Ackermann function: Difference between revisions

From Rosetta Code
Line 3,722: Line 3,722:
   
 
local fn Ackerman( m as NSInteger, n as NSInteger ) as NSInteger
 
local fn Ackerman( m as NSInteger, n as NSInteger ) as NSInteger
NSInteger result
+
NSInteger result
  +
 
select
+
select
case m == 0 : result = ( n + 1 )
+
case m == 0 : result = ( n + 1 )
case n == 0 : result = fn Ackerman( ( m - 1 ), 1 )
+
case n == 0 : result = fn Ackerman( ( m - 1 ), 1 )
case else : result = fn Ackerman( ( m - 1 ), fn Ackerman( m, ( n - 1 ) ) )
+
case else : result = fn Ackerman( ( m - 1 ), fn Ackerman( m, ( n - 1 ) ) )
end select
+
end select
 
end fn = result
 
end fn = result
   

Revision as of 10:13, 20 September 2022

Task
Ackermann function
You are encouraged to solve this task according to the task description, using any language you may know.

The Ackermann function is a classic example of a recursive function, notable especially because it is not a primitive recursive function. It grows very quickly in value, as does the size of its call tree.


The Ackermann function is usually defined as follows:


Its arguments are never negative and it always terminates.


Task

Write a function which returns the value of . Arbitrary precision is preferred (since the function grows so quickly), but not required.


See also



11l

Translation of: Python
F ack2(m, n) -> Int
   R I m == 0 {(n + 1)
 } E I m == 1 {(n + 2)
 } E I m == 2 {(2 * n + 3)
 } E I m == 3 {(8 * (2 ^ n - 1) + 5)
 } E I n == 0 {ack2(m - 1, 1)
 } E ack2(m - 1, ack2(m, n - 1))

print(ack2(0, 0))
print(ack2(3, 4))
print(ack2(4, 1))
Output:
1
125
65533

360 Assembly

Translation of: AWK

The OS/360 linkage is a bit tricky with the S/360 basic instruction set. To simplify, the program is recursive not reentrant.

*        Ackermann function        07/09/2015
&LAB     XDECO &REG,&TARGET
.*-----------------------------------------------------------------*
.*       THIS MACRO DISPLAYS THE REGISTER CONTENTS AS A TRUE       *
.*       DECIMAL VALUE. XDECO IS NOT PART OF STANDARD S360 MACROS! *
*------------------------------------------------------------------*
         AIF   (T'&REG EQ 'O').NOREG
         AIF   (T'&TARGET EQ 'O').NODEST
&LAB     B     I&SYSNDX               BRANCH AROUND WORK AREA
W&SYSNDX DS    XL8                    CONVERSION WORK AREA
I&SYSNDX CVD   &REG,W&SYSNDX          CONVERT TO DECIMAL
         MVC   &TARGET,=XL12'402120202020202020202020'
         ED    &TARGET,W&SYSNDX+2     MAKE FIELD PRINTABLE
         BC    2,*+12                 BYPASS NEGATIVE
         MVI   &TARGET+12,C'-'        INSERT NEGATIVE SIGN
         B     *+8                    BYPASS POSITIVE
         MVI   &TARGET+12,C'+'        INSERT POSITIVE SIGN
         MEXIT
.NOREG   MNOTE 8,'INPUT REGISTER OMITTED'
         MEXIT
.NODEST  MNOTE 8,'TARGET FIELD OMITTED'
         MEXIT
         MEND
ACKERMAN CSECT
         USING  ACKERMAN,R12       r12 : base register
         LR     R12,R15            establish base register
         ST     R14,SAVER14A       save r14
         LA     R4,0               m=0
LOOPM    CH     R4,=H'3'           do m=0 to 3
         BH     ELOOPM
         LA     R5,0               n=0
LOOPN    CH     R5,=H'8'           do n=0 to 8         
         BH     ELOOPN
         LR     R1,R4              m
         LR     R2,R5              n
         BAL    R14,ACKER          r1=acker(m,n)
         XDECO  R1,PG+19
         XDECO  R4,XD
         MVC    PG+10(2),XD+10
         XDECO  R5,XD
         MVC    PG+13(2),XD+10
         XPRNT  PG,44              print buffer
         LA     R5,1(R5)           n=n+1
         B      LOOPN
ELOOPN   LA     R4,1(R4)           m=m+1
         B      LOOPM
ELOOPM   L      R14,SAVER14A       restore r14
         BR     R14                return to caller
SAVER14A DS     F                  static save r14
PG       DC     CL44'Ackermann(xx,xx) = xxxxxxxxxxxx'
XD       DS     CL12
ACKER    CNOP   0,4                function r1=acker(r1,r2)
         LR     R3,R1              save argument r1 in r3
         LR     R9,R10             save stackptr (r10) in r9 temp
         LA     R1,STACKLEN        amount of storage required
         GETMAIN RU,LV=(R1)        allocate storage for stack
         USING  STACK,R10          make storage addressable
         LR     R10,R1             establish stack addressability
         ST     R14,SAVER14B       save previous r14
         ST     R9,SAVER10B        save previous r10
         LR     R1,R3              restore saved argument r1
START    ST     R1,M               stack m
         ST     R2,N               stack n
IF1      C      R1,=F'0'           if m<>0
         BNE    IF2                then goto if2
         LR     R11,R2             n
         LA     R11,1(R11)         return n+1
         B      EXIT
IF2      C      R2,=F'0'           else if m<>0
         BNE    IF3                then goto if3
         BCTR   R1,0               m=m-1
         LA     R2,1               n=1
         BAL    R14,ACKER          r1=acker(m)
         LR     R11,R1             return acker(m-1,1)
         B      EXIT
IF3      BCTR   R2,0               n=n-1
         BAL    R14,ACKER          r1=acker(m,n-1)
         LR     R2,R1              acker(m,n-1)
         L      R1,M               m
         BCTR   R1,0               m=m-1         
         BAL    R14,ACKER          r1=acker(m-1,acker(m,n-1))
         LR     R11,R1             return acker(m-1,1)
EXIT     L      R14,SAVER14B       restore r14
         L      R9,SAVER10B        restore r10 temp
         LA     R0,STACKLEN        amount of storage to free
         FREEMAIN A=(R10),LV=(R0)  free allocated storage
         LR     R1,R11             value returned
         LR     R10,R9             restore r10
         BR     R14                return to caller
         LTORG
         DROP   R12                base no longer needed
STACK    DSECT                     dynamic area
SAVER14B DS     F                  saved r14
SAVER10B DS     F                  saved r10
M        DS     F                  m
N        DS     F                  n
STACKLEN EQU    *-STACK
         YREGS  
         END    ACKERMAN
Output:
Ackermann( 0, 0) =            1
Ackermann( 0, 1) =            2
Ackermann( 0, 2) =            3
Ackermann( 0, 3) =            4
Ackermann( 0, 4) =            5
Ackermann( 0, 5) =            6
Ackermann( 0, 6) =            7
Ackermann( 0, 7) =            8
Ackermann( 0, 8) =            9
Ackermann( 1, 0) =            2
Ackermann( 1, 1) =            3
Ackermann( 1, 2) =            4
Ackermann( 1, 3) =            5
Ackermann( 1, 4) =            6
Ackermann( 1, 5) =            7
Ackermann( 1, 6) =            8
Ackermann( 1, 7) =            9
Ackermann( 1, 8) =           10
Ackermann( 2, 0) =            3
Ackermann( 2, 1) =            5
Ackermann( 2, 2) =            7
Ackermann( 2, 3) =            9
Ackermann( 2, 4) =           11
Ackermann( 2, 5) =           13
Ackermann( 2, 6) =           15
Ackermann( 2, 7) =           17
Ackermann( 2, 8) =           19
Ackermann( 3, 0) =            5
Ackermann( 3, 1) =           13
Ackermann( 3, 2) =           29
Ackermann( 3, 3) =           61
Ackermann( 3, 4) =          125
Ackermann( 3, 5) =          253
Ackermann( 3, 6) =          509
Ackermann( 3, 7) =         1021
Ackermann( 3, 8) =         2045

68000 Assembly

This implementation is based on the code shown in the computerphile episode in the youtube link at the top of this page (time index 5:00).

;
; Ackermann function for Motorola 68000 under AmigaOs 2+ by Thorham
;
; Set stack space to 60000 for m = 3, n = 5.
;
; The program will print the ackermann values for the range m = 0..3, n = 0..5
;
_LVOOpenLibrary equ -552
_LVOCloseLibrary equ -414
_LVOVPrintf equ -954

m equ 3 ; Nr of iterations for the main loop.
n equ 5 ; Do NOT set them higher, or it will take hours to complete on
        ; 68k, not to mention that the stack usage will become astronomical.
        ; Perhaps n can be a little higher... If you do increase the ranges
        ; then don't forget to increase the stack size.

execBase=4

start
    move.l  execBase,a6

    lea     dosName,a1
    moveq   #36,d0
    jsr     _LVOOpenLibrary(a6)
    move.l  d0,dosBase
    beq     exit

    move.l  dosBase,a6
    lea     printfArgs,a2

    clr.l   d3 ; m
.loopn
    clr.l   d4 ; n
.loopm
    bsr     ackermann

    move.l  d3,0(a2)
    move.l  d4,4(a2)
    move.l  d5,8(a2)
    move.l  #outString,d1
    move.l  a2,d2
    jsr     _LVOVPrintf(a6)

    addq.l  #1,d4
    cmp.l   #n,d4
    ble     .loopm

    addq.l  #1,d3
    cmp.l   #m,d3
    ble     .loopn

exit
    move.l  execBase,a6
    move.l  dosBase,a1
    jsr     _LVOCloseLibrary(a6)
    rts
;
; ackermann function
;
; in:
;
; d3 = m
; d4 = n
;
; out:
;
; d5 = ans
;
ackermann
    move.l  d3,-(sp)
    move.l  d4,-(sp)

    tst.l   d3
    bne     .l1
    move.l  d4,d5
    addq.l  #1,d5
    bra     .return
.l1
    tst.l   d4
    bne     .l2
    subq.l  #1,d3
    moveq   #1,d4
    bsr     ackermann
    bra     .return
.l2
    subq.l  #1,d4
    bsr     ackermann
    move.l  d5,d4
    subq.l  #1,d3
    bsr     ackermann

.return
    move.l  (sp)+,d4
    move.l  (sp)+,d3
    rts
;
; variables
;
dosBase
    dc.l    0

printfArgs
    dcb.l   3
;
; strings
;
dosName
    dc.b    "dos.library",0

outString
    dc.b    "ackermann (%ld,%ld) is: %ld",10,0

8080 Assembly

This function does 16-bit math. The test code prints a table of ack(m,n) for m ∊ [0,4) and n ∊ [0,9), on a real 8080 this takes a little over two minutes.

	org	100h
	jmp	demo
	;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
	;;;	ACK(M,N); DE=M, HL=N, return value in HL.
ack:	mov	a,d	; M=0?
	ora	e
	jnz	ackm
	inx	h	; If so, N+1.
	ret
ackm:	mov	a,h	; N=0?
	ora	l
	jnz	ackmn
	lxi	h,1	; If so, N=1,
	dcx	d	; N-=1,
	jmp	ack	; A(M,N) - tail recursion
ackmn:	push	d	; M>0 and N>0: store M on the stack
	dcx	h	; N-=1
	call	ack	; N = ACK(M,N-1)
	pop	d	; Restore previous M
	dcx	d	; M-=1
	jmp	ack	; A(M,N) - tail recursion
	;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
	;;;	Print table of ack(m,n)
MMAX:	equ	4	; Size of table to print. Note that math is done in
NMAX:	equ	9	; 16 bits. 
demo:	lhld	6	; Put stack pointer at top of available memory
	sphl
	lxi	b,0	; let B,C hold 8-bit M and N.
acknum:	xra	a	; Set high bit of M and N to zero
	mov	d,a	; DE = B (M)
	mov	e,b
	mov	h,a	; HL = C (N)
	mov	l,c
	call	ack	; HL = ack(DE,HL)
	call	prhl	; Print the number
	inr	c	; N += 1
	mvi	a,NMAX	; Time for next line?
	cmp	c
	jnz	acknum	; If not, print next number
	push	b	; Otherwise, save BC
	mvi	c,9	; Print newline
	lxi	d,nl
	call	5
	pop	b	; Restore BC
	mvi	c,0	; Set N to 0
	inr	b	; M += 1
	mvi	a,MMAX	; Time to stop?
	cmp	b
	jnz	acknum	; If not, print next number
	rst	0
	;;;	Print HL as ASCII number.
prhl:	push	h	; Save all registers
	push	d
	push	b
	lxi	b,pnum	; Store pointer to num string on stack
	push	b
	lxi	b,-10	; Divisor
prdgt:	lxi	d,-1
prdgtl:	inx	d	; Divide by 10 through trial subtraction
	dad	b
	jc	prdgtl
	mvi	a,'0'+10
	add	l	; L = remainder - 10
	pop	h	; Get pointer from stack
	dcx	h	; Store digit
	mov	m,a
	push	h	; Put pointer back on stack
	xchg		; Put quotient in HL
	mov	a,h	; Check if zero 
	ora	l
	jnz	prdgt	; If not, next digit
	pop	d	; Get pointer and put in DE
	mvi	c,9	; CP/M print string
	call	5
	pop	b	; Restore registers 
	pop	d
	pop	h
	ret
	db	'*****'	; Placeholder for number
pnum:	db	9,'$'
nl:	db	13,10,'$'
Output:
1       2       3       4       5       6       7       8       9
2       3       4       5       6       7       8       9       10
3       5       7       9       11      13      15      17      19
5       13      29      61      125     253     509     1021    2045

8086 Assembly

This code does 16-bit math just like the 8080 version.

	cpu	8086
	bits	16
	org	100h
section	.text
	jmp	demo
	;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
	;;;	ACK(M,N); DX=M, AX=N, return value in AX.
ack:	and	dx,dx	; N=0?
	jnz	.m
	inc	ax	; If so, return N+1
	ret
.m:	and	ax,ax	; M=0?
	jnz	.mn
	mov	ax,1	; If so, N=1,
	dec	dx	; M -= 1
	jmp	ack	; ACK(M-1,1) - tail recursion
.mn:	push	dx	; Keep M on the stack
	dec	ax	; N-=1
	call	ack	; N = ACK(M,N-1)
	pop	dx	; Restore M
	dec	dx	; M -= 1
	jmp	ack	; ACK(M-1,ACK(M,N-1)) - tail recursion
	;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
	;;;	Print table of ack(m,n)
MMAX:	equ	4	; Size of table to print. Noe that math is done
NMAX:	equ	9	; in 16 bits.
demo:	xor	si,si	; Let SI hold M,
	xor	di,di	; and DI hold N.
acknum:	mov	dx,si	; Calculate ack(M,N)
	mov	ax,di
	call	ack
	call	prax	; Print number
	inc	di	; N += 1
	cmp	di,NMAX	; Row done?
	jb	acknum	; If not, print next number on row
	xor	di,di	; Otherwise, N=0,
	inc	si	; M += 1
	mov	dx,nl	; Print newline
	call	prstr
	cmp	si,MMAX	; Done?
	jb	acknum	; If not, start next row
	ret		; Otherwise, stop.
	;;;	Print AX as ASCII number.
prax:	mov	bx,pnum	; Pointer to number string
	mov	cx,10	; Divisor
.dgt:	xor	dx,dx	; Divide AX by ten
	div	cx
	add	dl,'0'	; DX holds remainder - add ASCII 0
	dec	bx	; Move pointer backwards
	mov	[bx],dl	; Save digit in string
	and	ax,ax	; Are we done yet?
	jnz	.dgt	; If not, next digit
	mov	dx,bx	; Tell DOS to print the string
prstr:	mov	ah,9
	int	21h
	ret
section	.data
	db	'*****'	; Placeholder for ASCII number
pnum:	db	9,'$'
nl:	db	13,10,'$'
Output:
1       2       3       4       5       6       7       8       9
2       3       4       5       6       7       8       9       10
3       5       7       9       11      13      15      17      19
5       13      29      61      125     253     509     1021    2045

8th

\ Ackermann function, illustrating use of "memoization".

\ Memoization is a technique whereby intermediate computed values are stored
\ away against later need.  It is particularly valuable when calculating those
\ values is time or resource intensive, as with the Ackermann function.

\ make the stack much bigger so this can complete!
100000 stack-size

\ This is where memoized values are stored:
{} var, dict

\ Simple accessor words
: dict! \ "key" val --
  dict @ -rot m:! drop ;

: dict@ \ "key" -- val
  dict @ swap m:@ nip ;

defer: ack1

\ We just jam the string representation of the two numbers together for a key:
: makeKey  \ m n -- m n key
	2dup >s swap >s s:+ ;

: ack2 \ m n -- A
	makeKey dup
	dict@ null?
	if \ can't find key in dict
		\ m n key null
		drop \ m n key
		-rot \ key m n
		ack1 \ key A
		tuck \ A key A
		dict! \ A
	else \ found value
		\ m n key value
		>r drop 2drop r>
	then ;

: ack \ m n -- A
	over not 
	if
		nip n:1+
	else
		dup not
		if
			drop n:1- 1 ack2
		else
			over swap n:1- ack2
			swap n:1- swap ack2
		then
	then ;

' ack is ack1

: ackOf \ m n --
        2dup
        "Ack(" . swap . ", " . . ") = " . ack . cr ;


0 0 ackOf
0 4 ackOf
1 0 ackOf
1 1 ackOf
2 1 ackOf
2 2 ackOf
3 1 ackOf
3 3 ackOf
4 0 ackOf

\ this last requires a very large data stack.  So start 8th with a parameter '-k 100000'
4 1 ackOf

bye
The output:
Ack(0, 0) = 1
Ack(0, 4) = 5
Ack(1, 0) = 2
Ack(1, 1) = 3
Ack(2, 1) = 5
Ack(2, 2) = 7
Ack(3, 1) = 13
Ack(3, 3) = 61
Ack(4, 0) = 13
Ack(4, 1) = 65533

AArch64 Assembly

Works with: as version Raspberry Pi 3B version Buster 64 bits
or android 64 bits with application Termux
/* ARM assembly AARCH64 Raspberry PI 3B or android 64 bits */
/*  program ackermann64.s   */ 

/*******************************************/
/* Constantes file                         */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly*/
.include "../includeConstantesARM64.inc"
.equ MMAXI,   4
.equ NMAXI,   10

/*********************************/
/* Initialized data              */
/*********************************/
.data
sMessResult:        .asciz "Result for @  @  : @ \n"
szMessError:        .asciz "Overflow !!.\n"
szCarriageReturn:   .asciz "\n"
 
/*********************************/
/* UnInitialized data            */
/*********************************/
.bss
sZoneConv:        .skip 24
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main 
main:                           // entry of program 
    mov x3,#0
    mov x4,#0
1:
    mov x0,x3
    mov x1,x4
    bl ackermann
    mov x5,x0
    mov x0,x3
    ldr x1,qAdrsZoneConv        // else display odd message
    bl conversion10             // call decimal conversion
    ldr x0,qAdrsMessResult
    ldr x1,qAdrsZoneConv        // insert value conversion in message
    bl strInsertAtCharInc
    mov x6,x0
    mov x0,x4
    ldr x1,qAdrsZoneConv        // else display odd message
    bl conversion10             // call decimal conversion
    mov x0,x6
    ldr x1,qAdrsZoneConv        // insert value conversion in message
    bl strInsertAtCharInc
    mov x6,x0
    mov x0,x5
    ldr x1,qAdrsZoneConv        // else display odd message
    bl conversion10             // call decimal conversion
    mov x0,x6
    ldr x1,qAdrsZoneConv        // insert value conversion in message
    bl strInsertAtCharInc
    bl affichageMess
    add x4,x4,#1
    cmp x4,#NMAXI
    blt 1b
    mov x4,#0
    add x3,x3,#1
    cmp x3,#MMAXI
    blt 1b
100:                            // standard end of the program 
    mov x0, #0                  // return code
    mov x8, #EXIT               // request to exit program
    svc #0                      // perform the system call
 
qAdrszCarriageReturn:     .quad szCarriageReturn
qAdrsMessResult:          .quad sMessResult
qAdrsZoneConv:            .quad sZoneConv
/***************************************************/
/*     fonction ackermann              */
/***************************************************/
// x0 contains a number m
// x1 contains a number n
// x0 return résult
ackermann:
    stp x1,lr,[sp,-16]!         // save  registres
    stp x2,x3,[sp,-16]!         // save  registres
    cmp x0,0
    beq 5f
    mov x3,-1
    csel x0,x3,x0,lt               // error
    blt 100f
    cmp x1,#0
    csel x0,x3,x0,lt               // error
    blt 100f
    bgt 1f
    sub x0,x0,#1
    mov x1,#1
    bl ackermann
    b 100f
1:
    mov x2,x0
    sub x1,x1,#1
    bl ackermann
    mov x1,x0
    sub x0,x2,#1
    bl ackermann
    b 100f
5:
    adds x0,x1,#1
    bcc 100f
    ldr x0,qAdrszMessError
    bl affichageMess
    mov x0,-1
100:

    ldp x2,x3,[sp],16         // restaur des  2 registres
    ldp x1,lr,[sp],16         // restaur des  2 registres
    ret
qAdrszMessError:        .quad szMessError

/********************************************************/
/*        File Include fonctions                        */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
Output:
Result for 0  0  : 1
Result for 0  1  : 2
Result for 0  2  : 3
Result for 0  3  : 4
Result for 0  4  : 5
Result for 0  5  : 6
Result for 0  6  : 7
Result for 0  7  : 8
Result for 0  8  : 9
Result for 0  9  : 10
Result for 1  0  : 2
Result for 1  1  : 3
Result for 1  2  : 4
Result for 1  3  : 5
Result for 1  4  : 6
Result for 1  5  : 7
Result for 1  6  : 8
Result for 1  7  : 9
Result for 1  8  : 10
Result for 1  9  : 11
Result for 2  0  : 3
Result for 2  1  : 5
Result for 2  2  : 7
Result for 2  3  : 9
Result for 2  4  : 11
Result for 2  5  : 13
Result for 2  6  : 15
Result for 2  7  : 17
Result for 2  8  : 19
Result for 2  9  : 21
Result for 3  0  : 5
Result for 3  1  : 13
Result for 3  2  : 29
Result for 3  3  : 61
Result for 3  4  : 125
Result for 3  5  : 253
Result for 3  6  : 509
Result for 3  7  : 1021
Result for 3  8  : 2045
Result for 3  9  : 4093

ABAP

REPORT  zhuberv_ackermann.

CLASS zcl_ackermann DEFINITION.
  PUBLIC SECTION.
    CLASS-METHODS ackermann IMPORTING m TYPE i
                                      n TYPE i
                            RETURNING value(v) TYPE i.
ENDCLASS.            "zcl_ackermann DEFINITION


CLASS zcl_ackermann IMPLEMENTATION.

  METHOD: ackermann.

    DATA: lv_new_m TYPE i,
          lv_new_n TYPE i.

    IF m = 0.
      v = n + 1.
    ELSEIF m > 0 AND n = 0.
      lv_new_m = m - 1.
      lv_new_n = 1.
      v = ackermann( m = lv_new_m n = lv_new_n ).
    ELSEIF m > 0 AND n > 0.
      lv_new_m = m - 1.

      lv_new_n = n - 1.
      lv_new_n = ackermann( m = m n = lv_new_n ).

      v = ackermann( m = lv_new_m n = lv_new_n ).
    ENDIF.

  ENDMETHOD.                    "ackermann

ENDCLASS.                    "zcl_ackermann IMPLEMENTATION


PARAMETERS: pa_m TYPE i,
            pa_n TYPE i.

DATA: lv_result TYPE i.

START-OF-SELECTION.
  lv_result = zcl_ackermann=>ackermann( m = pa_m n = pa_n ).
  WRITE: / lv_result.

Action!

Action! language does not support recursion. Therefore an iterative approach with a stack has been proposed.

DEFINE MAXSIZE="1000"
CARD ARRAY stack(MAXSIZE)
CARD stacksize=[0]

BYTE FUNC IsEmpty()
  IF stacksize=0 THEN
    RETURN (1)
  FI
RETURN (0)

PROC Push(BYTE v)
  IF stacksize=maxsize THEN
    PrintE("Error: stack is full!")
    Break()
  FI
  stack(stacksize)=v
  stacksize==+1
RETURN

BYTE FUNC Pop()
  IF IsEmpty() THEN
    PrintE("Error: stack is empty!")
    Break()
  FI
  stacksize==-1
RETURN (stack(stacksize))

CARD FUNC Ackermann(CARD m,n)
  Push(m)
  WHILE IsEmpty()=0
  DO
    m=Pop()
    IF m=0 THEN
      n==+1
    ELSEIF n=0 THEN
      n=1
      Push(m-1)
    ELSE
      n==-1
      Push(m-1)
      Push(m)
    FI
  OD
RETURN (n)

PROC Main()
  CARD m,n,res

  FOR m=0 TO 3
  DO
    FOR n=0 TO 4
    DO
      res=Ackermann(m,n)
      PrintF("Ack(%U,%U)=%U%E",m,n,res)
    OD
  OD
RETURN
Output:

Screenshot from Atari 8-bit computer

Ack(0,0)=1
Ack(0,1)=2
Ack(0,2)=3
Ack(0,3)=4
Ack(0,4)=5
Ack(1,0)=2
Ack(1,1)=3
Ack(1,2)=4
Ack(1,3)=5
Ack(1,4)=6
Ack(2,0)=3
Ack(2,1)=5
Ack(2,2)=7
Ack(2,3)=9
Ack(2,4)=11
Ack(3,0)=5
Ack(3,1)=13
Ack(3,2)=29
Ack(3,3)=61
Ack(3,4)=125

ActionScript

public function ackermann(m:uint, n:uint):uint
{
    if (m == 0)
    {
        return n + 1;
    }
    if (n == 0)
    {
        return ackermann(m - 1, 1);
    }		

    return ackermann(m - 1, ackermann(m, n - 1));
}

Ada

with Ada.Text_IO;  use Ada.Text_IO;

procedure Test_Ackermann is
   function Ackermann (M, N : Natural) return Natural is
   begin
      if M = 0 then
         return N + 1;
      elsif N = 0 then
         return Ackermann (M - 1, 1);
      else
         return Ackermann (M - 1, Ackermann (M, N - 1));
      end if;
   end Ackermann;
begin
   for M in 0..3 loop
      for N in 0..6 loop
         Put (Natural'Image (Ackermann (M, N)));
      end loop;
      New_Line;
   end loop;
end Test_Ackermann;

The implementation does not care about arbitrary precision numbers because the Ackermann function does not only grow, but also slow quickly, when computed recursively.

Output:
the first 4x7 Ackermann's numbers
 1 2 3 4 5 6 7
 2 3 4 5 6 7 8
 3 5 7 9 11 13 15
 5 13 29 61 125 253 509

Agda

Works with: Agda version 2.5.2
open import Data.Nat
open import Data.Nat.Show
open import IO
  
module Ackermann where

ack :  ->  -> 
ack zero n = n + 1
ack (suc m) zero = ack m 1
ack (suc m) (suc n) = ack m (ack (suc m) n)
      
main = run (putStrLn (show (ack 3 9)))

Note the unicode ℕ characters, they can be input in emacs agda mode using "\bN". Running in bash:

agda --compile Ackermann.agda
./Ackermann
Output:
4093

ALGOL 60

Works with: A60
begin 
   integer procedure ackermann(m,n);value m,n;integer m,n;
   ackermann:=if m=0 then n+1
         else if n=0 then ackermann(m-1,1)
                     else ackermann(m-1,ackermann(m,n-1));
   integer m,n;
   for m:=0 step 1 until 3 do begin
      for n:=0 step 1 until 6 do
         outinteger(1,ackermann(m,n));
      outstring(1,"\n")
   end
end
Output:
 1  2  3  4  5  6  7
 2  3  4  5  6  7  8
 3  5  7  9  11  13  15
 5  13  29  61  125  253  509

ALGOL 68

Translation of: Ada
Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
PROC test ackermann = VOID: 
BEGIN
   PROC ackermann = (INT m, n)INT:
   BEGIN
      IF m = 0 THEN
         n + 1
      ELIF n = 0 THEN
         ackermann (m - 1, 1)
      ELSE
         ackermann (m - 1, ackermann (m, n - 1))
      FI
   END # ackermann #;

   FOR m FROM 0 TO 3 DO
      FOR n FROM 0 TO 6 DO
         print(ackermann (m, n))
      OD;
      new line(stand out)
   OD
END # test ackermann #;
test ackermann
Output:
         +1         +2         +3         +4         +5         +6         +7
         +2         +3         +4         +5         +6         +7         +8
         +3         +5         +7         +9        +11        +13        +15
         +5        +13        +29        +61       +125       +253       +509

ALGOL W

Translation of: ALGOL 60
begin 
   integer procedure ackermann( integer value m,n ) ;
       if m=0 then n+1
       else if n=0 then ackermann(m-1,1)
                   else ackermann(m-1,ackermann(m,n-1));
   for m := 0 until 3 do begin
      write( ackermann( m, 0 ) );
      for n := 1 until 6 do writeon( ackermann( m, n ) );
   end for_m
end.
Output:
             1               2               3               4               5               6               7
             2               3               4               5               6               7               8
             3               5               7               9              11              13              15
             5              13              29              61             125             253             509

APL

Works with: Dyalog APL
ackermann{
     0=1⍵:1+2
     0=2⍵:∇(¯1+1)1
     (¯1+1),(1),¯1+2
 }

AppleScript

on ackermann(m, n)
    if m is equal to 0 then return n + 1
    if n is equal to 0 then return ackermann(m - 1, 1)
    return ackermann(m - 1, ackermann(m, n - 1))
end ackermann

Argile

Works with: Argile version 1.0.0
use std

for each (val nat n) from 0 to 6
  for each (val nat m) from 0 to 3
    print "A("m","n") = "(A m n)

.:A <nat m, nat n>:. -> nat
  return (n+1)				if m == 0
  return (A (m - 1) 1)			if n == 0
  A (m - 1) (A m (n - 1))

ARM Assembly

Works with: as version Raspberry Pi
or android 32 bits with application Termux
/* ARM assembly Raspberry PI  or android 32 bits */
/*  program ackermann.s   */ 

/* REMARK 1 : this program use routines in a include file 
   see task Include a file language arm assembly 
   for the routine affichageMess conversion10 
   see at end of this program the instruction include */
/* for constantes see task include a file in arm assembly */
/************************************/
/* Constantes                       */
/************************************/
.include "../constantes.inc"
.equ MMAXI,   4
.equ NMAXI,   10

/*********************************/
/* Initialized data              */
/*********************************/
.data
sMessResult:        .asciz "Result for @  @  : @ \n"
szMessError:        .asciz "Overflow !!.\n"
szCarriageReturn:   .asciz "\n"
 
/*********************************/
/* UnInitialized data            */
/*********************************/
.bss
sZoneConv:        .skip 24
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main 
main:                           @ entry of program 
    mov r3,#0
    mov r4,#0
1:
    mov r0,r3
    mov r1,r4
    bl ackermann
    mov r5,r0
    mov r0,r3
    ldr r1,iAdrsZoneConv        @ else display odd message
    bl conversion10             @ call decimal conversion
    ldr r0,iAdrsMessResult
    ldr r1,iAdrsZoneConv        @ insert value conversion in message
    bl strInsertAtCharInc
    mov r6,r0
    mov r0,r4
    ldr r1,iAdrsZoneConv        @ else display odd message
    bl conversion10             @ call decimal conversion
    mov r0,r6
    ldr r1,iAdrsZoneConv        @ insert value conversion in message
    bl strInsertAtCharInc
    mov r6,r0
    mov r0,r5
    ldr r1,iAdrsZoneConv        @ else display odd message
    bl conversion10             @ call decimal conversion
    mov r0,r6
    ldr r1,iAdrsZoneConv        @ insert value conversion in message
    bl strInsertAtCharInc
    bl affichageMess
    add r4,#1
    cmp r4,#NMAXI
    blt 1b
    mov r4,#0
    add r3,#1
    cmp r3,#MMAXI
    blt 1b
100:                            @ standard end of the program 
    mov r0, #0                  @ return code
    mov r7, #EXIT               @ request to exit program
    svc #0                      @ perform the system call
 
iAdrszCarriageReturn:     .int szCarriageReturn
iAdrsMessResult:          .int sMessResult
iAdrsZoneConv:            .int sZoneConv
/***************************************************/
/*     fonction ackermann              */
/***************************************************/
// r0 contains a number m
// r1 contains a number n
// r0 return résult
ackermann:
    push {r1-r2,lr}             @ save  registers 
    cmp r0,#0
    beq 5f
    movlt r0,#-1               @ error
    blt 100f
    cmp r1,#0
    movlt r0,#-1               @ error
    blt 100f
    bgt 1f
    sub r0,r0,#1
    mov r1,#1
    bl ackermann
    b 100f
1:
    mov r2,r0
    sub r1,r1,#1
    bl ackermann
    mov r1,r0
    sub r0,r2,#1
    bl ackermann
    b 100f
5:
    adds r0,r1,#1
    bcc 100f
    ldr r0,iAdrszMessError
    bl affichageMess
    bkpt
100:
    pop {r1-r2,lr}             @ restaur registers
    bx lr                      @ return
iAdrszMessError:        .int szMessError
/***************************************************/
/*      ROUTINES INCLUDE                           */
/***************************************************/
.include "../affichage.inc"
Output:
Result for 0            0            : 1
Result for 0            1            : 2
Result for 0            2            : 3
Result for 0            3            : 4
Result for 0            4            : 5
Result for 0            5            : 6
Result for 0            6            : 7
Result for 0            7            : 8
Result for 0            8            : 9
Result for 0            9            : 10
Result for 1            0            : 2
Result for 1            1            : 3
Result for 1            2            : 4
Result for 1            3            : 5
Result for 1            4            : 6
Result for 1            5            : 7
Result for 1            6            : 8
Result for 1            7            : 9
Result for 1            8            : 10
Result for 1            9            : 11
Result for 2            0            : 3
Result for 2            1            : 5
Result for 2            2            : 7
Result for 2            3            : 9
Result for 2            4            : 11
Result for 2            5            : 13
Result for 2            6            : 15
Result for 2            7            : 17
Result for 2            8            : 19
Result for 2            9            : 21
Result for 3            0            : 5
Result for 3            1            : 13
Result for 3            2            : 29
Result for 3            3            : 61
Result for 3            4            : 125
Result for 3            5            : 253
Result for 3            6            : 509
Result for 3            7            : 1021
Result for 3            8            : 2045
Result for 3            9            : 4093

Arturo

ackermann: function [m,n][
    (m=0)? -> n+1 [
        (n=0)? -> ackermann m-1 1
               -> ackermann m-1 ackermann m n-1
    ]
]

loop 0..3 'a [
    loop 0..4 'b [
        print ["ackermann" a b "=>" ackermann a b]
    ]
]
Output:
ackermann 0 0 => 1 
ackermann 0 1 => 2 
ackermann 0 2 => 3 
ackermann 0 3 => 4 
ackermann 0 4 => 5 
ackermann 1 0 => 2 
ackermann 1 1 => 3 
ackermann 1 2 => 4 
ackermann 1 3 => 5 
ackermann 1 4 => 6 
ackermann 2 0 => 3 
ackermann 2 1 => 5 
ackermann 2 2 => 7 
ackermann 2 3 => 9 
ackermann 2 4 => 11 
ackermann 3 0 => 5 
ackermann 3 1 => 13 
ackermann 3 2 => 29 
ackermann 3 3 => 61 
ackermann 3 4 => 125

ATS

fun ackermann
  {m,n:nat} .<m,n>.
  (m: int m, n: int n): Nat =
  case+ (m, n) of
  | (0, _) => n+1
  | (_, 0) =>> ackermann (m-1, 1)
  | (_, _) =>> ackermann (m-1, ackermann (m, n-1))
// end of [ackermann]

AutoHotkey

A(m, n) {
  If (m > 0) && (n = 0)
    Return A(m-1,1)
  Else If (m > 0) && (n > 0)
    Return A(m-1,A(m, n-1))
  Else If (m=0)
    Return n+1
}

; Example:
MsgBox, % "A(1,2) = " A(1,2)

AutoIt

Classical version

Func Ackermann($m, $n)
    If ($m = 0) Then
        Return $n+1
    Else
        If ($n = 0) Then
            Return Ackermann($m-1, 1)
        Else
            return Ackermann($m-1, Ackermann($m, $n-1))
        EndIf
    EndIf
EndFunc

Classical + cache implementation

This version works way faster than the classical one: Ackermann(3, 5) runs in 12,7 ms, while the classical version takes 402,2 ms.

Global $ackermann[2047][2047] ; Set the size to whatever you want
Func Ackermann($m, $n)
	If ($ackermann[$m][$n] <> 0) Then
		Return $ackermann[$m][$n]
	Else
		If ($m = 0) Then
			$return = $n + 1
		Else
			If ($n = 0) Then
				$return = Ackermann($m - 1, 1)
			Else
				$return = Ackermann($m - 1, Ackermann($m, $n - 1))
			EndIf
		EndIf
		$ackermann[$m][$n] = $return
		Return $return
	EndIf
EndFunc   ;==>Ackermann

AWK

function ackermann(m, n) 
{
  if ( m == 0 ) { 
    return n+1
  }
  if ( n == 0 ) { 
    return ackermann(m-1, 1)
  }
  return ackermann(m-1, ackermann(m, n-1))
}

BEGIN {
  for(n=0; n < 7; n++) {
    for(m=0; m < 4; m++) {
      print "A(" m "," n ") = " ackermann(m,n)
    }
  }
}

Babel

main: 
        {((0 0) (0 1) (0 2)
        (0 3) (0 4) (1 0)
        (1 1) (1 2) (1 3)
        (1 4) (2 0) (2 1)
        (2 2) (2 3) (3 0)
        (3 1) (3 2) (4 0)) 
    
        { dup
        "A(" << { %d " " . << } ... ") = " <<
    reverse give 
    ack 
    %d cr << } ... }

ack!: 
    { dup zero?
        { <-> dup zero?
            { <-> 
                cp
                1 -
                <- <- 1 - ->
                ack -> 
                ack }
            { <->
                1 - 
                <- 1 -> 
                ack }
        if }
        { zap 1 + }
    if }

zero?!: { 0 = }
Output:
A(0 0 ) = 1
A(0 1 ) = 2
A(0 2 ) = 3
A(0 3 ) = 4
A(0 4 ) = 5
A(1 0 ) = 2
A(1 1 ) = 3
A(1 2 ) = 4
A(1 3 ) = 5
A(1 4 ) = 6
A(2 0 ) = 3
A(2 1 ) = 5
A(2 2 ) = 7
A(2 3 ) = 9
A(3 0 ) = 5
A(3 1 ) = 13
A(3 2 ) = 29
A(4 0 ) = 13

BASIC

Applesoft BASIC

100 DIM R%(2900),M(2900),N(2900)
110 FOR M = 0 TO 3
120     FOR N = 0 TO 4
130         GOSUB 200"ACKERMANN
140         PRINT "ACK("M","N") = "ACK
150 NEXT N, M
160 END 

200 M(SP) = M
210 N(SP) = N

REM A(M - 1, A(M, N - 1))
220 IF M > 0 AND N > 0 THEN N = N - 1 : R%(SP) = 0 : SP = SP + 1 : GOTO 200

REM A(M - 1, 1)
230 IF M > 0 THEN M = M - 1 : N = 1 : R%(SP) = 1 : SP = SP + 1 : GOTO 200

REM N + 1
240 ACK = N + 1

REM RETURN
250 M = M(SP) : N = N(SP) : IF SP = 0 THEN RETURN
260 FOR SP = SP - 1 TO 0 STEP -1 : IF R%(SP) THEN M = M(SP) : N = N(SP) : NEXT SP : SP = 0 : RETURN
270 M = M - 1 : N = ACK : R%(SP) = 1 : SP = SP + 1 : GOTO 200

BASIC256

dim stack(5000, 3)	# BASIC-256 lacks functions (as of ver. 0.9.6.66)
stack[0,0] = 3		# M
stack[0,1] = 7		# N
lev = 0 

gosub ackermann
print "A("+stack[0,0]+","+stack[0,1]+") = "+stack[0,2]
end

ackermann:
	if stack[lev,0]=0 then
		stack[lev,2] = stack[lev,1]+1
		return
	end if
	if stack[lev,1]=0 then 
		lev = lev+1
		stack[lev,0] = stack[lev-1,0]-1
		stack[lev,1] = 1
		gosub ackermann
		stack[lev-1,2] = stack[lev,2]
		lev = lev-1
		return
	end if
	lev = lev+1
	stack[lev,0] = stack[lev-1,0]
	stack[lev,1] = stack[lev-1,1]-1
	gosub ackermann
	stack[lev,0] = stack[lev-1,0]-1
	stack[lev,1] = stack[lev,2]
	gosub ackermann
	stack[lev-1,2] = stack[lev,2]
	lev = lev-1
	return
Output:
 A(3,7) = 1021
# BASIC256 since 0.9.9.1 supports functions
for m = 0 to 3
   for n = 0 to 4
      print m + " " + n + " " + ackermann(m,n)
   next n
next m
end

function ackermann(m,n)
   if m = 0 then
      ackermann = n+1
   else
      if n = 0 then
         ackermann = ackermann(m-1,1)
      else
         ackermann = ackermann(m-1,ackermann(m,n-1))
      endif
   end if
end function
Output:
0 0 1
0 1 2
0 2 3
0 3 4
0 4 5
1 0 2
1 1 3
1 2 4
1 3 5
1 4 6
2 0 3
2 1 5
2 2 7
2 3 9
2 4 11
3 0 5
3 1 13
3 2 29
3 3 61
3 4 125

BBC BASIC

      PRINT FNackermann(3, 7)
      END
      
      DEF FNackermann(M%, N%)
      IF M% = 0 THEN = N% + 1
      IF N% = 0 THEN = FNackermann(M% - 1, 1)
      = FNackermann(M% - 1, FNackermann(M%, N%-1))

QuickBasic

Works with: QuickBasic version 4.5

BASIC runs out of stack space very quickly. The call ack(3, 4) gives a stack error.

DECLARE FUNCTION ack! (m!, n!)

FUNCTION ack (m!, n!)
       IF m = 0 THEN ack = n + 1

       IF m > 0 AND n = 0 THEN
               ack = ack(m - 1, 1)
       END IF
       IF m > 0 AND n > 0 THEN
               ack = ack(m - 1, ack(m, n - 1))
       END IF
END FUNCTION

Batch File

Had trouble with this, so called in the gurus at StackOverflow. Thanks to Patrick Cuff for pointing out where I was going wrong.

::Ackermann.cmd
@echo off
set depth=0
:ack
if %1==0 goto m0
if %2==0 goto n0

:else
set /a n=%2-1
set /a depth+=1
call :ack %1 %n%
set t=%errorlevel%
set /a depth-=1
set /a m=%1-1
set /a depth+=1
call :ack %m% %t%
set t=%errorlevel%
set /a depth-=1
if %depth%==0 ( exit %t% ) else ( exit /b %t% )

:m0
set/a n=%2+1
if %depth%==0 ( exit %n% ) else ( exit /b %n% )

:n0
set /a m=%1-1
set /a depth+=1
call :ack %m% 1
set t=%errorlevel%
set /a depth-=1
if %depth%==0 ( exit %t% ) else ( exit /b %t% )

Because of the exit statements, running this bare closes one's shell, so this test routine handles the calling of Ackermann.cmd

::Ack.cmd
@echo off
cmd/c ackermann.cmd %1 %2
echo Ackermann(%1, %2)=%errorlevel%

A few test runs:

D:\Documents and Settings\Bruce>ack 0 4
Ackermann(0, 4)=5

D:\Documents and Settings\Bruce>ack 1 4
Ackermann(1, 4)=6

D:\Documents and Settings\Bruce>ack 2 4
Ackermann(2, 4)=11

D:\Documents and Settings\Bruce>ack 3 4
Ackermann(3, 4)=125

bc

Requires a bc that supports long names and the print statement.

Works with: OpenBSD bc
Works with: GNU bc
define ack(m, n) {
   if ( m == 0 ) return (n+1);
   if ( n == 0 ) return (ack(m-1, 1));
   return (ack(m-1, ack(m, n-1)));
}

for (n=0; n<7; n++) {
  for (m=0; m<4; m++) {
     print "A(", m, ",", n, ") = ", ack(m,n), "\n"; 
  }
}
quit

BCPL

GET "libhdr"

LET ack(m, n) = m=0 -> n+1,
                n=0 -> ack(m-1, 1),
                ack(m-1, ack(m, n-1))

LET start() = VALOF
{ FOR i = 0 TO 6 FOR m = 0 TO 3 DO
    writef("ack(%n, %n) = %n*n", m, n, ack(m,n))
  RESULTIS 0
}

beeswax

Iterative slow version:

                         >M?f@h@gMf@h3yzp            if m>0 and n>0 => replace m,n with m-1,m,n-1
                    >@h@g'b?1f@h@gM?f@hzp            if m>0 and n=0 => replace m,n with m-1,1
_ii>Ag~1?~Lpz1~2h@g'd?g?Pfzp                         if m=0         => replace m,n with n+1
           >I;
     b                     <            <

A functional and recursive realization of the version above. Functions are realized by direct calls of functions via jumps (instruction J) to the entry points of two distinct functions:

1st function _ii (input function) with entry point at (row,col) = (4,1)

2nd function Ag~1.... (Ackermann function) with entry point at (row,col) = (1,1)

Each block of 1FJ or 1fFJ in the code is a call of the Ackermann function itself.

Ag~1?Lp1~2@g'p?g?Pf1FJ                               Ackermann function.  if m=0 => run Ackermann function (m, n+1)
      xI;    x@g'p??@Mf1fFJ                                               if m>0 and n=0 => run Ackermann (m-1,1)
                 xM@~gM??f~f@f1FJ                                         if m>0 and n>0 => run Ackermann(m,Ackermann(m-1,n-1))
_ii1FJ                                               input function. Input m,n, then execute Ackermann(m,n)

Highly optimized and fast version, returns A(4,1)/A(5,0) almost instantaneously:

                             >Mf@Ph?@g@2h@Mf@Php     if m>4 and n>0 => replace m,n with m-1,m,n-1  
                 >~4~L#1~2hg'd?1f@hgM?f2h      p     if m>4 and n=0 => replace m,n with m-1,1
                #            q      <                                                   /n+3 times  \
                #X~4K#?2Fg?PPP>@B@M"pb               if m=4         => replace m,n with 2^(2^(....)))-3
                 # >~3K#?g?PPP~2BMMp>@MMMp           if m=3         => replace m,n with 2^(n+3)-3
_ii>Ag~1?~Lpz1~2h@gX'#?g?P      p  M                 if m=0         => replace m,n with n+1
    z      I       ~>~1K#?g?PP  p                    if m=1         => replace m,n with n+2
     f     ;       >2K#?g?~2.PPPp                    if m=2         => replace m,n with 2n+3
   z  b                         <  <     <
   d                                           <

Higher values than A(4,1)/(5,0) lead to UInt64 wraparound, support for numbers bigger than 2^64-1 is not implemented in these solutions.

Befunge

Befunge-93

Translation of: ERRE

Since Befunge-93 doesn't have recursive capabilities we need to use an iterative algorithm.

&>&>vvg0>#0\#-:#1_1v
@v:\<vp0    0:-1<\+<
^>00p>:#^_$1+\:#^_$.

Befunge-98

Works with: CCBI version 2.1
r[1&&{0
>v
 j
u>.@ 
1>  \:v
^  v:\_$1+
\^v_$1\1-
u^>1-0fp:1-\0fg101-

The program reads two integers (first m, then n) from command line, idles around funge space, then outputs the result of the Ackerman function. Since the latter is calculated truly recursively, the execution time becomes unwieldy for most m>3.

BQN

A ← {
  A 0‿n: n+1;
  A m‿0: A (m-1)‿1;
  A m‿n: A (m-1)‿(A m‿(n-1))
}

Example usage:

    A 0‿3
4
    A 1‿4
6
    A 2‿4
11
    A 3‿4
125

Bracmat

Three solutions are presented here. The first one is a purely recursive version, only using the formulas at the top of the page. The value of A(4,1) cannot be computed due to stack overflow. It can compute A(3,9) (4093), but probably not A(3,10)

( Ack
=   m n
  .   !arg:(?m,?n)
    & ( !m:0&!n+1
      | !n:0&Ack$(!m+-1,1)
      | Ack$(!m+-1,Ack$(!m,!n+-1))
      )
);

The second version is a purely non-recursive solution that easily can compute A(4,1). The program uses a stack for Ackermann function calls that are to be evaluated, but that cannot be computed given the currently known function values - the "known unknowns". The currently known values are stored in a hash table. The Hash table also contains incomplete Ackermann function calls, namely those for which the second argument is not known yet - "the unknown unknowns". These function calls are associated with "known unknowns" that are going to provide the value of the second argument. As soon as such an associated known unknown becomes known, the unknown unknown becomes a known unknown and is pushed onto the stack.

Although all known values are stored in the hash table, the converse is not true: an element in the hash table is either a "known known" or an "unknown unknown" associated with an "known unknown".

  ( A
  =     m n value key eq chain
      , find insert future stack va val
    .   ( chain
        =   key future skey
          .   !arg:(?key.?future)
            & str$!key:?skey
            & (cache..insert)$(!skey..!future)
            & 
        )
      & (find=.(cache..find)$(str$!arg))
      & ( insert
        =   key value future v futureeq futurem skey
          .   !arg:(?key.?value)
            & str$!key:?skey
            & (   (cache..find)$!skey:(?key.?v.?future)
                & (cache..remove)$!skey
                & (cache..insert)$(!skey.!value.)
                & (   !future:(?futurem.?futureeq)
                    & (!futurem,!value.!futureeq)
                  | 
                  )
              | (cache..insert)$(!skey.!value.)&
              )
        )
      & !arg:(?m,?n)
      & !n+1:?value
      & :?eq:?stack
      &   whl
        ' ( (!m,!n):?key
          &     (   find$!key:(?.#%?value.?future)
                  & insert$(!eq.!value) !future
                |   !m:0
                  & !n+1:?value
                  & ( !eq:&insert$(!key.!value)
                    |   insert$(!key.!value) !stack:?stack
                      & insert$(!eq.!value)
                    )
                |   !n:0
                  &   (!m+-1,1.!key)
                      (!eq:|(!key.!eq))
                |   find$(!m,!n+-1):(?.?val.?)
                  & (   !val:#%
                      & (   find$(!m+-1,!val):(?.?va.?)
                          & !va:#%
                          & insert$(!key.!va)
                        |   (!m+-1,!val.!eq)
                            (!m,!n.!eq)
                        )
                    | 
                    )
                |   chain$(!m,!n+-1.!m+-1.!key)
                  &   (!m,!n+-1.)
                      (!eq:|(!key.!eq))
                )
                !stack
            : (?m,?n.?eq) ?stack
          )
      & !value
  )
& new$hash:?cache
Some results:
A$(0,0):1
A$(3,13):65533
A$(3,14):131069
A$(4,1):65533

The last solution is a recursive solution that employs some extra formulas, inspired by the Common Lisp solution further down.

( AckFormula
=   m n
  .   !arg:(?m,?n)
    & ( !m:0&!n+1
      | !m:1&!n+2
      | !m:2&2*!n+3
      | !m:3&2^(!n+3)+-3
      | !n:0&AckFormula$(!m+-1,1)
      | AckFormula$(!m+-1,AckFormula$(!m,!n+-1))
      )
)
Some results:
AckFormula$(4,1):65533
AckFormula$(4,2):2003529930406846464979072351560255750447825475569751419265016973.....22087777506072339445587895905719156733

The last computation costs about 0,03 seconds.

Brat

ackermann = { m, n |
	when { m == 0 } { n + 1 }
		{ m > 0 && n == 0 } { ackermann(m - 1, 1) }
		{ m > 0 && n > 0 } { ackermann(m - 1, ackermann(m, n - 1)) }
}

p ackermann 3, 4  #Prints 125

C

Straightforward implementation per Ackermann definition:

#include <stdio.h>

int ackermann(int m, int n)
{
        if (!m) return n + 1;
        if (!n) return ackermann(m - 1, 1);
        return ackermann(m - 1, ackermann(m, n - 1));
}

int main()
{
        int m, n;
        for (m = 0; m <= 4; m++)
                for (n = 0; n < 6 - m; n++)
                        printf("A(%d, %d) = %d\n", m, n, ackermann(m, n));

        return 0;
}
Output:
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(4, 0) = 13
A(4, 1) = 65533

Ackermann function makes a lot of recursive calls, so the above program is a bit naive. We need to be slightly less naive, by doing some simple caching:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int m_bits, n_bits;
int *cache;

int ackermann(int m, int n)
{
        int idx, res;
        if (!m) return n + 1;

        if (n >= 1<<n_bits) {
                printf("%d, %d\n", m, n);
                idx = 0;
        } else {
                idx = (m << n_bits) + n;
                if (cache[idx]) return cache[idx];
        }

        if (!n) res = ackermann(m - 1, 1);
        else    res = ackermann(m - 1, ackermann(m, n - 1));

        if (idx) cache[idx] = res;
        return res;
}
int main()
{
        int m, n;

        m_bits = 3;
        n_bits = 20;  /* can save n values up to 2**20 - 1, that's 1 meg */
        cache = malloc(sizeof(int) * (1 << (m_bits + n_bits)));
        memset(cache, 0, sizeof(int) * (1 << (m_bits + n_bits)));

        for (m = 0; m <= 4; m++)
                for (n = 0; n < 6 - m; n++)
                        printf("A(%d, %d) = %d\n", m, n, ackermann(m, n));

        return 0;
}
Output:
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(4, 0) = 13
A(4, 1) = 65533

Whee. Well, with some extra work, we calculated one more n value, big deal, right? But see, A(4, 2) = A(3, A(4, 1)) = A(3, 65533) = A(2, A(3, 65532)) = ... you can see how fast it blows up. In fact, no amount of caching will help you calculate large m values; on the machine I use A(4, 2) segfaults because the recursions run out of stack space--not a whole lot I can do about it. At least it runs out of stack space quickly, unlike the first solution...

A couple of alternative approaches...

/* Thejaka Maldeniya */

#include <conio.h>

unsigned long long HR(unsigned int n, unsigned long long a, unsigned long long b) {
	// (Internal) Recursive Hyperfunction: Perform a Hyperoperation...

	unsigned long long r = 1;

	while(b--)
		r = n - 3 ? HR(n - 1, a, r) : /* Exponentiation */ r * a;

	return r;
}

unsigned long long H(unsigned int n, unsigned long long a, unsigned long long b) {
	// Hyperfunction (Recursive-Iterative-O(1) Hybrid): Perform a Hyperoperation...

	switch(n) {
		case 0:
			// Increment
			return ++b;
		case 1:
			// Addition
			return a + b;
		case 2:
			// Multiplication
			return a * b;
	}

	return HR(n, a, b);
}

unsigned long long APH(unsigned int m, unsigned int n) {
	// Ackermann-Péter Function (Recursive-Iterative-O(1) Hybrid)
	return H(m, 2, n + 3) - 3;
}

unsigned long long * p = 0;

unsigned long long APRR(unsigned int m, unsigned int n) {
	if (!m) return ++n;

	unsigned long long r = p ? p[m] : APRR(m - 1, 1);

	--m;
	while(n--)
		r = APRR(m, r);

	return r;
}

unsigned long long APRA(unsigned int m, unsigned int n) {
	return
		m ?
			n ?
				APRR(m, n)
				: p ? p[m] : APRA(--m, 1)
			: ++n
		;
}

unsigned long long APR(unsigned int m, unsigned int n) {
	unsigned long long r = 0;

	// Allocate
	p = (unsigned long long *) malloc(sizeof(unsigned long long) * (m + 1));

	// Initialize
	for(; r <= m; ++r)
		p[r] = r ? APRA(r - 1, 1) : APRA(r, 0);

	// Calculate
	r = APRA(m, n);

	// Free
	free(p);

	return r;
}

unsigned long long AP(unsigned int m, unsigned int n) {
	return APH(m, n);
	return APR(m, n);
}

int main(int n, char ** a) {
	unsigned int M, N;

	if (n != 3) {
		printf("Usage: %s <m> <n>\n", *a);
		return 1;
	}

	printf("AckermannPeter(%u, %u) = %llu\n", M = atoi(a[1]), N = atoi(a[2]), AP(M, N));

	//printf("\nPress any key...");
	//getch();
	return 0;
}

A couple of more iterative techniques...

/* Thejaka Maldeniya */

#include <conio.h>

unsigned long long HI(unsigned int n, unsigned long long a, unsigned long long b) {
	// Hyperfunction (Iterative): Perform a Hyperoperation...

	unsigned long long *I, r = 1;
	unsigned int N = n - 3;

	if (!N)
		// Exponentiation
		while(b--)
			r *= a;
	else if(b) {
		n -= 2;

		// Allocate
		I = (unsigned long long *) malloc(sizeof(unsigned long long) * n--);

		// Initialize
		I[n] = b;

		// Calculate
		for(;;) {
			if(I[n]) {
				--I[n];
				if (n)
					I[--n] = r, r = 1;
				else
					r *= a;
			} else
				for(;;)
					if (n == N)
						goto a;
					else if(I[++n])
						break;
		}
a:

		// Free
		free(I);
	}

	return r;
}

unsigned long long H(unsigned int n, unsigned long long a, unsigned long long b) {
	// Hyperfunction (Iterative-O(1) Hybrid): Perform a Hyperoperation...

	switch(n) {
		case 0:
			// Increment
			return ++b;
		case 1:
			// Addition
			return a + b;
		case 2:
			// Multiplication
			return a * b;
	}

	return HI(n, a, b);
}

unsigned long long APH(unsigned int m, unsigned int n) {
	// Ackermann-Péter Function (Recursive-Iterative-O(1) Hybrid)
	return H(m, 2, n + 3) - 3;
}

unsigned long long * p = 0;

unsigned long long APIA(unsigned int m, unsigned int n) {
	if (!m) return ++n;

	// Initialize
	unsigned long long *I, r = p ? p[m] : APIA(m - 1, 1);
	unsigned int M = m;

	if (n) {
		// Allocate
		I = (unsigned long long *) malloc(sizeof(unsigned long long) * (m + 1));

		// Initialize
		I[m] = n;

		// Calculate
		for(;;) {
			if(I[m]) {
				if (m)
					--I[m], I[--m] = r, r = p ? p[m] : APIA(m - 1, 1);
				else
					r += I[m], I[m] = 0;
			} else
				for(;;)
					if (m == M)
						goto a;
					else if(I[++m])
						break;
		}
a:

		// Free
		free(I);
	}

	return r;
}

unsigned long long API(unsigned int m, unsigned int n) {
	unsigned long long r = 0;

	// Allocate
	p = (unsigned long long *) malloc(sizeof(unsigned long long) * (m + 1));

	// Initialize
	for(; r <= m; ++r)
		p[r] = r ? APIA(r - 1, 1) : APIA(r, 0);

	// Calculate
	r = APIA(m, n);

	// Free
	free(p);

	return r;
}

unsigned long long AP(unsigned int m, unsigned int n) {
	return APH(m, n);
	return API(m, n);
}

int main(int n, char ** a) {
	unsigned int M, N;

	if (n != 3) {
		printf("Usage: %s <m> <n>\n", *a);
		return 1;
	}

	printf("AckermannPeter(%u, %u) = %llu\n", M = atoi(a[1]), N = atoi(a[2]), AP(M, N));

	//printf("\nPress any key...");
	//getch();
	return 0;
}

A few further tweaks/optimizations may be possible.

C#

Basic Version

using System;
class Program
{
    public static long Ackermann(long m, long n)
    {
        if(m > 0)
        {
            if (n > 0)
                return Ackermann(m - 1, Ackermann(m, n - 1));
            else if (n == 0)
                return Ackermann(m - 1, 1);
        }
        else if(m == 0)
        {
            if(n >= 0) 
                return n + 1;
        }

        throw new System.ArgumentOutOfRangeException();
    }
    
    static void Main()
    {
        for (long m = 0; m <= 3; ++m)
        {
            for (long n = 0; n <= 4; ++n)
            {
                Console.WriteLine("Ackermann({0}, {1}) = {2}", m, n, Ackermann(m, n));
            }
        }
    }
}
Output:
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125

Efficient Version

using System;
using System.Numerics;
using System.IO;
using System.Diagnostics;

namespace Ackermann_Function
{
    class Program
    {
        static void Main(string[] args)
        {
            int _m = 0;
            int _n = 0;
            Console.Write("m = ");
            try
            {
                _m = Convert.ToInt32(Console.ReadLine());
            }
            catch (Exception)
            {
                Console.WriteLine("Please enter a number.");
            }
            Console.Write("n = ");
            try
            {
                _n = Convert.ToInt32(Console.ReadLine());
            }
            catch (Exception)
            {
                Console.WriteLine("Please enter a number.");
            }
            //for (long m = 0; m <= 10; ++m)
            //{
            //    for (long n = 0; n <= 10; ++n)
            //    {
            //        DateTime now = DateTime.Now;
            //        Console.WriteLine("Ackermann({0}, {1}) = {2}", m, n, Ackermann(m, n));
            //        Console.WriteLine("Time taken:{0}", DateTime.Now - now);
            //    }
            //}

            DateTime now = DateTime.Now;
            Console.WriteLine("Ackermann({0}, {1}) = {2}", _m, _n, Ackermann(_m, _n));
            Console.WriteLine("Time taken:{0}", DateTime.Now - now);
            File.WriteAllText("number.txt", Ackermann(_m, _n).ToString());
            Process.Start("number.txt");
            Console.ReadKey();
        }
        public class OverflowlessStack<T>
        {
            internal sealed class SinglyLinkedNode
            {
                private const int ArraySize = 2048;
                T[] _array;
                int _size;
                public SinglyLinkedNode Next;
                public SinglyLinkedNode()
                {
                    _array = new T[ArraySize];
                }
                public bool IsEmpty { get { return _size == 0; } }
                public SinglyLinkedNode Push(T item)
                {
                    if (_size == ArraySize - 1)
                    {
                        SinglyLinkedNode n = new SinglyLinkedNode();
                        n.Next = this;
                        n.Push(item);
                        return n;
                    }
                    _array[_size++] = item;
                    return this;
                }
                public T Pop()
                {
                    return _array[--_size];
                }
            }
            private SinglyLinkedNode _head = new SinglyLinkedNode();

            public T Pop()
            {
                T ret = _head.Pop();
                if (_head.IsEmpty && _head.Next != null)
                    _head = _head.Next;
                return ret;
            }
            public void Push(T item)
            {
                _head = _head.Push(item);
            }
            public bool IsEmpty
            {
                get { return _head.Next == null && _head.IsEmpty; }
            }
        }
        public static BigInteger Ackermann(BigInteger m, BigInteger n)
        {
            var stack = new OverflowlessStack<BigInteger>();
            stack.Push(m);
            while (!stack.IsEmpty)
            {
                m = stack.Pop();
            skipStack:
                if (m == 0)
                    n = n + 1;
                else if (m == 1)
                    n = n + 2;
                else if (m == 2)
                    n = n * 2 + 3;
                else if (n == 0)
                {
                    --m;
                    n = 1;
                    goto skipStack;
                }
                else
                {
                    stack.Push(m - 1);
                    --n;
                    goto skipStack;
                }
            }
            return n;
        }
    }
}

Possibly the most efficient implementation of Ackermann in C#. It successfully runs Ack(4,2) when executed in Visual Studio. Don't forget to add a reference to System.Numerics.

C++

Basic version

#include <iostream>

unsigned int ackermann(unsigned int m, unsigned int n) {
  if (m == 0) {
    return n + 1;
  }
  if (n == 0) {
    return ackermann(m - 1, 1);
  }
  return ackermann(m - 1, ackermann(m, n - 1));
}

int main() {
  for (unsigned int m = 0; m < 4; ++m) {
    for (unsigned int n = 0; n < 10; ++n) {
      std::cout << "A(" << m << ", " << n << ") = " << ackermann(m, n) << "\n";
    }
  }
}

Efficient version

Translation of: D

C++11 with boost's big integer type. Compile with:

g++ -std=c++11 -I /path/to/boost ackermann.cpp.
#include <iostream>
#include <sstream>
#include <string>
#include <boost/multiprecision/cpp_int.hpp>

using big_int = boost::multiprecision::cpp_int;

big_int ipow(big_int base, big_int exp) {
  big_int result(1);
  while (exp) {
    if (exp & 1) {
      result *= base;
    }
    exp >>= 1;
    base *= base;
  }
  return result;
}

big_int ackermann(unsigned m, unsigned n) {
  static big_int (*ack)(unsigned, big_int) =
      [](unsigned m, big_int n)->big_int {
    switch (m) {
    case 0:
      return n + 1;
    case 1:
      return n + 2;
    case 2:
      return 3 + 2 * n;
    case 3:
      return 5 + 8 * (ipow(big_int(2), n) - 1);
    default:
      return n == 0 ? ack(m - 1, big_int(1)) : ack(m - 1, ack(m, n - 1));
    }
  };
  return ack(m, big_int(n));
}

int main() {
  for (unsigned m = 0; m < 4; ++m) {
    for (unsigned n = 0; n < 10; ++n) {
      std::cout << "A(" << m << ", " << n << ") = " << ackermann(m, n) << "\n";
    }
  }

  std::cout << "A(4, 1) = " << ackermann(4, 1) << "\n";

  std::stringstream ss;
  ss << ackermann(4, 2);
  auto text = ss.str();
  std::cout << "A(4, 2) = (" << text.length() << " digits)\n"
            << text.substr(0, 80) << "\n...\n"
            << text.substr(text.length() - 80) << "\n";
}
<pre>
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(0, 6) = 7
A(0, 7) = 8
A(0, 8) = 9
A(0, 9) = 10
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(1, 5) = 7
A(1, 6) = 8
A(1, 7) = 9
A(1, 8) = 10
A(1, 9) = 11
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(2, 4) = 11
A(2, 5) = 13
A(2, 6) = 15
A(2, 7) = 17
A(2, 8) = 19
A(2, 9) = 21
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(3, 3) = 61
A(3, 4) = 125
A(3, 5) = 253
A(3, 6) = 509
A(3, 7) = 1021
A(3, 8) = 2045
A(3, 9) = 4093
A(4, 1) = 65533
A(4, 2) = (19729 digits)
2003529930406846464979072351560255750447825475569751419265016973710894059556311
...
4717124577965048175856395072895337539755822087777506072339445587895905719156733

Chapel

proc A(m:int, n:int):int {
        if m == 0 then
                return n + 1;
        else if n == 0 then
                return A(m - 1, 1);
        else
                return A(m - 1, A(m, n - 1));
}

Clay

ackermann(m, n) {
    if(m == 0)
      return n + 1;
    if(n == 0)
      return ackermann(m - 1, 1);

    return ackermann(m - 1, ackermann(m, n - 1));
}

CLIPS

Functional solution

(deffunction ackerman
  (?m ?n)
  (if (= 0 ?m)
    then (+ ?n 1)
    else (if (= 0 ?n)
      then (ackerman (- ?m 1) 1)
      else (ackerman (- ?m 1) (ackerman ?m (- ?n 1)))
    )
  )
)
Example usage:
CLIPS> (ackerman 0 4)
5
CLIPS> (ackerman 1 4)
6
CLIPS> (ackerman 2 4)
11
CLIPS> (ackerman 3 4)
125

Fact-based solution

(deffacts solve-items
  (solve 0 4)
  (solve 1 4)
  (solve 2 4)
  (solve 3 4)
)

(defrule acker-m-0
  ?compute <- (compute 0 ?n)
  =>
  (retract ?compute)
  (assert (ackerman 0 ?n (+ ?n 1)))
)

(defrule acker-n-0-pre
  (compute ?m&:(> ?m 0) 0)
  (not (ackerman =(- ?m 1) 1 ?))
  =>
  (assert (compute (- ?m 1) 1))
)

(defrule acker-n-0
  ?compute <- (compute ?m&:(> ?m 0) 0)
  (ackerman =(- ?m 1) 1 ?val)
  =>
  (retract ?compute)
  (assert (ackerman ?m 0 ?val))
)

(defrule acker-m-n-pre-1
  (compute ?m&:(> ?m 0) ?n&:(> ?n 0))
  (not (ackerman ?m =(- ?n 1) ?))
  =>
  (assert (compute ?m (- ?n 1)))
)

(defrule acker-m-n-pre-2
  (compute ?m&:(> ?m 0) ?n&:(> ?n 0))
  (ackerman ?m =(- ?n 1) ?newn)
  (not (ackerman =(- ?m 1) ?newn ?))
  =>
  (assert (compute (- ?m 1) ?newn))
)

(defrule acker-m-n
  ?compute <- (compute ?m&:(> ?m 0) ?n&:(> ?n 0))
  (ackerman ?m =(- ?n 1) ?newn)
  (ackerman =(- ?m 1) ?newn ?val)
  =>
  (retract ?compute)
  (assert (ackerman ?m ?n ?val))
)

(defrule acker-solve
  (solve ?m ?n)
  (not (compute ?m ?n))
  (not (ackerman ?m ?n ?))
  =>
  (assert (compute ?m ?n))
)

(defrule acker-solved
  ?solve <- (solve ?m ?n)
  (ackerman ?m ?n ?result)
  =>
  (retract ?solve)
  (printout t "A(" ?m "," ?n ") = " ?result crlf)
)

When invoked, each required A(m,n) needed to solve the requested (solve ?m ?n) facts gets generated as its own fact. Below shows the invocation of the above, as well as an excerpt of the final facts list. Regardless of how many input (solve ?m ?n) requests are made, each possible A(m,n) is only solved once.

CLIPS> (reset)
CLIPS> (facts)
f-0     (initial-fact)
f-1     (solve 0 4)
f-2     (solve 1 4)
f-3     (solve 2 4)
f-4     (solve 3 4)
For a total of 5 facts.
CLIPS> (run)
A(3,4) = 125
A(2,4) = 11
A(1,4) = 6
A(0,4) = 5
CLIPS> (facts)
f-0     (initial-fact)
f-15    (ackerman 0 1 2)
f-16    (ackerman 1 0 2)
f-18    (ackerman 0 2 3)
...
f-632   (ackerman 1 123 125)
f-633   (ackerman 2 61 125)
f-634   (ackerman 3 4 125)
For a total of 316 facts.
CLIPS>

Clojure

(defn ackermann [m n] 
  (cond (zero? m) (inc n)
        (zero? n) (ackermann (dec m) 1)
        :else (ackermann (dec m) (ackermann m (dec n)))))

CLU

% Ackermann function
ack = proc (m, n: int) returns (int)
    if     m=0 then return(n+1)
    elseif n=0 then return(ack(m-1, 1))
    else            return(ack(m-1, ack(m, n-1)))
    end
end ack

% Print a table of ack( 0..3, 0..8 )
start_up = proc ()
    po: stream := stream$primary_output()
    
    for m: int in int$from_to(0, 3) do
        for n: int in int$from_to(0, 8) do
            stream$putright(po, int$unparse(ack(m,n)), 8)
        end
        stream$putl(po, "")
    end
end start_up
Output:
       1       2       3       4       5       6       7       8       9
       2       3       4       5       6       7       8       9      10
       3       5       7       9      11      13      15      17      19
       5      13      29      61     125     253     509    1021    2045

COBOL

       IDENTIFICATION DIVISION.
       PROGRAM-ID. Ackermann.

       DATA DIVISION.
       LINKAGE SECTION.
       01  M          USAGE UNSIGNED-LONG.
       01  N          USAGE UNSIGNED-LONG.

       01  Return-Val USAGE UNSIGNED-LONG.

       PROCEDURE DIVISION USING M N Return-Val.
           EVALUATE M ALSO N
               WHEN 0 ALSO ANY
                   ADD 1 TO N GIVING Return-Val

               WHEN NOT 0 ALSO 0
                   SUBTRACT 1 FROM M
                   CALL "Ackermann" USING BY CONTENT M BY CONTENT 1
                       BY REFERENCE Return-Val

               WHEN NOT 0 ALSO NOT 0
                   SUBTRACT 1 FROM N
                   CALL "Ackermann" USING BY CONTENT M BY CONTENT N
                       BY REFERENCE Return-Val
                       
                   SUBTRACT 1 FROM M
                   CALL "Ackermann" USING BY CONTENT M
                       BY CONTENT Return-Val BY REFERENCE Return-Val
           END-EVALUATE

           GOBACK
           .

CoffeeScript

ackermann = (m, n) ->
  if m is 0 then n + 1
  else if m > 0 and n is 0 then ackermann m - 1, 1
  else ackermann m - 1, ackermann m, n - 1

Comal

0010 //
0020 // Ackermann function
0030 //
0040 FUNC a#(m#,n#)
0050   IF m#=0 THEN RETURN n#+1
0060   IF n#=0 THEN RETURN a#(m#-1,1)
0070   RETURN a#(m#-1,a#(m#,n#-1))
0080 ENDFUNC a#
0090 //
0100 // Print table of Ackermann values
0110 //
0120 ZONE 5
0130 FOR m#:=0 TO 3 DO
0140   FOR n#:=0 TO 4 DO PRINT a#(m#,n#),
0150   PRINT
0160 ENDFOR m#
0170 END
Output:
1    2    3    4    5
2    3    4    5    6
3    5    7    9    11
5    13   29   61   125

Common Lisp

(defun ackermann (m n)
  (cond ((zerop m) (1+ n))
        ((zerop n) (ackermann (1- m) 1))
        (t         (ackermann (1- m) (ackermann m (1- n))))))

More elaborately:

(defun ackermann (m n)
  (case m ((0) (1+ n))
    ((1) (+ 2 n))
    ((2) (+ n n 3))
    ((3) (- (expt 2 (+ 3 n)) 3))
    (otherwise (ackermann (1- m) (if (zerop n) 1 (ackermann m (1- n)))))))

(loop for m from 0 to 4 do
      (loop for n from (- 5 m) to (- 6 m) do
	    (format t "A(~d, ~d) = ~d~%" m n (ackermann m n))))
Output:
A(0, 5) = 6

A(0, 6) = 7 A(1, 4) = 6 A(1, 5) = 7 A(2, 3) = 9 A(2, 4) = 11 A(3, 2) = 29 A(3, 3) = 61 A(4, 1) = 65533

A(4, 2) = 2003529930 <... skipping a few digits ...> 56733

Component Pascal

BlackBox Component Builder

MODULE NpctAckerman;
 
IMPORT  StdLog;
 
VAR     
	m,n: INTEGER;
 
PROCEDURE Ackerman (x,y: INTEGER):INTEGER;
 
BEGIN
  IF    x = 0  THEN  RETURN  y + 1
  ELSIF y = 0  THEN  RETURN  Ackerman (x - 1 , 1)
  ELSE
    RETURN  Ackerman (x - 1 , Ackerman (x , y - 1))
  END
END Ackerman;
 
PROCEDURE Do*;
BEGIN
  FOR  m := 0  TO  3  DO
    FOR  n := 0  TO  6  DO
      StdLog.Int (Ackerman (m, n));StdLog.Char (' ')
    END;
    StdLog.Ln
  END;
  StdLog.Ln
END Do;

END NpctAckerman.

Execute: ^Q NpctAckerman.Do

<pre>
 1  2  3  4  5  6  7 
 2  3  4  5  6  7  8 
 3  5  7  9  11  13  15 
 5  13  29  61  125  253  509 

Coq

Require Import Arith.
Fixpoint A m := fix A_m n :=
  match m with
    | 0 => n + 1
    | S pm =>
      match n with
        | 0 => A pm 1
        | S pn => A pm (A_m pn)
      end
  end.
Require Import Utf8.

Section FOLD.
  Context {A: Type} (f: A  A) (a: A).
  Fixpoint fold (n: nat) : A :=
    match n with
    | O => a
    | S n' => f (fold n')
    end.
End FOLD.

Definition ackermann : nat  nat  nat :=
  fold  g, fold g (g (S O))) S.

Crystal

Translation of: Ruby
def ack(m, n)
  if m == 0
    n + 1
  elsif n == 0
    ack(m-1, 1)
  else
    ack(m-1, ack(m, n-1))
  end
end

#Example:
(0..3).each do |m|
  puts (0..6).map { |n| ack(m, n) }.join(' ')
end
Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509

D

Basic version

ulong ackermann(in ulong m, in ulong n) pure nothrow @nogc {
    if (m == 0)
        return n + 1;
    if (n == 0)
        return ackermann(m - 1, 1);
    return ackermann(m - 1, ackermann(m, n - 1));
}

void main() {
    assert(ackermann(2, 4) == 11);
}

More Efficient Version

Translation of: Mathematica
import std.stdio, std.bigint, std.conv;

BigInt ipow(BigInt base, BigInt exp) pure nothrow {
    auto result = 1.BigInt;
    while (exp) {
        if (exp & 1)
            result *= base;
        exp >>= 1;
        base *= base;
    }

    return result;
}

BigInt ackermann(in uint m, in uint n) pure nothrow
out(result) {
    assert(result >= 0);
} body {
    static BigInt ack(in uint m, in BigInt n) pure nothrow {
        switch (m) {
            case 0: return n + 1;
            case 1: return n + 2;
            case 2: return 3 + 2 * n;
            //case 3: return 5 + 8 * (2 ^^ n - 1);
            case 3: return 5 + 8 * (ipow(2.BigInt, n) - 1);
            default: return (n == 0) ?
                        ack(m - 1, 1.BigInt) :
                        ack(m - 1, ack(m, n - 1));
        }
    }

    return ack(m, n.BigInt);
}

void main() {
    foreach (immutable m; 1 .. 4)
        foreach (immutable n; 1 .. 9)
            writefln("ackermann(%d, %d): %s", m, n, ackermann(m, n));
    writefln("ackermann(4, 1): %s", ackermann(4, 1));

    immutable a = ackermann(4, 2).text;
    writefln("ackermann(4, 2)) (%d digits):\n%s...\n%s",
             a.length, a[0 .. 94], a[$ - 96 .. $]);
}
Output:
ackermann(1, 1): 3
ackermann(1, 2): 4
ackermann(1, 3): 5
ackermann(1, 4): 6
ackermann(1, 5): 7
ackermann(1, 6): 8
ackermann(1, 7): 9
ackermann(1, 8): 10
ackermann(2, 1): 5
ackermann(2, 2): 7
ackermann(2, 3): 9
ackermann(2, 4): 11
ackermann(2, 5): 13
ackermann(2, 6): 15
ackermann(2, 7): 17
ackermann(2, 8): 19
ackermann(3, 1): 13
ackermann(3, 2): 29
ackermann(3, 3): 61
ackermann(3, 4): 125
ackermann(3, 5): 253
ackermann(3, 6): 509
ackermann(3, 7): 1021
ackermann(3, 8): 2045
ackermann(4, 1): 65533
ackermann(4, 2)) (19729 digits):
2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880...
699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156733

Dart

no caching, the implementation takes ages even for A(4,1)

int A(int m, int n) => m==0 ? n+1 : n==0 ? A(m-1,1) : A(m-1,A(m,n-1));

main() {
  print(A(0,0));
  print(A(1,0));
  print(A(0,1));
  print(A(2,2));
  print(A(2,3));
  print(A(3,3));
  print(A(3,4));
  print(A(3,5));
  print(A(4,0));
}

Dc

This needs a modern Dc with r (swap) and # (comment). It easily can be adapted to an older Dc, but it will impact readability a lot.

[               # todo: n 0 -- n+1 and break 2 levels
  + 1 +         # n+1
  q
] s1

[               # todo: m 0 -- A(m-1,1) and break 2 levels
  + 1 -         # m-1
  1             # m-1 1
  lA x          # A(m-1,1)
  q
] s2

[               # todo: m n -- A(m,n)
  r d 0=1       # n m(!=0)
  r d 0=2       # m(!=0) n(!=0)
  Sn            # m(!=0)
  d 1 - r       # m-1 m
  Ln 1 -        # m-1 m n-1
  lA x          # m-1 A(m,n-1)
  lA x          # A(m-1,A(m,n-1))
] sA

3 9 lA x f
Output:
4093

Delphi

function Ackermann(m,n:Int64):Int64;
begin
    if m = 0 then
        Result := n + 1
    else if n = 0 then
        Result := Ackermann(m-1, 1)
    else
        Result := Ackermann(m-1, Ackermann(m, n - 1));
end;

Draco

/* Ackermann function */
proc ack(word m, n) word:
    if   m=0 then n+1
    elif n=0 then ack(m-1, 1)
    else          ack(m-1, ack(m, n-1))
    fi
corp;

/* Write a table of Ackermann values */
proc nonrec main() void:
    byte m, n;
    for m from 0 upto 3 do
        for n from 0 upto 8 do
            write(ack(m,n) : 5)
        od;
        writeln()
    od
corp
Output:
    1    2    3    4    5    6    7    8    9
    2    3    4    5    6    7    8    9   10
    3    5    7    9   11   13   15   17   19
    5   13   29   61  125  253  509 1021 2045

DWScript

function Ackermann(m, n : Integer) : Integer;
begin
    if m = 0 then
        Result := n+1
    else if n = 0 then
        Result := Ackermann(m-1, 1)
    else Result := Ackermann(m-1, Ackermann(m, n-1));
end;

Dylan

define method ack(m == 0, n :: <integer>)
   n + 1
end;
define method ack(m :: <integer>, n :: <integer>)
   ack(m - 1, if (n == 0) 1 else ack(m, n - 1) end)
end;

E

def A(m, n) {
    return if (m <=> 0)          { n+1              } \
      else if (m > 0 && n <=> 0) { A(m-1, 1)        } \
      else                       { A(m-1, A(m,n-1)) }
}

EasyLang

func ackerm m n . r .
  if m = 0
    r = n + 1
  elif n = 0
    call ackerm m - 1 1 r
  else
    call ackerm m n - 1 h
    call ackerm m - 1 h r
  .
.
call ackerm 3 6 r
print r

Egel

def ackermann =
    [ 0 N -> N + 1
    | M 0 -> ackermann (M - 1) 1
    | M N -> ackermann (M - 1) (ackermann M (N - 1)) ]

Eiffel

Example code

Test of Example code

class
	ACKERMAN_EXAMPLE

feature -- Basic Operations

	ackerman (m, n: NATURAL): NATURAL
			-- Recursively compute the n-th term of a series.
		require
			non_negative_m: m >= 0
			non_negative_n: n >= 0
		do
			if m = 0 then
				Result := n + 1
			elseif m > 0 and n = 0 then
				Result := ackerman (m - 1, 1)
			elseif m > 0 and n > 0 then
				Result := ackerman (m - 1, ackerman (m, n - 1))
			else
				check invalid_arg_state: False end
			end
		end

end

Ela

ack 0 n = n+1
ack m 0 = ack (m - 1) 1
ack m n = ack (m - 1) <| ack m <| n - 1

Elena

ELENA 4.x :

import extensions;

ackermann(m,n)
{
    if(n < 0 || m < 0)
    {
        InvalidArgumentException.raise()
    };
    
    m =>
       0 { ^n + 1 }
       : {
              n => 
                0 { ^ackermann(m - 1,1) }
                : { ^ackermann(m - 1,ackermann(m,n-1)) }
           }
}

public program()
{
    for(int i:=0, i <= 3, i += 1)
    {
        for(int j := 0, j <= 5, j += 1)
        {
            console.printLine("A(",i,",",j,")=",ackermann(i,j))
        }
    };

    console.readChar()
}
Output:
A(0,0)=1
A(0,1)=2
A(0,2)=3
A(0,3)=4
A(0,4)=5
A(0,5)=6
A(1,0)=2
A(1,1)=3
A(1,2)=4
A(1,3)=5
A(1,4)=6
A(1,5)=7
A(2,0)=3
A(2,1)=5
A(2,2)=7
A(2,3)=9
A(2,4)=11
A(2,5)=13
A(3,0)=5
A(3,1)=13
A(3,2)=29
A(3,3)=61
A(3,4)=125
A(3,5)=253

Elixir

defmodule Ackermann do
  def ack(0, n), do: n + 1 
  def ack(m, 0), do: ack(m - 1, 1)
  def ack(m, n), do: ack(m - 1, ack(m, n - 1))
end

Enum.each(0..3, fn m ->
  IO.puts Enum.map_join(0..6, " ", fn n -> Ackermann.ack(m, n) end)
end)
Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509

Emacs Lisp

(defun ackermann (m n)
  (cond ((zerop m) (1+ n))
	((zerop n) (ackermann (1- m) 1))
	(t         (ackermann (1- m)
			      (ackermann m (1- n))))))

Erlang

-module(ackermann).
-export([ackermann/2]).

ackermann(0, N) -> 
  N+1;
ackermann(M, 0) -> 
  ackermann(M-1, 1);
ackermann(M, N) when M > 0 andalso N > 0 ->
  ackermann(M-1, ackermann(M, N-1)).

ERRE

Iterative version, using a stack. First version used various GOTOs statement, now removed and substituted with the new ERRE control statements.

PROGRAM ACKERMAN

!
! computes Ackermann function
! (second version for rosettacode.org)
!

!$INTEGER

DIM STACK[10000]

!$INCLUDE="PC.LIB"

PROCEDURE ACK(M,N->N)
  LOOP
    CURSOR_SAVE(->CURX%,CURY%) 
    LOCATE(8,1)
    PRINT("Livello Stack:";S;"  ")
    LOCATE(CURY%,CURX%)
    IF M<>0 THEN
       IF N<>0 THEN
           STACK[S]=M
           S+=1
           N-=1
        ELSE
           M-=1
           N+=1
        END IF
        CONTINUE LOOP
     ELSE
        N+=1
        S-=1
    END IF
    IF S<>0 THEN
        M=STACK[S]
        M-=1
        CONTINUE LOOP
      ELSE
        EXIT PROCEDURE
    END IF
  END LOOP
END PROCEDURE

BEGIN
   PRINT(CHR$(12);)
   FOR X=0 TO 3 DO
     FOR Y=0 TO 9 DO
        S=1
        ACK(X,Y->ANS)
        PRINT(ANS;)
     END FOR
     PRINT
   END FOR
END PROGRAM

Prints a list of Ackermann function values: from A(0,0) to A(3,9). Uses a stack to avoid overflow. Formating options to make this pretty are available, but for this example only basic output is used.

 1  2  3  4  5  6  7  8  9  10
 2  3  4  5  6  7  8  9  10  11
 3  5  7  9  11  13  15  17  19  21
 5  13  29  61  125  253  509  1021  2045  4093


Stack Level: 1

Euler Math Toolbox

>M=zeros(1000,1000);
>function map A(m,n) ...
$  global M;
$  if m==0 then return n+1; endif;
$  if n==0 then return A(m-1,1); endif;
$  if m<=cols(M) and n<=cols(M) then
$    M[m,n]=A(m-1,A(m,n-1));
$    return M[m,n];
$  else return A(m-1,A(m,n-1));
$  endif;
$endfunction
>shortestformat; A((0:3)',0:5)
         1         2         3         4         5         6 
         2         3         4         5         6         7 
         3         5         7         9        11        13 
         5        13        29        61       125       253

Euphoria

This is based on the VBScript example.

function ack(atom m, atom n)
    if m = 0 then 
        return n + 1
    elsif m > 0 and n = 0 then
        return ack(m - 1, 1)
    else
        return ack(m - 1, ack(m, n - 1))
    end if
end function

for i = 0 to 3 do
    for j = 0 to 6 do
        printf( 1, "%5d", ack( i, j ) )
    end for
    puts( 1, "\n" )
end for

Ezhil

நிரல்பாகம் அகெர்மன்(முதலெண், இரண்டாமெண்)

  @((முதலெண் < 0) || (இரண்டாமெண் < 0)) ஆனால்

    பின்கொடு -1

  முடி

  @(முதலெண் == 0) ஆனால்

    பின்கொடு இரண்டாமெண்+1

  முடி

  @((முதலெண் > 0) && (இரண்டாமெண் == 00)) ஆனால்

    பின்கொடு அகெர்மன்(முதலெண் - 1, 1)

  முடி

  பின்கொடு அகெர்மன்(முதலெண் - 1, அகெர்மன்(முதலெண், இரண்டாமெண் - 1))

முடி

 = int(உள்ளீடு("ஓர் எண்ணைத் தாருங்கள், அது பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாக இருக்கலாம்: "))
 = int(உள்ளீடு("அதேபோல் இன்னோர் எண்ணைத் தாருங்கள், இதுவும் பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாகவோ இருக்கலாம்: "))

விடை = அகெர்மன்(, )

@(விடை < 0) ஆனால்

  பதிப்பி "தவறான எண்களைத் தந்துள்ளீர்கள்!"

இல்லை

  பதிப்பி "நீங்கள் தந்த எண்களுக்கான அகர்மென் மதிப்பு: ", விடை

முடி

F#

The following program implements the Ackermann function in F# but is not tail-recursive and so runs out of stack space quite fast.

let rec ackermann m n = 
    match m, n with
    | 0, n -> n + 1
    | m, 0 -> ackermann (m - 1) 1
    | m, n -> ackermann (m - 1) ackermann m (n - 1)

do
    printfn "%A" (ackermann (int fsi.CommandLineArgs.[1]) (int fsi.CommandLineArgs.[2]))

Transforming this into continuation passing style avoids limited stack space by permitting tail-recursion.

let ackermann M N =
    let rec acker (m, n, k) =
        match m,n with
            | 0, n -> k(n + 1)
            | m, 0 -> acker ((m - 1), 1, k)
            | m, n -> acker (m, (n - 1), (fun x -> acker ((m - 1), x, k)))
    acker (M, N, (fun x -> x))

Factor

USING: kernel math locals combinators ;
IN: ackermann

:: ackermann ( m n -- u ) 
    { 
        { [ m 0 = ] [ n 1 + ] } 
        { [ n 0 = ] [ m 1 - 1 ackermann ] } 
        [ m 1 - m n 1 - ackermann ackermann ] 
    } cond ;

Falcon

function ackermann( m, n )
 if m == 0:  return( n + 1 )
 if n == 0:  return( ackermann( m - 1, 1 ) )
 return( ackermann( m - 1, ackermann( m, n - 1 ) ) )
end

for M in [ 0:4 ]
 for N in [ 0:7 ]
   >> ackermann( M, N ), " "
 end
 >
end

The above will output the below. Formating options to make this pretty are available, but for this example only basic output is used.

1 2 3 4 5 6 7 
2 3 4 5 6 7 8 
3 5 7 9 11 13 15 
5 13 29 61 125 253 509 

FALSE

[$$[%
  \$$[%
     1-\$@@a;!  { i j -> A(i-1, A(i, j-1)) }
  1]?0=[
     %1         { i 0 -> A(i-1, 1) }
   ]?
  \1-a;!
1]?0=[
  %1+           { 0 j -> j+1 }
 ]?]a: { j i }

3 3 a;! .  { 61 }

Fantom

class Main
{
  // assuming m,n are positive
  static Int ackermann (Int m, Int n)
  {
    if (m == 0)
      return n + 1
    else if (n == 0)
      return ackermann (m - 1, 1)
    else
      return ackermann (m - 1, ackermann (m, n - 1))
  }

  public static Void main ()
  {
    (0..3).each |m|
    {
      (0..6).each |n|
      {
        echo ("Ackerman($m, $n) = ${ackermann(m, n)}")
      }
    }
  }
}
Output:
Ackerman(0, 0) = 1
Ackerman(0, 1) = 2
Ackerman(0, 2) = 3
Ackerman(0, 3) = 4
Ackerman(0, 4) = 5
Ackerman(0, 5) = 6
Ackerman(0, 6) = 7
Ackerman(1, 0) = 2
Ackerman(1, 1) = 3
Ackerman(1, 2) = 4
Ackerman(1, 3) = 5
Ackerman(1, 4) = 6
Ackerman(1, 5) = 7
Ackerman(1, 6) = 8
Ackerman(2, 0) = 3
Ackerman(2, 1) = 5
Ackerman(2, 2) = 7
Ackerman(2, 3) = 9
Ackerman(2, 4) = 11
Ackerman(2, 5) = 13
Ackerman(2, 6) = 15
Ackerman(3, 0) = 5
Ackerman(3, 1) = 13
Ackerman(3, 2) = 29
Ackerman(3, 3) = 61
Ackerman(3, 4) = 125
Ackerman(3, 5) = 253
Ackerman(3, 6) = 509

FBSL

Mixed-language solution using pure FBSL, Dynamic Assembler, and Dynamic C layers of FBSL v3.5 concurrently. The following is a single script; the breaks are caused by switching between RC's different syntax highlighting schemes:

#APPTYPE CONSOLE

TestAckermann()

PAUSE

SUB TestAckermann()
	FOR DIM m = 0 TO 3
		FOR DIM n = 0 TO 10
			PRINT AckermannF(m, n), " ";
		NEXT
		PRINT
	NEXT
END SUB

FUNCTION AckermannF(m AS INTEGER, n AS INTEGER) AS INTEGER
	IF NOT m THEN RETURN n + 1
	IF NOT n THEN RETURN AckermannA(m - 1, 1)
	RETURN AckermannC(m - 1, AckermannF(m, n - 1))
END FUNCTION

DYNC AckermannC(m AS INTEGER, n AS INTEGER) AS INTEGER
	int Ackermann(int m, int n)
	{
		if (!m) return n + 1;
		if (!n) return Ackermann(m - 1, 1);
		return Ackermann(m - 1, Ackermann(m, n - 1));
	}
	
	int main(int m, int n)
	{
		return Ackermann(m, n);
	}
END DYNC

DYNASM AckermannA(m AS INTEGER, n AS INTEGER) AS INTEGER
	ENTER 0, 0
	INVOKE Ackermann, m, n
	LEAVE
	RET
	
	@Ackermann
	ENTER 0, 0

	.IF DWORD PTR [m] .THEN
		JMP @F
	.ENDIF
	MOV EAX, n
	INC EAX
	JMP xit

	@@
	.IF DWORD PTR [n] .THEN
		JMP @F
	.ENDIF
	MOV EAX, m
	DEC EAX
	INVOKE Ackermann, EAX, 1
	JMP xit

	@@
	MOV EAX, n
	DEC EAX
	INVOKE Ackermann, m, EAX
	MOV ECX, m
	DEC ECX
	INVOKE Ackermann, ECX, EAX
	
	@xit
	LEAVE
	RET 8
END DYNASM
Output:
1 2 3 4 5 6 7 8 9 10 11
2 3 4 5 6 7 8 9 10 11 12
3 5 7 9 11 13 15 17 19 21 23
5 13 29 61 125 253 509 1021 2045 4093 8189

Press any key to continue...

Fermat

Func A(m,n) = if m = 0 then n+1 else if n = 0 then A(m-1,1) else A(m-1,A(m,n-1)) fi fi.;
A(3,8)
Output:
     2045

Forth

: acker ( m n -- u )
	over 0= IF  nip 1+ EXIT  THEN
	swap 1- swap ( m-1 n -- )
	dup  0= IF  1+  recurse EXIT  THEN
	1- over 1+ swap recurse recurse ;
Example of use:
FORTH> 0 0 acker . 1  ok
FORTH> 3 4 acker . 125  ok

An optimized version:

: ackermann                            ( m n -- u ) 
  over                                 ( case statement) 
  0 over = if drop nip 1+     else
  1 over = if drop nip 2 +    else
  2 over = if drop nip 2* 3 + else
  3 over = if drop swap 5 + swap lshift 3 - else
    drop swap 1- swap dup
    if
      1- over 1+ swap recurse recurse exit
    else
      1+ recurse exit                  \ allow tail recursion
    then
  then then then then
;

Fortran

Works with: Fortran version 90 and later
PROGRAM EXAMPLE  
  IMPLICIT NONE
 
  INTEGER :: i, j
 
  DO i = 0, 3
    DO j = 0, 6
       WRITE(*, "(I10)", ADVANCE="NO") Ackermann(i, j)
    END DO
    WRITE(*,*)
  END DO
 
CONTAINS
 
  RECURSIVE FUNCTION Ackermann(m, n) RESULT(ack)
    INTEGER :: ack, m, n

    IF (m == 0) THEN
      ack = n + 1
    ELSE IF (n == 0) THEN
      ack = Ackermann(m - 1, 1)
    ELSE
      ack = Ackermann(m - 1, Ackermann(m, n - 1))
    END IF
  END FUNCTION Ackermann

END PROGRAM EXAMPLE

Free Pascal

See #Delphi or #Pascal.

FreeBASIC

' version 28-10-2016
' compile with: fbc -s console
' to do A(4, 2) the stack size needs to be increased
' compile with: fbc -s console -t 2000

Function ackerman (m As Long, n As Long) As Long

    If m = 0 Then ackerman = n +1

    If m > 0 Then
        If n = 0 Then
            ackerman = ackerman(m -1, 1)
        Else
            If n > 0 Then
                ackerman = ackerman(m -1, ackerman(m, n -1))
            End If
        End If
    End If
End Function

' ------=< MAIN >=------

Dim As Long m, n
Print

For m = 0 To 4
    Print Using "###"; m;
    For n = 0 To 10
        ' A(4, 1) or higher will run out of stack memory (default 1M)
        ' change n = 1 to n = 2 to calculate A(4, 2), increase stack!
        If m = 4 And n = 1 Then Exit For 
        Print Using "######"; ackerman(m, n);
    Next
    Print
Next

' empty keyboard buffer
While InKey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
  0     1     2     3     4     5     6     7     8     9    10    11
  1     2     3     4     5     6     7     8     9    10    11    12
  2     3     5     7     9    11    13    15    17    19    21    23
  3     5    13    29    61   125   253   509  1021  2045  4093  8189
  4    13

FunL

def
  ackermann( 0, n ) = n + 1
  ackermann( m, 0 ) = ackermann( m - 1, 1 )
  ackermann( m, n ) = ackermann( m - 1, ackermann(m, n - 1) )

for m <- 0..3, n <- 0..4
  printf( 'Ackermann( %d, %d ) = %d\n', m, n, ackermann(m, n) )
Output:
Ackermann( 0, 0 ) = 1
Ackermann( 0, 1 ) = 2
Ackermann( 0, 2 ) = 3
Ackermann( 0, 3 ) = 4
Ackermann( 0, 4 ) = 5
Ackermann( 1, 0 ) = 2
Ackermann( 1, 1 ) = 3
Ackermann( 1, 2 ) = 4
Ackermann( 1, 3 ) = 5
Ackermann( 1, 4 ) = 6
Ackermann( 2, 0 ) = 3
Ackermann( 2, 1 ) = 5
Ackermann( 2, 2 ) = 7
Ackermann( 2, 3 ) = 9
Ackermann( 2, 4 ) = 11
Ackermann( 3, 0 ) = 5
Ackermann( 3, 1 ) = 13
Ackermann( 3, 2 ) = 29
Ackermann( 3, 3 ) = 61
Ackermann( 3, 4 ) = 125

Futhark

fun ackermann(m: int, n: int): int =
  if m == 0 then n + 1
  else if n == 0 then ackermann(m-1, 1)
  else ackermann(m - 1, ackermann(m, n-1))

FutureBasic

include "NSLog.incl"

local fn Ackerman( m as NSInteger, n as NSInteger ) as NSInteger
  NSInteger result
  
  select
    case m == 0 : result = ( n + 1 )
    case n == 0 : result = fn Ackerman( ( m - 1 ), 1 )
    case else   : result = fn Ackerman( ( m - 1 ), fn Ackerman( m, ( n - 1 ) ) )
  end select
end fn = result

NSInteger          m, n
CFMutableStringRef mutStr

mutStr = fn StringWithCapacity( 0 )

for m = 0 to 3
  for n = 0 to 9
    StringAppendString( mutStr, fn StringWithFormat( @"fn Ackerman( %ld, %ld ) = %ld\n", m, n, fn Ackerman( m, n ) ) )
  next
next

NSLog( @"%@", mutStr )

HandleEvents

Output:

fn Ackerman( 0, 0 ) = 1
fn Ackerman( 0, 1 ) = 2
fn Ackerman( 0, 2 ) = 3
fn Ackerman( 0, 3 ) = 4
fn Ackerman( 0, 4 ) = 5
fn Ackerman( 0, 5 ) = 6
fn Ackerman( 0, 6 ) = 7
fn Ackerman( 0, 7 ) = 8
fn Ackerman( 0, 8 ) = 9
fn Ackerman( 0, 9 ) = 10
fn Ackerman( 1, 0 ) = 2
fn Ackerman( 1, 1 ) = 3
fn Ackerman( 1, 2 ) = 4
fn Ackerman( 1, 3 ) = 5
fn Ackerman( 1, 4 ) = 6
fn Ackerman( 1, 5 ) = 7
fn Ackerman( 1, 6 ) = 8
fn Ackerman( 1, 7 ) = 9
fn Ackerman( 1, 8 ) = 10
fn Ackerman( 1, 9 ) = 11
fn Ackerman( 2, 0 ) = 3
fn Ackerman( 2, 1 ) = 5
fn Ackerman( 2, 2 ) = 7
fn Ackerman( 2, 3 ) = 9
fn Ackerman( 2, 4 ) = 11
fn Ackerman( 2, 5 ) = 13
fn Ackerman( 2, 6 ) = 15
fn Ackerman( 2, 7 ) = 17
fn Ackerman( 2, 8 ) = 19
fn Ackerman( 2, 9 ) = 21
fn Ackerman( 3, 0 ) = 5
fn Ackerman( 3, 1 ) = 13
fn Ackerman( 3, 2 ) = 29
fn Ackerman( 3, 3 ) = 61
fn Ackerman( 3, 4 ) = 125
fn Ackerman( 3, 5 ) = 253
fn Ackerman( 3, 6 ) = 509
fn Ackerman( 3, 7 ) = 1021
fn Ackerman( 3, 8 ) = 2045
fn Ackerman( 3, 9 ) = 4093

Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

In this page you can see the program(s) related to this task and their results.

Gambas

Public Function Ackermann(m As Float, n As Float) As Float
  If m = 0 Then
    Return n + 1
  End If
  If n = 0 Then
    Return Ackermann(m - 1, 1)
  End If
  Return Ackermann(m - 1, Ackermann(m, n - 1))
End

Public Sub Main()
  Dim m, n As Float
  For m = 0 To 3
    For n = 0 To 4
      Print "Ackermann("; m; ", "; n; ") = "; Ackermann(m, n)
    Next
  Next
End

GAP

ack := function(m, n)
  if m = 0 then
    return n + 1;
  elif (m > 0) and (n = 0) then
    return ack(m - 1, 1);
  elif (m > 0) and (n > 0) then
    return ack(m - 1, ack(m, n - 1));
  else
    return fail;
  fi;
end;

Genyris

def A (m n)
   cond
      (equal? m 0)
          + n 1
      (equal? n 0) 
          A (- m 1) 1
      else
          A (- m 1)
             A m (- n 1)

GML

Define a script resource named ackermann and paste this code inside:

///ackermann(m,n)
var m, n;
m = argument0;
n = argument1;
if(m=0)
    {
    return (n+1)
    }
else if(n == 0)
    {
    return (ackermann(m-1,1,1))
    }
else
    {
    return (ackermann(m-1,ackermann(m,n-1,2),1))
    }

gnuplot

A (m, n) = m == 0 ? n + 1 : n == 0 ? A (m - 1, 1) : A (m - 1, A (m, n - 1))
print A (0, 4)
print A (1, 4)
print A (2, 4)
print A (3, 4)
Output:
5
6
11
stack overflow

Go

Classic version

func Ackermann(m, n uint) uint {
	switch 0 {
	case m:
		return n + 1
	case n:
		return Ackermann(m - 1, 1)
	}
	return Ackermann(m - 1, Ackermann(m, n - 1))
}

Expanded version

func Ackermann2(m, n uint) uint {
  switch {
    case m == 0:
      return n + 1
    case m == 1:
      return n + 2
    case m == 2:
      return 2*n + 3
    case m == 3:
      return 8 << n - 3
    case n == 0:
      return Ackermann2(m - 1, 1)
  }
  return Ackermann2(m - 1, Ackermann2(m, n - 1))
}

Expanded version with arbitrary precision

package main

import (
	"fmt"
	"math/big"
	"math/bits" // Added in Go 1.9
)

var one = big.NewInt(1)
var two = big.NewInt(2)
var three = big.NewInt(3)
var eight = big.NewInt(8)

func Ackermann2(m, n *big.Int) *big.Int {
	if m.Cmp(three) <= 0 {
		switch m.Int64() {
		case 0:
			return new(big.Int).Add(n, one)
		case 1:
			return new(big.Int).Add(n, two)
		case 2:
			r := new(big.Int).Lsh(n, 1)
			return r.Add(r, three)
		case 3:
			if nb := n.BitLen(); nb > bits.UintSize {
				// n is too large to represent as a
				// uint for use in the Lsh method.
				panic(TooBigError(nb))

				// If we tried to continue anyway, doing
				// 8*2^n-3 as bellow, we'd use hundreds
				// of megabytes and lots of CPU time
				// without the Exp call even returning.
				r := new(big.Int).Exp(two, n, nil)
				r.Mul(eight, r)
				return r.Sub(r, three)
			}
			r := new(big.Int).Lsh(eight, uint(n.Int64()))
			return r.Sub(r, three)
		}
	}
	if n.BitLen() == 0 {
		return Ackermann2(new(big.Int).Sub(m, one), one)
	}
	return Ackermann2(new(big.Int).Sub(m, one),
		Ackermann2(m, new(big.Int).Sub(n, one)))
}

type TooBigError int

func (e TooBigError) Error() string {
	return fmt.Sprintf("A(m,n) had n of %d bits; too large", int(e))
}

func main() {
	show(0, 0)
	show(1, 2)
	show(2, 4)
	show(3, 100)
	show(3, 1e6)
	show(4, 1)
	show(4, 2)
	show(4, 3)
}

func show(m, n int64) {
	defer func() {
		// Ackermann2 could/should have returned an error
		// instead of a panic. But here's how to recover
		// from the panic, and report "expected" errors.
		if e := recover(); e != nil {
			if err, ok := e.(TooBigError); ok {
				fmt.Println("Error:", err)
			} else {
				panic(e)
			}
		}
	}()

	fmt.Printf("A(%d, %d) = ", m, n)
	a := Ackermann2(big.NewInt(m), big.NewInt(n))
	if a.BitLen() <= 256 {
		fmt.Println(a)
	} else {
		s := a.String()
		fmt.Printf("%d digits starting/ending with: %s...%s\n",
			len(s), s[:20], s[len(s)-20:],
		)
	}
}
Output:
A(0, 0) = 1
A(1, 2) = 4
A(2, 4) = 11
A(3, 100) = 10141204801825835211973625643005
A(3, 1000000) = 301031 digits starting/ending with: 79205249834367186005...39107225301976875005
A(4, 1) = 65533
A(4, 2) = 19729 digits starting/ending with: 20035299304068464649...45587895905719156733
A(4, 3) = Error: A(m,n) had n of 65536 bits; too large

Golfscript

{
  :_n; :_m;
  _m 0= {_n 1+}
        {_n 0= {_m 1- 1 ack}
               {_m 1- _m _n 1- ack ack}
               if}
        if
}:ack;

Groovy

def ack ( m, n ) {
    assert m >= 0 && n >= 0 : 'both arguments must be non-negative'
    m == 0 ? n + 1 : n == 0 ? ack(m-1, 1) : ack(m-1, ack(m, n-1))
}

Test program:

def ackMatrix = (0..3).collect { m -> (0..8).collect { n -> ack(m, n) } }
ackMatrix.each { it.each { elt -> printf "%7d", elt }; println() }
Output:
      1      2      3      4      5      6      7      8      9
      2      3      4      5      6      7      8      9     10
      3      5      7      9     11     13     15     17     19
      5     13     29     61    125    253    509   1021   2045

Note: In the default groovyConsole configuration for WinXP, "ack(4,1)" caused a stack overflow error!

Hare

use fmt;

fn ackermann(m: u64, n: u64) u64 = {
	if (m == 0) {
		return n + 1;
	};
	if (n == 0) {
		return ackermann(m - 1, 1);
	};
	return ackermann(m - 1, ackermann(m, n - 1));
};

export fn main() void = {
	for (let m = 0u64; m < 4; m += 1) {
		for (let n = 0u64; n < 10; n += 1) {
			fmt::printfln("A({}, {}) = {}", m, n, ackermann(m, n))!;
		};
		fmt::println()!;
	};
};

Haskell

ack :: Int -> Int -> Int
ack 0 n = succ n
ack m 0 = ack (pred m) 1
ack m n = ack (pred m) (ack m (pred n))

main :: IO ()
main = mapM_ print $ uncurry ack <$> [(0, 0), (3, 4)]
Output:
1
125

Generating a list instead:

import Data.List (mapAccumL)

-- everything here are [Int] or [[Int]], which would overflow
-- * had it not overrun the stack first *
ackermann = iterate ack [1..] where
	ack a = s where
		s = snd $ mapAccumL f (tail a) (1 : zipWith (-) s (1:s))
	f a b = (aa, head aa) where aa = drop b a

main = mapM_ print $ map (\n -> take (6 - n) $ ackermann !! n) [0..5]

Haxe

class RosettaDemo
{
    static public function main()
    {
        Sys.print(ackermann(3, 4));
    }

    static function ackermann(m : Int, n : Int)
    {
        if (m == 0)
        {
            return n + 1;
        }
        else if (n == 0)
        {
            return ackermann(m-1, 1);
        }
        return ackermann(m-1, ackermann(m, n-1));
    }
}

Hoon

|=  [m=@ud n=@ud]
?:  =(m 0)
  +(n)
?:  =(n 0)
  $(n 1, m (dec m))
$(m (dec m), n $(n (dec n)))

Icon and Unicon

Taken from the public domain Icon Programming Library's acker in memrfncs, written by Ralph E. Griswold.

procedure acker(i, j)
   static memory

   initial {
      memory := table()
      every memory[0 to 100] := table()
      }

   if i = 0 then return j + 1

   if j = 0 then /memory[i][j] := acker(i - 1, 1)
   else /memory[i][j] := acker(i - 1, acker(i, j - 1))

   return memory[i][j]

end

procedure main()
   every m := 0 to 3 do {
      every n := 0 to 8 do {
         writes(acker(m, n) || " ")
         }
      write()
      }
end
Output:
1 2 3 4 5 6 7 8 9 
2 3 4 5 6 7 8 9 10 
3 5 7 9 11 13 15 17 19 
5 13 29 61 125 253 509 1021 2045

Idris

A : Nat -> Nat -> Nat
A Z n = S n
A (S m) Z = A m (S Z)
A (S m) (S n) = A m (A (S m) n)

Ioke

Translation of: Clojure
ackermann = method(m,n,
  cond(
    m zero?, n succ,
    n zero?, ackermann(m pred, 1),
    ackermann(m pred, ackermann(m, n pred)))
)

J

As posted at the J wiki

ack=: c1`c1`c2`c3 @. (#.@,&*) M.
c1=: >:@]                        NB. if 0=x, 1+y
c2=: <:@[ ack 1:                 NB. if 0=y, (x-1) ack 1
c3=: <:@[ ack [ ack <:@]         NB. else,   (x-1) ack x ack y-1
Example use:
   0 ack 3
4
   1 ack 3
5
   2 ack 3
9
   3 ack 3
61

J's stack was too small for me to compute 4 ack 1.

Alternative Primitive Recursive Version

This version works by first generating verbs (functions) and then applying them to compute the rows of the related Buck function; then the Ackermann function is obtained in terms of the Buck function. It uses extended precision to be able to compute 4 Ack 2.

The Ackermann function derived in this fashion is primitive recursive. This is possible because in J (as in some other languages) functions, or representations of them, are first-class values.

   Ack=. 3 -~ [ ({&(2 4$'>:  2x&+') ::(,&'&1'&'2x&*'@:(-&2))"0@:[ 128!:2 ]) 3 + ]
Example use:
   0 1 2 3 Ack 0 1 2 3 4 5 6 7
1  2  3  4   5   6   7    8
2  3  4  5   6   7   8    9
3  5  7  9  11  13  15   17
5 13 29 61 125 253 509 1021
   
   3 4 Ack 0 1 2
 5    13                                                                                                                                                                                                                                                        ...
13 65533 2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348101038234342907263181822949382118812668869506364761547029165041871916351587966347219442930927982084309104855990570159318959639524863372367203002916...
   
   4 # @: ": @: Ack 2 NB. Number of digits of 4 Ack 2
19729
   
   5 Ack 0
65533

A structured derivation of Ack follows:

   o=. @: NB. Composition of verbs (functions) 
   x=. o[ NB. Composing the left noun (argument)
   
   (rows2up=. ,&'&1'&'2x&*') o i. 4 
2x&*      
2x&*&1    
2x&*&1&1  
2x&*&1&1&1
   NB. 2's multiplication, exponentiation, tetration, pentation, etc.
    
   0 1 2 (BuckTruncated=. (rows2up  x apply ]) f.) 0 1 2 3 4 5
0 2 4  6     8                                                                                                                                                                                                                                                  ...
1 2 4  8    16                                                                                                                                                                                                                                                  ...
1 2 4 16 65536 2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348101038234342907263181822949382118812668869506364761547029165041871916351587966347219442930927982084309104855990570159318959639524863372367203...
   NB. Buck truncated function (missing the first two rows)
   
   BuckTruncated NB. Buck truncated function-level code
,&'&1'&'2x&*'@:[ 128!:2 ]
   
   (rows01=. {&('>:',:'2x&+')) 0 1 NB. The missing first two rows
>:  
2x&+
   
   Buck=. (rows01 :: (rows2up o (-&2)))"0 x apply ]
   
   (Ack=. (3 -~ [ Buck 3 + ])f.)  NB. Ackermann function-level code
3 -~ [ ({&(2 4$'>:  2x&+') ::(,&'&1'&'2x&*'@:(-&2))"0@:[ 128!:2 ]) 3 + ]

Java

import java.math.BigInteger;

public static BigInteger ack(BigInteger m, BigInteger n) {
    return m.equals(BigInteger.ZERO)
            ? n.add(BigInteger.ONE)
            : ack(m.subtract(BigInteger.ONE),
                        n.equals(BigInteger.ZERO) ? BigInteger.ONE : ack(m, n.subtract(BigInteger.ONE)));
}
Works with: Java version 8+
@FunctionalInterface
public interface FunctionalField<FIELD extends Enum<?>> {
  public Object untypedField(FIELD field);

  @SuppressWarnings("unchecked")
  public default <VALUE> VALUE field(FIELD field) {
    return (VALUE) untypedField(field);
  }
}
import java.util.function.BiFunction;
import java.util.function.Function;
import java.util.function.Predicate;
import java.util.function.UnaryOperator;
import java.util.stream.Stream;

public interface TailRecursive {
  public static <INPUT, INTERMEDIARY, OUTPUT> Function<INPUT, OUTPUT> new_(Function<INPUT, INTERMEDIARY> toIntermediary, UnaryOperator<INTERMEDIARY> unaryOperator, Predicate<INTERMEDIARY> predicate, Function<INTERMEDIARY, OUTPUT> toOutput) {
    return input ->
      $.new_(
        Stream.iterate(
          toIntermediary.apply(input),
          unaryOperator
        ),
        predicate,
        toOutput
      )
    ;
  }

  public static <INPUT1, INPUT2, INTERMEDIARY, OUTPUT> BiFunction<INPUT1, INPUT2, OUTPUT> new_(BiFunction<INPUT1, INPUT2, INTERMEDIARY> toIntermediary, UnaryOperator<INTERMEDIARY> unaryOperator, Predicate<INTERMEDIARY> predicate, Function<INTERMEDIARY, OUTPUT> toOutput) {
    return (input1, input2) ->
      $.new_(
        Stream.iterate(
          toIntermediary.apply(input1, input2),
          unaryOperator
        ),
        predicate,
        toOutput
      )
    ;
  }

  public enum $ {
    $$;

    private static <INTERMEDIARY, OUTPUT> OUTPUT new_(Stream<INTERMEDIARY> stream, Predicate<INTERMEDIARY> predicate, Function<INTERMEDIARY, OUTPUT> function) {
      return stream
        .filter(predicate)
        .map(function)
        .findAny()
        .orElseThrow(RuntimeException::new)
      ;
    }
  }
}
import java.math.BigInteger;
import java.util.Stack;
import java.util.function.BinaryOperator;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public interface Ackermann {
  public static Ackermann new_(BigInteger number1, BigInteger number2, Stack<BigInteger> stack, boolean flag) {
    return $.new_(number1, number2, stack, flag);
  }
  public static void main(String... arguments) {
    $.main(arguments);
  }
  public BigInteger number1();
  public BigInteger number2();

  public Stack<BigInteger> stack();

  public boolean flag();

  public enum $ {
    $$;

    private static final BigInteger ZERO = BigInteger.ZERO;
    private static final BigInteger ONE = BigInteger.ONE;
    private static final BigInteger TWO = BigInteger.valueOf(2);
    private static final BigInteger THREE = BigInteger.valueOf(3);
    private static final BigInteger FOUR = BigInteger.valueOf(4);

    private static Ackermann new_(BigInteger number1, BigInteger number2, Stack<BigInteger> stack, boolean flag) {
      return (FunctionalAckermann) field -> {
        switch (field) {
          case number1: return number1;
          case number2: return number2;
          case stack: return stack;
          case flag: return flag;
          default: throw new UnsupportedOperationException(
            field instanceof Field
              ? "Field checker has not been updated properly."
              : "Field is not of the correct type."
          );
        }
      };
    }

    private static final BinaryOperator<BigInteger> ACKERMANN = 
      TailRecursive.new_(
        (BigInteger number1, BigInteger number2) ->
          new_(
            number1,
            number2,
            Stream.of(number1).collect(
              Collectors.toCollection(Stack::new)
            ),
            false
          )
        ,
        ackermann -> {
          BigInteger number1 = ackermann.number1();
          BigInteger number2 = ackermann.number2();
          Stack<BigInteger> stack = ackermann.stack();
          if (!stack.empty() && !ackermann.flag()) {
            number1 = stack.pop();
          }
          switch (number1.intValue()) {
            case 0:
              return new_(
                number1,
                number2.add(ONE),
                stack,
                false
              );
            case 1:
              return new_(
                number1,
                number2.add(TWO),
                stack,
                false
              );
            case 2:
              return new_(
                number1,
                number2.multiply(TWO).add(THREE),
                stack,
                false
              );
            default:
              if (ZERO.equals(number2)) {
                return new_(
                  number1.subtract(ONE),
                  ONE,
                  stack,
                  true
                );
              } else {
                stack.push(number1.subtract(ONE));
                return new_(
                  number1,
                  number2.subtract(ONE),
                  stack,
                  true
                );
              }
          }
        },
        ackermann -> ackermann.stack().empty(),
        Ackermann::number2
      )::apply
    ;

    private static void main(String... arguments) {
      System.out.println(ACKERMANN.apply(FOUR, TWO));
    }

    private enum Field {
      number1,
      number2,
      stack,
      flag
    }

    @FunctionalInterface
    private interface FunctionalAckermann extends FunctionalField<Field>, Ackermann {
      @Override
      public default BigInteger number1() {
        return field(Field.number1);
      }

      @Override
      public default BigInteger number2() {
        return field(Field.number2);
      }

      @Override
      public default Stack<BigInteger> stack() {
        return field(Field.stack);
      }

      @Override
      public default boolean flag() {
        return field(Field.flag);
      }
    }
  }
}

Template:Iterative version

/*
 * Source https://stackoverflow.com/a/51092690/5520417
 */

package matematicas;

import java.math.BigInteger;
import java.util.HashMap;
import java.util.Stack;

/**
 * @author rodri
 *
 */

public class IterativeAckermannMemoryOptimization extends Thread {

  /**
   * Max percentage of free memory that the program will use. Default is 10% since
   * the majority of the used devices are mobile and therefore it is more likely
   * that the user will have more opened applications at the same time than in a
   * desktop device
   */
  private static Double SYSTEM_MEMORY_LIMIT_PERCENTAGE = 0.1;

  /**
   * Attribute of the type IterativeAckermann
   */
  private IterativeAckermann iterativeAckermann;

  /**
   * @param iterativeAckermann
   */
  public IterativeAckermannMemoryOptimization(IterativeAckermann iterativeAckermann) {
    super();
    this.iterativeAckermann = iterativeAckermann;
  }

  /**
   * @return
   */
  public IterativeAckermann getIterativeAckermann() {
    return iterativeAckermann;
  }

  /**
   * @param iterativeAckermann
   */
  public void setIterativeAckermann(IterativeAckermann iterativeAckermann) {
    this.iterativeAckermann = iterativeAckermann;
  }

  public static Double getSystemMemoryLimitPercentage() {
    return SYSTEM_MEMORY_LIMIT_PERCENTAGE;
  }

  /**
   * Principal method of the thread. Checks that the memory used doesn't exceed or
   * equal the limit, and informs the user when that happens.
   */
  @Override
  public void run() {
    String operating_system = System.getProperty("os.name").toLowerCase();
    if ( operating_system.equals("windows") || operating_system.equals("linux") || operating_system.equals("macintosh") ) {
      SYSTEM_MEMORY_LIMIT_PERCENTAGE = 0.25;
    }

    while ( iterativeAckermann.getConsumed_heap() >= SYSTEM_MEMORY_LIMIT_PERCENTAGE * Runtime.getRuntime().freeMemory() ) {
      try {
        wait();
      }
      catch ( InterruptedException e ) {
        // TODO Auto-generated catch block
        e.printStackTrace();
      }
    }
    if ( ! iterativeAckermann.isAlive() )
      iterativeAckermann.start();
    else
      notifyAll();

  }

}


public class IterativeAckermann extends Thread {

  /*
   * Adjust parameters conveniently
   */
  /**
   * 
   */
  private static final int HASH_SIZE_LIMIT = 636;

  /**
   * 
   */
  private BigInteger m;

  /**
   * 
   */
  private BigInteger n;

  /**
   * 
   */
  private Integer hash_size;

  /**
   * 
   */
  private Long consumed_heap;

  /**
   * @param m
   * @param n
   * @param invalid
   * @param invalid2
   */
  public IterativeAckermann(BigInteger m, BigInteger n, Integer invalid, Long invalid2) {
    super();
    this.m = m;
    this.n = n;
    this.hash_size = invalid;
    this.consumed_heap = invalid2;
  }

  /**
   * 
   */
  public IterativeAckermann() {
    // TODO Auto-generated constructor stub
    super();
    m = null;
    n = null;
    hash_size = 0;
    consumed_heap = 0l;
  }

  /**
   * @return
   */
  public static BigInteger getLimit() {
    return LIMIT;
  }

  /**
   * @author rodri
   *
   * @param <T1>
   * @param <T2>
   */
  /**
   * @author rodri
   *
   * @param <T1>
   * @param <T2>
   */
  static class Pair<T1, T2> {

    /**
     * 
     */
    /**
     * 
     */
    T1 x;

    /**
     * 
     */
    /**
     * 
     */
    T2 y;

    /**
     * @param x_
     * @param y_
     */
    /**
     * @param x_
     * @param y_
     */
    Pair(T1 x_, T2 y_) {
      x = x_;
      y = y_;
    }

    /**
     *
     */
    /**
     *
     */
    @Override
    public int hashCode() {
      return x.hashCode() ^ y.hashCode();
    }

    /**
     *
     */
    /**
     *
     */
    @Override
    public boolean equals(Object o_) {

      if ( o_ == null ) {
        return false;
      }
      if ( o_.getClass() != this.getClass() ) {
        return false;
      }
      Pair<?, ?> o = (Pair<?, ?>) o_;
      return x.equals(o.x) && y.equals(o.y);
    }
  }

  /**
   * 
   */
  private static final BigInteger LIMIT = new BigInteger("6");

  /**
   * @param m
   * @param n
   * @return
   */

  /**
   *
   */
  @Override
  public void run() {
    while ( hash_size >= HASH_SIZE_LIMIT ) {
      try {
        this.wait();
      }
      catch ( InterruptedException e ) {
        // TODO Auto-generated catch block
        e.printStackTrace();
      }
    }
    for ( BigInteger i = BigInteger.ZERO; i.compareTo(LIMIT) == - 1; i = i.add(BigInteger.ONE) ) {
      for ( BigInteger j = BigInteger.ZERO; j.compareTo(LIMIT) == - 1; j = j.add(BigInteger.ONE) ) {
        IterativeAckermann iterativeAckermann = new IterativeAckermann(i, j, null, null);
        System.out.printf("Ackmermann(%d, %d) = %d\n", i, j, iterativeAckermann.iterative_ackermann(i, j));

      }
    }
  }

  /**
   * @return
   */
  public BigInteger getM() {
    return m;
  }

  /**
   * @param m
   */
  public void setM(BigInteger m) {
    this.m = m;
  }

  /**
   * @return
   */
  public BigInteger getN() {
    return n;
  }

  /**
   * @param n
   */
  public void setN(BigInteger n) {
    this.n = n;
  }

  /**
   * @return
   */
  public Integer getHash_size() {
    return hash_size;
  }

  /**
   * @param hash_size
   */
  public void setHash_size(Integer hash_size) {
    this.hash_size = hash_size;
  }

  /**
   * @return
   */
  public Long getConsumed_heap() {
    return consumed_heap;
  }

  /**
   * @param consumed_heap
   */
  public void setConsumed_heap(Long consumed_heap) {
    this.consumed_heap = consumed_heap;
  }

  /**
   * @param m
   * @param n
   * @return
   */
  public BigInteger iterative_ackermann(BigInteger m, BigInteger n) {
    if ( m.compareTo(BigInteger.ZERO) != - 1 && m.compareTo(BigInteger.ZERO) != - 1 )
      try {
        HashMap<Pair<BigInteger, BigInteger>, BigInteger> solved_set = new HashMap<Pair<BigInteger, BigInteger>, BigInteger>(900000);
        Stack<Pair<BigInteger, BigInteger>> to_solve = new Stack<Pair<BigInteger, BigInteger>>();
        to_solve.push(new Pair<BigInteger, BigInteger>(m, n));

        while ( ! to_solve.isEmpty() ) {
          Pair<BigInteger, BigInteger> head = to_solve.peek();
          if ( head.x.equals(BigInteger.ZERO) ) {
            solved_set.put(head, head.y.add(BigInteger.ONE));
            to_solve.pop();
          }
          else if ( head.y.equals(BigInteger.ZERO) ) {
            Pair<BigInteger, BigInteger> next = new Pair<BigInteger, BigInteger>(head.x.subtract(BigInteger.ONE), BigInteger.ONE);
            BigInteger result = solved_set.get(next);
            if ( result == null ) {
              to_solve.push(next);
            }
            else {
              solved_set.put(head, result);
              to_solve.pop();
            }
          }
          else {
            Pair<BigInteger, BigInteger> next0 = new Pair<BigInteger, BigInteger>(head.x, head.y.subtract(BigInteger.ONE));
            BigInteger result0 = solved_set.get(next0);
            if ( result0 == null ) {
              to_solve.push(next0);
            }
            else {
              Pair<BigInteger, BigInteger> next = new Pair<BigInteger, BigInteger>(head.x.subtract(BigInteger.ONE), result0);
              BigInteger result = solved_set.get(next);
              if ( result == null ) {
                to_solve.push(next);
              }
              else {
                solved_set.put(head, result);
                to_solve.pop();
              }
            }
          }
        }
        this.hash_size = solved_set.size();
        System.out.println("Hash Size: " + hash_size);
        consumed_heap = (Runtime.getRuntime().totalMemory() / (1024 * 1024));
        System.out.println("Consumed Heap: " + consumed_heap + "m");
        setHash_size(hash_size);
        setConsumed_heap(consumed_heap);
        return solved_set.get(new Pair<BigInteger, BigInteger>(m, n));

      }
      catch ( OutOfMemoryError e ) {
        // TODO: handle exception
        e.printStackTrace();
      }
    throw new IllegalArgumentException("The arguments must be non-negative integers.");
  }

  /**
   * @param args
   */
  /**
   * @param args
   */
  public static void main(String[] args) {
    IterativeAckermannMemoryOptimization iterative_ackermann_memory_optimization = new IterativeAckermannMemoryOptimization(
        new IterativeAckermann());
    iterative_ackermann_memory_optimization.start();
  }
}

JavaScript

ES5

function ack(m, n) {
 return m === 0 ? n + 1 : ack(m - 1, n === 0  ? 1 : ack(m, n - 1));
}

Eliminating Tail Calls

function ack(M,N) {
  for (; M > 0; M--) {
    N = N === 0 ? 1 : ack(M,N-1);
  }
  return N+1;
}

Iterative, With Explicit Stack

function stackermann(M, N) {
  const stack = [];
  for (;;) {
    if (M === 0) {
      N++;
      if (stack.length === 0) return N;
      const r = stack[stack.length-1];
      if (r[1] === 1) stack.length--;
      else r[1]--;
      M = r[0];
    } else if (N === 0) {
      M--;
      N = 1;
    } else {
      M--
      stack.push([M, N]);
      N = 1;
    }
  }
}

Stackless Iterative

#!/usr/bin/env nodejs
function ack(M, N){
	const next = new Float64Array(M + 1);
	const goal = new Float64Array(M + 1).fill(1, 0, M);
	const n = N + 1;

	// This serves as a sentinel value;
	// next[M] never equals goal[M] == -1,
	// so we don't need an extra check for
	// loop termination below.
	goal[M] = -1;

	let v;
	do {
		v = next[0] + 1;
		let m = 0;
		while (next[m] === goal[m]) {
			goal[m] = v;
			next[m++]++;
		}
		next[m]++;
	} while (next[M] !== n);
	return v;
}
var args = process.argv;
console.log(ack(parseInt(args[2]), parseInt(args[3])));
Output:
> time ./ack.js 4 1            
65533
./ack.js 4 1  0,48s user 0,03s system 100% cpu 0,505 total ; AMD FX-8350 @ 4 GHz

ES6

(() => {
    'use strict';

    // ackermann :: Int -> Int -> Int
    const ackermann = m => n => {
        const go = (m, n) =>
            0 === m ? (
                succ(n)
            ) : go(pred(m), 0 === n ? (
                1
            ) : go(m, pred(n)));
        return go(m, n);
    };

    // TEST -----------------------------------------------
    const main = () => console.log(JSON.stringify(
        [0, 1, 2, 3].map(
            flip(ackermann)(3)
        )
    ));


    // GENERAL FUNCTIONS ----------------------------------

    // flip :: (a -> b -> c) -> b -> a -> c
    const flip = f =>
        x => y => f(y)(x);

    // pred :: Enum a => a -> a
    const pred = x => x - 1;

    // succ :: Enum a => a -> a
    const succ = x => 1 + x;


    // MAIN ---
    return main();
})();
Output:
[4,5,9,61]

Joy

DEFINE ack == [[[pop null] popd succ] 
[[null] pop pred 1 ack] 
[[dup pred swap] dip pred ack ack]] 
cond.

another using a combinator:

DEFINE ack == [[[pop null] [popd succ]] 
[[null] [pop pred 1] []] 
[[[dup pred swap] dip pred] [] []]] 
condnestrec.

jq

Works with: jq version 1.4

Without Memoization

# input: [m,n]
def ack:
  .[0] as $m | .[1] as $n
  | if $m == 0 then $n + 1
    elif $n == 0 then [$m-1, 1] | ack
    else [$m-1, ([$m, $n-1 ] | ack)] | ack
    end ;

Example:

range(0;5) as $i
| range(0; if $i > 3 then 1 else 6 end) as $j
| "A(\($i),\($j)) = \( [$i,$j] | ack )"
Output:
# jq -n -r -f ackermann.jq
A(0,0) = 1
A(0,1) = 2
A(0,2) = 3
A(0,3) = 4
A(0,4) = 5
A(0,5) = 6
A(1,0) = 2
A(1,1) = 3
A(1,2) = 4
A(1,3) = 5
A(1,4) = 6
A(1,5) = 7
A(2,0) = 3
A(2,1) = 5
A(2,2) = 7
A(2,3) = 9
A(2,4) = 11
A(2,5) = 13
A(3,0) = 5
A(3,1) = 13
A(3,2) = 29
A(3,3) = 61
A(3,4) = 125
A(3,5) = 253
A(4,0) = 13

With Memoization and Optimization

# input: [m,n, cache]
# output [value, updatedCache]
def ack:

  # input: [value,cache]; output: [value, updatedCache]
  def cache(key): .[1] += { (key): .[0] };
  
  def pow2: reduce range(0; .) as $i (1; .*2);
 
  .[0] as $m | .[1] as $n | .[2] as $cache
  | if   $m == 0 then [$n + 1, $cache]
    elif $m == 1 then [$n + 2, $cache]
    elif $m == 2 then [2 * $n + 3, $cache]
    elif $m == 3 then [8 * ($n|pow2) - 3, $cache]
    else
    (.[0:2]|tostring) as $key
    | $cache[$key] as $value
    | if $value then [$value, $cache]
      elif $n == 0 then
        ([$m-1, 1, $cache] | ack)
        | cache($key)
      else
        ([$m, $n-1, $cache ] | ack) 
        | [$m-1, .[0], .[1]] | ack
        | cache($key)
      end
    end;

def A(m;n): [m,n,{}] | ack | .[0];
Example:
A(4,1)
Output:
65533

Jsish

From javascript entry.

/* Ackermann function, in Jsish */

function ack(m, n) {
 return m === 0 ? n + 1 : ack(m - 1, n === 0  ? 1 : ack(m, n - 1));
}

if (Interp.conf('unitTest')) {
    Interp.conf({maxDepth:4096});
;    ack(1,3);
;    ack(2,3);
;    ack(3,3);
;    ack(1,5);
;    ack(2,5);
;    ack(3,5);
}

/*
=!EXPECTSTART!=
ack(1,3) ==> 5
ack(2,3) ==> 9
ack(3,3) ==> 61
ack(1,5) ==> 7
ack(2,5) ==> 13
ack(3,5) ==> 253
=!EXPECTEND!=
*/
Output:
prompt$ jsish --U Ackermann.jsi
ack(1,3) ==> 5
ack(2,3) ==> 9
ack(3,3) ==> 61
ack(1,5) ==> 7
ack(2,5) ==> 13
ack(3,5) ==> 253

Julia

function ack(m,n)
    if m == 0
        return n + 1
    elseif n == 0
        return ack(m-1,1)
    else
        return ack(m-1,ack(m,n-1))
    end
end

One-liner:

ack2(m::Integer, n::Integer) = m == 0 ? n + 1 : n == 0 ? ack2(m - 1, 1) : ack2(m - 1, ack2(m, n - 1))

Using memoization, source:

using Memoize
@memoize ack3(m::Integer, n::Integer) = m == 0 ? n + 1 : n == 0 ? ack3(m - 1, 1) : ack3(m - 1, ack3(m, n - 1))

Benchmarking:

julia> @time ack2(4,1)
elapsed time: 71.98668457 seconds (96 bytes allocated)
65533

julia> @time ack3(4,1)
elapsed time: 0.49337724 seconds (30405308 bytes allocated)
65533

K

See the K wiki

ack:{:[0=x;y+1;0=y;_f[x-1;1];_f[x-1;_f[x;y-1]]]}
ack[2;2]

Kdf9 Usercode

V6; W0;
YS26000;
RESTART; J999; J999;
PROGRAM;                   (main program);
   V1 = B1212121212121212; (radix 10 for FRB);
   V2 = B2020202020202020; (high bits for decimal digits);
   V3 = B0741062107230637; ("A[3,"  in Flexowriter code);
   V4 = B0727062200250007; ("7] = " in Flexowriter code);
   V5 = B7777777777777777;

      ZERO; NOT; =M1;      (Q1 := 0/0/-1);
      SETAYS0; =M2; I2=2;  (Q2 := 0/2/AYS0: M2 is the stack pointer);
      SET 3; =RC7;         (Q7 := 3/1/0: C7 = m);
      SET 7; =RC8;         (Q8 := 7/1/0: C8 = n);
   JSP1;                   (call Ackermann function);
      V1; REV; FRB;        (convert result to base 10);
      V2; OR;              (convert decimal digits to characters);
      V5; REV;
      SHLD+24; =V5; ERASE; (eliminate leading zeros);
      SETAV5; =RM9;
      SETAV3; =I9;
      POAQ9;               (write result to Flexowriter);

999;  ZERO; OUT;           (terminate run);

P1; (To compute A[m, n]);

   99;
      J1C7NZ;           (to 1 if m ± 0);
         I8; =+C8;      (n := n + 1);
         C8;            (result to NEST);
      EXIT 1;           (return);
   *1;
      J2C8NZ;           (to 2 if n ± 0);
         I8; =C8;       (n := 1);
         DC7;           (m := m - 1);
      J99;              (tail recursion for A[m-1, 1]);
   *2;
         LINK; =M0M2;   (push return address);
         C7; =M0M2QN;   (push m);
         DC8;           (n := n - 1);
      JSP1;             (full recursion for A[m, n-1]);
         =C8;           (n := A[m, n-1]);
         M1M2; =C7;     (m := top of stack);
         DC7;           (m := m - 1);
         M-I2;          (pop stack);
         M0M2; =LINK;   (return address := top of stack);
      J99;              (tail recursion for A[m-1, A[m, n-1]]);

FINISH;

Klingphix

:ack
    %n !n %m !m
 
    $m 0 ==
    ( [$n 1 +]
      [$n 0 ==
        ( [$m 1 - 1 ack]
          [$m 1 - $m $n 1 - ack ack]
        ) if
      ]
    ) if
;
 
3 6 ack print nl
msec print

" " input

Klong

ack::{:[0=x;y+1:|0=y;.f(x-1;1);.f(x-1;.f(x;y-1))]}
ack(2;2)

Kotlin

tailrec fun A(m: Long, n: Long): Long {
    require(m >= 0L) { "m must not be negative" }
    require(n >= 0L) { "n must not be negative" }
    if (m == 0L) {
        return n + 1L
    }
    if (n == 0L) {
        return A(m - 1L, 1L)
    }
    return A(m - 1L, A(m, n - 1L))
}

inline fun<T> tryOrNull(block: () -> T): T? = try { block() } catch (e: Throwable) { null }

const val N = 10L
const val M = 4L

fun main() {
    (0..M)
        .map { it to 0..N }
        .map { (m, Ns) -> (m to Ns) to Ns.map { n -> tryOrNull { A(m, n) } } }
        .map { (input, output) -> "A(${input.first}, ${input.second})" to output.map { it?.toString() ?: "?" } }
        .map { (input, output) -> "$input = $output" }
        .forEach(::println)
}
Output:
A(0, 0..10) = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
A(1, 0..10) = [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
A(2, 0..10) = [3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23]
A(3, 0..10) = [5, 13, 29, 61, 125, 253, 509, 1021, 2045, 4093, 8189]
A(4, 0..10) = [13, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?]

Lambdatalk

{def ack
 {lambda {:m :n}
  {if {= :m 0}
   then {+ :n 1}
   else {if {= :n 0}
   then {ack {- :m 1} 1}
   else {ack {- :m 1} {ack :m {- :n 1}}}}}}}
-> ack

{S.map {ack 0} {S.serie 0 300000}}  // 2090ms
{S.map {ack 1} {S.serie 0 500}}     // 2038ms
{S.map {ack 2} {S.serie 0 70}}      // 2100ms
{S.map {ack 3} {S.serie 0 6}}       // 1800ms

{ack 2 700}     // 8900ms
-> 1403

{ack 3 7}       // 6000ms
-> 1021

{ack 4 1}       // too much
-> ???

Lasso

#!/usr/bin/lasso9
 
define ackermann(m::integer, n::integer) => {
  if(#m == 0) => {
    return ++#n
  else(#n == 0)
    return ackermann(--#m, 1)
  else
    return ackermann(#m-1, ackermann(#m, --#n))
  }
}

with x in generateSeries(1,3),
     y in generateSeries(0,8,2)
do stdoutnl(#x+', '#y+': ' + ackermann(#x, #y))
Output:
1, 0: 2
1, 2: 4
1, 4: 6
1, 6: 8
1, 8: 10
2, 0: 3
2, 2: 7
2, 4: 11
2, 6: 15
2, 8: 19
3, 0: 5
3, 2: 29
3, 4: 125
3, 6: 509
3, 8: 2045

LFE

(defun ackermann
  ((0 n) (+ n 1))
  ((m 0) (ackermann (- m 1) 1))
  ((m n) (ackermann (- m 1) (ackermann m (- n 1)))))

Liberty BASIC

Print Ackermann(1, 2)

    Function Ackermann(m, n)
        Select Case
            Case (m < 0) Or (n < 0)
                Exit Function
            Case (m = 0)
                Ackermann = (n + 1)
            Case (m > 0) And (n = 0)
                Ackermann = Ackermann((m - 1), 1)
            Case (m > 0) And (n > 0)
                Ackermann = Ackermann((m - 1), Ackermann(m, (n - 1)))
        End Select
    End Function

LiveCode

function ackermann m,n
    switch
        Case m = 0
            return n + 1
        Case (m > 0 And n = 0)
            return ackermann((m - 1), 1)
        Case (m > 0 And n > 0)
            return ackermann((m - 1), ackermann(m, (n - 1)))
    end switch
end ackermann

to ack :i :j
  if :i = 0 [output :j+1]
  if :j = 0 [output ack :i-1 1]
  output ack :i-1 ack :i :j-1
end

Logtalk

ack(0, N, V) :-
    !,
    V is N + 1.
ack(M, 0, V) :-
    !,
    M2 is M - 1,
    ack(M2, 1, V).
ack(M, N, V) :-
    M2 is M - 1,
    N2 is N - 1,
    ack(M, N2, V2),
    ack(M2, V2, V).

LOLCODE

Translation of: C
HAI 1.3

HOW IZ I ackermann YR m AN YR n
    NOT m, O RLY?
        YA RLY, FOUND YR SUM OF n AN 1
    OIC

    NOT n, O RLY?
        YA RLY, FOUND YR I IZ ackermann YR DIFF OF m AN 1 AN YR 1 MKAY
    OIC

    FOUND YR I IZ ackermann YR DIFF OF m AN 1 AN YR...
     I IZ ackermann YR m AN YR DIFF OF n AN 1 MKAY MKAY
IF U SAY SO

IM IN YR outer UPPIN YR m TIL BOTH SAEM m AN 5
    IM IN YR inner UPPIN YR n TIL BOTH SAEM n AN DIFF OF 6 AN m
        VISIBLE "A(" m ", " n ") = " I IZ ackermann YR m AN YR n MKAY
    IM OUTTA YR inner
IM OUTTA YR outer

KTHXBYE

Lua

function ack(M,N)
    if M == 0 then return N + 1 end
    if N == 0 then return ack(M-1,1) end
    return ack(M-1,ack(M, N-1))
end

Stackless iterative solution with multiple precision, fast

 
#!/usr/bin/env luajit
local gmp = require 'gmp' ('libgmp')
local mpz, 		z_mul, 		z_add, 		z_add_ui, 		z_set_d = 
	gmp.types.z, gmp.z_mul,	gmp.z_add,	gmp.z_add_ui, 	gmp.z_set_d
local z_cmp, 	z_cmp_ui, 		z_init_d, 			z_set=
	gmp.z_cmp,	gmp.z_cmp_ui, 	gmp.z_init_set_d, 	gmp.z_set
local printf = gmp.printf

local function ack(i,n)
	local nxt=setmetatable({},  {__index=function(t,k) local z=mpz() z_init_d(z, 0) t[k]=z return z end})
	local goal=setmetatable({}, {__index=function(t,k) local o=mpz() z_init_d(o, 1) t[k]=o return o end})
	goal[i]=mpz() z_init_d(goal[i], -1)
	local v=mpz() z_init_d(v, 0) 
	local ic
	local END=n+1
	local ntmp,gtmp
	repeat
		ic=0
		ntmp,gtmp=nxt[ic], goal[ic]
		z_add_ui(v, ntmp, 1)
		while z_cmp(ntmp, gtmp) == 0 do
			z_set(gtmp,v)
			z_add_ui(ntmp, ntmp, 1)
			nxt[ic], goal[ic]=ntmp, gtmp
			ic=ic+1
			ntmp,gtmp=nxt[ic], goal[ic]
		end
		z_add_ui(ntmp, ntmp, 1)
		nxt[ic]=ntmp
	until z_cmp_ui(nxt[i], END) == 0
	return v
end

if #arg<1 then
	print("Ackermann: "..arg[0].." <num1> [num2]")
else
	printf("%Zd\n", ack(tonumber(arg[1]), arg[2] and tonumber(arg[2]) or 0))
end
Output:
> time ./ackermann_iter.lua 4 1
65533
./ackermann_iter.lua 4 1  0,01s user 0,01s system 95% cpu 0,015 total // AMD FX-8350@4 GHz
> time ./ackermann.lua 3 10                                              ⏎
8189
./ackermann.lua 3 10  0,22s user 0,00s system 98% cpu 0,222 total // recursive solution
> time ./ackermann_iter.lua 3 10
8189
./ackermann_iter.lua 3 10  0,00s user 0,00s system 92% cpu 0,009 total

Lucid

ack(m,n)
 where
  ack(m,n) = if m eq 0 then n+1
                       else if n eq 0 then ack(m-1,1)
                                      else ack(m-1, ack(m, n-1)) fi
                       fi;
 end

Luck

function ackermann(m: int, n: int): int = (
   if m==0 then n+1
   else if n==0 then ackermann(m-1,1)
   else ackermann(m-1,ackermann(m,n-1))
)

M2000 Interpreter

Module Checkit {
      Def ackermann(m,n) =If(m=0-> n+1, If(n=0-> ackermann(m-1,1), ackermann(m-1,ackermann(m,n-1))))
      For m = 0 to 3 {For n = 0 to 4 {Print m;" ";n;" ";ackermann(m,n)}}
}
Checkit


Module Checkit {
      Module Inner (ack) {
            For m = 0 to 3 {For n = 0 to 4 {Print m;" ";n;" ";ack(m,n)}} 
      }
      Inner lambda (m,n) ->If(m=0-> n+1, If(n=0-> lambda(m-1,1),lambda(m-1,lambda(m,n-1))))
}
Checkit

M4

define(`ack',`ifelse($1,0,`incr($2)',`ifelse($2,0,`ack(decr($1),1)',`ack(decr($1),ack($1,decr($2)))')')')dnl
ack(3,3)
Output:
61 

MAD

While MAD supports function calls, it does not handle recursion automatically. There is support for a stack, but the programmer has to set it up himself (by defining an array to reserve memory, then making it the stack using the SET LIST) command. Values have to be pushed and popped from it by hand (using SAVE and RESTORE), and for a function to be reentrant, even the return address has to be kept.

On top of this, all variables are global throughout the program (there is no scope), and argument passing is done by reference, meaning that even once the stack is set up, arguments cannot be passed in the normal way. To define a function that takes arguments, one would have to declare a helper function that then passes the arguments to the recursive function via the stack or the global variables. The following program demonstrates this.

            NORMAL MODE IS INTEGER
            DIMENSION LIST(3000)
            SET LIST TO LIST
          
            INTERNAL FUNCTION(DUMMY)
            ENTRY TO ACKH.
LOOP        WHENEVER M.E.0
                FUNCTION RETURN N+1
            OR WHENEVER N.E.0
                M=M-1
                N=1
                TRANSFER TO LOOP
            OTHERWISE
                SAVE RETURN
                SAVE DATA M
                N=N-1
                N=ACKH.(0)
                RESTORE DATA M
                RESTORE RETURN
                M=M-1
                TRANSFER TO LOOP
            END OF CONDITIONAL
            ERROR RETURN
            END OF FUNCTION
            
            INTERNAL FUNCTION(MM,NN)
            ENTRY TO ACK.
            M=MM
            N=NN
            FUNCTION RETURN ACKH.(0)
            END OF FUNCTION
            
            THROUGH SHOW, FOR I=0, 1, I.G.3
            THROUGH SHOW, FOR J=0, 1, J.G.8
SHOW        PRINT FORMAT ACKF,I,J,ACK.(I,J)
            
            VECTOR VALUES ACKF = $4HACK(,I1,1H,,I1,4H) = ,I4*$
            END OF PROGRAM
Output:
ACK(0,0) =    1
ACK(0,1) =    2
ACK(0,2) =    3
ACK(0,3) =    4
ACK(0,4) =    5
ACK(0,5) =    6
ACK(0,6) =    7
ACK(0,7) =    8
ACK(0,8) =    9
ACK(1,0) =    2
ACK(1,1) =    3
ACK(1,2) =    4
ACK(1,3) =    5
ACK(1,4) =    6
ACK(1,5) =    7
ACK(1,6) =    8
ACK(1,7) =    9
ACK(1,8) =   10
ACK(2,0) =    3
ACK(2,1) =    5
ACK(2,2) =    7
ACK(2,3) =    9
ACK(2,4) =   11
ACK(2,5) =   13
ACK(2,6) =   15
ACK(2,7) =   17
ACK(2,8) =   19
ACK(3,0) =    5
ACK(3,1) =   13
ACK(3,2) =   29
ACK(3,3) =   61
ACK(3,4) =  125
ACK(3,5) =  253
ACK(3,6) =  509
ACK(3,7) = 1021
ACK(3,8) = 2045



Maple

Strictly by the definition given above, we can code this as follows.

Ackermann := proc( m :: nonnegint, n :: nonnegint )
  option remember; # optional automatic memoization
  if m = 0 then
    n + 1
  elif n = 0 then
    thisproc( m - 1, 1 )
  else
    thisproc( m - 1, thisproc( m, n - 1 ) )
  end if
end proc:
In Maple, the keyword
thisproc
refers to the currently executing procedure (closure) and is used when writing recursive procedures. (You could also use the name of the procedure, Ackermann in this case, but then a concurrently executing task or thread could re-assign that name while the recursive procedure is executing, resulting in an incorrect result.)

To make this faster, you can use known expansions for small values of . (See Wikipedia:Ackermann function)

Ackermann := proc( m :: nonnegint, n :: nonnegint )
  option remember; # optional automatic memoization
  if m = 0 then
    n + 1
  elif m = 1 then
    n + 2
  elif m = 2 then
    2 * n + 3
  elif m = 3 then
    8 * 2^n - 3
  elif n = 0 then
    thisproc( m - 1, 1 )
  else
    thisproc( m - 1, thisproc( m, n - 1 ) )
  end if
end proc:

This makes it possible to compute Ackermann( 4, 1 ) and Ackermann( 4, 2 ) essentially instantly, though Ackermann( 4, 3 ) is still out of reach.

To compute Ackermann( 1, i ) for i from 1 to 10 use

> map2( Ackermann, 1, [seq]( 1 .. 10 ) );
               [3, 4, 5, 6, 7, 8, 9, 10, 11, 12]

To get the first 10 values for m = 2 use

> map2( Ackermann, 2, [seq]( 1 .. 10 ) );
               [5, 7, 9, 11, 13, 15, 17, 19, 21, 23]

For Ackermann( 4, 2 ) we get a very long number with

> length( Ackermann( 4, 2 ) );
                      19729

digits.

Mathcad

Mathcad is a non-text-based programming environment. The equation below is an approximation of the way that it is entered (and) displayed on a Mathcad worksheet. The worksheet is available at https://community.ptc.com/t5/PTC-Mathcad/Rosetta-Code-Ackermann-Function/m-p/750117#M197410

This particular version of Ackermann's function was created in Mathcad Prime Express 7.0, a free version of Mathcad Prime 7.0 with restrictions (such as no programming or symbolics). All Prime Express numbers are complex. There is a recursion depth limit of about 4,500.

A(m,n):=if(m=0,n+1,if(n=0,A(m-1,1),A(m-1,A(m,n-1))))

The worksheet also contains an explictly-calculated version of Ackermann's function that calls the tetration function na.

na(a,n):=if(n=0,1,ana(a,n-1))


aerror(m,n):=error(format("cannot compute a({0},{1})",m,n))

a(m,n):=if(m=0,n+1,if(m=1,n+2,if(m=2,2n+3,if(m=3,2^(n+3)-3,aerror(m,n)))))

a(m,n):=if(m=4,na(2,n+3)-3,a(m,n)

Mathematica / Wolfram Language

Two possible implementations would be:

$RecursionLimit=Infinity
Ackermann1[m_,n_]:=
 If[m==0,n+1,
  If[ n==0,Ackermann1[m-1,1],
   Ackermann1[m-1,Ackermann1[m,n-1]]
  ]
 ]

 Ackermann2[0,n_]:=n+1;
 Ackermann2[m_,0]:=Ackermann1[m-1,1];
 Ackermann2[m_,n_]:=Ackermann1[m-1,Ackermann1[m,n-1]]

Note that the second implementation is quite a bit faster, as doing 'if' comparisons is slower than the built-in pattern matching algorithms. Examples:

Flatten[#,1]&@Table[{"Ackermann2["<>ToString[i]<>","<>ToString[j]<>"] =",Ackermann2[i,j]},{i,3},{j,8}]//Grid

gives back:

Ackermann2[1,1] =	3
Ackermann2[1,2] =	4
Ackermann2[1,3] =	5
Ackermann2[1,4] =	6
Ackermann2[1,5] =	7
Ackermann2[1,6] =	8
Ackermann2[1,7] =	9
Ackermann2[1,8] =	10
Ackermann2[2,1] =	5
Ackermann2[2,2] =	7
Ackermann2[2,3] =	9
Ackermann2[2,4] =	11
Ackermann2[2,5] =	13
Ackermann2[2,6] =	15
Ackermann2[2,7] =	17
Ackermann2[2,8] =	19
Ackermann2[3,1] =	13
Ackermann2[3,2] =	29
Ackermann2[3,3] =	61
Ackermann2[3,4] =	125
Ackermann2[3,5] =	253
Ackermann2[3,6] =	509
Ackermann2[3,7] =	1021
Ackermann2[3,8] =	2045

If we would like to calculate Ackermann[4,1] or Ackermann[4,2] we have to optimize a little bit:

Clear[Ackermann3]
$RecursionLimit=Infinity;
Ackermann3[0,n_]:=n+1;
Ackermann3[1,n_]:=n+2;
Ackermann3[2,n_]:=3+2n;
Ackermann3[3,n_]:=5+8 (2^n-1);
Ackermann3[m_,0]:=Ackermann3[m-1,1];
Ackermann3[m_,n_]:=Ackermann3[m-1,Ackermann3[m,n-1]]

Now computing Ackermann[4,1] and Ackermann[4,2] can be done quickly (<0.01 sec): Examples 2:

Ackermann3[4, 1]
Ackermann3[4, 2]

gives back:

65533
2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880........699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156733

Ackermann[4,2] has 19729 digits, several thousands of digits omitted in the result above for obvious reasons. Ackermann[5,0] can be computed also quite fast, and is equal to 65533. Summarizing Ackermann[0,n_], Ackermann[1,n_], Ackermann[2,n_], and Ackermann[3,n_] can all be calculated for n>>1000. Ackermann[4,0], Ackermann[4,1], Ackermann[4,2] and Ackermann[5,0] are only possible now. Maybe in the future we can calculate higher Ackermann numbers efficiently and fast. Although showing the results will always be a problem.

MATLAB

function A = ackermannFunction(m,n)
    if m == 0
        A = n+1;
    elseif (m > 0) && (n == 0)
        A = ackermannFunction(m-1,1);
    else
        A = ackermannFunction( m-1,ackermannFunction(m,n-1) );
    end
end

Maxima

ackermann(m, n) := if integerp(m) and integerp(n) then ackermann[m, n] else 'ackermann(m, n)$

ackermann[m, n] := if m = 0 then n + 1
                   elseif m = 1 then 2 + (n + 3) - 3
                   elseif m = 2 then 2 * (n + 3) - 3
                   elseif m = 3 then 2^(n + 3) - 3
                   elseif n = 0 then ackermann[m - 1, 1]
                   else ackermann[m - 1, ackermann[m, n - 1]]$

tetration(a, n) := if integerp(n) then block([b: a], for i from 2 thru n do b: a^b, b) else 'tetration(a, n)$

/* this should evaluate to zero */
ackermann(4, n) - (tetration(2, n + 3) - 3);
subst(n = 2, %);
ev(%, nouns);

MAXScript

Use with caution. Will cause a stack overflow for m > 3.

fn ackermann m n =
(
    if m == 0 then
    (
        return n + 1
    )
    else if n == 0 then
    (
        ackermann (m-1) 1
    )
    else
    (
        ackermann (m-1) (ackermann m (n-1))
    )
)

Mercury

This is the Ackermann function with some (obvious) elements elided. The ack/3 predicate is implemented in terms of the ack/2 function. The ack/2 function is implemented in terms of the ack/3 predicate. This makes the code both more concise and easier to follow than would otherwise be the case. The integer type is used instead of int because the problem statement stipulates the use of bignum integers if possible.

:- func ack(integer, integer) = integer.
ack(M, N) = R :- ack(M, N, R).

:- pred ack(integer::in, integer::in, integer::out) is det.
ack(M, N, R) :-
	( ( M < integer(0)  
	  ; N < integer(0) ) -> throw(bounds_error)
	; M = integer(0)     -> R = N + integer(1)
	; N = integer(0)     -> ack(M - integer(1), integer(1), R)
	;                       ack(M - integer(1), ack(M, N - integer(1)), R) ).

min

Works with: min version 0.19.3
(
  :n :m
  (
    ((m 0 ==) (n 1 +))
    ((n 0 ==) (m 1 - 1 ackermann))
    ((true) (m 1 - m n 1 - ackermann ackermann))
  ) case
) :ackermann

MiniScript

ackermann = function(m, n)
    if m == 0 then return n+1
    if n == 0 then return ackermann(m - 1, 1)
    return ackermann(m - 1, ackermann(m, n - 1))
end function
 
for m in range(0, 3)
    for n in range(0, 4)
        print "(" + m + ", " + n + "): " + ackermann(m, n)
    end for
end for

МК-61/52

П1	<->	П0	ПП	06	С/П	ИП0	x=0	13	ИП1
1	+	В/О	ИП1	x=0	24	ИП0	1	П1	-
П0	ПП	06	В/О	ИП0	П2	ИП1	1	-	П1
ПП	06	П1	ИП2	1	-	П0	ПП	06	В/О

ML/I

ML/I loves recursion, but runs out of its default amount of storage with larger numbers than those tested here!

Program

MCSKIP "WITH" NL
"" Ackermann function
"" Will overflow when it reaches implementation-defined signed integer limit
MCSKIP MT,<>
MCINS %.
MCDEF ACK WITHS ( , )
AS <MCSET T1=%A1.
MCSET T2=%A2.
MCGO L1 UNLESS T1 EN 0
%%T2.+1.MCGO L0
%L1.MCGO L2 UNLESS T2 EN 0
ACK(%%T1.-1.,1)MCGO L0
%L2.ACK(%%T1.-1.,ACK(%T1.,%%T2.-1.))>
"" Macro ACK now defined, so try it out
a(0,0) => ACK(0,0)
a(0,1) => ACK(0,1)
a(0,2) => ACK(0,2)
a(0,3) => ACK(0,3)
a(0,4) => ACK(0,4)
a(0,5) => ACK(0,5)
a(1,0) => ACK(1,0)
a(1,1) => ACK(1,1)
a(1,2) => ACK(1,2)
a(1,3) => ACK(1,3)
a(1,4) => ACK(1,4)
a(2,0) => ACK(2,0)
a(2,1) => ACK(2,1)
a(2,2) => ACK(2,2)
a(2,3) => ACK(2,3)
a(3,0) => ACK(3,0)
a(3,1) => ACK(3,1)
a(3,2) => ACK(3,2)
a(4,0) => ACK(4,0)
Output:
a(0,0) => 1
a(0,1) => 2
a(0,2) => 3
a(0,3) => 4
a(0,4) => 5
a(0,5) => 6
a(1,0) => 2
a(1,1) => 3
a(1,2) => 4
a(1,3) => 5
a(1,4) => 6
a(2,0) => 3
a(2,1) => 5
a(2,2) => 7
a(2,3) => 9
a(3,0) => 5
a(3,1) => 13
a(3,2) => 29
a(4,0) => 13

mLite

fun ackermann( 0, n ) = n + 1 
	| ( m, 0 ) = ackermann( m - 1, 1 )
	| ( m, n ) = ackermann( m - 1, ackermann(m, n - 1) )

Test code providing tuples from (0,0) to (3,8)

fun jota x = map (fn x = x-1) ` iota x

fun test_tuples (x, y) = append_map (fn a = map (fn b = (b, a)) ` jota x) ` jota y

map ackermann (test_tuples(4,9))

Result

[1, 2, 3, 5, 2, 3, 5, 13, 3, 4, 7, 29, 4, 5, 9, 61, 5, 6, 11, 125, 6, 7, 13, 253, 7, 8, 15, 509, 8, 9, 17, 1021, 9, 10, 19, 2045]

Modula-2

MODULE ackerman;

IMPORT  ASCII, NumConv, InOut;

VAR     m, n            : LONGCARD;
        string          : ARRAY [0..19] OF CHAR;
        OK              : BOOLEAN;

PROCEDURE Ackerman (x, y   : LONGCARD) : LONGCARD;

BEGIN
  IF    x = 0  THEN  RETURN  y + 1
  ELSIF y = 0  THEN  RETURN  Ackerman (x - 1 , 1)
  ELSE
    RETURN  Ackerman (x - 1 , Ackerman (x , y - 1))
  END
END Ackerman;

BEGIN
  FOR  m := 0  TO  3  DO
    FOR  n := 0  TO  6  DO
      NumConv.Num2Str (Ackerman (m, n), 10, string, OK);
      IF  OK  THEN
        InOut.WriteString (string)
      ELSE
        InOut.WriteString ("* Error in number * ")
      END;
      InOut.Write (ASCII.HT)
    END;
    InOut.WriteLn
  END;
  InOut.WriteLn
END ackerman.
Output:
jan@Beryllium:~/modula/rosetta$ ackerman

1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15

5 13 29 61 125 253 509

Modula-3

The type CARDINAL is defined in Modula-3 as [0..LAST(INTEGER)], in other words, it can hold all positive integers.

MODULE Ack EXPORTS Main;

FROM IO IMPORT Put;
FROM Fmt IMPORT Int;

PROCEDURE Ackermann(m, n: CARDINAL): CARDINAL =
  BEGIN
    IF m = 0 THEN 
      RETURN n + 1;
    ELSIF n = 0 THEN
      RETURN Ackermann(m - 1, 1);
    ELSE
      RETURN Ackermann(m - 1, Ackermann(m, n - 1));
    END;
  END Ackermann;

BEGIN
  FOR m := 0 TO 3 DO
    FOR n := 0 TO 6 DO
      Put(Int(Ackermann(m, n)) & " ");
    END;
    Put("\n");
  END;
END Ack.
Output:
1 2 3 4 5 6 7 
2 3 4 5 6 7 8 
3 5 7 9 11 13 15 
5 13 29 61 125 253 509 

MUMPS

Ackermann(m,n)	;
	If m=0 Quit n+1
	If m>0,n=0 Quit $$Ackermann(m-1,1)
	If m>0,n>0 Quit $$Ackermann(m-1,$$Ackermann(m,n-1))
	Set $Ecode=",U13-Invalid parameter for Ackermann: m="_m_", n="_n_","

Write $$Ackermann(1,8) ; 10
Write $$Ackermann(2,8) ; 19
Write $$Ackermann(3,5) ; 253

Neko

/**
 Ackermann recursion, in Neko
 Tectonics:
    nekoc ackermann.neko
    neko ackermann 4 0
*/
ack = function(x,y) {
   if (x == 0) return y+1;
   if (y == 0) return ack(x-1,1);
   return ack(x-1, ack(x,y-1));
};

var arg1 = $int($loader.args[0]);
var arg2 = $int($loader.args[1]);

/* If not given, or negative, default to Ackermann(3,4) */
if (arg1 == null || arg1 < 0) arg1 = 3;
if (arg2 == null || arg2 < 0) arg2 = 4;

try
   $print("Ackermann(", arg1, ",", arg2, "): ", ack(arg1,arg2), "\n")
catch problem
   $print("Ackermann(", arg1, ",", arg2, "): ", problem, "\n")
Output:
prompt$ nekoc ackermann.neko
prompt$ neko ackermann.n 3 4
Ackermann(3,4): 125

prompt$ time neko ackermann.n 4 1
Ackermann(4,1): Stack Overflow

real    0m31.475s
user    0m31.460s
sys     0m0.012s

prompt$ time neko ackermann 3 10
Ackermann(3,10): 8189

real    0m1.865s
user    0m1.862s
sys     0m0.004s

Nemerle

In Nemerle, we can state the Ackermann function as a lambda. By using pattern-matching, our definition strongly resembles the mathematical notation.

using System;
using Nemerle.IO;


def ackermann(m, n) {
    def A = ackermann;
    match(m, n) {
        | (0, n) => n + 1
        | (m, 0) when m > 0 => A(m - 1, 1)
        | (m, n) when m > 0 && n > 0 => A(m - 1, A(m, n - 1))
        | _ => throw Exception("invalid inputs");
    }
}


for(mutable m = 0; m < 4; m++) {
    for(mutable n = 0; n < 5; n++) {
        print("ackermann($m, $n) = $(ackermann(m, n))\n");
    }
}

A terser version using implicit match (which doesn't use the alias A internally):

def ackermann(m, n) {
    | (0, n) => n + 1
    | (m, 0) when m > 0 => ackermann(m - 1, 1)
    | (m, n) when m > 0 && n > 0 => ackermann(m - 1, ackermann(m, n - 1))
    | _ => throw Exception("invalid inputs");
}

Or, if we were set on using the A notation, we could do this:

def ackermann = {
    def A(m, n) {
        | (0, n) => n + 1
        | (m, 0) when m > 0 => A(m - 1, 1)
        | (m, n) when m > 0 && n > 0 => A(m - 1, A(m, n - 1))
        | _ => throw Exception("invalid inputs");
    }
    A
}

NetRexx

/* NetRexx */
options replace format comments java crossref symbols binary

numeric digits 66

parse arg j_ k_ .
if j_ = '' | j_ = '.' | \j_.datatype('w') then j_ = 3
if k_ = '' | k_ = '.' | \k_.datatype('w') then k_ = 5

loop m_ = 0 to j_
  say
  loop n_ = 0 to k_
    say 'ackermann('m_','n_') =' ackermann(m_, n_).right(5)
    end n_
  end m_
return

method ackermann(m, n) public static
  select
    when m = 0 then rval = n + 1
    when n = 0 then rval = ackermann(m - 1, 1)
    otherwise       rval = ackermann(m - 1, ackermann(m, n - 1))
    end
  return rval

NewLISP

#! /usr/local/bin/newlisp

(define (ackermann m n)
  (cond ((zero? m) (inc n))
        ((zero? n) (ackermann (dec m) 1))
        (true (ackermann (- m 1) (ackermann m (dec n))))))
In case of stack overflow error, you have to start your program with a proper "-s <value>" flag
as "newlisp -s 100000 ./ackermann.lsp".
See http://www.newlisp.org/newlisp_manual.html#stack_size

Nim

from strutils import parseInt

proc ackermann(m, n: int64): int64 =
  if m == 0:
    result = n + 1
  elif n == 0:
    result = ackermann(m - 1, 1)
  else:
    result = ackermann(m - 1, ackermann(m, n - 1))

proc getNumber(): int =
  try:
    result = stdin.readLine.parseInt
  except ValueError:
    echo "An integer, please!"
    result = getNumber()
  if result < 0:
    echo "Please Enter a non-negative Integer: "
    result = getNumber()

echo "First non-negative Integer please: "
let first = getNumber()
echo "Second non-negative Integer please: "
let second = getNumber()
echo "Result: ", $ackermann(first, second)

Nit

Source: the official Nit’s repository.

# Task: Ackermann function
#
# A simple straightforward recursive implementation.
module ackermann_function

fun ack(m, n: Int): Int
do
	if m == 0 then return n + 1
	if n == 0 then return ack(m-1,1)
	return ack(m-1, ack(m, n-1))
end

for m in [0..3] do
	for n in [0..6] do
		print ack(m,n)
	end
	print ""
end

Output:

1
2
3
4
5
6
7

2
3
4
5
6
7
8

3
5
7
9
11
13
15

5
13
29
61
125
253
509

Oberon-2

MODULE ackerman;

IMPORT  Out;

VAR     m, n    : INTEGER;

PROCEDURE Ackerman (x, y   : INTEGER) : INTEGER;

BEGIN
  IF    x = 0  THEN  RETURN  y + 1
  ELSIF y = 0  THEN  RETURN  Ackerman (x - 1 , 1)
  ELSE
    RETURN  Ackerman (x - 1 , Ackerman (x , y - 1))
  END
END Ackerman;

BEGIN
  FOR  m := 0  TO  3  DO
    FOR  n := 0  TO  6  DO
      Out.Int (Ackerman (m, n), 10);
      Out.Char (9X)
    END;
    Out.Ln
  END;
  Out.Ln
END ackerman.

Objeck

Translation of: C# – C sharp
class Ackermann {
  function : Main(args : String[]) ~ Nil {
    for(m := 0; m <= 3; ++m;) {
      for(n := 0; n <= 4; ++n;) {
        a := Ackermann(m, n);
        if(a > 0) {
          "Ackermann({$m}, {$n}) = {$a}"->PrintLine();
        };
      };
    };
  }
  
  function : Ackermann(m : Int, n : Int) ~ Int {
    if(m > 0) {
      if (n > 0) {
        return Ackermann(m - 1, Ackermann(m, n - 1));
      }
      else if (n = 0) {
        return Ackermann(m - 1, 1);
      };
    }
    else if(m = 0) {
      if(n >= 0) { 
        return n + 1;
      };
    };
    
    return -1;
  }
}
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125

OCaml

let rec a m n =
  if m=0 then (n+1) else
  if n=0 then (a (m-1) 1) else
  (a (m-1) (a m (n-1)))

or:

let rec a = function
  | 0, n -> (n+1)
  | m, 0 -> a(m-1, 1)
  | m, n -> a(m-1, a(m, n-1))

with memoization using an hash-table:

let h = Hashtbl.create 4001

let a m n =
  try Hashtbl.find h (m, n)
  with Not_found ->
    let res = a (m, n) in
    Hashtbl.add h (m, n) res;
    (res)

taking advantage of the memoization we start calling small values of m and n in order to reduce the recursion call stack:

let a m n =
  for _m = 0 to m do
    for _n = 0 to n do
      ignore(a _m _n);
    done;
  done;
  (a m n)

Arbitrary precision

With arbitrary-precision integers (Big_int module):

open Big_int
let one  = unit_big_int
let zero = zero_big_int
let succ = succ_big_int
let pred = pred_big_int
let eq = eq_big_int

let rec a m n =
  if eq m zero then (succ n) else
  if eq n zero then (a (pred m) one) else
  (a (pred m) (a m (pred n)))

compile with:

ocamlopt -o acker nums.cmxa acker.ml

Tail-Recursive

Here is a tail-recursive version:

let rec find_option h v =
  try Some(Hashtbl.find h v)
  with Not_found -> None

let rec a bounds caller todo m n =
  match m, n with
  | 0, n ->
      let r = (n+1) in
      ( match todo with
        | [] -> r
        | (m,n)::todo ->
            List.iter (fun k ->
              if not(Hashtbl.mem bounds k)
              then Hashtbl.add bounds k r) caller;
            a bounds [] todo m n )

  | m, 0 ->
      a bounds caller todo (m-1) 1

  | m, n ->
      match find_option bounds (m, n-1) with
      | Some a_rec ->
          let caller = (m,n)::caller in
          a bounds caller todo (m-1) a_rec
      | None ->
          let todo = (m,n)::todo
          and caller = [(m, n-1)] in
          a bounds caller todo m (n-1)

let a = a (Hashtbl.create 42 (* arbitrary *) ) [] [] ;;

This one uses the arbitrary precision, the tail-recursion, and the optimisation explain on the Wikipedia page about (m,n) = (3,_).

open Big_int
let one  = unit_big_int
let zero = zero_big_int
let succ = succ_big_int
let pred = pred_big_int
let add = add_big_int
let sub = sub_big_int
let eq = eq_big_int
let three = succ(succ one)
let power = power_int_positive_big_int

let eq2 (a1,a2) (b1,b2) =
  (eq a1 b1) && (eq a2 b2)

module H = Hashtbl.Make
  (struct
     type t = Big_int.big_int * Big_int.big_int
     let equal = eq2
     let hash (x,y) = Hashtbl.hash
       (Big_int.string_of_big_int x ^ "," ^
          Big_int.string_of_big_int y)
       (* probably not a very good hash function *)
   end)

let rec find_option h v =
  try Some (H.find h v)
  with Not_found -> None

let rec a bounds caller todo m n =
  let may_tail r =
    let k = (m,n) in
    match todo with
    | [] -> r
    | (m,n)::todo ->
        List.iter (fun k ->
                     if not (H.mem bounds k)
                     then H.add bounds k r) (k::caller);
        a bounds [] todo m n
  in
  match m, n with
  | m, n when eq m zero ->
      let r = (succ n) in
      may_tail r
 
  | m, n when eq n zero ->
      let caller = (m,n)::caller in
      a bounds caller todo (pred m) one
 
  | m, n when eq m three ->
      let r = sub (power 2 (add n three)) three in
      may_tail r

  | m, n ->
      match find_option bounds (m, pred n) with
      | Some a_rec ->
          let caller = (m,n)::caller in
          a bounds caller todo (pred m) a_rec
      | None ->
          let todo = (m,n)::todo in
          let caller = [(m, pred n)] in
          a bounds caller todo m (pred n)
 
let a = a (H.create 42 (* arbitrary *)) [] [] ;;

let () =
  let m, n =
    try
      big_int_of_string Sys.argv.(1),
      big_int_of_string Sys.argv.(2)
    with _ ->
      Printf.eprintf "usage: %s <int> <int>\n" Sys.argv.(0);
      exit 1
  in
  let r = a m n in
  Printf.printf "(a %s %s) = %s\n"
      (string_of_big_int m)
      (string_of_big_int n)
      (string_of_big_int r);
;;

Octave

function r = ackerman(m, n)
  if ( m == 0 )
    r = n + 1;
  elseif ( n == 0 )
    r = ackerman(m-1, 1);
  else
    r = ackerman(m-1, ackerman(m, n-1));
  endif
endfunction

for i = 0:3
  disp(ackerman(i, 4));
endfor

Oforth

: A( m n -- p )
   m ifZero: [ n 1+ return ]
   m 1- n ifZero: [ 1 ] else: [ A( m, n 1- ) ] A
;

Ol

; simple version
(define (A m n)
    (cond
        ((= m 0) (+ n 1))
        ((= n 0) (A (- m 1) 1))
        (else (A (- m 1) (A m (- n 1))))))

(print "simple version (A 3 6): " (A 3 6))

; smart (lazy) version
(define (ints-from n)
   (cons* n (delay (ints-from (+ n 1)))))

(define (knuth-up-arrow a n b)
  (let loop ((n n) (b b))
    (cond ((= b 0) 1)
          ((= n 1) (expt a b))
          (else    (loop (- n 1) (loop n (- b 1)))))))

(define (A+ m n)
   (define (A-stream)
      (cons*
         (ints-from 1) ;; m = 0
         (ints-from 2) ;; m = 1
         ;; m = 2
         (lmap (lambda (n)
                  (+ (* 2 (+ n 1)) 1))
            (ints-from 0))
         ;; m = 3
         (lmap (lambda (n)
               (- (knuth-up-arrow 2 (- m 2) (+ n 3)) 3))
            (ints-from 0))
         ;; m = 4...
         (delay (ldrop (A-stream) 3))))
   (llref (llref (A-stream) m) n))

(print "extended version (A 3 6): " (A+ 3 6))
Output:
simple version (A 3 6): 509
extended version (A 3 6): 509

OOC

ack: func (m: Int, n: Int) -> Int {
  if (m == 0) {
    n + 1 
  } else if (n == 0) {
    ack(m - 1, 1)
  } else {
    ack(m - 1, ack(m, n - 1)) 
  }
}

main: func {
  for (m in 0..4) {
    for (n in 0..10) {
      "ack(#{m}, #{n}) = #{ack(m, n)}" println()
    }   
  }
}

ooRexx

loop m = 0 to 3
    loop n = 0 to 6
        say "Ackermann("m", "n") =" ackermann(m, n)
    end
end

::routine ackermann
  use strict arg m, n
  -- give us some precision room
  numeric digits 10000
  if m = 0 then return n + 1
  else if n = 0 then return ackermann(m - 1, 1)
  else return ackermann(m - 1, ackermann(m, n - 1))
Output:
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(0, 5) = 6
Ackermann(0, 6) = 7
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(1, 5) = 7
Ackermann(1, 6) = 8
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(2, 5) = 13
Ackermann(2, 6) = 15
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125
Ackermann(3, 5) = 253
Ackermann(3, 6) = 509

Order

#include <order/interpreter.h>

#define ORDER_PP_DEF_8ack ORDER_PP_FN(    \
8fn(8X, 8Y,                               \
    8cond((8is_0(8X), 8inc(8Y))           \
          (8is_0(8Y), 8ack(8dec(8X), 1))  \
          (8else, 8ack(8dec(8X), 8ack(8X, 8dec(8Y)))))))

ORDER_PP(8to_lit(8ack(3, 4)))      // 125

Oz

Oz has arbitrary precision integers.

declare

  fun {Ack M N}
     if     M == 0 then N+1
     elseif N == 0 then {Ack M-1 1}
     else               {Ack M-1 {Ack M N-1}}
     end
  end

in

  {Show {Ack 3 7}}

PARI/GP

Naive implementation.

A(m,n)={
  if(m,
    if(n,
      A(m-1, A(m,n-1))
    ,
      A(m-1,1)
    )
  ,
    n+1
  )
};

Pascal

Program Ackerman;

function ackermann(m, n: Integer) : Integer;
begin
   if m = 0 then
      ackermann := n+1
   else if n = 0 then
      ackermann := ackermann(m-1, 1)
   else
      ackermann := ackermann(m-1, ackermann(m, n-1));
end;

var
   m, n	: Integer;

begin
   for n := 0 to 6 do
      for m := 0 to 3 do
	 WriteLn('A(', m, ',', n, ') = ', ackermann(m,n));
end.

Perl

We memoize calls to A to make A(2, n) and A(3, n) feasible for larger values of n.

{
    my @memo;
    sub A {
        my( $m, $n ) = @_;
        $memo[ $m ][ $n ] and return $memo[ $m ][ $n ];
        $m or return $n + 1;
        return $memo[ $m ][ $n ] = (
            $n
               ? A( $m - 1, A( $m, $n - 1 ) )
               : A( $m - 1, 1 )
        );
    }
}

An implementation using the conditional statements 'if', 'elsif' and 'else':

sub A {
    my ($m, $n) = @_;
    if    ($m == 0) { $n + 1 }
    elsif ($n == 0) { A($m - 1, 1) }
    else            { A($m - 1, A($m, $n - 1)) }
}

An implementation using ternary chaining:

sub A {
  my ($m, $n) = @_;
  $m == 0 ? $n + 1 :
  $n == 0 ? A($m - 1, 1) :
            A($m - 1, A($m, $n - 1))
}

Adding memoization and extra terms:

use Memoize;  memoize('ack2');
use bigint try=>"GMP";

sub ack2 {
   my ($m, $n) = @_;
   $m == 0 ? $n + 1 :
   $m == 1 ? $n + 2 :
   $m == 2 ? 2*$n + 3 :
   $m == 3 ? 8 * (2**$n - 1) + 5 :
   $n == 0 ? ack2($m-1, 1)
           : ack2($m-1, ack2($m, $n-1));
}
print "ack2(3,4) is ", ack2(3,4), "\n";
print "ack2(4,1) is ", ack2(4,1), "\n";
print "ack2(4,2) has ", length(ack2(4,2)), " digits\n";
Output:
ack2(3,4) is 125
ack2(4,1) is 65533
ack2(4,2) has 19729 digits

An optimized version, which uses @_ as a stack, instead of recursion. Very fast.

use strict;
use warnings;
use Math::BigInt;

use constant two => Math::BigInt->new(2);

sub ack {
	my $n = pop;
	while( @_ ) {
		my $m = pop;
		if( $m > 3 ) {
			push @_, (--$m) x $n;
			push @_, reverse 3 .. --$m;
			$n = 13;
		} elsif( $m == 3 ) {
			if( $n < 29 ) {
				$n = ( 1 << ( $n + 3 ) ) - 3;
			} else {
				$n = two ** ( $n + 3 ) - 3;
			}
		} elsif( $m == 2 ) {
			$n = 2 * $n + 3;
		} elsif( $m >= 0 ) {
			$n = $n + $m + 1;
		} else {
			die "negative m!";
		}
	}
	$n;
}
 
print "ack(3,4) is ", ack(3,4), "\n";
print "ack(4,1) is ", ack(4,1), "\n";
print "ack(4,2) has ", length(ack(4,2)), " digits\n";

Phix

native version

function ack(integer m, integer n)
    if m=0 then
        return n+1
    elsif m=1 then
        return n+2
    elsif m=2 then
        return 2*n+3
    elsif m=3 then
        return power(2,n+3)-3
    elsif m>0 and n=0 then
        return ack(m-1,1)
    else
        return ack(m-1,ack(m,n-1))
    end if
end function
 
constant limit = 23,
         fmtlens = {1,2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8}
 
atom t0 = time()
printf(1,"   0")
for j=1 to limit do
    string fmt = sprintf(" %%%dd",fmtlens[j+1])
    printf(1,fmt,j)
end for
printf(1,"\n")
for i=0 to 5 do
    printf(1,"%d:",i)
    for j=0 to iff(i>=4?5-i:limit) do
        string fmt = sprintf(" %%%dd",fmtlens[j+1])
        printf(1,fmt,{ack(i,j)})
    end for
    printf(1,"\n")
end for
Output:
   0  1  2  3   4   5   6    7    8    9   10    11    12    13     14     15     16      17      18      19      20       21       22       23
0: 1  2  3  4   5   6   7    8    9   10   11    12    13    14     15     16     17      18      19      20      21       22       23       24
1: 2  3  4  5   6   7   8    9   10   11   12    13    14    15     16     17     18      19      20      21      22       23       24       25
2: 3  5  7  9  11  13  15   17   19   21   23    25    27    29     31     33     35      37      39      41      43       45       47       49
3: 5 13 29 61 125 253 509 1021 2045 4093 8189 16381 32765 65533 131069 262141 524285 1048573 2097149 4194301 8388605 16777213 33554429 67108861
4: 13 65533
5: 65533

ack(4,2) and above fail with power function overflow. ack(3,100) will get you an answer, but only accurate to 16 or so digits.

gmp version

Translation of: Go
Library: Phix/mpfr
-- demo\rosetta\Ackermann.exw
include mpfr.e

procedure ack(integer m, mpz n)
    if m=0 then                 
        mpz_add_ui(n, n, 1)                     -- return n+1
    elsif m=1 then
        mpz_add_ui(n, n, 2)                     -- return n+2
    elsif m=2 then
        mpz_mul_si(n, n, 2)
        mpz_add_ui(n, n, 3)                     -- return 2*n+3
    elsif m=3 then
        if not mpz_fits_integer(n) then
            -- As per Go: 2^MAXINT would most certainly run out of memory.
            -- (think about it: a million digits is fine but pretty daft; 
            --  however a billion digits requires > addressable memory.)
            integer bn = mpz_sizeinbase(n, 2)
            throw(sprintf("A(m,n) had n of %d bits; too large",bn))
        end if
        integer ni = mpz_get_integer(n)
        mpz_set_si(n, 8)
        mpz_mul_2exp(n, n, ni) -- (n:=8*2^ni)
        mpz_sub_ui(n, n, 3)                     -- return power(2,n+3)-3
    elsif mpz_cmp_si(n,0)=0 then
        mpz_set_si(n, 1)
        ack(m-1,n)                              -- return ack(m-1,1)
    else
        mpz_sub_ui(n, n, 1)
        ack(m,n)
        ack(m-1,n)                              -- return ack(m-1,ack(m,n-1))
    end if
end procedure

constant limit = 23,
         fmtlens = {1,2,2,2,3,3,3,4,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8},
         extras = {{3,100},{3,1e6},{4,2},{4,3}}

procedure ackermann_tests()
    atom t0 = time()
    atom n = mpz_init()
    printf(1,"   0")
    for j=1 to limit do
        string fmt = sprintf(" %%%dd",fmtlens[j+1])
        printf(1,fmt,j)
    end for
    printf(1,"\n")
    for i=0 to 5 do
        printf(1,"%d:",i)
        for j=0 to iff(i>=4?5-i:limit) do
            mpz_set_si(n, j)
            ack(i,n)
            string fmt = sprintf(" %%%ds",fmtlens[j+1])
            printf(1,fmt,{mpz_get_str(n)})
        end for
        printf(1,"\n")
    end for
    printf(1,"\n")
    for i=1 to length(extras) do
        integer {em, en} = extras[i]
        mpz_set_si(n, en)
        string res
        try
            ack(em,n)
            res = mpz_get_str(n)
            integer lr = length(res)
            if lr>50 then
                res[21..-21] = "..."
                res &= sprintf(" (%d digits)",lr)
            end if
        catch e
            -- ack(4,3), ack(5,1) and ack(6,0) all fail,
            --                   just as they should do
            res = "***ERROR***: "&e[E_USER]
        end try
        printf(1,"ack(%d,%d) %s\n",{em,en,res})
    end for     
    n = mpz_free(n)
    printf(1,"\n")
    printf(1,"ackermann_tests completed (%s)\n\n",{elapsed(time()-t0)})
end procedure

ackermann_tests()
Output:
   0  1  2  3   4   5   6    7    8    9   10    11    12    13     14     15     16      17      18      19      20       21       22       23
0: 1  2  3  4   5   6   7    8    9   10   11    12    13    14     15     16     17      18      19      20      21       22       23       24
1: 2  3  4  5   6   7   8    9   10   11   12    13    14    15     16     17     18      19      20      21      22       23       24       25
2: 3  5  7  9  11  13  15   17   19   21   23    25    27    29     31     33     35      37      39      41      43       45       47       49
3: 5 13 29 61 125 253 509 1021 2045 4093 8189 16381 32765 65533 131069 262141 524285 1048573 2097149 4194301 8388605 16777213 33554429 67108861
4: 13 65533
5: 65533

ack(3,100) 10141204801825835211973625643005
ack(3,1000000) 79205249834367186005...39107225301976875005 (301031 digits)
ack(4,2) 20035299304068464649...45587895905719156733 (19729 digits)
ack(4,3) ***ERROR***: A(m,n) had n of 65536 bits; too large

ackermann_tests completed (0.2s)

Phixmonti

def ack
    var n var m
    
    m 0 == if
        n 1 +
    else
        n 0 == if
            m 1 - 1 ack
        else
            m 1 - m n 1 - ack ack
        endif
    endif
enddef

3 6 ack print nl

PHP

function ackermann( $m , $n )
{
    if ( $m==0 )
    {
        return $n + 1;
    }
    elseif ( $n==0 )
    {
        return ackermann( $m-1 , 1 );
    }
    return ackermann( $m-1, ackermann( $m , $n-1 ) );
}

echo ackermann( 3, 4 );
// prints 125

Picat

go =>
    foreach(M in 0..3)
        println([m=M,[a(M,N) : N in 0..16]])
    end,
    nl,
    printf("a2(4,1): %d\n", a2(4,1)),
    nl,
    time(check_larger(3,10000)),
    nl,
    time(check_larger(4,2)),
    nl.

% Using a2/2 and chop off large output
check_larger(M,N) => 
    printf("a2(%d,%d): ", M,N),
    A = a2(M,N).to_string,
    Len = A.len,
    if Len < 50 then
      println(A)
    else 
      println(A[1..20] ++ ".." ++ A[Len-20..Len])
    end,
    println(digits=Len).

% Plain tabled (memoized) version with guards
table
a(0, N) = N+1 => true.
a(M, 0) = a(M-1,1), M > 0 => true.
a(M, N) = a(M-1,a(M, N-1)), M > 0, N > 0 => true.

% Faster and pure function version (no guards). 
% (Based on Python example.)
table
a2(0,N) = N + 1.
a2(1,N) = N + 2.
a2(2,N) = 2*N + 3.
a2(3,N) = 8*(2**N - 1) + 5.
a2(M,N) = cond(N == 0,a2(M-1, 1), a2(M-1, a2(M, N-1))).
Output:
[m = 0,[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17]]
[m = 1,[2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]]
[m = 2,[3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35]]
[m = 3,[5,13,29,61,125,253,509,1021,2045,4093,8189,16381,32765,65533,131069,262141,524285]]

a2(4,1): 65533

a2(3,10000): 15960504935046067079..454194438340773675005
digits = 3012
CPU time 0.02 seconds.

a2(4,2): 20035299304068464649..445587895905719156733
digits = 19729
CPU time 0.822 seconds.

PicoLisp

(de ack (X Y)
   (cond
      ((=0 X) (inc Y))
      ((=0 Y) (ack (dec X) 1))
      (T (ack (dec X) (ack X (dec Y)))) ) )

Piet

Rendered as wikitable:

ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww
ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww

This is a naive implementation that does not use any optimization. Find the explanation at [[1]]. Computing the Ackermann function for (4,1) is possible, but takes quite a while because the stack grows very fast to large dimensions.

Example output:

   ? 3
   ? 5
   253

Pike

int main(){
   write(ackermann(3,4) + "\n");
}
 
int ackermann(int m, int n){
   if(m == 0){
      return n + 1;
   } else if(n == 0){
      return ackermann(m-1, 1);
   } else {
      return ackermann(m-1, ackermann(m, n-1));
   }
}

PL/I

Ackerman: procedure (m, n) returns (fixed (30)) recursive;
   declare (m, n) fixed (30);
   if m = 0 then return (n+1);
   else if m > 0 & n = 0 then return (Ackerman(m-1, 1));
   else if m > 0 & n > 0 then return (Ackerman(m-1, Ackerman(m, n-1)));
   return (0);
end Ackerman;

PL/SQL

DECLARE

  FUNCTION ackermann(pi_m IN NUMBER,
                     pi_n IN NUMBER) RETURN NUMBER IS
  BEGIN
    IF pi_m = 0 THEN
      RETURN pi_n + 1;
    ELSIF pi_n = 0 THEN
      RETURN ackermann(pi_m - 1, 1);
    ELSE
      RETURN ackermann(pi_m - 1, ackermann(pi_m, pi_n - 1));
    END IF;
  END ackermann;

BEGIN
  FOR n IN 0 .. 6 LOOP
    FOR m IN 0 .. 3 LOOP
      dbms_output.put_line('A(' || m || ',' || n || ') = ' || ackermann(m, n));
    END LOOP;
  END LOOP;
END;
Output:
A(0,0) = 1
A(1,0) = 2
A(2,0) = 3
A(3,0) = 5
A(0,1) = 2
A(1,1) = 3
A(2,1) = 5
A(3,1) = 13
A(0,2) = 3
A(1,2) = 4
A(2,2) = 7
A(3,2) = 29
A(0,3) = 4
A(1,3) = 5
A(2,3) = 9
A(3,3) = 61
A(0,4) = 5
A(1,4) = 6
A(2,4) = 11
A(3,4) = 125
A(0,5) = 6
A(1,5) = 7
A(2,5) = 13
A(3,5) = 253
A(0,6) = 7
A(1,6) = 8
A(2,6) = 15
A(3,6) = 509

PostScript

/ackermann{
/n exch def
/m exch def %PostScript takes arguments in the reverse order as specified in the function definition
m 0 eq{
n 1 add
}if
m 0 gt n 0 eq and
{
m 1 sub 1 ackermann
}if
m 0 gt n 0 gt and{
m 1 sub m n 1 sub ackermann ackermann
}if
}def
Library: initlib
/A {
[/.m /.n] let
{
    {.m 0 eq} {.n succ} is?
    {.m 0 gt .n 0 eq and} {.m pred 1 A} is?
    {.m 0 gt .n 0 gt and} {.m pred .m .n pred A A} is?
} cond
end}.

Potion

ack = (m, n):
  if (m == 0): n + 1
. elsif (n == 0): ack(m - 1, 1)
. else: ack(m - 1, ack(m, n - 1)).
.

4 times(m):
  7 times(n):
    ack(m, n) print
    " " print.
  "\n" print.

PowerBASIC

FUNCTION PBMAIN () AS LONG
    DIM m AS QUAD, n AS QUAD

    m = ABS(VAL(INPUTBOX$("Enter a whole number.")))
    n = ABS(VAL(INPUTBOX$("Enter another whole number.")))

    MSGBOX STR$(Ackermann(m, n))
END FUNCTION

FUNCTION Ackermann (m AS QUAD, n AS QUAD) AS QUAD
    IF 0 = m THEN
        FUNCTION = n + 1
    ELSEIF 0 = n THEN
        FUNCTION = Ackermann(m - 1, 1)
    ELSE    ' m > 0; n > 0
        FUNCTION = Ackermann(m - 1, Ackermann(m, n - 1))
    END IF
END FUNCTION

PowerShell

Translation of: PHP
function ackermann ([long] $m, [long] $n) {
    if ($m -eq 0) {
        return $n + 1
    }
    
    if ($n -eq 0) {
        return (ackermann ($m - 1) 1)
    }
    
    return (ackermann ($m - 1) (ackermann $m ($n - 1)))
}

Building an example table (takes a while to compute, though, especially for the last three numbers; also it fails with the last line in Powershell v1 since the maximum recursion depth is only 100 there):

foreach ($m in 0..3) {
    foreach ($n in 0..6) {
        Write-Host -NoNewline ("{0,5}" -f (ackermann $m $n))
    }
    Write-Host
}
Output:
    1    2    3    4    5    6    7
    2    3    4    5    6    7    8
    3    5    7    9   11   13   15
    5   13   29   61  125  253  509

A More "PowerShelly" Way

function Get-Ackermann ([int64]$m, [int64]$n)
{
    if ($m -eq 0)
    {
        return $n + 1
    }
 
    if ($n -eq 0)
    {
        return Get-Ackermann ($m - 1) 1
    }
 
    return (Get-Ackermann ($m - 1) (Get-Ackermann $m ($n - 1)))
}

Save the result to an array (for possible future use?), then display it using the Format-Wide cmdlet:

$ackermann = 0..3 | ForEach-Object {$m = $_; 0..6 | ForEach-Object {Get-Ackermann $m  $_}}

$ackermann | Format-Wide {"{0,3}" -f $_} -Column 7 -Force
Output:
  1                   2                  3                  4                  5                  6                  7               
  2                   3                  4                  5                  6                  7                  8               
  3                   5                  7                  9                 11                 13                 15               
  5                  13                 29                 61                125                253                509               

Processing

int ackermann(int m, int n) {
  if (m == 0)
    return n + 1;
  else if (m > 0 && n == 0)
    return ackermann(m - 1, 1);
  else
    return ackermann( m - 1, ackermann(m, n - 1) );
}

// Call function to produce output:
// the first 4x7 Ackermann numbers
void setup() {
  for (int m=0; m<4; m++) {
    for (int n=0; n<7; n++) {
      print(ackermann(m, n), " ");
    }
    println();
  }
}
Output:
1  2  3  4  5  6  7
2  3  4  5  6  7  8
3  5  7  9  11  13  15
5  13  29  61  125  253  509

Processing Python mode

Python is not very adequate for deep recursion, so even setting sys.setrecursionlimit(1000000000) if m = 5 it throws 'maximum recursion depth exceeded'

from __future__ import print_function

def setup():
    for m in range(4):
        for n in range(7):
            print("{} ".format(ackermann(m, n)), end = "")
        print()
    # print('finished')

def ackermann(m, n):
    if m == 0:
        return n + 1
    elif m > 0 and n == 0:
        return ackermann(m - 1, 1)
    else:
        return ackermann(m - 1, ackermann(m, n - 1))

Processing.R

Processing.R may exceed its stack depth at ~n==6 and returns null.

setup <- function() {
  for (m in 0:3) {
    for (n in 0:4) {
      stdout$print(paste(ackermann(m, n), " "))
    }
    stdout$println("")
  }
}

ackermann <- function(m, n) {
  if ( m == 0 ) {
    return(n+1)
  } else if ( n == 0 ) {
    ackermann(m-1, 1)
  } else {
    ackermann(m-1, ackermann(m, n-1))
  }
}
Output:
1  2  3  4  5  
2  3  4  5  6  
3  5  7  9  11  
5  13  29  61  125

Prolog

Works with: SWI Prolog
:- table ack/3. % memoization reduces the execution time of ack(4,1,X) from several
                % minutes to about one second on a typical desktop computer.
ack(0, N, Ans) :- Ans is N+1.
ack(M, 0, Ans) :- M>0, X is M-1, ack(X, 1, Ans).
ack(M, N, Ans) :- M>0, N>0, X is M-1, Y is N-1, ack(M, Y, Ans2), ack(X, Ans2, Ans).

"Pure" Prolog Version (Uses Peano arithmetic instead of is/2):

ack(0,N,s(N)).
ack(s(M),0,P):- ack(M,s(0),P).
ack(s(M),s(N),P):- ack(s(M),N,S), ack(M,S,P).

% Peano's first axiom in Prolog is that s(0) AND s(s(N)):- s(N)
% Thanks to this we don't need explicit N > 0 checks.
% Nor explicit arithmetic operations like X is M-1.
% Recursion and unification naturally decrement s(N) to N.
% But: Prolog clauses are relations and cannot be replaced by their result, like functions.
% Because of this we do need an extra argument to hold the output of the function.
% And we also need an additional call to the function in the last clause.

% Example input/output:
% ?- ack(s(0),s(s(0)),P).
% P = s(s(s(s(0)))) ;
% false.

Pure

A 0 n = n+1;
A m 0 = A (m-1) 1 if m > 0;
A m n = A (m-1) (A m (n-1)) if m > 0 && n > 0;

Pure Data

#N canvas 741 265 450 436 10;
#X obj 83 111 t b l;
#X obj 115 163 route 0;
#X obj 115 185 + 1;
#X obj 83 380 f;
#X obj 161 186 swap;
#X obj 161 228 route 0;
#X obj 161 250 - 1;
#X obj 161 208 pack;
#X obj 115 314 t f f;
#X msg 161 272 \$1 1;
#X obj 115 142 t l;
#X obj 207 250 swap;
#X obj 273 271 - 1;
#X obj 207 272 t f f;
#X obj 207 298 - 1;
#X obj 207 360 pack;
#X obj 239 299 pack;
#X obj 83 77 inlet;
#X obj 83 402 outlet;
#X connect 0 0 3 0;
#X connect 0 1 10 0;
#X connect 1 0 2 0;
#X connect 1 1 4 0;
#X connect 2 0 8 0;
#X connect 3 0 18 0;
#X connect 4 0 7 0;
#X connect 4 1 7 1;
#X connect 5 0 6 0;
#X connect 5 1 11 0;
#X connect 6 0 9 0;
#X connect 7 0 5 0;
#X connect 8 0 3 1;
#X connect 8 1 15 1;
#X connect 9 0 10 0;
#X connect 10 0 1 0;
#X connect 11 0 13 0;
#X connect 11 1 12 0;
#X connect 12 0 16 1;
#X connect 13 0 14 0;
#X connect 13 1 16 0;
#X connect 14 0 15 0;
#X connect 15 0 10 0;
#X connect 16 0 10 0;
#X connect 17 0 0 0;

PureBasic

Procedure.q Ackermann(m, n)
  If m = 0
    ProcedureReturn n + 1
  ElseIf  n = 0
    ProcedureReturn Ackermann(m - 1, 1)
  Else
    ProcedureReturn Ackermann(m - 1, Ackermann(m, n - 1))
  EndIf
EndProcedure

Debug Ackermann(3,4)

Purity

data Iter = f => FoldNat <const $f One, $f> 
data Ackermann = FoldNat <const Succ, Iter>

Python

Python: Explicitly recursive

Works with: Python version 2.5
def ack1(M, N):
   return (N + 1) if M == 0 else (
      ack1(M-1, 1) if N == 0 else ack1(M-1, ack1(M, N-1)))

Another version:

from functools import lru_cache

@lru_cache(None)
def ack2(M, N):
    if M == 0:
        return N + 1
    elif N == 0:
        return ack2(M - 1, 1)
    else:
        return ack2(M - 1, ack2(M, N - 1))
Example of use:
>>> import sys
>>> sys.setrecursionlimit(3000)
>>> ack1(0,0)
1
>>> ack1(3,4)
125
>>> ack2(0,0)
1
>>> ack2(3,4)
125

From the Mathematica ack3 example:

def ack2(M, N):
   return (N + 1)   if M == 0 else (
          (N + 2)   if M == 1 else (
          (2*N + 3) if M == 2 else (
          (8*(2**N - 1) + 5) if M == 3 else (
          ack2(M-1, 1) if N == 0 else ack2(M-1, ack2(M, N-1))))))

Results confirm those of Mathematica for ack(4,1) and ack(4,2)

Python: Without recursive function calls

The heading is more correct than saying the following is iterative as an explicit stack is used to replace explicit recursive function calls. I don't think this is what Comp. Sci. professors mean by iterative.

from collections import deque

def ack_ix(m, n):
    "Paddy3118's iterative with optimisations on m"

    stack = deque([])
    stack.extend([m, n])

    while  len(stack) > 1:
        n, m = stack.pop(), stack.pop()

        if   m == 0:
            stack.append(n + 1)
        elif m == 1:
            stack.append(n + 2)
        elif m == 2:
            stack.append(2*n + 3)
        elif m == 3:
            stack.append(2**(n + 3) - 3)
        elif n == 0:
            stack.extend([m-1, 1])
        else:
            stack.extend([m-1, m, n-1])

    return stack[0]
Output:

(From an ipython shell)

In [26]: %time a_4_2 = ack_ix(4, 2)
Wall time: 0 ns

In [27]: # How big is the answer?

In [28]: float(a_4_2)
Traceback (most recent call last):

  File "<ipython-input-28-af4ad951eff8>", line 1, in <module>
    float(a_4_2)

OverflowError: int too large to convert to float


In [29]: # How many decimal digits in the answer?

In [30]: len(str(a_4_2))
Out[30]: 19729

Quackery

                           forward is ackermann ( m n --> r )
  [ over 0 = iff
      [ nip 1 + ] done
    dup 0 = iff
      [ drop 1 - 1
        ackermann ] done
     over 1 - unrot 1 -
     ackermann ackermann ]   resolves ackermann ( m n --> r )

  3 10 ackermann echo

Output:

8189

R

ackermann <- function(m, n) {
  if ( m == 0 ) {
    n+1
  } else if ( n == 0 ) {
    ackermann(m-1, 1)
  } else {
    ackermann(m-1, ackermann(m, n-1))
  }
}
for ( i in 0:3 ) {
  print(ackermann(i, 4))
}

Racket

#lang racket
(define (ackermann m n)
  (cond [(zero? m) (add1 n)]
        [(zero? n) (ackermann (sub1 m) 1)]
        [else (ackermann (sub1 m) (ackermann m (sub1 n)))]))

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.03
sub A(Int $m, Int $n) {
    if    $m == 0 { $n + 1 } 
    elsif $n == 0 { A($m - 1, 1) }
    else          { A($m - 1, A($m, $n - 1)) }
}

An implementation using multiple dispatch:

multi sub A(0,      Int $n) { $n + 1                   }
multi sub A(Int $m, 0     ) { A($m - 1, 1)             }
multi sub A(Int $m, Int $n) { A($m - 1, A($m, $n - 1)) }

Note that in either case, Int is defined to be arbitrary precision in Raku.

Here's a caching version of that, written in the sigilless style, with liberal use of Unicode, and the extra optimizing terms to make A(4,2) possible:

proto A(Int \𝑚, Int \𝑛) { (state @)[𝑚][𝑛] //= {*} }

multi A(0,      Int \𝑛) { 𝑛 + 1 }
multi A(1,      Int \𝑛) { 𝑛 + 2 }
multi A(2,      Int \𝑛) { 3 + 2 * 𝑛 }
multi A(3,      Int \𝑛) { 5 + 8 * (2 ** 𝑛 - 1) }

multi A(Int \𝑚, 0     ) { A(𝑚 - 1, 1) }
multi A(Int \𝑚, Int \𝑛) { A(𝑚 - 1, A(𝑚, 𝑛 - 1)) }

# Testing:
say A(4,1);
say .chars, " digits starting with ", .substr(0,50), "..." given A(4,2);
Output:
65533
19729 digits starting with 20035299304068464649790723515602557504478254755697...

REBOL

ackermann: func [m n] [
    case [
        m = 0 [n + 1]
        n = 0 [ackermann m - 1 1]
        true [ackermann m - 1 ackermann m n - 1]
    ]
]

ReScript

let _m = Sys.argv[2]
let _n = Sys.argv[3]

let m = int_of_string(_m)
let n = int_of_string(_n)

let rec a = (m, n) =>
  switch (m, n) {
  | (0, n) => (n+1)
  | (m, 0) => a(m-1, 1)
  | (m, n) => a(m-1, a(m, n-1))
  }

Js.log("ackermann(" ++ _m ++ ", " ++ _n ++ ") = "
    ++ string_of_int(a(m, n)))
Output:
$ bsc acker.res > acker.bs.js
$ node acker.bs.js 2 3
ackermann(2, 3) = 9
$ node acker.bs.js 3 4
ackermann(3, 4) = 125

REXX

no optimization

/*REXX program  calculates and displays  some values for the  Ackermann function.       */
            /*╔════════════════════════════════════════════════════════════════════════╗
              ║  Note:  the Ackermann function  (as implemented here)  utilizes deep   ║
              ║         recursive and is limited by the largest number that can have   ║
              ║         "1"  (unity) added to a number  (successfully and accurately). ║
              ╚════════════════════════════════════════════════════════════════════════╝*/
high=24
         do     j=0  to 3;                    say
             do k=0  to high % (max(1, j))
             call tell_Ack  j, k
             end   /*k*/
         end       /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell_Ack:  parse arg mm,nn;   calls=0            /*display an echo message to terminal. */
           #=right(nn,length(high))
           say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high),
                                      left('', 12)     'calls='right(calls, high)
           return
/*──────────────────────────────────────────────────────────────────────────────────────*/
ackermann: procedure expose calls                /*compute value of Ackermann function. */
           parse arg m,n;   calls=calls+1
           if m==0  then return n+1
           if n==0  then return ackermann(m-1, 1)
                         return ackermann(m-1, ackermann(m, n-1) )
output   when using the internal default input:
Ackermann(0, 0)=                       1              calls=                       1
Ackermann(0, 1)=                       2              calls=                       1
Ackermann(0, 2)=                       3              calls=                       1
Ackermann(0, 3)=                       4              calls=                       1
Ackermann(0, 4)=                       5              calls=                       1
Ackermann(0, 5)=                       6              calls=                       1
Ackermann(0, 6)=                       7              calls=                       1
Ackermann(0, 7)=                       8              calls=                       1
Ackermann(0, 8)=                       9              calls=                       1
Ackermann(0, 9)=                      10              calls=                       1
Ackermann(0,10)=                      11              calls=                       1
Ackermann(0,11)=                      12              calls=                       1
Ackermann(0,12)=                      13              calls=                       1
Ackermann(0,13)=                      14              calls=                       1
Ackermann(0,14)=                      15              calls=                       1
Ackermann(0,15)=                      16              calls=                       1
Ackermann(0,16)=                      17              calls=                       1
Ackermann(0,17)=                      18              calls=                       1
Ackermann(0,18)=                      19              calls=                       1
Ackermann(0,19)=                      20              calls=                       1
Ackermann(0,20)=                      21              calls=                       1
Ackermann(0,21)=                      22              calls=                       1
Ackermann(0,22)=                      23              calls=                       1
Ackermann(0,23)=                      24              calls=                       1
Ackermann(0,24)=                      25              calls=                       1

Ackermann(1, 0)=                       2              calls=                       2
Ackermann(1, 1)=                       3              calls=                       4
Ackermann(1, 2)=                       4              calls=                       6
Ackermann(1, 3)=                       5              calls=                       8
Ackermann(1, 4)=                       6              calls=                      10
Ackermann(1, 5)=                       7              calls=                      12
Ackermann(1, 6)=                       8              calls=                      14
Ackermann(1, 7)=                       9              calls=                      16
Ackermann(1, 8)=                      10              calls=                      18
Ackermann(1, 9)=                      11              calls=                      20
Ackermann(1,10)=                      12              calls=                      22
Ackermann(1,11)=                      13              calls=                      24
Ackermann(1,12)=                      14              calls=                      26
Ackermann(1,13)=                      15              calls=                      28
Ackermann(1,14)=                      16              calls=                      30
Ackermann(1,15)=                      17              calls=                      32
Ackermann(1,16)=                      18              calls=                      34
Ackermann(1,17)=                      19              calls=                      36
Ackermann(1,18)=                      20              calls=                      38
Ackermann(1,19)=                      21              calls=                      40
Ackermann(1,20)=                      22              calls=                      42
Ackermann(1,21)=                      23              calls=                      44
Ackermann(1,22)=                      24              calls=                      46
Ackermann(1,23)=                      25              calls=                      48
Ackermann(1,24)=                      26              calls=                      50

Ackermann(2, 0)=                       3              calls=                       5
Ackermann(2, 1)=                       5              calls=                      14
Ackermann(2, 2)=                       7              calls=                      27
Ackermann(2, 3)=                       9              calls=                      44
Ackermann(2, 4)=                      11              calls=                      65
Ackermann(2, 5)=                      13              calls=                      90
Ackermann(2, 6)=                      15              calls=                     119
Ackermann(2, 7)=                      17              calls=                     152
Ackermann(2, 8)=                      19              calls=                     189
Ackermann(2, 9)=                      21              calls=                     230
Ackermann(2,10)=                      23              calls=                     275
Ackermann(2,11)=                      25              calls=                     324
Ackermann(2,12)=                      27              calls=                     377

Ackermann(3, 0)=                       5              calls=                      15
Ackermann(3, 1)=                      13              calls=                     106
Ackermann(3, 2)=                      29              calls=                     541
Ackermann(3, 3)=                      61              calls=                    2432
Ackermann(3, 4)=                     125              calls=                   10307
Ackermann(3, 5)=                     253              calls=                   42438
Ackermann(3, 6)=                     509              calls=                  172233
Ackermann(3, 7)=                    1021              calls=                  693964
Ackermann(3, 8)=                    2045              calls=                 2785999

optimized for m ≤ 2

/*REXX program  calculates and displays  some values for the  Ackermann function.       */
high=24
         do     j=0  to 3;                    say
             do k=0  to high % (max(1, j))
             call tell_Ack  j, k
             end   /*k*/
         end       /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell_Ack:  parse arg mm,nn;   calls=0            /*display an echo message to terminal. */
           #=right(nn,length(high))
           say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high),
                                      left('', 12)     'calls='right(calls, high)
           return
/*──────────────────────────────────────────────────────────────────────────────────────*/
ackermann: procedure expose calls                /*compute value of Ackermann function. */
           parse arg m,n;   calls=calls+1
           if m==0  then return n + 1
           if n==0  then return ackermann(m-1, 1)
           if m==2  then return n + 3 + n
                         return ackermann(m-1, ackermann(m, n-1) )
output   when using the internal default input:
Ackermann(0, 0)=                       1              calls=         1
Ackermann(0, 1)=                       2              calls=         1
Ackermann(0, 2)=                       3              calls=         1
Ackermann(0, 3)=                       4              calls=         1
Ackermann(0, 4)=                       5              calls=         1
Ackermann(0, 5)=                       6              calls=         1
Ackermann(0, 6)=                       7              calls=         1
Ackermann(0, 7)=                       8              calls=         1
Ackermann(0, 8)=                       9              calls=         1
Ackermann(0, 9)=                      10              calls=         1
Ackermann(0,10)=                      11              calls=         1
Ackermann(0,11)=                      12              calls=         1
Ackermann(0,12)=                      13              calls=         1
Ackermann(0,13)=                      14              calls=         1
Ackermann(0,14)=                      15              calls=         1
Ackermann(0,15)=                      16              calls=         1
Ackermann(0,16)=                      17              calls=         1
Ackermann(0,17)=                      18              calls=         1
Ackermann(0,18)=                      19              calls=         1
Ackermann(0,19)=                      20              calls=         1
Ackermann(0,20)=                      21              calls=         1
Ackermann(0,21)=                      22              calls=         1
Ackermann(0,22)=                      23              calls=         1
Ackermann(0,23)=                      24              calls=         1
Ackermann(0,24)=                      25              calls=         1

Ackermann(1, 0)=                       2              calls=         2
Ackermann(1, 1)=                       3              calls=         4
Ackermann(1, 2)=                       4              calls=         6
Ackermann(1, 3)=                       5              calls=         8
Ackermann(1, 4)=                       6              calls=        10
Ackermann(1, 5)=                       7              calls=        12
Ackermann(1, 6)=                       8              calls=        14
Ackermann(1, 7)=                       9              calls=        16
Ackermann(1, 8)=                      10              calls=        18
Ackermann(1, 9)=                      11              calls=        20
Ackermann(1,10)=                      12              calls=        22
Ackermann(1,11)=                      13              calls=        24
Ackermann(1,12)=                      14              calls=        26
Ackermann(1,13)=                      15              calls=        28
Ackermann(1,14)=                      16              calls=        30
Ackermann(1,15)=                      17              calls=        32
Ackermann(1,16)=                      18              calls=        34
Ackermann(1,17)=                      19              calls=        36
Ackermann(1,18)=                      20              calls=        38
Ackermann(1,19)=                      21              calls=        40
Ackermann(1,20)=                      22              calls=        42
Ackermann(1,21)=                      23              calls=        44
Ackermann(1,22)=                      24              calls=        46
Ackermann(1,23)=                      25              calls=        48
Ackermann(1,24)=                      26              calls=        50

Ackermann(2, 0)=                       3              calls=         5
Ackermann(2, 1)=                       5              calls=         1
Ackermann(2, 2)=                       7              calls=         1
Ackermann(2, 3)=                       9              calls=         1
Ackermann(2, 4)=                      11              calls=         1
Ackermann(2, 5)=                      13              calls=         1
Ackermann(2, 6)=                      15              calls=         1
Ackermann(2, 7)=                      17              calls=         1
Ackermann(2, 8)=                      19              calls=         1
Ackermann(2, 9)=                      21              calls=         1
Ackermann(2,10)=                      23              calls=         1
Ackermann(2,11)=                      25              calls=         1
Ackermann(2,12)=                      27              calls=         1

Ackermann(3, 0)=                       5              calls=         2
Ackermann(3, 1)=                      13              calls=         4
Ackermann(3, 2)=                      29              calls=         6
Ackermann(3, 3)=                      61              calls=         8
Ackermann(3, 4)=                     125              calls=        10
Ackermann(3, 5)=                     253              calls=        12
Ackermann(3, 6)=                     509              calls=        14
Ackermann(3, 7)=                    1021              calls=        16
Ackermann(3, 8)=                    2045              calls=        18

optimized for m ≤ 4

This REXX version takes advantage that some of the lower numbers for the Ackermann function have direct formulas.

If the   numeric digits 100   were to be increased to   20000,   then the value of   Ackermann(4,2)  
(the last line of output)   would be presented with the full   19,729   decimal digits.

/*REXX program  calculates and displays  some values for the  Ackermann function.       */
numeric digits 100                               /*use up to 100 decimal digit integers.*/
                       /*╔═════════════════════════════════════════════════════════════╗
                         ║ When REXX raises a number to an integer power  (via the  ** ║
                         ║ operator,  the power can be positive, zero, or negative).   ║
                         ║ Ackermann(5,1)   is a bit impractical to calculate.         ║
                         ╚═════════════════════════════════════════════════════════════╝*/
high=24
         do     j=0  to 4;                   say
             do k=0  to high % (max(1, j))
             call tell_Ack  j, k
             if j==4 & k==2  then leave          /*there's no sense in going overboard. */
             end   /*k*/
         end       /*j*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell_Ack:  parse arg mm,nn;   calls=0            /*display an echo message to terminal. */
           #=right(nn,length(high))
           say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high),
                                      left('', 12)     'calls='right(calls, high)
           return
/*──────────────────────────────────────────────────────────────────────────────────────*/
ackermann: procedure expose calls                /*compute value of Ackermann function. */
           parse arg m,n;   calls=calls+1
           if m==0  then return n + 1
           if m==1  then return n + 2
           if m==2  then return n + 3 + n
           if m==3  then return 2**(n+3) - 3
           if m==4  then do; #=2                 /* [↓]  Ugh!  ···  and still more ughs.*/
                                      do (n+3)-1 /*This is where the heavy lifting is.  */
                                      #=2**#
                                      end
                             return #-3
                         end
           if n==0  then return ackermann(m-1, 1)
                         return ackermann(m-1, ackermann(m, n-1) )

Output note:   none of the numbers shown below use recursion to compute.

output   when using the internal default input:
Ackermann(0, 0)=                       1              calls=         1
Ackermann(0, 1)=                       2              calls=         1
Ackermann(0, 2)=                       3              calls=         1
Ackermann(0, 3)=                       4              calls=         1
Ackermann(0, 4)=                       5              calls=         1
Ackermann(0, 5)=                       6              calls=         1
Ackermann(0, 6)=                       7              calls=         1
Ackermann(0, 7)=                       8              calls=         1
Ackermann(0, 8)=                       9              calls=         1
Ackermann(0, 9)=                      10              calls=         1
Ackermann(0,10)=                      11              calls=         1
Ackermann(0,11)=                      12              calls=         1
Ackermann(0,12)=                      13              calls=         1
Ackermann(0,13)=                      14              calls=         1
Ackermann(0,14)=                      15              calls=         1
Ackermann(0,15)=                      16              calls=         1
Ackermann(0,16)=                      17              calls=         1
Ackermann(0,17)=                      18              calls=         1
Ackermann(0,18)=                      19              calls=         1
Ackermann(0,19)=                      20              calls=         1
Ackermann(0,20)=                      21              calls=         1
Ackermann(0,21)=                      22              calls=         1
Ackermann(0,22)=                      23              calls=         1
Ackermann(0,23)=                      24              calls=         1
Ackermann(0,24)=                      25              calls=         1

Ackermann(1, 0)=                       2              calls=         1
Ackermann(1, 1)=                       3              calls=         1
Ackermann(1, 2)=                       4              calls=         1
Ackermann(1, 3)=                       5              calls=         1
Ackermann(1, 4)=                       6              calls=         1
Ackermann(1, 5)=                       7              calls=         1
Ackermann(1, 6)=                       8              calls=         1
Ackermann(1, 7)=                       9              calls=         1
Ackermann(1, 8)=                      10              calls=         1
Ackermann(1, 9)=                      11              calls=         1
Ackermann(1,10)=                      12              calls=         1
Ackermann(1,11)=                      13              calls=         1
Ackermann(1,12)=                      14              calls=         1
Ackermann(1,13)=                      15              calls=         1
Ackermann(1,14)=                      16              calls=         1
Ackermann(1,15)=                      17              calls=         1
Ackermann(1,16)=                      18              calls=         1
Ackermann(1,17)=                      19              calls=         1
Ackermann(1,18)=                      20              calls=         1
Ackermann(1,19)=                      21              calls=         1
Ackermann(1,20)=                      22              calls=         1
Ackermann(1,21)=                      23              calls=         1
Ackermann(1,22)=                      24              calls=         1
Ackermann(1,23)=                      25              calls=         1
Ackermann(1,24)=                      26              calls=         1

Ackermann(2, 0)=                       3              calls=         1
Ackermann(2, 1)=                       5              calls=         1
Ackermann(2, 2)=                       7              calls=         1
Ackermann(2, 3)=                       9              calls=         1
Ackermann(2, 4)=                      11              calls=         1
Ackermann(2, 5)=                      13              calls=         1
Ackermann(2, 6)=                      15              calls=         1
Ackermann(2, 7)=                      17              calls=         1
Ackermann(2, 8)=                      19              calls=         1
Ackermann(2, 9)=                      21              calls=         1
Ackermann(2,10)=                      23              calls=         1
Ackermann(2,11)=                      25              calls=         1
Ackermann(2,12)=                      27              calls=         1

Ackermann(3, 0)=                       5              calls=         1
Ackermann(3, 1)=                      13              calls=         1
Ackermann(3, 2)=                      29              calls=         1
Ackermann(3, 3)=                      61              calls=         1
Ackermann(3, 4)=                     125              calls=         1
Ackermann(3, 5)=                     253              calls=         1
Ackermann(3, 6)=                     509              calls=         1
Ackermann(3, 7)=                    1021              calls=         1
Ackermann(3, 8)=                    2045              calls=         1

Ackermann(4, 0)=                      13              calls=         1
Ackermann(4, 1)=                   65533              calls=         1
Ackermann(4, 2)=89506130880933368E+19728              calls=         1

Ring

Translation of: C#
for m = 0 to 3
        for n = 0 to 4
                see "Ackermann(" + m + ", " + n + ") = " + Ackermann(m, n) + nl
         next
next

func Ackermann m, n
        if m > 0
           if n > 0
                return Ackermann(m - 1, Ackermann(m, n - 1))
            but n = 0
                return Ackermann(m - 1, 1)
            ok 
        but m = 0
            if n >= 0 
                return n + 1
            ok
        ok
Raise("Incorrect Numerical input !!!")
Output:
Ackermann(0, 0) = 1
Ackermann(0, 1) = 2
Ackermann(0, 2) = 3
Ackermann(0, 3) = 4
Ackermann(0, 4) = 5
Ackermann(1, 0) = 2
Ackermann(1, 1) = 3
Ackermann(1, 2) = 4
Ackermann(1, 3) = 5
Ackermann(1, 4) = 6
Ackermann(2, 0) = 3
Ackermann(2, 1) = 5
Ackermann(2, 2) = 7
Ackermann(2, 3) = 9
Ackermann(2, 4) = 11
Ackermann(3, 0) = 5
Ackermann(3, 1) = 13
Ackermann(3, 2) = 29
Ackermann(3, 3) = 61
Ackermann(3, 4) = 125

Risc-V

the basic recursive function, because memorization and other improvements would blow the clarity.

ackermann: 	#x: a1, y: a2, return: a0
beqz a1, npe #case m = 0
beqz a2, mme #case m > 0 & n = 0
addi sp, sp, -8 #case m > 0 & n > 0
sw ra, 4(sp)
sw a1, 0(sp)
addi a2, a2, -1
jal ackermann
lw a1, 0(sp)
addi a1, a1, -1
mv a2, a0
jal ackermann
lw t0, 4(sp)
addi sp, sp, 8
jr t0, 0
npe:
addi a0, a2, 1
jr ra, 0
mme:
addi sp, sp, -4
sw ra, 0(sp)
addi a1, a1, -1
li a2, 1
jal ackermann
lw t0, 0(sp)
addi sp, sp, 4
jr t0, 0

Ruby

Translation of: Ada
def ack(m, n)
  if m == 0
    n + 1
  elsif n == 0
    ack(m-1, 1)
  else
    ack(m-1, ack(m, n-1))
  end
end

Example:

(0..3).each do |m|
  puts (0..6).map { |n| ack(m, n) }.join(' ')
end
Output:
 1 2 3 4 5 6 7 
 2 3 4 5 6 7 8 
 3 5 7 9 11 13 15 
 5 13 29 61 125 253 509

Run BASIC

print ackermann(1, 2)
 
function ackermann(m, n)
   if (m = 0)             then ackermann = (n + 1)
   if (m > 0) and (n = 0) then ackermann = ackermann((m - 1), 1)
   if (m > 0) and (n > 0) then ackermann = ackermann((m - 1), ackermann(m, (n - 1)))
end function

Rust

fn ack(m: isize, n: isize) -> isize {
    if m == 0 {
        n + 1
    } else if n == 0 {
        ack(m - 1, 1)
    } else {
        ack(m - 1, ack(m, n - 1))
    }
}

fn main() {
    let a = ack(3, 4);
    println!("{}", a); // 125
}

Or:

fn ack(m: u64, n: u64) -> u64 {
	match (m, n) {
		(0, n) => n + 1,
		(m, 0) => ack(m - 1, 1),
		(m, n) => ack(m - 1, ack(m, n - 1)),
	}
}

Sather

class MAIN is

  ackermann(m, n:INT):INT
    pre m >= 0 and n >= 0
  is
    if m = 0 then return n + 1; end;
    if n = 0 then return ackermann(m-1, 1); end;
    return ackermann(m-1, ackermann(m, n-1));
  end;

  main is
    n, m :INT;
    loop n := 0.upto!(6);
      loop m := 0.upto!(3);
        #OUT + "A(" + m + ", " + n + ") = " + ackermann(m, n) + "\n";
      end;
    end; 
  end;
end;

Instead of INT, the class INTI could be used, even though we need to use a workaround since in the GNU Sather v1.2.3 compiler the INTI literals are not implemented yet.

class MAIN is

  ackermann(m, n:INTI):INTI is
    zero ::= 0.inti; -- to avoid type conversion each time
    one  ::= 1.inti;
    if m = zero then return n + one; end;
    if n = zero then return ackermann(m-one, one); end;
    return ackermann(m-one, ackermann(m, n-one));
  end;

  main is
    n, m :INT;
    loop n := 0.upto!(6);
      loop m := 0.upto!(3);
        #OUT + "A(" + m + ", " + n + ") = " + ackermann(m.inti, n.inti) + "\n";
      end;
    end; 
  end;
end;

Scala

def ack(m: BigInt, n: BigInt): BigInt = {
  if (m==0) n+1
  else if (n==0) ack(m-1, 1)
  else ack(m-1, ack(m, n-1))
}
Example:
scala> for ( m <- 0 to 3; n <- 0 to 6 ) yield ack(m,n)
res0: Seq.Projection[BigInt] = RangeG(1, 2, 3, 4, 5, 6, 7, 2, 3, 4, 5, 6, 7, 8, 3, 5, 7, 9, 11, 13, 15, 5, 13, 29, 61, 125, 253, 509)

Memoized version using a mutable hash map:

val ackMap = new mutable.HashMap[(BigInt,BigInt),BigInt]
def ackMemo(m: BigInt, n: BigInt): BigInt = {
  ackMap.getOrElseUpdate((m,n), ack(m,n))
}

Scheme

(define (A m n)
    (cond
        ((= m 0) (+ n 1))
        ((= n 0) (A (- m 1) 1))
        (else (A (- m 1) (A m (- n 1))))))

An improved solution that uses a lazy data structure, streams, and defines Knuth up-arrows to calculate iterative exponentiation:

(define (A m n)
  (letrec ((A-stream
    (cons-stream
      (ints-from 1) ;; m = 0
      (cons-stream
        (ints-from 2) ;; m = 1
        (cons-stream
          ;; m = 2
          (stream-map (lambda (n)
                        (1+ (* 2 (1+ n))))
                      (ints-from 0))
          (cons-stream
            ;; m = 3
            (stream-map (lambda (n)
                          (- (knuth-up-arrow 2 (- m 2) (+ n 3)) 3))
                        (ints-from 0))
             ;; m = 4...
            (stream-tail A-stream 3)))))))
    (stream-ref (stream-ref A-stream m) n)))

(define (ints-from n)
  (letrec ((ints-rec (cons-stream n (stream-map 1+ ints-rec))))
    ints-rec))

(define (knuth-up-arrow a n b)
  (let loop ((n n) (b b))
    (cond ((= b 0) 1)
          ((= n 1) (expt a b))
          (else    (loop (-1+ n) (loop n (-1+ b)))))))

Scilab

clear
function acker=ackermann(m,n)
    global calls
    calls=calls+1
    if m==0 then     acker=n+1
    else
        if n==0 then acker=ackermann(m-1,1)
                else acker=ackermann(m-1,ackermann(m,n-1))
        end
    end
endfunction
function printacker(m,n)
    global calls
    calls=0
    printf('ackermann(%d,%d)=',m,n)
    printf('%d  calls=%d\n',ackermann(m,n),calls)
endfunction
maxi=3; maxj=6
for i=0:maxi
   for j=0:maxj
       printacker(i,j)
   end
end
Output:
ackermann(0,0)=1  calls=1
ackermann(0,1)=2  calls=1
ackermann(0,2)=3  calls=1
ackermann(0,3)=4  calls=1
ackermann(0,4)=5  calls=1
ackermann(0,5)=6  calls=1
ackermann(0,6)=7  calls=1
ackermann(1,0)=2  calls=2
ackermann(1,1)=3  calls=4
ackermann(1,2)=4  calls=6
ackermann(1,3)=5  calls=8
ackermann(1,4)=6  calls=10
ackermann(1,5)=7  calls=12
ackermann(1,6)=8  calls=14
ackermann(2,0)=3  calls=5
ackermann(2,1)=5  calls=14
ackermann(2,2)=7  calls=27
ackermann(2,3)=9  calls=44
ackermann(2,4)=11  calls=65
ackermann(2,5)=13  calls=90
ackermann(2,6)=15  calls=119
ackermann(3,0)=5  calls=15
ackermann(3,1)=13  calls=106
ackermann(3,2)=29  calls=541
ackermann(3,3)=61  calls=2432
ackermann(3,4)=125  calls=10307
ackermann(3,5)=253  calls=42438
ackermann(3,6)=509  calls=172233

Seed7

const func integer: ackermann (in integer: m, in integer: n) is func
  result
    var integer: result is 0;
  begin
    if m = 0 then
      result := succ(n);
    elsif n = 0 then
      result := ackermann(pred(m), 1);
    else
      result := ackermann(pred(m), ackermann(m, pred(n)));
    end if;
  end func;

Original source: [2]

SETL

program ackermann;

(for m in [0..3])
  print(+/ [rpad('' + ack(m, n), 4): n in [0..6]]);
end;

proc ack(m, n);
  return {[0,n+1]}(m) ? ack(m-1, {[0,1]}(n) ? ack(m, n-1));
end proc;

end program;

Shen

(define ack
  0 N -> (+ N 1)
  M 0 -> (ack (- M 1) 1)
  M N -> (ack (- M 1)
              (ack M (- N 1))))

Sidef

func A(m, n) {
    m == 0 ? (n + 1)
           : (n == 0 ? (A(m - 1, 1))
                     : (A(m - 1, A(m, n - 1))));
}

Alternatively, using multiple dispatch:

func A((0), n) { n + 1 }
func A(m, (0)) { A(m - 1, 1) }
func A(m,  n)  { A(m-1, A(m, n-1)) }

Calling the function:

say A(3, 2);     # prints: 29

Simula

as modified by R. Péter and R. Robinson:

 BEGIN
    INTEGER procedure
    Ackermann(g, p); SHORT INTEGER g, p;
        Ackermann:= IF g = 0 THEN p+1
            ELSE Ackermann(g-1, IF p = 0 THEN 1
                         ELSE Ackermann(g, p-1));

    INTEGER g, p;
    FOR p := 0 STEP 3 UNTIL 13 DO BEGIN
    	g := 4 - p/3;
        outtext("Ackermann("); outint(g, 0);
        outchar(','); outint(p, 2); outtext(") = ");
        outint(Ackermann(g, p), 0); outimage
    END
END
Output:
Ackermann(4, 0) = 13
Ackermann(3, 3) = 61
Ackermann(2, 6) = 15
Ackermann(1, 9) = 11
Ackermann(0,12) = 13

Slate

m@(Integer traits) ackermann: n@(Integer traits)
[
  m isZero
    ifTrue: [n + 1]
    ifFalse:
      [n isZero
	 ifTrue: [m - 1 ackermann: n]
	 ifFalse: [m - 1 ackermann: (m ackermann: n - 1)]]
].

Smalltalk

|ackermann|
ackermann := [ :n :m |
  (n = 0) ifTrue: [ (m + 1) ]
          ifFalse: [
           (m = 0) ifTrue: [ ackermann value: (n-1) value: 1 ]
                   ifFalse: [
                        ackermann value: (n-1)
                                  value: ( ackermann value: n
                                                     value: (m-1) )
                   ]
          ]
].

(ackermann value: 0 value: 0) displayNl.
(ackermann value: 3 value: 4) displayNl.

SmileBASIC

DEF ACK(M,N)
 IF M==0 THEN
  RETURN N+1
 ELSEIF M>0 AND N==0 THEN
  RETURN ACK(M-1,1)
 ELSE
  RETURN ACK(M-1,ACK(M,N-1))
 ENDIF
END

SNOBOL4

Works with: Macro Spitbol

Both Snobol4+ and CSnobol stack overflow, at ack(3,3) and ack(3,4), respectively.

define('ack(m,n)') :(ack_end)
ack     ack = eq(m,0) n + 1 :s(return)
        ack = eq(n,0) ack(m - 1,1) :s(return)
        ack = ack(m - 1,ack(m,n - 1)) :(return)
ack_end

*       # Test and display ack(0,0) .. ack(3,6)
L1      str = str ack(m,n) ' '
        n = lt(n,6) n + 1 :s(L1)
        output = str; str = ''
        n = 0; m = lt(m,3) m + 1 :s(L1)
end
Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509

SNUSP

   /==!/==atoi=@@@-@-----#
   |   |                          Ackermann function
   |   |       /=========\!==\!====\  recursion:
$,@/>,@/==ack=!\?\<+#    |   |     |   A(0,j) -> j+1
 j   i           \<?\+>-@/#  |     |   A(i,0) -> A(i-1,1)
                    \@\>@\->@/@\<-@/#  A(i,j) -> A(i-1,A(i,j-1))
                      |  |     |
            #      #  |  |     |             /+<<<-\  
            /-<<+>>\!=/  \=====|==!/========?\>>>=?/<<#
            ?      ?           |   \<<<+>+>>-/
            \>>+<<-/!==========/
            #      #

One could employ tail recursion elimination by replacing "@/#" with "/" in two places above.

SPAD

NNI ==> NonNegativeInteger

A:(NNI,NNI) -> NNI

A(m,n) ==
  m=0 => n+1
  m>0 and n=0 => A(m-1,1)
  m>0 and n>0 => A(m-1,A(m,n-1))
  
-- Example  
matrix [[A(i,j) for i in 0..3] for j in 0..3]
Output:

        +1  2  3  5 +
        |           |
        |2  3  5  13|
   (1)  |           |
        |3  4  7  29|
        |           |
        +4  5  9  61+
                                             Type: Matrix(NonNegativeInteger)

SQL PL

Works with: Db2 LUW
version 9.7 or higher.

With SQL PL:

--#SET TERMINATOR @

SET SERVEROUTPUT ON@

CREATE OR REPLACE FUNCTION ACKERMANN(
  IN M SMALLINT,
  IN N BIGINT
 ) RETURNS BIGINT
 BEGIN
  DECLARE RET BIGINT;
  DECLARE STMT STATEMENT;

  IF (M = 0) THEN
   SET RET = N + 1;
  ELSEIF (N = 0) THEN
   PREPARE STMT FROM 'SET ? = ACKERMANN(? - 1, 1)';
   EXECUTE STMT INTO RET USING M;
  ELSE
   PREPARE STMT FROM 'SET ? = ACKERMANN(? - 1, ACKERMANN(?, ? - 1))';
   EXECUTE STMT INTO RET USING M, M, N;
  END IF;
  RETURN RET;
 END @
 
BEGIN
 DECLARE M SMALLINT DEFAULT 0;
 DECLARE N SMALLINT DEFAULT 0;
 DECLARE MAX_LEVELS CONDITION FOR SQLSTATE '54038';
 DECLARE CONTINUE HANDLER FOR MAX_LEVELS BEGIN END;

 WHILE (N <= 6) DO
  WHILE (M <= 3) DO
   CALL DBMS_OUTPUT.PUT_LINE('ACKERMANN(' || M || ', ' || N || ') = ' || ACKERMANN(M, N));
   SET M = M + 1;
  END WHILE;
  SET M = 0;
  SET N = N + 1;
 END WHILE;
END @

Output:

db2 -td@
db2 => CREATE OR REPLACE FUNCTION ACKERMANN(
...
db2 (cont.) => END @
DB20000I  The SQL command completed successfully.
db2 => BEGIN
db2 (cont.) => END
...
DB20000I  The SQL command completed successfully.

ACKERMANN(0, 0) = 1
ACKERMANN(1, 0) = 2
ACKERMANN(2, 0) = 3
ACKERMANN(3, 0) = 5
ACKERMANN(0, 1) = 2
ACKERMANN(1, 1) = 3
ACKERMANN(2, 1) = 5
ACKERMANN(3, 1) = 13
ACKERMANN(0, 2) = 3
ACKERMANN(1, 2) = 4
ACKERMANN(2, 2) = 7
ACKERMANN(3, 2) = 29
ACKERMANN(0, 3) = 4
ACKERMANN(1, 3) = 5
ACKERMANN(2, 3) = 9
ACKERMANN(3, 3) = 61
ACKERMANN(0, 4) = 5
ACKERMANN(1, 4) = 6
ACKERMANN(2, 4) = 11
ACKERMANN(0, 5) = 6
ACKERMANN(1, 5) = 7
ACKERMANN(2, 5) = 13
ACKERMANN(0, 6) = 7
ACKERMANN(1, 6) = 8
ACKERMANN(2, 6) = 15

The maximum levels of cascade calls in Db2 are 16, and in some cases when executing the Ackermann function, it arrives to this limit (SQL0724N). Thus, the code catches the exception and continues with the next try.

Standard ML

fun a (0, n) = n+1
  | a (m, 0) = a (m-1, 1)
  | a (m, n) = a (m-1, a (m, n-1))

Stata

mata
function ackermann(m,n) {
	if (m==0) {
		return(n+1)
	} else if (n==0) {
		return(ackermann(m-1,1))
	} else {
		return(ackermann(m-1,ackermann(m,n-1)))
	}
}

for (i=0; i<=3; i++) printf("%f\n",ackermann(i,4))
5
6
11
125
end

Swift

func ackerman(m:Int, n:Int) -> Int {
    if m == 0 {
        return n+1
    } else if n == 0 {
        return ackerman(m-1, 1)
    } else {
        return ackerman(m-1, ackerman(m, n-1))
    }
}

Tcl

Simple

Translation of: Ruby
proc ack {m n} {
    if {$m == 0} {
        expr {$n + 1}
    } elseif {$n == 0} {
        ack [expr {$m - 1}] 1
    } else {
        ack [expr {$m - 1}] [ack $m [expr {$n - 1}]]
    }
}

With Tail Recursion

With Tcl 8.6, this version is preferred (though the language supports tailcall optimization, it does not apply it automatically in order to preserve stack frame semantics):

proc ack {m n} {
    if {$m == 0} {
        expr {$n + 1}
    } elseif {$n == 0} {
        tailcall ack [expr {$m - 1}] 1
    } else {
        tailcall ack [expr {$m - 1}] [ack $m [expr {$n - 1}]]
    }
}

To Infinity… and Beyond!

If we want to explore the higher reaches of the world of Ackermann's function, we need techniques to really cut the amount of computation being done.

Works with: Tcl version 8.6
package require Tcl 8.6

# A memoization engine, from http://wiki.tcl.tk/18152
oo::class create cache {
    filter Memoize
    variable ValueCache
    method Memoize args {
        # Do not filter the core method implementations
        if {[lindex [self target] 0] eq "::oo::object"} {
            return [next {*}$args]
        }

        # Check if the value is already in the cache
        set key [self target],$args
        if {[info exist ValueCache($key)]} {
            return $ValueCache($key)
        }

        # Compute value, insert into cache, and return it
        return [set ValueCache($key) [next {*}$args]]
    }
    method flushCache {} {
        unset ValueCache
        # Skip the cacheing
        return -level 2 ""
    }
}

# Make an object, attach the cache engine to it, and define ack as a method
oo::object create cached
oo::objdefine cached {
    mixin cache
    method ack {m n} {
        if {$m==0} {
            expr {$n+1}
        } elseif {$m==1} {
            # From the Mathematica version
            expr {$m+2}
        } elseif {$m==2} {
            # From the Mathematica version
            expr {2*$n+3}
        } elseif {$m==3} {
            # From the Mathematica version
            expr {8*(2**$n-1)+5}
        } elseif {$n==0} {
            tailcall my ack [expr {$m-1}] 1
        } else {
            tailcall my ack [expr {$m-1}] [my