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
- Conway chained arrow notation for the Ackermann function.
11l
<lang 11l>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))</lang>
- Output:
1 125 65533
360 Assembly
The OS/360 linkage is a bit tricky with the S/360 basic instruction set. To simplify, the program is recursive not reentrant. <lang 360asm>* Ackermann function 07/09/2015 &LAB XDECO ®,&TARGET .*-----------------------------------------------------------------* .* THIS MACRO DISPLAYS THE REGISTER CONTENTS AS A TRUE * .* DECIMAL VALUE. XDECO IS NOT PART OF STANDARD S360 MACROS! *
- ------------------------------------------------------------------*
AIF (T'® 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 ®,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</lang>
- 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).
<lang 68000devpac>;
- 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</lang>
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.
<lang 8080asm> 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,'$' nl: db 13,10,'$'</lang>
- 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.
<lang asm> 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,'$' nl: db 13,10,'$'</lang>
- 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
<lang Forth> \ 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</lang>
- 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
<lang AArch64 Assembly> /* 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" </lang>
- 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
<lang 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.
</lang>
Action!
Action! language does not support recursion. Therefore an iterative approach with a stack has been proposed. <lang Action!>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</lang>
- 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
<lang 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));
}</lang>
Ada
<lang 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;</lang> 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
<lang agda> open import Data.Nat open import Data.Nat.Show open import IO
module Ackermann where
ack : ℕ -> ℕ -> ℕ 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))) </lang>
Note the unicode ℕ characters, they can be input in emacs agda mode using "\bN". Running in bash:
<lang bash> agda --compile Ackermann.agda ./Ackermann </lang>
- Output:
4093
ALGOL 60
<lang algol60>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 </lang>
- 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
<lang algol68>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</lang>
- 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
<lang algolw>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.</lang>
- 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
<lang APL>ackermann←{
0=1⊃⍵:1+2⊃⍵
0=2⊃⍵:∇(¯1+1⊃⍵)1
∇(¯1+1⊃⍵),∇(1⊃⍵),¯1+2⊃⍵
}</lang>
AppleScript
<lang 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</lang>
Argile
<lang Argile>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))</lang>
ARM Assembly
<lang ARM Assembly> /* 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" </lang>
- 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
<lang rebol>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]
]
]</lang>
- 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
<lang 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]</lang>
AutoHotkey
<lang 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)</lang>
AutoIt
Classical version
<lang autoit>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</lang>
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.
<lang autoit>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</lang>
AWK
<lang 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)
}
}
}</lang>
Babel
<lang 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 = } </lang>
- 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
<lang 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</lang>
BASIC256
<lang 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</lang>
- Output:
A(3,7) = 1021
<lang basic256># 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</lang>
- 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
<lang bbcbasic> 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))</lang>
QuickBasic
BASIC runs out of stack space very quickly. The call ack(3, 4) gives a stack error. <lang qbasic>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</lang>
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. <lang dos>::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% )</lang>
Because of the exit statements, running this bare closes one's shell, so this test routine handles the calling of Ackermann.cmd
<lang dos>::Ack.cmd
@echo off
cmd/c ackermann.cmd %1 %2
echo Ackermann(%1, %2)=%errorlevel%</lang>
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.
<lang 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</lang>
BCPL
<lang 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
}</lang>
beeswax
Iterative slow version:
<lang beeswax>
>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 < <
</lang>
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.
<lang beeswax>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)</lang>
Highly optimized and fast version, returns A(4,1)/A(5,0) almost instantaneously: <lang beeswax>
>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 <
</lang>
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
Since Befunge-93 doesn't have recursive capabilities we need to use an iterative algorithm. <lang befunge>&>&>vvg0>#0\#-:#1_1v @v:\<vp0 0:-1<\+< ^>00p>:#^_$1+\:#^_$. </lang>
Befunge-98
<lang befunge>r[1&&{0 >v
j
u>.@ 1> \:v ^ v:\_$1+ \^v_$1\1- u^>1-0fp:1-\0fg101-</lang> The program reads two integers (first m, then n) from command line, idles around funge space, then outputs the result of the Ackerman function. Since the latter is calculated truly recursively, the execution time becomes unwieldy for most m>3.
BQN
<lang BQN>A ← {
A 0‿n: n+1; A m‿0: A (m-1)‿1; A m‿n: A (m-1)‿(A m‿(n-1))
}</lang> Example usage: <lang> A 0‿3 4
A 1‿4
6
A 2‿4
11
A 3‿4
125</lang>
Bracmat
Three solutions are presented here. The first one is a purely recursive version, only using the formulas at the top of the page. The value of A(4,1) cannot be computed due to stack overflow. It can compute A(3,9) (4093), but probably not A(3,10) <lang bracmat>( Ack = m n
. !arg:(?m,?n)
& ( !m:0&!n+1
| !n:0&Ack$(!m+-1,1)
| Ack$(!m+-1,Ack$(!m,!n+-1))
)
);</lang> 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". <lang bracmat> ( 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$hash:?cache</lang>
- Some results:
A$(0,0):1 A$(3,13):65533 A$(3,14):131069 A$(4,1):65533
The last solution is a recursive solution that employs some extra formulas, inspired by the Common Lisp solution further down. <lang bracmat>( AckFormula = m n
. !arg:(?m,?n)
& ( !m:0&!n+1
| !m:1&!n+2
| !m:2&2*!n+3
| !m:3&2^(!n+3)+-3
| !n:0&AckFormula$(!m+-1,1)
| AckFormula$(!m+-1,AckFormula$(!m,!n+-1))
)
)</lang>
- Some results:
AckFormula$(4,1):65533 AckFormula$(4,2):2003529930406846464979072351560255750447825475569751419265016973.....22087777506072339445587895905719156733
The last computation costs about 0,03 seconds.
Brat
<lang 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</lang>
C
Straightforward implementation per Ackermann definition: <lang C>#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;
}</lang>
- 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: <lang C>#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;
}</lang>
- 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...
C#
Basic Version
<lang csharp>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));
}
}
}
}</lang>
- 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
<lang csharp> 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;
}
}
} </lang> 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
<lang cpp>#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";
}
}
} </lang>
Efficient version
C++11 with boost's big integer type. Compile with:
g++ -std=c++11 -I /path/to/boost ackermann.cpp.
<lang 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";
}</lang>
<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
<lang 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));
}</lang>
Clay
<lang 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));
}</lang>
CLIPS
Functional solution <lang clips>(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)))
)
)
)</lang>
- 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 <lang clips>(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)
)</lang> 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
<lang clojure>(defn ackermann [m n]
(cond (zero? m) (inc n)
(zero? n) (ackermann (dec m) 1)
:else (ackermann (dec m) (ackermann m (dec n)))))</lang>
CLU
<lang 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 </lang>
- 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
<lang 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
.</lang>
CoffeeScript
<lang 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</lang>
Comal
<lang 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</lang>
- Output:
1 2 3 4 5 2 3 4 5 6 3 5 7 9 11 5 13 29 61 125
Common Lisp
<lang lisp>(defun ackermann (m n)
(cond ((zerop m) (1+ n))
((zerop n) (ackermann (1- m) 1))
(t (ackermann (1- m) (ackermann m (1- n))))))</lang>
More elaborately: <lang lisp>(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))))</lang>
- Output:
A(0, 5) = 6A(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 <lang oberon2> 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.
</lang>
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
<lang 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.</lang>
<lang coq>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 (λ g, fold g (g (S O))) S.
</lang>
Crystal
<lang 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 </lang>
- 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
<lang d>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);
}</lang>
More Efficient Version
<lang d>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 .. $]);
}</lang>
- 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) <lang dart>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));
}</lang>
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.
<lang dc>[ # 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</lang>
- Output:
4093
Delphi
<lang 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;</lang>
Draco
<lang 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</lang>
- 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
<lang delphi>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;</lang>
Dylan
<lang 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;</lang>
E
<lang 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)) }
}</lang>
EasyLang
<lang>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</lang>
Egel
<lang Egel> def ackermann =
[ 0 N -> N + 1 | M 0 -> ackermann (M - 1) 1 | M N -> ackermann (M - 1) (ackermann M (N - 1)) ]
</lang>
Eiffel
<lang Eiffel> note description: "Example of Ackerman function" synopsis: "[ The EIS link below (in Eiffel Studio) will launch in either an in-IDE browser or and external browser (your choice). The protocol informs Eiffel Studio about what program to use to open the `src' reference, which can be URI, PDF, or DOC. See second EIS for more information. ]" EIS: "name=Ackermann_function", "protocol=URI", "tag=rosetta_code", "src=http://rosettacode.org/wiki/Ackermann_function" EIS: "name=eis_protocols", "protocol=URI", "tag=eiffel_docs", "src=https://docs.eiffel.com/book/eiffelstudio/protocols"
class APPLICATION
create make
feature {NONE} -- Initialization
make do print ("%N A(0,0):" + ackerman (0, 0).out) print ("%N A(1,0):" + ackerman (1, 0).out) print ("%N A(0,1):" + ackerman (0, 1).out) print ("%N A(1,1):" + ackerman (1, 1).out) print ("%N A(2,0):" + ackerman (2, 0).out) print ("%N A(2,1):" + ackerman (2, 1).out) print ("%N A(2,2):" + ackerman (2, 2).out) print ("%N A(0,2):" + ackerman (0, 2).out) print ("%N A(1,2):" + ackerman (1, 2).out) print ("%N A(3,3):" + ackerman (3, 3).out) print ("%N A(3,4):" + ackerman (3, 4).out) end
feature -- Access
ackerman (m, n: NATURAL): NATURAL do if m = 0 then Result := n + 1 elseif n = 0 then Result := ackerman (m - 1, 1) else Result := ackerman (m - 1, ackerman (m, n - 1)) end end end </lang>
Ela
<lang ela>ack 0 n = n+1 ack m 0 = ack (m - 1) 1 ack m n = ack (m - 1) <| ack m <| n - 1</lang>
Elena
ELENA 4.x : <lang elena>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()
}</lang>
- 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
<lang 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)</lang>
- 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
<lang lisp>(defun ackermann (m n)
(cond ((zerop m) (1+ n))
((zerop n) (ackermann (1- m) 1)) (t (ackermann (1- m) (ackermann m (1- n))))))</lang>
Erlang
<lang 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)).
</lang>
ERRE
Iterative version, using a stack. First version used various GOTOs statement, now removed and substituted with the new ERRE control statements.
<lang erre> 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 </lang>
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
<lang 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
</lang>
Euphoria
This is based on the VBScript example. <lang Euphoria>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</lang>
Ezhil
<lang 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
அ = int(உள்ளீடு("ஓர் எண்ணைத் தாருங்கள், அது பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாக இருக்கலாம்: ")) ஆ = int(உள்ளீடு("அதேபோல் இன்னோர் எண்ணைத் தாருங்கள், இதுவும் பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாகவோ இருக்கலாம்: "))
விடை = அகெர்மன்(அ, ஆ)
@(விடை < 0) ஆனால்
பதிப்பி "தவறான எண்களைத் தந்துள்ளீர்கள்!"
இல்லை
பதிப்பி "நீங்கள் தந்த எண்களுக்கான அகர்மென் மதிப்பு: ", விடை
முடி </lang>
F#
The following program implements the Ackermann function in F# but is not tail-recursive and so runs out of stack space quite fast. <lang fsharp>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]))</lang>
Transforming this into continuation passing style avoids limited stack space by permitting tail-recursion. <lang fsharp>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))</lang>
Factor
<lang 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 ;</lang>
Falcon
<lang 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</lang> 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
<lang 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 }</lang>
Fantom
<lang 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)}")
}
}
}
}</lang>
- 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: <lang qbasic>#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</lang><lang C> 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); }</lang><lang qbasic> END DYNC
DYNASM AckermannA(m AS INTEGER, n AS INTEGER) AS INTEGER</lang><lang asm> 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 </lang><lang qbasic>END DYNASM</lang>
- 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
<lang 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)</lang>
- Output:
2045
Forth
<lang 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 ;</lang>
- Example of use:
FORTH> 0 0 acker . 1 ok FORTH> 3 4 acker . 125 ok
An optimized version: <lang forth>: 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
- </lang>
Fortran
<lang fortran>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</lang>
Free Pascal
FreeBASIC
<lang 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</lang>
- 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
<lang 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) )</lang>
- 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
<lang 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))
</lang>
FutureBasic
<lang 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 </lang>
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
<lang 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</lang>
GAP
<lang 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;</lang>
Genyris
<lang 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)</lang>
GML
Define a script resource named ackermann and paste this code inside: <lang GML>///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))
}</lang>
gnuplot
<lang 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)</lang>
- Output:
5 6 11 stack overflow
Go
Classic version
<lang go>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)) }</lang>
Expanded version
<lang go>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))
}</lang>
Expanded version with arbitrary precision
<lang go>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:], ) } }</lang>
- 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
<lang golfscript>{
:_n; :_m;
_m 0= {_n 1+}
{_n 0= {_m 1- 1 ack}
{_m 1- _m _n 1- ack ack}
if}
if
}:ack;</lang>
Groovy
<lang 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))
}</lang> Test program: <lang groovy>def ackMatrix = (0..3).collect { m -> (0..8).collect { n -> ack(m, n) } } ackMatrix.each { it.each { elt -> printf "%7d", elt }; println() }</lang>
- 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!
Haskell
<lang 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)]</lang>
- Output:
1 125
Generating a list instead: <lang haskell>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]</lang>
Haxe
<lang 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));
}
}</lang>
Icon and Unicon
Taken from the public domain Icon Programming Library's acker in memrfncs, written by Ralph E. Griswold. <lang Icon>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</lang>
- 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
<lang 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)</lang>
Ioke
<lang ioke>ackermann = method(m,n,
cond( m zero?, n succ, n zero?, ackermann(m pred, 1), ackermann(m pred, ackermann(m, n pred)))
)</lang>
J
As posted at the J wiki <lang j>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</lang>
- Example use:
<lang j> 0 ack 3 4
1 ack 3
5
2 ack 3
9
3 ack 3
61</lang> 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. <lang j> Ack=. 3 -~ [ ({&(2 4$'>: 2x&+') ::(,&'&1'&'2x&*'@:(-&2))"0@:[ 128!:2 ]) 3 + ]</lang>
- Example use:
<lang j> 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 </lang>
A structured derivation of Ack follows:
<lang j> 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 + ]</lang>
Java
<lang 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)));
}</lang>
<lang java5>@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);
}
}</lang> <lang java5>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)
;
}
}
}</lang> <lang java5>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);
}
}
}
}</lang> Template:Iterative version <lang java5>/*
* 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();
}
}
</lang>
JavaScript
ES5
<lang javascript>function ack(m, n) {
return m === 0 ? n + 1 : ack(m - 1, n === 0 ? 1 : ack(m, n - 1));
}</lang>
Eliminating Tail Calls
<lang javascript>function ack(M,N) {
for (; M > 0; M--) {
N = N === 0 ? 1 : ack(M,N-1);
}
return N+1;
}</lang>
Iterative, With Explicit Stack
<lang javascript>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;
}
}
}</lang>
Stackless Iterative
<lang javascript>#!/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])));</lang>
- 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
<lang javascript>(() => {
'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();
})();</lang>
- Output:
[4,5,9,61]
Joy
From here <lang joy>DEFINE ack == [ [ [pop null] popd succ ]
[ [null] pop pred 1 ack ]
[ [dup pred swap] dip pred ack ack ] ]
cond.</lang>
another using a combinator <lang joy>DEFINE ack == [ [ [pop null] [popd succ] ] [ [null] [pop pred 1] [] ] [ [[dup pred swap] dip pred] [] [] ] ]
condnestrec.</lang>
Whenever there are two definitions with the same name, the last one is the one that is used, when invoked.
jq
Without Memoization
<lang jq># 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 ;</lang>
Example: <lang jq>range(0;5) as $i | range(0; if $i > 3 then 1 else 6 end) as $j | "A(\($i),\($j)) = \( [$i,$j] | ack )"</lang>
- Output:
<lang sh># 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</lang>
With Memoization and Optimization
<lang jq># 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($key)
end
end;
def A(m;n): [m,n,{}] | ack | .[0]; </lang> Example:<lang jq>A(4,1)</lang>
- Output:
<lang sh>65533</lang>
Jsish
From javascript entry. <lang javascript>/* Ackermann function, in Jsish */
function ack(m, n) {
return m === 0 ? n + 1 : ack(m - 1, n === 0 ? 1 : ack(m, n - 1));
}
if (Interp.conf('unitTest')) {
Interp.conf({maxDepth:4096});
- ack(1,3);
- ack(2,3);
- ack(3,3);
- ack(1,5);
- ack(2,5);
- ack(3,5);
}
/*
!EXPECTSTART!
ack(1,3) ==> 5 ack(2,3) ==> 9 ack(3,3) ==> 61 ack(1,5) ==> 7 ack(2,5) ==> 13 ack(3,5) ==> 253
!EXPECTEND!
- /</lang>
- Output:
prompt$ 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
<lang 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</lang>
One-liner: <lang julia>ack2(m::Integer, n::Integer) = m == 0 ? n + 1 : n == 0 ? ack2(m - 1, 1) : ack2(m - 1, ack2(m, n - 1))</lang>
Using memoization, source: <lang julia>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))</lang>
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 <lang k>ack:{:[0=x;y+1;0=y;_f[x-1;1];_f[x-1;_f[x;y-1]]]} ack[2;2]</lang>
Kdf9 Usercode
<lang 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;</lang>
Klingphix
<lang>:ack
%n !n %m !m
$m 0 ==
( [$n 1 +]
[$n 0 ==
( [$m 1 - 1 ack]
[$m 1 - $m $n 1 - ack ack]
) if
]
) if
3 6 ack print nl msec print
" " input</lang>
Klong
<lang k> ack::{:[0=x;y+1:|0=y;.f(x-1;1);.f(x-1;.f(x;y-1))]} ack(2;2)</lang>
Kotlin
<lang scala> tailrec fun A(m: Long, n: Long): Long {
require(m >= 0L) { "m must not be negative" }
require(n >= 0L) { "n must not be negative" }
if (m == 0L) {
return n + 1L
}
if (n == 0L) {
return A(m - 1L, 1L)
}
return A(m - 1L, A(m, n - 1L))
}
inline fun<T> tryOrNull(block: () -> T): T? = try { block() } catch (e: Throwable) { null }
const val N = 10L const val M = 4L
fun main() {
(0..M)
.map { it to 0..N }
.map { (m, Ns) -> (m to Ns) to Ns.map { n -> tryOrNull { A(m, n) } } }
.map { (input, output) -> "A(${input.first}, ${input.second})" to output.map { it?.toString() ?: "?" } }
.map { (input, output) -> "$input = $output" }
.forEach(::println)
} </lang>
- 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
<lang Scheme> {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 -> ??? </lang>
Lasso
<lang 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)) </lang>
- 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
<lang lisp>(defun ackermann
((0 n) (+ n 1)) ((m 0) (ackermann (- m 1) 1)) ((m n) (ackermann (- m 1) (ackermann m (- n 1)))))</lang>
Liberty BASIC
<lang lb>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</lang>
LiveCode
<lang 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</lang>
Logo
<lang logo>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</lang>
Logtalk
<lang 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).</lang>
LOLCODE
<lang LOLCODE>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</lang>
Lua
<lang 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</lang>
Stackless iterative solution with multiple precision, fast
<lang lua>
- !/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 </lang>
- 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
<lang 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</lang>
Luck
<lang 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))
)</lang>
M2000 Interpreter
<lang 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 </lang>
M4
<lang 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)</lang>
- 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.
<lang MAD> 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
</lang>
- 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. <lang Maple> 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: </lang> In Maple, the keyword <lang Maple>thisproc</lang> 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) <lang Maple> 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:
</lang>
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 <lang Maple> > map2( Ackermann, 1, [seq]( 1 .. 10 ) );
[3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
</lang> To get the first 10 values for m = 2 use <lang Maple> > map2( Ackermann, 2, [seq]( 1 .. 10 ) );
[5, 7, 9, 11, 13, 15, 17, 19, 21, 23]
</lang> For Ackermann( 4, 2 ) we get a very long number with <lang Maple> > length( Ackermann( 4, 2 ) );
19729
</lang> 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.
<lang Mathcad>A(m,n):=if(m=0,n+1,if(n=0,A(m-1,1),A(m-1,A(m,n-1))))</lang>
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: <lang Mathematica>$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]]</lang>
Note that the second implementation is quite a bit faster, as doing 'if' comparisons is slower than the built-in pattern matching algorithms. Examples: <lang Mathematica>Flatten[#,1]&@Table[{"Ackermann2["<>ToString[i]<>","<>ToString[j]<>"] =",Ackermann2[i,j]},{i,3},{j,8}]//Grid</lang> gives back: <lang Mathematica>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</lang> If we would like to calculate Ackermann[4,1] or Ackermann[4,2] we have to optimize a little bit: <lang Mathematica>Clear[Ackermann3] $RecursionLimit=Infinity; Ackermann3[0,n_]:=n+1; Ackermann3[1,n_]:=n+2; Ackermann3[2,n_]:=3+2n; Ackermann3[3,n_]:=5+8 (2^n-1); Ackermann3[m_,0]:=Ackermann3[m-1,1]; Ackermann3[m_,n_]:=Ackermann3[m-1,Ackermann3[m,n-1]]</lang> Now computing Ackermann[4,1] and Ackermann[4,2] can be done quickly (<0.01 sec): Examples 2: <lang Mathematica>Ackermann3[4, 1] Ackermann3[4, 2]</lang> gives back:
Ackermann[4,2] has 19729 digits, several thousands of digits omitted in the result above for obvious reasons. Ackermann[5,0] can be computed also quite fast, and is equal to 65533. Summarizing Ackermann[0,n_], Ackermann[1,n_], Ackermann[2,n_], and Ackermann[3,n_] can all be calculated for n>>1000. Ackermann[4,0], Ackermann[4,1], Ackermann[4,2] and Ackermann[5,0] are only possible now. Maybe in the future we can calculate higher Ackermann numbers efficiently and fast. Although showing the results will always be a problem.
MATLAB
<lang MATLAB>function A = ackermannFunction(m,n)
if m == 0
A = n+1;
elseif (m > 0) && (n == 0)
A = ackermannFunction(m-1,1);
else
A = ackermannFunction( m-1,ackermannFunction(m,n-1) );
end
end</lang>
Maxima
<lang maxima>ackermann(m, n) := if integerp(m) and integerp(n) then ackermann[m, n] else 'ackermann(m, n)$
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);</lang>
MAXScript
Use with caution. Will cause a stack overflow for m > 3. <lang maxscript>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))
)
)</lang>
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.
<lang mercury>:- 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) ).</lang>
min
<lang min>(
:n :m ( ((m 0 ==) (n 1 +)) ((n 0 ==) (m 1 - 1 ackermann)) ((true) (m 1 - m n 1 - ackermann ackermann)) ) case
) :ackermann</lang>
MiniScript
<lang 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</lang>
МК-61/52
<lang>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</lang>
ML/I
ML/I loves recursion, but runs out of its default amount of storage with larger numbers than those tested here!
Program
<lang ML/I>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)</lang>
- Output:
<lang ML/I>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</lang>
mLite
<lang haskell>fun ackermann( 0, n ) = n + 1 | ( m, 0 ) = ackermann( m - 1, 1 ) | ( m, n ) = ackermann( m - 1, ackermann(m, n - 1) )</lang> Test code providing tuples from (0,0) to (3,8) <lang haskell>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))</lang> 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
<lang modula2>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.</lang>
- Output:
jan@Beryllium:~/modula/rosetta$ ackerman1 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. <lang modula3>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.</lang>
- 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
<lang 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</lang>
Neko
<lang ActionScript>/**
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
$print("Ackermann(", arg1, ",", arg2, "): ", problem, "\n")</lang>
- Output:
prompt$ 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. <lang Nemerle> 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");
}
}
</lang>
A terser version using implicit match (which doesn't use the alias A internally):
<lang Nemerle>
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");
}
</lang>
Or, if we were set on using the A notation, we could do this:
<lang Nemerle>
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
} </lang>
NetRexx
<lang 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
</lang>
NewLISP
<lang 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))))))
</lang>
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
<lang 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: ", $ackermann(first, second) </lang>
Nit
Source: the official Nit’s repository.
<lang nit># Task: Ackermann function
- A simple straightforward recursive implementation.
module ackermann_function
fun ack(m, n: Int): Int do if m == 0 then return n + 1 if n == 0 then return ack(m-1,1) return ack(m-1, ack(m, n-1)) end
for m in [0..3] do for n in [0..6] do print ack(m,n) end print "" end</lang>
Output:
1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509
Oberon-2
<lang oberon2>MODULE ackerman;
IMPORT Out;
VAR m, n : INTEGER;
PROCEDURE Ackerman (x, y : INTEGER) : INTEGER;
BEGIN
IF x = 0 THEN RETURN y + 1 ELSIF y = 0 THEN RETURN Ackerman (x - 1 , 1) ELSE RETURN Ackerman (x - 1 , Ackerman (x , y - 1)) END
END Ackerman;
BEGIN
FOR m := 0 TO 3 DO
FOR n := 0 TO 6 DO
Out.Int (Ackerman (m, n), 10);
Out.Char (9X)
END;
Out.Ln
END;
Out.Ln
END ackerman.</lang>
Objeck
<lang objeck>class Ackermann {
function : Main(args : String[]) ~ Nil {
for(m := 0; m <= 3; ++m;) {
for(n := 0; n <= 4; ++n;) {
a := Ackermann(m, n);
if(a > 0) {
"Ackermann({$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;
}
}</lang>
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
<lang 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)))</lang>
or: <lang ocaml>let rec a = function
| 0, n -> (n+1) | m, 0 -> a(m-1, 1) | m, n -> a(m-1, a(m, n-1))</lang>
with memoization using an hash-table: <lang ocaml>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)</lang>
taking advantage of the memoization we start calling small values of m and n in order to reduce the recursion call stack: <lang ocaml>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)</lang>
Arbitrary precision
With arbitrary-precision integers (Big_int module): <lang ocaml>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)))</lang>
compile with:
ocamlopt -o acker nums.cmxa acker.ml
Tail-Recursive
Here is a tail-recursive version: <lang ocaml>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 *) ) [] [] ;;</lang> This one uses the arbitrary precision, the tail-recursion, and the optimisation explain on the Wikipedia page about (m,n) = (3,_). <lang ocaml>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);
- </lang>
Octave
<lang 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</lang>
Oforth
<lang Oforth>: A( m n -- p )
m ifZero: [ n 1+ return ] m 1- n ifZero: [ 1 ] else: [ A( m, n 1- ) ] A
- </lang>
OOC
<lang 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()
}
}
} </lang>
ooRexx
<lang 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))
</lang>
- 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
<lang c>#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</lang>
Oz
Oz has arbitrary precision integers. <lang oz>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}}</lang>
PARI/GP
Naive implementation. <lang parigp>A(m,n)={
if(m,
if(n,
A(m-1, A(m,n-1))
,
A(m-1,1)
)
,
n+1
)
};</lang>
Pascal
<lang 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.</lang>
Perl
We memoize calls to A to make A(2, n) and A(3, n) feasible for larger values of n. <lang perl>{
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( $m - 1, 1 )
);
}
}</lang>
An implementation using the conditional statements 'if', 'elsif' and 'else': <lang perl>sub A {
my ($m, $n) = @_;
if ($m == 0) { $n + 1 }
elsif ($n == 0) { A($m - 1, 1) }
else { A($m - 1, A($m, $n - 1)) }
}</lang>
An implementation using ternary chaining: <lang perl>sub A {
my ($m, $n) = @_;
$m == 0 ? $n + 1 :
$n == 0 ? A($m - 1, 1) :
A($m - 1, A($m, $n - 1))
}</lang>
Adding memoization and extra terms: <lang perl>use Memoize; memoize('ack2'); use bigint try=>"GMP";
sub ack2 {
my ($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";</lang>
- 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.
<lang Perl>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";
</lang>
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
-- 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
<lang 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</lang>
PHP
<lang 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</lang>
Picat
<lang 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))). </lang>
- 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
<lang PicoLisp>(de ack (X Y)
(cond
((=0 X) (inc Y))
((=0 Y) (ack (dec X) 1))
(T (ack (dec X) (ack X (dec Y)))) ) )</lang>
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
<lang 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));
}
}</lang>
PL/I
<lang 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;</lang>
PL/SQL
<lang PLSQL>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; </lang>
- 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
<lang 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</lang>
<lang postscript>/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}.</lang>
Potion
<lang 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.</lang>
PowerBASIC
<lang 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</lang>
PowerShell
<lang powershell>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)))
}</lang> 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): <lang powershell>foreach ($m in 0..3) {
foreach ($n in 0..6) {
Write-Host -NoNewline ("{0,5}" -f (ackermann $m $n))
}
Write-Host
}</lang>
- Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509
A More "PowerShelly" Way
<lang PowerShell> function Get-Ackermann ([int64]$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)))
}
</lang>
Save the result to an array (for possible future use?), then display it using the Format-Wide cmdlet:
<lang PowerShell>
$ackermann = 0..3 | ForEach-Object {$m = $_; 0..6 | ForEach-Object {Get-Ackermann $m $_}}
$ackermann | Format-Wide {"{0,3}" -f $_} -Column 7 -Force </lang>
- Output:
1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509
Processing
<lang java>int ackermann(int m, int n) {
if (m == 0) return n + 1; else if (m > 0 && n == 0) return ackermann(m - 1, 1); else return ackermann( m - 1, ackermann(m, n - 1) );
}
// Call function to produce output: // the first 4x7 Ackermann numbers void setup() {
for (int m=0; m<4; m++) {
for (int n=0; n<7; n++) {
print(ackermann(m, n), " ");
}
println();
}
}</lang>
- Output:
1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509
Processing Python mode
Python is not very adequate for deep recursion, so even setting sys.setrecursionlimit(1000000000) if m = 5 it throws 'maximum recursion depth exceeded'
<lang python>from __future__ import print_function
def setup():
for m in range(4):
for n in range(7):
print("{} ".format(ackermann(m, n)), end = "")
print()
# print('finished')
def ackermann(m, n):
if m == 0:
return n + 1
elif m > 0 and n == 0:
return ackermann(m - 1, 1)
else:
return ackermann(m - 1, ackermann(m, n - 1))</lang>
Processing.R
Processing.R may exceed its stack depth at ~n==6 and returns null.
<lang r>setup <- function() {
for (m in 0:3) {
for (n in 0:4) {
stdout$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))
}
}</lang>
- Output:
1 2 3 4 5 2 3 4 5 6 3 5 7 9 11 5 13 29 61 125
Prolog
<lang 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).</lang>
Pure
<lang 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;</lang>
Pure Data
<lang 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;</lang>
PureBasic
<lang 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)</lang>
Purity
<lang Purity>data Iter = f => FoldNat <const $f One, $f> data Ackermann = FoldNat <const Succ, Iter></lang>
Python
Python: Explicitly recursive
<lang python>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)))</lang>
Another version: <lang python>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))</lang>
- Example of use:
<lang python>>>> import sys >>> sys.setrecursionlimit(3000) >>> ack1(0,0) 1 >>> ack1(3,4) 125 >>> ack2(0,0) 1 >>> ack2(3,4) 125</lang> From the Mathematica ack3 example: <lang python>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))))))</lang>
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.
<lang python>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]</lang>
- 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
<lang 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</lang>
Output:
8189
R
<lang 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))
}
}</lang> <lang R>for ( i in 0:3 ) {
print(ackermann(i, 4))
}</lang>
Racket
<lang 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)))]))
</lang>
Raku
(formerly Perl 6)
<lang perl6>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)) }
}</lang> An implementation using multiple dispatch: <lang perl6>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)) }</lang> 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: <lang perl6>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);</lang>
- 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
<lang 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)))</lang>
- 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
<lang rexx>/*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) )</lang>
- 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
<lang rexx>/*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) )</lang>
- 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.
<lang rexx>/*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) )</lang>
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
<lang ring>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 !!!") </lang>
- 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. <lang Risc-V>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 </lang>
Ruby
<lang 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</lang> Example: <lang ruby>(0..3).each do |m|
puts (0..6).map { |n| ack(m, n) }.join(' ')
end</lang>
- 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
<lang runbasic>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</lang>
Rust
<lang 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
} </lang>
Or:
<lang rust> 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)), } } </lang>
Sather
<lang 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;</lang>
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.
<lang sather>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;</lang>
Scala
<lang 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))
}</lang>
- 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: <lang scala>val ackMap = new mutable.HashMap[(BigInt,BigInt),BigInt] def ackMemo(m: BigInt, n: BigInt): BigInt = {
ackMap.getOrElseUpdate((m,n), ack(m,n))
}</lang>
Scheme
<lang scheme>(define (A m n)
(cond
((= m 0) (+ n 1))
((= n 0) (A (- m 1) 1))
(else (A (- m 1) (A m (- n 1))))))</lang>
An improved solution that uses a lazy data structure, streams, and defines Knuth up-arrows to calculate iterative exponentiation:
<lang scheme>(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)))))))</lang>
Scilab
<lang>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</lang>
- 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
<lang 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;</lang>
Original source: [2]
SETL
<lang 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;</lang>
Shen
<lang shen>(define ack
0 N -> (+ N 1)
M 0 -> (ack (- M 1) 1)
M N -> (ack (- M 1)
(ack M (- N 1))))</lang>
Sidef
<lang ruby>func A(m, n) {
m == 0 ? (n + 1)
: (n == 0 ? (A(m - 1, 1))
: (A(m - 1, A(m, n - 1))));
}</lang>
Alternatively, using multiple dispatch: <lang ruby>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)) }</lang>
Calling the function: <lang ruby>say A(3, 2); # prints: 29</lang>
Simula
as modified by R. Péter and R. Robinson: <lang Simula> 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</lang>
- Output:
Ackermann(4, 0) = 13 Ackermann(3, 3) = 61 Ackermann(2, 6) = 15 Ackermann(1, 9) = 11 Ackermann(0,12) = 13
Slate
<lang 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)]] ].</lang>
Smalltalk
<lang 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.</lang>
SmileBASIC
<lang 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</lang>
SNOBOL4
Both Snobol4+ and CSnobol stack overflow, at ack(3,3) and ack(3,4), respectively. <lang SNOBOL4>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</lang>
- 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
<lang 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))
| | |
# # | | | /+<<<-\
/-<<+>>\!=/ \=====|==!/========?\>>>=?/<<#
? ? | \<<<+>+>>-/
\>>+<<-/!==========/
# #</lang>
One could employ tail recursion elimination by replacing "@/#" with "/" in two places above.
SPAD
<lang 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] </lang>
- Output:
+1 2 3 5 +
| |
|2 3 5 13|
(1) | |
|3 4 7 29|
| |
+4 5 9 61+
Type: Matrix(NonNegativeInteger)
SQL PL
version 9.7 or higher.
With SQL PL: <lang 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 @ </lang> 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
<lang sml>fun a (0, n) = n+1
| a (m, 0) = a (m-1, 1) | a (m, n) = a (m-1, a (m, n-1))</lang>
Stata
<lang 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</lang>
Swift
<lang 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))
}
}</lang>
Tcl
Simple
<lang tcl>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}]]
}
}</lang>
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): <lang tcl>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}]]
}
}</lang>
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.
<lang tcl>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 ack $m [expr {$n-1}]]
}
}
}
- Some small tweaks...
interp recursionlimit {} 100000 interp alias {} ack {} cacheable ack</lang> But even with all this, you still run into problems calculating as that's kind-of large…
TI-83 BASIC
This program assumes the variables N and M are the arguments of the function, and that the list L1 is empty. It stores the result in the system variable ANS. (Program names can be no longer than 8 characters, so I had to truncate the function's name.)
<lang ti83b>PROGRAM:ACKERMAN
- If not(M
- Then
- N+1→N
- Return
- Else
- If not(N
- Then
- 1→N
- M-1→M
- prgmACKERMAN
- Else
- N-1→N
- M→L1(1+dim(L1
- prgmACKERMAN
- Ans→N
- L1(dim(L1))-1→M
- dim(L1)-1→dim(L1
- prgmACKERMAN
- End
- End</lang>
Here is a handler function that makes the previous function easier to use. (You can name it whatever you want.)
<lang ti83b>PROGRAM:AHANDLER
- 0→dim(L1
- Prompt M
- Prompt N
- prgmACKERMAN
- Disp Ans</lang>
TI-89 BASIC
<lang ti89b>Define A(m,n) = when(m=0, n+1, when(n=0, A(m-1,1), A(m-1, A(m, n-1))))</lang>
TorqueScript
<lang TorqueScript>function ackermann(%m,%n) {
if(%m==0)
return %n+1;
if(%m>0&&%n==0)
return ackermann(%m-1,1);
if(%m>0&&%n>0)
return ackermann(%m-1,ackermann(%m,%n-1));
}</lang>
TSE SAL
<lang TSESAL>// library: math: get: ackermann: recursive <description></description> <version>1.0.0.0.5</version> <version control></version control> (filenamemacro=getmaare.s) [kn, ri, tu, 27-12-2011 14:46:59] INTEGER PROC FNMathGetAckermannRecursiveI( INTEGER mI, INTEGER nI )
IF ( mI == 0 ) RETURN( nI + 1 ) ENDIF IF ( nI == 0 ) RETURN( FNMathGetAckermannRecursiveI( mI - 1, 1 ) ) ENDIF RETURN( FNMathGetAckermannRecursiveI( mI - 1, FNMathGetAckermannRecursiveI( mI, nI - 1 ) ) )
END
PROC Main() STRING s1[255] = "2" STRING s2[255] = "3" IF ( NOT ( Ask( "math: get: ackermann: recursive: m = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF IF ( NOT ( Ask( "math: get: ackermann: recursive: n = ", s2, _EDIT_HISTORY_ ) ) AND ( Length( s2 ) > 0 ) ) RETURN() ENDIF
Message( FNMathGetAckermannRecursiveI( Val( s1 ), Val( s2 ) ) ) // gives e.g. 9
END</lang>
TXR
with memoization.
<lang txrlisp>(defmacro defmemofun (name (. args) . body)
(let ((hash (gensym "hash-"))
(argl (gensym "args-"))
(hent (gensym "hent-"))
(uniq (copy-str "uniq")))
^(let ((,hash (hash :equal-based)))
(defun ,name (,*args)
(let* ((,argl (list ,*args))
(,hent (inhash ,hash ,argl ,uniq)))
(if (eq (cdr ,hent) ,uniq)
(set (cdr ,hent) (block ,name (progn ,*body)))
(cdr ,hent)))))))
(defmemofun ack (m n)
(cond ((= m 0) (+ n 1)) ((= n 0) (ack (- m 1) 1)) (t (ack (- m 1) (ack m (- n 1))))))
(each ((i (range 0 3)))
(each ((j (range 0 4))) (format t "ack(~a, ~a) = ~a\n" i j (ack i j))))</lang>
- Output:
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
UNIX Shell
<lang bash>ack() {
local m=$1 local n=$2 if [ $m -eq 0 ]; then echo -n $((n+1)) elif [ $n -eq 0 ]; then ack $((m-1)) 1 else ack $((m-1)) $(ack $m $((n-1))) fi
}</lang> Example: <lang bash>for ((m=0;m<=3;m++)); do
for ((n=0;n<=6;n++)); do ack $m $n echo -n " " done echo
done</lang>
- Output:
1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509
Ursala
Anonymous recursion is the usual way of doing things like this. <lang Ursala>#import std
- import nat
ackermann =
~&al^?\successor@ar ~&ar?(
^R/~&f ^/predecessor@al ^|R/~& ^|/~& predecessor, ^|R/~& ~&\1+ predecessor@l)</lang>
test program for the first 4 by 7 numbers: <lang Ursala>#cast %nLL
test = block7 ackermann*K0 iota~~/4 7</lang>
- Output:
< <1,2,3,4,5,6,7>, <2,3,4,5,6,7,8>, <3,5,7,9,11,13,15>, <5,13,29,61,125,253,509>>
V
<lang v>[ack
[ [pop zero?] [popd succ]
[zero?] [pop pred 1 ack]
[true] [[dup pred swap] dip pred ack ack ]
] when].</lang>
using destructuring view <lang v>[ack
[ [pop zero?] [ [m n : [n succ]] view i]
[zero?] [ [m n : [m pred 1 ack]] view i]
[true] [ [m n : [m pred m n pred ack ack]] view i]
] when].</lang>
Vala
<lang vala>uint64 ackermann(uint64 m, uint64 n) {
if (m == 0) return n + 1; if (n == 0) return ackermann(m - 1, 1); return ackermann(m - 1, ackermann(m, n - 1));
}
void main () {
for (uint64 m = 0; m < 4; ++m) {
for (uint64 n = 0; n < 10; ++n) {
print(@"A($m,$n) = $(ackermann(m,n))\n");
}
}
}</lang>
- 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(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
VBA
<lang vb>Private Function Ackermann_function(m As Variant, n As Variant) As Variant
Dim result As Variant
Debug.Assert m >= 0
Debug.Assert n >= 0
If m = 0 Then
result = CDec(n + 1)
Else
If n = 0 Then
result = Ackermann_function(m - 1, 1)
Else
result = Ackermann_function(m - 1, Ackermann_function(m, n - 1))
End If
End If
Ackermann_function = CDec(result)
End Function Public Sub main()
Debug.Print " n=",
For j = 0 To 7
Debug.Print j,
Next j
Debug.Print
For i = 0 To 3
Debug.Print "m=" & i,
For j = 0 To 7
Debug.Print Ackermann_function(i, j),
Next j
Debug.Print
Next i
End Sub</lang>
- Output:
n= 0 1 2 3 4 5 6 7 m=0 1 2 3 4 5 6 7 8 m=1 2 3 4 5 6 7 8 9 m=2 3 5 7 9 11 13 15 17 m=3 5 13 29 61 125 253 509 1021
VBScript
Based on BASIC version. Uncomment all the lines referring to depth and see just how deep the recursion goes.
- Implementation
<lang vb>option explicit '~ dim depth function ack(m, n) '~ wscript.stdout.write depth & " " if m = 0 then '~ depth = depth + 1 ack = n + 1 '~ depth = depth - 1 elseif m > 0 and n = 0 then '~ depth = depth + 1 ack = ack(m - 1, 1) '~ depth = depth - 1 '~ elseif m > 0 and n > 0 then else '~ depth = depth + 1 ack = ack(m - 1, ack(m, n - 1)) '~ depth = depth - 1 end if
end function</lang>
- Invocation
<lang vb>wscript.echo ack( 1, 10 ) '~ depth = 0 wscript.echo ack( 2, 1 ) '~ depth = 0 wscript.echo ack( 4, 4 )</lang>
- Output:
12 5 C:\foo\ackermann.vbs(16, 3) Microsoft VBScript runtime error: Out of stack space: 'ack'
Visual Basic
<lang vb> Option Explicit Dim calls As Long Sub main()
Const maxi = 4
Const maxj = 9
Dim i As Long, j As Long
For i = 0 To maxi
For j = 0 To maxj
Call print_acker(i, j)
Next j
Next i
End Sub 'main Sub print_acker(m As Long, n As Long)
calls = 0
Debug.Print "ackermann("; m; ","; n; ")=";
Debug.Print ackermann(m, n), "calls="; calls
End Sub 'print_acker Function ackermann(m As Long, n As Long) As Long
calls = calls + 1
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 If
End If
End Function 'ackermann</lang>
- 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( 0 , 7 )= 8 calls= 1 ackermann( 0 , 8 )= 9 calls= 1 ackermann( 0 , 9 )= 10 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( 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( 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 ackermann( 3 , 9 )= 4093 calls= 11164370 ackermann( 4 , 0 )= 13 calls= 107 ackermann( 4 , 1 )= out of stack space
Vlang
<lang go>fn ackermann(m int, n int ) int {
if m == 0 {
return n + 1
}
else if n == 0 {
return ackermann(m - 1, 1)
}
return ackermann(m - 1, ackermann(m, n - 1) )
}
fn main() {
for m := 0; m <= 4; m++ {
for n := 0; n < ( 6 - m ); n++ {
println('Ackermann($m, $n) = ${ackermann(m, n)}')
}
}
} </lang>
- 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(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(3, 0) = 5 Ackermann(3, 1) = 13 Ackermann(3, 2) = 29 Ackermann(4, 0) = 13 Ackermann(4, 1) = 65533
Wart
<lang wart>def (ackermann m n)
(if m=0
n+1
n=0
(ackermann m-1 1)
:else
(ackermann m-1 (ackermann m n-1)))</lang>
WDTE
<lang WDTE>let memo a m n => true { == m 0 => + n 1; == n 0 => a (- m 1) 1; true => a (- m 1) (a m (- n 1)); };</lang>
Wren
<lang ecmascript>// To use recursion definition and declaration must be on separate lines var Ackermann Ackermann = Fn.new {|m, n|
if (m == 0) return n + 1 if (n == 0) return Ackermann.call(m - 1, 1) return Ackermann.call(m - 1, Ackermann.call(m, n - 1))
}
var pairs = [ [1, 3], [2, 3], [3, 3], [1, 5], [2, 5], [3, 5] ] for (pair in pairs) {
var p1 = pair[0]
var p2 = pair[1]
System.print("A[%(p1), %(p2)] = %(Ackermann.call(p1, p2))")
}</lang>
- Output:
A[1, 3] = 5 A[2, 3] = 9 A[3, 3] = 61 A[1, 5] = 7 A[2, 5] = 13 A[3, 5] = 253
X86 Assembly
<lang asm> section .text
global _main _main:
mov eax, 3 ;m mov ebx, 4 ;n call ack ;returns number in ebx ret
ack:
cmp eax, 0
je M0 ;if M == 0
cmp ebx, 0
je N0 ;if N == 0
dec ebx ;else N-1
push eax ;save M
call ack1 ;ack(m,n) -> returned in ebx so no further instructions needed
pop eax ;restore M
dec eax ;M - 1
call ack1 ;return ack(m-1,ack(m,n-1))
ret
M0:
inc ebx ;return n + 1
ret
N0:
dec eax
inc ebx ;ebx always 0: inc -> ebx = 1
call ack1 ;return ack(M-1,1)
ret
</lang>
XLISP
<lang lisp>(defun ackermann (m n)
(cond
((= m 0) (+ n 1))
((= n 0) (ackermann (- m 1) 1))
(t (ackermann (- m 1) (ackermann m (- n 1))))))</lang>
Test it: <lang lisp>(print (ackermann 3 9))</lang> Output (after a very perceptible pause):
4093
That worked well. Test it again: <lang lisp>(print (ackermann 4 1))</lang> Output (after another pause):
Abort: control stack overflow happened in: #<Code ACKERMANN>
XPL0
<lang XPL0>include c:\cxpl\codes;
func Ackermann(M, N); int 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)); ]; \Ackermann
int M, N; [for M:= 0 to 3 do
[for N:= 0 to 7 do
[IntOut(0, Ackermann(M, N)); ChOut(0,9\tab\)];
CrLf(0);
];
]</lang> Recursion overflows the stack if either M or N is extended by a single count.
- Output:
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
XSLT
The following named template calculates the Ackermann function: <lang xml>
<xsl:template name="ackermann"> <xsl:param name="m"/> <xsl:param name="n"/>
<xsl:choose>
<xsl:when test="$m = 0">
<xsl:value-of select="$n+1"/>
</xsl:when>
<xsl:when test="$n = 0">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="'1'"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:variable name="p">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m"/>
<xsl:with-param name="n" select="$n - 1"/>
</xsl:call-template>
</xsl:variable>
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="$p"/>
</xsl:call-template>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</lang>
Here it is as part of a template <lang xml> <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:template match="arguments">
<xsl:for-each select="args">
<xsl:value-of select="m"/>, <xsl:value-of select="n"/>:
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="m"/>
<xsl:with-param name="n" select="n"/>
</xsl:call-template>
</xsl:for-each> </xsl:template>
<xsl:template name="ackermann"> <xsl:param name="m"/> <xsl:param name="n"/>
<xsl:choose>
<xsl:when test="$m = 0">
<xsl:value-of select="$n+1"/>
</xsl:when>
<xsl:when test="$n = 0">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="'1'"/>
</xsl:call-template>
</xsl:when>
<xsl:otherwise>
<xsl:variable name="p">
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m"/>
<xsl:with-param name="n" select="$n - 1"/>
</xsl:call-template>
</xsl:variable>
<xsl:call-template name="ackermann">
<xsl:with-param name="m" select="$m - 1"/>
<xsl:with-param name="n" select="$p"/>
</xsl:call-template>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet> </lang>
Which will transform this input <lang xml> <?xml version="1.0" ?> <?xml-stylesheet type="text/xsl" href="ackermann.xslt"?> <arguments>
<args> <m>0</m> <n>0</n> </args> <args> <m>0</m> <n>1</n> </args> <args> <m>0</m> <n>2</n> </args> <args> <m>0</m> <n>3</n> </args> <args> <m>0</m> <n>4</n> </args> <args> <m>0</m> <n>5</n> </args> <args> <m>0</m> <n>6</n> </args> <args> <m>0</m> <n>7</n> </args> <args> <m>0</m> <n>8</n> </args> <args> <m>1</m> <n>0</n> </args> <args> <m>1</m> <n>1</n> </args> <args> <m>1</m> <n>2</n> </args> <args> <m>1</m> <n>3</n> </args> <args> <m>1</m> <n>4</n> </args> <args> <m>1</m> <n>5</n> </args> <args> <m>1</m> <n>6</n> </args> <args> <m>1</m> <n>7</n> </args> <args> <m>1</m> <n>8</n> </args> <args> <m>2</m> <n>0</n> </args> <args> <m>2</m> <n>1</n> </args> <args> <m>2</m> <n>2</n> </args> <args> <m>2</m> <n>3</n> </args> <args> <m>2</m> <n>4</n> </args> <args> <m>2</m> <n>5</n> </args> <args> <m>2</m> <n>6</n> </args> <args> <m>2</m> <n>7</n> </args> <args> <m>2</m> <n>8</n> </args> <args> <m>3</m> <n>0</n> </args> <args> <m>3</m> <n>1</n> </args> <args> <m>3</m> <n>2</n> </args> <args> <m>3</m> <n>3</n> </args> <args> <m>3</m> <n>4</n> </args> <args> <m>3</m> <n>5</n> </args> <args> <m>3</m> <n>6</n> </args> <args> <m>3</m> <n>7</n> </args> <args> <m>3</m> <n>8</n> </args>
</arguments> </lang>
into this output
0, 0: 1 0, 1: 2 0, 2: 3 0, 3: 4 0, 4: 5 0, 5: 6 0, 6: 7 0, 7: 8 0, 8: 9 1, 0: 2 1, 1: 3 1, 2: 4 1, 3: 5 1, 4: 6 1, 5: 7 1, 6: 8 1, 7: 9 1, 8: 10 2, 0: 3 2, 1: 5 2, 2: 7 2, 3: 9 2, 4: 11 2, 5: 13 2, 6: 15 2, 7: 17 2, 8: 19 3, 0: 5 3, 1: 13 3, 2: 29 3, 3: 61 3, 4: 125 3, 5: 253 3, 6: 509 3, 7: 1021 3, 8: 2045
Yabasic
<lang Yabasic>sub 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))
end sub
print ack(3, 4) </lang>
What smart code can get. Fast as lightning!
<lang Yabasic>sub ack(m, 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 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 sub
sub Ackermann()
local i, j
for i=0 to 3
for j=0 to 10
print ack(i,j) using "#####";
next
print
next
print "ack(4,1) ";: print ack(4,1) using "#####"
end sub
Ackermann()</lang>
Yorick
<lang yorick>func ack(m, n) {
if(m == 0)
return n + 1;
else if(n == 0)
return ack(m - 1, 1);
else
return ack(m - 1, ack(m, n - 1));
}</lang> Example invocation: <lang yorick>for(m = 0; m <= 3; m++) {
for(n = 0; n <= 6; n++)
write, format="%d ", ack(m, n);
write, "";
}</lang>
- 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
Z80 Assembly
This function does 16-bit math. Sjasmplus syntax, CP/M executable. <lang z80> OPT --syntax=abf : OUTPUT "ackerman.com"
ORG $100 jr demo_start
- --------------------------------------------------------------------------------------------------------------------
- entry
- ackermann_fn
- input
- bc = m, hl = n
- output
- hl = A(m,n) (16bit only)
ackermann_fn.inc_n:
inc hl
ackermann_fn:
inc hl ld a,c or b ret z ; m == 0 -> return n+1 ; m > 0 case ; bc = m, hl = n+1 dec bc dec hl ; m-1, n restored ld a,l or h jr z,.inc_n ; n == 0 -> return A(m-1, 1) ; m > 0, n > 0 ; bc = m-1, hl = n push bc inc bc dec hl call ackermann_fn ; n = A(m, n-1) pop bc jp ackermann_fn ; return A(m-1,A(m, n-1))
- --------------------------------------------------------------------------------------------------------------------
- helper functions for demo printing 4x9 table
print_str:
push bc push hl ld c,9
.call_cpm:
call 5 pop hl pop bc ret
print_hl:
ld b,' ' ld e,b call print_char ld de,-10000 call extract_digit ld de,-1000 call extract_digit ld de,-100 call extract_digit ld de,-10 call extract_digit ld a,l
print_digit:
ld b,'0' add a,b ld e,a
print_char:
push bc push hl ld c,2 jr print_str.call_cpm
extract_digit:
ld a,-1
.digit_loop:
inc a add hl,de jr c,.digit_loop sbc hl,de or a jr nz,print_digit ld e,b jr print_char
- --------------------------------------------------------------------------------------------------------------------
demo_start: ; do m: [0,4) cross n: [0,9) table
ld bc,0
.loop_m:
ld hl,0 ; bc = m, hl = n = 0 ld de,txt_m_is call print_str ld a,c or '0' ld e,a call print_char ld e,':' call print_char
.loop_n:
push bc push hl call ackermann_fn call print_hl pop hl pop bc inc hl ld a,l cp 9 jr c,.loop_n ld de,crlf call print_str inc bc ld a,c cp 4 jr c,.loop_m rst 0 ; return to CP/M
txt_m_is: db "m=$" crlf: db 10,13,'$'</lang>
- Output:
m=0: 1 2 3 4 5 6 7 8 9 m=1: 2 3 4 5 6 7 8 9 10 m=2: 3 5 7 9 11 13 15 17 19 m=3: 5 13 29 61 125 253 509 1021 2045
ZED
Source -> http://ideone.com/53FzPA Compiled -> http://ideone.com/OlS7zL <lang zed>(A) m n comment: (=) m 0 (add1) n
(A) m n comment: (=) n 0 (A) (sub1) m 1
(A) m n comment:
- true
(A) (sub1) m (A) m (sub1) n
(add1) n comment:
- true
(003) "+" n 1
(sub1) n comment:
- true
(003) "-" n 1
(=) n1 n2 comment:
- true
(003) "=" n1 n2</lang>
Zig
<lang zig>pub fn ack(m: u64, n: u64) u64 {
if (m == 0) return n + 1; if (n == 0) return ack(m - 1, 1); return ack(m - 1, ack(m, n - 1));
}
pub fn main() !void {
const stdout = @import("std").io.getStdOut().writer();
var m: u8 = 0;
while (m <= 3) : (m += 1) {
var n: u8 = 0;
while (n <= 8) : (n += 1)
try stdout.print("{d:>8}", .{ack(m, n)});
try stdout.print("\n", .{});
}
}</lang>
- 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
ZX Spectrum Basic
<lang zxbasic>10 DIM s(2000,3) 20 LET s(1,1)=3: REM M 30 LET s(1,2)=7: REM N 40 LET lev=1 50 GO SUB 100 60 PRINT "A(";s(1,1);",";s(1,2);") = ";s(1,3) 70 STOP 100 IF s(lev,1)=0 THEN LET s(lev,3)=s(lev,2)+1: RETURN 110 IF s(lev,2)=0 THEN LET lev=lev+1: LET s(lev,1)=s(lev-1,1)-1: LET s(lev,2)=1: GO SUB 100: LET s(lev-1,3)=s(lev,3): LET lev=lev-1: RETURN 120 LET lev=lev+1 130 LET s(lev,1)=s(lev-1,1) 140 LET s(lev,2)=s(lev-1,2)-1 150 GO SUB 100 160 LET s(lev,1)=s(lev-1,1)-1 170 LET s(lev,2)=s(lev,3) 180 GO SUB 100 190 LET s(lev-1,3)=s(lev,3) 200 LET lev=lev-1 210 RETURN </lang>
- Output:
A(3,7) = 1021
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