# Towers of Hanoi

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Solve the   Towers of Hanoi   problem with recursion.

## 11l

Translation of: Python
F hanoi(ndisks, startPeg = 1, endPeg = 3) -> N
I ndisks
hanoi(ndisks - 1, startPeg, 6 - startPeg - endPeg)
print(‘Move disk #. from peg #. to peg #.’.format(ndisks, startPeg, endPeg))
hanoi(ndisks - 1, 6 - startPeg - endPeg, endPeg)

hanoi(ndisks' 3)
Output:
Move disk 1 from peg 1 to peg 3
Move disk 2 from peg 1 to peg 2
Move disk 1 from peg 3 to peg 2
Move disk 3 from peg 1 to peg 3
Move disk 1 from peg 2 to peg 1
Move disk 2 from peg 2 to peg 3
Move disk 1 from peg 1 to peg 3


## 360 Assembly

Translation of: PL/I
*        Towers of Hanoi           08/09/2015
HANOITOW CSECT
USING  HANOITOW,R12       r12 : base register
LR     R12,R15            establish base register
ST     R14,SAVE14         save r14
BEGIN    LH     R2,=H'4'           n <===
L      R3,=C'123 '        stating position
BAL    R14,MOVE           r1=move(m,n)
RETURN   L      R14,SAVE14         restore r14
SAVE14   DS     F                  static save r14
PG       DC     CL44'xxxxxxxxxxxx Move disc from pole X to pole Y'
NN       DC     F'0'
POLEX    DS     F                  current poles
POLEN    DS     F                  new poles
*        ....   recursive          subroutine move(n, poles)  [r2,r3]
MOVE     LR     R10,R11            save stackptr (r11) in r10 temp
LA     R1,STACKLEN        amount of storage required
GETMAIN RU,LV=(R1)        allocate storage for stack
ST     R14,SAVE14M        save previous r14
ST     R10,SAVE11M        save previous r11
LR     R1,R5              restore saved argument r5
BEGINM   STM    R2,R3,STACK        push arguments to stack
ST     R3,POLEX
CH     R2,=H'1'           if n<>1
BNE    RECURSE            then goto recurse
L      R1,NN
LA     R1,1(R1)           nn=nn+1
ST     R1,NN
XDECO  R1,PG              nn
MVC    PG+33(1),POLEX+0   from
MVC    PG+43(1),POLEX+1   to
XPRNT  PG,44              print "move disk from to"
B      RETURNM
RECURSE  L      R2,N               n
BCTR   R2,0               n=n-1
MVC    POLEN+0(1),POLES+0 from
MVC    POLEN+1(1),POLES+2 via
MVC    POLEN+2(1),POLES+1 to
L      R3,POLEN           new poles
BAL    R14,MOVE           call move(n-1,from,via,to)
LA     R2,1               n=1
MVC    POLEN,POLES
L      R3,POLEN           new poles
BAL    R14,MOVE           call move(1,from,to,via)
L      R2,N               n
BCTR   R2,0               n=n-1
MVC    POLEN+0(1),POLES+2 via
MVC    POLEN+1(1),POLES+1 to
MVC    POLEN+2(1),POLES+0 from
L      R3,POLEN           new poles
BAL    R14,MOVE           call move(n-1,via,to,from)
RETURNM  LM     R2,R3,STACK        pull arguments from stack
LR     R1,R11             current stack
L      R14,SAVE14M        restore r14
L      R11,SAVE11M        restore r11
LA     R0,STACKLEN        amount of storage to free
FREEMAIN A=(R1),LV=(R0)   free allocated storage
LTORG
DROP   R12                base no longer needed
STACKDS  DSECT                     dynamic area
SAVE14M  DS     F                  saved r14
SAVE11M  DS     F                  saved r11
STACK    DS     0F                 stack
N        DS     F                  r2 n
POLES    DS     F                  r3 poles
STACKLEN EQU    *-STACKDS
YREGS
END    HANOITOW
Output:
           1 Move disc from pole 1 to pole 3
2 Move disc from pole 1 to pole 2
3 Move disc from pole 3 to pole 2
4 Move disc from pole 1 to pole 3
5 Move disc from pole 2 to pole 1
6 Move disc from pole 2 to pole 3
7 Move disc from pole 1 to pole 3
8 Move disc from pole 1 to pole 2
9 Move disc from pole 3 to pole 2
10 Move disc from pole 3 to pole 1
11 Move disc from pole 2 to pole 1
12 Move disc from pole 3 to pole 2
13 Move disc from pole 1 to pole 3
14 Move disc from pole 1 to pole 2
15 Move disc from pole 3 to pole 2


## 6502 Assembly

Works with: Commodore

This should work on any Commodore 8-bit computer; just set temp to an appropriate zero-page location.

temp   = $FB ; this works on a VIC-20 or C-64; adjust for other machines. Need two bytes zero-page space unused by the OS. ; kernal print-char routine chrout =$FFD2

; Main Towers of Hanoi routine. To call, load the accumulator with the number of disks to move,
; the X register with the source peg (1-3), and the Y register with the target peg.

hanoi:   cmp #$00 ; do nothing if the number of disks to move is zero bne nonzero rts nonzero: pha ; save registers on stack txa pha tya pha pha ; and make room for the spare peg number ; Parameters are now on the stack at these offsets: count =$0104
source = $0103 target =$0102
spare  = $0101 ; compute spare rod number (6 - source - dest) tsx lda #6 sec sbc source, x sec sbc target, x sta spare, x ; prepare for first recursive call tay ; target is the spare peg tsx lda source, x ; source is the same sta temp ; we're using X to access the stack, so save its value here for now lda count, x ; move count - 1 disks sec sbc #1 ldx temp ; now load X for call ; and recurse jsr hanoi ; restore X and Y for print call tsx ldy target, x lda source, x tax ; print instructions to move the last disk jsr print_move ; prepare for final recursive call tsx lda spare, x ; source is now spare sta temp lda target, x ; going to the original target tay lda count, x ; and again moving count-1 disks sec sbc #1 ldx temp jsr hanoi ; pop our stack frame, restore registers, and return pla pla tay pla tax pla rts ; constants for printing prelude: .asciiz "MOVE DISK FROM " interlude: .asciiz " TO " postlude: .byte 13,0 ; print instructions: move disk from (X) to (Y) print_move: pha txa pha tya pha ; Parameters are now on the stack at these offsets: from =$0102
to   = $0101 lda #<prelude ldx #>prelude jsr print_string tsx lda from,x clc adc #$30
jsr chrout
lda #<interlude
ldx #>interlude
jsr print_string
tsx
lda to,x
clc
adc #$30 jsr chrout lda #<postlude ldx #>postlude jsr print_string pla tay pla tax pla rts ; utility routine: print null-terminated string at address AX print_string: sta temp stx temp+1 ldy #0 loop: lda (temp),y beq done_print jsr chrout iny bne loop done_print: rts Output: MOVE DISK FROM 1 TO 2 MOVE DISK FROM 1 TO 3 MOVE DISK FROM 2 TO 3 MOVE DISK FROM 1 TO 2 MOVE DISK FROM 3 TO 1 MOVE DISK FROM 3 TO 2 MOVE DISK FROM 1 TO 2 MOVE DISK FROM 1 TO 3 MOVE DISK FROM 2 TO 3 MOVE DISK FROM 2 TO 1 MOVE DISK FROM 3 TO 1 MOVE DISK FROM 2 TO 3 MOVE DISK FROM 1 TO 2 MOVE DISK FROM 1 TO 3 MOVE DISK FROM 2 TO 3 ## 8080 Assembly  org 100h lhld 6 ; Top of CP/M usable memory sphl ; Put the stack there lxi b,0401h ; Set up first arguments to move() lxi d,0203h call move ; move(4, 1, 2, 3) rst 0 ; quit program ;;; Move B disks from C via D to E. move: dcr b ; One fewer disk in next iteration jz mvout ; If this was the last disk, print move and stop push b ; Otherwise, save registers, push d mov a,d ; First recursive call mov d,e mov e,a call move ; move(B-1, C, E, D) pop d ; Restore registers pop b call mvout ; Print current move mov a,c ; Second recursive call mov c,d mov d,a jmp move ; move(B-1, D, C, E) - tail call optimization ;;; Print move, saving registers. mvout: push b ; Save registers on stack push d mov a,c ; Store 'from' as ASCII digit in 'from' space adi '0' sta out1 mov a,e ; Store 'to' as ASCII digit in 'to' space adi '0' sta out2 lxi d,outstr mvi c,9 ; CP/M call to print the string call 5 pop d ; Restore register contents pop b ret ;;; Move output with placeholder for pole numbers outstr: db 'Move disk from pole ' out1: db '* to pole ' out2: db '*',13,10,'$'
Output:
Move disk from pole 1 to pole 2
Move disk from pole 1 to pole 3
Move disk from pole 2 to pole 3
Move disk from pole 1 to pole 2
Move disk from pole 3 to pole 1
Move disk from pole 3 to pole 2
Move disk from pole 1 to pole 2
Move disk from pole 1 to pole 3
Move disk from pole 2 to pole 3
Move disk from pole 2 to pole 1
Move disk from pole 3 to pole 1
Move disk from pole 2 to pole 3
Move disk from pole 1 to pole 2
Move disk from pole 1 to pole 3
Move disk from pole 2 to pole 3


## 8086 Assembly

	cpu	8086
bits	16
org	100h
section	.text
mov	bx,0402h	; Set up first arguments to move()
mov	cx,0103h	; Registers chosen s.t. CX contains output
;;;	Move BH disks from CH via BL to CL
move:	dec	bh		; One fewer disk in next iteration
jz	.out		; If this was last disk, just print move
push	bx		; Save the registers for a recursive call
push	cx
xchg	bl,cl		; Swap the 'to' and 'via' registers
call	move		; move(BH, CH, CL, BL)
pop	cx		; Restore the registers from the stack
pop	bx
call	.out		; Print the move
xchg	ch,bl		; Swap the 'from' and 'via' registers
jmp	move		; move(BH, BL, CH, CL)
;;;	Print the move
.out:	mov	ax,'00'		; Add ASCII 0 to both 'from' and 'to'
add	ax,cx		; in one 16-bit operation
mov	[out1],ah	; Store 'from' field in output
mov	[out2],al	; Store 'to' field in output
mov	dx,outstr	; MS-DOS system call to print string
mov	ah,9
int	21h
ret
section	.data
outstr:	db	'Move disk from pole '
out1:	db	'* to pole '
out2:	db	'*',13,10,'$'  Output: Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3  ## 8th 5 var, disks var sa var sb var sc : save sc ! sb ! sa ! disks ! ; : get sa @ sb @ sc @ ; : get2 get swap ; : hanoi save disks @ not if ;; then disks @ get disks @ n:1- get2 hanoi save cr " move a ring from " . sa @ . " to " . sb @ . disks @ n:1- get2 rot hanoi ; " Tower of Hanoi, with " . disks @ . " rings: " . disks @ 1 2 3 hanoi cr bye  ## ActionScript public function move(n:int, from:int, to:int, via:int):void { if (n > 0) { move(n - 1, from, via, to); trace("Move disk from pole " + from + " to pole " + to); move(n - 1, via, to, from); } }  ## Ada with Ada.Text_Io; use Ada.Text_Io; procedure Towers is type Pegs is (Left, Center, Right); procedure Hanoi (Ndisks : Natural; Start_Peg : Pegs := Left; End_Peg : Pegs := Right; Via_Peg : Pegs := Center) is begin if Ndisks > 0 then Hanoi(Ndisks - 1, Start_Peg, Via_Peg, End_Peg); Put_Line("Move disk" & Natural'Image(Ndisks) & " from " & Pegs'Image(Start_Peg) & " to " & Pegs'Image(End_Peg)); Hanoi(Ndisks - 1, Via_Peg, End_Peg, Start_Peg); end if; end Hanoi; begin Hanoi(4); end Towers;  ## Agena move := proc(n::number, src::number, dst::number, via::number) is if n > 0 then move(n - 1, src, via, dst) print(src & ' to ' & dst) move(n - 1, via, dst, src) fi end move(4, 1, 2, 3) ## ALGOL 60 begin procedure movedisk(n, f, t); integer n, f, t; begin outstring (1, "Move disk from"); outinteger(1, f); outstring (1, "to"); outinteger(1, t); outstring (1, "\n"); end; procedure dohanoi(n, f, t, u); integer n, f, t, u; begin if n < 2 then movedisk(1, f, t) else begin dohanoi(n - 1, f, u, t); movedisk(1, f, t); dohanoi(n - 1, u, t, f); end; end; dohanoi(4, 1, 2, 3); outstring(1,"Towers of Hanoi puzzle completed!") end Output: Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Move disk from 1 to 3 Move disk from 2 to 1 Move disk from 2 to 3 Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Move disk from 3 to 1 Move disk from 2 to 1 Move disk from 3 to 2 Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Towers of Hanoi puzzle completed! ## ALGOL 68 PROC move = (INT n, from, to, via) VOID: IF n > 0 THEN move(n - 1, from, via, to); printf(($"Move disk from pole "g" to pole "gl$, from, to)); move(n - 1, via, to, from) FI ; main: ( move(4, 1,2,3) ) COMMENT Disk number is also printed in this code (works with a68g): COMMENT PROC move = (INT n, from, to, via) VOID: IF n > 0 THEN move(n - 1, from, via, to); printf(($"Move disk "g"     from pole "g"     to pole "gl$, n, from, to)); move(n - 1, via, to, from) FI ; main: ( move(4, 1,2,3) ) ## ALGOL-M begin procedure move(n, src, via, dest); integer n; string(1) src, via, dest; begin if n > 0 then begin move(n-1, src, dest, via); write("Move disk from pole "); writeon(src); writeon(" to pole "); writeon(dest); move(n-1, via, src, dest); end; end; move(4, "1", "2", "3"); end Output: Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 ## ALGOL W ### Recursive Following Agena, Algol 68, AmigaE... begin procedure move ( integer value n, from, to, via ) ; if n > 0 then begin move( n - 1, from, via, to ); write( i_w := 1, s_w := 0, "Move disk from peg: ", from, " to peg: ", to ); move( n - 1, via, to, from ) end move ; move( 4, 1, 2, 3 ) end.  ### Iterative Translation of: Tiny BASIC begin % iterative towers of hanoi - translated from Tiny Basic % integer d, n; while begin writeon( "How many disks? " ); read( d ); d < 1 or d > 10 end do begin end; n := 1; while d not = 0 do begin d := d - 1; n := 2 * n end; for x := 1 until n - 1 do begin integer s, t; % Algol W has the necessary bit and modulo operators so these are used here % % instead of implementing them via subroutines % s := number( bitstring( x ) and bitstring( x - 1 ) ) rem 3; t := ( number( bitstring( x ) or bitstring( x - 1 ) ) + 1 ) rem 3; write( i_w := 1, s_w := 0, "Move disc on peg ", s + 1, " to peg ", t + 1 ) end end.  ## AmigaE PROC move(n, from, to, via) IF n > 0 move(n-1, from, via, to) WriteF('Move disk from pole \d to pole \d\n', from, to) move(n-1, via, to, from) ENDIF ENDPROC PROC main() move(4, 1,2,3) ENDPROC ## Amazing Hopper #include <hopper.h> #proto hanoi(_X_,_Y_,_Z_,_W_) main: get arg number (2,discos) {discos}!neg? do{fail=0,mov(fail),{"I need a positive (or zero) number here, not: ",fail}println,exit(0)} pos? do{ _hanoi( discos, "A", "B", "C" ) } exit(0) .locals hanoi(discos,inicio,aux,fin) iif( {discos}eqto(1), {inicio, "->", fin, "\n"};print, _hanoi({discos}minus(1), inicio,fin,aux);\ {inicio, "->", fin, "\n"};print;\ _hanoi({discos}minus(1), aux, inicio, fin)) back Output: For 4 discs: A->B A->C B->C A->B C->A C->B A->B A->C B->C B->A C->A B->C A->B A->C B->C  ## APL Works with: Dyalog APL hanoi←{ move←{ n from to via←⍵ n≤0:⍬ l←∇(n-1) from via to r←∇(n-1) via to from l,(⊂from to),r } '⊂Move disk from pole ⊃,I1,⊂ to pole ⊃,I1'⎕FMT↑move ⍵ }  Output:  hanoi 4 1 2 3 Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 ## AppleScript --------------------- TOWERS OF HANOI -------------------- -- hanoi :: Int -> (String, String, String) -> [(String, String)] on hanoi(n, abc) script go on |λ|(n, {x, y, z}) if n > 0 then set m to n - 1 |λ|(m, {x, z, y}) & ¬ {{x, y}} & |λ|(m, {z, y, x}) else {} end if end |λ| end script go's |λ|(n, abc) end hanoi --------------------------- TEST ------------------------- on run unlines(map(intercalate(" -> "), ¬ hanoi(3, {"left", "right", "mid"}))) end run -------------------- GENERIC FUNCTIONS ------------------- -- intercalate :: String -> [String] -> String on intercalate(delim) script on |λ|(xs) set {dlm, my text item delimiters} to ¬ {my text item delimiters, delim} set s to xs as text set my text item delimiters to dlm s end |λ| end script end intercalate -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: First-class m => (a -> b) -> m (a -> b) on mReturn(f) if class of f is script then f else script property |λ| : f end script end if end mReturn -- map :: (a -> b) -> [a] -> [b] on map(f, xs) tell mReturn(f) set lng to length of xs set lst to {} repeat with i from 1 to lng set end of lst to |λ|(item i of xs, i, xs) end repeat return lst end tell end map -- unlines :: [String] -> String on unlines(xs) set {dlm, my text item delimiters} to ¬ {my text item delimiters, linefeed} set str to xs as text set my text item delimiters to dlm str end unlines  Output: left -> right left -> mid right -> mid left -> right mid -> left mid -> right left -> right More illustratively: (I've now eliminated the recursive |move|() handler's tail calls. So it's now only called 2 ^ (n - 1) times as opposed to 2 ^ (n + 1) - 1 with full recursion. The maximum call depth of n is only reached once, whereas with full recursion, the maximum depth was n + 1 and this was reached 2 ^ n times.) on hanoi(n, source, target) set t1 to tab & "tower 1: " & tab set t2 to tab & "tower 2: " & tab set t3 to tab & "tower 3: " & tab script o property m : 0 property tower1 : {} property tower2 : {} property tower3 : {} property towerRefs : {a reference to tower1, a reference to tower2, a reference to tower3} property process : missing value on |move|(n, source, target) set aux to 6 - source - target repeat with n from n to 2 by -1 -- Tail call elimination repeat. |move|(n - 1, source, aux) set end of item target of my towerRefs to n tell item source of my towerRefs to set its contents to reverse of rest of its reverse set m to m + 1 set end of my process to ¬ {(m as text) & ". move disc " & n & (" from tower " & source) & (" to tower " & target & ":"), ¬ t1 & tower1, ¬ t2 & tower2, ¬ t3 & tower3} tell source set source to aux set aux to it end tell end repeat -- Specific code for n = 1: set end of item target of my towerRefs to 1 tell item source of my towerRefs to set its contents to reverse of rest of its reverse set m to m + 1 set end of my process to ¬ {(m as text) & ". move disc 1 from tower " & source & (" to tower " & target & ":"), ¬ t1 & tower1, ¬ t2 & tower2, ¬ t3 & tower3} end |move| end script repeat with i from n to 1 by -1 set end of item source of o's towerRefs to i end repeat set astid to AppleScript's text item delimiters set AppleScript's text item delimiters to ", " set o's process to {"Starting with " & n & (" discs on tower " & (source & ":")), ¬ t1 & o's tower1, t2 & o's tower2, t3 & o's tower3} if (n > 0) then tell o to |move|(n, source, target) set end of o's process to "That's it!" set AppleScript's text item delimiters to linefeed set process to o's process as text set AppleScript's text item delimiters to astid return process end hanoi -- Test: set numberOfDiscs to 3 set sourceTower to 1 set destinationTower to 2 hanoi(numberOfDiscs, sourceTower, destinationTower)  Output: "Starting with 3 discs on tower 1: tower 1: 3, 2, 1 tower 2: tower 3: 1. move disc 1 from tower 1 to tower 2: tower 1: 3, 2 tower 2: 1 tower 3: 2. move disc 2 from tower 1 to tower 3: tower 1: 3 tower 2: 1 tower 3: 2 3. move disc 1 from tower 2 to tower 3: tower 1: 3 tower 2: tower 3: 2, 1 4. move disc 3 from tower 1 to tower 2: tower 1: tower 2: 3 tower 3: 2, 1 5. move disc 1 from tower 3 to tower 1: tower 1: 1 tower 2: 3 tower 3: 2 6. move disc 2 from tower 3 to tower 2: tower 1: 1 tower 2: 3, 2 tower 3: 7. move disc 1 from tower 1 to tower 2: tower 1: tower 2: 3, 2, 1 tower 3: That's it!" ## ARM Assembly .text .global _start _start: mov r0,#4 @ 4 disks, mov r1,#1 @ from pole 1, mov r2,#2 @ via pole 2, mov r3,#3 @ to pole 3. bl move mov r0,#0 @ Exit to Linux afterwards mov r7,#1 swi #0 @@@ Move r0 disks from r1 via r2 to r3 move: subs r0,r0,#1 @ One fewer disk in next iteration beq show @ If last disk, just print move push {r0-r3,lr} @ Save all the registers incl. link register eor r2,r2,r3 @ Swap the 'to' and 'via' registers eor r3,r2,r3 eor r2,r2,r3 bl move @ Recursive call pop {r0-r3} @ Restore all the registers except LR bl show @ Show current move eor r1,r1,r3 @ Swap the 'to' and 'via' registers eor r3,r1,r3 eor r1,r1,r3 pop {lr} @ Restore link register b move @ Tail call @@@ Show move show: push {r0-r3,lr} @ Save all the registers add r1,r1,#'0 @ Write the source pole ldr lr,=spole strb r1,[lr] add r3,r3,#'0 @ Write the destination pole ldr lr,=dpole strb r3,[lr] mov r0,#1 @ 1 = stdout ldr r1,=moves @ Pointer to string ldr r2,=mlen @ Length of string mov r7,#4 @ 4 = Linux write syscall swi #0 @ Print the move pop {r0-r3,pc} @ Restore all the registers and return .data moves: .ascii "Move disk from pole " spole: .ascii "* to pole " dpole: .ascii "*\n" mlen = . - moves  Output: Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 3 to pole 1 Move disk from pole 1 to pole 2 Move disk from pole 2 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 3 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 2 to pole 3 Move disk from pole 3 to pole 1 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 3 to pole 1 ## Arturo Translation of: D hanoi: function [n f dir via][ if n>0 [ hanoi n-1 f via dir print ["Move disk" n "from" f "to" dir] hanoi n-1 via dir f ] ] hanoi 3 'L 'M 'R  Output: Move disk 1 from L to M Move disk 2 from L to R Move disk 1 from M to R Move disk 3 from L to M Move disk 1 from R to L Move disk 2 from R to M Move disk 1 from L to M ## AutoHotkey move(n, from, to, via) ;n = # of disks, from = start pole, to = end pole, via = remaining pole { if (n = 1) { msgbox , Move disk from pole %from% to pole %to% } else { move(n-1, from, via, to) move(1, from, to, via) move(n-1, via, to, from) } } move(64, 1, 3, 2)  ## AutoIt Func move($n, $from,$to, $via) If ($n = 1) Then
ConsoleWrite(StringFormat("Move disk from pole "&$from&" To pole "&$to&"\n"))
Else
move($n - 1,$from, $via,$to)
move(1, $from,$to, $via) move($n - 1, $via,$to, $from) EndIf EndFunc move(4, 1,2,3)  ## AWK Translation of: Logo $ awk 'func hanoi(n,f,t,v){if(n>0){hanoi(n-1,f,v,t);print(f,"->",t);hanoi(n-1,v,t,f)}}
BEGIN{hanoi(4,"left","middle","right")}'

Output:
left -> right
left -> middle
right -> middle
left -> right
middle -> left
middle -> right
left -> right
left -> middle
right -> middle
right -> left
middle -> left
right -> middle
left -> right
left -> middle
right -> middle

## BASIC

### Using a Subroutine

Works with: FreeBASIC
Works with: RapidQ
SUB move (n AS Integer, fromPeg AS Integer, toPeg AS Integer, viaPeg AS Integer)
IF n>0 THEN
move n-1, fromPeg, viaPeg, toPeg
PRINT "Move disk from "; fromPeg; " to "; toPeg
move n-1, viaPeg, toPeg, fromPeg
END IF
END SUB

move 4,1,2,3


### Using GOSUBs

Here's an example of implementing recursion in an old BASIC that only has global variables:

Works with: Applesoft BASIC
Works with: Commodore BASIC
Works with: GW-BASIC
10 DEPTH=4: REM SHOULD EQUAL NUMBER OF DISKS
20 DIM N(DEPTH), F(DEPTH), T(DEPTH), V(DEPTH): REM STACK PER PARAMETER
30 SP = 0:    REM STACK POINTER
50 F(SP) = 1: REM ON PEG 1
60 T(SP) = 2: REM MOVE TO PEG 2
70 V(SP) = 3: REM VIA PEG 3
80 GOSUB 100
90 END
99 REM MOVE SUBROUTINE
100 IF N(SP) = 0 THEN RETURN
110 OS = SP:           REM STORE STACK POINTER
120 SP = SP + 1:       REM INCREMENT STACK POINTER
130 N(SP) = N(OS) - 1: REM MOVE N-1 DISCS
140 F(SP) = F(OS)    : REM FROM START PEG
150 T(SP) = V(OS)    : REM TO VIA PEG
160 V(SP) = T(OS)    : REM VIA TO PEG
170 GOSUB 100
180 OS = SP - 1: REM OS WILL HAVE CHANGED
190 PRINT "MOVE DISC FROM"; F(OS); "TO"; T(OS)
200 N(SP) = N(OS) - 1: REM MOVE N-1 DISCS
210 F(SP) = V(OS)    : REM FROM VIA PEG
220 T(SP) = T(OS)    : REM TO DEST PEG
230 V(SP) = F(OS)    : REM VIA FROM PEG
240 GOSUB 100
250 SP = SP - 1      : REM RESTORE STACK POINTER FOR CALLER
260 RETURN


### Using binary method

Works with: Commodore BASIC

Very fast version in BASIC V2 on Commodore C-64

   10 DEF FNM3(X)=X-INT(X/3)*3:REM MODULO 3
20 N=4:GOSUB 100
30 END
99 REM HANOI
100 :FOR M=1 TO 2^N-1
110 ::PRINT MID$(STR$(M),2);":",FNM3(M AND M-1)+1;"TO";FNM3((M OR M-1)+1)+1
130 :NEXT M
140 RETURN

Output:
1:     1 TO 3
2:     1 TO 2
3:     3 TO 2
4:     1 TO 3
5:     2 TO 1
6:     2 TO 3
7:     1 TO 3
8:     1 TO 2
9:     3 TO 2
10:    3 TO 1
11:    2 TO 1
12:    3 TO 2
13:    1 TO 3
14:    1 TO 2
15:    3 TO 2 

## BASIC256

call move(4,1,2,3)
print "Towers of Hanoi puzzle completed!"
end

subroutine move (n, fromPeg, toPeg, viaPeg)
if n>0 then
call move(n-1, fromPeg, viaPeg, toPeg)
print "Move disk from "+fromPeg+" to "+toPeg
call move(n-1, viaPeg, toPeg, fromPeg)
end if
end subroutine
Output:
Move disk from 1 to 3
Move disk from 1 to 2
Move disk from 3 to 2
Move disk from 1 to 3
Move disk from 2 to 1
Move disk from 2 to 3
Move disk from 1 to 3
Move disk from 1 to 2
Move disk from 3 to 2
Move disk from 3 to 1
Move disk from 2 to 1
Move disk from 3 to 2
Move disk from 1 to 3
Move disk from 1 to 2
Move disk from 3 to 2
Towers of Hanoi puzzle completed!


## Batch File

@echo off
setlocal enabledelayedexpansion

%==The main thing==%
%==First param - Number of disks==%
%==Second param - Start pole==%
%==Third param - End pole==%
%==Fourth param - Helper pole==%
call :move 4 START END HELPER
echo.
pause
exit /b 0

%==The "function"==%
:move
setlocal
set n=%1
set from=%2
set to=%3
set via=%4

if %n% gtr 0 (
set /a x=!n!-1
call :move !x! %from% %via% %to%
echo Move top disk from pole %from% to pole %to%.
call :move !x! %via% %to% %from%
)
exit /b 0

Output:
Move top disk from pole START to pole HELPER.
Move top disk from pole START to pole END.
Move top disk from pole HELPER to pole END.
Move top disk from pole START to pole HELPER.
Move top disk from pole END to pole START.
Move top disk from pole END to pole HELPER.
Move top disk from pole START to pole HELPER.
Move top disk from pole START to pole END.
Move top disk from pole HELPER to pole END.
Move top disk from pole HELPER to pole START.
Move top disk from pole END to pole START.
Move top disk from pole HELPER to pole END.
Move top disk from pole START to pole HELPER.
Move top disk from pole START to pole END.
Move top disk from pole HELPER to pole END.

Press any key to continue . . .

## BBC BASIC

      DIM Disc$(13),Size%(3) FOR disc% = 1 TO 13 Disc$(disc%) = STRING$(disc%," ")+STR$disc%+STRING$(disc%," ") IF disc%>=10 Disc$(disc%) = MID$(Disc$(disc%),2)
Disc$(disc%) = CHR$17+CHR$(128+disc%-(disc%>7))+Disc$(disc%)+CHR$17+CHR$128
NEXT disc%

MODE 3
OFF
ndiscs% = 13
FOR n% = ndiscs% TO 1 STEP -1
PROCput(n%,1)
NEXT
INPUT TAB(0,0) "Press Enter to start" dummy$PRINT TAB(0,0) SPC(20); PROChanoi(ndiscs%,1,2,3) VDU 30 END DEF PROChanoi(a%,b%,c%,d%) IF a%=0 ENDPROC PROChanoi(a%-1,b%,d%,c%) PROCtake(a%,b%) PROCput(a%,c%) PROChanoi(a%-1,d%,c%,b%) ENDPROC DEF PROCput(disc%,peg%) PRINTTAB(13+26*(peg%-1)-disc%,20-Size%(peg%))Disc$(disc%);
Size%(peg%) = Size%(peg%)+1
ENDPROC

DEF PROCtake(disc%,peg%)
Size%(peg%) = Size%(peg%)-1
PRINTTAB(13+26*(peg%-1)-disc%,20-Size%(peg%))STRING$(2*disc%+1," "); ENDPROC  ## BCPL get "libhdr" let start() be move(4, 1, 2, 3) and move(n, src, via, dest) be if n > 0 do$(  move(n-1, src, dest, via)
writef("Move disk from pole %N to pole %N*N", src, dest);
move(n-1, via, src, dest)
$) Output: Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 ## Befunge This is loosely based on the Python sample. The number of disks is specified by the first integer on the stack (the initial character 4 in the example below). If you want the program to prompt the user for that value, you can replace the 4 with a & (the read integer command). 48*2+1>#v_:!#@_0" ksid evoM">:#,_$:8/:.v
>8v8:<$#<+9-+*2%3\*3/3:,+55.+1%3:$_,#!>#:<
: >/!#^_:0\:8/1-8vv,_$8%:3/1+.>0" gep ot"^ ^++3-%3\*2/3:%8\*<>:^:"from peg "0\*8-1<  Output: Move disk 1 from peg 1 to peg 2 Move disk 2 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 Move disk 3 from peg 1 to peg 2 Move disk 1 from peg 3 to peg 1 Move disk 2 from peg 3 to peg 2 Move disk 1 from peg 1 to peg 2 Move disk 4 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 Move disk 2 from peg 2 to peg 1 Move disk 1 from peg 3 to peg 1 Move disk 3 from peg 2 to peg 3 Move disk 1 from peg 1 to peg 2 Move disk 2 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 ## BQN Based on: APL Move ← { 𝕩⊑⊸≤0 ? ⟨⟩; 𝕊 n‿from‿to‿via: l ← 𝕊 ⟨n-1, from, via, to⟩ r ← 𝕊 ⟨n-1, via, to, from⟩ l∾(<from‿to)∾r } {"Move disk from pole "∾(•Fmt 𝕨)∾" to pole "∾•Fmt 𝕩}´˘>Move 4‿1‿2‿3 ┌─ ╵"Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2" ┘  ## Bracmat ( ( move = n from to via . !arg:(?n,?from,?to,?via) & ( !n:>0 & move$(!n+-1,!from,!via,!to)
& out$("Move disk from pole " !from " to pole " !to) & move$(!n+-1,!via,!to,!from)
|
)
)
& move$(4,1,2,3) ); Output: Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 2 ## Brainf*** [ This implementation is recursive and uses a stack, consisting of frames that are 8 bytes long. The layout is as follows: Byte Description 0 recursion flag (the program stops if the flag is zero) 1 the step which is currently executed 4 means a call to move(a, c, b, n - 1) 3 means a call to move(a, b, c, 1) 2 means a call to move(b, a, c, n - 1) 1 prints the source and dest pile 2 flag to check whether the current step has already been done or if it still must be executed 3 the step which will be executed in the next loop 4 the source pile 5 the helper pile 6 the destination pile 7 the number of disks to move The first stack frame (0 0 0 0 0 0 0 0) is used to abort the recursion. ] >>>>>>>> These are the parameters for the program (1 4 1 0 'a 'b 'c 5) +>++++>+>> >>>>++++++++[<++++++++++++>-]< [<<<+>+>+>-]<<<+>++>+++>+++++> <<<<<<<< [> while (recurse) [- if (step gt 0) >[-]+< todo = 1 [- if (step gt 1) [- if (step gt 2) [- if (step gt 3) >>+++<< next = 3 >-< todo = 0 >>>>>>[>+>+<<-]>[<+>-]> n dup - [[-] if (sub(n 1) gt 0) <+>>>++++> push (1 0 0 4) copy and push a <<<<<<<<[>>>>>>>>+>+ <<<<<<<<<-]>>>>>>>> >[<<<<<<<<<+>>>>>>>>>-]< > copy and push c <<<<<<<[>>>>>>>+>+ <<<<<<<<-]>>>>>>> >[<<<<<<<<+>>>>>>>>-]< > copy and push b <<<<<<<<<[>>>>>>>>>+>+ <<<<<<<<<<-]>>>>>>>>> >[<<<<<<<<<<+>>>>>>>>>>-]< > copy n and push sub(n 1) <<<<<<<<[>>>>>>>>+>+ <<<<<<<<<-]>>>>>>>> >[<<<<<<<<<+>>>>>>>>>-]< - >> ] <<<<<<<< ] >[-< if ((step gt 2) and todo) >>++<< next = 2 >>>>>>> +>>>+> push 1 0 0 1 a b c 1 <<<<<<<<[>>>>>>>>+>+ <<<<<<<<<-]>>>>>>>> >[<<<<<<<<<+>>>>>>>>>-]< > a <<<<<<<<[>>>>>>>>+>+ <<<<<<<<<-]>>>>>>>> >[<<<<<<<<<+>>>>>>>>>-]< > b <<<<<<<<[>>>>>>>>+>+ <<<<<<<<<-]>>>>>>>> >[<<<<<<<<<+>>>>>>>>>-]< > c + >> >]< ] >[-< if ((step gt 1) and todo) >>>>>>[>+>+<<-]>[<+>-]> n dup - [[-] if (n sub 1 gt 0) <+>>>++++> push (1 0 0 4) copy and push b <<<<<<<[>>>>>>>+ <<<<<<<-]>>>>>>> >[<<<<<<<<+>>>>>>>>-]< > copy and push a <<<<<<<<<[>>>>>>>>>+ <<<<<<<<<-]>>>>>>>>> >[<<<<<<<<<<+>>>>>>>>>>-]< > copy and push c <<<<<<<<[>>>>>>>>+ <<<<<<<<-]>>>>>>>> >[<<<<<<<<<+>>>>>>>>>-]< > copy n and push sub(n 1) <<<<<<<<[>>>>>>>>+>+ <<<<<<<<<-]>>>>>>>> >[<<<<<<<<<+>>>>>>>>>-]< - >> ] <<<<<<<< >]< ] >[-< if ((step gt 0) and todo) >>>>>>> >++++[<++++++++>-]< >>++++++++[<+++++++++>-]<++++ >>++++++++[<++++++++++++>-]<+++++ >>+++++++++[<++++++++++++>-]<+++ <<< >.+++++++>.++.--.<<. >>-.+++++.----.<<. >>>.<---.+++.>+++.+.+.<.<<. >.>--.+++++.---.++++. -------.+++.<<. >>>++.-------.-.<<<. >+.>>+++++++.---.-----.<<<. <<<<.>>>>. >>----.>++++++++.<+++++.<<. >.>>.---.-----.<<<. <<.>>++++++++++++++. >>>[-]<[-]<[-]<[-] +++++++++++++.---.[-] <<<<<<< >]< >>[<<+>>-]<< step = next ] return with clear stack frame <[-]>[-]>[-]>[-]>[-]>[-]>[-]>[-]<<<<<< <<<<<<<< >>[<<+>>-]<< step = next < ]  ## C #include <stdio.h> void move(int n, int from, int via, int to) { if (n > 1) { move(n - 1, from, to, via); printf("Move disk from pole %d to pole %d\n", from, to); move(n - 1, via, from, to); } else { printf("Move disk from pole %d to pole %d\n", from, to); } } int main() { move(4, 1,2,3); return 0; }  Animate it for fun: #include <stdio.h> #include <stdlib.h> #include <unistd.h> typedef struct { int *x, n; } tower; tower *new_tower(int cap) { tower *t = calloc(1, sizeof(tower) + sizeof(int) * cap); t->x = (int*)(t + 1); return t; } tower *t; int height; void text(int y, int i, int d, const char *s) { printf("\033[%d;%dH", height - y + 1, (height + 1) * (2 * i + 1) - d); while (d--) printf("%s", s); } void add_disk(int i, int d) { t[i]->x[t[i]->n++] = d; text(t[i]->n, i, d, "=="); usleep(100000); fflush(stdout); } int remove_disk(int i) { int d = t[i]->x[--t[i]->n]; text(t[i]->n + 1, i, d, " "); return d; } void move(int n, int from, int to, int via) { if (!n) return; move(n - 1, from, via, to); add_disk(to, remove_disk(from)); move(n - 1, via, to, from); } int main(int c, char *v[]) { puts("\033[H\033[J"); if (c <= 1 || (height = atoi(v)) <= 0) height = 8; for (c = 0; c < 3; c++) t[c] = new_tower(height); for (c = height; c; c--) add_disk(0, c); move(height, 0, 2, 1); text(1, 0, 1, "\n"); return 0; }  ## C# public void move(int n, int from, int to, int via) { if (n == 1) { System.Console.WriteLine("Move disk from pole " + from + " to pole " + to); } else { move(n - 1, from, via, to); move(1, from, to, via); move(n - 1, via, to, from); } }  ## C++ Works with: g++ void move(int n, int from, int to, int via) { if (n == 1) { std::cout << "Move disk from pole " << from << " to pole " << to << std::endl; } else { move(n - 1, from, via, to); move(1, from, to, via); move(n - 1, via, to, from); } }  ## Clojure ### Side-Effecting Solution (defn towers-of-hanoi [n from to via] (when (pos? n) (towers-of-hanoi (dec n) from via to) (printf "Move from %s to %s\n" from to) (recur (dec n) via to from)))  ### Lazy Solution (defn towers-of-hanoi [n from to via] (when (pos? n) (lazy-cat (towers-of-hanoi (dec n) from via to) (cons [from '-> to] (towers-of-hanoi (dec n) via to from)))))  ## CLU move = proc (n, from, via, to: int) po: stream := stream$primary_output()
if n > 0 then
move(n-1, from, to, via)
stream$putl(po, "Move disk from pole " || int$unparse(from)
|| " to pole "
|| int$unparse(to)) move(n-1, via, from, to) end end move start_up = proc () move(4, 1, 2, 3) end start_up Output: Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 ## COBOL Translation of: C Works with: OpenCOBOL version 2.0  >>SOURCE FREE IDENTIFICATION DIVISION. PROGRAM-ID. towers-of-hanoi. PROCEDURE DIVISION. CALL "move-disk" USING 4, 1, 2, 3 . END PROGRAM towers-of-hanoi. IDENTIFICATION DIVISION. PROGRAM-ID. move-disk RECURSIVE. DATA DIVISION. LINKAGE SECTION. 01 n PIC 9 USAGE COMP. 01 from-pole PIC 9 USAGE COMP. 01 to-pole PIC 9 USAGE COMP. 01 via-pole PIC 9 USAGE COMP. PROCEDURE DIVISION USING n, from-pole, to-pole, via-pole. IF n > 0 SUBTRACT 1 FROM n CALL "move-disk" USING CONTENT n, from-pole, via-pole, to-pole DISPLAY "Move disk from pole " from-pole " to pole " to-pole CALL "move-disk" USING CONTENT n, via-pole, to-pole, from-pole END-IF . END PROGRAM move-disk.   IDENTIFICATION DIVISION. PROGRAM-ID. towers-of-hanoi. PROCEDURE DIVISION. CALL "move-disk" USING 4, 1, 2, 3 . END PROGRAM towers-of-hanoi. IDENTIFICATION DIVISION. PROGRAM-ID. move-disk RECURSIVE. DATA DIVISION. LINKAGE SECTION. 01 n PIC 9 USAGE COMP. 01 from-pole PIC 9 USAGE COMP. 01 to-pole PIC 9 USAGE COMP. 01 via-pole PIC 9 USAGE COMP. PROCEDURE DIVISION USING n, from-pole, to-pole, via-pole. IF n > 0 SUBTRACT 1 FROM n CALL "move-disk" USING CONTENT n, from-pole, via-pole, to-pole ADD 1 TO n DISPLAY "Move disk number "n " from pole " from-pole " to pole " to-pole SUBTRACT 1 FROM n CALL "move-disk" USING CONTENT n, via-pole, to-pole, from-pole END-IF . END PROGRAM move-disk.  ### ANSI-74 solution Early versions of COBOL did not have recursion. There are no locally-scoped variables and the call of a procedure does not have to use a stack to save return state. Recursion would cause undefined results. It is therefore necessary to use an iterative algorithm. This solution is an adaptation of Kolar's Hanoi Tower algorithm no. 1. Works with: CIS COBOL version 4.2 Works with: GnuCOBOL version 3.0-rc1.0  IDENTIFICATION DIVISION. PROGRAM-ID. ITERATIVE-TOWERS-OF-HANOI. AUTHOR. SOREN ROUG. DATE-WRITTEN. 2019-06-28. ENVIRONMENT DIVISION. CONFIGURATION SECTION. SOURCE-COMPUTER. LINUX. OBJECT-COMPUTER. KAYPRO4. INPUT-OUTPUT SECTION. FILE-CONTROL. DATA DIVISION. WORKING-STORAGE SECTION. 77 NUM-DISKS PIC 9 VALUE 4. 77 N1 PIC 9 COMP. 77 N2 PIC 9 COMP. 77 FROM-POLE PIC 9 COMP. 77 TO-POLE PIC 9 COMP. 77 VIA-POLE PIC 9 COMP. 77 FP-TMP PIC 9 COMP. 77 TO-TMP PIC 9 COMP. 77 P-TMP PIC 9 COMP. 77 TMP-P PIC 9 COMP. 77 I PIC 9 COMP. 77 DIV PIC 9 COMP. 01 STACKNUMS. 05 NUMSET OCCURS 3 TIMES. 10 DNUM PIC 9 COMP. 01 GAMESET. 05 POLES OCCURS 3 TIMES. 10 STACK OCCURS 10 TIMES. 15 POLE PIC 9 USAGE COMP. PROCEDURE DIVISION. HANOI. DISPLAY "TOWERS OF HANOI PUZZLE WITH ", NUM-DISKS, " DISKS.". ADD NUM-DISKS, 1 GIVING N1. ADD NUM-DISKS, 2 GIVING N2. MOVE 1 TO DNUM (1). MOVE N1 TO DNUM (2), DNUM (3). MOVE N1 TO POLE (1, N1), POLE (2, N1), POLE (3, N1). MOVE 1 TO POLE (1, N2). MOVE 2 TO POLE (2, N2). MOVE 3 TO POLE (3, N2). MOVE 1 TO I. PERFORM INIT-PUZZLE UNTIL I = N1. MOVE 1 TO FROM-POLE. DIVIDE 2 INTO NUM-DISKS GIVING DIV. MULTIPLY 2 BY DIV. IF DIV NOT = NUM-DISKS PERFORM INITODD ELSE PERFORM INITEVEN. PERFORM MOVE-DISK UNTIL DNUM (3) NOT > 1. DISPLAY "TOWERS OF HANOI PUZZLE COMPLETED!". STOP RUN. INIT-PUZZLE. MOVE I TO POLE (1, I). MOVE 0 TO POLE (2, I), POLE (3, I). ADD 1 TO I. INITEVEN. MOVE 2 TO TO-POLE. MOVE 3 TO VIA-POLE. INITODD. MOVE 3 TO TO-POLE. MOVE 2 TO VIA-POLE. MOVE-DISK. MOVE DNUM (FROM-POLE) TO FP-TMP. MOVE POLE (FROM-POLE, FP-TMP) TO I. DISPLAY "MOVE DISK FROM ", POLE (FROM-POLE, N2), " TO ", POLE (TO-POLE, N2). ADD 1 TO DNUM (FROM-POLE). MOVE VIA-POLE TO TMP-P. SUBTRACT 1 FROM DNUM (TO-POLE). MOVE DNUM (TO-POLE) TO TO-TMP. MOVE I TO POLE (TO-POLE, TO-TMP). DIVIDE 2 INTO I GIVING DIV. MULTIPLY 2 BY DIV. IF I NOT = DIV PERFORM MOVE-TO-VIA ELSE PERFORM MOVE-FROM-VIA. MOVE-TO-VIA. MOVE TO-POLE TO VIA-POLE. MOVE DNUM (FROM-POLE) TO FP-TMP. MOVE DNUM (TMP-P) TO P-TMP. IF POLE (FROM-POLE, FP-TMP) > POLE (TMP-P, P-TMP) PERFORM MOVE-FROM-TO ELSE MOVE TMP-P TO TO-POLE. MOVE-FROM-TO. MOVE FROM-POLE TO TO-POLE. MOVE TMP-P TO FROM-POLE. MOVE DNUM (FROM-POLE) TO FP-TMP. MOVE DNUM (TMP-P) TO P-TMP. MOVE-FROM-VIA. MOVE FROM-POLE TO VIA-POLE. MOVE TMP-P TO FROM-POLE.  ## CoffeeScript hanoi = (ndisks, start_peg=1, end_peg=3) -> if ndisks staging_peg = 1 + 2 + 3 - start_peg - end_peg hanoi(ndisks-1, start_peg, staging_peg) console.log "Move disk #{ndisks} from peg #{start_peg} to #{end_peg}" hanoi(ndisks-1, staging_peg, end_peg) hanoi(4)  ## Common Lisp (defun move (n from to via) (cond ((= n 1) (format t "Move from ~A to ~A.~%" from to)) (t (move (- n 1) from via to) (format t "Move from ~A to ~A.~%" from to) (move (- n 1) via to from))))  ## D ### Recursive Version import std.stdio; void hanoi(in int n, in char from, in char to, in char via) { if (n > 0) { hanoi(n - 1, from, via, to); writefln("Move disk %d from %s to %s", n, from, to); hanoi(n - 1, via, to, from); } } void main() { hanoi(3, 'L', 'M', 'R'); }  Output: Move disk 1 from L to M Move disk 2 from L to R Move disk 1 from M to R Move disk 3 from L to M Move disk 1 from R to L Move disk 2 from R to M Move disk 1 from L to M ### Fast Iterative Version // Code found and then improved by Glenn C. Rhoads, // then some more by M. Kolar (2000). void main(in string[] args) { import core.stdc.stdio, std.conv, std.typetuple; immutable size_t n = (args.length > 1) ? args.to!size_t : 3; size_t p = [(1 << n) - 1, 0, 0]; // Show the start configuration of the pegs. '|'.putchar; foreach_reverse (immutable i; 1 .. n + 1) printf(" %d", i); "\n|\n|".puts; foreach (immutable size_t x; 1 .. (1 << n)) { { immutable size_t i1 = x & (x - 1); immutable size_t fr = (i1 + i1 / 3) & 3; immutable size_t i2 = (x | (x - 1)) + 1; immutable size_t to = (i2 + i2 / 3) & 3; size_t j = 1; for (size_t w = x; !(w & 1); w >>= 1, j <<= 1) {} // Now j is not the number of the disk to move, // it contains the single bit to be moved: p[fr] &= ~j; p[to] |= j; } // Show the current configuration of pegs. foreach (immutable size_t k; TypeTuple!(0, 1, 2)) { "\n|".printf; size_t j = 1 << n; foreach_reverse (immutable size_t w; 1 .. n + 1) { j >>= 1; if (j & p[k]) printf(" %zd", w); } } '\n'.putchar; } }  Output: | 3 2 1 | | | 3 2 | | 1 | 3 | 2 | 1 | 3 | 2 1 | | | 2 1 | 3 | 1 | 2 | 3 | 1 | | 3 2 | | | 3 2 1  ## Dart main() { moveit(from,to) { print("move${from} ---> ${to}"); } hanoi(height,toPole,fromPole,usePole) { if (height>0) { hanoi(height-1,usePole,fromPole,toPole); moveit(fromPole,toPole); hanoi(height-1,toPole,usePole,fromPole); } } hanoi(3,3,1,2); }  The same as above, with optional static type annotations and styled according to http://www.dartlang.org/articles/style-guide/ main() { String say(String from, String to) => "$from ---> $to"; hanoi(int height, int toPole, int fromPole, int usePole) { if (height > 0) { hanoi(height - 1, usePole, fromPole, toPole); print(say(fromPole.toString(), toPole.toString())); hanoi(height - 1, toPole, usePole, fromPole); } } hanoi(3, 3, 1, 2); }  Output: move 1 ---> 3 move 1 ---> 2 move 3 ---> 2 move 1 ---> 3 move 2 ---> 1 move 2 ---> 3 move 1 ---> 3  ## Dc From Here  [ # move(from, to) n # print from [ --> ]n # print " --> " p # print to\n sw # p doesn't pop, so get rid of the value ]sm [ # init(n) sw # tuck n away temporarily 9 # sentinel as bottom of stack lw # bring n back 1 # "from" tower's label 3 # "to" tower's label 0 # processed marker ]si [ # Move() lt # push to lf # push from lmx # call move(from, to) ]sM [ # code block <d> ln # push n lf # push from lt # push to 1 # push processed marker 1 ln # push n 1 # push 1 - # n - 1 lf # push from ll # push left 0 # push processed marker 0 ]sd [ # code block <e> ln # push n 1 # push 1 - # n - 1 ll # push left lt # push to 0 # push processed marker 0 ]se [ # code block <x> ln 1 =M ln 1 !=d ]sx [ # code block <y> lMx lex ]sy [ # quit() q # exit the program ]sq [ # run() d 9 =q # if stack empty, quit() sp # processed st # to sf # from sn # n 6 # lf # - # lt # - # 6 - from - to sl # lp 0 =x # lp 0 !=y # lrx # loop ]sr 5lix # init(n) lrx # run()  ## Delphi See Pascal. ## Draco proc move(byte n, src, via, dest) void: if n>0 then move(n-1, src, dest, via); writeln("Move disk from pole ",src," to pole ",dest); move(n-1, via, src, dest) fi corp proc nonrec main() void: move(4, 1, 2, 3) corp Output: Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 ## Dyalect Translation of: Swift func hanoi(n, a, b, c) { if n > 0 { hanoi(n - 1, a, c, b) print("Move disk from \(a) to \(c)") hanoi(n - 1, b, a, c) } } hanoi(4, "A", "B", "C") Output: Move disk from A to B Move disk from A to C Move disk from B to C Move disk from A to B Move disk from C to A Move disk from C to B Move disk from A to B Move disk from A to C Move disk from B to C Move disk from B to A Move disk from C to A Move disk from B to C Move disk from A to B Move disk from A to C Move disk from B to C ## E def move(out, n, fromPeg, toPeg, viaPeg) { if (n.aboveZero()) { move(out, n.previous(), fromPeg, viaPeg, toPeg) out.println(Move disk$n from $fromPeg to$toPeg.)
move(out, n.previous(), viaPeg, toPeg, fromPeg)
}
}

move(stdout, 4, def left {}, def right {}, def middle {})

## EasyLang

func hanoi n src dst aux . .
if n >= 1
call hanoi n - 1 src aux dst
print "Move " & src & " to " & dst
call hanoi n - 1 aux dst src
.
.
call hanoi 5 1 2 3


## EDSAC order code

The Wikipedia article on EDSAC says "recursive calls were forbidden", and this is true if the standard "Wheeler jump" is used. In the Wheeler jump, the caller (in effect) passes the return address to the subroutine, which uses that address to manufacture a "link order", i.e. a jump back to the caller. This link order is normally stored at a fixed location in the subroutine, so if the subroutine were to call itself then the original link order would be overwritten and lost. However, it is easy enough to make a subroutine save its link orders in a stack, so that it can be called recursively, as the Rosetta Code task requires.

The program has a maximum of 9 discs, so as to simplify the printout of the disc numbers. Discs are numbered 1, 2, 3, ... in increasing order of size. The program could be speeded up by shortening the messages, which at present take up most of the runtime.

[Towers of Hanoi task for Rosetta Code.]
[EDSAC program, Initial Orders 2.]

T100K   [load program at location 100 (arbitrary)]
GK
[Number of discs, in the address field]
   P3F     [<--- edit here, value 1..9]
[Letters to represent the rods]
   LF      [left]
   CF      [centre]
   RF      [right]

[Main routine. Enter with acc = 0]
   T1F     [1F := 0]
   A5@     [initialize recursive subroutine]
G104@
A@      [number of discs]
T1F     [pass to subroutines]
A1@     [source rod]
T4F     [pass to subroutines]
A3@     [target rod]
T5F     [pass to subroutines]
   A13@    [call subroutine to write header ]
G18@
   A15@    [call recursive subroutine to write moves ]
G104@
ZF      [stop]

[Input:  1F = number of discs (in the address field)]
[4F = letter for starting rod]
[5F = letter for ending rod]
[Output: None. 1F, 4F, 5F must be preserved.]
   A3F     [plant return link as usual]
T35@
A1F     [number of discs]
L512F   [shift 11 left to make output char]
T39@    [plant in message]
A4F     [starting rod]
T53@    [plant in message]
A5F     [ending rod]
T58@    [plant in message]
A36@    [O order for first char of message]
E30@    [skip next order (code for 'O' is positive)]
   A37@    [restore acc after test below]
   U31@    [plant order to write next character]
   OF      [(planted) write next character]
A2F     [inc address in previous order]
S37@    [finished yet?]
G29@    [if not, loop back]
   ZF      [(planted) exit with acc = 0]
   O38@    [O order for start of message]
   O61@    [O order for exclusive end of message]
   #F
   PFK2048F!FDFIFSFCFSF!FFFRFOFMF!F
   PF!FTFOF!F
   PF@F&F


[Subroutine to write a move of one disc.]
[Input:  1F = disc number 1..9 (in the address field)]
[4F = letter for source rod]
[5F = letter for target rod]
[Output: None. 1F, 4F, 5F must be preserved.]
[Condensed to save space; very similar to previous subroutine.]
   A3FT78@A1FL512FT88@ A4FT96@A5FT101@A79@E73@
   A80@
   U74@
   OFA2FS80@G72@
   ZF      [(planted) exit with acc = 0]
   O81@
   O104@
   K2048FMFOFVFEF!F#F
   PFK2048F!FFFRFOFMF!F
   PF!FTFOF!F
   PF@F&F


[Recursive subroutine to move discs 1..n, where 1 <= n <= 9.]
[Call with n = 0 to initialize.]
[Input:  1F = n (in the address field)]
[4F = letter for source rod]
[5F = letter for target rod]
[Output: None. 1F, 4F, 5F must be preserved.]
   A3F     [plant link as usual ]
T167@
[The link will be saved in a stack if recursive calls are required.]
G115@   [jump if n > 0]
[Here if n = 0. Initialize; no recursive calls.]
A169@   [initialize push order to start of stack]
T122@
A1@     [find total of the codes for the rod letters]
A2@
A3@
T168@   [store for future use]
[Here with acc = -n in address field]
G120@   [jump if n > 1]
[Here if n = 1. Just write the move; no recursive calls.]
   A117@   [call write subroutine]
G61@
[Here if n > 1. Recursive calls are required.]
   TF      [clear acc]
   TF      [(planted) push link order onto stack]
A122@   [inc address in previous order]
A2F
T122@
[First recursive call. Modify parameters 1F and 5F; 4F stays the same]
S2F     [make n - 1]
T1F     [pass as parameter]
A168@   [get 3rd rod, neither source nor target]
S4F
S5F
T5F
   A133@   [recursive call]
G104@
[Returned, restore parameters]
A1F
A2F
T1F
A168@
S4F
S5F
T5F
[Write move of largest disc]
   A142@
G61@
[Second recursive call. Modify parameters 1F and 4F; 5F stays the same]
[Condensed to save space; very similar to first recursice call.]
A1FS2FT1FA168@S4FS5FT4F
   A151@G104@A1FA2FT1FA168@S4FS5FT4F
A122@   [dec address in push order]
S2F
U122@
A170@   [make A order with same address]
T165@   [plant in code]
   AF      [(planted) pop return link from stack]
T167@   [plant in code]
[Constants]
   PF      [(planted) sum of letters for rods]
   T171@   [T order for start of stack]
   MF      [add to T order to make A order, same address]
[Stack: placed at end of program, grows into free space.]

E4Z     [define entry point]
PF      [acc = 0 on entry]
[end]
Output:
3 DISCS FROM L TO R
MOVE 1 FROM L TO R
MOVE 2 FROM L TO C
MOVE 1 FROM R TO C
MOVE 3 FROM L TO R
MOVE 1 FROM C TO L
MOVE 2 FROM C TO R
MOVE 1 FROM L TO R


## Eiffel

class
APPLICATION

create
make

feature {NONE} -- Initialization

make
do
move (4, "A", "B", "C")
end

feature -- Towers of Hanoi

move (n: INTEGER; frm, to, via: STRING)
require
n > 0
do
if n = 1 then
print ("Move disk from pole " + frm + " to pole " + to + "%N")
else
move (n - 1, frm, via, to)
move (1, frm, to, via)
move (n - 1, via, to, frm)
end
end
end


## Ela

open monad io
:::IO

//Functional approach
hanoi 0 _ _ _ = []
hanoi n a b c = hanoi (n - 1) a c b ++ [(a,b)] ++ hanoi (n - 1) c b a

hanoiIO n = mapM_ f $hanoi n 1 2 3 where f (x,y) = putStrLn$ "Move " ++ show x ++ " to " ++ show y

hanoiM n = hanoiM' n 1 2 3 where
hanoiM' 0 _ _ _ = return ()
hanoiM' n a b c = do
hanoiM' (n - 1) a c b
putStrLn $"Move " ++ show a ++ " to " ++ show b hanoiM' (n - 1) c b a ## Elena ELENA 4.x : move = (n,from,to,via) { if (n == 1) { console.printLine("Move disk from pole ",from," to pole ",to) } else { move(n-1,from,via,to); move(1,from,to,via); move(n-1,via,to,from) } }; ## Elixir defmodule RC do def hanoi(n) when 0<n and n<10, do: hanoi(n, 1, 2, 3) defp hanoi(1, f, _, t), do: move(f, t) defp hanoi(n, f, u, t) do hanoi(n-1, f, t, u) move(f, t) hanoi(n-1, u, f, t) end defp move(f, t), do: IO.puts "Move disk from #{f} to #{t}" end RC.hanoi(3)  Output: Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Move disk from 1 to 3 Move disk from 2 to 1 Move disk from 2 to 3 Move disk from 1 to 3  ## Emacs Lisp Translation of: Common Lisp (defun move (n from to via) (if (= n 1) (message "Move from %S to %S" from to) (move (- n 1) from via to) (message "Move from %S to %S" from to) (move (- n 1) via to from)))  ## Erlang move(1, F, T, _V) -> io:format("Move from ~p to ~p~n", [F, T]); move(N, F, T, V) -> move(N-1, F, V, T), move(1 , F, T, V), move(N-1, V, T, F).  ## ERRE !----------------------------------------------------------- ! HANOI.R : solve tower of Hanoi puzzle using a recursive ! modified algorithm. !----------------------------------------------------------- PROGRAM HANOI !$INTEGER

!VAR I,J,MOSSE,NUMBER

PROCEDURE PRINTMOVE
LOCAL SOURCE$,DEST$
MOSSE=MOSSE+1
CASE I OF
1-> SOURCE$="Left" END -> 2-> SOURCE$="Center" END ->
3-> SOURCE$="Right" END -> END CASE CASE J OF 1-> DEST$="Left" END ->
2-> DEST$="Center" END -> 3-> DEST$="Right" END ->
END CASE
PRINT("I move a disk from ";SOURCE$;" to ";DEST$)
END PROCEDURE

PROCEDURE MOVE
IF NUMBER<>0 THEN
NUMBER=NUMBER-1
J=6-I-J
MOVE
J=6-I-J
PRINTMOVE
I=6-I-J
MOVE
I=6-I-J
NUMBER=NUMBER+1
END IF
END PROCEDURE

BEGIN
MAXNUM=12
MOSSE=0
PRINT(CHR$(12);TAB(25);"--- TOWERS OF HANOI ---") REPEAT PRINT("Number of disks ";) INPUT(NUMBER) UNTIL NUMBER>1 AND NUMBER<=MAXNUM PRINT PRINT("For ";NUMBER;"disks the total number of moves is";2^NUMBER-1) I=1 ! number of source pole J=3 ! number of destination pole MOVE END PROGRAM Output:  --- TOWER OF HANOI --- Number of disks ? 3 For 3 disks the total number of moves is 7 I move a disk from Left to Right I move a disk from Left to Center I move a disk from Right to Center I move a disk from Left to Right I move a disk from Center to Left I move a disk from Center to Right I move a disk from Left to Right  ## Excel ### LAMBDA With the names HANOI and SHOWHANOI bound to the following lambdas in the Excel worksheet Name Manager: SHOWHANOI =LAMBDA(n, FILTERP( LAMBDA(x, "" <> x) )( HANOI(n)("left")("right")("mid") ) ) HANOI =LAMBDA(n, LAMBDA(l, LAMBDA(r, LAMBDA(m, IF(0 = n, "", LET( next, n - 1, APPEND( APPEND( HANOI(next)(l)(m)(r) )( CONCAT(l, " -> ", r) ) )( HANOI(next)(m)(r)(l) ) ) ) ) ) ) )  And assuming that these generic lambdas are also bound to the following names in Name Manager: APPEND =LAMBDA(xs, LAMBDA(ys, LET( nx, ROWS(xs), rowIndexes, SEQUENCE(nx + ROWS(ys)), colIndexes, SEQUENCE( 1, MAX(COLUMNS(xs), COLUMNS(ys)) ), IF( rowIndexes <= nx, INDEX(xs, rowIndexes, colIndexes), INDEX(ys, rowIndexes - nx, colIndexes) ) ) ) ) FILTERP =LAMBDA(p, LAMBDA(xs, FILTER(xs, p(xs)) ) )  In the output below, the expression in B2 defines an array of strings which additionally populate the following cells. Output:  =SHOWHANOI(A2) fx A B 1 Disks Steps 2 3 left -> right 3 left -> mid 4 right -> mid 5 left -> right 6 mid -> left 7 mid -> right 8 left -> right ## Ezhil # (C) 2013 Ezhil Language Project # Tower of Hanoi – recursive solution நிரல்பாகம் ஹோனாய்(வட்டுகள், முதல்அச்சு, இறுதிஅச்சு,வட்டு) @(வட்டுகள் == 1 ) ஆனால் பதிப்பி “வட்டு ” + str(வட்டு) + “ஐ \t (” + str(முதல்அச்சு) + “ —> ” + str(இறுதிஅச்சு)+ “) அச்சிற்கு நகர்த்துக.” இல்லை @( ["இ", "அ", "ஆ"] இல் அச்சு ) ஒவ்வொன்றாக @( (முதல்அச்சு != அச்சு) && (இறுதிஅச்சு != அச்சு) ) ஆனால் நடு = அச்சு முடி முடி # solve problem for n-1 again between src and temp pegs ஹோனாய்(வட்டுகள்-1, முதல்அச்சு,நடு,வட்டுகள்-1) # move largest disk from src to destination ஹோனாய்(1, முதல்அச்சு, இறுதிஅச்சு,வட்டுகள்) # solve problem for n-1 again between different pegs ஹோனாய்(வட்டுகள்-1, நடு, இறுதிஅச்சு,வட்டுகள்-1) முடி முடி ஹோனாய்(4,”அ”,”ஆ”,0)  ## F# #light let rec hanoi num start finish = match num with | 0 -> [ ] | _ -> let temp = (6 - start - finish) (hanoi (num-1) start temp) @ [ start, finish ] @ (hanoi (num-1) temp finish) [<EntryPoint>] let main args = (hanoi 4 1 2) |> List.iter (fun pair -> match pair with | a, b -> printf "Move disc from %A to %A\n" a b) 0  ## Factor USING: formatting kernel locals math ; IN: rosettacode.hanoi : move ( from to -- ) "%d->%d\n" printf ; :: hanoi ( n from to other -- ) n 0 > [ n 1 - from other to hanoi from to move n 1 - other to from hanoi ] when ;  In the REPL: ( scratchpad ) 3 1 3 2 hanoi 1->3 1->2 3->2 1->3 2->1 2->3 1->3 ## FALSE ["Move disk from "$!\" to "$!\" "]p: { to from } [n;0>[n;1-n: @\ h;! @\ p;! \@ h;! \@ n;1+n:]?]h: { via to from } 4n:["right"]["middle"]["left"]h;!%%% ## Fermat Func Hanoi( n, f, t, v ) = if n = 0 then !''; else Hanoi(n - 1, f, v, t); !f;!' -> ';!t;!', '; Hanoi(n - 1, v, t, f) fi. Output: 1 -> 3, 1 -> 2, 3 -> 2, 1 -> 3, 2 -> 1, 2 -> 3, 1 -> 3, 1 -> 2, 3 -> 2, 3 -> 1, 2 -> 1, 3 -> 2, 1 -> 3, 1 -> 2, 3 -> 2, ## FOCAL 01.10 S N=4;S S=1;S V=2;S T=3 01.20 D 2 01.30 Q 02.02 S N(D)=N(D)-1;I (N(D)),2.2,2.04 02.04 S D=D+1 02.06 S N(D)=N(D-1);S S(D)=S(D-1) 02.08 S T(D)=V(D-1);S V(D)=T(D-1) 02.10 D 2 02.12 S D=D-1 02.14 D 3 02.16 S A=S(D);S S(D)=V(D);S V(D)=A 02.18 G 2.02 02.20 D 3 03.10 T %1,"MOVE DISK FROM POLE",S(D) 03.20 T " TO POLE",T(D),! Output: MOVE DISK FROM POLE= 1 TO POLE= 2 MOVE DISK FROM POLE= 1 TO POLE= 3 MOVE DISK FROM POLE= 2 TO POLE= 3 MOVE DISK FROM POLE= 1 TO POLE= 2 MOVE DISK FROM POLE= 3 TO POLE= 1 MOVE DISK FROM POLE= 3 TO POLE= 2 MOVE DISK FROM POLE= 1 TO POLE= 2 MOVE DISK FROM POLE= 1 TO POLE= 3 MOVE DISK FROM POLE= 2 TO POLE= 3 MOVE DISK FROM POLE= 2 TO POLE= 1 MOVE DISK FROM POLE= 3 TO POLE= 1 MOVE DISK FROM POLE= 2 TO POLE= 3 MOVE DISK FROM POLE= 1 TO POLE= 2 MOVE DISK FROM POLE= 1 TO POLE= 3 MOVE DISK FROM POLE= 2 TO POLE= 3 ## Forth With locals: CREATE peg1 ," left " CREATE peg2 ," middle " CREATE peg3 ," right " : .$   COUNT TYPE ;
: MOVE-DISK
LOCALS| via to from n |
n 1 =
IF   CR ." Move disk from " from .$." to " to .$
ELSE n 1- from via to RECURSE
1    from to via RECURSE
n 1- via to from RECURSE
THEN ;


Without locals, executable pegs:

: left   ." left" ;
: right  ." right" ;
: middle ." middle" ;

: move-disk ( v t f n -- v t f )
dup 0= if drop exit then
1-       >R
rot swap R@ ( t v f n-1 ) recurse
rot swap
2dup cr ." Move disk from " execute ."  to " execute
swap rot R> ( f t v n-1 ) recurse
swap rot ;
: hanoi ( n -- )
1 max >R ['] right ['] middle ['] left R> move-disk drop drop drop ;


## Fortran

Works with: Fortran version 90 and later
PROGRAM TOWER

CALL Move(4, 1, 2, 3)

CONTAINS

RECURSIVE SUBROUTINE Move(ndisks, from, to, via)
INTEGER, INTENT (IN) :: ndisks, from, to, via

IF (ndisks == 1) THEN
WRITE(*, "(A,I1,A,I1)") "Move disk from pole ", from, " to pole ", to
ELSE
CALL Move(ndisks-1, from, via, to)
CALL Move(1, from, to, via)
CALL Move(ndisks-1, via, to, from)
END IF
END SUBROUTINE Move

END PROGRAM TOWER

PROGRAM TOWER2

CALL Move(4, 1, 2, 3)

CONTAINS

RECURSIVE SUBROUTINE Move(ndisks, from, via, to)
INTEGER, INTENT (IN) :: ndisks, from, via, to

IF (ndisks > 1) THEN
CALL Move(ndisks-1, from, to, via)
WRITE(*, "(A,I1,A,I1,A,I1)") "Move disk ", ndisks, "  from pole ", from, " to pole ", to
Call Move(ndisks-1,via,from,to)
ELSE
WRITE(*, "(A,I1,A,I1,A,I1)") "Move disk ", ndisks, "  from pole ", from, " to pole ", to
END IF
END SUBROUTINE Move

END PROGRAM TOWER2


## FreeBASIC

' FB 1.05.0 Win64

Sub move(n As Integer, from As Integer, to_ As Integer, via As Integer)
If n > 0 Then
move(n - 1, from, via, to_)
Print "Move disk"; n; " from pole"; from; " to pole"; to_
move(n - 1, via, to_, from)
End If
End Sub

Print "Three disks" : Print
move 3, 1, 2, 3
Print
Print "Four disks" : Print
move 4, 1, 2, 3
Print "Press any key to quit"
Sleep

Output:
Three disks

Move disk 1 from pole 1 to pole 2
Move disk 2 from pole 1 to pole 3
Move disk 1 from pole 2 to pole 3
Move disk 3 from pole 1 to pole 2
Move disk 1 from pole 3 to pole 1
Move disk 2 from pole 3 to pole 2
Move disk 1 from pole 1 to pole 2

Four disks

Move disk 1 from pole 1 to pole 3
Move disk 2 from pole 1 to pole 2
Move disk 1 from pole 3 to pole 2
Move disk 3 from pole 1 to pole 3
Move disk 1 from pole 2 to pole 1
Move disk 2 from pole 2 to pole 3
Move disk 1 from pole 1 to pole 3
Move disk 4 from pole 1 to pole 2
Move disk 1 from pole 3 to pole 2
Move disk 2 from pole 3 to pole 1
Move disk 1 from pole 2 to pole 1
Move disk 3 from pole 3 to pole 2
Move disk 1 from pole 1 to pole 3
Move disk 2 from pole 1 to pole 2
Move disk 1 from pole 3 to pole 2


## Frink

/** Set up the recursive call for n disks */
hanoi[n] := hanoi[n, 1, 3, 2]

/** The recursive call. */
hanoi[n, source, target, aux] :=
{
if n > 0
{
hanoi[n-1, source, aux, target]
println["Move from $source to$target"]
hanoi[n-1, aux, target, source]
}
}

hanoi

## FutureBasic

window 1, @"Towers of Hanoi", ( 0, 0, 300, 300 )

void local fn Move( n as long, fromPeg as long, toPeg as long, viaPeg as long )
if n > 0
fn Move( n-1, fromPeg, viaPeg, toPeg )
print "Move disk from "; fromPeg; " to "; toPeg
fn Move( n-1, viaPeg, toPeg, fromPeg )
end if
end fn

fn Move( 4, 1, 2, 3 )
print
print "Towers of Hanoi puzzle solved."

HandleEvents

Output:

Move disk from  1 to  3
Move disk from  1 to  2
Move disk from  3 to  2
Move disk from  1 to  3
Move disk from  2 to  1
Move disk from  2 to  3
Move disk from  1 to  3
Move disk from  1 to  2
Move disk from  3 to  2
Move disk from  3 to  1
Move disk from  2 to  1
Move disk from  3 to  2
Move disk from  1 to  3
Move disk from  1 to  2
Move disk from  3 to  2

Towers of Hanoi puzzle solved.


## 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.

## GAP

Hanoi := function(n)
local move;
move := function(n, a, b, c) # from, through, to
if n = 1 then
Print(a, " -> ", c, "\n");
else
move(n - 1, a, c, b);
move(1, a, b, c);
move(n - 1, b, a, c);
fi;
end;
move(n, "A", "B", "C");
end;

Hanoi(1);
# A -> C

Hanoi(2);
# A -> B
# A -> C
# B -> C

Hanoi(3);
# A -> C
# A -> B
# C -> B
# A -> C
# B -> A
# B -> C
# A -> C


## Go

package main

import "fmt"

// a towers of hanoi solver just has one method, play
type solver interface {
play(int)
}

func main() {
var t solver    // declare variable of solver type
t = new(towers) // type towers must satisfy solver interface
t.play(4)
}

// towers is example of type satisfying solver interface
type towers struct {
// an empty struct.  some other solver might fill this with some
// data representation, maybe for algorithm validation, or maybe for
// visualization.
}

// play is sole method required to implement solver type
func (t *towers) play(n int) {
// drive recursive solution, per task description
t.moveN(n, 1, 2, 3)
}

// recursive algorithm
func (t *towers) moveN(n, from, to, via int) {
if n > 0 {
t.moveN(n-1, from, via, to)
t.move1(from, to)
t.moveN(n-1, via, to, from)
}
}

// example function prints actions to screen.
// enhance with validation or visualization as needed.
func (t *towers) move1(from, to int) {
fmt.Println("move disk from rod", from, "to rod", to)
}


In other words:

package main

import "fmt"

func main() {
move(3, "A", "B", "C")
}

func move(n uint64, a, b, c string) {
if n > 0 {
move(n-1, a, c, b)
fmt.Println("Move disk from " + a + " to " + c)
move(n-1, b, a, c)
}
}


## Groovy

Unlike most solutions here this solution manipulates more-or-less actual stacks of more-or-less actual rings.

def tail = { list, n ->  def m = list.size(); list.subList([m - n, 0].max(),m) }

final STACK = [A:[],B:[],C:[]].asImmutable()

def report = { it -> }
def check = { it -> }

def moveRing = { from, to ->  to << from.pop(); report(); check(to) }

def moveStack
moveStack = { from, to, using = STACK.values().find { !(it.is(from) || it.is(to)) } ->
if (!from) return
def n = from.size()
moveStack(tail(from, n-1), using, to)
moveRing(from, to)
moveStack(tail(using, n-1), to, from)
}


Test program:

enum Ring {
S('°'), M('o'), L('O'), XL('( )');
private sym
private Ring(sym) { this.sym=sym }
String toString() { sym }
}

report = { STACK.each { k, v ->  println "${k}:${v}" }; println() }
check = { to -> assert to == ([] + to).sort().reverse() }

Ring.values().reverseEach { STACK.A << it }
report()
check(STACK.A)
moveStack(STACK.A, STACK.C)

Output:
A: [( ), O, o, °]
B: []
C: []

A: [( ), O, o]
B: [°]
C: []

A: [( ), O]
B: [°]
C: [o]

A: [( ), O]
B: []
C: [o, °]

A: [( )]
B: [O]
C: [o, °]

A: [( ), °]
B: [O]
C: [o]

A: [( ), °]
B: [O, o]
C: []

A: [( )]
B: [O, o, °]
C: []

A: []
B: [O, o, °]
C: [( )]

A: []
B: [O, o]
C: [( ), °]

A: [o]
B: [O]
C: [( ), °]

A: [o, °]
B: [O]
C: [( )]

A: [o, °]
B: []
C: [( ), O]

A: [o]
B: [°]
C: [( ), O]

A: []
B: [°]
C: [( ), O, o]

A: []
B: []
C: [( ), O, o, °]

Most of the programs on this page use an imperative approach (i.e., print out movements as side effects during program execution). Haskell favors a purely functional approach, where you would for example return a (lazy) list of movements from a to b via c:

hanoi :: Integer -> a -> a -> a -> [(a, a)]
hanoi 0 _ _ _ = []
hanoi n a b c = hanoi (n-1) a c b ++ [(a,b)] ++ hanoi (n-1) c b a


You can also do the above with one tail-recursion call:

hanoi :: Integer -> a -> a -> a -> [(a, a)]

hanoi n a b c = hanoiToList n a b c []
where
hanoiToList 0 _ _ _ l = l
hanoiToList n a b c l = hanoiToList (n-1) a c b ((a, b) : hanoiToList (n-1) c b a l)


One can use this function to produce output, just like the other programs:

hanoiIO n = mapM_ f $hanoi n 1 2 3 where f (x,y) = putStrLn$ "Move " ++ show x ++ " to " ++ show y


or, instead, one can of course also program imperatively, using the IO monad directly:

hanoiM :: Integer -> IO ()
hanoiM n = hanoiM' n 1 2 3 where
hanoiM' 0 _ _ _ = return ()
hanoiM' n a b c = do
hanoiM' (n-1) a c b
putStrLn $"Move " ++ show a ++ " to " ++ show b hanoiM' (n-1) c b a  or, defining it as a monoid, and adding some output: -------------------------- HANOI ------------------------- hanoi :: Int -> String -> String -> String -> [(String, String)] hanoi 0 _ _ _ = mempty hanoi n l r m = hanoi (n - 1) l m r <> [(l, r)] <> hanoi (n - 1) m r l --------------------------- TEST ------------------------- main :: IO () main = putStrLn$ showHanoi 5

------------------------- DISPLAY ------------------------
showHanoi :: Int -> String
showHanoi n =
unlines $fmap ( \(from, to) -> concat [justifyRight 5 ' ' from, " -> ", to] ) (hanoi n "left" "right" "mid") justifyRight :: Int -> Char -> String -> String justifyRight n c = (drop . length) <*> (replicate n c <>)  Output:  left -> right left -> mid right -> mid left -> right mid -> left mid -> right left -> right left -> mid right -> mid right -> left mid -> left right -> mid left -> right left -> mid right -> mid left -> right mid -> left mid -> right left -> right mid -> left right -> mid right -> left mid -> left mid -> right left -> right left -> mid right -> mid left -> right mid -> left mid -> right left -> right ## HolyC Translation of: C U0 Move(U8 n, U8 from, U8 to, U8 via) { if (n > 0) { Move(n - 1, from, via, to); Print("Move disk from pole %d to pole %d\n", from, to); Move(n - 1, via, to, from); } } Move(4, 1, 2, 3); ## Icon and Unicon The following is based on a solution in the Unicon book. procedure main(arglist) hanoi(arglist) | stop("Usage: hanoi n\n\rWhere n is the number of disks to move.") end #procedure hanoi(n:integer, needle1:1, needle2:2) # unicon shorthand for icon code 1,2,3 below procedure hanoi(n, needle1, needle2) #: solve towers of hanoi by moving n disks from needle 1 to needle2 via other local other n := integer(0 < n) | runerr(n,101) # 1 ensure integer (this also ensures it's positive too) /needle1 := 1 # 2 default /needle2 := 2 # 3 default if n = 1 then write("Move disk from ", needle1, " to ", needle2) else { other := 6 - needle1 - needle2 # clever but somewhat un-iconish way to find other hanoi(n-1, needle1, other) write("Move disk from ", needle1, " to ", needle2) hanoi(n-1, other, needle2) } return end  ## Inform 7 Hanoi is a room. A post is a kind of supporter. A post is always fixed in place. The left post, the middle post, and the right post are posts in Hanoi. A disk is a kind of supporter. The red disk is a disk on the left post. The orange disk is a disk on the red disk. The yellow disk is a disk on the orange disk. The green disk is a disk on the yellow disk. Definition: a disk is topmost if nothing is on it. When play begins: move 4 disks from the left post to the right post via the middle post. To move (N - number) disk/disks from (FP - post) to (TP - post) via (VP - post): if N > 0: move N - 1 disks from FP to VP via TP; say "Moving a disk from [FP] to [TP]..."; let D be a random topmost disk enclosed by FP; if a topmost disk (called TD) is enclosed by TP, now D is on TD; otherwise now D is on TP; move N - 1 disks from VP to TP via FP.  ## Io hanoi := method(n, from, to, via, if (n == 1) then ( writeln("Move from ", from, " to ", to) ) else ( hanoi(n - 1, from, via, to ) hanoi(1 , from, to , via ) hanoi(n - 1, via , to , from) ) )  ## Ioke  = method(n, f, u, t, if(n < 2, "#{f} --> #{t}" println, H(n - 1, f, t, u) "#{f} --> #{t}" println H(n - 1, u, f, t) ) ) hanoi = method(n, H(n, 1, 2, 3) )  ## IS-BASIC 100 PROGRAM "Hanoi.bas" 110 CALL HANOI(4,1,3,2) 120 DEF HANOI(DISK,FRO,TO,WITH) 130 IF DISK>0 THEN 140 CALL HANOI(DISK-1,FRO,WITH,TO) 150 PRINT "Move disk";DISK;"from";FRO;"to";TO 160 CALL HANOI(DISK-1,WITH,TO,FRO) 170 END IF 180 END DEF ## J Solutions H =: i.@,&2  (({&0 2 1,0 2,{&1 0 2)@$:@<:) @. *    NB. tacit using anonymous recursion

Example use:
   H 3
0 2
0 1
2 1
0 2
1 2
1 0
2 0


The result is a 2-column table; a row i,j is interpreted as: move a disk (the top disk) from peg i to peg j . Or, using explicit rather than implicit code:

H1=: monad define                                   NB. explicit equivalent of H
if. y do.
({&0 2 1 , 0 2 , {&1 0 2) H1 y-1
else.
i.0 2
end.
)


The usage here is the same:

   H1 2
0 1
0 2
1 2
Alternative solution

If a textual display is desired, similar to some of the other solutions here (counting from 1 instead of 0, tracking which disk is on the top of the stack, and of course formatting the result for a human reader instead of providing a numeric result):

hanoi=: monad define
moves=. H y
disks=.  $~ ((],[,])$:@<:) @.* y
('move disk ';' from peg ';' to peg ');@,."1 ":&.>disks,.1+moves
)

Demonstration:
   hanoi 3
move disk 1 from peg 1 to peg 3
move disk 2 from peg 1 to peg 2
move disk 1 from peg 3 to peg 2
move disk 3 from peg 1 to peg 3
move disk 1 from peg 2 to peg 1
move disk 2 from peg 2 to peg 3
move disk 1 from peg 1 to peg 3


## Java

public void move(int n, int from, int to, int via) {
if (n == 1) {
System.out.println("Move disk from pole " + from + " to pole " + to);
} else {
move(n - 1, from, via, to);
move(1, from, to, via);
move(n - 1, via, to, from);
}
}


Where n is the number of disks to move and from, to, and via are the poles.

Example use:
move(3, 1, 2, 3);

Output:
Move disk from pole 1 to pole 2
Move disk from pole 1 to pole 3
Move disk from pole 2 to pole 3
Move disk from pole 1 to pole 2
Move disk from pole 3 to pole 1
Move disk from pole 3 to pole 2
Move disk from pole 1 to pole 2


## JavaScript

### ES5

function move(n, a, b, c) {
if (n > 0) {
move(n-1, a, c, b);
console.log("Move disk from " + a + " to " + c);
move(n-1, b, a, c);
}
}
move(4, "A", "B", "C");


Or, as a functional expression, rather than a statement with side effects:

(function () {

// hanoi :: Int -> String -> String -> String -> [[String, String]]
function hanoi(n, a, b, c) {
return n ? hanoi(n - 1, a, c, b)
.concat([
[a, b]
])
.concat(hanoi(n - 1, c, b, a)) : [];
}

return hanoi(3, 'left', 'right', 'mid')
.map(function (d) {
return d + ' -> ' + d;
});
})();

Output:
["left -> right", "left -> mid",
"right -> mid", "left -> right",
"mid -> left", "mid -> right",
"left -> right"]


### ES6

(() => {
"use strict";

// ----------------- TOWERS OF HANOI -----------------

// hanoi :: Int -> String -> String ->
// String -> [[String, String]]
const hanoi = n =>
(a, b, c) => {
const go = hanoi(n - 1);

return Boolean(n) ? [
...go(a, c, b),
...[
[a, b]
],
...go(c, b, a)
] : [];
};

// ---------------------- TEST -----------------------
return hanoi(3)("left", "right", "mid")
.map(d => ${d} ->${d})
.join("\n");
})();

Output:
left -> right
left -> mid
right -> mid
left -> right
mid -> left
mid -> right
left -> right

## Joy

DEFINE hanoi == [[rolldown] infra] dip
[[[null] [pop pop] ]
[[dup2 [[rotate] infra] dip pred]
[[dup rest put] dip
[[swap] infra] dip pred]
[]]]
condnestrec.

Using it (5 is the number of disks.)

[source destination temp] 5 hanoi.

## jq

Works with: jq version 1.4

The algorithm used here is used elsewhere on this page but it is worthwhile pointing out that it can also be read as a proof that:

(a) move(n;"A";"B";"C") will logically succeed for n>=0; and

(b) move(n;"A";"B";"C") will generate the sequence of moves, assuming sufficient computing resources.

The proof of (a) is by induction:

• As explained in the comments, the algorithm establishes that move(n;x;y;z) is possible for all n>=0 and distinct x,y,z if move(n-1;x;y;z)) is possible;
• Since move(0;x;y;z) evidently succeeds, (a) is established by induction.

The truth of (b) follows from the fact that the algorithm emits an instruction of what to do when moving a single disk.

# n is the number of disks to move from From to To
def move(n; From; To; Via):
if n > 0 then
# move all but the largest at From to Via (according to the rules):
move(n-1; From; Via; To),
# ... so the largest disk at From is now free to move to its final destination:
"Move disk from \(From) to \(To)",
# Move the remaining disks at Via to To:
move(n-1; Via; To; From)
else empty
end;

Example:

move(5; "A"; "B"; "C")


## Jsish

From Javascript ES5 entry.

/* Towers of Hanoi, in Jsish */

function move(n, a, b, c) {
if (n > 0) {
move(n-1, a, c, b);
puts("Move disk from " + a + " to " + c);
move(n-1, b, a, c);
}
}

if (Interp.conf('unitTest')) move(4, "A", "B", "C");

/*
=!EXPECTSTART!=
Move disk from A to B
Move disk from A to C
Move disk from B to C
Move disk from A to B
Move disk from C to A
Move disk from C to B
Move disk from A to B
Move disk from A to C
Move disk from B to C
Move disk from B to A
Move disk from C to A
Move disk from B to C
Move disk from A to B
Move disk from A to C
Move disk from B to C
=!EXPECTEND!=
*/

Output:
prompt$jsish -u towersOfHanoi.jsi [PASS] towersOfHanoi.jsi ## Julia Translation of: R function solve(n::Integer, from::Integer, to::Integer, via::Integer) if n == 1 println("Move disk from$from to $to") else solve(n - 1, from, via, to) solve(1, from, to, via) solve(n - 1, via, to, from) end end solve(4, 1, 2, 3)  Output: Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Move disk from 1 to 3 Move disk from 2 to 1 Move disk from 2 to 3 Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Move disk from 3 to 1 Move disk from 2 to 1 Move disk from 3 to 2 Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2  ## K  h:{[n;a;b;c]if[n>0;_f[n-1;a;c;b];0:,//$($n,":",$a,"->",$b,"\n");_f[n-1;c;b;a]]} h[4;1;2;3] 1:1->3 2:1->2 1:3->2 3:1->3 1:2->1 2:2->3 1:1->3 4:1->2 1:3->2 2:3->1 1:2->1 3:3->2 1:1->3 2:1->2 1:3->2 The disk to move in the i'th step is the same as the position of the leftmost 1 in the binary representation of 1..2^n.  s:();{[n;a;b;c]if[n>0;_f[n-1;a;c;b];s,:n;_f[n-1;c;b;a]]}[4;1;2;3];s 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 1_{*1+&|x}'a:(2_vs!_2^4) 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 ## Klingphix Translation of: MiniScript include ..\Utilitys.tlhy :moveDisc %B !B %C !C %A !A %n !n { n A C B }$n [
$n 1 -$A $B$C moveDisc
( "Move disc " $n " from pole "$A " to pole " $C ) lprint nl$n 1 - $B$C $A moveDisc ] if ; { Move disc 3 from pole 1 to pole 3, with pole 2 as spare } 3 1 3 2 moveDisc " " input Output: Move disc 1 from pole 1 to pole 3 Move disc 2 from pole 1 to pole 2 Move disc 1 from pole 3 to pole 2 Move disc 3 from pole 1 to pole 3 Move disc 1 from pole 2 to pole 1 Move disc 2 from pole 2 to pole 3 Move disc 1 from pole 1 to pole 3 ## Kotlin // version 1.1.0 class Hanoi(disks: Int) { private var moves = 0 init { println("Towers of Hanoi with$disks disks:\n")
move(disks, 'L', 'C', 'R')
println("\nCompleted in $moves moves\n") } private fun move(n: Int, from: Char, to: Char, via: Char) { if (n > 0) { move(n - 1, from, via, to) moves++ println("Move disk$n from $from to$to")
move(n - 1, via, to, from)
}
}
}

fun main(args: Array<String>) {
Hanoi(3)
Hanoi(4)
}

Output:
Towers of Hanoi with 3 disks:

Move disk 1 from L to C
Move disk 2 from L to R
Move disk 1 from C to R
Move disk 3 from L to C
Move disk 1 from R to L
Move disk 2 from R to C
Move disk 1 from L to C

Completed in 7 moves

Towers of Hanoi with 4 disks:

Move disk 1 from L to R
Move disk 2 from L to C
Move disk 1 from R to C
Move disk 3 from L to R
Move disk 1 from C to L
Move disk 2 from C to R
Move disk 1 from L to R
Move disk 4 from L to C
Move disk 1 from R to C
Move disk 2 from R to L
Move disk 1 from C to L
Move disk 3 from R to C
Move disk 1 from L to R
Move disk 2 from L to C
Move disk 1 from R to C

Completed in 15 moves


## lambdatalk

(Following NewLisp, PicoLisp, Racket, Scheme)

{def move
{lambda {:n :from :to :via}
{if {<= :n 0}
then >
else {move {- :n 1} :from :via :to}
move disk :n from :from to :to {br}
{move {- :n 1} :via :to :from} }}}
-> move
{move 4 A B C}
> move disk 1 from A to C
> move disk 2 from A to B
> move disk 1 from C to B
> move disk 3 from A to C
> move disk 1 from B to A
> move disk 2 from B to C
> move disk 1 from A to C
> move disk 4 from A to B
> move disk 1 from C to B
> move disk 2 from C to A
> move disk 1 from B to A
> move disk 3 from C to B
> move disk 1 from A to C
> move disk 2 from A to B
> move disk 1 from C to B


## Lasso

#!/usr/bin/lasso9

define towermove(
disks::integer,
a,b,c
) => {
if(#disks > 0) => {
towermove(#disks - 1, #a, #c, #b )
stdoutnl("Move disk from " + #a + " to " + #c)
towermove(#disks - 1, #b, #a, #c )
}
}

towermove((integer($argv -> second || 3)), "A", "B", "C")  Called from command line: ./towers  Output: Move disk from A to C Move disk from A to B Move disk from C to B Move disk from A to C Move disk from B to A Move disk from B to C Move disk from A to C Called from command line: ./towers 4  Output: Move disk from A to B Move disk from A to C Move disk from B to C Move disk from A to B Move disk from C to A Move disk from C to B Move disk from A to B Move disk from A to C Move disk from B to C Move disk from B to A Move disk from C to A Move disk from B to C Move disk from A to B Move disk from A to C Move disk from B to C ## Liberty BASIC This looks much better with a GUI interface.  source$ ="A"
via$="B" target$ ="C"

call hanoi 4, source$, target$, via$' ie call procedure to move legally 4 disks from peg A to peg C via peg B wait sub hanoi numDisks, source$, target$, via$
if numDisks =0 then
exit sub
else
call hanoi numDisks -1, source$, via$, target$print " Move disk "; numDisks; " from peg "; source$; " to peg "; target$call hanoi numDisks -1, via$, target$, source$
end if
end sub

end

## Lingo

on hanoi (n, a, b, c)
if n > 0 then
hanoi(n-1, a, c, b)
put "Move disk from" && a && "to" && c
hanoi(n-1, b, a, c)
end if
end
hanoi(3, "A", "B", "C")
-- "Move disk from A to C"
-- "Move disk from A to B"
-- "Move disk from C to B"
-- "Move disk from A to C"
-- "Move disk from B to A"
-- "Move disk from B to C"
-- "Move disk from A to C"

## Logo

to move :n :from :to :via
if :n = 0 [stop]
move :n-1 :from :via :to
(print [Move disk from] :from [to] :to)
move :n-1 :via :to :from
end
move 4 "left "middle "right

## Logtalk

:- object(hanoi).

:- public(run/1).
:- mode(run(+integer), one).
:- info(run/1, [
comment is 'Solves the towers of Hanoi problem for the specified number of disks.',
argnames is ['Disks']]).

run(Disks) :-
move(Disks, left, middle, right).

move(1, Left, _, Right):-
!,
report(Left, Right).
move(Disks, Left, Aux, Right):-
Disks2 is Disks - 1,
move(Disks2, Left, Right, Aux),
report(Left, Right),
move(Disks2, Aux, Left, Right).

report(Pole1, Pole2):-
write('Move a disk from '),
writeq(Pole1),
write(' to '),
writeq(Pole2),
write('.'),
nl.

:- end_object.


## LOLCODE

HAI 1.2

HOW IZ I HANOI YR N AN YR SRC AN YR DST AN YR VIA
BTW VISIBLE SMOOSH "HANOI N=" N " SRC=" SRC " DST=" DST " VIA=" VIA MKAY
BOTH SAEM N AN 0, O RLY?
YA RLY
BTW VISIBLE "Done."
GTFO
NO WAI
I HAS A LOWER ITZ DIFF OF N AN 1
I IZ HANOI YR LOWER AN YR SRC AN YR VIA AN YR DST MKAY
VISIBLE SMOOSH "Move disc " N " from " SRC " to " DST MKAY
I IZ HANOI YR LOWER AN YR VIA AN YR DST AN YR SRC MKAY
OIC
IF U SAY SO

I IZ HANOI YR 4 AN YR 1 AN YR 2 AN YR 3  MKAY

KTHXBYE

## Lua

function move(n, src, dst, via)
if n > 0 then
move(n - 1, src, via, dst)
print(src, 'to', dst)
move(n - 1, via, dst, src)
end
end

move(4, 1, 2, 3)

function move(n, src, via, dst)
if n > 0 then
move(n - 1, src, dst, via)
print('Disk ',n,' from ' ,src, 'to', dst)
move(n - 1, via, src, dst)

end
end

move(4, 1, 2, 3)


### Hanoi Iterative

#!/usr/bin/env luajit
local function printf(fmt, ...) io.write(string.format(fmt, ...)) end
local runs=0
local function move(tower, from, to)
if #tower[from]==0
or (#tower[to]>0
and tower[from][#tower[from]]>tower[to][#tower[to]]) then
to,from=from,to
end
if #tower[from]>0 then
tower[to][#tower[to]+1]=tower[from][#tower[from]]
tower[from][#tower[from]]=nil

io.write(tower[to][#tower[to]],":",from, "→", to, " ")
end
end

local function hanoi(n)
local src,dst,via={},{},{}
local tower={src,dst,via}
for i=1,n do src[i]=n-i+1 end
local one,nxt,lst
if n%2==1 then -- odd
one,nxt,lst=1,2,3
else
one,nxt,lst=1,3,2
end
--repeat
::loop::
move(tower, one, nxt)
if #dst==n then return end
move(tower, one, lst)
one,nxt,lst=nxt,lst,one
goto loop
--until false
end

local num=arg and tonumber(arg) or 4

hanoi(num)

Output:
> ./hanoi_iter.lua 5
1:1→2 2:1→3 1:2→3 3:1→2 1:3→1 2:3→2 1:1→2 4:1→3 1:2→3 2:2→1 1:3→1 3:2→3 1:1→2 2:1→3 1:2→3 5:1→2 1:3→1 2:3→2 1:1→2 3:3→1 1:2→3 2:2→1 1:3→1 4:3→2 1:1→2 2:1→3 1:2→3 3:1→2 1:3→1 2:3→2 1:1→2


### Hanoi Bitwise Fast

#!/usr/bin/env luajit
-- binary solution
local bit=require"bit"
local band,bor=bit.band,bit.bor
local function hanoi(n)
local even=(n-1)%2
for m=1,2^n-1 do
io.write(m,":",band(m,m-1)%3+1, "→", (bor(m,m-1)+1)%3+1, " ")
end
end

local num=arg and tonumber(arg) or 4

hanoi(num)

Output:
> ./hanoi_bit.lua 4
1:1→3 2:1→2 3:3→2 4:1→3 5:2→1 6:2→3 7:1→3 8:1→2 9:3→2 10:3→1 11:2→1 12:3→2 13:1→3 14:1→2 15:3→2
> time ./hanoi_bit.lua 30 >/dev/null  ; on AMD FX-8350 @ 4 GHz
./hanoi_bit.lua 30 > /dev/null  297,40s user 1,39s system 99% cpu 4:59,01 total


## M2000 Interpreter

Translation of: FreeBasic
Module Hanoi {
Rem HANOI TOWERS
Print "Three disks" : Print
move(3, 1, 2, 3)
Print
Print "Four disks" : Print
move(4, 1, 2, 3)

Sub move(n, from, to, via)
If n <=0 Then Exit Sub
move(n - 1, from, via, to)
Print "Move disk"; n; " from pole"; from; " to pole"; to
move(n - 1, via, to, from)
End Sub
}
Hanoi
Output:

same as in FreeBasic

            NORMAL MODE IS INTEGER
DIMENSION LIST(100)
SET LIST TO LIST

VECTOR VALUES MOVFMT =
0  $20HMOVE DISK FROM POLE ,I1,S1,8HTO POLE ,I1*$

INTERNAL FUNCTION(DUMMY)
ENTRY TO MOVE.
LOOP        NUM = NUM - 1
WHENEVER NUM.E.0
PRINT FORMAT MOVFMT,FROM,DEST
OTHERWISE
SAVE RETURN
SAVE DATA NUM,FROM,VIA,DEST
TEMP=DEST
DEST=VIA
VIA=TEMP
MOVE.(0)
RESTORE DATA NUM,FROM,VIA,DEST
RESTORE RETURN
PRINT FORMAT MOVFMT,FROM,DEST
TEMP=FROM
FROM=VIA
VIA=TEMP
TRANSFER TO LOOP
END OF CONDITIONAL
FUNCTION RETURN
END OF FUNCTION

NUM  = 4
FROM = 1
VIA  = 2
DEST = 3
MOVE.(0)

END OF PROGRAM
Output:
MOVE DISK FROM POLE 1 TO POLE 2
MOVE DISK FROM POLE 1 TO POLE 3
MOVE DISK FROM POLE 2 TO POLE 3
MOVE DISK FROM POLE 1 TO POLE 2
MOVE DISK FROM POLE 3 TO POLE 1
MOVE DISK FROM POLE 3 TO POLE 2
MOVE DISK FROM POLE 1 TO POLE 2
MOVE DISK FROM POLE 1 TO POLE 3
MOVE DISK FROM POLE 2 TO POLE 3
MOVE DISK FROM POLE 2 TO POLE 1
MOVE DISK FROM POLE 3 TO POLE 1
MOVE DISK FROM POLE 2 TO POLE 3
MOVE DISK FROM POLE 1 TO POLE 2
MOVE DISK FROM POLE 1 TO POLE 3
MOVE DISK FROM POLE 2 TO POLE 3

## Maple

Hanoi := proc(n::posint,a,b,c)
if n = 1 then
printf("Move disk from tower %a to tower %a.\n",a,c);
else
Hanoi(n-1,a,c,b);
Hanoi(1,a,b,c);
Hanoi(n-1,b,a,c);
fi;
end:

printf("Moving 2 disks from tower A to tower C using tower B.\n");
Hanoi(2,A,B,C);
Output:

Moving 2 disks from tower A to tower C using tower B.

Move disk from tower A to tower B.

Move disk from tower A to tower C.

Move disk from tower B to tower C.

## Mathematica/Wolfram Language

Hanoi[0, from_, to_, via_] := Null
Hanoi[n_Integer, from_, to_, via_] := (Hanoi[n-1, from, via, to];Print["Move disk from pole ", from, " to ", to, "."];Hanoi[n-1, via, to, from])


## MATLAB

This is a direct translation from the Python example given in the Wikipedia entry for the Tower of Hanoi puzzle.

function towerOfHanoi(n,A,C,B)
if (n~=0)
towerOfHanoi(n-1,A,B,C);
disp(sprintf('Move plate %d from tower %d to tower %d',[n A C]));
towerOfHanoi(n-1,B,C,A);
end
end

Sample output:
towerOfHanoi(3,1,3,2)
Move plate 1 from tower 1 to tower 3
Move plate 2 from tower 1 to tower 2
Move plate 1 from tower 3 to tower 2
Move plate 3 from tower 1 to tower 3
Move plate 1 from tower 2 to tower 1
Move plate 2 from tower 2 to tower 3
Move plate 1 from tower 1 to tower 3

## MiniScript

moveDisc = function(n, A, C, B)
if n == 0 then return
moveDisc n-1, A, B, C
print "Move disc " + n + " from pole " + A + " to pole " + C
moveDisc n-1, B, C, A
end function

// Move disc 3 from pole 1 to pole 3, with pole 2 as spare
moveDisc 3, 1, 3, 2

Output:
Move disc 1 from pole 1 to pole 3
Move disc 2 from pole 1 to pole 2
Move disc 1 from pole 3 to pole 2
Move disc 3 from pole 1 to pole 3
Move disc 1 from pole 2 to pole 1
Move disc 2 from pole 2 to pole 3
Move disc 1 from pole 1 to pole 3

## MIPS Assembly

# Towers of Hanoi
# MIPS assembly implementation (tested with MARS)
# Source: https://stackoverflow.com/questions/50382420/hanoi-towers-recursive-solution-using-mips/50383530#50383530

.data
prompt: .asciiz "Enter a number: "
part1: .asciiz "\nMove disk "
part2: .asciiz " from rod "
part3: .asciiz " to rod "

.text
.globl main
main:
li $v0, 4 # print string la$a0,  prompt
syscall
li $v0, 5 # read integer syscall # parameters for the routine add$a0, $v0,$zero # move to $a0 li$a1, 'A'
li $a2, 'B' li$a3, 'C'

jal hanoi           # call hanoi routine

li $v0, 10 # exit syscall hanoi: #save in stack addi$sp, $sp, -20 sw$ra, 0($sp) sw$s0, 4($sp) sw$s1, 8($sp) sw$s2, 12($sp) sw$s3, 16($sp) add$s0, $a0,$zero
add $s1,$a1, $zero add$s2, $a2,$zero
add $s3,$a3, $zero addi$t1, $zero, 1 beq$s0, $t1, output recur1: addi$a0, $s0, -1 add$a1, $s1,$zero
add $a2,$s3, $zero add$a3, $s2,$zero
jal hanoi

j output

recur2:

addi $a0,$s0, -1
add $a1,$s3, $zero add$a2, $s2,$zero
add $a3,$s1, $zero jal hanoi exithanoi: lw$ra, 0($sp) # restore registers from stack lw$s0, 4($sp) lw$s1, 8($sp) lw$s2, 12($sp) lw$s3, 16($sp) addi$sp, $sp, 20 # restore stack pointer jr$ra

output:

li $v0, 4 # print string la$a0,  part1
syscall
li $v0, 1 # print integer add$a0, $s0,$zero
syscall
li $v0, 4 # print string la$a0,  part2
syscall
li $v0, 11 # print character add$a0, $s1,$zero
syscall
li $v0, 4 # print string la$a0,  part3
syscall
li $v0, 11 # print character add$a0, $s2,$zero
syscall

beq $s0,$t1, exithanoi
j recur2

## МК-61/52

^	2	x^y	П0	<->	2	/	{x}	x#0	16
3	П3	2	П2	БП	20	3	П2	2	П3
1	П1	ПП	25	КППB	ПП	28	КППA	ПП	31
КППB	ПП	34	КППA	ИП1	ИП3	КППC	ИП1	ИП2	КППC
ИП3	ИП2	КППC	ИП1	ИП3	КППC	ИП2	ИП1	КППC	ИП2
ИП3	КППC	ИП1	ИП3	КППC	В/О	ИП1	ИП2	БП	62
ИП2	ИП1	КППC	ИП1	ИП2	ИП3	П1	->	П3	->
П2	В/О	1	0	/	+	С/П	КИП0	ИП0	x=0
89	3	3	1	ИНВ	^	ВП	2	С/П	В/О


Instruction: РA = 56; РB = 60; РC = 72; N В/О С/П, where 2 <= N <= 7.

## Modula-2

MODULE Towers;
FROM FormatString IMPORT FormatString;

PROCEDURE Move(n,from,to,via : INTEGER);
VAR buf : ARRAY[0..63] OF CHAR;
BEGIN
IF n>0 THEN
Move(n-1, from, via, to);
FormatString("Move disk %i from pole %i to pole %i\n", buf, n, from, to);
WriteString(buf);
Move(n-1, via, to, from)
END
END Move;

BEGIN
Move(3, 1, 3, 2);

END Towers.


## Modula-3

MODULE Hanoi EXPORTS Main;

FROM IO IMPORT Put;
FROM Fmt IMPORT Int;

PROCEDURE doHanoi(n, from, to, using: INTEGER) =
BEGIN
IF n > 0 THEN
doHanoi(n - 1, from, using, to);
Put("move " & Int(from) & " --> " & Int(to) & "\n");
doHanoi(n - 1, using, to, from);
END;
END doHanoi;

BEGIN
doHanoi(4, 1, 2, 3);
END Hanoi.


## Monte

def move(n, fromPeg, toPeg, viaPeg):
if (n > 0):
move(n.previous(), fromPeg, viaPeg, toPeg)

## Oz

declare
proc {TowersOfHanoi N From To Via}
if N > 0 then
{TowersOfHanoi N-1 From Via To}
{System.showInfo "Move from "#From#" to "#To}
{TowersOfHanoi N-1 Via To From}
end
end
in
{TowersOfHanoi 4 left middle right}

## PARI/GP

Translation of: Python
\\ Towers of Hanoi
\\ 8/19/2016 aev
\\ Where: n - number of disks, sp - start pole, ep - end pole.
HanoiTowers(n,sp,ep)={
if(n!=0,
HanoiTowers(n-1,sp,6-sp-ep);
print("Move disk ", n, " from pole ", sp," to pole ", ep);
HanoiTowers(n-1,6-sp-ep,ep);
);
}
\\ Testing n=3:
HanoiTowers(3,1,3);
Output:
> HanoiTower(3,1,3);
Move disk 1 from pole 1 to pole 3
Move disk 2 from pole 1 to pole 2
Move disk 1 from pole 3 to pole 2
Move disk 3 from pole 1 to pole 3
Move disk 1 from pole 2 to pole 1
Move disk 2 from pole 2 to pole 3
Move disk 1 from pole 1 to pole 3


## Pascal

Works with: Free Pascal version 2.0.4

I think it is standard pascal, except for the constant array "strPole". I am not sure if constant arrays are part of the standard. However, as far as I know, they are a "de facto" standard in every compiler.

program Hanoi;
type
TPole = (tpLeft, tpCenter, tpRight);
const
strPole:array[TPole] of string=('left','center','right');

procedure MoveStack (const Ndisks : integer; const Origin,Destination,Auxiliary:TPole);
begin
if Ndisks >0 then begin
MoveStack(Ndisks - 1, Origin,Auxiliary, Destination );
Writeln('Move disk ',Ndisks ,' from ',strPole[Origin],' to ',strPole[Destination]);
MoveStack(Ndisks - 1, Auxiliary, Destination, origin);
end;
end;

begin
MoveStack(4,tpLeft,tpCenter,tpRight);
end.


A little longer, but clearer for my taste

program Hanoi;
type
TPole = (tpLeft, tpCenter, tpRight);
const
strPole:array[TPole] of string=('left','center','right');

procedure MoveOneDisk(const DiskNum:integer; const Origin,Destination:TPole);
begin
Writeln('Move disk ',DiskNum,' from ',strPole[Origin],' to ',strPole[Destination]);
end;

procedure MoveStack (const Ndisks : integer; const Origin,Destination,Auxiliary:TPole);
begin
if Ndisks =1 then
MoveOneDisk(1,origin,Destination)
else begin
MoveStack(Ndisks - 1, Origin,Auxiliary, Destination );
MoveOneDisk(Ndisks,origin,Destination);
MoveStack(Ndisks - 1, Auxiliary, Destination, origin);
end;
end;

begin
MoveStack(4,tpLeft,tpCenter,tpRight);
end.


## Perl

sub hanoi {
my ($n,$from, $to,$via) = (@_, 1, 2, 3);

if ($n == 1) { print "Move disk from pole$from to pole $to.\n"; } else { hanoi($n - 1, $from,$via, $to); hanoi(1,$from, $to,$via);
hanoi($n - 1,$via, $to,$from);
};
};


## Phix

constant poles = {"left","middle","right"}
enum               left,  middle,  right

sequence disks
integer moves

procedure showpegs(integer src, integer dest)
string desc = sprintf("%s to %s:",{poles[src],poles[dest]})
disks[dest] &= disks[src][$] disks[src] = disks[src][1..$-1]
for i=1 to length(disks) do
desc = ""
end for
printf(1,"\n")
moves += 1
end procedure

procedure hanoir(integer n, src=left, dest=right, via=middle)
if n>0 then
hanoir(n-1, src, via, dest)
showpegs(src,dest)
hanoir(n-1, via, dest, src)
end if
end procedure

procedure hanoi(integer n)
disks = {reverse(tagset(n)),{},{}}
moves = 0
hanoir(n)
printf(1,"completed in %d moves\n",{moves})
end procedure

hanoi(3) -- (output of 4,5,6 also shown)

Output:
left to right:   | 3 2
|
| 1

left to middle:  | 3
| 2
| 1

right to middle: | 3
| 2 1
|

left to right:   |
| 2 1
| 3

middle to left:  | 1
| 2
| 3

middle to right: | 1
|
| 3 2

left to right:   |
|
| 3 2 1

completed in 7 moves

left to middle:  | 4 3 2
| 1
|

left to right:   | 4 3
| 1
| 2

middle to right: | 4 3
|
| 2 1

...

left to middle:  | 2
| 1
| 4 3

left to right:   |
| 1
| 4 3 2

middle to right: |
|
| 4 3 2 1

completed in 15 moves

left to right:   | 5 4 3 2
|
| 1

left to middle:  | 5 4 3
| 2
| 1

right to middle: | 5 4 3
| 2 1
|

...

middle to left:  | 1
| 2
| 5 4 3

middle to right: | 1
|
| 5 4 3 2

left to right:   |
|
| 5 4 3 2 1

completed in 31 moves

left to middle:  | 6 5 4 3 2
| 1
|

left to right:   | 6 5 4 3
| 1
| 2

middle to right: | 6 5 4 3
|
| 2 1

...

left to middle:  | 2
| 1
| 6 5 4 3

left to right:   |
| 1
| 6 5 4 3 2

middle to right: |
|
| 6 5 4 3 2 1

completed in 63 moves


## PHL

Translation of: C
module hanoi;

extern printf;

@Void move(@Integer n, @Integer from, @Integer to, @Integer via) [
if (n > 0) {
move(n - 1, from, via, to);
printf("Move disk from pole %d to pole %d\n", from, to);
move(n - 1, via, to, from);
}
]

@Integer main [
move(4, 1,2,3);
return 0;
]

## PHP

Translation of: Java
function move($n,$from,$to,$via) {
if ($n === 1) { print("Move disk from pole$from to pole $to"); } else { move($n-1,$from,$via,$to); move(1,$from,$to,$via);
move($n-1,$via,$to,$from);
}
}


## Picat

main =>
hanoi(3, left, center, right).

hanoi(0, _From, _To, _Via) => true.
hanoi(N, From, To, Via) =>
hanoi(N - 1, From, Via, To),
printf("Move disk %w from pole %w to pole %w\n", N, From, To),
hanoi(N - 1, Via, To, From).
Output:
Move disk 1 from pole left to pole center
Move disk 2 from pole left to pole right
Move disk 1 from pole center to pole right
Move disk 3 from pole left to pole center
Move disk 1 from pole right to pole left
Move disk 2 from pole right to pole center
Move disk 1 from pole left to pole center
count=7, theoretical=7

### Fast counting

main =>
hanoi(64).

hanoi(N) =>
printf("N=%d\n", N),
Count = move(N, left, center, right) ,
printf("count=%w, theoretical=%w\n", Count, 2**N-1).

table
move(0, _From, _To, _Via) = 0.
move(N, From, To, Via) = Count =>
Count1 = move(N - 1, From, Via, To),
Count2 = move(N - 1, Via, To, From),
Count = Count1+Count2+1.
Output:
N=64
count=18446744073709551615, theoretical=18446744073709551615

## PicoLisp

(de move (N A B C)  # Use: (move 3 'left 'center 'right)
(unless (=0 N)
(move (dec N) A C B)
(println 'Move 'disk 'from A 'to B)
(move (dec N) C B A) ) )

## PL/I

Translation of: Fortran
tower: proc options (main);

call Move (4,1,2,3);

Move: procedure (ndiscs, from, to, via) recursive;
declare (ndiscs, from, to, via) fixed binary;

if ndiscs = 1 then
put skip edit ('Move disc from pole ', trim(from), ' to pole ',
trim(to) ) (a);
else
do;
call Move (ndiscs-1, from, via, to);
call Move (1, from, to, via);
call Move (ndiscs-1, via, to, from);
end;
end Move;

end tower;

## PL/M

Translation of: Tiny BASIC

Iterative solution as PL/M doesn't do recursion.

Works with: 8080 PL/M Compiler
... under CP/M (or an emulator)
100H: /* ITERATIVE TOWERS OF HANOI; TRANSLATED FROM TINY BASIC (VIA ALGOL W) */

/* CP/M BDOS SYSTEM CALL                                                  */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
/* I/O ROUTINES                                                           */
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END; PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;

DECLARE ( D, N, X, S, T ) ADDRESS;
/* FIXED NUMBER OF DISCS: 4 */
N = 1;
DO D = 1 TO 4;
N = N + N;
END;
DO X = 1 TO N - 1;
/* AS IN ALGOL W, WE CAN USE PL/M'S BIT ABD MOD OPERATORS             */
S =   ( X AND ( X - 1 ) )       MOD 3;
T = ( ( X OR  ( X - 1 ) ) + 1 ) MOD 3;
CALL PR$STRING( .'MOVE DISC ON PEG$' );
CALL PR$CHAR( '1' + S ); CALL PR$STRING( .' TO PEG $' ); CALL PR$CHAR( '1' + T );
CALL PR$STRING( .( 0DH, 0AH, '$' ) );
END;
EOF
Output:
MOVE DISC ON PEG 1 TO PEG 3
MOVE DISC ON PEG 1 TO PEG 2
MOVE DISC ON PEG 3 TO PEG 2
MOVE DISC ON PEG 1 TO PEG 3
MOVE DISC ON PEG 2 TO PEG 1
MOVE DISC ON PEG 2 TO PEG 3
MOVE DISC ON PEG 1 TO PEG 3
MOVE DISC ON PEG 1 TO PEG 2
MOVE DISC ON PEG 3 TO PEG 2
MOVE DISC ON PEG 3 TO PEG 1
MOVE DISC ON PEG 2 TO PEG 1
MOVE DISC ON PEG 3 TO PEG 2
MOVE DISC ON PEG 1 TO PEG 3
MOVE DISC ON PEG 1 TO PEG 2
MOVE DISC ON PEG 3 TO PEG 2


## Plain TeX

\newcount\hanoidepth
\def\hanoi#1{%
\hanoidepth = #1
\move abc
}%
\def\move#1#2#3{%
\ifnum \hanoidepth > 0
\move #1#3#2
\fi
Move the upper disk from pole #1 to pole #3.\par
\ifnum \hanoidepth > 0
\move#2#1#3
\fi
}

\hanoi{5}
\end


## Pop11

define hanoi(n, src, dst, via);
if n > 0 then
hanoi(n - 1, src, via, dst);
'Move disk ' >< n >< ' from ' >< src >< ' to ' >< dst >< '.' =>
hanoi(n - 1, via, dst, src);
endif;
enddefine;

hanoi(4, "left", "middle", "right");

## PostScript

A million-page document, each page showing one move.

%!PS-Adobe-3.0
%%BoundingBox: 0 0 300 300

/plate {
dup s mul neg 2 div 0 rmoveto
dup s mul 0 rlineto
0 th rlineto
s neg mul 0 rlineto
closepath gsave .5 setgray fill grestore 0 setgray stroke
} def

/drawtower {
0 1 2 { /x exch def /y 0 def
tower x get {
dup 0 gt { x y plate /y y 1 add def } {pop} ifelse
} forall
} for showpage
} def

/apop { [ exch aload pop /last exch def ] last } def
/apush{ [ 3 1 roll aload pop counttomark -1 roll ] } def

/hanoi {
0 dict begin /from /mid /to /h 5 -1 2 { -1 roll def } for
h 1 eq {
tower from get apop tower to get apush
tower to 3 -1 roll put
tower from 3 -1 roll put
drawtower
} {
/h h 1 sub def
from to mid h hanoi
from mid to 1 hanoi
mid from to h hanoi
} ifelse
end
} def

/n 12 def
/s 90 n div def
/th 180 n div def
/tower [ [n 1 add -1 2 { } for ] [] [] ] def

drawtower 0 1 2 n hanoi

%%EOF


## PowerShell

Works with: PowerShell version 4.0
function hanoi($n,$a,  $b,$c) {
if($n -eq 1) { "$a -> $c" } else{ hanoi ($n - 1) $a$c $b hanoi 1$a $b$c
hanoi ($n - 1)$b $a$c
}
}
hanoi 3 "A" "B" "C"


Output:

A -> C
A -> B
C -> B
A -> C
B -> A
B -> C
A -> C


## Prolog

From Programming in Prolog by W.F. Clocksin & C.S. Mellish

hanoi(N) :- move(N,left,center,right).

move(0,_,_,_) :- !.
move(N,A,B,C) :-
M is N-1,
move(M,A,C,B),
inform(A,B),
move(M,C,B,A).

inform(X,Y) :- write([move,a,disk,from,the,X,pole,to,Y,pole]), nl.


Using DCGs and separating core logic from IO

hanoi(N, Src, Aux, Dest, Moves-NMoves) :-
NMoves is 2^N - 1,
length(Moves, NMoves),
phrase(move(N, Src, Aux, Dest), Moves).

move(1, Src, _, Dest) --> !,
[Src->Dest].

move(2, Src, Aux, Dest) --> !,
[Src->Aux,Src->Dest,Aux->Dest].

move(N, Src, Aux, Dest) -->
{ succ(N0, N) },
move(N0, Src, Dest, Aux),
move(1, Src, Aux, Dest),
move(N0, Aux, Src, Dest).


## PureBasic

Algorithm according to http://en.wikipedia.org/wiki/Towers_of_Hanoi

Procedure Hanoi(n, A.s, C.s, B.s)
If n
Hanoi(n-1, A, B, C)
PrintN("Move the plate from "+A+" to "+C)
Hanoi(n-1, B, C, A)
EndIf
EndProcedure


Full program

Procedure Hanoi(n, A.s, C.s, B.s)
If n
Hanoi(n-1, A, B, C)
PrintN("Move the plate from "+A+" to "+C)
Hanoi(n-1, B, C, A)
EndIf
EndProcedure

If OpenConsole()
Define n=3
PrintN("Moving "+Str(n)+" pegs."+#CRLF$) Hanoi(n,"Left Peg","Middle Peg","Right Peg") PrintN(#CRLF$+"Press ENTER to exit."): Input()
EndIf

Output:
Moving 3 pegs.

Move the plate from Left Peg to Middle Peg
Move the plate from Left Peg to Right Peg
Move the plate from Middle Peg to Right Peg
Move the plate from Left Peg to Middle Peg
Move the plate from Right Peg to Left Peg
Move the plate from Right Peg to Middle Peg
Move the plate from Left Peg to Middle Peg

Press ENTER to exit.


## Python

### Recursive

def hanoi(ndisks, startPeg=1, endPeg=3):
if ndisks:
hanoi(ndisks-1, startPeg, 6-startPeg-endPeg)
print(f"Move disk {ndisks} from peg {startPeg} to peg {endPeg}")
hanoi(ndisks-1, 6-startPeg-endPeg, endPeg)

hanoi(4)

Output:
for ndisks=2
Move disk 1 from peg 1 to peg 2
Move disk 2 from peg 1 to peg 3
Move disk 1 from peg 2 to peg 3


Or, separating the definition of the data from its display:

Works with: Python version 3.7
'''Towers of Hanoi'''

# hanoi :: Int -> String -> String -> String -> [(String, String)]
def hanoi(n):
'''A list of (from, to) label pairs,
where a, b and c are labels for each of the
three Hanoi tower positions.'''
def go(n, a, b, c):
p = n - 1
return (
go(p, a, c, b) + [(a, b)] + go(p, c, b, a)
) if 0 < n else []
return lambda a: lambda b: lambda c: go(n, a, b, c)

# TEST ----------------------------------------------------
if __name__ == '__main__':

# fromTo :: (String, String) -> String
def fromTo(xy):
'''x -> y'''
x, y = xy
return x.rjust(5, ' ') + ' -> ' + y

print(__doc__ + ':\n\n' + '\n'.join(
map(fromTo, hanoi(4)('left')('right')('mid'))
))

Output:
Towers of Hanoi:

left -> mid
left -> right
mid -> right
left -> mid
right -> left
right -> mid
left -> mid
left -> right
mid -> right
mid -> left
right -> left
mid -> right
left -> mid
left -> right
mid -> right

### Graphic

Refactoring the version above to recursively generate a simple visualisation:

Works with: Python version 3.7
'''Towers of Hanoi'''

from itertools import accumulate, chain, repeat
from inspect import signature
import operator

# hanoi :: Int -> [(Int, Int)]
def hanoi(n):
'''A list of index pairs, representing disk moves
between indexed Hanoi positions.
'''
def go(n, a, b, c):
p = n - 1
return (
go(p, a, c, b) + [(a, b)] + go(p, c, b, a)
) if 0 < n else []
return go(n, 0, 2, 1)

# hanoiState :: ([Int],[Int],[Int], String) -> (Int, Int) ->
#               ([Int],[Int],[Int], String)
def hanoiState(tpl, ab):
'''A new Hanoi tower state'''
a, b = ab
xs, ys = tpl[a], tpl[b]

w = 3 * (2 + (2 * max(map(max, filter(len, tpl[:-1])))))

def delta(i):
return tpl[i] if i not in ab else xs[1:] if (
i == a
) else [xs] + ys

tkns = moveName(('left', 'mid', 'right'))(ab)
caption = ' '.join(tkns)
return tuple(map(delta, [0, 1, 2])) + (
(caption if tkns != 'mid' else caption.rjust(w, ' ')),
)

# showHanoi :: ([Int],[Int],[Int], String) -> String
def showHanoi(tpl):
'''Captioned string representation of an updated Hanoi tower state.'''

def fullHeight(n):
return lambda xs: list(repeat('', n - len(xs))) + xs

mul = curry(operator.mul)
lt = curry(operator.lt)
rods = fmap(fmap(mul('__')))(
list(tpl[0:3])
)
h = max(map(len, rods))
w = 2 + max(
map(
compose(max)(fmap(len)),
filter(compose(lt(0))(len), rods)
)
)
xs = fmap(concat)(
transpose(fmap(
compose(fmap(center(w)(' ')))(
fullHeight(h)
)
)(rods))
)
return tpl + '\n\n' + unlines(xs) + '\n' + ('___' * w)

# moveName :: (String, String, String) -> (Int, Int) -> [String]
def moveName(labels):
'''(from, to) index pair represented as an a -> b string.'''
def go(ab):
a, b = ab
return [labels[a], ' to ', labels[b]] if a < b else [
labels[b], ' from ', labels[a]
]
return lambda ab: go(ab)

# TEST ----------------------------------------------------
def main():
'''Visualisation of a Hanoi tower sequence for N discs.
'''
n = 3
print('Hanoi sequence for ' + str(n) + ' disks:\n')
print(unlines(
fmap(showHanoi)(
scanl(hanoiState)(
(enumFromTo(1)(n), [], [], '')
)(hanoi(n))
)
))

# GENERIC -------------------------------------------------

# center :: Int -> Char -> String -> String
def center(n):
'''String s padded with c to approximate centre,
fitting in but not truncated to width n.'''
return lambda c: lambda s: s.center(n, c)

# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
def compose(g):
'''Right to left function composition.'''
return lambda f: lambda x: g(f(x))

# concat :: [[a]] -> [a]
# concat :: [String] -> String
def concat(xs):
'''The concatenation of all the elements
in a list or iterable.'''

def f(ys):
zs = list(chain(*ys))
return ''.join(zs) if isinstance(ys, str) else zs

return (
f(xs) if isinstance(xs, list) else (
chain.from_iterable(xs)
)
) if xs else []

# curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
'''A curried function derived
from an uncurried function.'''
if 1 < len(signature(f).parameters):
return lambda x: lambda y: f(x, y)
else:
return f

# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))

# fmap :: (a -> b) -> [a] -> [b]
def fmap(f):
'''fmap over a list.
f lifted to a function over a list.
'''
return lambda xs: list(map(f, xs))

# scanl :: (b -> a -> b) -> b -> [a] -> [b]
def scanl(f):
'''scanl is like reduce, but returns a succession of
intermediate values, building from the left.
'''
return lambda a: lambda xs: (
accumulate(chain([a], xs), f)
)

# showLog :: a -> IO String
def showLog(*s):
'''Arguments printed with
intercalated arrows.'''
print(
' -> '.join(map(str, s))
)

# transpose :: Matrix a -> Matrix a
def transpose(m):
'''The rows and columns of the argument transposed.
(The matrix containers and rows can be lists or tuples).
'''
if m:
inner = type(m)
z = zip(*m)
return (type(m))(
map(inner, z) if tuple != inner else z
)
else:
return m

# unlines :: [String] -> String
def unlines(xs):
'''A single string derived by the intercalation
of a list of strings with the newline character.
'''
return '\n'.join(xs)

# TEST ----------------------------------------------------
if __name__ == '__main__':
main()

Hanoi sequence for 3 disks:

__
____
______
________________________
left  to  right

____
______            __
________________________
left  to  mid

______   ____     __
________________________
mid  from  right

__
______   ____
________________________
left  to  right

__
____   ______
________________________
left  from  mid

__     ____   ______
________________________
mid  to  right

____
__            ______
________________________
left  to  right

__
____
______
________________________

### Library: VPython

There is a 3D hanoi-game in the examples that come with VPython, and at github.

## Quackery

  [ stack ]                     is rings    (     --> [ )

[ rings share
depth share -
8 * times sp
emit sp emit sp
say 'move' cr ]             is echomove ( c c -->   )

[ dup rings put
depth put
char a char b char c
[ swap decurse
rot 2dup echomove
decurse
swap rot ]
3 times drop
depth release
rings release ]             is hanoi    (   n --> n )

say 'How to solve a three ring Towers of Hanoi puzzle:' cr cr
3 hanoi cr
Output:
How to solve a three ring Towers of Hanoi puzzle:

a c move
a b move
c b move
a c move
b a move
b c move
a c move
a b move
c b move
c a move
b a move
c b move
a c move
a b move
c b move



## Quite BASIC

'This is implemented on the Quite BASIC website
'http://www.quitebasic.com/prj/puzzle/towers-of-hanoi/

1000 REM Towers of Hanoi
1010 REM Quite BASIC Puzzle Project
1020 CLS
1030 PRINT "Towers of Hanoi"
1040 PRINT
1050 PRINT "This is a recursive solution for seven discs."
1060 PRINT
1070 PRINT "See the REM statements in the program if you didn't think that recursion was possible in classic BASIC!"
1080 REM Yep, recursive GOSUB calls works in Quite BASIC!
1090 REM However, to actually write useful recursive algorithms, it helps to have variable scoping and parameters to subroutines -- something classic BASIC is lacking.  In this case we have only one "parameter" -- the variable N.  And subroutines are always called with N-1.  This is lucky for us because we can keep track of the value by decrementing it when we enter subroutines and incrementing it back when we exit.
1100 REM If we had subroutine parameters we could have written a single subroutine for moving discs from peg P to peg Q where P and Q were subroutine parameters, but no such luck.  Instead we have to write six different subroutines for moving from peg to peg.  See Subroutines 4000, 5000, 6000, 7000, 8000, and 9000.
1110 REM ===============================
2000 REM A, B, and C are arrays holding the discs
2010 REM We refer to the corresponding pegs as peg A, B, and C
2020 ARRAY A
2030 ARRAY B
2040 ARRAY C
2050 REM Fill peg A with seven discs
2060 FOR I = 0 TO 6
2070 LET A[I] = 7 - I
2080 NEXT I
2090 REM X, Y, Z hold the number of discs on pegs A, B, and C
2100 LET X = 7
2110 LET Y = 0
2120 LET Z = 0
2130 REM Disc colors
2140 ARRAY P
2150 LET P = "cyan"
2160 LET P = "blue"
2170 LET P = "green"
2180 LET P = "yellow"
2190 LET P = "magenta"
2200 LET P = "orange"
2210 LET P = "red"
2220 REM Draw initial position -- all discs on the A peg
2230 FOR I = 0 TO 6
2240 FOR J = 8 - A[I] TO 8 + A[I]
2250 PLOT J, I, P[A[I]]
2260 NEXT J
2270 NEXT I
2280 REM N is the number of discs to move
2290 LET N = 7
2320 REM Move all discs from peg A to peg B
2310 GOSUB 6000
2320 END
3000 REM The subroutines 3400, 3500, 3600, 3700, 3800, 3900
3010 REM handle the drawing of the discs on the canvas as we
3020 REM move discs from one peg to another.
3030 REM These subroutines also update the variables X, Y, and Z
3040 REM which hold the number of discs on each peg.
3050 REM ==============================
3400 REM Subroutine -- Remove disc from peg A
3410 LET X = X - 1
3420 FOR I = 8 - A[X] TO 8 + A[X]
3430 PLOT I, X, "gray"
3440 NEXT I
3450 RETURN
3500 REM Subroutine -- Add disc to peg A
3510 FOR I = 8 - A[X] TO 8 + A[X]
3520 PLOT I, X, P[A[X]]
3530 NEXT I
3540 LET X = X + 1
3550 PAUSE 400 * (5 - LEVEL) + 10
3560 RETURN
3600 REM Subroutine -- Remove disc from peg B
3610 LET Y = Y - 1
3620 FOR I = 24 - B[Y] TO 24 + B[Y]
3630 PLOT I, Y, "gray"
3640 NEXT I
3650 RETURN
3700 REM Subroutine -- Add disc to peg B
3710 FOR I = 24 - B[Y] TO 24 + B[Y]
3720 PLOT I, Y, P[B[Y]]
3730 NEXT I
3740 LET Y = Y + 1
3750 PAUSE 400 * (5 - LEVEL) + 10
3760 RETURN
3800 REM Subroutine -- Remove disc from peg C
3810 LET Z = Z - 1
3820 FOR I = 40 - C[Z] TO 40 + C[Z]
3830 PLOT I, Z, "gray"
3840 NEXT I
3850 RETURN
3900 REM Subroutine -- Add disc to peg C
3910 FOR I = 40 - C[Z] TO 40 + C[Z]
3920 PLOT I, Z, P[C[Z]]
3930 NEXT I
3940 LET Z = Z + 1
3950 PAUSE 400 * (5 - LEVEL) + 10
3960 RETURN
4000 REM ======================================
4010 REM Recursive Subroutine -- move N discs from peg B to peg A
4020 REM First move N-1 discs from peg B to peg C
4030 LET N = N - 1
4040 IF N <> 0 THEN GOSUB 9000
4050 REM Then move one disc from peg B to peg A
4060 GOSUB 3600
4070 LET A[X] = B[Y]
4080 GOSUB 3500
4090 REM And finally move N-1 discs from peg C to peg A
4100 IF N <> 0 THEN GOSUB 5000
4110 REM Restore N before returning
4120 LET N = N + 1
4130 RETURN
5000 REM ======================================
5010 REM Recursive Subroutine -- Move N discs from peg C to peg A
5020 REM First move N-1 discs from peg C to peg B
5030 LET N = N - 1
5040 IF N <> 0 THEN GOSUB 8000
5050 REM Then move one disc from peg C to peg A
5060 GOSUB 3800
5070 LET A[X] = C[Z]
5080 GOSUB 3500
5090 REM And finally move N-1 discs from peg B to peg A
5100 IF N <> 0 THEN GOSUB 4000
5120 REM Restore N before returning
5130 LET N = N + 1
5140 RETURN
6000 REM ======================================
6000 REM Recursive Subroutine -- Move N discs from peg A to peg B
6010 REM First move N-1 discs from peg A to peg C
6020 LET N = N - 1
6030 IF N <> 0 THEN GOSUB 7000
6040 REM Then move one disc from peg A to peg B
6050 GOSUB 3400
6060 LET B[Y] = A[X]
6070 GOSUB 3700
6090 REM And finally move N-1 discs from peg C to peg B
6100 IF N <> 0 THEN GOSUB 8000
6110 REM Restore N before returning
6120 LET N = N + 1
6130 RETURN
7000 REM ======================================
7010 REM Recursive Subroutine -- Move N discs from peg A to peg C
7020 REM First move N-1 discs from peg A to peg B
7030 LET N = N - 1
7040 IF N <> 0 THEN GOSUB 6000
7050 REM Then move one disc from peg A to peg C
7060 GOSUB 3400
7070 LET C[Z] = A[X]
7080 GOSUB 3900
7090 REM And finally move N-1 discs from peg B to peg C
7100 IF N <> 0 THEN GOSUB 9000
7110 REM Restore N before returning
7120 LET N = N + 1
7130 RETURN
8000 REM ======================================
8010 REM Recursive Subroutine -- Move N discs from peg C to peg B
8020 REM First move N-1 discs from peg C to peg A
8030 LET N = N - 1
8040 IF N <> 0 THEN GOSUB 5000
8050 REM Then move one disc from peg C to peg B
8060 GOSUB 3800
8070 LET B[Y] = C[Z]
8080 GOSUB 3700
8090 REM And finally move N-1 discs from peg A to peg B
8100 IF N <> 0 THEN GOSUB 6000
8110 REM Restore N before returning
8120 LET N = N + 1
8130 RETURN
9000 REM ======================================
9010 REM Recursive Subroutine -- Move N discs from peg B to peg C
9020 REM First move N-1 discs from peg B to peg A
9030 LET N = N - 1
9040 IF N <> 0 THEN GOSUB 4000
9050 REM Then move one disc from peg B to peg C
9060 GOSUB 3600
9070 LET C[Z] = B[Y]
9080 GOSUB 3900
9090 REM And finally move N-1 discs from peg A to peg C
9100 IF N <> 0 THEN GOSUB 7000
9110 REM Restore N before returning
9120 LET N = N + 1
9130 RETURN

## R

Translation of: Octave
hanoimove <- function(ndisks, from, to, via) {
if (ndisks == 1) {
cat("move disk from", from, "to", to, "\n")
} else {
hanoimove(ndisks - 1, from, via, to)
hanoimove(1, from, to, via)
hanoimove(ndisks - 1, via, to, from)
}
}

hanoimove(4, 1, 2, 3)


## Racket

#lang racket
(define (hanoi n a b c)
(when (> n 0)
(hanoi (- n 1) a c b)
(printf "Move ~a to ~a\n" a b)
(hanoi (- n 1) c b a)))
(hanoi 4 'left 'middle 'right)


## Raku

(formerly Perl 6)

subset Peg of Int where 1|2|3;

multi hanoi (0,      Peg $a, Peg$b,     Peg $c) { } multi hanoi (Int$n, Peg $a = 1, Peg$b = 2, Peg $c = 3) { hanoi$n - 1, $a,$c, $b; say "Move$a to $b."; hanoi$n - 1, $c,$b, $a; }  ## Rascal Translation of: Python public void hanoi(ndisks, startPeg, endPeg){ if(ndisks>0){ hanoi(ndisks-1, startPeg, 6 - startPeg - endPeg); println("Move disk <ndisks> from peg <startPeg> to peg <endPeg>"); hanoi(ndisks-1, 6 - startPeg - endPeg, endPeg); } } Output: rascal>hanoi(4,1,3) Move disk 1 from peg 1 to peg 2 Move disk 2 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 Move disk 3 from peg 1 to peg 2 Move disk 1 from peg 3 to peg 1 Move disk 2 from peg 3 to peg 2 Move disk 1 from peg 1 to peg 2 Move disk 4 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 Move disk 2 from peg 2 to peg 1 Move disk 1 from peg 3 to peg 1 Move disk 3 from peg 2 to peg 3 Move disk 1 from peg 1 to peg 2 Move disk 2 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 ok ## Raven Translation of: Python define hanoi use ndisks, startpeg, endpeg ndisks 0 > if 6 startpeg - endpeg - startpeg ndisks 1 - hanoi endpeg startpeg ndisks "Move disk %d from peg %d to peg %d\n" print endpeg 6 startpeg - endpeg - ndisks 1 - hanoi define dohanoi use ndisks # startpeg=1, endpeg=3 3 1 ndisks hanoi # 4 disks 4 dohanoi Output: raven hanoi.rv Move disk 1 from peg 1 to peg 2 Move disk 2 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 Move disk 3 from peg 1 to peg 2 Move disk 1 from peg 3 to peg 1 Move disk 2 from peg 3 to peg 2 Move disk 1 from peg 1 to peg 2 Move disk 4 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3 Move disk 2 from peg 2 to peg 1 Move disk 1 from peg 3 to peg 1 Move disk 3 from peg 2 to peg 3 Move disk 1 from peg 1 to peg 2 Move disk 2 from peg 1 to peg 3 Move disk 1 from peg 2 to peg 3  ## REBOL REBOL [ Title: "Towers of Hanoi" URL: http://rosettacode.org/wiki/Towers_of_Hanoi ] hanoi: func [ {Begin moving the golden disks from one pole to the next. Note: when last disk moved, the world will end.} disks [integer!] "Number of discs on starting pole." /poles "Name poles." from to via ][ if disks = 0 [return] if not poles [from: 'left to: 'middle via: 'right] hanoi/poles disks - 1 from via to print [from "->" to] hanoi/poles disks - 1 via to from ] hanoi 4  Output: left -> right left -> middle right -> middle left -> right middle -> left middle -> right left -> right left -> middle right -> middle right -> left middle -> left right -> middle left -> right left -> middle right -> middle ## Retro [[User:Wodan58|Wodan58]] ([[User talk:Wodan58|talk]]) { 'Num 'From 'To 'Via } [ var ] a:for-each :set !Via !To !From !Num ; :display @To @From 'Move_a_ring_from_%n_to_%n\n s:format s:put ; :hanoi (num,from,to,via-) set @Num n:-zero? [ @Num @From @To @Via @Num n:dec @From @Via @To hanoi set display @Num n:dec @Via @To @From hanoi ] if ; #3 #1 #3 #2 hanoi nl [[User:Wodan58|Wodan58]] ([[User talk:Wodan58|talk]]) ## REXX ### simple text moves /*REXX program displays the moves to solve the Tower of Hanoi (with N disks). */ parse arg N . /*get optional number of disks from CL.*/ if N=='' | N=="," then N=3 /*Not specified? Then use the default.*/ #= 0 /*#: the number of disk moves (so far)*/ z= 2**N - 1 /*Z: " " " minimum # of moves.*/ call mov 1, 3, N /*move the top disk, then recurse ··· */ say /* [↓] Display the minimum # of moves.*/ say 'The minimum number of moves to solve a ' N"─disk Tower of Hanoi is " z exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ mov: procedure expose # z; parse arg @1,@2,@3; L= length(z) if @3==1 then do; #= # + 1 /*bump the (disk) move counter by one. */ say 'step' right(#, L)": move disk on tower" @1 '───►' @2 end else do; call mov @1, 6 -@1 -@2, @3 -1 call mov @1, @2, 1 call mov 6 - @1 - @2, @2, @3 -1 end return /* [↑] this subroutine uses recursion.*/  output when using the default input: step 1: move disk on tower 1 ───► 3 step 2: move disk on tower 1 ───► 2 step 3: move disk on tower 3 ───► 2 step 4: move disk on tower 1 ───► 3 step 5: move disk on tower 2 ───► 1 step 6: move disk on tower 2 ───► 3 step 7: move disk on tower 1 ───► 3 The minimum number of moves to solve a 3-disk Tower of Hanoi is 7  output when the following was entered (to solve with four disks): 4 step 1: move disk on tower 1 ───► 2 step 2: move disk on tower 1 ───► 3 step 3: move disk on tower 2 ───► 3 step 4: move disk on tower 1 ───► 2 step 5: move disk on tower 3 ───► 1 step 6: move disk on tower 3 ───► 2 step 7: move disk on tower 1 ───► 2 step 8: move disk on tower 1 ───► 3 step 9: move disk on tower 2 ───► 3 step 10: move disk on tower 2 ───► 1 step 11: move disk on tower 3 ───► 1 step 12: move disk on tower 2 ───► 3 step 13: move disk on tower 1 ───► 2 step 14: move disk on tower 1 ───► 3 step 15: move disk on tower 2 ───► 3 The minimum number of moves to solve a 4-disk Tower of Hanoi is 15  ### pictorial moves This REXX version pictorially shows (via ASCII art) the moves for solving the Town of Hanoi. Quite a bit of code has been dedicated to showing a "picture" of the towers with the disks, and the movement of the disk (for each move). "Coloring" of the disks is attempted with dithering. In addition, it shows each move in a countdown manner (the last move is marked as #1). It may not be obvious from the pictorial display of the moves, but whenever a disk is moved from one tower to another, it is always the top disk that is moved (to the target tower). Also, since the pictorial showing of the moves may be voluminous (especially for a larger number of disks), the move counter is started with the maximum and is the count shown is decremented so the viewer can see how many moves are left to display. /*REXX program displays the moves to solve the Tower of Hanoi (with N disks). */ parse arg N . /*get optional number of disks from CL.*/ if N=='' | N=="," then N=3 /*Not specified? Then use the default.*/ sw= 80; wp= sw%3 - 1; blanks= left('', wp) /*define some default REXX variables. */ c.1= sw % 3 % 2 /* [↑] SW: assume default Screen Width*/ c.2= sw % 2 - 1 /* ◄─── C.1 C.2 C.2 are the positions*/ c.3= sw - 2 - c.1 /* of the 3 columns.*/ #= 0; z= 2**N - 1; moveK= z /*#moves; min# of moves; where to move.*/ @abc= 'abcdefghijklmnopqrstuvwxyN' /*dithering chars when many disks used.*/ ebcdic= ('f2'x==2) /*determine if EBCDIC or ASCII machine.*/ if ebcdic then do; bar= 'bf'x; ar= "df"x; dither= 'db9f9caf'x; down= "9a"x tr= 'bc'x; bl= "ab"x; br= 'bb'x; vert= "fa"x; tl= 'ac'x end else do; bar= 'c4'x; ar= "10"x; dither= 'b0b1b2db'x; down= "19"x tr= 'bf'x; bl= "c0"x; br= 'd9'x; vert= "b3"x; tl= 'da'x end verts= vert || vert; Tcorners= tl || tr; box = left(dither, 1) downs= down || down; Bcorners= bl || br; boxChars= dither || @abc$.= 0;         $.1= N; k= N; kk= k + k do j=1 for N; @.3.j= blanks; @.2.j= blanks; @.1.j= center( copies(box, kk), wp) if N<=length(boxChars) then @.1.j= translate( @.1.j, , substr( boxChars, kk%2, 1), box) kk= kk - 2 end /*j*/ /*populate the tower of Hanoi spindles.*/ call showTowers; call mov 1,3,N; say say 'The minimum number of moves to solve a ' N"-disk Tower of Hanoi is " z exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ dsk: parse arg from dest; #= # + 1; pp= if from==1 then do; pp= overlay(bl, pp, c.1) pp= overlay(bar, pp, c.1+1, c.dest-c.1-1, bar) || tr end if from==2 then do if dest==1 then do; pp= overlay(tl, pp, c.1) pp= overlay(bar, pp, c.1+1, c.2-c.1-1,bar)||br end if dest==3 then do; pp= overlay(bl, pp, c.2) pp= overlay(bar, pp, c.2+1, c.3-c.2-1,bar)||tr end end if from==3 then do; pp= overlay(br, pp, c.3) pp= overlay(bar, pp, c.dest+1, c.3-c.dest-1, bar) pp= overlay(tl, pp, c.dest) end say translate(pp, downs, Bcorners || Tcorners || bar); say overlay(moveK, pp, 1) say translate(pp, verts, Tcorners || Bcorners || bar) say translate(pp, downs, Tcorners || Bcorners || bar); moveK= moveK - 1$.from= $.from - 1;$.dest= $.dest + 1; _f=$.from + 1;           _t= $.dest @.dest._t= @.from._f; @.from._f= blanks; call showTowers return /*──────────────────────────────────────────────────────────────────────────────────────*/ mov: if arg(3)==1 then call dsk arg(1) arg(2) else do; call mov arg(1), 6 -arg(1) -arg(2), arg(3) -1 call mov arg(1), arg(2), 1 call mov 6 -arg(1) -arg(2), arg(2), arg(3) -1 end /* [↑] The MOV subroutine is recursive, */ return /*it uses no variables, is uses BIFs instead*/ /*──────────────────────────────────────────────────────────────────────────────────────*/ showTowers: do j=N by -1 for N; _=@.1.j @.2.j @.3.j; if _\='' then say _; end; return  output when using the default input:  ░░ ▒▒▒▒ ▓▓▓▓▓▓ ↓ 7 └───────────────────────────────────────────────────┐ │ ↓ ▒▒▒▒ ▓▓▓▓▓▓ ░░ ↓ 6 └─────────────────────────┐ │ ↓ ▓▓▓▓▓▓ ▒▒▒▒ ░░ ↓ 5 ┌─────────────────────────┘ │ ↓ ░░ ▓▓▓▓▓▓ ▒▒▒▒ ↓ 4 └───────────────────────────────────────────────────┐ │ ↓ ░░ ▒▒▒▒ ▓▓▓▓▓▓ ↓ 3 ┌─────────────────────────┘ │ ↓ ░░ ▒▒▒▒ ▓▓▓▓▓▓ ↓ 2 └─────────────────────────┐ │ ↓ ▒▒▒▒ ░░ ▓▓▓▓▓▓ ↓ 1 └───────────────────────────────────────────────────┐ │ ↓ ░░ ▒▒▒▒ ▓▓▓▓▓▓ The minimum number of moves to solve a 3-disk Tower of Hanoi is 7  ## Ring move(4, 1, 2, 3) func move n, src, dst, via if n > 0 move(n - 1, src, via, dst) see "" + src + " to " + dst + nl move(n - 1, via, dst, src) ok ## Ruby ### version 1 def move(num_disks, start=0, target=1, using=2) if num_disks == 1 @towers[target] << @towers[start].pop puts "Move disk from #{start} to #{target} : #{@towers}" else move(num_disks-1, start, using, target) move(1, start, target, using) move(num_disks-1, using, target, start) end end n = 5 @towers = [[*1..n].reverse, [], []] move(n)  Output: Move disk from 0 to 1 : [[5, 4, 3, 2], , []] Move disk from 0 to 2 : [[5, 4, 3], , ] Move disk from 1 to 2 : [[5, 4, 3], [], [2, 1]] Move disk from 0 to 1 : [[5, 4], , [2, 1]] Move disk from 2 to 0 : [[5, 4, 1], , ] Move disk from 2 to 1 : [[5, 4, 1], [3, 2], []] Move disk from 0 to 1 : [[5, 4], [3, 2, 1], []] Move disk from 0 to 2 : [, [3, 2, 1], ] Move disk from 1 to 2 : [, [3, 2], [4, 1]] Move disk from 1 to 0 : [[5, 2], , [4, 1]] Move disk from 2 to 0 : [[5, 2, 1], , ] Move disk from 1 to 2 : [[5, 2, 1], [], [4, 3]] Move disk from 0 to 1 : [[5, 2], , [4, 3]] Move disk from 0 to 2 : [, , [4, 3, 2]] Move disk from 1 to 2 : [, [], [4, 3, 2, 1]] Move disk from 0 to 1 : [[], , [4, 3, 2, 1]] Move disk from 2 to 0 : [, , [4, 3, 2]] Move disk from 2 to 1 : [, [5, 2], [4, 3]] Move disk from 0 to 1 : [[], [5, 2, 1], [4, 3]] Move disk from 2 to 0 : [, [5, 2, 1], ] Move disk from 1 to 2 : [, [5, 2], [4, 1]] Move disk from 1 to 0 : [[3, 2], , [4, 1]] Move disk from 2 to 0 : [[3, 2, 1], , ] Move disk from 2 to 1 : [[3, 2, 1], [5, 4], []] Move disk from 0 to 1 : [[3, 2], [5, 4, 1], []] Move disk from 0 to 2 : [, [5, 4, 1], ] Move disk from 1 to 2 : [, [5, 4], [2, 1]] Move disk from 0 to 1 : [[], [5, 4, 3], [2, 1]] Move disk from 2 to 0 : [, [5, 4, 3], ] Move disk from 2 to 1 : [, [5, 4, 3, 2], []] Move disk from 0 to 1 : [[], [5, 4, 3, 2, 1], []]  ### version 2 # solve(source, via, target) # Example: # solve([5, 4, 3, 2, 1], [], []) # Note this will also solve randomly placed disks, # "place all disk in target with legal moves only". def solve(*towers) # total number of disks disks = towers.inject(0){|sum, tower| sum+tower.length} x=0 # sequence number p towers # initial trace # have we solved the puzzle yet? while towers.last.length < disks do x+=1 # assume the next step from = (x&x-1)%3 to = ((x|(x-1))+1)%3 # can we actually take from tower? if top = towers[from].last bottom = towers[to].last # is the move legal? if !bottom || bottom > top # ok, do it! towers[to].push(towers[from].pop) p towers # trace end end end end solve([5, 4, 3, 2, 1], [], [])  Output: [[5, 4, 3, 2, 1], [], []] [[5, 4, 3, 2], [], ] [[5, 4, 3], , ] [[5, 4, 3], [2, 1], []] [[5, 4], [2, 1], ] [[5, 4, 1], , ] [[5, 4, 1], [], [3, 2]] [[5, 4], [], [3, 2, 1]] [, , [3, 2, 1]] [, [4, 1], [3, 2]] [[5, 2], [4, 1], ] [[5, 2, 1], , ] [[5, 2, 1], [4, 3], []] [[5, 2], [4, 3], ] [, [4, 3, 2], ] [, [4, 3, 2, 1], []] [[], [4, 3, 2, 1], ] [, [4, 3, 2], ] [, [4, 3], [5, 2]] [[], [4, 3], [5, 2, 1]] [, , [5, 2, 1]] [, [4, 1], [5, 2]] [[3, 2], [4, 1], ] [[3, 2, 1], , ] [[3, 2, 1], [], [5, 4]] [[3, 2], [], [5, 4, 1]] [, , [5, 4, 1]] [, [2, 1], [5, 4]] [[], [2, 1], [5, 4, 3]] [, , [5, 4, 3]] [, [], [5, 4, 3, 2]] [[], [], [5, 4, 3, 2, 1]]  ## Run BASIC a = move(4, "1", "2", "3") function move(n, a$, b$, c$)
if n > 0 then
a = move(n-1, a$, c$, b$) print "Move disk from " ; a$ ; " to " ; c$a = move(n-1, b$, a$, c$)
end if
end function
Move disk from 1 to 3
Move disk from 1 to 2
Move disk from 3 to 2
Move disk from 1 to 3
Move disk from 2 to 1
Move disk from 2 to 3
Move disk from 1 to 3
Move disk from 1 to 2
Move disk from 3 to 2
Move disk from 3 to 1
Move disk from 2 to 1
Move disk from 3 to 2
Move disk from 1 to 3
Move disk from 1 to 2
Move disk from 3 to 2

## Rust

Translation of: C
fn move_(n: i32, from: i32, to: i32, via: i32) {
if n > 0 {
move_(n - 1, from, via, to);
println!("Move disk from pole {} to pole {}", from, to);
move_(n - 1, via, to, from);
}
}

fn main() {
move_(4, 1,2,3);
}


## SASL

Copied from SAL manual, Appendix II, answer (3)

hanoi 8 ‘abc"
WHERE
hanoi 0 (a,b,c,) = ()
hanoi n ( a,b,c) = hanoi (n-1) (a,c,b) ,
‘move a disc from " , a , ‘ to " , b , NL ,
hanoi (n-1) (c,b,a)
?

## Sather

Translation of: Fortran
class MAIN is

move(ndisks, from, to, via:INT) is
if ndisks = 1 then
#OUT + "Move disk from pole " + from + " to pole " + to + "\n";
else
move(ndisks-1, from, via, to);
move(1, from, to, via);
move(ndisks-1, via, to, from);
end;
end;

main is
move(4, 1, 2, 3);
end;
end;

## Scala

def move(n: Int, from: Int, to: Int, via: Int) : Unit = {
if (n == 1) {
Console.println("Move disk from pole " + from + " to pole " + to)
} else {
move(n - 1, from, via, to)
move(1, from, to, via)
move(n - 1, via, to, from)
}
}


This next example is from http://gist.github.com/66925 it is a translation to Scala of a Prolog solution and solves the problem at compile time

object TowersOfHanoi {
import scala.reflect.Manifest

def simpleName(m:Manifest[_]):String = {
val name = m.toString
name.substring(name.lastIndexOf('$')+1) } trait Nat final class _0 extends Nat final class Succ[Pre<:Nat] extends Nat type _1 = Succ[_0] type _2 = Succ[_1] type _3 = Succ[_2] type _4 = Succ[_3] case class Move[N<:Nat,A,B,C]() implicit def move0[A,B,C](implicit a:Manifest[A],b:Manifest[B]):Move[_0,A,B,C] = { System.out.println("Move from "+simpleName(a)+" to "+simpleName(b));null } implicit def moveN[P<:Nat,A,B,C](implicit m1:Move[P,A,C,B],m2:Move[_0,A,B,C],m3:Move[P,C,B,A]) :Move[Succ[P],A,B,C] = null def run[N<:Nat,A,B,C](implicit m:Move[N,A,B,C]) = null case class Left() case class Center() case class Right() def main(args:Array[String]){ run[_2,Left,Right,Center] } }  ## Scheme Recursive Process (define (towers-of-hanoi n from to spare) (define (print-move from to) (display "Move[") (display from) (display ", ") (display to) (display "]") (newline)) (cond ((= n 0) "done") (else (towers-of-hanoi (- n 1) from spare to) (print-move from to) (towers-of-hanoi (- n 1) spare to from)))) (towers-of-hanoi 3 "A" "B" "C")  Output: Move[A, B] Move[A, C] Move[B, C] Move[A, B] Move[C, A] Move[C, B] Move[A, B] "done" ## Seed7 const proc: hanoi (in integer: disk, in string: source, in string: dest, in string: via) is func begin if disk > 0 then hanoi(pred(disk), source, via, dest); writeln("Move disk " <& disk <& " from " <& source <& " to " <& dest); hanoi(pred(disk), via, dest, source); end if; end func; ## Sidef Translation of: Perl func hanoi(n, from=1, to=2, via=3) { if (n == 1) { say "Move disk from pole #{from} to pole #{to}."; } else { hanoi(n-1, from, via, to); hanoi( 1, from, to, via); hanoi(n-1, via, to, from); } } hanoi(4);  ## SNOBOL4 * # Note: count is global define('hanoi(n,src,trg,tmp)') :(hanoi_end) hanoi hanoi = eq(n,0) 1 :s(return) hanoi(n - 1, src, tmp, trg) count = count + 1 output = count ': Move disc from ' src ' to ' trg hanoi(n - 1, tmp, trg, src) :(return) hanoi_end * # Test with 4 discs hanoi(4,'A','C','B') end Output: 1: Move disc from A to B 2: Move disc from A to C 3: Move disc from B to C 4: Move disc from A to B 5: Move disc from C to A 6: Move disc from C to B 7: Move disc from A to B 8: Move disc from A to C 9: Move disc from B to C 10: Move disc from B to A 11: Move disc from C to A 12: Move disc from B to C 13: Move disc from A to B 14: Move disc from A to C 15: Move disc from B to C ## Standard ML  fun hanoi(0, a, b, c) = [] | hanoi(n, a, b, c) = hanoi(n-1, a, c, b) @ [(a,b)] @ hanoi(n-1, c, b, a);  ## Stata function hanoi(n, a, b, c) { if (n>0) { hanoi(n-1, a, c, b) printf("Move from %f to %f\n", a, b) hanoi(n-1, c, b, a) } } hanoi(3, 1, 2, 3) Move from 1 to 2 Move from 1 to 3 Move from 2 to 3 Move from 1 to 2 Move from 3 to 1 Move from 3 to 2 Move from 1 to 2  ## Swift Translation of: JavaScript func hanoi(n:Int, a:String, b:String, c:String) { if (n > 0) { hanoi(n - 1, a, c, b) println("Move disk from \(a) to \(c)") hanoi(n - 1, b, a, c) } } hanoi(4, "A", "B", "C")  Swift 2.1 func hanoi(n:Int, a:String, b:String, c:String) { if (n > 0) { hanoi(n - 1, a: a, b: c, c: b) print("Move disk from \(a) to \(c)") hanoi(n - 1, a: b, b: a, c: c) } } hanoi(4, a:"A", b:"B", c:"C")  ## Tcl The use of interp alias shown is a sort of closure: keep track of the number of moves required interp alias {} hanoi {} do_hanoi 0 proc do_hanoi {count n {from A} {to C} {via B}} { if {$n == 1} {
interp alias {} hanoi {} do_hanoi [incr count]
puts "$count: move from$from to $to" } else { incr n -1 hanoi$n $from$via $to hanoi 1$from $to$via
hanoi $n$via $to$from
}
}

hanoi 4

Output:
1: move from A to B
2: move from A to C
3: move from B to C
4: move from A to B
5: move from C to A
6: move from C to B
7: move from A to B
8: move from A to C
9: move from B to C
10: move from B to A
11: move from C to A
12: move from B to C
13: move from A to B
14: move from A to C
15: move from B to C

## TI-83 BASIC

TI-83 BASIC lacks recursion, so technically this task is impossible, however here is a version that uses an iterative method.

PROGRAM:TOHSOLVE
0→A
1→B
0→C
0→D
0→M
1→R
While A<1 or A>7
Input "No. of rings=?",A
End
randM(A+1,3)→[C]
[[1,2][1,3][2,3]]→[E]

Fill(0,[C])
For(I,1,A,1)
I?[C](I,1)
End
ClrHome
While [C](1,3)≠1 and [C](1,2)≠1

For(J,1,3)
For(I,1,A)
If [C](I,J)≠0:Then
Output(I+1,3J,[C](I,J))
End
End
End
While C=0
Output(1,3B," ")
1→I
[E](R,2)→J
While [C](I,J)=0 and I≤A
I+1→I
End
[C](I,J)→D
1→I
[E](R,1)→J
While [C](I,J)=0 and I≤A
I+1→I
End
If (D<[C](I,J) and D≠0) or [C](I,J)=0:Then
[E](R,2)→B
Else
[E](R,1)→B
End

1→I
While [C](I,B)=0 and I≤A
I+1→I
End
If I≤A:Then
[C](I,B)→C
0→[C](I,B)
Output(I+1,3B," ")
End
Output(1,3B,"V")
End

While C≠0
Output(1,3B," ")
If B=[E](R,2):Then
[E](R,1)→B
Else
[E](R,2)→B
End

1→I
While [C](I,B)=0 and I≤A
I+1→I
End
If [C](I,B)=0 or [C](I,B)>C:Then
C→[C](I-1,B)
0→C
M+1→M
End
End
Output(1,3B,"V")
R+1→R
If R=4:Then:1→R:End

End

## Tiny BASIC

Tiny BASIC does not have recursion, so only an iterative solution is possible... and it has no arrays, so actually keeping track of individual discs is not feasible.

But as if by magic, it turns out that the source and destination pegs on iteration number n are given by (n&n-1) mod 3 and ((n|n-1) + 1) mod 3 respectively, where & and | are the bitwise and and or operators. Line 40 onward is dedicated to implementing those bitwise operations, since Tiny BASIC hasn't got them natively.

 5  PRINT "How many disks?"
INPUT D
IF D < 1 THEN GOTO 5
IF D > 10 THEN GOTO 5
LET N = 1
10  IF D = 0 THEN GOTO 20
LET D = D - 1
LET N = 2*N
GOTO 10
20  LET X = 0
30  LET X = X + 1
IF X = N THEN END
GOSUB 40
LET S = S - 3*(S/3)
GOSUB 50
LET T = T + 1
LET T = T - 3*(T/3)
PRINT "Move disc on peg ",S+1," to peg ",T+1
GOTO 30
40  LET B = X - 1
LET A = X
LET S = 0
LET Z = 2048
45  LET C = 0
IF B >= Z THEN LET C = 1
IF A >= Z THEN LET C = C + 1
IF C = 2 THEN LET S = S + Z
IF A >= Z THEN LET A = A - Z
IF B >= Z THEN LET B = B - Z
LET Z = Z / 2
IF Z = 0 THEN RETURN
GOTO 45
50  LET B = X - 1
LET A = X
LET T = 0
LET Z = 2048
55  LET C = 0
IF B >= Z THEN LET C = 1
IF A >= Z THEN LET C = C + 1
IF C > 0 THEN LET T = T + Z
IF A >= Z THEN LET A = A - Z
IF B >= Z THEN LET B = B - Z
LET Z = Z / 2
IF Z = 0 THEN RETURN
GOTO 55
Output:

How many discs?
4
Move disc on peg 1 to peg 3
Move disc on peg 1 to peg 2
Move disc on peg 3 to peg 2
Move disc on peg 1 to peg 3
Move disc on peg 2 to peg 1
Move disc on peg 2 to peg 3
Move disc on peg 1 to peg 3
Move disc on peg 1 to peg 2
Move disc on peg 3 to peg 2
Move disc on peg 3 to peg 1
Move disc on peg 2 to peg 1
Move disc on peg 3 to peg 2
Move disc on peg 1 to peg 3
Move disc on peg 1 to peg 2
Move disc on peg 3 to peg 2



## Toka

value| sa sb sc n |
[ to sc to sb to sa to n ] is vars!
[ ( num from to via -- )
vars!
n 0 <>
[
n sa sb sc
n 1- sa sc sb recurse
vars!
." Move a ring from " sa . ." to " sb . cr
n 1- sc sb sa recurse
] ifTrue
] is hanoi

## True BASIC

Translation of: FreeBASIC
DECLARE SUB hanoi

SUB hanoi(n, desde , hasta, via)
IF n > 0 THEN
CALL hanoi(n - 1, desde, via, hasta)
PRINT "Mover disco"; n; "desde posición"; desde; "hasta posición"; hasta
CALL hanoi(n - 1, via, hasta, desde)
END IF
END SUB

PRINT "Tres discos"
PRINT
CALL hanoi(3, 1, 2, 3)
PRINT
PRINT "Cuatro discos"
PRINT
CALL hanoi(4, 1, 2, 3)
PRINT
PRINT "Pulsa un tecla para salir"
END


## TSE SAL

// library: program: run: towersofhanoi: recursive: sub <description></description> <version>1.0.0.0.0</version> <version control></version control> (filenamemacro=runprrsu.s) [kn, ri, tu, 07-02-2012 19:54:23]
PROC PROCProgramRunTowersofhanoiRecursiveSub( INTEGER totalDiskI, STRING fromS, STRING toS, STRING viaS, INTEGER bufferI )
IF ( totalDiskI == 0 )
RETURN()
ENDIF
PROCProgramRunTowersofhanoiRecursiveSub( totalDiskI - 1, fromS, viaS, toS, bufferI )
AddLine( Format( "Move disk", " ", totalDiskI, " ", "from peg", " ", "'", fromS, "'", " ", "to peg", " ", "'", toS, "'" ), bufferI )
PROCProgramRunTowersofhanoiRecursiveSub( totalDiskI - 1, viaS, toS, fromS, bufferI )
END

// library: program: run: towersofhanoi: recursive <description></description> <version>1.0.0.0.6</version> <version control></version control> (filenamemacro=runprtre.s) [kn, ri, tu, 07-02-2012 19:40:45]
PROC PROCProgramRunTowersofhanoiRecursive( INTEGER totalDiskI, STRING fromS, STRING toS, STRING viaS )
INTEGER bufferI = 0
PushPosition()
bufferI = CreateTempBuffer()
PopPosition()
PROCProgramRunTowersofhanoiRecursiveSub( totalDiskI, fromS, toS, viaS, bufferI )
GotoBufferId( bufferI )
END

PROC Main()
STRING s1 = "4"
IF ( NOT ( Ask( "program: run: towersofhanoi: recursive: totalDiskI = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIF
PROCProgramRunTowersofhanoiRecursive( Val( s1 ), "source", "target", "via" )
END

## uBasic/4tH

Translation of: C
Proc  _Move(4, 1,2,3)                  ' 4 disks, 3 poles
End

_Move Param(4)
If (a@ > 0) Then
Proc _Move (a@ - 1, b@, d@, c@)
Print "Move disk from pole ";b@;" to pole ";c@
Proc _Move (a@ - 1, d@, c@, b@)
EndIf
Return


## UNIX Shell

Works with: Bourne Again SHell
Works with: Korn Shell
Works with: Z Shell
function move {
typeset -i n=$1 typeset from=$2
typeset to=$3 typeset via=$4

if (( n )); then
move $(( n - 1 )) "$from" "$via" "$to"
echo "Move disk from pole $from to pole$to"
move $(( n - 1 )) "$via" "$to" "$from"
fi
}

move "$@"  A strict POSIX (or just really old) shell has no subprogram capability, but scripts are naturally reentrant, so: Works with: Bourne Shell Works with: Almquist Shell #!/bin/sh if [ "$1" -gt 0 ]; then
"$0" "expr$1 - 1" "$2" "$4" "$3" echo "Move disk from pole$2 to pole $3" "$0" "expr $1 - 1" "$4" "$3" "$2"
fi


Output from any of the above:

Output:
$hanoi 4 1 3 2 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 3 to pole 1 Move disk from pole 3 to pole 2 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 Move disk from pole 2 to pole 1 Move disk from pole 3 to pole 1 Move disk from pole 2 to pole 3 Move disk from pole 1 to pole 2 Move disk from pole 1 to pole 3 Move disk from pole 2 to pole 3 ## Ursala #import nat move = ~&al^& ^rlPlrrPCT/~&arhthPX ^|W/~& ^|G/predecessor ^/~&htxPC ~&zyxPC #show+ main = ^|T(~&,' -> '--)* move/4 <'start','end','middle'> Output: start -> middle start -> end middle -> end start -> middle end -> start end -> middle start -> middle start -> end middle -> end middle -> start end -> start middle -> end start -> middle start -> end middle -> end ## VBScript Derived from the BASIC256 version. Sub Move(n,fromPeg,toPeg,viaPeg) If n > 0 Then Move n-1, fromPeg, viaPeg, toPeg WScript.StdOut.Write "Move disk from " & fromPeg & " to " & toPeg WScript.StdOut.WriteBlankLines(1) Move n-1, viaPeg, toPeg, fromPeg End If End Sub Move 4,1,2,3 WScript.StdOut.Write("Towers of Hanoi puzzle completed!")  Output: Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Move disk from 1 to 3 Move disk from 2 to 1 Move disk from 2 to 3 Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Move disk from 3 to 1 Move disk from 2 to 1 Move disk from 3 to 2 Move disk from 1 to 3 Move disk from 1 to 2 Move disk from 3 to 2 Towers of Hanoi puzzle completed! ## Vedit macro language This implementation outputs the results in current edit buffer. #1=1; #2=2; #3=3; #4=4 // move 4 disks from 1 to 2 Call("MOVE_DISKS") Return // Move disks // #1 = from, #2 = to, #3 = via, #4 = number of disks // :MOVE_DISKS: if (#4 > 0) { Num_Push(1,4) #9=#2; #2=#3; #3=#9; #4-- // #1 to #3 via #2 Call("MOVE_DISKS") Num_Pop(1,4) Ins_Text("Move a disk from ") // move one disk Num_Ins(#1, LEFT+NOCR) Ins_Text(" to ") Num_Ins(#2, LEFT) Num_Push(1,4) #9=#1; #1=#3; #3 = #9; #4-- // #3 to #2 via #1 Call("MOVE_DISKS") Num_Pop(1,4) } Return ## Vim Script function TowersOfHanoi(n, from, to, via) if (a:n > 1) call TowersOfHanoi(a:n-1, a:from, a:via, a:to) endif echom("Move a disc from " . a:from . " to " . a:to) if (a:n > 1) call TowersOfHanoi(a:n-1, a:via, a:to, a:from) endif endfunction call TowersOfHanoi(4, 1, 3, 2) Output: Move a disc from 1 to 2 Move a disc from 1 to 3 Move a disc from 2 to 3 Move a disc from 1 to 2 Move a disc from 3 to 1 Move a disc from 3 to 2 Move a disc from 1 to 2 Move a disc from 1 to 3 Move a disc from 2 to 3 Move a disc from 2 to 1 Move a disc from 3 to 1 Move a disc from 2 to 3 Move a disc from 1 to 2 Move a disc from 1 to 3 Move a disc from 2 to 3 ## Visual Basic .NET Module TowersOfHanoi Sub MoveTowerDisks(ByVal disks As Integer, ByVal fromTower As Integer, ByVal toTower As Integer, ByVal viaTower As Integer) If disks > 0 Then MoveTowerDisks(disks - 1, fromTower, viaTower, toTower) System.Console.WriteLine("Move disk {0} from {1} to {2}", disks, fromTower, toTower) MoveTowerDisks(disks - 1, viaTower, toTower, fromTower) End If End Sub Sub Main() MoveTowerDisks(4, 1, 2, 3) End Sub End Module  ## VTL-2 VTL-2 doesn't have procedure parameters, so this stacks and unstacks the return line number and parameters as reuired. The "move" routune starts at line 2000, the routine at 4000 stacks the return line number and parameters for "move" and the routine at 5000 unstacks the return line number and parameters. 1000 N=4 1010 F=1 1020 T=2 1030 V=3 1040 S=0 1050 #=2000 1060 #=9999 2000 R=! 2010 #=N<1*2210 2020 #=4000 2030 N=N-1 2040 A=T 2050 T=V 2060 V=A 2070 #=2000 2080 #=5000 2090 ?="Move disk from peg: "; 2100 ?=F 2110 ?=" to peg: "; 2120 ?=T 2130 ?="" 2140 #=4000 2150 N=N-1 2160 A=F 2170 F=V 2180 V=A 2190 #=2000 2200 #=5000 2210 #=R 4000 S=S+1 4010 :S)=R 4020 S=S+1 4030 :S)=N 4040 S=S+1 4050 :S)=F 4060 S=S+1 4070 :S)=V 4080 S=S+1 4090 :S)=T 4100 #=! 5000 T=:S) 5010 S=S-1 5020 V=:S) 5030 S=S-1 5040 F=:S) 5050 S=S-1 5060 N=:S) 5070 S=S-1 5080 R=:S) 5090 S=S-1 5100 #=! Output: Move disk from peg: 1 to peg: 3 Move disk from peg: 1 to peg: 2 Move disk from peg: 3 to peg: 2 Move disk from peg: 1 to peg: 3 Move disk from peg: 2 to peg: 1 Move disk from peg: 2 to peg: 3 Move disk from peg: 1 to peg: 3 Move disk from peg: 1 to peg: 2 Move disk from peg: 3 to peg: 2 Move disk from peg: 3 to peg: 1 Move disk from peg: 2 to peg: 1 Move disk from peg: 3 to peg: 2 Move disk from peg: 1 to peg: 3 Move disk from peg: 1 to peg: 2 Move disk from peg: 3 to peg: 2  ## Wren Translation of: Kotlin class Hanoi { construct new(disks) { _moves = 0 System.print("Towers of Hanoi with %(disks) disks:\n") move(disks, "L", "C", "R") System.print("\nCompleted in %(_moves) moves\n") } move(n, from, to, via) { if (n > 0) { move(n - 1, from, via, to) _moves = _moves + 1 System.print("Move disk %(n) from %(from) to %(to)") move(n - 1, via, to, from) } } } Hanoi.new(3) Hanoi.new(4)  Output: Towers of Hanoi with 3 disks: Move disk 1 from L to C Move disk 2 from L to R Move disk 1 from C to R Move disk 3 from L to C Move disk 1 from R to L Move disk 2 from R to C Move disk 1 from L to C Completed in 7 moves Towers of Hanoi with 4 disks: Move disk 1 from L to R Move disk 2 from L to C Move disk 1 from R to C Move disk 3 from L to R Move disk 1 from C to L Move disk 2 from C to R Move disk 1 from L to R Move disk 4 from L to C Move disk 1 from R to C Move disk 2 from R to L Move disk 1 from C to L Move disk 3 from R to C Move disk 1 from L to R Move disk 2 from L to C Move disk 1 from R to C Completed in 15 moves  ## XPL0 code Text=12; proc MoveTower(Discs, From, To, Using); int Discs, From, To, Using; [if Discs > 0 then [MoveTower(Discs-1, From, Using, To); Text(0, "Move from "); Text(0, From); Text(0, " peg to "); Text(0, To); Text(0, " peg.^M^J"); MoveTower(Discs-1, Using, To, From); ]; ]; MoveTower(3, "left", "right", "center") Output: Move from left peg to right peg. Move from left peg to center peg. Move from right peg to center peg. Move from left peg to right peg. Move from center peg to left peg. Move from center peg to right peg. Move from left peg to right peg.  ## XQuery declare function local:hanoi($disk as xs:integer, $from as xs:integer,$to as xs:integer, $via as xs:integer) as element()* { if($disk > 0)
then (
local:hanoi($disk - 1,$from, $via,$to),
<move disk='{$disk}'><from>{$from}</from><to>{$to}</to></move>, local:hanoi($disk - 1, $via,$to, $from) ) else () }; <hanoi> { local:hanoi(4, 1, 2, 3) } </hanoi>  Output: <?xml version="1.0" encoding="UTF-8"?> <hanoi> <move disk="1"> <from>1</from> <to>3</to> </move> <move disk="2"> <from>1</from> <to>2</to> </move> <move disk="1"> <from>3</from> <to>2</to> </move> <move disk="3"> <from>1</from> <to>3</to> </move> <move disk="1"> <from>2</from> <to>1</to> </move> <move disk="2"> <from>2</from> <to>3</to> </move> <move disk="1"> <from>1</from> <to>3</to> </move> <move disk="4"> <from>1</from> <to>2</to> </move> <move disk="1"> <from>3</from> <to>2</to> </move> <move disk="2"> <from>3</from> <to>1</to> </move> <move disk="1"> <from>2</from> <to>1</to> </move> <move disk="3"> <from>3</from> <to>2</to> </move> <move disk="1"> <from>1</from> <to>3</to> </move> <move disk="2"> <from>1</from> <to>2</to> </move> <move disk="1"> <from>3</from> <to>2</to> </move> </hanoi>  ## XSLT <xsl:template name="hanoi"> <xsl:param name="n"/> <xsl:param name="from">left</xsl:param> <xsl:param name="to">middle</xsl:param> <xsl:param name="via">right</xsl:param> <xsl:if test="$n &gt; 0">
<xsl:call-template name="hanoi">
<xsl:with-param name="n"    select="$n - 1"/> <xsl:with-param name="from" select="$from"/>
<xsl:with-param name="to"   select="$via"/> <xsl:with-param name="via" select="$to"/>
</xsl:call-template>
<fo:block>
<xsl:text>Move disk from </xsl:text>
<xsl:value-of select="$from"/> <xsl:text> to </xsl:text> <xsl:value-of select="$to"/>
</fo:block>
<xsl:call-template name="hanoi">
<xsl:with-param name="n"    select="$n - 1"/> <xsl:with-param name="from" select="$via"/>
<xsl:with-param name="to"   select="$to"/> <xsl:with-param name="via" select="$from"/>
</xsl:call-template>
</xsl:if>
</xsl:template>

<xsl:call-template name="hanoi"><xsl:with-param name="n" select="4"/></xsl:call-template>


## Yabasic

sub hanoi(ndisks, startPeg, endPeg)
if ndisks then
hanoi(ndisks-1, startPeg, 6-startPeg-endPeg)
//print "Move disk ", ndisks, " from ", startPeg, " to ", endPeg
hanoi(ndisks-1, 6-startPeg-endPeg, endPeg)
end if
end sub

print "Hanoi 1 ellapsed ... ";

t1 = peek("millisrunning")
hanoi(22, 1, 3)
t2 = peek("millisrunning")
print t2-t1, " ms"

sub hanoi2(n, from, to_, via)
if n = 1 then
//print "Move from ", from, " to ", to_
else
hanoi2(n - 1, from, via , to_ )
hanoi2(1    , from, to_ , via )
hanoi2(n - 1, via , to_ , from)
end if
end sub

print "Hanoi 2 ellapsed ... ";
hanoi2(22, 1, 3, 2)
print peek("millisrunning") - t2, " ms"

## Zig

Translation of: C
const std = @import("std");

pub fn print(from: u32, to: u32) void {
std.log.info("Moving disk from rod {} to rod {}", .{ from, to });
}

pub fn move(n: u32, from: u32, via: u32, to: u32) void {
if (n > 1) {
move(n - 1, from, to, via);
print(from, to);
move(n - 1, via, from, to);
} else {
print(from, to);
}
}

pub fn main() !void {
move(4, 1, 2, 3);
}


## zkl

Translation of: C
fcn move(n, from,to,via){
if (n>0){
move(n-1, from,via,to);
println("Move disk from pole %d to pole %d".fmt(from, to));
move(n-1, via,to,from);
}
}
move(3, 1,2,3);
Output:
Move disk from pole 1 to pole 2
Move disk from pole 1 to pole 3
Move disk from pole 2 to pole 3
Move disk from pole 1 to pole 2
Move disk from pole 3 to pole 1
Move disk from pole 3 to pole 2
Move disk from pole 1 to pole 2
`