# Ackermann function

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

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

The Ackermann function is usually defined as follows:

${\displaystyle A(m,n)={\begin{cases}n+1&{\mbox{if }}m=0\\A(m-1,1)&{\mbox{if }}m>0{\mbox{ and }}n=0\\A(m-1,A(m,n-1))&{\mbox{if }}m>0{\mbox{ and }}n>0.\end{cases}}}$

Its arguments are never negative and it always terminates. Write a function which returns the value of ${\displaystyle A(m,n)}$. Arbitrary precision is preferred (since the function grows so quickly), but not required.

## 360 Assembly

Translation of: AWK

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

*        Ackermann function        07/09/2015&LAB     XDECO &REG,&TARGET.*-----------------------------------------------------------------*.*       THIS MACRO DISPLAYS THE REGISTER CONTENTS AS A TRUE       *.*       DECIMAL VALUE. XDECO IS NOT PART OF STANDARD S360 MACROS! **------------------------------------------------------------------*         AIF   (T'&REG EQ 'O').NOREG         AIF   (T'&TARGET EQ 'O').NODEST&LAB     B     I&SYSNDX               BRANCH AROUND WORK AREAW&SYSNDX DS    XL8                    CONVERSION WORK AREAI&SYSNDX CVD   &REG,W&SYSNDX          CONVERT TO DECIMAL         MVC   &TARGET,=XL12'402120202020202020202020'         ED    &TARGET,W&SYSNDX+2     MAKE FIELD PRINTABLE         BC    2,*+12                 BYPASS NEGATIVE         MVI   &TARGET+12,C'-'        INSERT NEGATIVE SIGN         B     *+8                    BYPASS POSITIVE         MVI   &TARGET+12,C'+'        INSERT POSITIVE SIGN         MEXIT.NOREG   MNOTE 8,'INPUT REGISTER OMITTED'         MEXIT.NODEST  MNOTE 8,'TARGET FIELD OMITTED'         MEXIT         MENDACKERMAN CSECT         USING  ACKERMAN,R12       r12 : base register         LR     R12,R15            establish base register         ST     R14,SAVER14A       save r14         LA     R4,0               m=0LOOPM    CH     R4,=H'3'           do m=0 to 3         BH     ELOOPM         LA     R5,0               n=0LOOPN    CH     R5,=H'8'           do n=0 to 8                  BH     ELOOPN         LR     R1,R4              m         LR     R2,R5              n         BAL    R14,ACKER          r1=acker(m,n)         XDECO  R1,PG+19         XDECO  R4,XD         MVC    PG+10(2),XD+10         XDECO  R5,XD         MVC    PG+13(2),XD+10         XPRNT  PG,44              print buffer         LA     R5,1(R5)           n=n+1         B      LOOPNELOOPN   LA     R4,1(R4)           m=m+1         B      LOOPMELOOPM   L      R14,SAVER14A       restore r14         BR     R14                return to callerSAVER14A DS     F                  static save r14PG       DC     CL44'Ackermann(xx,xx) = xxxxxxxxxxxx'XD       DS     CL12ACKER    CNOP   0,4                function r1=acker(r1,r2)         LR     R3,R1              save argument r1 in r3         LR     R9,R10             save stackptr (r10) in r9 temp         LA     R1,STACKLEN        amount of storage required         GETMAIN RU,LV=(R1)        allocate storage for stack         USING  STACK,R10          make storage addressable         LR     R10,R1             establish stack addressability         ST     R14,SAVER14B       save previous r14         ST     R9,SAVER10B        save previous r10         LR     R1,R3              restore saved argument r1START    ST     R1,M               stack m         ST     R2,N               stack nIF1      C      R1,=F'0'           if m<>0         BNE    IF2                then goto if2         LR     R11,R2             n         LA     R11,1(R11)         return n+1         B      EXITIF2      C      R2,=F'0'           else if m<>0         BNE    IF3                then goto if3         BCTR   R1,0               m=m-1         LA     R2,1               n=1         BAL    R14,ACKER          r1=acker(m)         LR     R11,R1             return acker(m-1,1)         B      EXITIF3      BCTR   R2,0               n=n-1         BAL    R14,ACKER          r1=acker(m,n-1)         LR     R2,R1              acker(m,n-1)         L      R1,M               m         BCTR   R1,0               m=m-1                  BAL    R14,ACKER          r1=acker(m-1,acker(m,n-1))         LR     R11,R1             return acker(m-1,1)EXIT     L      R14,SAVER14B       restore r14         L      R9,SAVER10B        restore r10 temp         LA     R0,STACKLEN        amount of storage to free         FREEMAIN A=(R10),LV=(R0)  free allocated storage         LR     R1,R11             value returned         LR     R10,R9             restore r10         BR     R14                return to caller         LTORG         DROP   R12                base no longer neededSTACK    DSECT                     dynamic areaSAVER14B DS     F                  saved r14SAVER10B DS     F                  saved r10M        DS     F                  mN        DS     F                  nSTACKLEN EQU    *-STACK         YREGS           END    ACKERMAN
Output:
Ackermann( 0, 0) =            1
Ackermann( 0, 1) =            2
Ackermann( 0, 2) =            3
Ackermann( 0, 3) =            4
Ackermann( 0, 4) =            5
Ackermann( 0, 5) =            6
Ackermann( 0, 6) =            7
Ackermann( 0, 7) =            8
Ackermann( 0, 8) =            9
Ackermann( 1, 0) =            2
Ackermann( 1, 1) =            3
Ackermann( 1, 2) =            4
Ackermann( 1, 3) =            5
Ackermann( 1, 4) =            6
Ackermann( 1, 5) =            7
Ackermann( 1, 6) =            8
Ackermann( 1, 7) =            9
Ackermann( 1, 8) =           10
Ackermann( 2, 0) =            3
Ackermann( 2, 1) =            5
Ackermann( 2, 2) =            7
Ackermann( 2, 3) =            9
Ackermann( 2, 4) =           11
Ackermann( 2, 5) =           13
Ackermann( 2, 6) =           15
Ackermann( 2, 7) =           17
Ackermann( 2, 8) =           19
Ackermann( 3, 0) =            5
Ackermann( 3, 1) =           13
Ackermann( 3, 2) =           29
Ackermann( 3, 3) =           61
Ackermann( 3, 4) =          125
Ackermann( 3, 5) =          253
Ackermann( 3, 6) =          509
Ackermann( 3, 7) =         1021
Ackermann( 3, 8) =         2045


## 68000 Assembly

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

;; Ackermann function for Motorola 68000 under AmigaOs 2+ by Thorham;; Set stack space to 60000 for m = 3, n = 5.;; The program will print the ackermann values for the range m = 0..3, n = 0..5;_LVOOpenLibrary equ -552_LVOCloseLibrary equ -414_LVOVPrintf equ -954 m equ 3 ; Nr of iterations for the main loop.n equ 5 ; Do NOT set them higher, or it will take hours to complete on        ; 68k, not to mention that the stack usage will become astronomical.        ; Perhaps n can be a little higher... If you do increase the ranges        ; then don't forget to increase the stack size. execBase=4 start    move.l  execBase,a6     lea     dosName,a1    moveq   #36,d0    jsr     _LVOOpenLibrary(a6)    move.l  d0,dosBase    beq     exit     move.l  dosBase,a6    lea     printfArgs,a2     clr.l   d3 ; m.loopn    clr.l   d4 ; n.loopm    bsr     ackermann     move.l  d3,0(a2)    move.l  d4,4(a2)    move.l  d5,8(a2)    move.l  #outString,d1    move.l  a2,d2    jsr     _LVOVPrintf(a6)     addq.l  #1,d4    cmp.l   #n,d4    ble     .loopm     addq.l  #1,d3    cmp.l   #m,d3    ble     .loopn exit    move.l  execBase,a6    move.l  dosBase,a1    jsr     _LVOCloseLibrary(a6)    rts;; ackermann function;; in:;; d3 = m; d4 = n;; out:;; d5 = ans;ackermann    move.l  d3,-(sp)    move.l  d4,-(sp)     tst.l   d3    bne     .l1    move.l  d4,d5    addq.l  #1,d5    bra     .return.l1    tst.l   d4    bne     .l2    subq.l  #1,d3    moveq   #1,d4    bsr     ackermann    bra     .return.l2    subq.l  #1,d4    bsr     ackermann    move.l  d5,d4    subq.l  #1,d3    bsr     ackermann .return    move.l  (sp)+,d4    move.l  (sp)+,d3    rts;; variables;dosBase    dc.l    0 printfArgs    dcb.l   3;; strings;dosName    dc.b    "dos.library",0 outString    dc.b    "ackermann (%ld,%ld) is: %ld",10,0

## 8th

 \ Ackermann function, illustrating use of "memoization". \ Memoization is a technique whereby intermediate computed values are stored\ away against later need.  It is particularly valuable when calculating those\ values is time or resource intensive, as with the Ackermann function. \ make the stack much bigger so this can complete!100000 stack-size \ This is where memoized values are stored:{} var, dict \ Simple accessor words: dict! \ "key" val --  dict @ -rot m:! drop ; : [email protected] \ "key" -- val  dict @ swap m:@ nip ; defer: ack1 \ We just jam the string representation of the two numbers together for a key:: makeKey  \ m n -- m n key	2dup >s swap >s s:+ ; : ack2 \ m n -- A	makeKey dup	[email protected] null?	if \ can't find key in dict		\ m n key null		drop \ m n key		-rot \ key m n		ack1 \ key A		tuck \ A key A		dict! \ A	else \ found value		\ m n key value		>r drop 2drop r>	then ; : ack \ m n -- A	over not 	if		nip n:1+	else		dup not		if			drop n:1- 1 ack2		else			over swap n:1- ack2			swap n:1- swap ack2		then	then ; ' ack is ack1 : ackOf \ m n --        2dup        "Ack(" . swap . ", " . . ") = " . ack . cr ;  0 0 ackOf0 4 ackOf1 0 ackOf1 1 ackOf2 1 ackOf2 2 ackOf3 1 ackOf3 3 ackOf4 0 ackOf \ this last requires a very large data stack.  So start 8th with a parameter '-k 100000'4 1 ackOf bye
The output:
Ack(0, 0) = 1
Ack(0, 4) = 5
Ack(1, 0) = 2
Ack(1, 1) = 3
Ack(2, 1) = 5
Ack(2, 2) = 7
Ack(3, 1) = 13
Ack(3, 3) = 61
Ack(4, 0) = 13
Ack(4, 1) = 65533


## ABAP

 REPORT  zhuberv_ackermann. CLASS zcl_ackermann DEFINITION.  PUBLIC SECTION.    CLASS-METHODS ackermann IMPORTING m TYPE i                                      n TYPE i                            RETURNING value(v) TYPE i.ENDCLASS.            "zcl_ackermann DEFINITION  CLASS zcl_ackermann IMPLEMENTATION.   METHOD: ackermann.     DATA: lv_new_m TYPE i,          lv_new_n TYPE i.     IF m = 0.      v = n + 1.    ELSEIF m > 0 AND n = 0.      lv_new_m = m - 1.      lv_new_n = 1.      v = ackermann( m = lv_new_m n = lv_new_n ).    ELSEIF m > 0 AND n > 0.      lv_new_m = m - 1.       lv_new_n = n - 1.      lv_new_n = ackermann( m = m n = lv_new_n ).       v = ackermann( m = lv_new_m n = lv_new_n ).    ENDIF.   ENDMETHOD.                    "ackermann ENDCLASS.                    "zcl_ackermann IMPLEMENTATION  PARAMETERS: pa_m TYPE i,            pa_n TYPE i. DATA: lv_result TYPE i. START-OF-SELECTION.  lv_result = zcl_ackermann=>ackermann( m = pa_m n = pa_n ).  WRITE: / lv_result.

## ActionScript

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

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

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

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

## Agda

Works with: Agda version 2.5.2
 open import Data.Natopen import Data.Nat.Showopen import IO module Ackermann where ack : ℕ -> ℕ -> ℕack zero n = n + 1ack (suc m) zero = ack m 1ack (suc m) (suc n) = ack m (ack (suc m) n) main = run (putStrLn (show (ack 3 9)))

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

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


## ALGOL 68

Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386
PROC test ackermann = VOID: BEGIN   PROC ackermann = (INT m, n)INT:   BEGIN      IF m = 0 THEN         n + 1      ELIF n = 0 THEN         ackermann (m - 1, 1)      ELSE         ackermann (m - 1, ackermann (m, n - 1))      FI   END # ackermann #;    FOR m FROM 0 TO 3 DO      FOR n FROM 0 TO 6 DO         print(ackermann (m, n))      OD;      new line(stand out)   ODEND # test ackermann #;test ackermann
Output:
         +1         +2         +3         +4         +5         +6         +7
+2         +3         +4         +5         +6         +7         +8
+3         +5         +7         +9        +11        +13        +15
+5        +13        +29        +61       +125       +253       +509


## APL

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

## AppleScript

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

## Argile

Works with: Argile version 1.0.0
use std for each (val nat n) from 0 to 6  for each (val nat m) from 0 to 3    print "A("m","n") = "(A m n) .:A <nat m, nat n>:. -> nat  return (n+1)				if m == 0  return (A (m - 1) 1)			if n == 0  A (m - 1) (A m (n - 1))

## ATS

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

## AutoHotkey

A(m, n) {  If (m > 0) && (n = 0)    Return A(m-1,1)  Else If (m > 0) && (n > 0)    Return A(m-1,A(m, n-1))  Else If (m=0)    Return n+1} ; Example:MsgBox, % "A(1,2) = " A(1,2)

## AutoIt

Func Ackermann($m,$n)    If ($m = 0) Then Return$n+1    Else        If ($n = 0) Then Return Ackermann($m-1, 1)        Else            return Ackermann($m-1, Ackermann($m, $n-1)) EndIf EndIfEndFunc ### Classical + cache implementation This version works way faster than the classical one: Ackermann(3, 5) runs in 12,7 ms, while the classical version takes 402,2 ms. Global$ackermann[2047][2047] ; Set the size to whatever you wantFunc Ackermann($m,$n)	If ($ackermann[$m][$n] <> 0) Then Return$ackermann[$m][$n]	Else		If ($m = 0) Then$return = $n + 1 Else If ($n = 0) Then				$return = Ackermann($m - 1, 1)			Else				$return = Ackermann($m - 1, Ackermann($m,$n - 1))			EndIf		EndIf		$ackermann[$m][$n] =$return		Return $return EndIfEndFunc ;==>Ackermann ## AWK function ackermann(m, n) { if ( m == 0 ) { return n+1 } if ( n == 0 ) { return ackermann(m-1, 1) } return ackermann(m-1, ackermann(m, n-1))} BEGIN { for(n=0; n < 7; n++) { for(m=0; m < 4; m++) { print "A(" m "," n ") = " ackermann(m,n) } }} ## Babel main: {((0 0) (0 1) (0 2) (0 3) (0 4) (1 0) (1 1) (1 2) (1 3) (1 4) (2 0) (2 1) (2 2) (2 3) (3 0) (3 1) (3 2) (4 0)) { dup "A(" << { %d " " . << } ... ") = " << reverse give ack %d cr << } ... } ack!: { dup zero? { <-> dup zero? { <-> cp 1 - <- <- 1 - -> ack -> ack } { <-> 1 - <- 1 -> ack } if } { zap 1 + } if } zero?!: { 0 = }  Output: A(0 0 ) = 1 A(0 1 ) = 2 A(0 2 ) = 3 A(0 3 ) = 4 A(0 4 ) = 5 A(1 0 ) = 2 A(1 1 ) = 3 A(1 2 ) = 4 A(1 3 ) = 5 A(1 4 ) = 6 A(2 0 ) = 3 A(2 1 ) = 5 A(2 2 ) = 7 A(2 3 ) = 9 A(3 0 ) = 5 A(3 1 ) = 13 A(3 2 ) = 29 A(4 0 ) = 13 ## BASIC Works with: QuickBasic version 4.5 BASIC runs out of stack space very quickly. The call ack(3, 4) gives a stack error. DECLARE FUNCTION ack! (m!, n!) FUNCTION ack (m!, n!) IF m = 0 THEN ack = n + 1 IF m > 0 AND n = 0 THEN ack = ack(m - 1, 1) END IF IF m > 0 AND n > 0 THEN ack = ack(m - 1, ack(m, n - 1)) END IFEND FUNCTION ## BASIC256 dim stack(5000, 3) # BASIC-256 lacks functions (as of ver. 0.9.6.66)stack[0,0] = 3 # Mstack[0,1] = 7 # Nlev = 0 gosub ackermannprint "A("+stack[0,0]+","+stack[0,1]+") = "+stack[0,2]end ackermann: if stack[lev,0]=0 then stack[lev,2] = stack[lev,1]+1 return end if if stack[lev,1]=0 then lev = lev+1 stack[lev,0] = stack[lev-1,0]-1 stack[lev,1] = 1 gosub ackermann stack[lev-1,2] = stack[lev,2] lev = lev-1 return end if lev = lev+1 stack[lev,0] = stack[lev-1,0] stack[lev,1] = stack[lev-1,1]-1 gosub ackermann stack[lev,0] = stack[lev-1,0]-1 stack[lev,1] = stack[lev,2] gosub ackermann stack[lev-1,2] = stack[lev,2] lev = lev-1 return Output:  A(3,7) = 1021  # BASIC256 since 0.9.9.1 supports functionsfor m = 0 to 3 for n = 0 to 4 print m + " " + n + " " + ackermann(m,n) next nnext mend function ackermann(m,n) if m = 0 then ackermann = n+1 else if n = 0 then ackermann = ackermann(m-1,1) else ackermann = ackermann(m-1,ackermann(m,n-1)) endif end ifend function Output: 0 0 1 0 1 2 0 2 3 0 3 4 0 4 5 1 0 2 1 1 3 1 2 4 1 3 5 1 4 6 2 0 3 2 1 5 2 2 7 2 3 9 2 4 11 3 0 5 3 1 13 3 2 29 3 3 61 3 4 125 ## Batch File Had trouble with this, so called in the gurus at StackOverflow. Thanks to Patrick Cuff for pointing out where I was going wrong. ::Ackermann.cmd@echo offset depth=0:ackif %1==0 goto m0if %2==0 goto n0 :elseset /a n=%2-1set /a depth+=1call :ack %1 %n%set t=%errorlevel%set /a depth-=1set /a m=%1-1set /a depth+=1call :ack %m% %t%set t=%errorlevel%set /a depth-=1if %depth%==0 ( exit %t% ) else ( exit /b %t% ) :m0set/a n=%2+1if %depth%==0 ( exit %n% ) else ( exit /b %n% ) :n0set /a m=%1-1set /a depth+=1call :ack %m% 1set t=%errorlevel%set /a depth-=1if %depth%==0 ( exit %t% ) else ( exit /b %t% ) Because of the exit statements, running this bare closes one's shell, so this test routine handles the calling of Ackermann.cmd ::Ack.cmd@echo offcmd/c ackermann.cmd %1 %2echo Ackermann(%1, %2)=%errorlevel% A few test runs: D:\Documents and Settings\Bruce>ack 0 4 Ackermann(0, 4)=5 D:\Documents and Settings\Bruce>ack 1 4 Ackermann(1, 4)=6 D:\Documents and Settings\Bruce>ack 2 4 Ackermann(2, 4)=11 D:\Documents and Settings\Bruce>ack 3 4 Ackermann(3, 4)=125 ## BBC BASIC  PRINT FNackermann(3, 7) END DEF FNackermann(M%, N%) IF M% = 0 THEN = N% + 1 IF N% = 0 THEN = FNackermann(M% - 1, 1) = FNackermann(M% - 1, FNackermann(M%, N%-1)) ## bc Requires a bc that supports long names and the print statement. Works with: OpenBSD bc Works with: GNU bc define ack(m, n) { if ( m == 0 ) return (n+1); if ( n == 0 ) return (ack(m-1, 1)); return (ack(m-1, ack(m, n-1)));} for (n=0; n<7; n++) { for (m=0; m<4; m++) { print "A(", m, ",", n, ") = ", ack(m,n), "\n"; }}quit ## BCPL GET "libhdr" LET ack(m, n) = m=0 -> n+1, n=0 -> ack(m-1, 1), ack(m-1, ack(m, n-1)) LET start() = VALOF{ FOR i = 0 TO 6 FOR m = 0 TO 3 DO writef("ack(%n, %n) = %n*n", m, n, ack(m,n)) RESULTIS 0} ## beeswax Iterative slow version:  >[email protected]@[email protected] if m>0 and n>0 => replace m,n with m-1,m,n-1 >@[email protected]'[email protected]@[email protected] if m>0 and n=0 => replace m,n with m-1,1_ii>[email protected]'d?g?Pfzp if m=0 => replace m,n with n+1 >I; b < <  A functional and recursive realization of the version above. Functions are realized by direct calls of functions via jumps (instruction J) to the entry points of two distinct functions: 1st function _ii (input function) with entry point at (row,col) = (4,1) 2nd function Ag~1.... (Ackermann function) with entry point at (row,col) = (1,1) Each block of 1FJ or 1fFJ in the code is a call of the Ackermann function itself. [email protected]'p?g?Pf1FJ Ackermann function. if m=0 => run Ackermann function (m, n+1) xI; [email protected]'[email protected] if m>0 and n=0 => run Ackermann (m-1,1) [email protected][email protected] if m>0 and n>0 => run Ackermann(m,Ackermann(m-1,n-1))_ii1FJ input function. Input m,n, then execute Ackermann(m,n) Highly optimized and fast version, returns A(4,1)/A(5,0) almost instantaneously:  >[email protected][email protected]@[email protected]@Php if m>4 and n>0 => replace m,n with m-1,m,n-1 >~4~L#1~2hg'[email protected]?f2h p if m>4 and n=0 => replace m,n with m-1,1 # q < /n+3 times \ #X~4K#?2Fg?PPP>@[email protected]"pb if m=4 => replace m,n with 2^(2^(....)))-3 # >~3K#?g?PPP~2BMMp>@MMMp if m=3 => replace m,n with 2^(n+3)-3_ii>[email protected]'#?g?P p M if m=0 => replace m,n with n+1 z I ~>~1K#?g?PP p if m=1 => replace m,n with n+2 f ; >2K#?g?~2.PPPp if m=2 => replace m,n with 2n+3 z b < < < d <  Higher values than A(4,1)/(5,0) lead to UInt64 wraparound, support for numbers bigger than 2^64-1 is not implemented in these solutions. ## Befunge ### Befunge-93 Translation of: ERRE Since Befunge-93 doesn't have recursive capabilities we need to use an iterative algorithm. &>&>vvg0>#0\#-:#1_1v@v:\<vp0 0:-1<\+<^>00p>:#^_$1+\:#^_$.  ### Befunge-98 Works with: CCBI version 2.1 r[1&&{0>v ju>[email protected] 1> \:v^ v:\_$1+\^v_$1\1-u^>1-0fp:1-\0fg101- The program reads two integers (first m, then n) from command line, idles around funge space, then outputs the result of the Ackerman function. Since the latter is calculated truly recursively, the execution time becomes unwieldy for most m>3. ## Bracmat Three solutions are presented here. The first one is a purely recursive version, only using the formulas at the top of the page. The value of A(4,1) cannot be computed due to stack overflow. It can compute A(3,9) (4093), but probably not A(3,10) ( Ack= m n . !arg:(?m,?n) & ( !m:0&!n+1 | !n:0&Ack$(!m+-1,1)      | Ack$(!m+-1,Ack$(!m,!n+-1))      ));

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

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

  ( A  =     m n value key eq chain      , find insert future stack va val    .   ( chain        =   key future skey          .   !arg:(?key.?future)            & str$!key:?skey & (cache..insert)$(!skey..!future)            &         )      & (find=.(cache..find)$(str$!arg))      & ( insert        =   key value future v futureeq futurem skey          .   !arg:(?key.?value)            & str$!key:?skey & ( (cache..find)$!skey:(?key.?v.?future)                & (cache..remove)$!skey & (cache..insert)$(!skey.!value.)                & (   !future:(?futurem.?futureeq)                    & (!futurem,!value.!futureeq)                  |                   )              | (cache..insert)$(!skey.!value.)& ) ) & !arg:(?m,?n) & !n+1:?value & :?eq:?stack & whl ' ( (!m,!n):?key & ( find$!key:(?.#%?value.?future)                  & insert$(!eq.!value) !future | !m:0 & !n+1:?value & ( !eq:&insert$(!key.!value)                    |   insert$(!key.!value) !stack:?stack & insert$(!eq.!value)                    )                |   !n:0                  &   (!m+-1,1.!key)                      (!eq:|(!key.!eq))                |   find$(!m,!n+-1):(?.?val.?) & ( !val:#% & ( find$(!m+-1,!val):(?.?va.?)                          & !va:#%                          & insert$(!key.!va) | (!m+-1,!val.!eq) (!m,!n.!eq) ) | ) | chain$(!m,!n+-1.!m+-1.!key)                  &   (!m,!n+-1.)                      (!eq:|(!key.!eq))                )                !stack            : (?m,?n.?eq) ?stack          )      & !value  )& new$hash:?cache Some results: A$(0,0):1
A$(3,13):65533 A$(3,14):131069
A$(4,1):65533  The last solution is a recursive solution that employs some extra formulas, inspired by the Common Lisp solution further down. ( AckFormula= m n . !arg:(?m,?n) & ( !m:0&!n+1 | !m:1&!n+2 | !m:2&2*!n+3 | !m:3&2^(!n+3)+-3 | !n:0&AckFormula$(!m+-1,1)      | AckFormula$(!m+-1,AckFormula$(!m,!n+-1))      ))
Some results:
AckFormula$(4,1):65533 AckFormula$(4,2):2003529930406846464979072351560255750447825475569751419265016973.....22087777506072339445587895905719156733


The last computation costs about 0,03 seconds.

## Brat

ackermann = { m, n |	when { m == 0 } { n + 1 }		{ m > 0 && n == 0 } { ackermann(m - 1, 1) }		{ m > 0 && n > 0 } { ackermann(m - 1, ackermann(m, n - 1)) }} p ackermann 3, 4  #Prints 125

## C

Straightforward implementation per Ackermann definition:

#include <stdio.h> int ackermann(int m, int n){        if (!m) return n + 1;        if (!n) return ackermann(m - 1, 1);        return ackermann(m - 1, ackermann(m, n - 1));} int main(){        int m, n;        for (m = 0; m <= 4; m++)                for (n = 0; n < 6 - m; n++)                        printf("A(%d, %d) = %d\n", m, n, ackermann(m, n));         return 0;}
Output:
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(4, 0) = 13
A(4, 1) = 65533

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

#include <stdio.h>#include <stdlib.h>#include <string.h> int m_bits, n_bits;int *cache; int ackermann(int m, int n){        int idx, res;        if (!m) return n + 1;         if (n >= 1<<n_bits) {                printf("%d, %d\n", m, n);                idx = 0;        } else {                idx = (m << n_bits) + n;                if (cache[idx]) return cache[idx];        }         if (!n) res = ackermann(m - 1, 1);        else    res = ackermann(m - 1, ackermann(m, n - 1));         if (idx) cache[idx] = res;        return res;}int main(){        int m, n;         m_bits = 3;        n_bits = 20;  /* can save n values up to 2**20 - 1, that's 1 meg */        cache = malloc(sizeof(int) * (1 << (m_bits + n_bits)));        memset(cache, 0, sizeof(int) * (1 << (m_bits + n_bits)));         for (m = 0; m <= 4; m++)                for (n = 0; n < 6 - m; n++)                        printf("A(%d, %d) = %d\n", m, n, ackermann(m, n));         return 0;}
Output:
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(4, 0) = 13
A(4, 1) = 65533

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

## C#

### Basic Version

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


### Efficient Version

 using System;using System.Numerics;using System.IO;using System.Diagnostics; namespace Ackermann_Function{    class Program    {        static void Main(string[] args)        {            int _m = 0;            int _n = 0;            Console.Write("m = ");            try            {                _m = Convert.ToInt32(Console.ReadLine());            }            catch (Exception)            {                Console.WriteLine("Please enter a number.");            }            Console.Write("n = ");            try            {                _n = Convert.ToInt32(Console.ReadLine());            }            catch (Exception)            {                Console.WriteLine("Please enter a number.");            }            //for (long m = 0; m <= 10; ++m)            //{            //    for (long n = 0; n <= 10; ++n)            //    {            //        DateTime now = DateTime.Now;            //        Console.WriteLine("Ackermann({0}, {1}) = {2}", m, n, Ackermann(m, n));            //        Console.WriteLine("Time taken:{0}", DateTime.Now - now);            //    }            //}             DateTime now = DateTime.Now;            Console.WriteLine("Ackermann({0}, {1}) = {2}", _m, _n, Ackermann(_m, _n));            Console.WriteLine("Time taken:{0}", DateTime.Now - now);            File.WriteAllText("number.txt", Ackermann(_m, _n).ToString());            Process.Start("number.txt");            Console.ReadKey();        }        public class OverflowlessStack<T>        {            internal sealed class SinglyLinkedNode            {                private const int ArraySize = 2048;                T[] _array;                int _size;                public SinglyLinkedNode Next;                public SinglyLinkedNode()                {                    _array = new T[ArraySize];                }                public bool IsEmpty { get { return _size == 0; } }                public SinglyLinkedNode Push(T item)                {                    if (_size == ArraySize - 1)                    {                        SinglyLinkedNode n = new SinglyLinkedNode();                        n.Next = this;                        n.Push(item);                        return n;                    }                    _array[_size++] = item;                    return this;                }                public T Pop()                {                    return _array[--_size];                }            }            private SinglyLinkedNode _head = new SinglyLinkedNode();             public T Pop()            {                T ret = _head.Pop();                if (_head.IsEmpty && _head.Next != null)                    _head = _head.Next;                return ret;            }            public void Push(T item)            {                _head = _head.Push(item);            }            public bool IsEmpty            {                get { return _head.Next == null && _head.IsEmpty; }            }        }        public static BigInteger Ackermann(BigInteger m, BigInteger n)        {            var stack = new OverflowlessStack<BigInteger>();            stack.Push(m);            while (!stack.IsEmpty)            {                m = stack.Pop();            skipStack:                if (m == 0)                    n = n + 1;                else if (m == 1)                    n = n + 2;                else if (m == 2)                    n = n * 2 + 3;                else if (n == 0)                {                    --m;                    n = 1;                    goto skipStack;                }                else                {                    stack.Push(m - 1);                    --n;                    goto skipStack;                }            }            return n;        }    }}

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

## C++

### Basic version

#include <iostream> unsigned int ackermann(unsigned int m, unsigned int n) {  if (m == 0) {    return n + 1;  }  if (n == 0) {    return ackermann(m - 1, 1);  }  return ackermann(m - 1, ackermann(m, n - 1));} int main() {  for (unsigned int m = 0; m < 4; ++m) {    for (unsigned int n = 0; n < 10; ++n) {      std::cout << "A(" << m << ", " << n << ") = " << ackermann(m, n) << "\n";    }  }}

### Efficient version

Translation of: D

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

g++ -std=c++11 -I /path/to/boost ackermann.cpp.

#include <iostream>#include <sstream>#include <string>#include <boost/multiprecision/cpp_int.hpp> using big_int = boost::multiprecision::cpp_int; big_int ipow(big_int base, big_int exp) {  big_int result(1);  while (exp) {    if (exp & 1) {      result *= base;    }    exp >>= 1;    base *= base;  }  return result;} big_int ackermann(unsigned m, unsigned n) {  static big_int (*ack)(unsigned, big_int) =      [](unsigned m, big_int n)->big_int {    switch (m) {    case 0:      return n + 1;    case 1:      return n + 2;    case 2:      return 3 + 2 * n;    case 3:      return 5 + 8 * (ipow(big_int(2), n) - 1);    default:      return n == 0 ? ack(m - 1, big_int(1)) : ack(m - 1, ack(m, n - 1));    }  };  return ack(m, big_int(n));} int main() {  for (unsigned m = 0; m < 4; ++m) {    for (unsigned n = 0; n < 10; ++n) {      std::cout << "A(" << m << ", " << n << ") = " << ackermann(m, n) << "\n";    }  }   std::cout << "A(4, 1) = " << ackermann(4, 1) << "\n";   std::stringstream ss;  ss << ackermann(4, 2);  auto text = ss.str();  std::cout << "A(4, 2) = (" << text.length() << " digits)\n"            << text.substr(0, 80) << "\n...\n"            << text.substr(text.length() - 80) << "\n";}
<pre>
A(0, 0) = 1
A(0, 1) = 2
A(0, 2) = 3
A(0, 3) = 4
A(0, 4) = 5
A(0, 5) = 6
A(0, 6) = 7
A(0, 7) = 8
A(0, 8) = 9
A(0, 9) = 10
A(1, 0) = 2
A(1, 1) = 3
A(1, 2) = 4
A(1, 3) = 5
A(1, 4) = 6
A(1, 5) = 7
A(1, 6) = 8
A(1, 7) = 9
A(1, 8) = 10
A(1, 9) = 11
A(2, 0) = 3
A(2, 1) = 5
A(2, 2) = 7
A(2, 3) = 9
A(2, 4) = 11
A(2, 5) = 13
A(2, 6) = 15
A(2, 7) = 17
A(2, 8) = 19
A(2, 9) = 21
A(3, 0) = 5
A(3, 1) = 13
A(3, 2) = 29
A(3, 3) = 61
A(3, 4) = 125
A(3, 5) = 253
A(3, 6) = 509
A(3, 7) = 1021
A(3, 8) = 2045
A(3, 9) = 4093
A(4, 1) = 65533
A(4, 2) = (19729 digits)
2003529930406846464979072351560255750447825475569751419265016973710894059556311
...
4717124577965048175856395072895337539755822087777506072339445587895905719156733


## Chapel

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

## Clay

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

## CLIPS

Functional solution

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


Fact-based solution

(deffacts solve-items  (solve 0 4)  (solve 1 4)  (solve 2 4)  (solve 3 4)) (defrule acker-m-0  ?compute <- (compute 0 ?n)  =>  (retract ?compute)  (assert (ackerman 0 ?n (+ ?n 1)))) (defrule acker-n-0-pre  (compute ?m&:(> ?m 0) 0)  (not (ackerman =(- ?m 1) 1 ?))  =>  (assert (compute (- ?m 1) 1))) (defrule acker-n-0  ?compute <- (compute ?m&:(> ?m 0) 0)  (ackerman =(- ?m 1) 1 ?val)  =>  (retract ?compute)  (assert (ackerman ?m 0 ?val))) (defrule acker-m-n-pre-1  (compute ?m&:(> ?m 0) ?n&:(> ?n 0))  (not (ackerman ?m =(- ?n 1) ?))  =>  (assert (compute ?m (- ?n 1)))) (defrule acker-m-n-pre-2  (compute ?m&:(> ?m 0) ?n&:(> ?n 0))  (ackerman ?m =(- ?n 1) ?newn)  (not (ackerman =(- ?m 1) ?newn ?))  =>  (assert (compute (- ?m 1) ?newn))) (defrule acker-m-n  ?compute <- (compute ?m&:(> ?m 0) ?n&:(> ?n 0))  (ackerman ?m =(- ?n 1) ?newn)  (ackerman =(- ?m 1) ?newn ?val)  =>  (retract ?compute)  (assert (ackerman ?m ?n ?val))) (defrule acker-solve  (solve ?m ?n)  (not (compute ?m ?n))  (not (ackerman ?m ?n ?))  =>  (assert (compute ?m ?n))) (defrule acker-solved  ?solve <- (solve ?m ?n)  (ackerman ?m ?n ?result)  =>  (retract ?solve)  (printout t "A(" ?m "," ?n ") = " ?result crlf))

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

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

## Clojure

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

## COBOL

       IDENTIFICATION DIVISION.       PROGRAM-ID. Ackermann.        DATA DIVISION.       LINKAGE SECTION.       01  M          USAGE UNSIGNED-LONG.       01  N          USAGE UNSIGNED-LONG.        01  Return-Val USAGE UNSIGNED-LONG.        PROCEDURE DIVISION USING M N Return-Val.           EVALUATE M ALSO N               WHEN 0 ALSO ANY                   ADD 1 TO N GIVING Return-Val                WHEN NOT 0 ALSO 0                   SUBTRACT 1 FROM M                   CALL "Ackermann" USING BY CONTENT M BY CONTENT 1                       BY REFERENCE Return-Val                WHEN NOT 0 ALSO NOT 0                   SUBTRACT 1 FROM N                   CALL "Ackermann" USING BY CONTENT M BY CONTENT N                       BY REFERENCE Return-Val                    SUBTRACT 1 FROM M                   CALL "Ackermann" USING BY CONTENT M                       BY CONTENT Return-Val BY REFERENCE Return-Val           END-EVALUATE            GOBACK           .

## CoffeeScript

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

## Common Lisp

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

More elaborately:

(defun ackermann (m n)  (case m ((0) (1+ n))    ((1) (+ 2 n))    ((2) (+ n n 3))    ((3) (- (expt 2 (+ 3 n)) 3))    (otherwise (ackermann (1- m) (if (zerop n) 1 (ackermann m (1- n))))))) (loop for m from 0 to 4 do      (loop for n from (- 5 m) to (- 6 m) do	    (format t "A(~d, ~d) = ~d~%" m n (ackermann m n))))
Output:
A(0, 5) = 6
A(0, 6) = 7
A(1, 4) = 6
A(1, 5) = 7
A(2, 3) = 9
A(2, 4) = 11
A(3, 2) = 29
A(3, 3) = 61
A(4, 1) = 65533

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

## Coq

Require Import Arith.Fixpoint A m := fix A_m n :=  match m with    | 0 => n + 1    | S pm =>      match n with        | 0 => A pm 1        | S pn => A pm (A_m pn)      end  end.
Require Import Utf8. Section FOLD.  Context {A: Type} (f: A → A) (a: A).  Fixpoint fold (n: nat) : A :=    match n with    | O => a    | S n' => f (fold n')    end.End FOLD. Definition ackermann : nat → nat → nat :=  fold (λ g, fold g (g (S O))) S.

## Component Pascal

BlackBox Component Builder

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

Execute: ^Q NpctAckerman.Do

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


## D

### Basic version

ulong ackermann(in ulong m, in ulong n) pure nothrow @nogc {    if (m == 0)        return n + 1;    if (n == 0)        return ackermann(m - 1, 1);    return ackermann(m - 1, ackermann(m, n - 1));} void main() {    assert(ackermann(2, 4) == 11);}

### More Efficient Version

Translation of: Mathematica
import std.stdio, std.bigint, std.conv; BigInt ipow(BigInt base, BigInt exp) pure nothrow {    auto result = 1.BigInt;    while (exp) {        if (exp & 1)            result *= base;        exp >>= 1;        base *= base;    }     return result;} BigInt ackermann(in uint m, in uint n) pure nothrowout(result) {    assert(result >= 0);} body {    static BigInt ack(in uint m, in BigInt n) pure nothrow {        switch (m) {            case 0: return n + 1;            case 1: return n + 2;            case 2: return 3 + 2 * n;            //case 3: return 5 + 8 * (2 ^^ n - 1);            case 3: return 5 + 8 * (ipow(2.BigInt, n) - 1);            default: return (n == 0) ?                        ack(m - 1, 1.BigInt) :                        ack(m - 1, ack(m, n - 1));        }    }     return ack(m, n.BigInt);} void main() {    foreach (immutable m; 1 .. 4)        foreach (immutable n; 1 .. 9)            writefln("ackermann(%d, %d): %s", m, n, ackermann(m, n));    writefln("ackermann(4, 1): %s", ackermann(4, 1));     immutable a = ackermann(4, 2).text;    writefln("ackermann(4, 2)) (%d digits):\n%s...\n%s",             a.length, a[0 .. 94], a[$- 96 ..$]);}
Output:
ackermann(1, 1): 3
ackermann(1, 2): 4
ackermann(1, 3): 5
ackermann(1, 4): 6
ackermann(1, 5): 7
ackermann(1, 6): 8
ackermann(1, 7): 9
ackermann(1, 8): 10
ackermann(2, 1): 5
ackermann(2, 2): 7
ackermann(2, 3): 9
ackermann(2, 4): 11
ackermann(2, 5): 13
ackermann(2, 6): 15
ackermann(2, 7): 17
ackermann(2, 8): 19
ackermann(3, 1): 13
ackermann(3, 2): 29
ackermann(3, 3): 61
ackermann(3, 4): 125
ackermann(3, 5): 253
ackermann(3, 6): 509
ackermann(3, 7): 1021
ackermann(3, 8): 2045
ackermann(4, 1): 65533
ackermann(4, 2)) (19729 digits):
2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880...
699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156733

## Dart

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

int A(int m, int n) => m==0 ? n+1 : n==0 ? A(m-1,1) : A(m-1,A(m,n-1)); main() {  print(A(0,0));  print(A(1,0));  print(A(0,1));  print(A(2,2));  print(A(2,3));  print(A(3,3));  print(A(3,4));  print(A(3,5));  print(A(4,0));}

## Delphi

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

## DWScript

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

## Dylan

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

## E

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

## Egel

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

## Eiffel

 note	description: "Example of Ackerman function"	synopsis: "[		The EIS link below (in Eiffel Studio) will launch in either an in-IDE browser or		and external browser (your choice). The protocol informs Eiffel Studio about what		program to use to open the src' reference, which can be URI, PDF, or DOC. See		second EIS for more information.		]"	EIS: "name=Ackermann_function", "protocol=URI", "tag=rosetta_code",		"src=http://rosettacode.org/wiki/Ackermann_function"	EIS: "name=eis_protocols", "protocol=URI", "tag=eiffel_docs",		"src=https://docs.eiffel.com/book/eiffelstudio/protocols" class	APPLICATION create	make feature {NONE} -- Initialization 	make		do			print ("%N A(0,0):" + ackerman (0, 0).out)			print ("%N A(1,0):" + ackerman (1, 0).out)			print ("%N A(0,1):" + ackerman (0, 1).out)			print ("%N A(1,1):" + ackerman (1, 1).out)			print ("%N A(2,0):" + ackerman (2, 0).out)			print ("%N A(2,1):" + ackerman (2, 1).out)			print ("%N A(2,2):" + ackerman (2, 2).out)			print ("%N A(0,2):" + ackerman (0, 2).out)			print ("%N A(1,2):" + ackerman (1, 2).out)			print ("%N A(3,3):" + ackerman (3, 3).out)			print ("%N A(3,4):" + ackerman (3, 4).out)		end feature -- Access 	ackerman (m, n: NATURAL): NATURAL		do			if m = 0 then				Result := n + 1			elseif n = 0 then				Result := ackerman (m - 1, 1)			else				Result := ackerman (m - 1, ackerman (m, n - 1))			end		endend 

## Ela

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

## Elena

ELENA 3.4 :

import extensions. ackermann(m,n)[    if((n < 0)||(m < 0))    [        InvalidArgumentException new; raise    ].     m =>       0 [ ^n + 1 ];       ! [              n =>                 0 [ ^ackermann(m - 1,1) ];                ! [ ^ackermann(m - 1,ackermann(m,n-1)) ]           ]] public program[    0 to:3 do(:i)    [        0 to:5 do(:j)        [            console printLine("A(",i,",",j,")=",ackermann(i,j))        ]    ].     console readChar]
Output:
A(0,0)=1
A(0,1)=2
A(0,2)=3
A(0,3)=4
A(0,4)=5
A(0,5)=6
A(1,0)=2
A(1,1)=3
A(1,2)=4
A(1,3)=5
A(1,4)=6
A(1,5)=7
A(2,0)=3
A(2,1)=5
A(2,2)=7
A(2,3)=9
A(2,4)=11
A(2,5)=13
A(3,0)=5
A(3,1)=13
A(3,2)=29
A(3,3)=61
A(3,4)=125
A(3,5)=253


## Elixir

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


## Erlang

 -module(ackermann).-export([ackermann/2]). ackermann(0, N) ->   N+1;ackermann(M, 0) ->   ackermann(M-1, 1);ackermann(M, N) when M > 0 andalso N > 0 ->  ackermann(M-1, ackermann(M, N-1)). 

## ERRE

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

## Ezhil

 நிரல்பாகம் அகெர்மன்(முதலெண், இரண்டாமெண்)   @((முதலெண் < 0) || (இரண்டாமெண் < 0)) ஆனால்     பின்கொடு -1   முடி   @(முதலெண் == 0) ஆனால்     பின்கொடு இரண்டாமெண்+1   முடி   @((முதலெண் > 0) && (இரண்டாமெண் == 00)) ஆனால்     பின்கொடு அகெர்மன்(முதலெண் - 1, 1)   முடி   பின்கொடு அகெர்மன்(முதலெண் - 1, அகெர்மன்(முதலெண், இரண்டாமெண் - 1)) முடி அ = int(உள்ளீடு("ஓர் எண்ணைத் தாருங்கள், அது பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாக இருக்கலாம்: "))ஆ = int(உள்ளீடு("அதேபோல் இன்னோர் எண்ணைத் தாருங்கள், இதுவும் பூஜ்ஜியமாகவோ, அதைவிடப் பெரியதாகவோ இருக்கலாம்: ")) விடை = அகெர்மன்(அ, ஆ) @(விடை < 0) ஆனால்   பதிப்பி "தவறான எண்களைத் தந்துள்ளீர்கள்!" இல்லை   பதிப்பி "நீங்கள் தந்த எண்களுக்கான அகர்மென் மதிப்பு: ", விடை முடி 

## F#

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

let rec ackermann m n =     match m, n with    | 0, n -> n + 1    | m, 0 -> ackermann (m - 1) 1    | m, n -> ackermann (m - 1) ackermann m (n - 1) do    printfn "%A" (ackermann (int fsi.CommandLineArgs.[1]) (int fsi.CommandLineArgs.[2]))

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

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

## Factor

USING: kernel math locals combinators ;IN: ackermann :: ackermann ( m n -- u )     {         { [ m 0 = ] [ n 1 + ] }         { [ n 0 = ] [ m 1 - 1 ackermann ] }         [ m 1 - m n 1 - ackermann ackermann ]     } cond ;

## Falcon

function ackermann( m, n ) if m == 0:  return( n + 1 ) if n == 0:  return( ackermann( m - 1, 1 ) ) return( ackermann( m - 1, ackermann( m, n - 1 ) ) )end for M in [ 0:4 ] for N in [ 0:7 ]   >> ackermann( M, N ), " " end >end

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

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


## FALSE

[$$[% \$$[%     1-\[email protected]@a;!  { i j -> A(i-1, A(i, j-1)) }  1]?0=[     %1         { i 0 -> A(i-1, 1) }   ]?  \1-a;!1]?0=[  %1+           { 0 j -> j+1 } ]?]a: { j i } 3 3 a;! .  { 61 }

## Haxe

class RosettaDemo{    static public function main()    {        Sys.print(ackermann(3, 4));    }     static function ackermann(m : Int, n : Int)    {        if (m == 0)        {            return n + 1;        }        else if (n == 0)        {            return ackermann(m-1, 1);        }        return ackermann(m-1, ackermann(m, n-1));    }}

## Icon and Unicon

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

procedure acker(i, j)   static memory    initial {      memory := table()      every memory[0 to 100] := table()      }    if i = 0 then return j + 1    if j = 0 then /memory[i][j] := acker(i - 1, 1)   else /memory[i][j] := acker(i - 1, acker(i, j - 1))    return memory[i][j] end procedure main()   every m := 0 to 3 do {      every n := 0 to 8 do {         writes(acker(m, n) || " ")         }      write()      }end
Output:
1 2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9 10
3 5 7 9 11 13 15 17 19
5 13 29 61 125 253 509 1021 2045

## Idris

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

## Ioke

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

## J

As posted at the J wiki

ack=: c1c1c2c3 @. (#[email protected],&*) M.c1=: >:@]                        NB. if 0=x, 1+yc2=: <:@[ ack 1:                 NB. if 0=y, (x-1) ack 1c3=: <:@[ ack [ ack <:@]         NB. else,   (x-1) ack x ack y-1
Example use:
   0 ack 34   1 ack 35   2 ack 39   3 ack 361

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

### Alternative Primitive Recursive Version

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

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

   Ack=. 3 -~ [ ({&(2 4$'>: 2x&+') ::(,&'&1'&'2x&*'@:(-&2))"0@:[ 128!:2 ]) 3 + ] Example use:  0 1 2 3 Ack 0 1 2 3 4 5 6 71 2 3 4 5 6 7 82 3 4 5 6 7 8 93 5 7 9 11 13 15 175 13 29 61 125 253 509 1021 3 4 Ack 0 1 2 5 13 ...13 65533 2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348101038234342907263181822949382118812668869506364761547029165041871916351587966347219442930927982084309104855990570159318959639524863372367203002916... 4 # @: ": @: Ack 2 NB. Number of digits of 4 Ack 219729 5 Ack 065533  A structured derivation of Ack follows:  o=. @: NB. Composition of verbs (functions) x=. o[ NB. Composing the left noun (argument) (rows2up=. ,&'&1'&'2x&*') o i. 4 2x&* 2x&*&1 2x&*&1&1 2x&*&1&1&1 NB. 2's multiplication, exponentiation, tetration, pentation, etc. 0 1 2 (BuckTruncated=. (rows2up x apply ]) f.) 0 1 2 3 4 50 2 4 6 8 ...1 2 4 8 16 ...1 2 4 16 65536 2003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880933348101038234342907263181822949382118812668869506364761547029165041871916351587966347219442930927982084309104855990570159318959639524863372367203... NB. Buck truncated function (missing the first two rows) BuckTruncated NB. Buck truncated function-level code,&'&1'&'2x&*'@:[ 128!:2 ] (rows01=. {&('>:',:'2x&+')) 0 1 NB. The missing first two rows>: 2x&+ Buck=. (rows01 :: (rows2up o (-&2)))"0 x apply ] (Ack=. (3 -~ [ Buck 3 + ])f.) NB. Ackermann function-level code3 -~ [ ({&(2 4$'>:  2x&+') ::(,&'&1'&'2x&*'@:(-&2))"0@:[ 128!:2 ]) 3 + ]

## Java

import java.math.BigInteger; public static BigInteger ack(BigInteger m, BigInteger n) {    return m.equals(BigInteger.ZERO)            ? n.add(BigInteger.ONE)            : ack(m.subtract(BigInteger.ONE),                        n.equals(BigInteger.ZERO) ? BigInteger.ONE : ack(m, n.subtract(BigInteger.ONE)));}
Works with: Java version 8+
@FunctionalInterfacepublic interface FunctionalField<FIELD extends Enum<?>> {  public Object untypedField(FIELD field);   @SuppressWarnings("unchecked")  public default <VALUE> VALUE field(FIELD field) {    return (VALUE) untypedField(field);  }}

Example:

range(0;5) as $i| range(0; if$i > 3 then 1 else 6 end) as $j| "A($$i),\(j)) = \( [i,j] | ack )" Output: # jq -n -r -f ackermann.jqA(0,0) = 1A(0,1) = 2A(0,2) = 3A(0,3) = 4A(0,4) = 5A(0,5) = 6A(1,0) = 2A(1,1) = 3A(1,2) = 4A(1,3) = 5A(1,4) = 6A(1,5) = 7A(2,0) = 3A(2,1) = 5A(2,2) = 7A(2,3) = 9A(2,4) = 11A(2,5) = 13A(3,0) = 5A(3,1) = 13A(3,2) = 29A(3,3) = 61A(3,4) = 125A(3,5) = 253A(4,0) = 13 ### With Memoization and Optimization # input: [m,n, cache]# output [value, updatedCache]def ack: # input: [value,cache]; output: [value, updatedCache] def cache(key): .[1] += { (key): .[0] }; def pow2: reduce range(0; .) as i (1; .*2); .[0] as m | .[1] as n | .[2] as cache | if m == 0 then [n + 1, cache] elif m == 1 then [n + 2, cache] elif m == 2 then [2 * n + 3, cache] elif m == 3 then [8 * (n|pow2) - 3, cache] else (.[0:2]|tostring) as key | cache[key] as value | if value then [value, cache] elif n == 0 then ([m-1, 1, cache] | ack) | cache(key) else ([m, n-1, cache ] | ack) | [m-1, .[0], .[1]] | ack | cache(key) end end; def A(m;n): [m,n,{}] | ack | .[0];  Example: A(4,1) Output: 65533 ## Julia function ack(m,n) if m == 0 return n + 1 elseif n == 0 return ack(m-1,1) else return ack(m-1,ack(m,n-1)) endend One-liner: ack2(m::Integer, n::Integer) = m == 0 ? n + 1 : n == 0 ? ack2(m - 1, 1) : ack2(m - 1, ack2(m, n - 1)) Using memoization, source: using Memoize@memoize ack3(m::Integer, n::Integer) = m == 0 ? n + 1 : n == 0 ? ack3(m - 1, 1) : ack3(m - 1, ack3(m, n - 1)) Benchmarking: julia> @time ack2(4,1) elapsed time: 71.98668457 seconds (96 bytes allocated) 65533 julia> @time ack3(4,1) elapsed time: 0.49337724 seconds (30405308 bytes allocated) 65533 ## K See the K wiki ack:{:[0=x;y+1;0=y;_f[x-1;1];_f[x-1;_f[x;y-1]]]}ack[2;2] ## Kdf9 Usercode V6; W0;YS26000;RESTART; J999; J999;PROGRAM; (main program); V1 = B1212121212121212; (radix 10 for FRB); V2 = B2020202020202020; (high bits for decimal digits); V3 = B0741062107230637; ("A[3," in Flexowriter code); V4 = B0727062200250007; ("7] = " in Flexowriter code); V5 = B7777777777777777; ZERO; NOT; =M1; (Q1 := 0/0/-1); SETAYS0; =M2; I2=2; (Q2 := 0/2/AYS0: M2 is the stack pointer); SET 3; =RC7; (Q7 := 3/1/0: C7 = m); SET 7; =RC8; (Q8 := 7/1/0: C8 = n); JSP1; (call Ackermann function); V1; REV; FRB; (convert result to base 10); V2; OR; (convert decimal digits to characters); V5; REV; SHLD+24; =V5; ERASE; (eliminate leading zeros); SETAV5; =RM9; SETAV3; =I9; POAQ9; (write result to Flexowriter); 999; ZERO; OUT; (terminate run); P1; (To compute A[m, n]); 99; J1C7NZ; (to 1 if m ± 0); I8; =+C8; (n := n + 1); C8; (result to NEST); EXIT 1; (return); *1; J2C8NZ; (to 2 if n ± 0); I8; =C8; (n := 1); DC7; (m := m - 1); J99; (tail recursion for A[m-1, 1]); *2; LINK; =M0M2; (push return address); C7; =M0M2QN; (push m); DC8; (n := n - 1); JSP1; (full recursion for A[m, n-1]); =C8; (n := A[m, n-1]); M1M2; =C7; (m := top of stack); DC7; (m := m - 1); M-I2; (pop stack); M0M2; =LINK; (return address := top of stack); J99; (tail recursion for A[m-1, A[m, n-1]]); FINISH; ## Klong  ack::{:[0=x;y+1:|0=y;.f(x-1;1);.f(x-1;.f(x;y-1))]}ack(2;2) ## Kotlin fun A(m: Long, n: Long): Long = when { m == 0L -> n + 1 m > 0L -> when { n == 0L -> A(m - 1, 1) n > 0L -> A(m - 1, A(m, n - 1)) else -> throw IllegalArgumentException("illegal n") } else -> throw IllegalArgumentException("illegal m")} fun main(args: Array<String>) { val M: Long = 4 val N: Long = 20 val r = 0..N for (m in 0..M) { print("\nA(m, r) =") var able = true r.forEach { try { if (able) { val a = A(m, it) print(" %6d".format(a)) } else print(" ?") } catch(e: Throwable) { print(" ?") able = false } } }} Output: A(0, 0..20) = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 A(1, 0..20) = 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 A(2, 0..20) = 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 A(3, 0..20) = 5 13 29 61 125 253 509 1021 2045 4093 8189 16381 32765 ? ? ? ? ? ? ? ? A(4, 0..20) = 13 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ## Lasso #!/usr/bin/lasso9 define ackermann(m::integer, n::integer) => { if(#m == 0) => { return ++#n else(#n == 0) return ackermann(--#m, 1) else return ackermann(#m-1, ackermann(#m, --#n)) }} with x in generateSeries(1,3), y in generateSeries(0,8,2)do stdoutnl(#x+', '#y+': ' + ackermann(#x, #y))  Output: 1, 0: 2 1, 2: 4 1, 4: 6 1, 6: 8 1, 8: 10 2, 0: 3 2, 2: 7 2, 4: 11 2, 6: 15 2, 8: 19 3, 0: 5 3, 2: 29 3, 4: 125 3, 6: 509 3, 8: 2045 ## LFE (defun ackermann ((0 n) (+ n 1)) ((m 0) (ackermann (- m 1) 1)) ((m n) (ackermann (- m 1) (ackermann m (- n 1))))) ## Liberty BASIC Print Ackermann(1, 2) Function Ackermann(m, n) Select Case Case (m < 0) Or (n < 0) Exit Function Case (m = 0) Ackermann = (n + 1) Case (m > 0) And (n = 0) Ackermann = Ackermann((m - 1), 1) Case (m > 0) And (n > 0) Ackermann = Ackermann((m - 1), Ackermann(m, (n - 1))) End Select End Function ## LiveCode function ackermann m,n switch Case m = 0 return n + 1 Case (m > 0 And n = 0) return ackermann((m - 1), 1) Case (m > 0 And n > 0) return ackermann((m - 1), ackermann(m, (n - 1))) end switchend ackermann ## Logo to ack :i :j if :i = 0 [output :j+1] if :j = 0 [output ack :i-1 1] output ack :i-1 ack :i :j-1end ## Logtalk ack(0, N, V) :- !, V is N + 1.ack(M, 0, V) :- !, M2 is M - 1, ack(M2, 1, V).ack(M, N, V) :- M2 is M - 1, N2 is N - 1, ack(M, N2, V2), ack(M2, V2, V). ## LOLCODE Translation of: C HAI 1.3 HOW IZ I ackermann YR m AN YR n NOT m, O RLY? YA RLY, FOUND YR SUM OF n AN 1 OIC NOT n, O RLY? YA RLY, FOUND YR I IZ ackermann YR DIFF OF m AN 1 AN YR 1 MKAY OIC FOUND YR I IZ ackermann YR DIFF OF m AN 1 AN YR... I IZ ackermann YR m AN YR DIFF OF n AN 1 MKAY MKAYIF U SAY SO IM IN YR outer UPPIN YR m TIL BOTH SAEM m AN 5 IM IN YR inner UPPIN YR n TIL BOTH SAEM n AN DIFF OF 6 AN m VISIBLE "A(" m ", " n ") = " I IZ ackermann YR m AN YR n MKAY IM OUTTA YR innerIM OUTTA YR outer KTHXBYE ## Lua function ack(M,N) if M == 0 then return N + 1 end if N == 0 then return ack(M-1,1) end return ack(M-1,ack(M, N-1))end ## Lucid ack(m,n) where ack(m,n) = if m eq 0 then n+1 else if n eq 0 then ack(m-1,1) else ack(m-1, ack(m, n-1)) fi fi; end ## Luck function ackermann(m: int, n: int): int = ( if m==0 then n+1 else if n==0 then ackermann(m-1,1) else ackermann(m-1,ackermann(m,n-1))) ## M2000 Interpreter  Module Checkit { Def ackermann(m,n) =If(m=0-> n+1, If(n=0-> ackermann(m-1,1), ackermann(m-1,ackermann(m,n-1)))) For m = 0 to 3 {For n = 0 to 4 {Print m;" ";n;" ";ackermann(m,n)}}}Checkit Module Checkit { Module Inner (ack) { For m = 0 to 3 {For n = 0 to 4 {Print m;" ";n;" ";ack(m,n)}} } Inner lambda (m,n) ->If(m=0-> n+1, If(n=0-> lambda(m-1,1),lambda(m-1,lambda(m,n-1))))}Checkit  ## M4 define(ack',ifelse(1,0,incr(2)',ifelse(2,0,ack(decr(1),1)',ack(decr(1),ack(1,decr(2)))')')')dnlack(3,3) Output: 61  ## Maple Strictly by the definition given above, we can code this as follows.  Ackermann := proc( m :: nonnegint, n :: nonnegint ) option remember; # optional automatic memoization if m = 0 then n + 1 elif n = 0 then thisproc( m - 1, 1 ) else thisproc( m - 1, thisproc( m, n - 1 ) ) end ifend proc:  In Maple, the keyword thisproc refers to the currently executing procedure (closure) and is used when writing recursive procedures. (You could also use the name of the procedure, Ackermann in this case, but then a concurrently executing task or thread could re-assign that name while the recursive procedure is executing, resulting in an incorrect result.) To make this faster, you can use known expansions for small values of ${\displaystyle m}$. (See Wikipedia:Ackermann function)  Ackermann := proc( m :: nonnegint, n :: nonnegint ) option remember; # optional automatic memoization if m = 0 then n + 1 elif m = 1 then n + 2 elif m = 2 then 2 * n + 3 elif m = 3 then 8 * 2^n - 3 elif n = 0 then thisproc( m - 1, 1 ) else thisproc( m - 1, thisproc( m, n - 1 ) ) end ifend proc:  This makes it possible to compute Ackermann( 4, 1 ) and Ackermann( 4, 2 ) essentially instantly, though Ackermann( 4, 3 ) is still out of reach. To compute Ackermann( 1, i ) for i from 1 to 10 use  > map2( Ackermann, 1, [seq]( 1 .. 10 ) ); [3, 4, 5, 6, 7, 8, 9, 10, 11, 12]  To get the first 10 values for m = 2 use  > map2( Ackermann, 2, [seq]( 1 .. 10 ) ); [5, 7, 9, 11, 13, 15, 17, 19, 21, 23]  For Ackermann( 4, 2 ) we get a very long number with  > length( Ackermann( 4, 2 ) ); 19729  digits. ## Mathematica / Wolfram Language Two possible implementations would be: RecursionLimit=InfinityAckermann1[m_,n_]:= If[m==0,n+1, If[ n==0,Ackermann1[m-1,1], Ackermann1[m-1,Ackermann1[m,n-1]] ] ] Ackermann2[0,n_]:=n+1; Ackermann2[m_,0]:=Ackermann1[m-1,1]; Ackermann2[m_,n_]:=Ackermann1[m-1,Ackermann1[m,n-1]] Note that the second implementation is quite a bit faster, as doing 'if' comparisons is slower than the built-in pattern matching algorithms. Examples: Flatten[#,1]&@Table[{"Ackermann2["<>ToString[i]<>","<>ToString[j]<>"] =",Ackermann2[i,j]},{i,3},{j,8}]//Grid gives back: Ackermann2[1,1] = 3Ackermann2[1,2] = 4Ackermann2[1,3] = 5Ackermann2[1,4] = 6Ackermann2[1,5] = 7Ackermann2[1,6] = 8Ackermann2[1,7] = 9Ackermann2[1,8] = 10Ackermann2[2,1] = 5Ackermann2[2,2] = 7Ackermann2[2,3] = 9Ackermann2[2,4] = 11Ackermann2[2,5] = 13Ackermann2[2,6] = 15Ackermann2[2,7] = 17Ackermann2[2,8] = 19Ackermann2[3,1] = 13Ackermann2[3,2] = 29Ackermann2[3,3] = 61Ackermann2[3,4] = 125Ackermann2[3,5] = 253Ackermann2[3,6] = 509Ackermann2[3,7] = 1021Ackermann2[3,8] = 2045 If we would like to calculate Ackermann[4,1] or Ackermann[4,2] we have to optimize a little bit: Clear[Ackermann3]RecursionLimit=Infinity;Ackermann3[0,n_]:=n+1;Ackermann3[1,n_]:=n+2;Ackermann3[2,n_]:=3+2n;Ackermann3[3,n_]:=5+8 (2^n-1);Ackermann3[m_,0]:=Ackermann3[m-1,1];Ackermann3[m_,n_]:=Ackermann3[m-1,Ackermann3[m,n-1]] Now computing Ackermann[4,1] and Ackermann[4,2] can be done quickly (<0.01 sec): Examples 2: Ackermann3[4, 1]Ackermann3[4, 2] gives back: 655332003529930406846464979072351560255750447825475569751419265016973710894059556311453089506130880........699146577530041384717124577965048175856395072895337539755822087777506072339445587895905719156733 Ackermann[4,2] has 19729 digits, several thousands of digits omitted in the result above for obvious reasons. Ackermann[5,0] can be computed also quite fast, and is equal to 65533. Summarizing Ackermann[0,n_], Ackermann[1,n_], Ackermann[2,n_], and Ackermann[3,n_] can all be calculated for n>>1000. Ackermann[4,0], Ackermann[4,1], Ackermann[4,2] and Ackermann[5,0] are only possible now. Maybe in the future we can calculate higher Ackermann numbers efficiently and fast. Although showing the results will always be a problem. ## MATLAB function A = ackermannFunction(m,n) if m == 0 A = n+1; elseif (m > 0) && (n == 0) A = ackermannFunction(m-1,1); else A = ackermannFunction( m-1,ackermannFunction(m,n-1) ); endend ## Maxima ackermann(m, n) := if integerp(m) and integerp(n) then ackermann[m, n] else 'ackermann(m, n) ackermann[m, n] := if m = 0 then n + 1 elseif m = 1 then 2 + (n + 3) - 3 elseif m = 2 then 2 * (n + 3) - 3 elseif m = 3 then 2^(n + 3) - 3 elseif n = 0 then ackermann[m - 1, 1] else ackermann[m - 1, ackermann[m, n - 1]] tetration(a, n) := if integerp(n) then block([b: a], for i from 2 thru n do b: a^b, b) else 'tetration(a, n) /* this should evaluate to zero */ackermann(4, n) - (tetration(2, n + 3) - 3);subst(n = 2, %);ev(%, nouns); ## MAXScript Use with caution. Will cause a stack overflow for m > 3. fn ackermann m n =( if m == 0 then ( return n + 1 ) else if n == 0 then ( ackermann (m-1) 1 ) else ( ackermann (m-1) (ackermann m (n-1)) )) ## МК-61/52 П1 <-> П0 ПП 06 С/П ИП0 x=0 13 ИП11 + В/О ИП1 x=0 24 ИП0 1 П1 -П0 ПП 06 В/О ИП0 П2 ИП1 1 - П1ПП 06 П1 ИП2 1 - П0 ПП 06 В/О ## mLite fun ackermann( 0, n ) = n + 1 | ( m, 0 ) = ackermann( m - 1, 1 ) | ( m, n ) = ackermann( m - 1, ackermann(m, n - 1) ) Test code providing tuples from (0,0) to (3,8) fun jota x = map (fn x = x-1)  iota x fun test_tuples (x, y) = append_map (fn a = map (fn b = (b, a))  jota x)  jota y map ackermann (test_tuples(4,9)) Result [1, 2, 3, 5, 2, 3, 5, 13, 3, 4, 7, 29, 4, 5, 9, 61, 5, 6, 11, 125, 6, 7, 13, 253, 7, 8, 15, 509, 8, 9, 17, 1021, 9, 10, 19, 2045] ## ML/I ML/I loves recursion, but runs out of its default amount of storage with larger numbers than those tested here! ### Program MCSKIP "WITH" NL"" Ackermann function"" Will overflow when it reaches implementation-defined signed integer limitMCSKIP MT,<>MCINS %.MCDEF ACK WITHS ( , )AS <MCSET T1=%A1.MCSET T2=%A2.MCGO L1 UNLESS T1 EN 0%%T2.+1.MCGO L0%L1.MCGO L2 UNLESS T2 EN 0ACK(%%T1.-1.,1)MCGO L0%L2.ACK(%%T1.-1.,ACK(%T1.,%%T2.-1.))>"" Macro ACK now defined, so try it outa(0,0) => ACK(0,0)a(0,1) => ACK(0,1)a(0,2) => ACK(0,2)a(0,3) => ACK(0,3)a(0,4) => ACK(0,4)a(0,5) => ACK(0,5)a(1,0) => ACK(1,0)a(1,1) => ACK(1,1)a(1,2) => ACK(1,2)a(1,3) => ACK(1,3)a(1,4) => ACK(1,4)a(2,0) => ACK(2,0)a(2,1) => ACK(2,1)a(2,2) => ACK(2,2)a(2,3) => ACK(2,3)a(3,0) => ACK(3,0)a(3,1) => ACK(3,1)a(3,2) => ACK(3,2)a(4,0) => ACK(4,0) Output: a(0,0) => 1a(0,1) => 2a(0,2) => 3a(0,3) => 4a(0,4) => 5a(0,5) => 6a(1,0) => 2a(1,1) => 3a(1,2) => 4a(1,3) => 5a(1,4) => 6a(2,0) => 3a(2,1) => 5a(2,2) => 7a(2,3) => 9a(3,0) => 5a(3,1) => 13a(3,2) => 29a(4,0) => 13 ## Mercury This is the Ackermann function with some (obvious) elements elided. The ack/3 predicate is implemented in terms of the ack/2 function. The ack/2 function is implemented in terms of the ack/3 predicate. This makes the code both more concise and easier to follow than would otherwise be the case. The integer type is used instead of int because the problem statement stipulates the use of bignum integers if possible. :- func ack(integer, integer) = integer.ack(M, N) = R :- ack(M, N, R). :- pred ack(integer::in, integer::in, integer::out) is det.ack(M, N, R) :- ( ( M < integer(0) ; N < integer(0) ) -> throw(bounds_error) ; M = integer(0) -> R = N + integer(1) ; N = integer(0) -> ack(M - integer(1), integer(1), R) ; ack(M - integer(1), ack(M, N - integer(1)), R) ). ## Modula-2 MODULE ackerman; IMPORT ASCII, NumConv, InOut; VAR m, n : LONGCARD; string : ARRAY [0..19] OF CHAR; OK : BOOLEAN; PROCEDURE Ackerman (x, y : LONGCARD) : LONGCARD; BEGIN IF x = 0 THEN RETURN y + 1 ELSIF y = 0 THEN RETURN Ackerman (x - 1 , 1) ELSE RETURN Ackerman (x - 1 , Ackerman (x , y - 1)) ENDEND Ackerman; BEGIN FOR m := 0 TO 3 DO FOR n := 0 TO 6 DO NumConv.Num2Str (Ackerman (m, n), 10, string, OK); IF OK THEN InOut.WriteString (string) ELSE InOut.WriteString ("* Error in number * ") END; InOut.Write (ASCII.HT) END; InOut.WriteLn END; InOut.WriteLnEND ackerman. Output: [email protected]:~/modula/rosetta ackerman 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509 ## Modula-3 The type CARDINAL is defined in Modula-3 as [0..LAST(INTEGER)], in other words, it can hold all positive integers. MODULE Ack EXPORTS Main; FROM IO IMPORT Put;FROM Fmt IMPORT Int; PROCEDURE Ackermann(m, n: CARDINAL): CARDINAL = BEGIN IF m = 0 THEN RETURN n + 1; ELSIF n = 0 THEN RETURN Ackermann(m - 1, 1); ELSE RETURN Ackermann(m - 1, Ackermann(m, n - 1)); END; END Ackermann; BEGIN FOR m := 0 TO 3 DO FOR n := 0 TO 6 DO Put(Int(Ackermann(m, n)) & " "); END; Put("\n"); END;END Ack. Output: 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509  ## MUMPS Ackermann(m,n) ; If m=0 Quit n+1 If m>0,n=0 Quit Ackermann(m-1,1) If m>0,n>0 Quit Ackermann(m-1,Ackermann(m,n-1)) Set Ecode=",U13-Invalid parameter for Ackermann: m="_m_", n="_n_"," Write Ackermann(1,8) ; 10Write Ackermann(2,8) ; 19Write Ackermann(3,5) ; 253 ## Nemerle In Nemerle, we can state the Ackermann function as a lambda. By using pattern-matching, our definition strongly resembles the mathematical notation.  using System;using Nemerle.IO; def ackermann(m, n) { def A = ackermann; match(m, n) { | (0, n) => n + 1 | (m, 0) when m > 0 => A(m - 1, 1) | (m, n) when m > 0 && n > 0 => A(m - 1, A(m, n - 1)) | _ => throw Exception("invalid inputs"); }} for(mutable m = 0; m < 4; m++) { for(mutable n = 0; n < 5; n++) { print("ackermann(m, n) = (ackermann(m, n))\n"); }}  A terser version using implicit match (which doesn't use the alias A internally):  def ackermann(m, n) { | (0, n) => n + 1 | (m, 0) when m > 0 => ackermann(m - 1, 1) | (m, n) when m > 0 && n > 0 => ackermann(m - 1, ackermann(m, n - 1)) | _ => throw Exception("invalid inputs");}  Or, if we were set on using the A notation, we could do this:  def ackermann = { def A(m, n) { | (0, n) => n + 1 | (m, 0) when m > 0 => A(m - 1, 1) | (m, n) when m > 0 && n > 0 => A(m - 1, A(m, n - 1)) | _ => throw Exception("invalid inputs"); } A}  ## NetRexx /* NetRexx */options replace format comments java crossref symbols binary numeric digits 66 parse arg j_ k_ .if j_ = '' | j_ = '.' | \j_.datatype('w') then j_ = 3if k_ = '' | k_ = '.' | \k_.datatype('w') then k_ = 5 loop m_ = 0 to j_ say loop n_ = 0 to k_ say 'ackermann('m_','n_') =' ackermann(m_, n_).right(5) end n_ end m_return method ackermann(m, n) public static select when m = 0 then rval = n + 1 when n = 0 then rval = ackermann(m - 1, 1) otherwise rval = ackermann(m - 1, ackermann(m, n - 1)) end return rval  ## NewLISP  #! /usr/local/bin/newlisp (define (ackermann m n) (cond ((zero? m) (inc n)) ((zero? n) (ackermann (dec m) 1)) (true (ackermann (- m 1) (ackermann m (dec n))))))  In case of stack overflow error, you have to start your program with a proper "-s <value>" flag as "newlisp -s 100000 ./ackermann.lsp". See http://www.newlisp.org/newlisp_manual.html#stack_size  ## Nim from strutils import parseInt proc ackermann(m, n: int64): int64 = if m == 0: result = n + 1 elif n == 0: result = ackermann(m - 1, 1) else: result = ackermann(m - 1, ackermann(m, n - 1)) proc getNumber(): int = try: result = stdin.readLine.parseInt except ValueError: echo "An integer, please!" result = getNumber() if result < 0: echo "Please Enter a non-negative Integer: " result = getNumber() echo "First non-negative Integer please: "let first = getNumber()echo "Second non-negative Integer please: "let second = getNumber()echo "Result: ", ackermann(first, second)  ## Nit Source: the official Nit’s repository. # Task: Ackermann function## A simple straightforward recursive implementation.module ackermann_function fun ack(m, n: Int): Intdo if m == 0 then return n + 1 if n == 0 then return ack(m-1,1) return ack(m-1, ack(m, n-1))end for m in [0..3] do for n in [0..6] do print ack(m,n) end print ""end Output: 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509  ## Objeck Translation of: C# – C sharp class Ackermann { function : Main(args : String[]) ~ Nil { for(m := 0; m <= 3; ++m;) { for(n := 0; n <= 4; ++n;) { a := Ackermann(m, n); if(a > 0) { "Ackermann({m}, {n}) = {a}"->PrintLine(); }; }; }; } function : Ackermann(m : Int, n : Int) ~ Int { if(m > 0) { if (n > 0) { return Ackermann(m - 1, Ackermann(m, n - 1)); } else if (n = 0) { return Ackermann(m - 1, 1); }; } else if(m = 0) { if(n >= 0) { return n + 1; }; }; return -1; }} Ackermann(0, 0) = 1 Ackermann(0, 1) = 2 Ackermann(0, 2) = 3 Ackermann(0, 3) = 4 Ackermann(0, 4) = 5 Ackermann(1, 0) = 2 Ackermann(1, 1) = 3 Ackermann(1, 2) = 4 Ackermann(1, 3) = 5 Ackermann(1, 4) = 6 Ackermann(2, 0) = 3 Ackermann(2, 1) = 5 Ackermann(2, 2) = 7 Ackermann(2, 3) = 9 Ackermann(2, 4) = 11 Ackermann(3, 0) = 5 Ackermann(3, 1) = 13 Ackermann(3, 2) = 29 Ackermann(3, 3) = 61 Ackermann(3, 4) = 125  ## OCaml let rec a m n = if m=0 then (n+1) else if n=0 then (a (m-1) 1) else (a (m-1) (a m (n-1))) or: let rec a = function | 0, n -> (n+1) | m, 0 -> a(m-1, 1) | m, n -> a(m-1, a(m, n-1)) with memoization using an hash-table: let h = Hashtbl.create 4001 let a m n = try Hashtbl.find h (m, n) with Not_found -> let res = a (m, n) in Hashtbl.add h (m, n) res; (res) taking advantage of the memoization we start calling small values of m and n in order to reduce the recursion call stack: let a m n = for _m = 0 to m do for _n = 0 to n do ignore(a _m _n); done; done; (a m n) ### Arbitrary precision With arbitrary-precision integers (Big_int module): open Big_intlet one = unit_big_intlet zero = zero_big_intlet succ = succ_big_intlet pred = pred_big_intlet eq = eq_big_int let rec a m n = if eq m zero then (succ n) else if eq n zero then (a (pred m) one) else (a (pred m) (a m (pred n))) compile with: ocamlopt -o acker nums.cmxa acker.ml  ### Tail-Recursive Here is a tail-recursive version: let rec find_option h v = try Some(Hashtbl.find h v) with Not_found -> None let rec a bounds caller todo m n = match m, n with | 0, n -> let r = (n+1) in ( match todo with | [] -> r | (m,n)::todo -> List.iter (fun k -> if not(Hashtbl.mem bounds k) then Hashtbl.add bounds k r) caller; a bounds [] todo m n ) | m, 0 -> a bounds caller todo (m-1) 1 | m, n -> match find_option bounds (m, n-1) with | Some a_rec -> let caller = (m,n)::caller in a bounds caller todo (m-1) a_rec | None -> let todo = (m,n)::todo and caller = [(m, n-1)] in a bounds caller todo m (n-1) let a = a (Hashtbl.create 42 (* arbitrary *) ) [] [] ;; This one uses the arbitrary precision, the tail-recursion, and the optimisation explain on the Wikipedia page about (m,n) = (3,_). open Big_intlet one = unit_big_intlet zero = zero_big_intlet succ = succ_big_intlet pred = pred_big_intlet add = add_big_intlet sub = sub_big_intlet eq = eq_big_intlet three = succ(succ one)let power = power_int_positive_big_int let eq2 (a1,a2) (b1,b2) = (eq a1 b1) && (eq a2 b2) module H = Hashtbl.Make (struct type t = Big_int.big_int * Big_int.big_int let equal = eq2 let hash (x,y) = Hashtbl.hash (Big_int.string_of_big_int x ^ "," ^ Big_int.string_of_big_int y) (* probably not a very good hash function *) end) let rec find_option h v = try Some (H.find h v) with Not_found -> None let rec a bounds caller todo m n = let may_tail r = let k = (m,n) in match todo with | [] -> r | (m,n)::todo -> List.iter (fun k -> if not (H.mem bounds k) then H.add bounds k r) (k::caller); a bounds [] todo m n in match m, n with | m, n when eq m zero -> let r = (succ n) in may_tail r | m, n when eq n zero -> let caller = (m,n)::caller in a bounds caller todo (pred m) one | m, n when eq m three -> let r = sub (power 2 (add n three)) three in may_tail r | m, n -> match find_option bounds (m, pred n) with | Some a_rec -> let caller = (m,n)::caller in a bounds caller todo (pred m) a_rec | None -> let todo = (m,n)::todo in let caller = [(m, pred n)] in a bounds caller todo m (pred n) let a = a (H.create 42 (* arbitrary *)) [] [] ;; let () = let m, n = try big_int_of_string Sys.argv.(1), big_int_of_string Sys.argv.(2) with _ -> Printf.eprintf "usage: %s <int> <int>\n" Sys.argv.(0); exit 1 in let r = a m n in Printf.printf "(a %s %s) = %s\n" (string_of_big_int m) (string_of_big_int n) (string_of_big_int r);;; ## Oberon-2 MODULE ackerman; IMPORT Out; VAR m, n : INTEGER; PROCEDURE Ackerman (x, y : INTEGER) : INTEGER; BEGIN IF x = 0 THEN RETURN y + 1 ELSIF y = 0 THEN RETURN Ackerman (x - 1 , 1) ELSE RETURN Ackerman (x - 1 , Ackerman (x , y - 1)) ENDEND Ackerman; BEGIN FOR m := 0 TO 3 DO FOR n := 0 TO 6 DO Out.Int (Ackerman (m, n), 10); Out.Char (9X) END; Out.Ln END; Out.LnEND ackerman. ## Octave function r = ackerman(m, n) if ( m == 0 ) r = n + 1; elseif ( n == 0 ) r = ackerman(m-1, 1); else r = ackerman(m-1, ackerman(m, n-1)); endifendfunction for i = 0:3 disp(ackerman(i, 4));endfor ## Oforth : A( m n -- p ) m ifZero: [ n 1+ return ] m 1- n ifZero: [ 1 ] else: [ A( m, n 1- ) ] A; ## OOC  ack: func (m: Int, n: Int) -> Int { if (m == 0) { n + 1 } else if (n == 0) { ack(m - 1, 1) } else { ack(m - 1, ack(m, n - 1)) }} main: func { for (m in 0..4) { for (n in 0..10) { "ack(#{m}, #{n}) = #{ack(m, n)}" println() } }}  ## ooRexx  loop m = 0 to 3 loop n = 0 to 6 say "Ackermann("m", "n") =" ackermann(m, n) endend ::routine ackermann use strict arg m, n -- give us some precision room numeric digits 10000 if m = 0 then return n + 1 else if n = 0 then return ackermann(m - 1, 1) else return ackermann(m - 1, ackermann(m, n - 1))  Output: Ackermann(0, 0) = 1 Ackermann(0, 1) = 2 Ackermann(0, 2) = 3 Ackermann(0, 3) = 4 Ackermann(0, 4) = 5 Ackermann(0, 5) = 6 Ackermann(0, 6) = 7 Ackermann(1, 0) = 2 Ackermann(1, 1) = 3 Ackermann(1, 2) = 4 Ackermann(1, 3) = 5 Ackermann(1, 4) = 6 Ackermann(1, 5) = 7 Ackermann(1, 6) = 8 Ackermann(2, 0) = 3 Ackermann(2, 1) = 5 Ackermann(2, 2) = 7 Ackermann(2, 3) = 9 Ackermann(2, 4) = 11 Ackermann(2, 5) = 13 Ackermann(2, 6) = 15 Ackermann(3, 0) = 5 Ackermann(3, 1) = 13 Ackermann(3, 2) = 29 Ackermann(3, 3) = 61 Ackermann(3, 4) = 125 Ackermann(3, 5) = 253 Ackermann(3, 6) = 509  ## Order #include <order/interpreter.h> #define ORDER_PP_DEF_8ack ORDER_PP_FN( \8fn(8X, 8Y, \ 8cond((8is_0(8X), 8inc(8Y)) \ (8is_0(8Y), 8ack(8dec(8X), 1)) \ (8else, 8ack(8dec(8X), 8ack(8X, 8dec(8Y))))))) ORDER_PP(8to_lit(8ack(3, 4))) // 125 ## Oz Oz has arbitrary precision integers. declare fun {Ack M N} if M == 0 then N+1 elseif N == 0 then {Ack M-1 1} else {Ack M-1 {Ack M N-1}} end end in {Show {Ack 3 7}} ## PARI/GP Naive implementation. A(m,n)={ if(m, if(n, A(m-1, A(m,n-1)) , A(m-1,1) ) , n+1 )}; ## Pascal Program Ackerman; function ackermann(m, n: Integer) : Integer;begin if m = 0 then ackermann := n+1 else if n = 0 then ackermann := ackermann(m-1, 1) else ackermann := ackermann(m-1, ackermann(m, n-1));end; var m, n : Integer; begin for n := 0 to 6 do for m := 0 to 3 do WriteLn('A(', m, ',', n, ') = ', ackermann(m,n));end. ## Perl We memoize calls to A to make A(2, n) and A(3, n) feasible for larger values of n. { my @memo; sub A { my( m, n ) = @_; memo[ m ][ n ] and return memo[ m ][ n ]; m or return n + 1; return memo[ m ][ n ] = ( n ? A( m - 1, A( m, n - 1 ) ) : A( m - 1, 1 ) ); }} An implementation using the conditional statements 'if', 'elsif' and 'else': sub A { my (m, n) = @_; if (m == 0) { n + 1 } elsif (n == 0) { A(m - 1, 1) } else { A(m - 1, A(m, n - 1)) }} An implementation using ternary chaining: sub A { my (m, n) = @_; m == 0 ? n + 1 : n == 0 ? A(m - 1, 1) : A(m - 1, A(m, n - 1))} Adding memoization and extra terms: use Memoize; memoize('ack2');use bigint try=>"GMP"; sub ack2 { my (m, n) = @_; m == 0 ? n + 1 : m == 1 ? n + 2 : m == 2 ? 2*n + 3 : m == 3 ? 8 * (2**n - 1) + 5 : n == 0 ? ack2(m-1, 1) : ack2(m-1, ack2(m, n-1));}print "ack2(3,4) is ", ack2(3,4), "\n";print "ack2(4,1) is ", ack2(4,1), "\n";print "ack2(4,2) has ", length(ack2(4,2)), " digits\n"; Output: ack2(3,4) is 125 ack2(4,1) is 65533 ack2(4,2) has 19729 digits An optimized version, which uses @_ as a stack, instead of recursion. Very fast. use strict;use warnings;use Math::BigInt; use constant two => Math::BigInt->new(2); sub ack { my n = pop; while( @_ ) { my m = pop; if( m > 3 ) { push @_, (--m) x n; push @_, reverse 3 .. --m; n = 13; } elsif( m == 3 ) { if( n < 29 ) { n = ( 1 << ( n + 3 ) ) - 3; } else { n = two ** ( n + 3 ) - 3; } } elsif( m == 2 ) { n = 2 * n + 3; } elsif( m >= 0 ) { n = n + m + 1; } else { die "negative m!"; } } n;} print "ack(3,4) is ", ack(3,4), "\n";print "ack(4,1) is ", ack(4,1), "\n";print "ack(4,2) has ", length(ack(4,2)), " digits\n";  ## Perl 6 Works with: Rakudo version 2018.03 sub A(Int m, Int n) { if m == 0 { n + 1 } elsif n == 0 { A(m - 1, 1) } else { A(m - 1, A(m, n - 1)) }} An implementation using multiple dispatch: multi sub A(0, Int n) { n + 1 }multi sub A(Int m, 0 ) { A(m - 1, 1) }multi sub A(Int m, Int n) { A(m - 1, A(m, n - 1)) } Note that in either case, Int is defined to be arbitrary precision in Perl 6. Here's a caching version of that, written in the sigilless style, with liberal use of Unicode, and the extra optimizing terms to make A(4,2) possible: proto A(Int \𝑚, Int \𝑛) { (state @)[𝑚][𝑛] //= {*} } multi A(0, Int \𝑛) { 𝑛 + 1 }multi A(1, Int \𝑛) { 𝑛 + 2 }multi A(2, Int \𝑛) { 3 + 2 * 𝑛 }multi A(3, Int \𝑛) { 5 + 8 * (2 ** 𝑛 - 1) } multi A(Int \𝑚, 0 ) { A(𝑚 - 1, 1) }multi A(Int \𝑚, Int \𝑛) { A(𝑚 - 1, A(𝑚, 𝑛 - 1)) } # Testing:say A(4,1);say .chars, " digits starting with ", .substr(0,50), "..." given A(4,2); Output: 65533 19729 digits starting with 20035299304068464649790723515602557504478254755697... ## Phix  ---- Ackermann.exw-- =============---- optimised. still no bignum library, so ack(4,2), which is power(2,65536)-3, which is-- apparently 19729 digits, and any above, are beyond (the CPU/FPU hardware) and this.-- (replaced ack(atom,atom) with ack(int,int) since the former fares no better.) function ack(integer m, integer n) if m=0 then return n+1 elsif m=1 then return n+2 elsif m=2 then return 2*n+3 elsif m=3 then return power(2,n+3)-3 elsif m>0 and n=0 then return ack(m-1,1) else return ack(m-1,ack(m,n-1)) end ifend function procedure Ackermann() for i=0 to 3 do for j=0 to 10 do printf(1,"%5d",ack(i,j)) end for puts(1,"\n") end for printf(1,"ack(4,1) %5d\n",ack(4,1))-- printf(1,"ack(4,2) %5d\n",ack(4,2)) -- power function overflow if getc(0) then end ifend procedure Ackermann()  Output:  1 2 3 4 5 6 7 8 9 10 11 2 3 4 5 6 7 8 9 10 11 12 3 5 7 9 11 13 15 17 19 21 23 5 13 29 61 125 253 509 1021 2045 4093 8189 ack(4,1) 65533  ## PHP function ackermann( m , n ){ if ( m==0 ) { return n + 1; } elseif ( n==0 ) { return ackermann( m-1 , 1 ); } return ackermann( m-1, ackermann( m , n-1 ) );} echo ackermann( 3, 4 );// prints 125 ## PicoLisp (de ack (X Y) (cond ((=0 X) (inc Y)) ((=0 Y) (ack (dec X) 1)) (T (ack (dec X) (ack X (dec Y)))) ) ) ## Piet Rendered as wikitable:  ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww ww This is a naive implementation that does not use any optimization. Find the explanation at [[1]]. Computing the Ackermann function for (4,1) is possible, but takes quite a while because the stack grows very fast to large dimensions. Example output:  ? 3 ? 5 253  ## Pike int main(){ write(ackermann(3,4) + "\n");} int ackermann(int m, int n){ if(m == 0){ return n + 1; } else if(n == 0){ return ackermann(m-1, 1); } else { return ackermann(m-1, ackermann(m, n-1)); }} ## PL/I Ackerman: procedure (m, n) returns (fixed (30)) recursive; declare (m, n) fixed (30); if m = 0 then return (n+1); else if m > 0 & n = 0 then return (Ackerman(m-1, 1)); else if m > 0 & n > 0 then return (Ackerman(m-1, Ackerman(m, n-1))); return (0);end Ackerman; ## PL/SQL DECLARE FUNCTION ackermann(pi_m IN NUMBER, pi_n IN NUMBER) RETURN NUMBER IS BEGIN IF pi_m = 0 THEN RETURN pi_n + 1; ELSIF pi_n = 0 THEN RETURN ackermann(pi_m - 1, 1); ELSE RETURN ackermann(pi_m - 1, ackermann(pi_m, pi_n - 1)); END IF; END ackermann; BEGIN FOR n IN 0 .. 6 LOOP FOR m IN 0 .. 3 LOOP DBMS_OUTPUT.put_line('A(' || m || ',' || n || ') = ' || ackermann(m, n)); END LOOP; END LOOP;END;  Output: A(0,0) = 1 A(1,0) = 2 A(2,0) = 3 A(3,0) = 5 A(0,1) = 2 A(1,1) = 3 A(2,1) = 5 A(3,1) = 13 A(0,2) = 3 A(1,2) = 4 A(2,2) = 7 A(3,2) = 29 A(0,3) = 4 A(1,3) = 5 A(2,3) = 9 A(3,3) = 61 A(0,4) = 5 A(1,4) = 6 A(2,4) = 11 A(3,4) = 125 A(0,5) = 6 A(1,5) = 7 A(2,5) = 13 A(3,5) = 253 A(0,6) = 7 A(1,6) = 8 A(2,6) = 15 A(3,6) = 509  ## PostScript /ackermann{/n exch def/m exch def %PostScript takes arguments in the reverse order as specified in the function definitionm 0 eq{n 1 add}ifm 0 gt n 0 eq and{m 1 sub 1 ackermann}ifm 0 gt n 0 gt and{m 1 sub m n 1 sub ackermann ackermann}if}def Library: initlib /A {[/.m /.n] let{ {.m 0 eq} {.n succ} is? {.m 0 gt .n 0 eq and} {.m pred 1 A} is? {.m 0 gt .n 0 gt and} {.m pred .m .n pred A A} is?} condend}. ## Potion ack = (m, n): if (m == 0): n + 1. elsif (n == 0): ack(m - 1, 1). else: ack(m - 1, ack(m, n - 1)).. 4 times(m): 7 times(n): ack(m, n) print " " print. "\n" print. ## PowerBASIC FUNCTION PBMAIN () AS LONG DIM m AS QUAD, n AS QUAD m = ABS(VAL(INPUTBOX("Enter a whole number."))) n = ABS(VAL(INPUTBOX("Enter another whole number."))) MSGBOX STR(Ackermann(m, n))END FUNCTION FUNCTION Ackermann (m AS QUAD, n AS QUAD) AS QUAD IF 0 = m THEN FUNCTION = n + 1 ELSEIF 0 = n THEN FUNCTION = Ackermann(m - 1, 1) ELSE ' m > 0; n > 0 FUNCTION = Ackermann(m - 1, Ackermann(m, n - 1)) END IFEND FUNCTION ## PowerShell Translation of: PHP function ackermann ([long] m, [long] n) { if (m -eq 0) { return n + 1 } if (n -eq 0) { return (ackermann (m - 1) 1) } return (ackermann (m - 1) (ackermann m (n - 1)))} Building an example table (takes a while to compute, though, especially for the last three numbers; also it fails with the last line in Powershell v1 since the maximum recursion depth is only 100 there): foreach (m in 0..3) { foreach (n in 0..6) { Write-Host -NoNewline ("{0,5}" -f (ackermann m n)) } Write-Host} Output:  1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509 ### A More "PowerShelly" Way  function Get-Ackermann ([int64]m, [int64]n){ if (m -eq 0) { return n + 1 } if (n -eq 0) { return Get-Ackermann (m - 1) 1 } return (Get-Ackermann (m - 1) (Get-Ackermann m (n - 1)))}  Save the result to an array (for possible future use?), then display it using the Format-Wide cmdlet:  ackermann = 0..3 | ForEach-Object {m = _; 0..6 | ForEach-Object {Get-Ackermann m _}} ackermann | Format-Wide {"{0,3}" -f _} -Column 7 -Force  Output:  1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509  ## Processing int ackermann(int m, n){ if (m == 0) return n + 1; else if (m > 0 && n == 0) return ackermann(m - 1, 1); else return ackermann( m - 1, ackermann(m, n - 1) );} ## Prolog Works with: SWI Prolog ack(0, N, Ans) :- Ans is N+1.ack(M, 0, Ans) :- M>0, X is M-1, ack(X, 1, Ans).ack(M, N, Ans) :- M>0, N>0, X is M-1, Y is N-1, ack(M, Y, Ans2), ack(X, Ans2, Ans). ## Pure A 0 n = n+1;A m 0 = A (m-1) 1 if m > 0;A m n = A (m-1) (A m (n-1)) if m > 0 && n > 0; ## PureBasic Procedure.q Ackermann(m, n) If m = 0 ProcedureReturn n + 1 ElseIf n = 0 ProcedureReturn Ackermann(m - 1, 1) Else ProcedureReturn Ackermann(m - 1, Ackermann(m, n - 1)) EndIfEndProcedure Debug Ackermann(3,4) ## Pure Data  #N canvas 741 265 450 436 10;#X obj 83 111 t b l;#X obj 115 163 route 0;#X obj 115 185 + 1;#X obj 83 380 f;#X obj 161 186 swap;#X obj 161 228 route 0;#X obj 161 250 - 1;#X obj 161 208 pack;#X obj 115 314 t f f;#X msg 161 272 \1 1;#X obj 115 142 t l;#X obj 207 250 swap;#X obj 273 271 - 1;#X obj 207 272 t f f;#X obj 207 298 - 1;#X obj 207 360 pack;#X obj 239 299 pack;#X obj 83 77 inlet;#X obj 83 402 outlet;#X connect 0 0 3 0;#X connect 0 1 10 0;#X connect 1 0 2 0;#X connect 1 1 4 0;#X connect 2 0 8 0;#X connect 3 0 18 0;#X connect 4 0 7 0;#X connect 4 1 7 1;#X connect 5 0 6 0;#X connect 5 1 11 0;#X connect 6 0 9 0;#X connect 7 0 5 0;#X connect 8 0 3 1;#X connect 8 1 15 1;#X connect 9 0 10 0;#X connect 10 0 1 0;#X connect 11 0 13 0;#X connect 11 1 12 0;#X connect 12 0 16 1;#X connect 13 0 14 0;#X connect 13 1 16 0;#X connect 14 0 15 0;#X connect 15 0 10 0;#X connect 16 0 10 0;#X connect 17 0 0 0; ## Purity data Iter = f => FoldNat <const f One, f> data Ackermann = FoldNat <const Succ, Iter> ## Python Works with: Python version 2.5 def ack1(M, N): return (N + 1) if M == 0 else ( ack1(M-1, 1) if N == 0 else ack1(M-1, ack1(M, N-1))) Another version: def ack2(M, N): if M == 0: return N + 1 elif N == 0: return ack2(M - 1, 1) else: return ack2(M - 1, ack2(M, N - 1)) Example of use: >>> import sys>>> sys.setrecursionlimit(3000)>>> ack1(0,0)1>>> ack1(3,4)125>>> ack2(0,0)1>>> ack2(3,4)125 From the Mathematica ack3 example: def ack2(M, N): return (N + 1) if M == 0 else ( (N + 2) if M == 1 else ( (2*N + 3) if M == 2 else ( (8*(2**N - 1) + 5) if M == 3 else ( ack2(M-1, 1) if N == 0 else ack2(M-1, ack2(M, N-1)))))) Results confirm those of Mathematica for ack(4,1) and ack(4,2) ## R ackermann <- function(m, n) { if ( m == 0 ) { n+1 } else if ( n == 0 ) { ackermann(m-1, 1) } else { ackermann(m-1, ackermann(m, n-1)) }} for ( i in 0:3 ) { print(ackermann(i, 4))} ## Racket  #lang racket(define (ackermann m n) (cond [(zero? m) (add1 n)] [(zero? n) (ackermann (sub1 m) 1)] [else (ackermann (sub1 m) (ackermann m (sub1 n)))]))  ## REBOL ackermann: func [m n] [ case [ m = 0 [n + 1] n = 0 [ackermann m - 1 1] true [ackermann m - 1 ackermann m n - 1] ] ] ## REXX ### no optimization /*REXX program calculates and displays some values for the Ackermann function. */ /*╔════════════════════════════════════════════════════════════════════════╗ ║ Note: the Ackermann function (as implemented here) utilizes deep ║ ║ recursive and is limited by the largest number that can have ║ ║ "1" (unity) added to a number (successfully and accurately). ║ ╚════════════════════════════════════════════════════════════════════════╝*/high=24 do j=0 to 3; say do k=0 to high % (max(1, j)) call tell_Ack j, k end /*k*/ end /*j*/exit /*stick a fork in it, we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/tell_Ack: parse arg mm,nn; calls=0 /*display an echo message to terminal. */ #=right(nn,length(high)) say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high), left('', 12) 'calls='right(calls, high) return/*──────────────────────────────────────────────────────────────────────────────────────*/ackermann: procedure expose calls /*compute value of Ackermann function. */ parse arg m,n; calls=calls+1 if m==0 then return n+1 if n==0 then return ackermann(m-1, 1) return ackermann(m-1, ackermann(m, n-1) ) output Ackermann(0, 0)= 1 calls= 1 Ackermann(0, 1)= 2 calls= 1 Ackermann(0, 2)= 3 calls= 1 Ackermann(0, 3)= 4 calls= 1 Ackermann(0, 4)= 5 calls= 1 Ackermann(0, 5)= 6 calls= 1 Ackermann(0, 6)= 7 calls= 1 Ackermann(0, 7)= 8 calls= 1 Ackermann(0, 8)= 9 calls= 1 Ackermann(0, 9)= 10 calls= 1 Ackermann(0,10)= 11 calls= 1 Ackermann(0,11)= 12 calls= 1 Ackermann(0,12)= 13 calls= 1 Ackermann(0,13)= 14 calls= 1 Ackermann(0,14)= 15 calls= 1 Ackermann(0,15)= 16 calls= 1 Ackermann(0,16)= 17 calls= 1 Ackermann(0,17)= 18 calls= 1 Ackermann(0,18)= 19 calls= 1 Ackermann(0,19)= 20 calls= 1 Ackermann(0,20)= 21 calls= 1 Ackermann(0,21)= 22 calls= 1 Ackermann(0,22)= 23 calls= 1 Ackermann(0,23)= 24 calls= 1 Ackermann(0,24)= 25 calls= 1 Ackermann(1, 0)= 2 calls= 2 Ackermann(1, 1)= 3 calls= 4 Ackermann(1, 2)= 4 calls= 6 Ackermann(1, 3)= 5 calls= 8 Ackermann(1, 4)= 6 calls= 10 Ackermann(1, 5)= 7 calls= 12 Ackermann(1, 6)= 8 calls= 14 Ackermann(1, 7)= 9 calls= 16 Ackermann(1, 8)= 10 calls= 18 Ackermann(1, 9)= 11 calls= 20 Ackermann(1,10)= 12 calls= 22 Ackermann(1,11)= 13 calls= 24 Ackermann(1,12)= 14 calls= 26 Ackermann(1,13)= 15 calls= 28 Ackermann(1,14)= 16 calls= 30 Ackermann(1,15)= 17 calls= 32 Ackermann(1,16)= 18 calls= 34 Ackermann(1,17)= 19 calls= 36 Ackermann(1,18)= 20 calls= 38 Ackermann(1,19)= 21 calls= 40 Ackermann(1,20)= 22 calls= 42 Ackermann(1,21)= 23 calls= 44 Ackermann(1,22)= 24 calls= 46 Ackermann(1,23)= 25 calls= 48 Ackermann(1,24)= 26 calls= 50 Ackermann(2, 0)= 3 calls= 5 Ackermann(2, 1)= 5 calls= 14 Ackermann(2, 2)= 7 calls= 27 Ackermann(2, 3)= 9 calls= 44 Ackermann(2, 4)= 11 calls= 65 Ackermann(2, 5)= 13 calls= 90 Ackermann(2, 6)= 15 calls= 119 Ackermann(2, 7)= 17 calls= 152 Ackermann(2, 8)= 19 calls= 189 Ackermann(2, 9)= 21 calls= 230 Ackermann(2,10)= 23 calls= 275 Ackermann(2,11)= 25 calls= 324 Ackermann(2,12)= 27 calls= 377 Ackermann(3, 0)= 5 calls= 15 Ackermann(3, 1)= 13 calls= 106 Ackermann(3, 2)= 29 calls= 541 Ackermann(3, 3)= 61 calls= 2432 Ackermann(3, 4)= 125 calls= 10307 Ackermann(3, 5)= 253 calls= 42438 Ackermann(3, 6)= 509 calls= 172233 Ackermann(3, 7)= 1021 calls= 693964 Ackermann(3, 8)= 2045 calls= 2785999  ### optimized for m ≤ 2 /*REXX program calculates and displays some values for the Ackermann function. */high=24 do j=0 to 3; say do k=0 to high % (max(1, j)) call tell_Ack j, k end /*k*/ end /*j*/exit /*stick a fork in it, we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/tell_Ack: parse arg mm,nn; calls=0 /*display an echo message to terminal. */ #=right(nn,length(high)) say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high), left('', 12) 'calls='right(calls, high) return/*──────────────────────────────────────────────────────────────────────────────────────*/ackermann: procedure expose calls /*compute value of Ackermann function. */ parse arg m,n; calls=calls+1 if m==0 then return n + 1 if n==0 then return ackermann(m-1, 1) if m==2 then return n + 3 + n return ackermann(m-1, ackermann(m, n-1) ) output Ackermann(0, 0)= 1 calls= 1 Ackermann(0, 1)= 2 calls= 1 Ackermann(0, 2)= 3 calls= 1 Ackermann(0, 3)= 4 calls= 1 Ackermann(0, 4)= 5 calls= 1 Ackermann(0, 5)= 6 calls= 1 Ackermann(0, 6)= 7 calls= 1 Ackermann(0, 7)= 8 calls= 1 Ackermann(0, 8)= 9 calls= 1 Ackermann(0, 9)= 10 calls= 1 Ackermann(0,10)= 11 calls= 1 Ackermann(0,11)= 12 calls= 1 Ackermann(0,12)= 13 calls= 1 Ackermann(0,13)= 14 calls= 1 Ackermann(0,14)= 15 calls= 1 Ackermann(0,15)= 16 calls= 1 Ackermann(0,16)= 17 calls= 1 Ackermann(0,17)= 18 calls= 1 Ackermann(0,18)= 19 calls= 1 Ackermann(0,19)= 20 calls= 1 Ackermann(0,20)= 21 calls= 1 Ackermann(0,21)= 22 calls= 1 Ackermann(0,22)= 23 calls= 1 Ackermann(0,23)= 24 calls= 1 Ackermann(0,24)= 25 calls= 1 Ackermann(1, 0)= 2 calls= 2 Ackermann(1, 1)= 3 calls= 4 Ackermann(1, 2)= 4 calls= 6 Ackermann(1, 3)= 5 calls= 8 Ackermann(1, 4)= 6 calls= 10 Ackermann(1, 5)= 7 calls= 12 Ackermann(1, 6)= 8 calls= 14 Ackermann(1, 7)= 9 calls= 16 Ackermann(1, 8)= 10 calls= 18 Ackermann(1, 9)= 11 calls= 20 Ackermann(1,10)= 12 calls= 22 Ackermann(1,11)= 13 calls= 24 Ackermann(1,12)= 14 calls= 26 Ackermann(1,13)= 15 calls= 28 Ackermann(1,14)= 16 calls= 30 Ackermann(1,15)= 17 calls= 32 Ackermann(1,16)= 18 calls= 34 Ackermann(1,17)= 19 calls= 36 Ackermann(1,18)= 20 calls= 38 Ackermann(1,19)= 21 calls= 40 Ackermann(1,20)= 22 calls= 42 Ackermann(1,21)= 23 calls= 44 Ackermann(1,22)= 24 calls= 46 Ackermann(1,23)= 25 calls= 48 Ackermann(1,24)= 26 calls= 50 Ackermann(2, 0)= 3 calls= 5 Ackermann(2, 1)= 5 calls= 1 Ackermann(2, 2)= 7 calls= 1 Ackermann(2, 3)= 9 calls= 1 Ackermann(2, 4)= 11 calls= 1 Ackermann(2, 5)= 13 calls= 1 Ackermann(2, 6)= 15 calls= 1 Ackermann(2, 7)= 17 calls= 1 Ackermann(2, 8)= 19 calls= 1 Ackermann(2, 9)= 21 calls= 1 Ackermann(2,10)= 23 calls= 1 Ackermann(2,11)= 25 calls= 1 Ackermann(2,12)= 27 calls= 1 Ackermann(3, 0)= 5 calls= 2 Ackermann(3, 1)= 13 calls= 4 Ackermann(3, 2)= 29 calls= 6 Ackermann(3, 3)= 61 calls= 8 Ackermann(3, 4)= 125 calls= 10 Ackermann(3, 5)= 253 calls= 12 Ackermann(3, 6)= 509 calls= 14 Ackermann(3, 7)= 1021 calls= 16 Ackermann(3, 8)= 2045 calls= 18  ### optimized for m ≤ 4 This REXX version takes advantage that some of the lower numbers for the Ackermann function have direct formulas. If the numeric digits 100 were to be increased to 20000, then the value of Ackermann(4,2) (the last line of output) would be presented with the full 19,729 decimal digits. /*REXX program calculates and displays some values for the Ackermann function. */numeric digits 100 /*use up to 100 decimal digit integers.*/ /*╔═════════════════════════════════════════════════════════════╗ ║ When REXX raises a number to an integer power (via the ** ║ ║ operator, the power can be positive, zero, or negative). ║ ║ Ackermann(5,1) is a bit impractical to calculate. ║ ╚═════════════════════════════════════════════════════════════╝*/high=24 do j=0 to 4; say do k=0 to high % (max(1, j)) call tell_Ack j, k if j==4 & k==2 then leave /*there's no sense in going overboard. */ end /*k*/ end /*j*/exit /*stick a fork in it, we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/tell_Ack: parse arg mm,nn; calls=0 /*display an echo message to terminal. */ #=right(nn,length(high)) say 'Ackermann('mm", "#')='right(ackermann(mm, nn), high), left('', 12) 'calls='right(calls, high) return/*──────────────────────────────────────────────────────────────────────────────────────*/ackermann: procedure expose calls /*compute value of Ackermann function. */ parse arg m,n; calls=calls+1 if m==0 then return n + 1 if m==1 then return n + 2 if m==2 then return n + 3 + n if m==3 then return 2**(n+3) - 3 if m==4 then do; #=2 /* [↓] Ugh! ··· and still more ughs.*/ do (n+3)-1 /*This is where the heavy lifting is. */ #=2**# end return #-3 end if n==0 then return ackermann(m-1, 1) return ackermann(m-1, ackermann(m, n-1) ) Output note: none of the numbers shown below use recursion to compute. output Ackermann(0, 0)= 1 calls= 1 Ackermann(0, 1)= 2 calls= 1 Ackermann(0, 2)= 3 calls= 1 Ackermann(0, 3)= 4 calls= 1 Ackermann(0, 4)= 5 calls= 1 Ackermann(0, 5)= 6 calls= 1 Ackermann(0, 6)= 7 calls= 1 Ackermann(0, 7)= 8 calls= 1 Ackermann(0, 8)= 9 calls= 1 Ackermann(0, 9)= 10 calls= 1 Ackermann(0,10)= 11 calls= 1 Ackermann(0,11)= 12 calls= 1 Ackermann(0,12)= 13 calls= 1 Ackermann(0,13)= 14 calls= 1 Ackermann(0,14)= 15 calls= 1 Ackermann(0,15)= 16 calls= 1 Ackermann(0,16)= 17 calls= 1 Ackermann(0,17)= 18 calls= 1 Ackermann(0,18)= 19 calls= 1 Ackermann(0,19)= 20 calls= 1 Ackermann(0,20)= 21 calls= 1 Ackermann(0,21)= 22 calls= 1 Ackermann(0,22)= 23 calls= 1 Ackermann(0,23)= 24 calls= 1 Ackermann(0,24)= 25 calls= 1 Ackermann(1, 0)= 2 calls= 1 Ackermann(1, 1)= 3 calls= 1 Ackermann(1, 2)= 4 calls= 1 Ackermann(1, 3)= 5 calls= 1 Ackermann(1, 4)= 6 calls= 1 Ackermann(1, 5)= 7 calls= 1 Ackermann(1, 6)= 8 calls= 1 Ackermann(1, 7)= 9 calls= 1 Ackermann(1, 8)= 10 calls= 1 Ackermann(1, 9)= 11 calls= 1 Ackermann(1,10)= 12 calls= 1 Ackermann(1,11)= 13 calls= 1 Ackermann(1,12)= 14 calls= 1 Ackermann(1,13)= 15 calls= 1 Ackermann(1,14)= 16 calls= 1 Ackermann(1,15)= 17 calls= 1 Ackermann(1,16)= 18 calls= 1 Ackermann(1,17)= 19 calls= 1 Ackermann(1,18)= 20 calls= 1 Ackermann(1,19)= 21 calls= 1 Ackermann(1,20)= 22 calls= 1 Ackermann(1,21)= 23 calls= 1 Ackermann(1,22)= 24 calls= 1 Ackermann(1,23)= 25 calls= 1 Ackermann(1,24)= 26 calls= 1 Ackermann(2, 0)= 3 calls= 1 Ackermann(2, 1)= 5 calls= 1 Ackermann(2, 2)= 7 calls= 1 Ackermann(2, 3)= 9 calls= 1 Ackermann(2, 4)= 11 calls= 1 Ackermann(2, 5)= 13 calls= 1 Ackermann(2, 6)= 15 calls= 1 Ackermann(2, 7)= 17 calls= 1 Ackermann(2, 8)= 19 calls= 1 Ackermann(2, 9)= 21 calls= 1 Ackermann(2,10)= 23 calls= 1 Ackermann(2,11)= 25 calls= 1 Ackermann(2,12)= 27 calls= 1 Ackermann(3, 0)= 5 calls= 1 Ackermann(3, 1)= 13 calls= 1 Ackermann(3, 2)= 29 calls= 1 Ackermann(3, 3)= 61 calls= 1 Ackermann(3, 4)= 125 calls= 1 Ackermann(3, 5)= 253 calls= 1 Ackermann(3, 6)= 509 calls= 1 Ackermann(3, 7)= 1021 calls= 1 Ackermann(3, 8)= 2045 calls= 1 Ackermann(4, 0)= 13 calls= 1 Ackermann(4, 1)= 65533 calls= 1 Ackermann(4, 2)=89506130880933368E+19728 calls= 1  ## Ring Translation of: C# for m = 0 to 3 for n = 0 to 4 see "Ackermann(" + m + ", " + n + ") = " + Ackermann(m, n) + nl nextnext func Ackermann m, n if m > 0 if n > 0 return Ackermann(m - 1, Ackermann(m, n - 1)) but n = 0 return Ackermann(m - 1, 1) ok but m = 0 if n >= 0 return n + 1 ok okRaise("Incorrect Numerical input !!!")  Output: Ackermann(0, 0) = 1 Ackermann(0, 1) = 2 Ackermann(0, 2) = 3 Ackermann(0, 3) = 4 Ackermann(0, 4) = 5 Ackermann(1, 0) = 2 Ackermann(1, 1) = 3 Ackermann(1, 2) = 4 Ackermann(1, 3) = 5 Ackermann(1, 4) = 6 Ackermann(2, 0) = 3 Ackermann(2, 1) = 5 Ackermann(2, 2) = 7 Ackermann(2, 3) = 9 Ackermann(2, 4) = 11 Ackermann(3, 0) = 5 Ackermann(3, 1) = 13 Ackermann(3, 2) = 29 Ackermann(3, 3) = 61 Ackermann(3, 4) = 125 ## Ruby Translation of: Ada def ack(m, n) if m == 0 n + 1 elsif n == 0 ack(m-1, 1) else ack(m-1, ack(m, n-1)) endend Example: (0..3).each do |m| puts (0..6).map { |n| ack(m, n) }.join(' ')end Output:  1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509 ## Run BASIC print ackermann(1, 2) function ackermann(m, n) if (m = 0) then ackermann = (n + 1) if (m > 0) and (n = 0) then ackermann = ackermann((m - 1), 1) if (m > 0) and (n > 0) then ackermann = ackermann((m - 1), ackermann(m, (n - 1)))end function ## Rust fn ack(m: isize, n: isize) -> isize { if m == 0 { n + 1 } else if n == 0 { ack(m - 1, 1) } else { ack(m - 1, ack(m, n - 1)) }} fn main() { let a = ack(3, 4); println!("{}", a); // 125}  Or:  fn ack(m: u64, n: u64) -> u64 { match (m, n) { (0, n) => n + 1, (m, 0) => ack(m - 1, 1), (m, n) => ack(m - 1, ack(m, n - 1)), }}  ## Sather class MAIN is ackermann(m, n:INT):INT pre m >= 0 and n >= 0 is if m = 0 then return n + 1; end; if n = 0 then return ackermann(m-1, 1); end; return ackermann(m-1, ackermann(m, n-1)); end; main is n, m :INT; loop n := 0.upto!(6); loop m := 0.upto!(3); #OUT + "A(" + m + ", " + n + ") = " + ackermann(m, n) + "\n"; end; end; end;end; Instead of INT, the class INTI could be used, even though we need to use a workaround since in the GNU Sather v1.2.3 compiler the INTI literals are not implemented yet. class MAIN is ackermann(m, n:INTI):INTI is zero ::= 0.inti; -- to avoid type conversion each time one ::= 1.inti; if m = zero then return n + one; end; if n = zero then return ackermann(m-one, one); end; return ackermann(m-one, ackermann(m, n-one)); end; main is n, m :INT; loop n := 0.upto!(6); loop m := 0.upto!(3); #OUT + "A(" + m + ", " + n + ") = " + ackermann(m.inti, n.inti) + "\n"; end; end; end;end; ## Scala def ack(m: BigInt, n: BigInt): BigInt = { if (m==0) n+1 else if (n==0) ack(m-1, 1) else ack(m-1, ack(m, n-1))} Example: scala> for ( m <- 0 to 3; n <- 0 to 6 ) yield ack(m,n)res0: Seq.Projection[BigInt] = RangeG(1, 2, 3, 4, 5, 6, 7, 2, 3, 4, 5, 6, 7, 8, 3, 5, 7, 9, 11, 13, 15, 5, 13, 29, 61, 125, 253, 509)  This example needs review by someone who knows about: Scala programming: Stream The following part of this "heavily recursive"??? is a way too complex solution and displays not the idiom of Scala. Memoization is by Stream (Last lesson of the Coursera course "Functional Programming Principles in Scala") which is much simpler and displays the power of the language. If you know Scala programming: Stream, please review this example and, as necessary, improve it or describe what should be done for someone who knows the language. I saw that this was one of the problems that needed an implementation in Scala, but when I made it, there was already one. But because ackerman is heavy recursive, I implemented it with memoization. So I still post it. Could still be done better, but that I leave to the reader. ;-) val maxDepth = 4 val ackMMap = scala.collection.mutable.Map[BigInt, BigInt]()val ackNMaps = Array.fill(maxDepth + 1) { scala.collection.mutable.Map[BigInt, BigInt]() } def ack(m: Int, n: BigInt): BigInt = { if ((m < 0) || (n < 0)) { throw new Exception("Negative parameters are not allowed: ack(%s, %s)".format(m, n)) } if (m > maxDepth) { throw new Exception("First parameter is greater as %s: ack(%s, %s)".format(maxDepth, m, n)) } val newM = m - 1 val newN = n - 1 if (m == 0) { n + 1 } else if (n == 0) { ackMMap.getOrElseUpdate(newM, ack(newM, 1)) } else { val createStep = 125 val index = m val mapCurrent = ackNMaps(index) val mapPrevious = ackNMaps(index - 1) val maxRecursion = 2 * createStep val nrOfElements : BigInt = if (mapCurrent.isEmpty) 0 else mapCurrent.max._1 if ((nrOfElements + maxRecursion) < n) { for (i <- nrOfElements + createStep to n by createStep) { mapCurrent.getOrElseUpdate(i, ack(m, i)) } } mapCurrent.getOrElseUpdate(n, { val ackVal = mapCurrent.getOrElseUpdate(newN, ack(m, newN)) mapPrevious.getOrElseUpdate(ackVal, ack(newM, ackVal)) }) }}  One important optimization is:  if ((nrOfElements + maxRecursion) < n) { for (i <- nrOfElements + createStep to n by createStep) { mapCurrent.getOrElseUpdate(i, ack(m, i)) } }  The recursion with ackermann can become very deep indeed. In this way it is still calculable without needing a very big stack. And because memoization is used, you circumvent a stack overflow without a (real) cost. ## Scheme (define (A m n) (cond ((= m 0) (+ n 1)) ((= n 0) (A (- m 1) 1)) (else (A (- m 1) (A m (- n 1)))))) ## Scilab clearfunction acker=ackermann(m,n) global calls calls=calls+1 if m==0 then acker=n+1 else if n==0 then acker=ackermann(m-1,1) else acker=ackermann(m-1,ackermann(m,n-1)) end endendfunctionfunction printacker(m,n) global calls calls=0 printf('ackermann(%d,%d)=',m,n) printf('%d calls=%d\n',ackermann(m,n),calls)endfunctionmaxi=3; maxj=6for i=0:maxi for j=0:maxj printacker(i,j) endend Output: ackermann(0,0)=1 calls=1 ackermann(0,1)=2 calls=1 ackermann(0,2)=3 calls=1 ackermann(0,3)=4 calls=1 ackermann(0,4)=5 calls=1 ackermann(0,5)=6 calls=1 ackermann(0,6)=7 calls=1 ackermann(1,0)=2 calls=2 ackermann(1,1)=3 calls=4 ackermann(1,2)=4 calls=6 ackermann(1,3)=5 calls=8 ackermann(1,4)=6 calls=10 ackermann(1,5)=7 calls=12 ackermann(1,6)=8 calls=14 ackermann(2,0)=3 calls=5 ackermann(2,1)=5 calls=14 ackermann(2,2)=7 calls=27 ackermann(2,3)=9 calls=44 ackermann(2,4)=11 calls=65 ackermann(2,5)=13 calls=90 ackermann(2,6)=15 calls=119 ackermann(3,0)=5 calls=15 ackermann(3,1)=13 calls=106 ackermann(3,2)=29 calls=541 ackermann(3,3)=61 calls=2432 ackermann(3,4)=125 calls=10307 ackermann(3,5)=253 calls=42438 ackermann(3,6)=509 calls=172233 ## Seed7 const func integer: ackermann (in integer: m, in integer: n) is func result var integer: result is 0; begin if m = 0 then result := succ(n); elsif n = 0 then result := ackermann(pred(m), 1); else result := ackermann(pred(m), ackermann(m, pred(n))); end if; end func; Original source: [2] ## SETL program ackermann; (for m in [0..3]) print(+/ [rpad('' + ack(m, n), 4): n in [0..6]]);end; proc ack(m, n); return {[0,n+1]}(m) ? ack(m-1, {[0,1]}(n) ? ack(m, n-1));end proc; end program; ## Shen (define ack 0 N -> (+ N 1) M 0 -> (ack (- M 1) 1) M N -> (ack (- M 1) (ack M (- N 1)))) ## Sidef func A(m, n) { m == 0 ? (n + 1) : (n == 0 ? (A(m - 1, 1)) : (A(m - 1, A(m, n - 1))));} Alternatively, using multiple dispatch: func A((0), n) { n + 1 }func A(m, (0)) { A(m - 1, 1) }func A(m, n) { A(m-1, A(m, n-1)) } Calling the function: say A(3, 2); # prints: 29 ## Simula as modified by R. Péter and R. Robinson:  BEGIN INTEGER procedure Ackermann(g, p); SHORT INTEGER g, p; Ackermann:= IF g = 0 THEN p+1 ELSE Ackermann(g-1, IF p = 0 THEN 1 ELSE Ackermann(g, p-1)); INTEGER g, p; FOR p := 0 STEP 3 UNTIL 13 DO BEGIN g := 4 - p/3; outtext("Ackermann("); outint(g, 0); outchar(','); outint(p, 2); outtext(") = "); outint(Ackermann(g, p), 0); outimage ENDEND Output: Ackermann(4, 0) = 13 Ackermann(3, 3) = 61 Ackermann(2, 6) = 15 Ackermann(1, 9) = 11 Ackermann(0,12) = 13  ## Slate [email protected](Integer traits) ackermann: [email protected](Integer traits)[ m isZero ifTrue: [n + 1] ifFalse: [n isZero ifTrue: [m - 1 ackermann: n] ifFalse: [m - 1 ackermann: (m ackermann: n - 1)]]]. ## Smalltalk |ackermann|ackermann := [ :n :m | (n = 0) ifTrue: [ (m + 1) ] ifFalse: [ (m = 0) ifTrue: [ ackermann value: (n-1) value: 1 ] ifFalse: [ ackermann value: (n-1) value: ( ackermann value: n value: (m-1) ) ] ]]. (ackermann value: 0 value: 0) displayNl.(ackermann value: 3 value: 4) displayNl. ## SNOBOL4 Works with: Macro Spitbol Both Snobol4+ and CSnobol stack overflow, at ack(3,3) and ack(3,4), respectively. define('ack(m,n)') :(ack_end)ack ack = eq(m,0) n + 1 :s(return) ack = eq(n,0) ack(m - 1,1) :s(return) ack = ack(m - 1,ack(m,n - 1)) :(return)ack_end * # Test and display ack(0,0) .. ack(3,6)L1 str = str ack(m,n) ' ' n = lt(n,6) n + 1 :s(L1) output = str; str = '' n = 0; m = lt(m,3) m + 1 :s(L1)end Output: 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509 ## SPAD  NNI ==> NonNegativeInteger A:(NNI,NNI) -> NNI A(m,n) == m=0 => n+1 m>0 and n=0 => A(m-1,1) m>0 and n>0 => A(m-1,A(m,n-1)) -- Example matrix [[A(i,j) for i in 0..3] for j in 0..3]  Output:  +1 2 3 5 + | | |2 3 5 13| (1) | | |3 4 7 29| | | +4 5 9 61+ Type: Matrix(NonNegativeInteger)  ## SNUSP  /==!/[email protected]@@[email protected]# | | Ackermann function | | /=========\!==\!====\ recursion:,@/>,@/==ack=!\?\<+# | | | A(0,j) -> j+1 j i \<?\+>[email protected]/# | | A(i,0) -> A(i-1,1) \@\>@\->@/@\<[email protected]/# A(i,j) -> A(i-1,A(i,j-1)) | | | # # | | | /+<<<-\ /-<<+>>\!=/ \=====|==!/========?\>>>=?/<<# ? ? | \<<<+>+>>-/ \>>+<<-/!==========/ # # One could employ tail recursion elimination by replacing "@/#" with "/" in two places above. ## Standard ML fun a (0, n) = n+1 | a (m, 0) = a (m-1, 1) | a (m, n) = a (m-1, a (m, n-1)) ## Stata matafunction ackermann(m,n) { if (m==0) { return(n+1) } else if (n==0) { return(ackermann(m-1,1)) } else { return(ackermann(m-1,ackermann(m,n-1))) }} for (i=0; i<=3; i++) printf("%f\n",ackermann(i,4))5611125end ## Swift func ackerman(m:Int, n:Int) -> Int { if m == 0 { return n+1 } else if n == 0 { return ackerman(m-1, 1) } else { return ackerman(m-1, ackerman(m, n-1)) }} ## Tcl ### Simple Translation of: Ruby proc ack {m n} { if {m == 0} { expr {n + 1} } elseif {n == 0} { ack [expr {m - 1}] 1 } else { ack [expr {m - 1}] [ack m [expr {n - 1}]] }} ### With Tail Recursion With Tcl 8.6, this version is preferred (though the language supports tailcall optimization, it does not apply it automatically in order to preserve stack frame semantics): proc ack {m n} { if {m == 0} { expr {n + 1} } elseif {n == 0} { tailcall ack [expr {m - 1}] 1 } else { tailcall ack [expr {m - 1}] [ack m [expr {n - 1}]] }} ### To Infinity… and Beyond! If we want to explore the higher reaches of the world of Ackermann's function, we need techniques to really cut the amount of computation being done. Works with: Tcl version 8.6 package require Tcl 8.6 # A memoization engine, from http://wiki.tcl.tk/18152oo::class create cache { filter Memoize variable ValueCache method Memoize args { # Do not filter the core method implementations if {[lindex [self target] 0] eq "::oo::object"} { return [next {*}args] } # Check if the value is already in the cache set key [self target],args if {[info exist ValueCache(key)]} { return ValueCache(key) } # Compute value, insert into cache, and return it return [set ValueCache(key) [next {*}args]] } method flushCache {} { unset ValueCache # Skip the cacheing return -level 2 "" }} # Make an object, attach the cache engine to it, and define ack as a methodoo::object create cachedoo::objdefine cached { mixin cache method ack {m n} { if {m==0} { expr {n+1} } elseif {m==1} { # From the Mathematica version expr {m+2} } elseif {m==2} { # From the Mathematica version expr {2*n+3} } elseif {m==3} { # From the Mathematica version expr {8*(2**n-1)+5} } elseif {n==0} { tailcall my ack [expr {m-1}] 1 } else { tailcall my ack [expr {m-1}] [my ack m [expr {n-1}]] } }} # Some small tweaks...interp recursionlimit {} 100000interp alias {} ack {} cacheable ack But even with all this, you still run into problems calculating ${\displaystyle {\mathit {ack}}(4,3)}$ as that's kind-of large… ## TSE SAL // library: math: get: ackermann: recursive <description></description> <version>1.0.0.0.5</version> <version control></version control> (filenamemacro=getmaare.s) [kn, ri, tu, 27-12-2011 14:46:59]INTEGER PROC FNMathGetAckermannRecursiveI( INTEGER mI, INTEGER nI ) IF ( mI == 0 ) RETURN( nI + 1 ) ENDIF IF ( nI == 0 ) RETURN( FNMathGetAckermannRecursiveI( mI - 1, 1 ) ) ENDIF RETURN( FNMathGetAckermannRecursiveI( mI - 1, FNMathGetAckermannRecursiveI( mI, nI - 1 ) ) )END PROC Main()STRING s1[255] = "2"STRING s2[255] = "3"IF ( NOT ( Ask( "math: get: ackermann: recursive: m = ", s1, _EDIT_HISTORY_ ) ) AND ( Length( s1 ) > 0 ) ) RETURN() ENDIFIF ( NOT ( Ask( "math: get: ackermann: recursive: n = ", s2, _EDIT_HISTORY_ ) ) AND ( Length( s2 ) > 0 ) ) RETURN() ENDIF Message( FNMathGetAckermannRecursiveI( Val( s1 ), Val( s2 ) ) ) // gives e.g. 9END ## TI-83 BASIC This program assumes the variables N and M are the arguments of the function, and that the list L1 is empty. It stores the result in the system variable ANS. (Program names can be no longer than 8 characters, so I had to truncate the function's name.) PROGRAM:ACKERMAN:If not(M:Then:N+1→N:Return:Else:If not(N:Then:1→N:M-1→M:prgmACKERMAN:Else:N-1→N:M→L1(1+dim(L1:prgmACKERMAN:Ans→N:L1(dim(L1))-1→M:dim(L1)-1→dim(L1:prgmACKERMAN:End:End Here is a handler function that makes the previous function easier to use. (You can name it whatever you want.) PROGRAM:AHANDLER:0→dim(L1:Prompt M:Prompt N:prgmACKERMAN:Disp Ans ## TI-89 BASIC Define A(m,n) = when(m=0, n+1, when(n=0, A(m-1,1), A(m-1, A(m, n-1)))) ## TorqueScript function ackermann(%m,%n){ if(%m==0) return %n+1; if(%m>0&&%n==0) return ackermann(%m-1,1); if(%m>0&&%n>0) return ackermann(%m-1,ackermann(%m,%n-1));} ## TXR Translation of: Scheme with memoization. (defmacro defmemofun (name (. args) . body) (let ((hash (gensym "hash-")) (argl (gensym "args-")) (hent (gensym "hent-")) (uniq (copy-str "uniq"))) ^(let ((,hash (hash :equal-based))) (defun ,name (,*args) (let* ((,argl (list ,*args)) (,hent (inhash ,hash ,argl ,uniq))) (if (eq (cdr ,hent) ,uniq) (set (cdr ,hent) (block ,name (progn ,*body))) (cdr ,hent))))))) (defmemofun ack (m n) (cond ((= m 0) (+ n 1)) ((= n 0) (ack (- m 1) 1)) (t (ack (- m 1) (ack m (- n 1)))))) (each ((i (range 0 3))) (each ((j (range 0 4))) (format t "ack(~a, ~a) = ~a\n" i j (ack i j)))) Output: ack(0, 0) = 1 ack(0, 1) = 2 ack(0, 2) = 3 ack(0, 3) = 4 ack(0, 4) = 5 ack(1, 0) = 2 ack(1, 1) = 3 ack(1, 2) = 4 ack(1, 3) = 5 ack(1, 4) = 6 ack(2, 0) = 3 ack(2, 1) = 5 ack(2, 2) = 7 ack(2, 3) = 9 ack(2, 4) = 11 ack(3, 0) = 5 ack(3, 1) = 13 ack(3, 2) = 29 ack(3, 3) = 61 ack(3, 4) = 125 ## UNIX Shell Works with: Bash ack() { local m=1 local n=2 if [ m -eq 0 ]; then echo -n ((n+1)) elif [ n -eq 0 ]; then ack ((m-1)) 1 else ack ((m-1)) (ack m ((n-1))) fi} Example: for ((m=0;m<=3;m++)); do for ((n=0;n<=6;n++)); do ack m n echo -n " " done echodone Output: 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 5 7 9 11 13 15 5 13 29 61 125 253 509 ## Ursala Anonymous recursion is the usual way of doing things like this. #import std#import nat ackermann = ~&al^?\[email protected] ~&ar?( ^R/~&f ^/[email protected] ^|R/~& ^|/~& predecessor, ^|R/~& ~&\1+ [email protected]) test program for the first 4 by 7 numbers: #cast %nLL test = block7 ackermann*K0 iota~~/4 7 Output: < <1,2,3,4,5,6,7>, <2,3,4,5,6,7,8>, <3,5,7,9,11,13,15>, <5,13,29,61,125,253,509>> ## V Translation of: Joy [ack [ [pop zero?] [popd succ] [zero?] [pop pred 1 ack] [true] [[dup pred swap] dip pred ack ack ] ] when]. using destructuring view [ack [ [pop zero?] [ [m n : [n succ]] view i] [zero?] [ [m n : [m pred 1 ack]] view i] [true] [ [m n : [m pred m n pred ack ack]] view i] ] when]. ## VBScript Based on BASIC version. Uncomment all the lines referring to depth and see just how deep the recursion goes. Implementation option explicit'~ dim depthfunction ack(m, n) '~ wscript.stdout.write depth & " " if m = 0 then '~ depth = depth + 1 ack = n + 1 '~ depth = depth - 1 elseif m > 0 and n = 0 then '~ depth = depth + 1 ack = ack(m - 1, 1) '~ depth = depth - 1 '~ elseif m > 0 and n > 0 then else '~ depth = depth + 1 ack = ack(m - 1, ack(m, n - 1)) '~ depth = depth - 1 end if end function Invocation wscript.echo ack( 1, 10 )'~ depth = 0wscript.echo ack( 2, 1 )'~ depth = 0wscript.echo ack( 4, 4 ) Output: 12 5 C:\foo\ackermann.vbs(16, 3) Microsoft VBScript runtime error: Out of stack space: 'ack'  ## Visual Basic Translation of: Rexx Works with: Visual Basic version VB6 Standard  Option ExplicitDim calls As LongSub main() Const maxi = 4 Const maxj = 9 Dim i As Long, j As Long For i = 0 To maxi For j = 0 To maxj Call print_acker(i, j) Next j Next iEnd Sub 'mainSub print_acker(m As Long, n As Long) calls = 0 Debug.Print "ackermann("; m; ","; n; ")="; Debug.Print ackermann(m, n), "calls="; callsEnd Sub 'print_ackerFunction ackermann(m As Long, n As Long) As Long calls = calls + 1 If m = 0 Then ackermann = n + 1 Else If n = 0 Then ackermann = ackermann(m - 1, 1) Else ackermann = ackermann(m - 1, ackermann(m, n - 1)) End If End IfEnd Function 'ackermann Output: ackermann( 0 , 0 )= 1 calls= 1 ackermann( 0 , 1 )= 2 calls= 1 ackermann( 0 , 2 )= 3 calls= 1 ackermann( 0 , 3 )= 4 calls= 1 ackermann( 0 , 4 )= 5 calls= 1 ackermann( 0 , 5 )= 6 calls= 1 ackermann( 0 , 6 )= 7 calls= 1 ackermann( 0 , 7 )= 8 calls= 1 ackermann( 0 , 8 )= 9 calls= 1 ackermann( 0 , 9 )= 10 calls= 1 ackermann( 1 , 0 )= 2 calls= 2 ackermann( 1 , 1 )= 3 calls= 4 ackermann( 1 , 2 )= 4 calls= 6 ackermann( 1 , 3 )= 5 calls= 8 ackermann( 1 , 4 )= 6 calls= 10 ackermann( 1 , 5 )= 7 calls= 12 ackermann( 1 , 6 )= 8 calls= 14 ackermann( 1 , 7 )= 9 calls= 16 ackermann( 1 , 8 )= 10 calls= 18 ackermann( 1 , 9 )= 11 calls= 20 ackermann( 2 , 0 )= 3 calls= 5 ackermann( 2 , 1 )= 5 calls= 14 ackermann( 2 , 2 )= 7 calls= 27 ackermann( 2 , 3 )= 9 calls= 44 ackermann( 2 , 4 )= 11 calls= 65 ackermann( 2 , 5 )= 13 calls= 90 ackermann( 2 , 6 )= 15 calls= 119 ackermann( 2 , 7 )= 17 calls= 152 ackermann( 2 , 8 )= 19 calls= 189 ackermann( 2 , 9 )= 21 calls= 230 ackermann( 3 , 0 )= 5 calls= 15 ackermann( 3 , 1 )= 13 calls= 106 ackermann( 3 , 2 )= 29 calls= 541 ackermann( 3 , 3 )= 61 calls= 2432 ackermann( 3 , 4 )= 125 calls= 10307 ackermann( 3 , 5 )= 253 calls= 42438 ackermann( 3 , 6 )= 509 calls= 172233 ackermann( 3 , 7 )= 1021 calls= 693964 ackermann( 3 , 8 )= 2045 calls= 2785999 ackermann( 3 , 9 )= 4093 calls= 11164370 ackermann( 4 , 0 )= 13 calls= 107 ackermann( 4 , 1 )= out of stack space ## Wart def (ackermann m n) (if m=0 n+1 n=0 (ackermann m-1 1) :else (ackermann m-1 (ackermann m n-1))) ## WDTE let memo a m n => switch true { == m 0 => + n 1; == n 0 => a (- m 1) 1; default => a (- m 1) (a m (- n 1));}; ## Wren  // To use recursion definition and declaration must be on separate linesvar Ackermann Ackermann = Fn.new {|m, n| if (m == 0) return n + 1 if (n == 0) return Ackermann.call(m - 1, 1) return Ackermann.call(m - 1, Ackermann.call(m, n - 1))}  ## XLISP (defun ackermann (m n) (cond ((= m 0) (+ n 1)) ((= n 0) (ackermann (- m 1) 1)) (t (ackermann (- m 1) (ackermann m (- n 1)))))) Test it: (print (ackermann 3 9)) Output (after a very perceptible pause): 4093 That worked well. Test it again: (print (ackermann 4 1)) Output (after another pause): Abort: control stack overflow happened in: #<Code ACKERMANN> ## XPL0 include c:\cxpl\codes; func Ackermann(M, N);int M, N;[if M=0 then return N+1; if N=0 then return Ackermann(M-1, 1);return Ackermann(M-1, Ackermann(M, N-1));]; \Ackermann int M, N;[for M:= 0 to 3 do [for N:= 0 to 7 do [IntOut(0, Ackermann(M, N)); ChOut(0,9\tab$$]; CrLf(0); ];] Recursion overflows the stack if either M or N is extended by a single count. Output: 1 2 3 4 5 6 7 8 2 3 4 5 6 7 8 9 3 5 7 9 11 13 15 17 5 13 29 61 125 253 509 1021  ## XSLT The following named template calculates the Ackermann function:  <xsl:template name="ackermann"> <xsl:param name="m"/> <xsl:param name="n"/> <xsl:choose> <xsl:when test="$m = 0">        <xsl:value-of select="$n+1"/> </xsl:when> <xsl:when test="$n = 0">        <xsl:call-template name="ackermann">          <xsl:with-param name="m" select="$m - 1"/> <xsl:with-param name="n" select="'1'"/> </xsl:call-template> </xsl:when> <xsl:otherwise> <xsl:variable name="p"> <xsl:call-template name="ackermann"> <xsl:with-param name="m" select="$m"/>            <xsl:with-param name="n" select="$n - 1"/> </xsl:call-template> </xsl:variable> <xsl:call-template name="ackermann"> <xsl:with-param name="m" select="$m - 1"/>          <xsl:with-param name="n" select="$p"/> </xsl:call-template> </xsl:otherwise> </xsl:choose> </xsl:template>  Here it is as part of a template  <?xml version="1.0" encoding="UTF-8"?><xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:template match="arguments"> <xsl:for-each select="args"> <div> <xsl:value-of select="m"/>, <xsl:value-of select="n"/>: <xsl:call-template name="ackermann"> <xsl:with-param name="m" select="m"/> <xsl:with-param name="n" select="n"/> </xsl:call-template> </div> </xsl:for-each> </xsl:template> <xsl:template name="ackermann"> <xsl:param name="m"/> <xsl:param name="n"/> <xsl:choose> <xsl:when test="$m = 0">        <xsl:value-of select="$n+1"/> </xsl:when> <xsl:when test="$n = 0">        <xsl:call-template name="ackermann">          <xsl:with-param name="m" select="$m - 1"/> <xsl:with-param name="n" select="'1'"/> </xsl:call-template> </xsl:when> <xsl:otherwise> <xsl:variable name="p"> <xsl:call-template name="ackermann"> <xsl:with-param name="m" select="$m"/>            <xsl:with-param name="n" select="$n - 1"/> </xsl:call-template> </xsl:variable> <xsl:call-template name="ackermann"> <xsl:with-param name="m" select="$m - 1"/>          <xsl:with-param name="n" select="\$p"/>        </xsl:call-template>      </xsl:otherwise>    </xsl:choose>  </xsl:template></xsl:stylesheet> 

Which will transform this input

 <?xml version="1.0" ?><?xml-stylesheet type="text/xsl" href="ackermann.xslt"?><arguments>  <args>    <m>0</m>    <n>0</n>  </args>  <args>    <m>0</m>    <n>1</n>  </args>  <args>    <m>0</m>    <n>2</n>  </args>  <args>    <m>0</m>    <n>3</n>  </args>  <args>    <m>0</m>    <n>4</n>  </args>  <args>    <m>0</m>    <n>5</n>  </args>  <args>    <m>0</m>    <n>6</n>  </args>  <args>    <m>0</m>    <n>7</n>  </args>  <args>    <m>0</m>    <n>8</n>  </args>  <args>    <m>1</m>    <n>0</n>  </args>  <args>    <m>1</m>    <n>1</n>  </args>  <args>    <m>1</m>    <n>2</n>  </args>  <args>    <m>1</m>    <n>3</n>  </args>  <args>    <m>1</m>    <n>4</n>  </args>  <args>    <m>1</m>    <n>5</n>  </args>  <args>    <m>1</m>    <n>6</n>  </args>  <args>    <m>1</m>    <n>7</n>  </args>  <args>    <m>1</m>    <n>8</n>  </args>  <args>    <m>2</m>    <n>0</n>  </args>  <args>    <m>2</m>    <n>1</n>  </args>  <args>    <m>2</m>    <n>2</n>  </args>  <args>    <m>2</m>    <n>3</n>  </args>  <args>    <m>2</m>    <n>4</n>  </args>  <args>    <m>2</m>    <n>5</n>  </args>  <args>    <m>2</m>    <n>6</n>  </args>  <args>    <m>2</m>    <n>7</n>  </args>  <args>    <m>2</m>    <n>8</n>  </args>  <args>    <m>3</m>    <n>0</n>  </args>  <args>    <m>3</m>    <n>1</n>  </args>  <args>    <m>3</m>    <n>2</n>  </args>  <args>    <m>3</m>    <n>3</n>  </args>  <args>    <m>3</m>    <n>4</n>  </args>  <args>    <m>3</m>    <n>5</n>  </args>  <args>    <m>3</m>    <n>6</n>  </args>  <args>    <m>3</m>    <n>7</n>  </args>  <args>    <m>3</m>    <n>8</n>  </args></arguments> 

into this output

0, 0: 1
0, 1: 2
0, 2: 3
0, 3: 4
0, 4: 5
0, 5: 6
0, 6: 7
0, 7: 8
0, 8: 9
1, 0: 2
1, 1: 3
1, 2: 4
1, 3: 5
1, 4: 6
1, 5: 7
1, 6: 8
1, 7: 9
1, 8: 10
2, 0: 3
2, 1: 5
2, 2: 7
2, 3: 9
2, 4: 11
2, 5: 13
2, 6: 15
2, 7: 17
2, 8: 19
3, 0: 5
3, 1: 13
3, 2: 29
3, 3: 61
3, 4: 125
3, 5: 253
3, 6: 509
3, 7: 1021
3, 8: 2045


## Yabasic

sub ack(M,N)    if M = 0 return N + 1    if N = 0 return ack(M-1,1)    return ack(M-1,ack(M, N-1))end sub print ack(3, 4) 

## Yorick

func ack(m, n) {    if(m == 0)        return n + 1;    else if(n == 0)        return ack(m - 1, 1);    else        return ack(m - 1, ack(m, n - 1));}

Example invocation:

for(m = 0; m <= 3; m++) {    for(n = 0; n <= 6; n++)        write, format="%d ", ack(m, n);    write, "";}
Output:
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 5 7 9 11 13 15
5 13 29 61 125 253 509

## ZED

Source -> http://ideone.com/53FzPA Compiled -> http://ideone.com/OlS7zL

(A) m ncomment:(=) m 0(add1) n (A) m ncomment:(=) n 0(A) (sub1) m 1 (A) m ncomment:#true(A) (sub1) m (A) m (sub1) n (add1) ncomment:#true(003) "+" n 1 (sub1) ncomment:#true(003) "-" n 1 (=) n1 n2comment:#true(003) "=" n1 n2

## ZX Spectrum Basic

Translation of: BASIC256
10 DIM s(2000,3)20 LET s(1,1)=3: REM M30 LET s(1,2)=7: REM N40 LET lev=150 GO SUB 10060 PRINT "A(";s(1,1);",";s(1,2);") = ";s(1,3)70 STOP 100 IF s(lev,1)=0 THEN LET s(lev,3)=s(lev,2)+1: RETURN 110 IF s(lev,2)=0 THEN LET lev=lev+1: LET s(lev,1)=s(lev-1,1)-1: LET s(lev,2)=1: GO SUB 100: LET s(lev-1,3)=s(lev,3): LET lev=lev-1: RETURN 120 LET lev=lev+1130 LET s(lev,1)=s(lev-1,1)140 LET s(lev,2)=s(lev-1,2)-1150 GO SUB 100160 LET s(lev,1)=s(lev-1,1)-1170 LET s(lev,2)=s(lev,3)180 GO SUB 100190 LET s(lev-1,3)=s(lev,3)200 LET lev=lev-1210 RETURN 
Output:
A(3,7) = 1021`