Ackermann function: Difference between revisions
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[ack |
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[ [pop zero?] [popd succ] |
[ [pop zero?] [popd succ] |
Revision as of 12:48, 25 September 2008
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
The Ackermann function is a classic recursive example in computer science. It is a function that grows very quickly (in its value and in the size of its call tree). It is defined as follows:
n+1 if m=0 A(m, n) = A(m-1, 1) if m>0 and n=0 A(m-1, A(m,n-1)) if m>0 and n>0
Its arguments are never negative and it always terminates. Write a function which returns the value of A(m, n). Arbitrary precision is preferred (since the funciton grows so quickly), but not required.
Java
<java>public static BigInteger ack(BigInteger m, BigInteger n){ if(m.equals(BigInteger.ZERO)) return n.add(BigInteger.ONE);
if(m.compareTo(BigInteger.ZERO) > 0 && n.equals(BigInteger.ZERO)) return ack(m.subtract(BigInteger.ONE), BigInteger.ONE);
if(m.compareTo(BigInteger.ZERO) > 0 && n.compareTo(BigInteger.ZERO) > 0) return ack(m.subtract(BigInteger.ONE), ack(m, n.subtract(BigInteger.ONE)));
return null; }</java>
Joy
From here
DEFINE ack == [ [ [pop null] popd succ ] [ [null] pop pred 1 ack ] [ [dup pred swap] dip pred ack ack ] ] cond.
another using a combinator
DEFINE ack == [ [ [0 =] [pop 1 +] ] [ [swap 0 =] [popd 1 - 1 swap] [] ] [ [dup rollup [1 -] dip] [swap 1 - ack] ] ] condlinrec.
Python
<python>Python 2.5.1 (r251:54863, May 2 2007, 08:46:07) [GCC 3.3.4 (pre 3.3.5 20040809)] on linux2 IDLE 1.2.1 >>> import sys
>>> def ack(M, N): return ( ( N + 1 ) if M == 0 else ( ( ack(M-1, 1) ) if N == 0 else ( ( ack(M-1, ack(M, N-1)) ) )))
>>> sys.setrecursionlimit(3000) >>> ack(0,0) 1 >>> ack(3,4) 125</python>
SNUSP
/==!/==atoi=@@@-@-----# | | Ackermann function | | /=========\!==\!====\ recursion: $,@/>,@/==ack=!\?\<+# | | | A(0,j) -> j+1 j i \<?\+>-@/# | | A(i,0) -> A(i-1,1) \@\>@\->@/@\<-@/# A(i,j) -> A(i-1,A(i,j-1)) | | | # # | | | /+<<<-\ /-<<+>>\!=/ \=====|==!/========?\>>>=?/<<# ? ? | \<<<+>+>>-/ \>>+<<-/!==========/ # #
V
[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].