Ackermann function: Difference between revisions
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A(m-1, A(m,n-1)) if m>0 and n>0 |
A(m-1, A(m,n-1)) if m>0 and n>0 |
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Its arguments are never negative and it always terminates. Write a function which returns the value of <tt>A(m, n)</tt>. Arbitrary precision is preferred (since the funciton grows so quickly), but not required. |
Its arguments are never negative and it always terminates. Write a function which returns the value of <tt>A(m, n)</tt>. Arbitrary precision is preferred (since the funciton grows so quickly), but not required. |
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=={{header|Forth}}== |
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: ack ( j i -- a ) |
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?dup if swap ?dup if 1- over recurse |
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else 1 |
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then swap 1- recurse |
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else 1+ then ; |
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=={{header|Java}}== |
=={{header|Java}}== |
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Revision as of 14:59, 25 September 2008
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.
Forth
: ack ( j i -- a )
?dup if swap ?dup if 1- over recurse
else 1
then swap 1- recurse
else 1+ then ;
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>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)))</python>
Example of use<python>>>> import sys >>> 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].