Mutual recursion: Difference between revisions
Content deleted Content added
Line 2,990:
echo implode(" ", $rb) . "\n";
?></lang>
=={{header|Picat}}==
Here are two approaches:
* a (tabled) function based: f/1 and f/2
* a (tabled) predicate based: male/2 and female/2 {{trans|Prolog}}
For small values (say N < 50) tabling is not really needed.
<lang Picat>go =>
N = 30,
println(func),
test_func(N),
println(pred),
test_pred(N),
nl.
nl.
% Testing the function based approach
test_func(N) =>
println([M : I in 0..N, male(I,M)]),
println([F : I in 0..N, female(I,F)]),
nl.
% Testing the predicate approach
test_pred(N) =>
println([M : I in 0..N, male(I,M)]),
println([F : I in 0..N, female(I,F)]),
nl.
% Function based
table
f(0) = 1.
f(N) = N - m(f(N-1)), N > 0 => true.
table
m(0) = 0.
m(N) = N - f(m(N-1)), N > 0 => true.
% predicate based
% translation of the Prolog solution
table
female(0,1).
female(N,F) :-
N>0,
N1 = N-1,
female(N1,R),
male(R, R1),
F = N-R1.
table
male(0,0).
male(N,F) :-
N>0,
N1 = N-1,
male(N1,R),
female(R, R1),
F = N-R1.</lang>
{{out}}
<pre>func
[0,0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12,12,13,14,14,15,16,16,17,17,18,19]
[1,1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12,13,13,14,14,15,16,16,17,17,18,19]
pred
[0,0,1,2,2,3,4,4,5,6,6,7,7,8,9,9,10,11,11,12,12,13,14,14,15,16,16,17,17,18,19]
[1,1,2,2,3,3,4,5,5,6,6,7,8,8,9,9,10,11,11,12,13,13,14,14,15,16,16,17,17,18,19]</pre>
For larger values, tabling is essential and then one can discern that the predicate based approach is a little faster.
Here are the times for testing N=1 000 000:
* func: 1.829s
* pred: 1.407s
=={{header|PicoLisp}}==
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