Continued fraction: Difference between revisions
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Pi: 3.141592654
</pre>
=={{header|Picat}}==
For Pi a test is added with a higher precision (200 -> 2000) to get a better result.
First a recursive version (based on Prolog's implementation):
<lang Picat>go =>
% square root 2
continued_fraction(200, sqrt_2_ab, V1),
printf("sqrt(2) = %w (diff: %0.15f)\n", V1, V1-sqrt(2)),
% napier
continued_fraction(200, napier_ab, V2),
printf("e = %w (diff: %0.15f)\n", V2, V2-math.e),
% pi
continued_fraction(200, pi_ab, V3),
printf("pi = %w (diff: %0.15f)\n", V3, V3-math.pi),
% get a better precision
continued_fraction(20000, pi_ab, V3b),
printf("pi = %w (diff: %0.15f)\n", V3b, V3b-math.pi),
nl.
continued_fraction(N, Compute_ab, V) ?=>
continued_fraction(N, Compute_ab, 0, V).
continued_fraction(0, Compute_ab, Temp, V) ?=>
call(Compute_ab, 0, A, _),
V = A + Temp.
continued_fraction(N, Compute_ab, Tmp, V) =>
call(Compute_ab, N, A, B),
Tmp1 = B / (A + Tmp),
N1 = N - 1,
continued_fraction(N1, Compute_ab, Tmp1, V).
% definitions for square root of 2
sqrt_2_ab(0, 1, 1).
sqrt_2_ab(_, 2, 1).
% definitions for napier
napier_ab(0, 2, _).
napier_ab(1, 1, 1).
napier_ab(N, N, V) :-
V is N - 1.
% definitions for pi
pi_ab(0, 3, _).
pi_ab(N, 6, V) :-
V is (2 * N - 1)*(2 * N - 1).</lang>
Output:
<pre>
sqrt(2) = 1.414213562373095 (diff: 0.000000000000000)
e = 2.718281828459046 (diff: 0.000000000000000)
pi = 3.141592622804847 (diff: -0.000000030784946)
pi = 3.141592653589762 (diff: -0.000000000000031)
</pre>
Then an iterative version (from Python's Fast Iterative version)
<lang Picat>continued_fraction_it(Fun, N) = Ret =>
Temp = 0.0,
foreach(I in N..-1..1)
[A,B] = apply(Fun,I),
Temp := B / (A + Temp)
end,
F = apply(Fun,0),
Ret = F[1] + Temp.
fsqrt2(N) = [cond(N > 0, 2, 1),1].
fnapier(N) = [cond(N > 0, N,2), cond(N>1,N-1,1)].
fpi(N) = [cond(N>0,6,3), (2*N-1) ** 2].</lang>
Which has exactly the same output as the recursive version.
=={{header|PicoLisp}}==
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