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Find palindromic numbers in both binary and ternary bases: Difference between revisions
Find palindromic numbers in both binary and ternary bases (view source)
Revision as of 08:53, 31 May 2022
, 1 year agoPicat constrain solver solution
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(Picat constrain solver solution) |
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2202021211210100110100002202101000110000220121210220000110001012022000010110010121121202022
</pre>
=={{header|Picat}}==
<lang Picat>
import sat.
to_num(List, Base, Num) =>
Len = length(List),
Num #= sum([List[I] * Base**(Len-I) : I in 1..Len]).
palindrom(S) =>
N = len(S),
Start :: 1..N, % start at the first non-zero position:
foreach(I in 1..N)
I1 #= max(1, min(N, N-(I-Start))), % I1 is the symmetry index partner of I (if relevant)
element(I1, S, S1), % S1 is the respective digit
I #< Start #=> S[I] #= 0, % skip leading 0´s
I #= Start #=> S[I] #> 0, % Start points to the first non-zero digit
I #>= Start #=> S[I] #= S1 % palindromic symmetry
end.
constrain(Max, B, X) =>
Len = floor(log(Max) / log(B)) + 1, % length of Max in Base B representation
Digits = new_list(Len), Digits :: 0..B-1,
to_num(Digits, B, X), % Digits show the Base B representation of X
palindrom(Digits).
main =>
N = 11, % maximum number of decimal digits for search, can be set freely
Max = 10**N - 1, % maximum number
X :: 2..Max,
constrain(Max, 2, X),
constrain(Max, 3, X),
Pnumbers = solve_all([X]),
foreach([Y] in [[0], [1]] ++ Pnumbers.sort()) % start with 0 and 1, then show solutions > 1
printf("%w %s %s%n", Y, to_radix_string(Y,2), to_radix_string(Y,3))
end.
</lang>
Output:
<pre>0 0 0
1 1 1
6643 1100111110011 100010001
1422773 101011011010110110101 2200021200022
5415589 10100101010001010100101 101012010210101
90396755477 1010100001100000100010000011000010101 22122022220102222022122</pre>
=={{header|PicoLisp}}==
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