Pascal's triangle/Puzzle: Difference between revisions
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Two-liner: x=5, y=13, z=8
</pre>
=={{header|Picat}}==
Below are three different approaches using constraint modelling (cp solver).
{{trans|Prolog}}
<lang Picat>import cp.
go =>
puzzle(T, X, Y,Z),
foreach(TT in T)
println(TT)
end,
println([x=X,y=Y,z=Z]),
nl,
fail, % are there any more solutions?
nl.
% Port of the Prolog solution
puzzle(Ts, X, Y, Z) :-
Ts = [ [151],
[_, _],
[40, _, _],
[_, _, _, _],
[X, 11, Y, 4, Z]],
Y #= X + Z,
triangle(Ts),
Vs = vars(Ts),
Vs :: 0..10000,
solve(Vs).
triangle([T|Ts]) :-
( Ts = [N|_] -> triangle_(T, N), triangle(Ts) ; true ).
triangle_([], _).
triangle_([T|Ts],[A,B|Rest]) :-
T #= A + B, triangle_(Ts, [B|Rest]).</lang>
{{out}}
<pre>[151]
[81,70]
[40,41,29]
[16,24,17,12]
[5,11,13,4,8]
[x = 5,y = 13,z = 8]</pre>
Here is a contraint model which calculates the positions in the triangle that should be added together.
<lang Picat>import cp.
puzzle2 =>
N = 5, % number of rows
Len = (N*(N+1)) div 2, % number of entries
% The triangle numbers for 1..N
T = [I*(I+1) div 2 : I in 1..N],
% The index of first number to use in addition
% create the indices of the numbers to add,
% i.e. Adds[I] + Adds[I+1]
Adds = new_list(T[N-1]),
Adds[1] := 2,
foreach(I in 2..T[N-1])
% "jump" of 2 when i-1 is a triangle number
if membchk(I-1,T) then
Adds[I] := Adds[I-1] + 2
else
Adds[I] := Adds[I-1] + 1
end
end,
% the pyramid
MaxVal = 10_000,
L = new_list(Len),
L :: 1..MaxVal,
% The clues.
L = [ 151,
_, _,
40, _, _,
_, _,_ , _ ,
X, 11, Y, 4, Z
],
% The sums
foreach(I in 1..T[N-1])
L[I] #= L[Adds[I]]+L[Adds[I]+1]
end,
% The extra constraint
Y #= X + Z,
solve(L),
println([x=X,y=Y,z=Z]),
fail, % check if there is another solution
nl.</lang>
{{out}}
<pre>[x = 5,y = 13,z = 8]</pre>
And then a version with hard coded constraints:
<lang Picat>import cp.
puzzle3 =>
% 1
% 2 3
% 4 5 6
% 7 8 9 10
%11 12 13 14 15
X = new_list(15),
X :: 0..10_000,
X[1] #= X[2]+X[3],
X[2] #= X[4]+X[5],
X[3] #= X[5]+X[6],
X[4] #= X[7]+X[8],
X[5] #= X[8]+X[9],
X[6] #= X[9]+X[10],
X[7] #= X[11]+X[12],
X[8] #= X[12]+X[13],
X[9] #= X[13]+X[14],
X[10] #= X[14]+X[15],
X[13] #= X[11] + X[15], % Y=X+Z,
% The hints
X[1] #= 151,
X[4] #= 40,
X[12] #= 11,
X[14] #= 4,
solve(X),
println([x=X[11],y=X[13],z=X[15]]),
fail,
nl.</lang>
{{out}}
<pre>[x = 5,y = 13,z = 8]</pre>
=={{header|PicoLisp}}==
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