The WAM: A Tutorial Reconstruction states that a query, p(Z, h(Z,W), f(W)), needs to be flattened using the following principles:
That being said, the query flattened form is:
X3=h(X2, X5), X4=f(X5), X1=p(X2, X3, X4);
I am lost with the definition of external variable, consider the following:
p(Z, h(Y, a(K, C), b(W)), f(W)).
Is Y an external variable? How should be the flattened form for this? From my understanding this would be the construction:
X1 = p(X2, X3, X4)
X2 = Z
X3 = h(X5, X6, X7)
X4 = f(X8)
X5 = Y
X6 = a(X7, X8)
X7 = K
X8 = C
X9 = b(X5)
But I am not sure, starting at X4 I got confused, should I have assigned the h inner values first?
You have the order the wrong way around: You are building terms before you have built their arguments. The text says to build the arguments before you build the outer terms. For example, you must build a(K, C) before you can build h(..., a(K, C), ...), and you must build that before you can build p(..., h(..., a(K, C), ...), ...). Here is one legal order:
X7 = K
X8 = C
X6 = a(X7, X8)
X5 = Y
X9 = b(X5)
X2 = Z
X3 = h(X5, X6, X7)
X4 = f(X8)
X1 = p(X2, X3, X4)
Related
I'm new to Prolog and I would like to convert given rules to Prolog language.
I have to define a predicate p(X1, Y1, X2, Y2) ,where X1, Y1 are integers, and X2 and Y2 are obtained from X1, Y1 following these rules:
if Y1 ≤ 0 and |X1|≤−Y1 then X2 is X1 + 1 and Y2 is Y1
else if X1 > 0 and |Y1| < X1 then X2 is X1, and Y2 is Y1+1
else if Y1 > 0 and −Y1 < X1 ≤ Y1 then X2 is X1−1, and Y2 is Y1
else X2 is X1 and Y2 is Y1−1
I'll provide a steer, here is rule 1:
p(X1, Y1, X2, Y2) :-
Y1 =< 0,
abs(X1) =< -Y1,
% Cut, to prevent processing alternatives
!,
X2 is X1 + 1,
Y2 is Y1.
Sample output in swi-prolog:
?- p(-3, -5, X2, Y2).
X2 = -2,
Y2 = -5.
So I recently started to learn Prolog and I had a question regarding making a predicate which gives u all the possible solutions (next moves) as the current spot is given. The best example is a maze. So this is my data which tells me 'w = white' and 'b = black' within a 5x5 maze:
grid([ [w, w, w, b, w],
[b ,b, w, w, w],
[w, w, w, b, w],
[w, b, b, b, b],
[w, w, w, w, w] ]).
I also implemented a predicated called white/1, which tells me if the given spot in the maze is white:
white(X/Y) :-
grid(M),
nth1(X, M, Line),
nth1(Y, Line, w).
What I now want to do is make a predicate that gives me all the possible moves within the maze. For example if I query:
?- move(5/2, NextState).
NextState = 4/2 ;
NextState = 5/1 ;
NextState = 5/3 ;
No
This is my code, but it gives false and I know it's completely wrong, but I don't know how to implement such a predicate:
move(5/5, _).
move(X/Y, NextState) :-
white(X/Y),
X1 is X + 1,
X1 =< 5,
move(X1/Y, NextState),
Y1 is Y + 1,
Y1 =< 5,
move(X/Y1, NextState),
X2 is X - 1,
X2 >= 5,
move(X2/Y, NextState),
Y2 is Y - 1,
Y2 >= 0,
move(X/Y2, NextState).
If someone could help me, I would really appreciate it! :)
EDIT:
move(X/Y, _) :-
white(X/Y).
move(X/Y, X/Y) :-
X1 is X + 1,
move(X/Y, X1/Y);
X2 is X - 1,
move(X/Y, X2/Y);
Y1 is Y + 1,
move(X/Y, X/Y1);
Y2 is Y - 1,
move(X/Y, X/Y2).
If I query it gives me:
?- move(3/3, NextState).
true ;
NextState = 3/3 ;
NextState = 3/3 ;
NextState = 3/3 ;
NextState = 3/3 ;
move(X/Y, X1/Y1, Xm/Ym) :-
X1 is X + 1, Y1 = Y, X1 =< Xm;
X1 is X - 1, Y1 = Y, X1 > 0;
X1 = X, Y1 is Y + 1, Y1 =< Ym;
X1 = X, Y1 is Y - 1, Y1 > 0.
For each line in the above predicate we have a new direction. The Xm/Ym are bounds of the maze.
| ?- move(5/2, X, 5/5).
X = 4/2 ? ;
X = 5/3 ? ;
X = 5/1
yes
You can remove Xm/Ym as an argument and just put 5 in the body. Or you can define a new predicate
move1(Current, Next) :- move(Current, Next, 5/5)
Instead of disjunctions, we can also write multiple clauses.
move(X/Y, X1/Y) :- X1 is X + 1, X1 =< 5.
move(X/Y, X1/Y) :- X1 is X - 1, X1 > 0.
move(X/Y, X/Y1) :- Y1 is Y + 1, Y1 =< 5.
move(X/Y, X/Y1) :- Y1 is Y - 1, Y1 > 0.
Both are equivalent and this is even clearer.
How can I find the result of an arithmetic expression composed of pluses and minuses without using 'is'?
For example, to find the result of 1+2+3+4-6, I do: X is 1+2+3+4-6 and I get: X = 4.
What I've tried so far:
eval(EXPR, EXPR) :-
integer(EXPR),
!.
eval(X+Y, RES) :-
eval(X, X1),
eval(Y, Y1),
plus(X1, Y1, RES).
eval(X-Y, RES) :-
eval(X, X1),
eval(Y, Y1),
Y2 = -Y1,
plus(X1, Y2, RES).
but when I try to compute an expression containing negative numbers, I get the error: `integer' expected, found `- number` (a compound).
How can I solve it?
Add plus(Y1,Y2,0) instead of Y2 = -Y1.
In the latter case, Y2 is a structure with functor - and Argument Y1, whereas in the former case, Y1 is a number.
Take Y1 = 3 for example. [Just for fun: try it with Y1 = -3]
?- Y1 = 3, Y2 = -Y1, Y2 =.. L.
Y1 = 3,
Y2 = - 3,
L = [-, 3].
?- Y1 = 3, plus(Y1,Y2,0), Y2 =.. L.
Y1 = 3,
Y2 = -3,
L = [-3].
However, Y2 is supposed to be an argument in the predicate plus/3, which does not allow structures as arguments.
Maybe you prefer this solution:
eval(X-Y, RES) :-
eval(X, X1),
eval(Y, Y1),
plus(Y1, RES, X1).
I am trying to compile a SWI-Prolog program but keep getting test is always true, var (sum) error on line 7. I cant figure out what this means. Can someone please help? This is a program I am hoping will eventually solve Latin squares. Thank you.
:- use_module(library(clpfd)).
magic_counter(Count) :-
findall(a, magic_1(_Soln), As),
length(As, Count).
magic_1(Soln) :-
Row1 = [W1, W2, W3],
Row2 = [X1, X2, X3],
Row3 = [Y1, Y2, Y3],
Row1 ins 1..3,
Row2 ins 1..3,
Row3 ins 1..3,
Sum #= 6,
all_different(Row1),
all_different(Row2),
all_different(Row3),
all_different([W1,X1,Y1]),
all_different([W2,X2,Y2]),
all_different([W3,X3,Y3]),
W1 + W2 + W3 #= Sum,
X1 + X2 + X3 #= Sum,
Y1 + Y2 + Y3 #= Sum,
W1 + X1 + Y1 #= Sum,
W2 + X2 + Y2 #= Sum,
W3 + X3 + Y3 #= Sum,
append(Row1,Row2,Row12),
append(Row12,Row3,Soln),
labeling([], Soln).
It's a warning, not an error.
I posted some time again a request on SWI-Prolog mailing list for this problem, since some existing code begun to raise this warning after a refinement to messages. Here is the answer from Jan.
I think you can ignore the warning, or disable it, but this seems not advisable.
If a row consists of three integers between 1 and 3, and these integers must be distinct, then the sum of such a row must be 6 (by definition). Stating that the sum of such a row equals 6 is therefore an idle constraint. The same reasoning applies to the 'columns' for which you assert similar disjointness constraints.
Edit: Even though the above reasoning is correct, this is not the origin of the warning. Carlo is right on this, it simply depends on the way in which the constraint library rewrites the constraints.
test1:-
L = [X],
L ins 1..2,
Y #= 2,
X #= Y.
test2:-
L = [X],
L ins 1..2,
X #= 2.
test1/0 gives the warning, test2/0 does not. Still, I find it difficult to see why the warning is given in the first place, i.e. what the rational behind it is. For example, here is the expansion of test1/0 (notice my comments):
:- use_module(library(clpfd)).
test1:-
A=[D],
A ins 1..2,
(
integer(B)
->
(
var(2)
->
2 is B
;
true
->
B=:=2
;
C is B,
clpfd:clpfd_equal(C, 2)
)
;
true
->
(
var(B) % This does not throw a warning.
->
B is 2
;
C is 2,
clpfd:clpfd_equal(B, C)
)
;
clpfd:clpfd_equal(B, 2)
),
(
integer(D)
->
(
var(B) % This throws a "Test is always true" warning.
->
B is D
;
integer(B)
->
D=:=B
;
E is D,
clpfd:clpfd_equal(E, B)
)
;
integer(B)
->
(
var(D)
->
D is B
;
E is B,
clpfd:clpfd_equal(D, E)
)
;
clpfd:clpfd_equal(D, B)
).
Given the equations
eqn1 = 5 x1 + 2 x2 + 3 x3 == 8
eqn2 = 4 x1 + 7 x2 + 9 x3 == 5
eqn3 = 6 x1 + x2 + 9 x3 == 2
how do I extract the coefficients of x1, x2, x3 to form a matrix?
I tried using CoefficientArrays but the output was given as a SparseArray.
Try Normal
(Normal[CoefficientArrays[{eqn1, eqn2, eqn3}, {x1, x2, x3}]][[2]]) // MatrixForm
I am not fond of Normal
Coefficient[# /. Equal[e_, _] -> e, {x1, x2, x3}] & /# {eqn1, eqn2, eqn3}
Shorter but not as clear:
Coefficient[First##, {x1, x2, x3}] & /# {eqn1, eqn2, eqn3}