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Currently working on an exercise where, given some predicate, the syntactic complexity has to be calculated. The syntactic complexity of some predicate is calculated as follows:
If the predicate is atomic or a function, its complexity is 2.
If the predicate is a variable, its complexity is 1.
For example, the syntactic complexity of loyalty(father(bob, Y), X) is worked out as follows:
loyalty = 2 (function)
father = 2 (function)
bob = 2 (atom)
Y = 1 (variable)
X = 1 (variable)
Total = 8
The approach taken was calculating such complexity if the predicate was in the form of a nested list, i.e. loyalty(father(bob, Y), X) = [loyalty, father, bob, Y, X], as follows:
complexity([], 0).
complexity([H|L], C) :- atomic(H), complexity(L, C1), C is C1+2.
complexity([H|L], C) :- var(H), complexity(L, C1), C is C1+1.
The remaining issue is converting the predicate to a flat list, as shown above. The ..= is useful, but its output is not complete, that is:
loyalty(father(bob, Y), X) ..= ["loyalty", "father(bob, Y)", "X"]
Any help would be appreciated.
You must apply =.. recursively as follows:
% term_to_list(+Term, -List)
term_to_list(Term, [Term]) :- var(Term), !.
term_to_list(Term, [Term]) :- atomic(Term), !.
term_to_list(Term, List) :-
compound(Term),
Term =.. Components,
maplist(term_to_list, Components, ListOfLists),
flatten(ListOfLists, List).
Example:
?- term_to_list(loyalty(father(bob, Y), X), L).
L = [loyalty, father, bob, Y, X].
Alternatively, you can define complexity/2 as follows:
% complexity(+Term, -Complexity)
complexity(Term, 1) :- var(Term), !.
complexity(Term, 2) :- atomic(Term), !.
complexity(Term, Complexity) :-
compound(Term),
Term =.. Components,
maplist(complexity, Components, Complexities),
sum_list(Complexities, Complexity).
Example:
?- complexity(loyalty(father(bob, Y), X), L).
L = 8.
Remark SWI-Prolog defines maplist/3 and sum_list/2 as follows:
maplist(Goal, List1, List2) :-
maplist_(List1, List2, Goal).
maplist_([], [], _).
maplist_([Elem1|Tail1], [Elem2|Tail2], Goal) :-
call(Goal, Elem1, Elem2),
maplist_(Tail1, Tail2, Goal).
sum_list(Xs, Sum) :-
sum_list(Xs, 0, Sum).
sum_list([], Sum, Sum).
sum_list([X|Xs], Sum0, Sum) :-
Sum1 is Sum0 + X,
sum_list(Xs, Sum1, Sum).
I have to define a predicate nshift/3 that shift a list N times either way.
Examples:
?- nshift(3,[a,b,c,d,e,f,g,h],Shifted).
Shifted = [d,e,f,g,h,a,b,c]
?-nshift(1,[1,2,3,4,5],Shifted).
Shifted = [2,3,4,5,1]
?-nshift(-2,[a,b,c,d,e,f,g,h],Shifted).
Shifted = [g,h,a,b,c,d,e,f]
I created a code that would take care of the first two examples but I'm having problem with the last example where the N time is -2. Can somebody help me.
My code:
my_shift([], []).
my_shift([H|T], L) :-
append(T, [H], L).
nshift(0, L, L) :- !.
nshift(N, L1, L2) :-
N1 is N-1,
my_shift(L1, L),
nshift(N1, L, L2).
I have this old code
rotate(right, L, [T|H]) :- append(H, [T], L).
rotate(left, [H|T], L) :- append(T, [H], L).
Then, I think you could adapt your nshift/3 testing if the first argument is < 0, something like
nshift(0, L, L) :- !.
nshift(N, L1, L2) :-
N < 0, rotate(right, L1, L), N1 is N+1, nshift(N1, L, L2).
nshift(N, L1, L2) :-
N > 0, rotate(left, L1, L), N1 is N-1, nshift(N1, L, L2).
As hinted in another answer, your type of shift is usually called rotate. Rotates with non-negative N can be written in a nicely declarative way as
naive_rotate(N, Xs, Ys) :-
length(Bs, N),
append(As, Bs, Xs),
append(Bs, As, Ys).
While this works, people will be quick to point out that its termination properties are poor: when you backtrack into rotate/3, i.e. ask for more solutions, it will not terminate. This can be addressed by adding redundant conditions on the list lengths, viz.
rotate(N, Xs, Ys) :-
same_length(Xs, Ys),
leq_length(Bs, Xs),
length(Bs, N),
append(As, Bs, Xs),
append(Bs, As, Ys).
same_length([], []).
same_length([_|Xs], [_|Ys]) :- same_length(Xs, Ys).
leq_length([], _).
leq_length([_|Xs], [_|Ys]) :- leq_length(Xs, Ys).
This now works nicely for various query patterns, e.g.
?- rotate(2, [a,b,c,d,e], Ys). % gives Ys = [d,e,a,b,c]
?- rotate(2, Xs, [a,b,c,d,e]). % gives Xs = [c,d,e,a,b]
?- rotate(N, [a,b,c,d,e], Ys). % 5 solutions
?- rotate(N, Xs, [a,b,c,d,e]). % 5 solutions
?- rotate(N, Xs, Ys). % many solutions
You can then write your original nshift/3 as
nshift(N, Xs, Ys) :-
( N>=0 -> rotate(N, Xs, Ys) ; M is -N, rotate(M, Ys, Xs) ).
I found this predicate for the calculation of all possible sums.
subset_sum(0,[],[]).
subset_sum(N,[_|Xs],L) :-
subset_sum(N,Xs,L).
subset_sum(N,[X|Xs],[X|Rest]) :-
R is N-X,
subset_sum(R,Xs,Rest).
Knowing that the division does not have the commutative property, how do I get the same result for the division?
This predicate only works for the division between the two elements and in order.
subset_div(1,[],[]).
subset_div(N,[_|Xs],L) :-
subset_div(N,Xs,L).
subset_div(N,[X|Xs],[X|Rest]) :-
R is X/N,
subset_div(R,Xs,Rest).
how you can get this result?
?-subset_div(20,[10,100,90,3,5],L).
L=[100,5].
?-subset_div(5,[10,4,59,200,12],L).
L=[200,10,4].
5= (200/10)/4 or 5 = (200/4)/10 but 5 \= (4/200)/10 or 5\= (10/4)/200
Thanks.
You can do it in terms of a product if you only care about left-associative solutions. Solutions when you can do, say [20 / (10 / 2) / 5] are harder, and would require a more complicated output format.
subset_prod(1, [], []).
subset_prod(N, [_|Xs], L) :-
subset_prod(N, Xs, L).
subset_prod(N, [X|Xs], [X|Rest]) :-
R is N/X,
subset_prod(R, Xs, Rest).
subset_div1(N, [X|Xs], [X|L]) :-
X1 is X / N,
integer(X1),
subset_prod(X1, Xs, L).
subset_div1(N, [_|Xs], L) :-
subset_div(N, Xs, L).
subset_div(N, L, M) :-
sort(L, L1),
reverse(L1, L2),
subset_div1(N, L2, M).
I am trying to write a program in Prolog to find a Latin Square of size N.
I have this right now:
delete(X, [X|T], T).
delete(X, [H|T], [H|S]) :-
delete(X, T, S).
permutation([], []).
permutation([H|T], R) :-
permutation(T, X),
delete(H, R, X).
latinSqaure([_]).
latinSquare([A,B|T], N) :-
permutation(A,B),
isSafe(A,B),
latinSquare([B|T]).
isSafe([], []).
isSafe([H1|T1], [H2|T2]) :-
H1 =\= H2,
isSafe(T1, T2).
using SWI-Prolog library:
:- module(latin_square, [latin_square/2]).
:- use_module(library(clpfd), [transpose/2]).
latin_square(N, S) :-
numlist(1, N, Row),
length(Rows, N),
maplist(copy_term(Row), Rows),
maplist(permutation, Rows, S),
transpose(S, T),
maplist(valid, T).
valid([X|T]) :-
memberchk(X, T), !, fail.
valid([_|T]) :- valid(T).
valid([_]).
test:
?- aggregate(count,S^latin_square(4,S),C).
C = 576.
edit your code, once corrected removing typos, it's a verifier, not a generator, but (as noted by ssBarBee in a deleted comment), it's flawed by missing test on not adjacent rows.
Here the corrected code
delete(X, [X|T], T).
delete(X, [H|T], [H|S]) :-
delete(X, T, S).
permutation([], []).
permutation([H|T], R):-
permutation(T, X),
delete(H, R, X).
latinSquare([_]).
latinSquare([A,B|T]) :-
permutation(A,B),
isSafe(A,B),
latinSquare([B|T]).
isSafe([], []).
isSafe([H1|T1], [H2|T2]) :-
H1 =\= H2,
isSafe(T1, T2).
and some test
?- latinSquare([[1,2,3],[2,3,1],[3,2,1]]).
false.
?- latinSquare([[1,2,3],[2,3,1],[3,1,2]]).
true .
?- latinSquare([[1,2,3],[2,3,1],[1,2,3]]).
true .
note the last test it's wrong, should give false instead.
Like #CapelliC, I recommend using CLP(FD) constraints for this, which are available in all serious Prolog systems.
In fact, consider using constraints more pervasively, to benefit from constraint propagation.
For example:
:- use_module(library(clpfd)).
latin_square(N, Rows, Vs) :-
length(Rows, N),
maplist(same_length(Rows), Rows),
maplist(all_distinct, Rows),
transpose(Rows, Cols),
maplist(all_distinct, Cols),
append(Rows, Vs),
Vs ins 1..N.
Example, counting all solutions for N = 4:
?- findall(., (latin_square(4,_,Vs),labeling([ff],Vs)), Ls), length(Ls, L).
L = 576,
Ls = [...].
The CLP(FD) version is much faster than the other version.
Notice that it is good practice to separate the core relation from the actual search with labeling/2. This lets you quickly see that the core relation terminates also for larger N:
?- latin_square(20, _, _), false.
false.
Thus, we directly see that this terminates, hence this plus any subsequent search with labeling/2 is guaranteed to find all solutions.
I have better solution, #CapelliC code takes very long time for squares with N length higher than 5.
:- use_module(library(clpfd)).
make_square(0,_,[]) :- !.
make_square(I,N,[Row|Rest]) :-
length(Row,N),
I1 is I - 1,
make_square(I1,N,Rest).
all_different_in_row([]) :- !.
all_different_in_row([Row|Rest]) :-
all_different(Row),
all_different_in_row(Rest).
all_different_in_column(Square) :-
transpose(Square,TSquare),
all_different_in_row(TSquare).
all_different_in_column1([[]|_]) :- !.
all_different_in_column1(Square) :-
maplist(column,Square,Column,Rest),
all_different(Column),
all_different_in_column1(Rest).
latin_square(N,Square) :-
make_square(N,N,Square),
append(Square,AllVars),
AllVars ins 1..N,
all_different_in_row(Square),
all_different_in_column(Square),
labeling([ff],AllVars).
I'm writing a prolog program that will check if two math expressions are actually the same. For example, if my math expression goal is: (a + b) + c then any of the following expressions are considered the same:
(a+b)+c
a+(b+c)
(b+a)+c
(c+a)+b
a+(c+b)
c+(a+b)
and other combinations
Certainly, I don't expect to check the combination of possible answers because the expression can be more complex than that.
Currently, this is my approach:
For example, if I want to check if a + b *c is the same with another expression such as c*b+a, then I store both expression recursively as binary expressions, and I should create a rule such as ValueOf that will give me the "value" of the first expression and the second expression. Then I just check if the "value" of both expression are the same, then I can say that both expression are the same. Problem is, because the content of the expression is not number, but identifier, I cannot use the prolog "is" keyword to get the value.
Any suggestion?
many thanks
% represent a + b * c
binExprID(binEx1).
hasLeftArg(binEx1, a).
hasRightArg(binEx1, binEx2).
hasOperator(binEx1, +).
binExprID(binEx2).
hasLeftArg(binEx2, b).
hasRightArg(binEx2, c).
hasOperator(binEx2, *).
% represent c * b + a
binExprID(binEx3).
hasLeftArg(binEx3, c).
hasRightArg(binEx3, b).
hasOperator(binEx3, *).
binExprID(binEx4).
hasLeftArg(binEx4, binEx3).
hasRightArg(binEx4, a).
hasOperator(binEx4, +).
goal:- valueOf(binEx1, V),
valueOf(binEx4, V).
Math expressions can be very complex, I presume you are referring to arithmetic instead. The normal form (I hope my wording is appropriate) is 'sum of monomials'.
Anyway, it's not an easy task to solve generally, and there is an ambiguity in your request: 2 expressions can be syntactically different (i.e. their syntax tree differ) but still have the same value. Obviously this is due to operations that leave unchanged the value, like adding/subtracting 0.
From your description, I presume that you are interested in 'evaluated' identity. Then you could normalize both expressions, before comparing for equality.
To evaluate syntactical identity, I would remove all parenthesis, 'distributing' factors over addends. The expression become a list of multiplicative terms. Essentially, we get a list of list, that can be sorted without changing the 'value'.
After the expression has been flattened, all multiplicative constants must be accumulated.
a simplified example:
a+(b+c)*5 will be [[1,a],[b,5],[c,5]] while a+5*(c+b) will be [[1,a],[5,c],[5,b]]
edit after some improvement, here is a very essential normalization procedure:
:- [library(apply)].
arith_equivalence(E1, E2) :-
normalize(E1, N),
normalize(E2, N).
normalize(E, N) :-
distribute(E, D),
sortex(D, N).
distribute(A, [[1, A]]) :- atom(A).
distribute(N, [[1, N]]) :- number(N).
distribute(X * Y, L) :-
distribute(X, Xn),
distribute(Y, Yn),
% distribute over factors
findall(Mono, (member(Xm, Xn), member(Ym, Yn), append(Xm, Ym, Mono)), L).
distribute(X + Y, L) :-
distribute(X, Xn),
distribute(Y, Yn),
append(Xn, Yn, L).
sortex(L, R) :-
maplist(msort, L, T),
maplist(accum, T, A),
sumeqfac(A, Z),
exclude(zero, Z, S),
msort(S, R).
accum(T2, [Total|Symbols]) :-
include(number, T2, Numbers),
foldl(mul, Numbers, 1, Total),
exclude(number, T2, Symbols).
sumeqfac([[N|F]|Fs], S) :-
select([M|F], Fs, Rs),
X is N+M,
!, sumeqfac([[X|F]|Rs], S).
sumeqfac([F|Fs], [F|Rs]) :-
sumeqfac(Fs, Rs).
sumeqfac([], []).
zero([0|_]).
mul(X, Y, Z) :- Z is X * Y.
Some test:
?- arith_equivalence(a+(b+c), (a+c)+b).
true .
?- arith_equivalence(a+b*c+0*77, c*b+a*1).
true .
?- arith_equivalence(a+a+a, a*3).
true .
I've used some SWI-Prolog builtin, like include/3, exclude/3, foldl/5, and msort/2 to avoid losing duplicates.
These are basic list manipulation builtins, easily implemented if your system doesn't have them.
edit
foldl/4 as defined in SWI-Prolog apply.pl:
:- meta_predicate
foldl(3, +, +, -).
foldl(Goal, List, V0, V) :-
foldl_(List, Goal, V0, V).
foldl_([], _, V, V).
foldl_([H|T], Goal, V0, V) :-
call(Goal, H, V0, V1),
foldl_(T, Goal, V1, V).
handling division
Division introduces some complexity, but this should be expected. After all, it introduces a full class of numbers: rationals.
Here are the modified predicates, but I think that the code will need much more debug. So I allegate also the 'unit test' of what this micro rewrite system can solve. Also note that I didn't introduce the negation by myself. I hope you can work out any required modification.
/* File: arith_equivalence.pl
Author: Carlo,,,
Created: Oct 3 2012
Purpose: answer to http://stackoverflow.com/q/12665359/874024
How to check if two math expressions are the same?
I warned that generalizing could be a though task :) See the edit.
*/
:- module(arith_equivalence,
[arith_equivalence/2,
normalize/2,
distribute/2,
sortex/2
]).
:- [library(apply)].
arith_equivalence(E1, E2) :-
normalize(E1, N),
normalize(E2, N), !.
normalize(E, N) :-
distribute(E, D),
sortex(D, N).
distribute(A, [[1, A]]) :- atom(A).
distribute(N, [[N]]) :- number(N).
distribute(X * Y, L) :-
distribute(X, Xn),
distribute(Y, Yn),
% distribute over factors
findall(Mono, (member(Xm, Xn), member(Ym, Yn), append(Xm, Ym, Mono)), L).
distribute(X / Y, L) :-
normalize(X, Xn),
normalize(Y, Yn),
divide(Xn, Yn, L).
distribute(X + Y, L) :-
distribute(X, Xn),
distribute(Y, Yn),
append(Xn, Yn, L).
sortex(L, R) :-
maplist(dsort, L, T),
maplist(accum, T, A),
sumeqfac(A, Z),
exclude(zero, Z, S),
msort(S, R).
dsort(L, S) :- is_list(L) -> msort(L, S) ; L = S.
divide([], _, []).
divide([N|Nr], D, [R|Rs]) :-
( N = [Nn|Ns],
D = [[Dn|Ds]]
-> Q is Nn/Dn, % denominator is monomial
remove_common(Ns, Ds, Ar, Br),
( Br = []
-> R = [Q|Ar]
; R = [Q|Ar]/[1|Br]
)
; R = [N/D] % no simplification available
),
divide(Nr, D, Rs).
remove_common(As, [], As, []) :- !.
remove_common([], Bs, [], Bs).
remove_common([A|As], Bs, Ar, Br) :-
select(A, Bs, Bt),
!, remove_common(As, Bt, Ar, Br).
remove_common([A|As], Bs, [A|Ar], Br) :-
remove_common(As, Bs, Ar, Br).
accum(T, [Total|Symbols]) :-
partition(number, T, Numbers, Symbols),
foldl(mul, Numbers, 1, Total), !.
accum(T, T).
sumeqfac([[N|F]|Fs], S) :-
select([M|F], Fs, Rs),
X is N+M,
!, sumeqfac([[X|F]|Rs], S).
sumeqfac([F|Fs], [F|Rs]) :-
sumeqfac(Fs, Rs).
sumeqfac([], []).
zero([0|_]).
mul(X, Y, Z) :- Z is X * Y.
:- begin_tests(arith_equivalence).
test(1) :-
arith_equivalence(a+(b+c), (a+c)+b).
test(2) :-
arith_equivalence(a+b*c+0*77, c*b+a*1).
test(3) :-
arith_equivalence(a+a+a, a*3).
test(4) :-
arith_equivalence((1+1)/x, 2/x).
test(5) :-
arith_equivalence(1/x+1, (1+x)/x).
test(6) :-
arith_equivalence((x+a)/(x*x), 1/x + a/(x*x)).
:- end_tests(arith_equivalence).
running the unit test:
?- run_tests(arith_equivalence).
% PL-Unit: arith_equivalence ...... done
% All 6 tests passed
true.