I'm trying to solve a linear relaxation of a problem I've already solved with a Python library in order to see if it behaves in the same way in Xpress Mosel.
One of the index sets I'm using is not the typical c=1..n but a set of sets, meaning I've taken the 1..n set and have created all the combinations of subsets possible (for example the set 1..3 creates the set of sets {{1},{2},{3},{1,2},{2,3},{1,2,3}}).
In one of my constraints, one of the indexes must run inside each one of those subsets.
The respective code in Python is as follows (using the Gurobi library):
cluster=[1,2,3,4,5,6]
cluster1=[]
for L in range(1,len(cluster)+1):
for subset in itertools.combinations(cluster, L):
clusters1.append(list(subset))
ConstraintA=LinExpr()
ConstraintB=LinExpr()
for i in range(len(nodes)):
for j in range(len(nodes)):
if i<j and A[i][j]==1:
for l in range(len(clusters1)):
ConstraintA+=z[i,j]
for h in clusters1[l]:
restricao2B+=(x[i][h]-x[j][h])
model.addConstr(ConstraintA,GRB.GREATER_EQUAL,ConstraintB)
ConstraintA=LinExpr()
ConstraintB=LinExpr()
(In case the code above is confusing, which I suspect it to be)The constraint I'm trying to write is:
z(i,j)>= sum_{h in C1}(x(i,h)-x(j,h)) forall C1 in C
in which the C1 is each of those subsets.
Is there a way to do this in Mosel?
You could use some Mosel code along these lines (however, independently of the language that you are using, please be aware that the calculated 'set of all subsets' very quickly grows in size with an increasing number of elements in the original set C, so this constraint formulation will not scale up well):
declarations
C: set of integer
CS: set of set of integer
z,x: array(I:range,J:range) of mpvar
end-declarations
C:=1..6
CS:=union(i in C) {{i}}
forall(j in 1..C.size-1)
forall(s in CS | s.size=j, i in C | i > max(k in s) k ) CS+={s+{i}}
forall(s in CS,i in I, j in J) z(i,j) >= sum(h in s) (x(i,h)-x(j,h))
Giving this some more thought, the following version working with lists in place of sets is more efficient (that is, faster):
uses "mmsystem"
declarations
C: set of integer
L: list of integer
CS: list of list of integer
z,x: array(I:range,J:range) of mpvar
end-declarations
C:=1..6
L:=list(C)
qsort(SYS_UP, L) ! Making sure L is ordered
CS:=union(i in L) [[i]]
forall(j in 1..L.size-1)
forall(s in CS | s.size=j, i in L | i > s.last ) CS+=[s+[i]]
forall(s in CS,i in I, j in J) z(i,j) >= sum(h in s) (x(i,h)-x(j,h))
I wrote the following function to perform SVD according to page 45 of 'the deep learning book' by Ian Goodfellow and co.
def SVD(A):
#A^T
AT = np.transpose(A)
#AA^T
AAT = A.dot(AT)
#A^TA
ATA = AT.dot(A)
#Left single values
LSV = np.linalg.eig(AAT)[1]
U = LSV #some values of U have the wrong sign
#Right single values
RSV = np.linalg.eig(ATA)[1]
V = RSV
V[:,0] = V[:,0] #V isnt arranged properly
values = np.sqrt(np.linalg.eig(ata)[0])
#descending order
values = np.sort(values)[::-1]
rows = A.shape[0]
columns = A.shape[1]
D = np.zeros((rows,columns))
np.fill_diagonal(D,values)
return U, D, V
However for any given matrix the results are not the same as using
np.linalg.svd(A)
and I have no idea why.
I tested my algorithm by saying
abs(UDV^T - A) < 0.0001
to check if it decomposed properly and it hasn't. The problem seems to lie with the V and U components but I can't see what's going wrong. D seems to be correct.
If anyone can see the problem it would be much appreciated.
I think you have a problem with the order of the eigenpairs that eig(ATA) and eig(AAT) return. The documentation of np.linalg.eig tells that no order is guaranteed. Replacing eig by eigh, which returns the eigenpairs in ascending order, should help. Also don't rearrange the values.
By the way, eigh is specific for symmetric matrices, such as the ones that you are passing, and will not return complex numbers if the original matrix is real.
We have this assignment for our Prolog course. After two months of one hour per week of Prolog, it is still an enigma to me, my thinking seems unable to adapt from procedural languages - yet.
There is a knowledge base containing predicates/functors with the same name and arities 1, 2 and 3.
The call form should be
search(functor_name, argument, S).
The answers should find all occurrences with this functor name and argument, regardless of arity.
The answers should be of the form:
S = functor_name(argument);
S = functor_name(argument,_);
S = functor_name(_,argument);
S = functor_name(argument,_,_);
S = functor_name(_,argument,_);
S = functor_name(_,_,argument);
false.
I have found out that I could use call to test if the entry in the knowledge base exists.
But call does not seem to work with a variable for the functor name. I am totally baffled, no idea how to use a variable for a functor name.
UPDATE:
My question has been partly answered.
My new code gives me true and false for arities 1, 2 and 3 (see below).
search(Person,Predicate) :-
ID = Person, Key = Predicate, current_functor(Key,1),
call(Key,ID)
; ID = Person, Key = Predicate, current_functor(Key,2),
(call(Key,ID,_);call(Key,_,ID))
; ID = Person, Key = Predicate, current_functor(Key,3),
(call(Key,ID,_,_);call(Key,_,ID,_);call(Key,_,_,ID)).
UPDATE2:
Another partial answer has come in. That one gives me S as a list of terms, but the "other" arguments are placeholders:
search2(Predicate, Arg, S) :-
( Arity = 2 ; Arity = 3 ; Arity = 4 ),
functor(S, Predicate, Arity),
S =.. [_,Predicate|Args],
member(Arg, Args).
The result is quite nice. Still missing: the Predicate should not be inside the brackets and the other arguments should be taken literally from the knowledge base, not written as placeholders. The current result looks like this:
?- search2(parent,lars,S).
S = parent(parent, lars) ;
S = parent(parent, lars, _G1575) ;
S = parent(parent, _G1574, lars) ;
S = parent(parent, lars, _G1575, _G1576) ;
S = parent(parent, _G1574, lars, _G1576) ;
S = parent(parent, _G1574, _G1575, lars).
I am giving up with this question, because the question was posed in the wrong way from the beginning. I should have asked more specifically - which I could not, because I am still no good in Prolog.
#false helped me most. I am accepting his answer.
There are two approaches here, one "traditional" (1970s) that implements literally what you want:
search(F, Arg, S) :-
( N = 1 ; N = 2 ; N = 3 ), % more compactly: between(1,3, N)
functor(S, F, N),
S =.. [_|Args], % more compactly: between(1,N, I), arg(I,S,Arg)
member(Arg, Args).
The other reconsiders the explicit construction of the goal. Actually, if you have a functor F, and arguments A1, A2, A3 you can immediately write the goal call(F, A1, A2, A3) without any use of functor/3 or (=..)/2.
There are many advantages of using call(F, A1, A2, A3) in place of Goal =.. [F, A1, A2, A3], call(Goal): In many situations it is cleaner, faster, and much easier to typecheck. Further, when using a module system, the handling of potential module qualifications for F will work seamlessly. Whereas (=..)/2 will have to handle all ugly details explicitly, that is more code, more errors.
search(F,A,call(F,A)).
search(F,A,call(F,A,_)).
search(F,A,call(F,_,A)).
search(F,A,call(F,A,_,_)).
search(F,A,call(F,_,A,_)).
search(F,A,call(F,_,_,A)).
If you want to shorten this, then rather construct call/N dynamically:
search(F, Arg, S) :-
( N = 2 ; N = 3 ; N = 4 ),
functor(S, call, N),
S =.. [_,F|Args],
member(Arg, Args).
Note that call needs an extra argument for the functor F!
You can use the "univ" operator, =.., to construct a goal dynamically:
?- F=member, X=1, L=[1,2,3], Goal =.. [F, X, L], call(Goal).
F = member,
X = 1,
L = [1, 2, 3],
Goal = member(1, [1, 2, 3]) .
Earlier, I asked a question on here asking for help with the conversion of a transition-graph of a finite automaton into a regular expression:
Understanding (and forming) the regular expression of this finite automaton
Thanks to the user Patrick87, I was able to find the help I was looking for. I also read the following link that he mentioned in his answer :
http://krchowdhary.com/toc/dfa-to-reg-exp.pdf
which explains three algorithmic methods to find regular expressions. Intuitively, I was attracted towards the Brzozowski Algebraic Method and tried to solve the FA that I had asked for help on in the previous post which is mentioned at the top.
The following are the characteristic equations that I have made for the FA. Please tell me and correct me if I am wrong and point me in the right direction!
R1 = bR2 + aR3
R2 = aR2 + bR4
R3 = aR3 + bR2 + λ
R4 = aR4 + bR3
Are these correct? If yes, then how how do I go about with the substitutions as every Ri will be in terms of Rj where i≠j.
Please help :D
I'll take a stab at this... We first reduce each equation so that we don't have any Ri's equal to expressions containing themselves...
R1 = bR2 + aR3
R2 = a*bR4
R3 = a*bR2 + a*
R4 = a*bR3
Now for the substitutions... first eliminate R4
R2 = a*ba*bR3
Now we get rid of R2
R1 = ba*ba*bR3 + aR3
R3 = a*ba*ba*bR3 + a*
We don't want R3 on its own RHS...
R3 = (a*ba*ba*b)*a*
Now eliminate R3
R1 = ba*ba*b(a*ba*ba*b)*a* + a(a*ba*ba*b)*a*
This looks right, and as such we can transform it to yours. You should try this exercise.