I'm wondering how to declare the constraint name using the ConstraintList's add() method. It seems not possible in a direct way. I found this alternative way to do that, but I can't manage to adapt the following code to it:
for i in i_list:
model.constr.add( sum(sum( (model.X[i,j,k] * get_coeff(j, k) kf get_coeff(j, k) > 0 else 0) for k in k_list) for j in get_j_list[i]) <= get_vilue(i) )
Any hint?
Related
I am using deff to define piecewise functions in Scilab. I use if-else statements to define the function in different intervals. All of them seem to work except for this one.
This is my approach:
deff('a=f(p)', ["if(q >= -2 & q <= -1) then"; "a=1"; "elseif(q >= 1 & q <= 2) then"; "a=-1"; "else"; "a=0"; "end";])
The same approach worked on similar functions such as this.
eff('a=f(p)', ["if((p > 0 & p <= 2)) then"; "a=p/2"; "else"; "a=0"; "end";])
I have defined q as so:
q = -3:Ts:3;
where Ts is some floating point number. What am I doing wrong?
You cannot use your function as-is when the input is a vector. A possible workaround is to use feval:
a = feval(q,f)
Another (better) solution is to code f in a vectorized manner:
function a = f(q)
a = 1*(q >= -2 & q <= -1) - 1*(q >= 1 & q <= 2)
end
the 1* are not necessary unless you need to change the values.
I'm having a real challenge declaring a 2D variable in Zimpl. (Parameters seem to work fine.)
The following is my MWE:
set I := {1 to 10};
set J := {1 to 5};
param A[I*J] := read InputFile as "n+";
var x[I] binary;
var s[J] binary; # this works but doesn't do what I need
var s2[I*J] binary; # this does what I need but doesn't work
minimize sum<i,j> in I*J with A[i,j] < 5: (s2[i,j] - x[i]) * A[i,j];
# this constraint compiles
subto constraint1:
forall <j> in J do sum <i> in I with A[i,j] < 5: x[i] <= 1 + s[j];
# this constraint does not compile
subto constraint2:
forall <j> in J do sum <i> in I with A[i,j] < 5: x[i] <= 1 + s2[i,j];
When try to create my lp file, I get
Error 133: Unknown symbol "i"
Does anyone have any insights into how I can get the second constraint to work? As far as I can tell this is identical to the implementation of the capacitated facility problem (Section 6.3) in the Zimpl user's manual.
Thanks in advance.
You have the sum over i on the left-hand side of the constraint, but then reference i on the right-hand side as well. Which value of i do you expect there?
What would work is
forall <j> in J do sum <i> in I with A[i,j] < 5: (x[i] - s2[i,j]) <= 1;
but I am not sure this is what you want to achieve.
Adding Leon's comment to make a more complete answer:
To add to what Gerald wrote, in ZIMPL sums always only consider the next variable, so you have to put parenthesis to make it work.
To add to what Gerald wrote, in ZIMPL sums always only consider the next variable, so you have to put parenthesis to make it work.
at minimize there is the name missing.
It should be minimize obj: sum ...
I'm using Julia to solve an integer program. My variables are of the form z[i,j], i in N and j in N and N=10and z[i,j] is a binary variable.
In the first half of the program, I have a set of solutions for which z[1,2]= 1 and z[1,3]=1 and all other variables are zero. Now, I need to pass these values to another set S in such a way that S={1,2,3}. I tried to code it in Julia, but I couldn't get it in the right way. The following is the what I've tried.Here, z_value is the way that I declare my variables z[i,j]. Can someone please help me to make it correct?
for i in N
for j in N
z_value = Pair(i,j)
if z_value == 1;
push!(S, Pair(i,j))
print(S)
end
end
end
Thanks, Michael and Stefan, I got required set S by rearranging the code as
for i in N
for j in N
if getvalue(z[i,j]) == 1
push!(S, i)
push!(S, j)
end
end
end
Thanks for your effort!!
I'm trying to make a solver that checks the block to make sure that no number repeats. Unfortunately, I can not get the correct logic on this and I'm not sure what I am doing incorrectly. Here is what I've got:
not quite sure why this is not working. Here's my code.
bool sudoku :: check_if_non_repeat(int r, int c, int v) //where r=row, c=column, v=value
Any idea why this is not working? I'm just getting infinite loops
if (!(j = brow && k == bcol))
Check that j=.... should be ==
I'm not sure what you tried to do, but I would do it like this:
bool sudoku :: check_if_non_repeat(int r, int c, int v) //where r=row, c=column, v=value
{
int brow = r/3;
int bcol = c/3;
for (int j = brow * 3; j < (brow * 3 + 3); j++)
for (int k = bcol * 3; k < (bcol * 3 + 3); k++)
if (sudoku_array[j][k] == v)
return true;
return false;
}
EDIT:
As noted below, the if statement need to be more complicated:
if ( sudoku_array[j][k] == v
&& v != 0
&& !(j == r && k == c))
return true;
I'm about to tell you about a different approach to the problem. I made a full solver a long while ago, and it basically used the opposite approach.
For each field, I had a std::bitset<9> which told me which values were still possible in that field. Each insertion would then update the other fields in the same row, column and box to remove that possibility, recursively filling out subsequent fields when any one of the them had one option left.
If it then tried to fill a number which was no longer allowed, then the last input given was no longer a valid number for that spot. That was also a far more thorough check than you're doing here: you won't be checking if you close off the last possibility for another field in the same row/column/box, let alone others.
I never did a couple planned optimizations, but even without them it outperformed (too quick to notice) my friend's solver (>50 seconds). Mostly because he had code like yours.
Hey, I wrote this (fortran) with the aim of finding the minimum spanning tree of a bunch of points (syscount of them). I know for a fact that this approach works, since i wrote it in javascript earlier today. js is slow though, and i wanted to see how much faster fortran would be!!
only problem is it's not working, i'm getting an annoying error;
prims.f95:72.43:
if((check == 1) .and. (path(nodesin(j))(k) < minpath)) then
1
Error: Expected a right parenthesis in expression at (1)
What the hell is that about?! the 43rd character on the line is the "h" of "path"
nodesin(1) = 1
do i = 1,syscount-1
pathstart = -1
pathend = -1
minpath = 2000
do j = 1,i
do k = 1, syscount
check = 1
do l = 1, i
if(nodesin(l) == k) then
check = 0
end if
end do
if((check == 1) .and. (path(nodesin(j))(k) < minpath)) then
minpath = path(nodesin(j))(k)
pathstart = nodesin(j)
pathend = k
end if
end do
end do
nodesin(i+1) = pathend
minpaths(i)(1) = pathstart
minpaths(i)(2) = pathend
end do
Also, i'm fairly new to fortran, so i have a few other questions;
can i use && instead of .and. ?
is there a versions of the for(object in list){} loop found in many other languages?
is there a verion of the php function in_array ? i.e. bool in_array(needle,haystack), and if there is, is there a better way of doing it than:
check = false
Asize = size(array)
do i = 1, Asize
if(array(i) == needle) then
check = true
end if
end do
then to using the check variable to see if it's there?
(I haven't posted anything on stackoverflow before. please don't get angry if i've broken loads of etiquette things!)
It looks like you have defined path and minpaths as two-dimensional arrays. Multi-dimensional arrays are accessed differently in Fortran when compared to C-like languages. In Fortran you separate the indices by commas within one set of parentheses.
I'm guessing by the use of these variables they are integer arrays. Here is how you access elements of those arrays (since you didn't share your variable declarations I am making up the shape of these arrays):
integer :: path(n1, n2)
integer :: minpaths(n3, 2)
your if statement should be:
if((check == 1) .and. (path(nodesin(j), k) < minpath)) then
your access to minpaths should be:
minpaths(i, 1) = pathstart
minpaths(i, 2) = pathend
Also, if you are not using IMPLICIT NONE I recommend you consider it. Not using it is dangerous, and you are using variable names that are close to each other (minpath and minpaths). You could save hours of hair pulling debugging by using IMPLICIT NONE.
While .EQ. can be replaced with ==, there is still only .AND.
For your code block to check whether a "variable is there", you can use "where" and have much shorter code!
In Fortran >= 90 statements and functions can operate on arrays so that explicit loops don't have to be used as frequently.
There is no for (object in list), but using the where statement can do something very similar.
Many of the intrinsic functions that act on arrays also take masks as optional arguments to selectively operate.
I suggest reading a book to learn about these features. I like the one by Metcalf, Reid and Cohen. In the meantime, the second Wikipedia article may help: http://en.wikipedia.org/wiki/Fortran_95_language_features