Anyone can help me to slove this gramma error?
I do know how to rewrite this code
constraint
forall(i in 1..LINENUMBER,j in 1..DAYSETNUMBER )
( PACKOUTCAP[i,j] -( sum(k in 1..j)(INPUTCAP[i,k] * rate[i,k,j-k+1])) >0 );
As mentioned in the comment: What is your full model (with some explanation what it should do). What error do you get? Which MiniZinc version?
The following model is your code with some fictive data added. It don't give any errors, and yield a lot of solutions.
int: LINENUMBER = 3;
int: DAYSETNUMBER = 3;
array[1..LINENUMBER,1..DAYSETNUMBER] of var 1..10: PACKOUTCAP;
array[1..LINENUMBER,1..DAYSETNUMBER] of var 1..10: INPUTCAP;
array[1..LINENUMBER,1..DAYSETNUMBER,1..LINENUMBER] of var 1..10: rate;
constraint
forall(i in 1..LINENUMBER,j in 1..DAYSETNUMBER )
( PACKOUTCAP[i,j] -( sum(k in 1..j)(INPUTCAP[i,k] * rate[i,k,j-k+1])) >0 );
Regarding cumulative sum, here's a simple model where the array y contains the cumulative sum of the values in the array x. Perhaps that help you?
int: n = 5;
array[1..n] of var 1..10: x;
array[1..n] of var 1..100: y; % cumulative sum
constraint
y[1] = x[1] /\
forall(i in 2..n) (
y[i] = y[i-1] + x[i]
)
;
Related
Any boys can help me write a function to caculate the n-th largest value in an array ?
for example :
array[1..10] of 1..200 : t=[30,20,30,40,50,50,70,10,90,100];
var int : nth_largest;
In general to determine the nth largest numbers you first have to determine the n-1 larger numbers. If there is only 1 or 2, then you might be able to do something smarter, but in general I think the best supported solution is to use the arg_sort function to determine the order of all elements, and then choose the nth largest one.
This can be written as the following function:
function var int: nth_largest(array[$$E] of $$V: x, var int: n) =
let {
any: perm = arg_sort(x);
} in x[perm[length(x)-n]];
In your fragment this can be used as:
array[1..10] of 1..200: t= [30,20,30,40,50,50,70,10,90,100];
var int: nth_largest ::output = nth_largest(t, 2);
function var int: nth_largest(array[$$E] of $$V: x, var int: n) =
let {
any: perm = arg_sort(x);
} in x[perm[length(x)-n]];
Although you might want to find a more meaningful name for the variable, since shadowing names of variables and functions is generally frowned upon.
Reading #Dekker1's solution, here's what I had in mind. Instead of sort(x,y) the function y = sort(x) is used.
include "globals.mzn";
int: n = 10;
int: nth = 3;
array[1..n] of 1..200: t = [30,20,30,40,50,50,70,10,90,100];
var 1..200: nth_largest;
constraint
nth_largest = sort(t)[n-nth+1]
;
Same idea using a function:
include "globals.mzn";
int: n = 10;
int: nth = 3;
array[1..n] of 1..200: t = [30,20,30,40,50,50,70,10,90,100];
var 1..200: nth_largest;
function var int: nth_largest_val(array[int] of var int: a, var int: m) =
sort(a)[length(a)-m+1]
;
constraint
nth_largest = nth_largest_val(t,nth)
;
Both give:
nth_largest = 70;
Update: Ignoring duplicates
Here is a variant which ignores duplicate values. nthis here a decision variable so we can see all solutions (and it makes it possible to "reverse" the constraint, see below).
include "globals.mzn";
int: n = 10;
var 1..n: nth; % : nth = 4;
array[1..n] of 1..200: t = [30,20,30,40,50,50,70,10,90,100];
var 1..200: nth_largest;
% Skipping duplicates
function var int: nth_largest_skip_dups(array[int] of var int: a, var int: nth) =
let {
var lb_array(a)..ub_array(a): v;
constraint
% count the number of larger values in the unique set
sum([ k > v | k in {a[j] | j in index_set(a) } ]) = nth -1
% ensure that v is in a
/\ exists(i in index_set(a)) (
v = a[i]
);
} in
v
;
constraint
nth_largest = nth_largest_skip_dups(t,nth)
;
output ["nth:\(nth)\nlargest: \(nth_largest)\n" ];
Here's the output:
nth:8
largest: 10
----------
nth:7
largest: 20
----------
nth:1
largest: 100
----------
nth:2
largest: 90
----------
nth:3
largest: 70
----------
nth:4
largest: 50
----------
nth:5
largest: 40
----------
nth:6
largest: 30
----------
Note about reversibility: If the following constraint is added the the constraint is reversed, i.e. we get what rank of the number 50:
constraint nth_largest = 50;
The output:
nth:4
largest: 50
A drawback is that it might be costly for large arrays, and it's better to use the first suggestion if you can ensure that the array has no duplicates.
I have the next MiniZinc code
array [0..5] of var 0..1: a;
constraint sum(a) = 3;
output [show(a)]
And I need to add two constraints to ensure that at most I have 3 contiguous 1's. I thought that adding constraint sum(a) = 3; helps but how can I add a constraint to make sure I have 3-contiguous 1s for example 111000 or 011100?
Another alternative is to constrain the distance between first and last 1:
int: n = 5;
set of int: N = 1..n;
array [N] of var 0..1: a;
var N: first = n + 1 - max(i in N)(i * a[n - i + 1]);
var N: last = max(i in N)(i * a[i]);
constraint sum(a) == 3;
constraint (last - first + 1) == 3;
output [show(a)]
Here is one way by constraining the sum of each block of 4 consecutive elements to 3:
array [0..5] of var 0..1: a;
int: MAX_CONTIGUOUS_ONES = 3;
constraint forall(start in 0..5-MAX_CONTIGUOUS_ONES)
(sum([a[i] | i in start..start+MAX_CONTIGUOUS_ONES]) <= MAX_CONTIGUOUS_ONES);
output [show(a)]
I have recently started studying minizinc, but I have got this strange behaviour in my program.
.dzn
n = 5;
c = 2;
.mzn
include "globals.mzn";
int: n;
int: c;
set of int: num_deliveries = 1..n-1;
int: headquarter = 1;
set of int: num_places = 1..n;
set of int: deliveries = 2..n;
set of int: couriers = 1..c;
set of int: num_max_deliveries = 1..n+2;
set of int: schedule_domain = 0..n;
int: first_place_idx = 1;
int: last_place_idx = n+2;
array[couriers,num_max_deliveries] of var schedule_domain: schedule;
array[1..2*n] of int: total = [schedule[i,j]|i,j in num_max_deliveries where i<=2 /\ j != first_place_idx /\ j!= last_place_idx];
output ["len_without_variable = \(length([ k | k in total where k != 0]))"];
var int: len_cleaned = length([ k | k in total where k != 0]);
output ["len_with_variable = \(len_cleaned)\n"];
In particular, from these lines of code I have different results, even if they are equal.
output ["len_without_variable = \(length([ k | k in total where k != 0]))"];
var int: len_cleaned = length([ k | k in total where k != 0]);
output ["len_with_variable = \(len_cleaned)\n"];
Why does it happen?
Here is one output of the model is this (with total added):
len_without_variable = 2
len_with_variable = 10
total: [3, 3, 0, 0, 0, 0, 0, 0, 0, 0]
To be honest, I'm not sure if the two calculations of the number of non zero elements in total should be the same or not. Perhaps this is a bug in how length operates on decision variables (with a where condition), and it should be 2 (in this example). Until this is settled, you should probably avoid using length like this.
The output of len_without_variable - the one defined in the output section - operates on the actual and known values of the solution. So this might not be the exact thing as using length in a constraint/variable definition.
To calculate the number of non zero values you can use sum instead:
var 0..n num_non_zeros = sum([ 1 | k in total where k != 0]);
However, the construct ... where k != 0 creates temporary variables which makes the model larger than necessary so it's better to use the following:
var 0..n num_non_zeros = sum([ total[i] != 0 | i in 1..n]);
I am working on this problem. I have two decision variables x and y. For same values in x, and corresponding same values in y, I have to count the occurences of this value in y.
I have tried to implement it. But I have doubts that the functionality is as expected.
My code
set of int: x = 1..4;
set of int: y = 1..3;
set of int: a = 1..10;
set of int: b = 1..5;
%Decision variables
array[x, y] of var a: recordA;
array[x, y] of var b: recordB;
constraint forall(i in x ) (
alldifferent([recordB[i,j] | j in y])
);
constraint forall(i in x ) (
alldifferent([recordA[i,j] | j in y])
);
%constraint forall(i,k in x, j in y where i<k /\ recordB[i,j]=recordB[k,j]) (
% forall(i,k in x,j in y where recordA[i,j]=recordA[k,j])(
% count(recordA, recordA[i,j])
%);
OR
%Maybe something like this
%constraint forall(i in x, j in y, m in b)(
% count(col(recordA, j),count(col(recordB,j),m))
% );
The idea is for every recorB and every recordA, to go through all the y (for all the x) and count how many values occur the same recordB at corresponding the same recordA.
For Example- for a value r1 in recordB->count of number of values r2 occuring in recordA.
In this case, recordB has three occurences of value 4 at recordB[1], recordB[6] and recordB[11]
correspondingly recordA has two occurrences of value 5 at recordA[1], recordA[6] but at recordA[11] the value is 1.
I want to count these occurrences of same values in recordA corresponding wuth recordB.
And each row of recordA is unique
And each row of recordB is unique
Thanks for suggestions
I want to find the smallest integer l that satisfies l^2 >= x, and mod(l,2)=0.
In the following example x=75, and hence l=10, since the previous even number doesn't fulfil the inequation: 8^2 <= 75 <= 10^2
I have tried this (ignoring the even-number requirement, which I can't to work):
syms l integer
eqn1 = l^2 >= 75;
% eqn2 = mod(l,2) == 0;
[sol_l, param, cond] = solve(eqn1, l, 'ReturnConditions', true);
But this does not give me anything helpful directly:
sol_l =
k
param =
k
cond =
(75^(1/2) <= k | k <= -75^(1/2)) & in(k, 'integer')
I would like to evaluate the conditions on the parameter and find the smallest value that satisfies the conditions.
Also, I would like to enforce the mod(l,2)=0 condition somehow, but I don't seem to get that work.
Using the solve for this task is like using a cannon to kill a mosquito. Actually, the answer of Lidia Parrilla is good and fast, although it can be simplified as follows:
l = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l = l + 1;
end
% if x = 75, then l = 10
But I would like to point out something that no one else noticed. The condition provided by the solve function for l^2 >= 75 is:
75^(1/2) <= k | k <= -75^(1/2)
and it's absolutely correct. Since l is being raised to the power of 2, and since a negative number raised to the power of 2 produces a positive number, the equation will always have two distinct solutions: a negative one and a positive one.
For x = 75, the solutions will be l = 10 and l = -10. So, if you want to find the smallest number (and a negative number is always smaller than a positive one), the right solution will be:
l = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l = l + 1;
end
l = l * -1;
If you want to return both solutions, the result will be:
l_pos = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l_pos = l_pos + 1;
end
l_neg = l_pos * -1;
l = [l_neg l_pos];
I guess the easiest solution without employing the inequality and solve function would be to find the exact solution to your equation l^2 >= x, and then finding the next even integer. The code would look like this:
x = 75;
y = ceil(sqrt(x)); %Ceil finds the next bigger integer
if(~mod(y,2)) %If it's even, we got the solution
sol = y;
else %If not, get the next integer
sol = y+1;
end
The previous code gives the correct solution to the provided example (x = 75; sol = 10)