National competition in programming math problem - c++
I encountered this problem practicing for an upcoming national competition. The problem goes as follows: You need to create a mixture of two ingredients being in relation to 1:1. You are given N different mixtures, each having its own weight Wi, and its relation in the mixture between the ingredients Mi, Ti (Each value, N, Wi, Mi, and Ti, will be less than 100). We need to find the biggest possible weight of the final mixture, keeping the relation to 1:1. We can take from each given mixture how much we want, we don't necessarily need to take the whole mixture, we can take some portion of it.
So with the given relation 1:1 in the final mixture, we know that we need to have an equal amount of weight from both ingredients possible. After that I need to know if I take K grams of some mixture, how much weight that is for ingredients A and B. So I came up with the following formula:
Let W be the weight in grams, and M and T be the relation between the ingredients respectively. If we want to take K (K <= W) grams we have the following:
Weight of ingredient A = M * (K / (M+T))
Weight of ingredient B = T * (K / (M+T))
#include <bits/stdc++.h>
using namespace std;
class state{
public:
int weight;
int A;
int B;
};
int n;
vector<state> arr;
double ans= 0;
void f(double weight_A, double weight_B, int idx){
if(weight_A == weight_B)
ans = max(ans, weight_A + weight_B);
if(idx >= n)
return;
int weight = arr[idx].weight, relA = arr[idx].A, relB = arr[idx].B;
for(int K = 0; K <= weight; K++){
f(weight_A + relA * (K * 1.0/(relA + relB)), weight_B + relB * (K * 1.0/(relA + relB)), idx+1);
}
}
int main(){
cin>>n;
for(int i = 0; i < n; i++){
state in;
cin>>in.weight>>in.A>>in.B;
arr.push_back(in);
}
f(0.0, 0.0, 0);
cout<<fixed<<setprecision(8);
cout<<ans<<endl;
}
The problem I encountered was that we don't necessarily need to take integer weights, some times to achieve the maximum possible weight of the final product we need to take decimal weights. Let's take a look at this example:
5
14 3 2
4 1 3
4 2 2
6 6 1
10 4 3
We have N = 5, and in each row are given 3 integers, Wi, Mi, and Ti. The weight of the ith mixture and its relation. My solution for this example gives 20.0000, and the correct solution for the above example is 20.85714286. Looking back my initial idea won't work because of the decimal numbers. I suppose there is some formula but I can't figure it out, can anyone help?
This is a Linear Programming problem, so you can solve it by constructing the problem in standard form, and then solve it with an optimization algorithm, like the simplex algorithm.
The objective is to maximize the quantity of medicine (from the original problem), that is the sum of quantities taken from each jar (I'll call the quantities x1, x2, ...).
The quantities are bounded to be lower than the weight Wi available in each jar.
The constraint is that the total amount of honey (first ingredient) is equal to the total amount of tahini (second ingredient). This would mean that:
sum(Mi/(Mi+Ti)*xi) = sum(Ti/(Mi+Ti)*xi)
You can take the second summation to the LHS and get:
sum((Mi-Ti)/(Mi+Ti)*xi) = 0
In order to get integer multipliers just multiply everything by the least common multiple of the denominators lcm(Mi+ti) and then divide by the gcd of the coefficients.
Using your example, the constraint would be:
(3-2)/(3+2) x1 + (1-3)/(1+3) x2 + (2-2)/(2+2) x3 + (6-1)/(6+1) x4 + (4-3)/(4+3) x5 = 0
that is
1/5 x1 -2/4 x2 + 0/4 x3 + 5/7 x4 + 1/7 x5 = 0
Multiply by the lcm(5,4,4,7,7)=140:
28 x1 -70 x2 + 0 x3 + 100 x4 + 20 x5 = 0
divide by 2:
14 x1 -35 x2 +0 x3 + 50 x4 + 10 x5 = 0
We are ready to solve the problem. Let's write it in CPLEX format:
maximize
quantity: x1 + x2 + x3 + x4 + x5
subject to
mix: 14 x1 -35 x2 +0 x3 + 50 x4 + 10 x5 = 0
bounds
x1 <= 14
x2 <= 4
x3 <= 4
x4 <= 6
x5 <= 10
end
Feed it to GLPK:
#include <stdio.h>
#include <stdlib.h>
#include <glpk.h>
int main(void)
{
glp_prob *P;
P = glp_create_prob();
glp_read_lp(P, NULL, "problem.cplex");
glp_adv_basis(P, 0);
glp_simplex(P, NULL);
glp_print_sol(P, "output.txt");
glp_delete_prob(P);
return 0;
}
And the output is:
Problem:
Rows: 1
Columns: 5
Non-zeros: 4
Status: OPTIMAL
Objective: quantity = 20.85714286 (MAXimum)
No. Row name St Activity Lower bound Upper bound Marginal
------ ------------ -- ------------- ------------- ------------- -------------
1 mix NS 0 0 = 0.0714286
No. Column name St Activity Lower bound Upper bound Marginal
------ ------------ -- ------------- ------------- ------------- -------------
1 x1 B 2.85714 0 14
2 x2 NU 4 0 4 3.5
3 x3 NU 4 0 4 1
4 x4 NL 0 0 6 -2.57143
5 x5 NU 10 0 10 0.285714
Karush-Kuhn-Tucker optimality conditions:
KKT.PE: max.abs.err = 0.00e+00 on row 0
max.rel.err = 0.00e+00 on row 0
High quality
KKT.PB: max.abs.err = 0.00e+00 on row 0
max.rel.err = 0.00e+00 on row 0
High quality
KKT.DE: max.abs.err = 0.00e+00 on column 0
max.rel.err = 0.00e+00 on column 0
High quality
KKT.DB: max.abs.err = 0.00e+00 on row 0
max.rel.err = 0.00e+00 on row 0
High quality
End of output
Of course given your input you should construct the problem in memory and feed it to the simplex algorithm without going through a file. Additionally, there's no need to get integer coefficients, it was just to allow a nicer problem formulation.
Related
Setting up an Minimizing the sum of absolute deviation linear programing problem in CPlex
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As Erwin Kalvelagen suggested, changing y1neg <= 0 to y1neg >= 0 was the answer since our our error factor in our constraint is y1pos - y1neg which we want to minimise.
Having negative value for non basic variable gives a non feasible solution in simplex method?
Objective function => x1 - 2x2 Subject to => x2 <= 5 x1 - x2 >= 2 x1 ,x2, x3 >= 0 Maximize? convert to standard form : Maximize -> -x1 + 2x2 Subject to -> x2 <= 5 -x1 + x2 <= -2 convert to slack form : Z = -x1 + 2x2 x3 = 5 - x2 x4 = -2 +x1 -x2 Basic solution (0,0,5,-2) Can I found optimal solution in here? If not why?
A many-to-one mapping in the natural domain using discrete input variables?
I would like to find a mapping f:X --> N, with multiple discrete natural variables X of varying dimension, where f produces a unique number between 0 to the multiplication of all dimensions. For example. Assume X = {a,b,c}, with dimensions |a| = 2, |b| = 3, |c| = 2. f should produce 0 to 12 (2*3*2). a b c | f(X) 0 0 0 | 0 0 0 1 | 1 0 1 0 | 2 0 1 1 | 3 0 2 0 | 4 0 2 1 | 5 1 0 0 | 6 1 0 1 | 7 1 1 0 | 8 1 1 1 | 9 1 2 0 | 10 1 2 1 | 11 This is easy when all dimensions are equal. Assume binary for example: f(a=1,b=0,c=1) = 1*2^2 + 0*2^1 + 1*2^0 = 5 Using this naively with varying dimensions we would get overlapping values: f(a=0,b=1,c=1) = 0*2^2 + 1*3^1 + 1*2^2 = 4 f(a=1,b=0,c=0) = 1*2^2 + 0*3^1 + 0*2^2 = 4 A computationally fast function is preferred as I intend to use/implement it in C++. Any help is appreciated!
Ok, the most important part here is math and algorythmics. You have variable dimensions of size (from least order to most one) d0, d1, ... ,dn. A tuple (x0, x1, ... , xn) with xi < di will represent the following number: x0 + d0 * x1 + ... + d0 * d1 * ... * dn-1 * xn In pseudo-code, I would write: result = 0 loop for i=n to 0 step -1 result = result * d[i] + x[i] To implement it in C++, my advice would be to create a class where the constructor would take the number of dimensions and the dimensions itself (or simply a vector<int> containing the dimensions), and a method that would accept an array or a vector of same size containing the values. Optionaly, you could control that no input value is greater than its dimension. A possible C++ implementation could be: class F { vector<int> dims; public: F(vector<int> d) : dims(d) {} int to_int(vector<int> x) { if (x.size() != dims.size()) { throw std::invalid_argument("Wrong size"); } int result = 0; for (int i = dims.size() - 1; i >= 0; i--) { if (x[i] >= dims[i]) { throw std::invalid_argument("Value >= dimension"); } result = result * dims[i] + x[i]; } return result; } };
Linear index upper triangular matrix
If I have the upper triangular portion of a matrix, offset above the diagonal, stored as a linear array, how can the (i,j) indices of a matrix element be extracted from the linear index of the array? For example, the linear array [a0, a1, a2, a3, a4, a5, a6, a7, a8, a9 is storage for the matrix 0 a0 a1 a2 a3 0 0 a4 a5 a6 0 0 0 a7 a8 0 0 0 0 a9 0 0 0 0 0 And we want to know the (i,j) index in the array corresponding to an offset in the linear matrix, without recursion. A suitable result, k2ij(int k, int n) -> (int, int) would satisfy, for example k2ij(k=0, n=5) = (0, 1) k2ij(k=1, n=5) = (0, 2) k2ij(k=2, n=5) = (0, 3) k2ij(k=3, n=5) = (0, 4) k2ij(k=4, n=5) = (1, 2) k2ij(k=5, n=5) = (1, 3) [etc]
The equations going from linear index to (i,j) index are i = n - 2 - floor(sqrt(-8*k + 4*n*(n-1)-7)/2.0 - 0.5) j = k + i + 1 - n*(n-1)/2 + (n-i)*((n-i)-1)/2 The inverse operation, from (i,j) index to linear index is k = (n*(n-1)/2) - (n-i)*((n-i)-1)/2 + j - i - 1 Verify in Python with: from numpy import triu_indices, sqrt n = 10 for k in range(n*(n-1)/2): i = n - 2 - int(sqrt(-8*k + 4*n*(n-1)-7)/2.0 - 0.5) j = k + i + 1 - n*(n-1)/2 + (n-i)*((n-i)-1)/2 assert np.triu_indices(n, k=1)[0][k] == i assert np.triu_indices(n, k=1)[1][k] == j for i in range(n): for j in range(i+1, n): k = (n*(n-1)/2) - (n-i)*((n-i)-1)/2 + j - i - 1 assert triu_indices(n, k=1)[0][k] == i assert triu_indices(n, k=1)[1][k] == j
First, let's renumber a[k] in opposite order. We'll get: 0 a9 a8 a7 a6 0 0 a5 a4 a3 0 0 0 a2 a1 0 0 0 0 a0 0 0 0 0 0 Then k2ij(k, n) will become k2ij(n - k, n). Now, the question is, how to calculate k2ij(k, n) in this new matrix. The sequence 0, 2, 5, 9 (indices of diagonal elements) corresponds to triangular numbers (after subtracting 1): a[n - i, n + 1 - i] = Ti - 1. Ti = i * (i + 1)/2, so if we know Ti, it's easy to solve this equation and get i (see formula in the linked wiki article, section "Triangular roots and tests for triangular numbers"). If k + 1 is not exactly a triangular number, the formula will still give you the useful result: after rounding it down, you'll get the highest value of i, for which Ti <= k, this value of i corresponds to the row index (counting from bottom), in which a[k] is located. To get the column (counting from right), you should simply calculate the value of Ti and subtract it: j = k + 1 - Ti. To be clear, these are not exacly i and j from your problem, you need to "flip" them. I didn't write the exact formula, but I hope that you got the idea, and it will now be trivial to find it after performing some boring but simple calculations.
The following is an implimentation in matlab, which can be easily transferred to another language, like C++. Here, we suppose the matrix has size m*m, ind is the index in the linear array. The only thing different is that here, we count the lower triangular part of the matrix column by column, which is analogus to your case (counting the upper triangular part row by row). function z= ind2lTra (ind, m) rvLinear = (m*(m-1))/2-ind; k = floor( (sqrt(1+8*rvLinear)-1)/2 ); j= rvLinear - k*(k+1)/2; z=[m-j, m-(k+1)];
For the records, this is the same function, but with one-based indexing, and in Julia: function iuppert(k::Integer,n::Integer) i = n - 1 - floor(Int,sqrt(-8*k + 4*n*(n-1) + 1)/2 - 0.5) j = k + i + ( (n-i+1)*(n-i) - n*(n-1) )÷2 return i, j end
Here is a more efficient formulation for k: k = (2 * n - 3 - i) * i / 2 + j - 1
In python 2: def k2ij(k, n): rows = 0 for t, cols in enumerate(xrange(n - 1, -1, -1)): rows += cols if k in xrange(rows): return (t, n - (rows - k)) return None
In python, the most efficient way is: array_size= 3 # make indices using k argument if you want above the diagonal u, v = np.triu_indices(n=array_size,k=1) # assuming linear indices above the diagonal i.e. 0 means (0,1) and not (0,0) linear_indices = [0,1] ijs = [(i,j) for (i,j) in zip(u[linear_indices], v[linear_indices])] ijs #[(0, 1), (0, 2)]
How can I write an if condition for my variable in GLPK?
Here is my full problem: Information: *Max. total investment: $125 *Pay-off is the sum of the units bought x pay-off/unit *Cost per investment: Buy-in cost + cost/unit x number of units if you buy at least one unit *The cost is sum of the costs per investment Constraints: *You may not invest in both 2 and 5. *You may invest in 1 only if you invest at least one of 2 and 3. *You must invest at least two of 3,4,5. *You may not invest more than max number of units. Problem: Maximize profit : pay-off - cost xi: # of units i ∈ {1,2,3,4,5} yi=1 if xi>0 else yi=0 cost = sum{i in I} buyInCost_i * yi + cost-unit_i*xi pay-off = sum{i in I} (pay-off/unit)_i*xi profit = pay-off - cost Maximize profit Subject to y2+y5 <= 1 y1<= y2+y3 y3+y4+y5 >= 2 x1<=5, x2<=4, x3<=5, x4<=7, x5<=3 cost<=125 Here is my question: For example I have this binary variable y yi=1 if xi>0 else yi=0 and i ∈ {1,2,3,4,5} I declared i as a data set set I; data; set I := 1 2 3 4 5; I don't know how to add if else condition to y variable in glpk. Can you please help me out? My modelling : set I; /*if x[i]>0 y[i]=1 else y[i]=0 ?????*/ var y{i in I}, binary; param a{i in I}; /* buy-in cost of investment i */ param b{i in I}; /* cost per unit of investment i */ param c{i in I}; /* pay-off per unit of investment i */ param d{i in I}; /* max number of units of investment i */ var x{i in I} >=0; /* Number of units that is bought of investment i */ var po := sum{i in I} c[i]*x[i]; var cost := sum{i in I} a[i]*y[i] + b[i]*x[i]; maximize profit: po-cost; s.t. c1: y[2]+y[5]<=1; s.t. c2: y[1]<y[2]+y[3]; s.t. c3: y[3]+y[4]+y[5]>=2; s.t. c4: x[1]<=5 x[2]<=4 x[3]<=5 x[4]<=7 x[5]<=3; s.t. c5: cost <=125; s.t. c6{i in I}: M * y[i] > x[i]; // if condition of y[i] set I := 1 2 3 4 5; param a := 1 25 2 35 3 28 4 20 5 40; param b := 1 5 2 7 3 6 4 4 5 8; param c := 1 15 2 25 3 17 4 13 5 18; param d := 1 5 2 4 3 5 4 7 5 3; param M := 10000; I am getting this syntax error: problem.mod:21: syntax error in variable statement Context: ...I } ; param d { i in I } ; var x { i in I } >= 0 ; var po := MathProg model processing error
You can't directly do that (there is no way to write 'directly' an if constraint in a LP). However, there are workarounds for this. For example, you can write: M * yi > xi where M is a large constant (greater than any value of xi). This way: if xi > 0, then the constraint is equivalent to yi > 0, that is yi == 1 since yi is binary (if M is large enough). if xi == 0, then the constraint is always verified, and yi will be equal to 0 since your objective is increasing with yi and you are minimizing. in both case, the constraint is equivalent to the if test.