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Hungarian Algorithm for Assignment Problem | Set 1 (Introduction)

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  • Difficulty Level : Hard
  • Last Updated : 20 Jul, 2022
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Let there be n agents and n tasks. Any agent can be assigned to perform any task, incurring some cost that may vary depending on the agent-task assignment. It is required to perform all tasks by assigning exactly one agent to each task and exactly one task to each agent in such a way that the total cost of the assignment is minimized. Example: You work as a manager for a chip manufacturer, and you currently have 3 people on the road meeting clients. Your salespeople are in Jaipur, Pune and Bangalore, and you want them to fly to three other cities: Delhi, Mumbai and Kerala. The table below shows the cost of airline tickets in INR between the cities: hungarian1 The question: where would you send each of your salespeople in order to minimize fair? Possible assignment: Cost = 11000 INR hungerain2 Other Possible assignment: Cost = 9500 INR and this is the best of the 3! possible assignments. hungarian4 Brute force solution is to consider every possible assignment implies a complexity of Ω(n!). The Hungarian algorithm, aka Munkres assignment algorithm, utilizes the following theorem for polynomial runtime complexity (worst case O(n3)) and guaranteed optimality: If a number is added to or subtracted from all of the entries of any one row or column of a cost matrix, then an optimal assignment for the resulting cost matrix is also an optimal assignment for the original cost matrix. We reduce our original weight matrix to contain zeros, by using the above theorem. We try to assign tasks to agents such that each agent is doing only one task and the penalty incurred in each case is zero. Core of the algorithm (assuming square matrix):

  1. For each row of the matrix, find the smallest element and subtract it from every element in its row.
  2. Do the same (as step 1) for all columns.
  3. Cover all zeros in the matrix using minimum number of horizontal and vertical lines.
  4. Test for Optimality: If the minimum number of covering lines is n, an optimal assignment is possible and we are finished. Else if lines are lesser than n, we haven’t found the optimal assignment, and must proceed to step 5.
  5. Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.

Try it before moving to see the solution

Explanation for above simple example:

 
Below is the cost matrix of example given in above diagrams.
 2500  4000  3500
 4000  6000  3500
 2000  4000  2500

Step 1: Subtract minimum of every row.
2500, 3500 and 2000 are subtracted from rows 1, 2 and 
3 respectively.

   0   1500  1000
  500  2500   0
   0   2000  500

Step 2: Subtract minimum of every column.
0, 1500 and 0 are subtracted from columns 1, 2 and 3 
respectively.

   0    0   1000
  500  1000   0
   0   500  500

Step 3: Cover all zeroes with minimum number of 
horizontal and vertical lines.


Step 4:  Since we need 3 lines to cover all zeroes,
we have found the optimal assignment. 
 2500  4000  3500
 4000  6000  3500
 2000  4000  2500

So the optimal cost is 4000 + 3500 + 2000 = 9500

  An example that doesn’t lead to optimal value in first attempt: In the above example, the first check for optimality did give us solution. What if we the number covering lines is less than n.

 
cost matrix:
 1500  4000  4500
 2000  6000  3500
 2000  4000  2500

Step 1: Subtract minimum of every row.
1500, 2000 and 2000 are subtracted from rows 1, 2 and 
3 respectively.

  0    2500  3000
  0    4000  1500
  0    2000   500

Step 2: Subtract minimum of every column.
0, 2000 and 500 are subtracted from columns 1, 2 and 3 
respectively.

  0     500  2500
  0    2000  1000 
  0      0      0 

Step 3: Cover all zeroes with minimum number of 
horizontal and vertical lines.


Step 4:  Since we only need 2 lines to cover all zeroes,
we have NOT found the optimal assignment. 

Step 5:  We subtract the smallest uncovered entry 
from all uncovered rows. Smallest entry is 500.
 -500    0   2000
 -500  1500   500
   0     0      0

Then we add the smallest entry to all covered columns, we get
   0     0   2000
   0   1500   500
  500    0      0

Now we return to Step 3:. Here we cover again using
lines. and go to Step 4:. Since we need 3 lines to 
cover, we found the optimal solution.
 1500  4000  4500
 2000  6000  3500
 2000  4000  2500

So the optimal cost is 4000 + 2000 + 2500 = 8500

C++




#include<bits/stdc++.h>
using namespace std;
 
 
class Solution {
  public:
   
    int cost[31][31]; //cost matrix
    int n, max_match; //n workers and n jobs
    int lx[31], ly[31]; //labels of X and Y parts
    int xy[31]; //xy[x] - vertex that is matched with x,
    int yx[31]; //yx[y] - vertex that is matched with y
    bool S[31], T[31]; //sets S and T in algorithm
    int slack[31]; //as in the algorithm description
    int slackx[31]; //slackx[y] such a vertex, that
    int prev_ious[31]; //array for memorizing alternating p
   
    void init_labels()
    {
        memset(lx, 0, sizeof(lx));
        memset(ly, 0, sizeof(ly));
        for (int x = 0; x < n; x++)
        for (int y = 0; y < n; y++)
        lx[x] = max(lx[x], cost[x][y]);
    }
     
      
    void update_labels()
    {
        int x, y;
        int delta = 99999999; //init delta as infinity
        for (y = 0; y < n; y++) //calculate delta using slack
            if (!T[y])
                delta = min(delta, slack[y]);
        for (x = 0; x < n; x++) //update X labels
            if (S[x])
                lx[x] -= delta;
        for (y = 0; y < n; y++) //update Y labels
            if (T[y])
                ly[y] += delta;
        for (y = 0; y < n; y++) //update slack array
            if (!T[y])
                slack[y] -= delta;
    }
     
     
    void add_to_tree(int x, int prev_iousx)
    //x - current vertex,prev_iousx - vertex from X before x in the alternating path,
    //so we add edges (prev_iousx, xy[x]), (xy[x], x)
    {
        S[x] = true; //add x to S
        prev_ious[x] = prev_iousx; //we need this when augmenting
        for (int y = 0; y < n; y++) //update slacks, because we add new vertex to S
            if (lx[x] + ly[y] - cost[x][y] < slack[y])
            {
                slack[y] = lx[x] + ly[y] - cost[x][y];
                slackx[y] = x;
            }
    }
     
     
     
    void augment() //main function of the algorithm
    {
        if (max_match == n) return; //check whether matching is already perfect
        int x, y, root; //just counters and root vertex
        int q[31], wr = 0, rd = 0; //q - queue for bfs, wr,rd - write and read
        //pos in queue
        memset(S, false, sizeof(S)); //init set S
        memset(T, false, sizeof(T)); //init set T
        memset(prev_ious, -1, sizeof(prev_ious)); //init set prev_ious - for the alternating tree
         
        for (x = 0; x < n; x++) //finding root of the tree
        {
            if (xy[x] == -1)
            {
                q[wr++] = root = x;
                prev_ious[x] = -2;
                S[x] = true;
                break;
            }
        }
         
        for (y = 0; y < n; y++) //initializing slack array
        {
            slack[y] = lx[root] + ly[y] - cost[root][y];
            slackx[y] = root;
        }
         
        //second part of augment() function
        while (true) //main cycle
        {
            while (rd < wr) //building tree with bfs cycle
            {
                x = q[rd++]; //current vertex from X part
                for (y = 0; y < n; y++) //iterate through all edges in equality graph
                    if (cost[x][y] == lx[x] + ly[y] && !T[y])
                    {
                        if (yx[y] == -1) break; //an exposed vertex in Y found, so
                                                //augmenting path exists!
                            T[y] = true; //else just add y to T,
                        q[wr++] = yx[y]; //add vertex yx[y], which is matched
                        //with y, to the queue
                        add_to_tree(yx[y], x); //add edges (x,y) and (y,yx[y]) to the tree
                    }
                if (y < n)
                    break; //augmenting path found!
            }
            if (y < n)
                break; //augmenting path found!
             
            update_labels(); //augmenting path not found, so improve labeling
             
            wr = rd = 0;
            for (y = 0; y < n; y++)
            //in this cycle we add edges that were added to the equality graph as a
            //result of improving the labeling, we add edge (slackx[y], y) to the tree if
            //and only if !T[y] && slack[y] == 0, also with this edge we add another one
            //(y, yx[y]) or augment the matching, if y was exposed
            if (!T[y] && slack[y] == 0)
            {
                if (yx[y] == -1) //exposed vertex in Y found - augmenting path exists!
                {
                    x = slackx[y];
                    break;
                }
                else
                {
                    T[y] = true; //else just add y to T,
                    if (!S[yx[y]])
                    {
                        q[wr++] = yx[y]; //add vertex yx[y], which is matched with
                        //y, to the queue
                        add_to_tree(yx[y], slackx[y]); //and add edges (x,y) and (y,
                        //yx[y]) to the tree
                    }
                }
            }
            if (y < n) break; //augmenting path found!
        }
         
        if (y < n) //we found augmenting path!
        {
            max_match++; //increment matching
            //in this cycle we inverse edges along augmenting path
            for (int cx = x, cy = y, ty; cx != -2; cx = prev_ious[cx], cy = ty)
            {
                ty = xy[cx];
                yx[cy] = cx;
                xy[cx] = cy;
            }
            augment(); //recall function, go to step 1 of the algorithm
        }
    }//end of augment() function
      
    int hungarian()
    {
        int ret = 0; //weight of the optimal matching
        max_match = 0; //number of vertices in current matching
        memset(xy, -1, sizeof(xy));
        memset(yx, -1, sizeof(yx));
        init_labels(); //step 0
        augment(); //steps 1-3
         
        for (int x = 0; x < n; x++) //forming answer there
            ret += cost[x][xy[x]];
     
        return ret;
    }
     
    int assignmentProblem(int Arr[], int N) {
         
        n = N;
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                cost[i][j] = -1*Arr[i*n+j];
                 
        int ans = -1 * hungarian();
         
        return ans;
    }
};
 
int main()
{
  int n=3;
  int Arr[3*3]={1500,4000,4500,2000,6000,3500,2000,4000,2500};   /*1500  4000  4500
                                                                   2000  6000  3500
                                                                   2000  4000  2500*/
  Solution ob;
  cout<<ob.assignmentProblem(Arr,n)<<endl;
}

Output

8500

In the next post, we will be discussing implementation of the above algorithm. The implementation requires more steps as we need to find minimum number of lines to cover all 0’s using a program. References: http://www.math.harvard.edu/archive/20_spring_05/handouts/assignment_overheads.pdf https://www.youtube.com/watch?v=dQDZNHwuuOY This article is contributed by Yash Varyani. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.


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