# Notebook 2.5 - The Hungarian Algorithm¶

The Hungarian Algorithm solves assignment problems where $n$ items must be assigned to $n$ elements.

This notebook will present a basic implementation of the Hungarian Algorithm and solve randomised problem instances.

We begin by importing numpy and generating an initial random problem instance of size $n \times n$. max_int defines the maximum values created by RNG (Random Number Generator).

In :
import numpy as np
import copy

max_int = 100
n = 5
cost_matrix = np.random.randint(max_int, size=(n,n))

print(cost_matrix)

[[60 65 75 32 93]
[95 41  6 33 46]
[18 11 32 65  5]
[13 40 69 41 58]
[26 28 54 88 10]]


## Part 1 - scipy¶

We can also use the scipy package to solve the assignment problem. The function linear_sum_assignment also uses the Hungarian algorithm. Let's see if the results are equal.

In :
# import scipy's linear_sum_assignment
from scipy.optimize import linear_sum_assignment

# execute the assignment
scp_assignment = linear_sum_assignment(cost_matrix)

# find the total cost
scp_total = 0
for i in range(len(scp_assignment)):
scp_total += cost_matrix[scp_assignment[i], scp_assignment[i]]
print(scp_total)

72


## Part 2 - Clean scipy implementation and figure¶

In this section, we implement a neater version of Part 1. We still use scipy to find the assignment, but we also produce readable results and a figure to accompany the final solution. The structure is separated into 4 functions:

1. run_assignment - the master function - it runs the linear_sum_assignment code and calls the suplementary functions to make the results more readable.
2. clean_assignment - scipy returns two arrays as a solution, here we combine then into a single array containing each assignment pair.
3. calc_costs - calculates the total assignment costs.
4. draw_network - draws a bipartite graph with the final assignment highlighted in red.
In :
def draw_network(cost_matrix, assignment):
import networkx as nx

x_diff = 10

y_min = 0
y_max = 5

G = nx.Graph()

for i in range(len(assignment)):

pos=nx.get_node_attributes(G,'pos')

for i in range(len(assignment)):
for j in range(len(assignment)):
val = [i, j]
if val in assignment:
c = 'r'
w = 4
else:
c = 'k'
w = 2

edges = G.edges()
colors = [G[u][v]['color'] for u,v in edges]
weights = [G[u][v]['weight'] for u,v in edges]
nx.draw(G,pos,with_labels=True, node_size=600, font_color='w', edge_color=colors, width=weights)

def calc_costs(cost_matrix, assignment):
total = 0
for a in assignment:
total += cost_matrix[a, a]

def clean_assignment(row, columns):
assignments = []
# create pairs
text = "The final assignment is "
for i in range(len(row)):
assignments.append([row[i], columns[i]])
if i > 0:
text += ", "
text += f"({row[i]}, {columns[i]})"
print(text)
return assignments

def run_assignment(cost_matrix):
row,columns = linear_sum_assignment(cost_matrix)
assignments = clean_assignment(row, columns)
total_cost = calc_costs(cost_matrix, assignments)
print(f"The total cost of the assignment is {total_cost}.")
draw_network(cost_matrix, assignments)


In :
run_assignment(cost_matrix)

The final assignment is (0, 3), (1, 2), (2, 1), (3, 0), (4, 4)
The total cost of the assignment is 72. ## Part 3 - Full implementation of the Hungarian Algorithm¶

This section presents a step-by-step implementation of the algorithm.

The algorithm has two stages, first we find the minimum value at each row, and then subtract that value to every element of the row. Then we repeat the process using the columns instead. We define this process in the function hungarian_step that has the cost_matrix as input.

In :
def hungarian_step(mat):
#The for-loop iterates through every column in the matrix so we subtract this value to every element of the row
for row_num in range(mat.shape):
mat[row_num] = mat[row_num] - np.min(mat[row_num])

#We repeat the process for the columns
for col_num in range(mat.shape):
mat[:,col_num] = mat[:,col_num] - np.min(mat[:,col_num])

return mat


The next step, while easy to carry out visually, becomes more difficult to code. We need to find the row containing the least number of zeros first.

The first step to do this is to define a function that finds the minimum number of rows to mark that contain a zero value, let's call this min_zeros. Let's assume that the matrix being input is boolean with True where 0 existed and False where non-zero.

Now, mark the column and row as False and repeat, saving the information where the last zero value was retrieved.

By repeating this process, we collect all zeros in the matrix.

In :
def min_zeros(zero_mat, mark_zero):
# min_row = [number of zeros, row index number]
min_row = [99999, -1]

for row_num in range(zero_mat.shape):
if np.sum(zero_mat[row_num] == True) > 0 and min_row > np.sum(zero_mat[row_num] == True):
min_row = [np.sum(zero_mat[row_num] == True), row_num]

# Marked the specific row and column as False
zero_index = np.where(zero_mat[min_row] == True)
mark_zero.append((min_row, zero_index))
zero_mat[min_row, :] = False
zero_mat[:, zero_index] = False


Now that we extracted the zeros in the matrix, we can mark the rows and columns to determine whether the hungarian algorithm is complete.

In :
def mark_matrix(mat):
#Transform the matrix to boolean matrix(0 = True, others = False)
cur_mat = mat
zero_bool_mat = (cur_mat == 0)
zero_bool_mat_copy = zero_bool_mat.copy()

#Recording possible answer positions by marked_zero
marked_zero = []
while (True in zero_bool_mat_copy):
min_zeros(zero_bool_mat_copy, marked_zero)

#Recording the row and column indexes seperately.
marked_zero_row = []
marked_zero_col = []
for i in range(len(marked_zero)):
marked_zero_row.append(marked_zero[i])
marked_zero_col.append(marked_zero[i])

# mark rows not containing zeros
non_marked_row = list(set(range(cur_mat.shape)) - set(marked_zero_row))

# mark columns with zeros
marked_cols = []
check_switch = True
while check_switch:
check_switch = False
for i in range(len(non_marked_row)):
row_array = zero_bool_mat[non_marked_row[i], :]
for j in range(row_array.shape):
if row_array[j] == True and j not in marked_cols:

marked_cols.append(j)
check_switch = True

for row_num, col_num in marked_zero:
if row_num not in non_marked_row and col_num in marked_cols:

non_marked_row.append(row_num)
check_switch = True

# mark rows with zeros
marked_rows = list(set(range(mat.shape)) - set(non_marked_row))

return(marked_zero, marked_rows, marked_cols)


In the case where marked rows and columns do not add up to $n$, we need to adjust the matrix to continue the search.

In :
def adjust_matrix(mat, cover_rows, cover_cols):
cur_mat = mat
non_zero_element = []

# find the minimum value of an element not in a marked column/row
for row in range(len(cur_mat)):
if row not in cover_rows:
for i in range(len(cur_mat[row])):
if i not in cover_cols:
non_zero_element.append(cur_mat[row][i])

min_num = min(non_zero_element)

# substract to all values not in a marked row/column
for row in range(len(cur_mat)):
if row not in cover_rows:
for i in range(len(cur_mat[row])):
if i not in cover_cols:
cur_mat[row, i] = cur_mat[row, i] - min_num
# add to all values in marked rows/column
for row in range(len(cover_rows)):
for col in range(len(cover_cols)):
cur_mat[cover_rows[row], cover_cols[col]] = cur_mat[cover_rows[row], cover_cols[col]] + min_num

return cur_mat


We can now put it all together into a single function:

In :
def hungarian_algorithm(cost_matrix):
n = cost_matrix.shape
cur_mat = copy.deepcopy(cost_matrix)

cur_mat = hungarian_step(cur_mat)

count_zero_lines = 0

while count_zero_lines < n:
ans_pos, marked_rows, marked_cols = mark_matrix(cur_mat)
count_zero_lines = len(marked_rows) + len(marked_cols)

if count_zero_lines < n:

return ans_pos

In :
assignment = hungarian_algorithm(cost_matrix)
print(f"The final assignment is: {assignment}")
print(cost_matrix)

The final assignment is: [(1, 4), (0, 3), (2, 0), (3, 2), (4, 1)]
[[1 8 2 0 0]
[3 2 3 7 0]
[1 6 4 8 8]
[9 6 0 4 5]
[7 0 1 6 6]]


We can now calculate the final cost of assignment.

In :
total = 0
for i in range(len(assignment)):
total += cost_matrix[assignment[i], assignment[i]]
print(f"The total cost of the assignment is {total}")

The total cost of the assignment is 1