PLEASE CODE IN PYTHON

Implement the algorithm for the closest pair of points (Section 5.3 in the textbook). Your program should input the coordinates of the points from (each of) the test files uploaded on D2L, and outputs: (1) the coordinates of a closest pair in the point-set, and (2) the distance between the two points in the closest pair.

Given a text file of pairs use the algorithm to find closest pair of points and the distance

textfile name: 10points.txt

Pairs in file like

1574 9098 7156 1322 511 4721 815 17928 1267 5292 666 405 889 14787 1184 2147 2560 18128 337 1320

Algorithm in textbook below

Def closest_pair(p):

n = len(p)

mergeSort(p,1,n)

return rec_cl_pair(p,1,n)

def rec_cl_pair(p,i,j):

if(j-i < 3):

mergeSort(p)

distance = dist(p[i],p[i+1])

closest_pair = (p[i],p[i+1])

if(j - i == 1):

return distance

if(dist(p[i+1], p[i+2]) < distance):

distance = dist(p[i+1], p[i+2])

closest_pair = (p[i+1], p[i+2])

if(dist(p[i], p[i+2]) < distance):

distance = dist(p[i], p[i+2])

closest_pair = (p[i], p[i+2])

return distance, closest_pair

k = (i+j)//2

l = p[k]

distanceL, closest_pairLeft = rec_cl_pair(p,i,k)

distanceR, closest_pairRight = rec_cl_pair(p,k+1,j)

distance = min(distanceL, distanceR)

if(distance == distanceR):

closest_pair = closest_pairRight

else:

closest_pair = closest_pairLeft

merge(p,i,k,j)

t = 0

for k in range(i, j):

if(p[k] > l - distance & p[k] < l + distance):

t += 1

v[t] = p(k)

for k in range(1, t-1):

for s in range(k+1, min(t, k+7)):

distance = min(distance, dist(v[k], v[s]))

return distance