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Create Rubidium2018.py
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kalwar authored Dec 8, 2018
1 parent ae4758c commit 781bd0e
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104 changes: 104 additions & 0 deletions Rubidium2018.py
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# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def closest_split_pair(p_x, p_y, delta, best_pair):
ln_x = len(p_x) # store length - quicker
mx_x = p_x[ln_x // 2][0] # select midpoint on x-sorted array

# Create a subarray of points not further than delta from
# midpoint on x-sorted array

s_y = [x for x in p_y if mx_x - delta <= x[0] <= mx_x + delta]

best = delta # assign best value to delta
ln_y = len(s_y) # store length of subarray for quickness
for i in range(ln_y - 1):
for j in range(i+1, min(i + 7, ln_y)):
p, q = s_y[i], s_y[j]
dst = dist(p, q)
if dst < best:
best_pair = p, q
best = dst
return best_pair[0], best_pair[1], best

def dist(p1, p2):
return max(abs(p1[0] - p2[0]), abs(p1[1] - p2[1]))

def brute(ax):
mi = dist(ax[0], ax[1])
p1 = ax[0]
p2 = ax[1]
ln_ax = len(ax)
if ln_ax == 2:
return p1, p2, mi
for i in range(ln_ax-1):
for j in range(i + 1, ln_ax):
if i != 0 and j != 1:
d = dist(ax[i], ax[j])
if d < mi: # Update min_dist and points
mi = d
p1, p2 = ax[i], ax[j]
return p1, p2, mi

def closest_pair(ax, ay):
ln_ax = len(ax) # It's quicker to assign variable
if ln_ax <= 3:
return brute(ax) # A call to bruteforce comparison
mid = ln_ax // 2 # Division without remainder, need int
Qx = ax[:mid] # Two-part split
Rx = ax[mid:]

# Determine midpoint on x-axis

midpoint = ax[mid][0]
Qy = list()
Ry = list()
for x in ay: # split ay into 2 arrays using midpoint
if x[0] <= midpoint:
Qy.append(x)
else:
Ry.append(x)

# Call recursively both arrays after split

(p1, q1, mi1) = closest_pair(Qx, Qy)
(p2, q2, mi2) = closest_pair(Rx, Ry)

# Determine smaller distance between points of 2 arrays

if mi1 <= mi2:
d = mi1
mn = (p1, q1)
else:
d = mi2
mn = (p2, q2)

# Call function to account for points on the boundary

(p3, q3, mi3) = closest_split_pair(ax, ay, d, mn)

# Determine smallest distance for the array

if d <= mi3:
return mn[0], mn[1], d
else:
return p3, q3, mi3

def solution(X, Y):
a = list(zip(X, Y))
ax = sorted(a, key=lambda x: x[0]) # Presorting x-wise
ay = sorted(a, key=lambda x: x[1]) # Presorting y-wise
(x1, y1), (x2, y2), mi = closest_pair(ax, ay) # Recursive D&C function

lo = 0
high = 10 ** 5
while lo < high:
mid = (lo + high + 1) // 2
if abs(x1 - x2) < 2 * mid and abs(y1 - y2) < 2 * mid:
high = mid - 1
else:
lo = mid
return lo

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