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simplified answer.78

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Hemanth21k 2025-07-26 23:10:48 +00:00
parent dbb54197e8
commit 44df2b0d3c
1 changed files with 11 additions and 14 deletions
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source/exercises100.ktx
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@ -1095,22 +1095,18 @@ P1 = np.random.uniform(-10,10,(10,2))
p = np.random.uniform(-10,10,( 1,2))
def distance_faster(P0,P1,p):
'''
Author: Hemanth Pasupuleti
Reference: https://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
---- Explainable solution - Slightly Faster as number of lines scale up exponentially ----
'''
v = P1- P0 # Shape: (n_lines,2), Compute: [(x2-x1) (y2-y1)]
v[:,[0,1]] = v[:,[1,0]] # Shape: (n_lines,2), Swap along axis to Compute: [(y2-y1) (x2-x1)]
v[:,1]*=-1 # Shape: (n_lines,2), Compute: [(y2-y1) -(x2-x1)]
norm = np.linalg.norm(v,axis=1) # Shape: (n_lines,), Compute: sqrt((x2-x1)**2 + (y2-y1)**2)
r = P0 - p # Shape: (n_lines,2), Compute: [(x1-x0) (y1-y0)]
# np.einsum('ij,ij->i',r,v) is equivalent to np.multiply(r,v)).sum(axis=1) which is scalar product of two matrices across axis 1.
#Author: Hemanth Pasupuleti
#Reference: https://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
v = P1- P0
v[:,[0,1]] = v[:,[1,0]]
v[:,1]*=-1
norm = np.linalg.norm(v,axis=1)
r = P0 - p
d = np.abs(np.einsum("ij,ij->i",r,v)) / norm
d = np.abs(np.einsum("ij,ij->i",r,v)) / norm # Shape: (n_lines,), Compute: d = |(x1-x0)*(y2-y1)-(y1-y0)*(x1-x0)|/sqrt((x2-x1)**2 + (y2-y1)**2)
return d
print(distance_faster(P0, P1, p))
##--------------- OR ---------------##
@ -1121,6 +1117,7 @@ def distance_slower(P0, P1, p):
U = U.reshape(len(U),1)
D = P0 + U*T - p
return np.sqrt((D**2).sum(axis=1))
print(distance_slower(P0, P1, p))
< q79