Q.78 solution expalined

Author: Hemanth21k

source/exercises100.ktx

@@ -1090,17 +1090,37 @@ Consider 2 sets of points P0,P1 describing lines (2d) and a point p, how to comp
 No hints provided...
 
 < a78
+P0 = np.random.uniform(-10,10,(10,2))
+P1 = np.random.uniform(-10,10,(10,2))
+p  = np.random.uniform(-10,10,( 1,2))
+
+def distance(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.
+
+    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(P0, P1, p))
+
+##--------------- OR ---------------##
 def distance(P0, P1, p):
     T = P1 - P0
     L = (T**2).sum(axis=1)
     U = -((P0[:,0]-p[...,0])*T[:,0] + (P0[:,1]-p[...,1])*T[:,1]) / L
     U = U.reshape(len(U),1)
     D = P0 + U*T - p
     return np.sqrt((D**2).sum(axis=1))
-
-P0 = np.random.uniform(-10,10,(10,2))
-P1 = np.random.uniform(-10,10,(10,2))
-p  = np.random.uniform(-10,10,( 1,2))
 print(distance(P0, P1, p))
 
 < q79