Shape Matching Experiments

Often, in Computer Vision tasks we have a model of an interesting shape and we want to know if this model matches with a target shape found through the analysis of the edges of an image. The target shape is never identical to the model we have, because it could be a representation of the model in a different pose or it could match only partially our model, so we have to find a proper transformation of the model in order to match the target. In this post we will see how to find the closest shape model to a target shape using the fmin function provided by Scipy.
We can represent a shape with a matrix like the following:

where each pair (x_i,y_i) is a "landmark" point of the shape and we can transform every generic point (x,y) using this function:

In this transformation we have that θ is a rotation angle, t_x and t_y are the translation along the x and the y axis respectively, while s is a scaling coefficient. Applying this transformation to every point of the shape described by X we get a new shape which is a transformation of the original one according to the parameters used in T. Given a matrix as defined above, the following Python function is able to apply the transformation to every point of the shape represented by the matrix:

import numpy as np
import pylab as pl
from scipy.optimize import fmin

def transform(X,theta,tx=0,ty=0,s=1):
 """ Performs scaling, rotation and translation
     on the set of points in the matrix X 
     
    Input:
      X, points (each row have to be a point [x, y])
      theta, rotation angle (in radiants)
      tx, translation along x
      ty, translation along y
      s, scaling coefficient
    Return:
      Y, transformed set of points
 """
 a = math.cos(theta)
 b = math.sin(theta)
 R = np.mat([[a*s, -b*s], [b*s, a*s]])
 Y = R*X # rotation and scaling
 Y[0,:] = Y[0,:]+tx # translation
 Y[1,:] = Y[1,:]+ty
 return Y

Now, we want to find the best pose (translation, scale and rotation) to match a model shape X to a target shape Y. We can solve this problem minimizing the sum of square distances between the points of X and the ones of Y, which means that we want to find

where T' is a function that applies T to all the point of the input shape. This task turns out to be pretty easy using the fmin function provided by Scipy:

def error_fun(p,X,Y):
 """ cost function to minimize """
 return np.linalg.norm(transform(X,p[0],p[1],p[2],p[3])-Y)

def match(X,Y,p0):
 """ Match the model X with the shape Y
     returns the set of parameters to transform
     X into Y and a set of partial solutions.
 """
 p_opt = fmin(error_fun, p0, args=(X,Y),retall=True)
 return p_opt[0],p_opt[1] # optimal solutions, other solutions

In the code above we have defined a cost function which measures how close the two shapes are and another function that minimizes the difference between the two shapes and returns the transfomation parameters. Now, we can try to match the trifoil shape starting from one of its transformations using the functions defined above:

# generating a shape to match
t = np.linspace(0,2*np.pi,50)
x = 2*np.cos(t)-.5*np.cos( 4*t )
y = 2*np.sin(t)-.5*np.sin( 4*t )
A = np.array([x,y]) # model shape
# transformation parameters
p = [-np.pi/2,.5,.8,1.5] # theta,tx,ty,s
Ar = transform(A,p[0],p[1],p[2],p[3]) # target shape
p0 = np.random.rand(4) # random starting parameters
p_opt, allsol = match(A,Ar,p0) # matching

It's interesting to visualize the error at each iteration:

pl.plot([error_fun(s,A,Ar) for s in allsol])
pl.show()

and the value of the parameters of the transformation at each iteration:

pl.plot(allsol)
pl.show()

From the graphs above we can observe that the the minimization process found its solution after 150 iterations. Indeed, after 150 iterations the error become very close to zero and there is almost no variation in the parameters. Now we can visualize the minimization process plotting some of the solutions in order to compare them with the target shape:

def plot_transform(A,p):
 Afound = transform(A,p[0],p[1],p[2],p[3])
 pl.plot(Afound[0,:].T,Afound[1,:].T,'-')

for k,i in enumerate(range(0,len(allsol),len(allsol)/15)):
 pl.subplot(4,4,k+1)
 pl.plot(Ar[0,:].T,Ar[1,:].T,'--',linewidth=2) # target shape
 plot_transform(A,allsol[i])
 pl.axis('equal')
pl.show()

In the graph above we can see that the initial solutions (the green ones) are very different from the shape we are trying to match (the dashed blue ones) and that during the process of minimization they get closer to the target shape until they fits it completely. In the example above we used two shapes of the same form. Let's see what happen when we have a starting shape that has a different form from the target model:

t = np.linspace(0,2*np.pi,50)
x = np.cos(t)
y = np.sin(t)
Ac = np.array([x,y]) # the circle (model shape)
p0 = np.random.rand(4) # random starting parameters
# the target shape is the trifoil again
p_opt, allsol = match(Ac,Ar,p0) # matching

In the example above we used a circle as model shape and a trifoil as target shape. Let's see what happened during the minimization process:

for k,i in enumerate(range(0,len(allsol),len(allsol)/15)):
 pl.subplot(4,4,k+1)
 pl.plot(Ar[0,:].T,Ar[1,:].T,'--',linewidth=2) # target shape
 plot_transform(Ac,allsol[i])
 pl.axis('equal')
pl.show()

In this case, where the model shape can't match the target shape completely, we see that the minimization process is able to find the circle that is closer to the trifoil.

Monte Carlo estimate for pi with numpy

In this post we will use a Monte Carlo method to approximate pi. The idea behind the method that we are going to see is the following:

Draw the unit square and the unit circle. Consider only the part of the circle inside the square and pick uniformly a large number of points at random over the square. Now, the unit circle has pi/4 the area of the square. So, it should be apparent that of the total number of points that hit within the square, the number of points that hit the circle quadrant is proportional to the area of that part. This gives a way to approximate pi/4 as the ratio between the number of points inside circle and the total number of points and multiplying it by 4 we have pi.

Let's see the python script that implements the method discussed above using the numpy's indexing facilities:

from pylab import plot,show,axis
from numpy import random,sqrt,pi

# scattering n points over the unit square
n = 1000000
p = random.rand(n,2)

# counting the points inside the unit circle
idx = sqrt(p[:,0]**2+p[:,1]**2) < 1

plot(p[idx,0],p[idx,1],'b.') # point inside
plot(p[idx==False,0],p[idx==False,1],'r.') # point outside
axis([-0.1,1.1,-0.1,1.1]) 
show()

# estimation of pi
print '%0.16f' % (sum(idx).astype('double')/n*4),'result'
print '%0.16f' % pi,'real pi'

The program will print the pi approximation on the standard out:

3.1457199999999998 result
3.1415926535897931 real pi

and will show a graph with the generated points:

Note that the lines of code used to estimate pi are just 3!

Fixed point iteration

A fixed point for a function is a point at which the value of the function does not change when the function is applied. More formally, x is a fixed point for a given function f if

and the fixed point iteration

converges to the a fixed point if f is continuous.
The following function implements the fixed point iteration algorithm:

from pylab import plot,show
from numpy import array,linspace,sqrt,sin
from numpy.linalg import norm

def fixedp(f,x0,tol=10e-5,maxiter=100):
 """ Fixed point algorithm """
 e = 1
 itr = 0
 xp = []
 while(e > tol and itr < maxiter):
  x = f(x0)      # fixed point equation
  e = norm(x0-x) # error at the current step
  x0 = x
  xp.append(x0)  # save the solution of the current step
  itr = itr + 1
 return x,xp

Let's find the fixed point of the square root funtion starting from x = 0.5 and plot the result

f = lambda x : sqrt(x)

x_start = .5
xf,xp = fixedp(f,x_start)

x = linspace(0,2,100)
y = f(x)
plot(x,y,xp,f(xp),'bo',
     x_start,f(x_start),'ro',xf,f(xf),'go',x,x,'k')
show()

The result of the program would appear as follows:

The red dot is the starting point, the blue ones are the sequence x_1,x_2,x_3,... and the green is the fixed point found.
In a similar way, we can compute the fixed point of function of multiple variables:

# 2 variables function
def g(x):
 x[0] = 1/4*(x[0]*x[0] + x[1]*x[1])
 x[1] = sin(x[0]+1)
 return array(x)

x,xf = fixedp(g,[0, 1])
print '   x =',x
print 'f(x) =',g(xf[len(xf)-1])

In this case g is a function of two variables and x is a vector, so the fixed point is a vector and the output is as follows:

   x = [ 0.          0.84147098]
f(x) = [ 0.          0.84147098]

Approximating pi

This script uses the following formula

to approximate the value of pi with a fixed number of correct digits.

import math
def pi(digits):
 k = 0
 pi = 0.0
 e = 1.0
 tol = pow(10,-digits) # is minimum error that we want to accept
 while e > tol:
  pi_old = pi
  pi += (2*pow(-1,k)*pow(3,.5-k))/(2*k+1)
  e = abs(pi-pi_old) # current error
  k += 1
  print '%0.16f %20.16f' % (pi,e)
 print '\nerror',e,'\niterations ',k
 print '%0.16f' % pi,'result'
 print '%0.16f' % math.pi,'real pi'

pi(8) # approximating pi with 8 correcet digits

During the execution you can see the current approximated value on the left column and the difference with the approximation at the previous step on the right column. At the end, the difference between the approximated value and the value provided by the math python module will be printed.

$ python pi.py 
3.4641016151377544   3.4641016151377544
3.0792014356780038   0.3849001794597506
3.1561814715699539   0.0769800358919501
3.1378528915956800   0.0183285799742738
3.1426047456630846   0.0047518540674045
3.1413087854628832   0.0012959602002014
3.1416743126988376   0.0003655272359544
3.1415687159417840   0.0001055967570536
3.1415997738115058   0.0000310578697218
3.1415905109380802   0.0000092628734256
3.1415933045030817   0.0000027935650015
3.1415924542876463   0.0000008502154354
3.1415927150203800   0.0000002607327336
3.1415926345473140   0.0000000804730660
3.1415926595217138   0.0000000249743999
3.1415926517339976   0.0000000077877162

error 7.78771624965e-09 
iterations  16
3.1415926517339976 result
3.1415926535897931 real pi