For an excellent tutorial on least squared fitting see : http://www.embedded.com/electrical-engineer-community/industry-blog/4027019/1/Why-all-the-math-

Author : Luke Whittlesey

email : [email protected]

In [1]:

```
# we need some ipython magic commands to
# 1. allow sympy to pretty print math equations inline
# 2. use pylab for inline graphing
%pylab inline
%load_ext sympy.interactive.ipythonprinting
```

In [2]:

```
def numerical_cubic_bezier_fn(control_pts, t):
"""
Numerically calculate an x,y point for the bezier curve at position t.
Parameters: control_pts -- list of 4 (x,y) control points
t -- 'time' value should be between 0 and 1
Return: (xpt, ypt) -- tuple of x,y data on the curve
"""
check = ((0 <= t) and (t <= 1))
assert check
# define the actual cubic bezier equation here
fn = lambda c,t : c[0]*(1 - t)**3 + c[1]*3*t*(1 - t)**2 + c[2]*3*t**2*(1 - t) + c[3]*t**3
xs = [x for x,y in control_pts]
ys = [y for x,y in control_pts]
# now calculate the x,y position from the bezier equation
xpt = fn(xs, t)
ypt = fn(ys, t)
return xpt, ypt
# just to make sure that I understand, lets create a list of numbers for a specific bezier curve
# arbitrarily define my 4 control points here
p0 = (0.7,0.2); p1 = (2.3,1.0); p2 = (7.2,0.8); p3 = (10.0,0.0)
# create a list of 1000 x,y points that traces over the length of this curve
points = [numerical_cubic_bezier_fn(control_pts=[p0,p1,p2,p3], t=t) for t in arange(0.0,1.0,1./1000)]
# separate the x's and y's so matplotlib can plot them
xs = [x for x,y in points]
ys = [y for x,y in points]
# plot the resulting curve and the control points
plot(xs, ys, 'b-')
hold(True)
for pt in [p0,p1,p2,p3]:
plot( pt[0], pt[1], 'ro')
hold(False)
t=title('example cubic bezier curve')
#legend(['curve','control pts'])
```

In [3]:

```
from sympy.printing.mathml import mathml
from sympy import Symbol, Matrix, summation, diff, expand, collect, preview, Sum, latex, Eq, factor
t = Symbol('t_i')
c0, c1, c2, c3 = [Symbol(name) for name in ['c0','c1','c2','c3']]
# the equation for a cubic bezier path b(t) over t in [0,1]
# Note: the c control points are (x,y) tuples, but we're going to compute
# the resulting x and y values independantly. x and y are independant of
# each other, but both are dependant on t which goes from 0 to 1
symbolic_cubic_bezier_fn = c0*(1 - t)**3 + c1*3*t*(1 - t)**2 + c2*3*t**2*(1 - t) + c3*t**3
print 'The cubic bezier function B(t) is:'
B = Symbol('B(t_i)')
Eq(B, symbolic_cubic_bezier_fn)
```

Out[3]:

In [4]:

```
d_i = Symbol('d_i')
# the error between what the 'best fit' bezier curve produces and the hand drawn point is defined as
error = symbolic_cubic_bezier_fn - d_i
e_i = Symbol('e_i')
# where symbolic_cubic_bezier_fn is bezier(t, control_points) and data_i is the i'th hand drawn point
Eq(e_i, error)
```

Out[4]:

In [5]:

```
# the least squares penalty function is the sum of all of the squared errors
penalty_fn = error**2
print 'The penalty function is :'
i = Symbol('i')
n = Symbol('n')
M = Symbol('M')
Eq(M, Sum( penalty_fn, (i,0,n) ))
```

Out[5]:

In [6]:

```
df_dc0 = collect( expand(penalty_fn.diff(c0), deep=True), [c0,c1,c2,c3])
print 'Partial derivative wrt c0 is:'
Eq(0, Sum(df_dc0, (i,0,n)) )
```

Out[6]:

In [7]:

```
df_dc1 = collect( expand(penalty_fn.diff(c1), deep=True), [c0,c1,c2,c3])
print 'Partial derivative wrt c1 is:'
Eq(0, Sum(df_dc1, (i,0,n)))
```

Out[7]:

In [8]:

```
df_dc2 = collect( expand(penalty_fn.diff(c2), deep=True), [c0,c1,c2,c3])
print 'Partial derivative wrt c2 is:'
Eq(0, Sum(df_dc2, (i,0,n)) )
```

Out[8]:

In [9]:

```
df_dc3 = collect( expand(penalty_fn.diff(c3), deep=True), [c0,c1,c2,c3])
print 'Partial derivative wrt c3 is:'
Eq(0, Sum(df_dc3, (i,0,n)) )
```

Out[9]:

In [10]:

```
def pull_arguments_by_var(eqn, variables):
"""
For a equation such as : a*2 + b*3*x + 5*x + 5
this function will return a dictionary of the arguments that are multiplied by each variable.
So if variables=[a,b], the dict returned would be:
a:[2],
b:[3*x],
'other':[5*x, 5]
Parameters: eqn -- a sympy equation that has already been term collected
variables -- a list of variables to search for arguments that are multiplied by each variable
Returns: vardict -- a dictionary containing the arguments for each variable and 'other'
"""
# try to find what is multiplied by the variables in variables and other
cdict = {}
for var in variables:
for arg in eqn.args:
if var in arg.atoms(Symbol):
cdict.setdefault(var, []).append(arg/var)
#print '%s : %s' %(c, arg)
for arg in eqn.args:
# test for non-cx arguments
if not any([(cc in arg) for cc in variables]):
cdict.setdefault('other', []).append(arg)
#print 'other : %s' %(arg)
return cdict
```

In [11]:

```
# from the 1st partial derivative, reshuffle the equations so the c's are on the left and the others are on the right
# also redistribute the sums
cvars = [c0,c1,c2,c3]
arg_dict = pull_arguments_by_var(eqn=df_dc0, variables=cvars)
leftside = c0*Sum(factor(arg_dict[c0][0]), (i,0,n)) + \
c1*Sum(factor(arg_dict[c1][0]), (i,0,n)) + \
c2*Sum(factor(arg_dict[c2][0]), (i,0,n)) + \
c3*Sum(factor(arg_dict[c3][0]), (i,0,n))
rightside = arg_dict['other'][0]
for arg in arg_dict['other'][1:]:
rightside = rightside + arg
print '1st Equation:'
Eq(leftside, -1*Sum(factor(rightside), (i,0,n)))
```

Out[11]:

In [12]:

```
# from the 2nd partial derivative, reshuffle the equations so the c's are on the left and the others are on the right
# also redistribute the sums
cvars = [c0,c1,c2,c3]
arg_dict = pull_arguments_by_var(eqn=df_dc1, variables=cvars)
leftside = c0*Sum(factor(arg_dict[c0][0]), (i,0,n)) + \
c1*Sum(factor(arg_dict[c1][0]), (i,0,n)) + \
c2*Sum(factor(arg_dict[c2][0]), (i,0,n)) + \
c3*Sum(factor(arg_dict[c3][0]), (i,0,n))
rightside = arg_dict['other'][0]
for arg in arg_dict['other'][1:]:
rightside = rightside + arg
print '2nd Equation:'
Eq(leftside, -1*Sum(factor(rightside), (i,0,n)))
```

Out[12]:

In [13]:

```
# from the 3rd partial derivative, reshuffle the equations so the c's are on the left and the others are on the right
# also redistribute the sums
cvars = [c0,c1,c2,c3]
arg_dict = pull_arguments_by_var(eqn=df_dc2, variables=cvars)
leftside = c0*Sum(factor(arg_dict[c0][0]), (i,0,n)) + \
c1*Sum(factor(arg_dict[c1][0]), (i,0,n)) + \
c2*Sum(factor(arg_dict[c2][0]), (i,0,n)) + \
c3*Sum(factor(arg_dict[c3][0]), (i,0,n))
rightside = arg_dict['other'][0]
for arg in arg_dict['other'][1:]:
rightside = rightside + arg
print '3rd Equation:'
Eq(leftside, -1*Sum(factor(rightside), (i,0,n)))
```

Out[13]:

In [14]:

```
# from the 4th partial derivative, reshuffle the equations so the c's are on the left and the others are on the right
# also redistribute the sums
cvars = [c0,c1,c2,c3]
arg_dict = pull_arguments_by_var(eqn=df_dc3, variables=cvars)
leftside = c0*Sum(factor(arg_dict[c0][0]), (i,0,n)) + \
c1*Sum(factor(arg_dict[c1][0]), (i,0,n)) + \
c2*Sum(factor(arg_dict[c2][0]), (i,0,n)) + \
c3*Sum(factor(arg_dict[c3][0]), (i,0,n))
rightside = arg_dict['other'][0]
for arg in arg_dict['other'][1:]:
rightside = rightside + arg
print '4th Equation:'
Eq(leftside, -1*Sum(factor(rightside), (i,0,n)))
```

Out[14]:

$\mathbf{A}$ is a 4x4 matrix of the summations listed on the left side of the equations

$\mathbf{x}$ is a 4x1 matrix containing the unknowns c0,c1,c2,c3

$\mathbf{b}$ is a 4x1 matrix of the summations listed on the right side of the equations

To solve is simply : $\mathbf{x = A^{-1} b}$

In [15]:

```
# build the actual matrices this time in a loop
cvars = [c0,c1,c2,c3]
a_vals = []
x_vals = cvars
b_vals = []
for eqn in [df_dc0, df_dc1, df_dc2, df_dc3]:
arg_dict = pull_arguments_by_var(eqn=eqn, variables=cvars)
for var in cvars:
a_vals.append( Sum(factor(arg_dict[var][0]), (i,0,n)) )
rightside = arg_dict['other'][0]
for arg in arg_dict['other'][1:]:
rightside = rightside + arg
b_vals.append( -1*Sum(factor(rightside), (i,0,n)) )
A = Matrix(4,4,a_vals)
x = Matrix(4,1,x_vals)
b = Matrix(4,1,b_vals)
```

In [16]:

```
print 'A = '
A
```

Out[16]:

In [17]:

```
print 'x = '
x
```

Out[17]:

In [18]:

```
print 'b = '
b
```

Out[18]:

In [19]:

```
from pylab import randn
# define the control points of a bezier curve we wish to generate
p0 = (0.7,0.2); p1 = (2.3,1.0); p2 = (7.2,0.8); p3 = (10.0,0.0)
# create a list of 1000 x,y points that traces over the length of this curve
points = [numerical_cubic_bezier_fn(control_pts=[p0,p1,p2,p3], t=t) for t in arange(0.0,1.0,1./1000)]
# separate the x's and y's so matplotlib can plot them; also add some noise
xs = [x+0.15*randn() for x,y in points]
ys = [y+0.15*randn() for x,y in points]
# plot the resulting 'hand drawn' curve
plot(xs, ys, 'b-')
t=title('noisy (hand drawn) curve')
```

We used $d_i$ to represent each of our drawing data points in sequence, but you might be confused on how we are going to use (x,y) points for each $d_i$. The answer is that we will compute only for x and then we will repeat the process only for y. This should give us the x portion of the control points and then the y portion of the control points. Note that x and y are orthogonal and so they are independant of each other, so we can get away with calculating them independantly.

For $t_i$, just imagine that someone is drawing the curve starting at one position and ending at another position in one continuous line. For this case we will assume it was drawn from the left and was finished on the right. $t_i$ is simply the time at which each data point was drawn. Remember that $t_i$ needs to go from 0 to 1 for the bezier equation to work. So, we assume that it took 1 second to draw the entire curve and each data point is drawn at the same relative speed. Of course time actually has nothing to do with anything, it's just a mental model of tying a $t_i$ to a $d_i$.

In [20]:

```
# we already have out d_i's in xs and ys from above
# make a t_i list
ts = arange(0.0, 1.0, 1.0/len(points))
# make the summation functions for A (16 of them)
A_fns = [None for i in range(16)]
A_fns[0] = lambda time_list, data_list : sum([2*(t_i - 1)**6 for t_i, d_i in zip(time_list, data_list)])
A_fns[1] = lambda time_list, data_list : sum([-6*t_i*(t_i - 1)**5 for t_i, d_i in zip(time_list, data_list)])
A_fns[2] = lambda time_list, data_list : sum([6*t_i**2*(t_i - 1)**4 for t_i, d_i in zip(time_list, data_list)])
A_fns[3] = lambda time_list, data_list : sum([-2*t_i**3*(t_i - 1)**3 for t_i, d_i in zip(time_list, data_list)])
A_fns[4] = lambda time_list, data_list : sum([-6*t_i*(t_i - 1)**5 for t_i, d_i in zip(time_list, data_list)])
A_fns[5] = lambda time_list, data_list : sum([18*t_i**2*(t_i - 1)**4 for t_i, d_i in zip(time_list, data_list)])
A_fns[6] = lambda time_list, data_list : sum([-18*t_i**3*(t_i - 1)**3 for t_i, d_i in zip(time_list, data_list)])
A_fns[7] = lambda time_list, data_list : sum([6*t_i**4*(t_i - 1)**2 for t_i, d_i in zip(time_list, data_list)])
A_fns[8] = lambda time_list, data_list : sum([6*t_i**2*(t_i - 1)**4 for t_i, d_i in zip(time_list, data_list)])
A_fns[9] = lambda time_list, data_list : sum([-18*t_i**3*(t_i - 1)**3 for t_i, d_i in zip(time_list, data_list)])
A_fns[10] = lambda time_list, data_list : sum([18*t_i**4*(t_i - 1)**2 for t_i, d_i in zip(time_list, data_list)])
A_fns[11] = lambda time_list, data_list : sum([-6*t_i**5*(t_i - 1) for t_i, d_i in zip(time_list, data_list)])
A_fns[12] = lambda time_list, data_list : sum([-2*t_i**3*(t_i - 1)**3 for t_i, d_i in zip(time_list, data_list)])
A_fns[13] = lambda time_list, data_list : sum([6*t_i**4*(t_i - 1)**2 for t_i, d_i in zip(time_list, data_list)])
A_fns[14] = lambda time_list, data_list : sum([-6*t_i**5*(t_i - 1) for t_i, d_i in zip(time_list, data_list)])
A_fns[15] = lambda time_list, data_list : sum([2*t_i**6 for t_i, d_i in zip(time_list, data_list)])
# make the summation functions for b (4 of them)
b_fns = [None for i in range(4)]
b_fns[0] = lambda time_list, data_list : -1.0 * sum([2*d_i*(t_i - 1)**3 for t_i, d_i in zip(time_list, data_list)])
b_fns[1] = lambda time_list, data_list : -1.0 * sum([-6*d_i*t_i*(t_i - 1)**2 for t_i, d_i in zip(time_list, data_list)])
b_fns[2] = lambda time_list, data_list : -1.0 * sum([6*d_i*t_i**2*(t_i - 1) for t_i, d_i in zip(time_list, data_list)])
b_fns[3] = lambda time_list, data_list : -1.0 * sum([-2*d_i*t_i**3 for t_i, d_i in zip(time_list, data_list)])
```

In [21]:

```
def solve_for_cs(time_series, data_series):
"""
Take an input series of t_i values and the corresponding d_i values,
compute the summation values that should go into the matrices and
solve for the 4 unknown variables.
Parameters: time_series -- t_i in increasing values
data_series -- d_i corresponding to each t_i
Returns: solution -- matrix containing the 4 solutions from solving the linear equations
"""
# compute the data we will put into matrix A
A_values = []
for fn in A_fns:
A_values.append(fn(time_series, data_series))
# fill the A matrix with data
A_numerical = Matrix(4,4, A_values)
# compute the data we will put into the b vector
b_values = []
for fn in b_fns:
b_values.append(fn(time_series, data_series))
# fill the b vector with data
b_numerical = Matrix(4,1, b_values)
# solve for the unknowns in vector x
x_numerical = A_numerical.inv() * b_numerical
#print 'A = \n', A_numerical
#print 'b = \n', b_numerical
#print 'x = \n', x_numerical
return x_numerical
```

In [22]:

```
# solve for the best fit in the x dimension
x_solutions = solve_for_cs(time_series=ts, data_series=xs)
# solve for the best fit in the y dimension
y_solutions = solve_for_cs(time_series=ts, data_series=ys)
# place the solutions into control points
best_fit_control_pts = [(x,y) for x,y in zip(x_solutions, y_solutions)]
```

In [23]:

```
# create a list of 1000 x,y points that traces over the length of our best fit curve
best_fit_points = [numerical_cubic_bezier_fn(control_pts=best_fit_control_pts, t=t) for t in arange(0.0,1.0,1./1000)]
# separate the x's and y's so matplotlib can plot them
best_fit_xs = [x for x,y in best_fit_points]
best_fit_ys = [y for x,y in best_fit_points]
# plot the original noisy curve
plot(xs, ys, 'b-')
hold(True)
# plot the best fit curve over top
plot(best_fit_xs, best_fit_ys, 'r-')
hold(False)
t=title('best fit cubic bezier curve')
# just for comparisons sake, print the 'clean' control points and then print the best
# fit control points taken from the noisy data
print 'Clean Control Points = ',
print [p0, p1, p2, p3]
print 'Best Fit Control Points = '
print best_fit_control_pts
```

In [23]:

```
```