# Tridiagonal Matrix Algorithm (TDMA) aka Thomas Algorithm, using Python with NumPy arrays

### Question

I found an implementation of the thomas algorithm or TDMA in MATLAB.

``````function x = TDMAsolver(a,b,c,d)
%a, b, c are the column vectors for the compressed tridiagonal matrix, d is the right vector
n = length(b); % n is the number of rows

% Modify the first-row coefficients
c(1) = c(1) / b(1);    % Division by zero risk.
d(1) = d(1) / b(1);    % Division by zero would imply a singular matrix.

for i = 2:n-1
temp = b(i) - a(i) * c(i-1);
c(i) = c(i) / temp;
d(i) = (d(i) - a(i) * d(i-1))/temp;
end

d(n) = (d(n) - a(n) * d(n-1))/( b(n) - a(n) * c(n-1));

% Now back substitute.
x(n) = d(n);
for i = n-1:-1:1
x(i) = d(i) - c(i) * x(i + 1);
end
end
``````

I need it in python using numpy arrays, here my first attempt at the algorithm in python.

``````import numpy

aa = (0.,8.,9.,3.,4.)
bb = (4.,5.,9.,4.,7.)
cc = (9.,4.,5.,7.,0.)
dd = (8.,4.,5.,9.,6.)

ary = numpy.array

a = ary(aa)
b = ary(bb)
c = ary(cc)
d = ary(dd)

n = len(b)## n is the number of rows

## Modify the first-row coefficients
c[0] = c[0]/ b[0]    ## risk of Division by zero.
d[0] = d[0]/ b[0]

for i in range(1,n,1):
temp = b[i] - a[i] * c[i-1]
c[i] = c[i]/temp
d[i] = (d[i] - a[i] * d[i-1])/temp

d[-1] = (d[-1] - a[-1] * d[-2])/( b[-1] - a[-1] * c[-2])

## Now back substitute.
x = numpy.zeros(5)
x[-1] = d[-1]
for i in range(-2, -n-1, -1):
x[i] = d[i] - c[i] * x[i + 1]
``````

They give different results, so what am I doing wrong?

1
2
1/4/2012 7:53:47 PM

I made this since none of the online implementations for python actually work. I've tested it against built-in matrix inversion and the results match.

Here a = Lower Diag, b = Main Diag, c = Upper Diag, d = solution vector

``````import numpy as np

def TDMA(a,b,c,d):
n = len(d)
w= np.zeros(n-1,float)
g= np.zeros(n, float)
p = np.zeros(n,float)

w[0] = c[0]/b[0]
g[0] = d[0]/b[0]

for i in range(1,n-1):
w[i] = c[i]/(b[i] - a[i-1]*w[i-1])
for i in range(1,n):
g[i] = (d[i] - a[i-1]*g[i-1])/(b[i] - a[i-1]*w[i-1])
p[n-1] = g[n-1]
for i in range(n-1,0,-1):
p[i-1] = g[i-1] - w[i-1]*p[i]
return p
``````

For an easy performance boost for large matrices, use numba! This code outperforms np.linalg.inv() in my tests:

``````import numpy as np
from numba import jit

@jit
def TDMA(a,b,c,d):
n = len(d)
w= np.zeros(n-1,float)
g= np.zeros(n, float)
p = np.zeros(n,float)

w[0] = c[0]/b[0]
g[0] = d[0]/b[0]

for i in range(1,n-1):
w[i] = c[i]/(b[i] - a[i-1]*w[i-1])
for i in range(1,n):
g[i] = (d[i] - a[i-1]*g[i-1])/(b[i] - a[i-1]*w[i-1])
p[n-1] = g[n-1]
for i in range(n-1,0,-1):
p[i-1] = g[i-1] - w[i-1]*p[i]
return p
``````
0
4/4/2017 6:27:12 PM