Question

This is a follow up question from a previous SO question. Now I have a bit which I have spread it into 8 bits. I have use Amro’s solution to spread the bit to 8 bits. Now I want an inverse way to convert the 8bits back to the single bit.

I have only managed to implement the inverse using for loop which take alot of time in the application.

Is there a faster way of doing it?

Answer 1

Since you are using the solution I suggested last time, lets say you have a matrix N-by-8 of these ‘bits’ where each row represent one 8-bit binary number. To convert to decimal in a vectorized way, its as simple as:

» M = randi([0 1], [5 8])      %# 5 random 8-bit numbers
M =
     1     0     1     0     1     0     1     1
     0     1     1     0     1     1     1     0
     1     1     0     1     1     0     1     1
     1     0     0     0     0     1     1     0
     1     0     0     1     0     1     1     0
» d = bin2dec( num2str(M) )
d =
   171
   110
   219
   134
   150

An alternative solution:

d = sum( bsxfun(@times, M, power(2,7:-1:0)), 2)