# Question related to vectorized matrix operation.

1 view (last 30 days)
Pappu Murthy on 22 Jan 2017
Edited: Pappu Murthy on 23 Jan 2017
i have two matrices A and B; A is say (Nx3) where N is rather large like 1000+... B is (nx2) where n is rather smaller of the order 10 or so.. but it does not matter it is smaller than A row size. i would like to compute
for i = 1:N
D = ( A(i,1)- B(1:n,1) )^2 + (A(i,2) - B(1:n,2)^2 )
end
for e.g. if N is 4, and n =2 then D = a [4x2] matrix...
I am able to perform this using for loops without much difficulty but would like to try using the vectorized Matrix operations instead for improving the performance and speed. Thanks in advance.

Stephen23 on 22 Jan 2017
Edited: Andrei Bobrov on 23 Jan 2017
>> A = randi(9,4,3)
A =
4 2 9
9 4 4
5 4 8
4 2 8
>> B = randi(9,3,2)
B =
3 2
6 4
3 1
>> D = bsxfun(@minus,A(:,1),B(:,1).').^2 + bsxfun(@minus,A(:,2),B(:,2).').^2
D =
1 8 2
40 9 45
8 1 13
1 8 2
Pappu Murthy on 23 Jan 2017
thanks to both of you after the corrections I am able to get the right answer and I found this answer to be the most direct and also verified that many times faster than my answer for large size matrices. Thanks again. I will now go ahead and accept the answer.

Andrei Bobrov on 23 Jan 2017
Edited: Andrei Bobrov on 23 Jan 2017
My variants:
For R2016b and later
>> C2 = squeeze(sum((A(:,1:2) - permute(B,[3,2,1])).^2,2))
C2 =
1 8 2
40 9 45
8 1 13
1 8 2
and with bsxfun:
>> C3 = squeeze(sum(bsxfun(@minus,A(:,1:2),permute(B,[3,2,1])).^2,2))
C3 =
1 8 2
40 9 45
8 1 13
1 8 2
Pappu Murthy on 23 Jan 2017
The first solution did not work in 2016a but worked ok in 2016b. Thanks for suggesting the alternative answers, although After testing a lot the original solution and your interpretation with bsxfun appear to be the best from speed point of view. Your first approach was a bit slower. I have tried on huge matrices to establish the speed correctly.