Transforming two matrices into one matrix using binary numbers

Question

A1 and A2 are two arrays of integers with the same dimensions 6000x2000.

I want to find a third matrix B by following these steps:

for i=1:1:6000
    for j=1:2000

        a1 = fliplr(dec2binvec(A1(i,j),14));  % Convert A1(i,j) to a binary vector
        a2 = fliplr(dec2binvec(A2(i,j),32));  % Convert A2(i,j) to a binary vector

        b = [a1 a2];

        B(i,j) = sum(b.*2.^(numel(b)-1:-1:0));  % Convert b to decimal


    end
end

my problem is the computation time to find B.

Is there a way to avoid loops in order to reduce the computation time ?

Example:

A1 = [2 3           A2 = [7 6
      4 5]                2 9]

A1(1,1) = 2 and A2(1,1) = 7

a1 = [0 0 0 1 0] (eg 5 bits) a2 = [0 0 0 1 1 1] (eg 6 bits)

b = [a1 a2] = [0 0 0 1 0 0 0 0 1 1 1]

B1(1,1) = sum(b.*2.^(numel(b)-1:-1:0)) = 135

Please post a small numeric example (say with 2x2 matrices), to clarify endianness etc — Luis Mendo, Feb 19 '15 at 22:21
`dec2binvec` returns least significant bit first, but `b.*2.^(numel(b)-1:-1:0)` uses it as highest significant bit first. Did you verify you really get what you want with this code? — Daniel, Feb 19 '15 at 22:30
@bzak: The example and your code don't match, your code flips the bits. — Daniel, Feb 19 '15 at 22:36
@Daniel: is there a function that returns highest significant bit first? — bzak, Feb 19 '15 at 22:42
@bzak: Matlab always uses lowest significant bit first for logical vectors and highest significant bit first for string representations. If you need something else, use `fliplr(dec2binvec(2,6))`. — Daniel, Feb 19 '15 at 22:46

score 4 · Answer 1 · answered Feb 19 '15 at 22:39

4

Using your example:

A1 = [2 3;           
      4 5];            
A2 = [7 6;
      2 9];
B=bitshift(A1,6)+A2

Output:

answered Feb 19 '15 at 22:39

beaker

16,331
3
32
49

Plus 1 for the `bitshift`. I love that function. – rayryeng Feb 20 '15 at 00:27
@rayryeng Thanks, I think for someone not used to working in binary it clarifies what multiplying by a power of 2 does to a number. – beaker Feb 20 '15 at 01:06
Yup it certainly does! – rayryeng Feb 20 '15 at 01:08

Luis Mendo · Answer 2 · 2015-02-19T22:59:35.297

1

If I understand your example correctly, you only need to bit-shift A1 (i.e. multiply by a power of 2):

M = 5; %// not used actually
N = 6;
B = A1 * 2^N + A2;

In your example, this gives

B =
   135   198
   258   329

edited Feb 19 '15 at 22:59

answered Feb 19 '15 at 22:38

Luis Mendo

110,752
13
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Daniel · Accepted Answer · 2015-03-02T15:55:23.963

1

I assume the example contains what you want. Simply use math ;)

A1 = [2 3;4,5]
A2=[7 6;2 9]
A1.*2^6+A2

Please be aware that doubles can hold up to 53 bits without precision loss. Recent versions of matlab support uint64. For even longer numbers check vpa, but vpa will result in slow code.

edited Mar 02 '15 at 15:55

answered Feb 19 '15 at 22:39

Daniel

36,610
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1

http://stackoverflow.com/questions/1848700/biggest-integer-that-can-be-stored-in-a-double – bzak Mar 02 '15 at 15:44

Transforming two matrices into one matrix using binary numbers

3 Answers3