Generating random numbers in specific binary range

Question

I have a matrix with decimal numbers and I know I can convert them to binary using this code:(IEEE 754 double-precision binary)

m = reshape(dec2bin(typecast(b(:),'uint8'),8).',1,[]);
b_recovered = reshape(typecast(uint8(bin2dec(reshape(m,8,[]).')),'double'),size(b));

Using this code, I think 8 last bits are most significant bits. I want to generate random decimal numbers and replace these elements as 8 last bits won't change after replacing with generated random numbers. I need to have a new number while keeping 8 last bits.

For example:

b=-1.12;
m=1110110001010001101110000001111010000101111010111111000110111111;

be replaced by:

 m=0000110000000000001110000000011010000001100010000010000110111111;

Which is equal to:

b=-1.337678432804527e-04

I know that I can generate random decimal numbers between two numbers but I'm not sure how to solve the above mentioned problem.

mymatrix(1:q,:)= value2 + (value2-value1).*rand(q,size(y_blk,2));

What are you actually trying to achieve here? Converting a number from decimal to binary, changing the first 56 bits and then converting it back to a decimal?? It seems like there'd be a shortcut depending on what you expect as an end result... — Wolfie
– Wolfie, Commented Mar 1, 2017 at 13:57
Generate integers between 0 and 255 (8bits) and convert those to binary. — mpaskov
– mpaskov, Commented Mar 1, 2017 at 14:15
@mpaskov It's not that simple...8 last bits are not equal to 0-255 — Hanna
– Hanna, Commented Mar 1, 2017 at 14:22
Misunderstood the question, you are keeping the last 8bits, and replacing the first 56, which are equal to 0-2^56. Updated the answer. — mpaskov
– mpaskov, Commented Mar 1, 2017 at 14:45
@Hanna is there something else that is missing from the answer? — mpaskov
– mpaskov, Commented Mar 1, 2017 at 16:23

mpaskov · Accepted Answer · 2017-03-01 15:10:43Z

1

Keeping the first 56 replacing the last 8

Generate an integer between 0 and 255 and convert it to binary. Replace the first/last 8 bits of you sequence with them.

b=-1.12;
m = reshape(dec2bin(typecast(b(:),'uint8'),8).',1,[]);
m_small = [dec2bin(randi(256)-1,8), m(9:end)]
m_large = [m(1:end-8),dec2bin(randi(256)-1,8)]
b_small = reshape(typecast(uint8(bin2dec(reshape(m_small,8,[]).')),'double'),size(b));
b_large = reshape(typecast(uint8(bin2dec(reshape(m_large,8,[]).')),'double'),size(b));

The value of b_small, hardly change indicating that the first 8bits are the least significant. b_large change is massive so the last 8 are the most significant bits.

Keeping the last 8 replacing the first 56

Misunderstood the question you want to keep the last 8 bits and replace the rest. In that case generate 56bits random data and add the last 8 that you have saved. Now as it turns out Matlab can only generate a max int of 2^53. Instead of generating one integer up to 2^56, generate 7 each one being 2^8, and concatenate them.

m_new = [reshape(dec2bin(randi(2^8,[7,1])-1,8),[1,56]), m(end-7:end)]
b_new = reshape(typecast(uint8(bin2dec(reshape(m_new,8,[]).')),'double'),size(b));

edited Mar 1, 2017 at 15:10

answered Mar 1, 2017 at 14:28

mpaskov

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2 Comments

Hanna Over a year ago

Can I keep just bit 57 and 58?

mpaskov Over a year ago

Yes, you would save m(end-7:end-5) and add dec2bin(randi(2^6)-1,6) to the end.

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