sha256 implementation in javascript

Question

I'm currently trying to implement sha256 from a scratch.

I'm implementing ch() function whose definition is ch(x,y,z) = (x&y) ^ (~x&z)

(&:indicate and gate, ^:indicate xor gate, ~:is negate)

I implemented like this:

function ch(x,y,z){
 (x&y) ^ (~x&z)
}

but after I implemented it, when I saw some other implementation such ashttp://point-at-infinity.org/jssha256/, the implementation is like this (below)

function SHA256_Ch(x, y, z) {
  return z ^ (x & (y ^ z));
}

what is this conversion?

can I obtain the same result from wikipedia's?

Could you tell me the path ?

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Thank you for answeing!

ch() stands for choose: x chooses y or z. When x is 0, z is chosen, and when x is 1, y is chosen

This is a critical sentence for me. I wanted to know how the desiner find second form ch function.

but I guess, if I obtain the view of your advise then, that is obvious.

if a designer of sha256 want to use ch function like it, that makes sense.when x is zero, y is 0, and z is left, because x and z is connected with xor gate,and x and y is connected with and gate.when x is one, y is y and z is nothing, because there are two z connected with xor gate;

I should start my guess from the perspective of the one who structured SHA256 algorythm.

Thanks for teaching!!

kouji

Dear mitch. I'm sorry. Althogh you helped me a lot. I was rude to you, because of my ignorance. I'll accept your graceful answer, as you suggest. Is it worked? Thank you for advises!! — kouji
– kouji, Commented Mar 15, 2016 at 14:19

Mitch Wheat · Accepted Answer · 2016-03-13 23:55:48Z

3

If you write out the truth table for x, y, z and the 2 expressions, you will see that they are identical.

In both cases (I've left out the intermediate results for brevity):

x  y  z  (x&y)^(~x&z)  z^(x&(y^z))
0  0  0  0             0
0  0  1  1             1
0  1  0  0             0  
0  1  1  1             1  
1  0  0  0             0  
1  0  1  0             0  
1  1  0  1             1  
1  1  1  1             1

ch() stands for choose: x chooses y or z. When x is 0, z is chosen, and when x is 1, y is chosen

The second form uses 1 less bitwise operation; but I'd guess that any speed improvement would be implementation specific.

edited Mar 13, 2016 at 23:55

answered Mar 13, 2016 at 6:53

Mitch Wheat

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