From: rab rub <lists@...>
Date: 2012-04-12T20:14:05+09:00
Subject: Re: Ruby openssl ECC help plz

Yes, what I am implementing requires me to treat the shared secret as an 
integer. I think I might be kind of close to figuring it out:

q = OpenSSL::PKey::EC.new("secp128r2")
q.generate_key

pub = OpenSSL::PKey::EC.new(q.group)
pub.public_key = q.public_key

a = pub.public_key.to_bn

ss = q.dh_compute_key(pub.public_key)

puts ss.unpack('I>*')


this puts a series of numbers such as the following:

2755907301
4161034086
221727499
1603331514

or

2041301904
2634281936
2269031273
1943826378

Just to see if this works, I have then been combining these into big 
integers, so far just by hand. My theory is that if the resulting 
integer is the same bit size as the curve, that it is an indication I 
may not be doing it incorrectly. And I am so close to almost not 
certainly doing it wrong, because sometimes it does make output that 
equals 128 bits when I convert it to binary, for example


275590730141610340862217274991603331514

but alas sometimes it is slightly more or less than 128 bits, as with


2041301904263428193622690312731943826378

which is 131 bits

regardless of the curve I use the output is either the correct size or 
near it, but I don't understand the variation.

ss.length always returns a number of bytes that when multiplied by 8 is 
the size of the curve, assuming that the curve consists of a number of 
bits that is divisible by 8 anyway.

I could be way off on this and doing it totally wrong, but since I hit a 
total dead end I figured that I might as well try doing things until I 
at least get the correct size integer and then try to confirm it as 
being correct further from there, but I can just get close.

thanks :).

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