Eddie
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 Jan4 accepted How does ECDH arrive on a shared secret? Jan4 comment How does ECDH arrive on a shared secret? To avoid continuing the back and forth in Comments, I created a Chat Room and have posted a few more questions there. Jan4 comment How does ECDH arrive on a shared secret? I see... ok, just to again make sure I understand.. and thanks to @StephenTouset's nugget of wisdom (thanks Stephen!). When Alice is dotting $g$ to itself $a$ times, Alice isn't really doing the "dot" operation $a$ times, but is somehow using a magical math formula that gets Alice $A$ much quicker than actually doing the "dot" $a$ times. So the final $S$ from Alice's perspective is $g$ dot $g$ $b$ times to get $B$ (supplied by Bob), then $B$ dot $B$ $a$ times. And from Bob's: $A$ dot $A$ $b$ times. Which ends up at the same point in the curve... the shared secret. Jan4 comment How does ECDH arrive on a shared secret? Let me just confirm I understand you...Everyone knows the starting point ($g$), everyone knows where Alice ended up ($A$), everyone knows where Bob ended up ($B$), and everyone knows the actual elliptic curve. But only Alice knows how many iterations of "dot" she did ($a$), and only Bob knows how many iterations he did ($b$). And once they know each other's $A$ and $B$, they then continue the other's "dot" operation. AKA, Alice takes $g$ and $B$ and "dots" it $a$ more times, and Bob takes $g$ and $A$ and "dots" it $b$ more times, and they both land at $S$, the shared secret... is that it? Jan4 revised How does ECDH arrive on a shared secret? added two