One explicit vote, one implicit vote for updating/clarifying the slides. I've created PIRK-69 to improve slide clarity.

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Unless this doesn't make sense (tell me) I'll mark PRs on this as WIPs until I've got some agreement from the community that the slides are clear enough. On Mon, Sep 19, 2016 at 3:31 PM, Ryan Carr <jryanc...@gmail.com> wrote: > Hey Walter / Tim, > > I just wanted to add I had some trouble similar to Tim's when trying to > understand the Wideskies paper. As a person without a background in group > theory/theoretical math trying to get my head around this stuff, it was > very difficult for me to even find with Google what the notations (Z/NZ) > and (Z/N^2 Z)^x (also called (Z/N^2 Z)* in the Wideskies/Paillier papers) > meant. Since these concepts are so central to how the algorithm works, I > think it would be really helpful if we had a footnote the first time those > are introduced defining that notation with links to a more in-depth > explanation, or at least a phrase that can be Googled to reliably find it. > > Thanks, > -Ryan Carr > > On Mon, Sep 19, 2016 at 1:50 PM Walter Ray-Dulany <raydul...@apache.org> > wrote: > > > Correction: > > > > ...bby the binomial theorem, (1+N)**N = 1 + N*N + other terms > divisible... > > > > I multiplied by N on the left when I ought to have exponentiated > > > > Walter > > > > On Mon, Sep 19, 2016 at 1:36 PM, Walter Ray-Dulany <raydul...@apache.org > > > > wrote: > > > > > Hi Tim, > > > > > > Apologies! It's disorienting at first, and most of all when one > actually > > > tries to sit down and do a real example. The version on the slides was > > not > > > written in one go, I assure you. > > > > > > Let's go through, and see what's not working. > > > > > > ************************** > > > > > > > I'm trying a very simple example. I'm going to choose, p = 3, q = 5 > > and > > > a message m = 42 > > > > > > Already we're in trouble. p and q are fine; but remember that the > > > plaintext space (let's call it P(N)) is the set of all integers in > Z/NZ; > > > that is, it is all numbers m > > > > > > 0 <= m < N > > > > > > You can see already that the m you chose is not in the plaintext space. > > > > > > Let's pick a new m to continue with; in this case, let's choose your m, > > > but mod 15 so that it lies in P(N). Thus, our new m going forward shall > > be > > > > > > m = 12 > > > > > > ************************** > > > > > > > I'm going to pick g = 240. I think it needs to be a multiple of N > that > > > is greater than N*N, correct? > > > > > > No, and this is important. g has to be an element of (Z/(N squared )Z)* > > of > > > order a nonzero multiple of N. That sentence is meaningless unless > you're > > > already embedded in the mathematics, so let's go through what it means, > > bit > > > by bit. > > > > > > g must be: > > > 1. *an element of (Z/(N squared)Z)**: everything but the outer * on the > > > right just means that 0 <= g < N*N; in this case that means 0 <= g < > 225. > > > The outer * on the right indicates that we only want to take a certain > > > special kind of g: one that is what we call a *unit* mod N*N; that is, > it > > > means that we require that there exist another element 0<= h < N*N such > > > that g*h = 1 mod N*N. In our current situation, N = p*q is a product of > > > primes, and so N*N = p**2 * q**2, and we can easily characterize G = > > (Z/(N > > > squared)Z)*: G = { 0<= g < N*N such that neither p nor q divide g}. So > as > > > long as we pick a g that does not have p or q as a factor, we're good > for > > > this condition (this also includes 0, so really all of my "0 <=" in > this > > > paragraph could have been "0 < "). Another way to characterize G is to > > say > > > that it is the set of integers less than N*N that are relatively prime > to > > > N*N. > > > > > > 2. *of order a nonzero multiple of N*: this is a little trickier. The > > > *order* of an element g of a finite group (which G is) is the least > > > integer k such that g^k = 1 in G. I'm not going to prove it here, but > it > > > turns out that every element of G has finite order (that is, if g is in > > G, > > > then there exists a finite non-zero k such that g^k = 1), and that it > is > > > less than or equal to the Carmichael number lambda(N*N). That takes > care > > of > > > what 'order' means, and, like I said, order is defined for all g in G. > > But! > > > We require a special order. Specifically, we only want g in G such that > > the > > > order of g is a non-zero multiple of N. We might ask whether we know > that > > > such always exists (a good question, since we require it), and we do! > > > Here's a quick proof of existence, one tied closely to Wideskies: > > > > > > * Take g = 1 + N (I'm going to prove, all at once, that 1+N is in G and > > > that it has an order that fits the bill). > > > * Consider g**N: by the binomial theorem, (1+N)*N = 1 + N*N + other > terms > > > divisible by N*N. This number is equivalent to 1 mod N*N. QED > > > > > > Ok, great, such g exist, and so we can require that we use one of them. > > > But you must be careful: you can't just choose any g in G off the > street > > > and expect it will satisfy our requirements. You chose g = 240, which > (1) > > > bigger than N*N, which isn't what we want, and (2) is divisible by N, > and > > > so even if we take 240 mod N*N, we still aren't in G, much less of the > > > 'right order' (turns out 240, being not relatively prime to N, can > never > > be > > > exponentiated to 1 mod N*N). For now, let's just take the standard > > > Wideskies g, g = 1 + N = 16. If you want to go through this with a > > > different g, give it a shot, but make sure it's got the right kind of > > order. > > > > > > ************************** > > > > > > > I'll pick zeta = 21. I think it needs to be greater than N. > > > > > > As in point 2, no. We require zeta to be in (Z/NZ)*, which, similar to > > the > > > above, means a number > > > > > > 0 < zeta < N such that zeta is a unit. > > > > > > You picked 21; if we take 21 mod N we get zeta = 6, which is not a unit > > > (in particular it is not relatively prime to p=3). Let's pick the next > > > number greater than 6 which is in (Z/NZ)*, which is > > > > > > zeta = 7. > > > > > > ************************** > > > > > > Let's see what we've got. > > > > > > ( (16**12)*(7**15) ) mod 225 = 208. > > > > > > I will leave it as an exercise to check that the decryption of 208 is > in > > > fact 12. > > > > > > ************************** > > > > > > Ok, that's all so far. If the above is still not computing (literally > or > > > metaphorically), I am available to converse one-on-one either over the > > > phone or some other medium (face time or what have you) and only too > > happy > > > to go over this more. > > > > > > Let me know, > > > > > > Walter > > > > > > > > > > > > On Mon, Sep 19, 2016 at 12:13 PM, Tim Ellison <t.p.elli...@gmail.com> > > > wrote: > > > > > >> On 17/09/16 06:42, wraydulany wrote: > > >> > [Pirk 67] - Add Slide Deck to the Website Documentation > Explaining > > >> the Mathematics of Pirk > > >> > > >> Argh! Walter you are driving me mad! :-) > > >> > > >> Thank you for putting up the slides, and I really want to grok this, > but > > >> I'm struggling to work through even the simplest of examples. > > >> I need the kindergarten version :-( > > >> > > >> What am I doing wrong? I'm trying a very simple example. I'm going > to > > >> choose, > > >> p = 3, q = 5 and a message m = 42 > > >> > > >> Looking at page 11 on the math_deck.pdf line by line > > >> > > >> line 2: > > >> N = 15. > > >> I'm going to pick g = 240. I think it needs to be a multiple of N > > >> that is greater than N*N, correct? > > >> > > >> line 3: > > >> I'll pick zeta = 21. I think it needs to be greater than N. > > >> > > >> Line 4: > > >> (240^42 x 21^15) mod 225 = 0 > > >> > > >> That's suspicious. > > >> ...and if I play with my free choices of m, g and zeta I always get > > zero. > > >> > > >> Help! > > >> > > >> I get the gist of the flow through the deck, but I want to work > through > > >> my own examples from top to bottom. > > >> > > >> Regards, > > >> Tim > > >> > > > > > > > > >