[bazzargh]

The typo is most likely the extra /, in (x/y)/(x^2+y^2) instead of (xy)/(x^2+y^2).

`2^x & 2^y ...is the & a bitwise operator...???? That would produce a unique ID? That would be very interesting, is that provable?`

Yes, & is bitwise and. It's just treating your players as a bit vector. It's not so much provable as a tautology, it is exactly the property that players x and y are present. It's not _useful_ tho because the field size you'd need to hold the bit vector is enormous.

As for the problem...it sounds bloom-filter adjacent (a bloom filter of players in a hand would give a single id with a low probability of collision for a set of players; you'd use this to accelerate exact checks), but also like an indexed many-to-many table might have done the job, but all depends on what the actual queries you needed to run were, I'm just idly speculating.

[noduerme]

At the time, at least, there was no way to index it for all 8 players involved in a hand. Each action taken would be indexed to the player that took it, and I'd need to sweep up adjacent actions for other players in each hand, but only the players who were consistently in lots of hands with that player. I've heard of bloom filters (now, not in 2012)... makes some sense. But the idea was to find some vector that made any set of players unique when running through a linear table, regardless of the order they presented in.

To that extent, I submit my solution as possibly being the best one.

I'm still a bit perplexed by why you say 2^x & 2^y is tautologically sound as a unique way to map f(x,y)==f(y,x), where x and y are nonequal integers. Throwing in the bitwise & makes it seem less safe to me. Why is that provably never replicable between any two pairs of integers?

[bazzargh]

I'm saying it's a tautology because it's just a binary representation of the set. Suppose we have 8 players, with x and y being 2 and 4: set the 2nd and 4th bits (ie 2^2 & 2^4) and you have 00001010.

But to lay it out: every positive integer is a sum of powers of 2. (this is obvious, since every number is a sum of 1s, ie 2^0). But also every number is a sum of _distinct_ powers of 2: if there are 2 identical powers 2^a+2^a in the sum, then they are replaced by 2^(a+1), this happens recursively until there are no more duplicated powers of 2.

It remains to show that each number has a unique binary representation, ie that there are no two numbers x=2^x1+2^x2+... and y=2^y1+2^y2+... that have the same sum, x=y, but from different powers. Suppose we have a smallest such number, and x1 y1 are the largest powers in each set. Then x1 != y1 because then we can subtract it from both numbers and get an _even smaller_ number that has distinct representations, a contradiction. Then either x1 < y1 or y1 < x1. Suppose without loss of generality that it's the first (we can just swap labels). then x<=2^(x1+1)-1 (just summing all powers of 2 from 1..x1) but y>=2^y1>=2^(x1+1)>x, a contradiction.

or, tl;dr just dealing with the case of 2 powers: we want to disprove that there exists a,b,c,d such that

2^a + 2^b = 2^c + 2^d, a>b, c>d, and (a,b) != (c,d).

Suppose a = c, then subtract 2^a from both sides and we have 2^b = 2^d, so b=d, a contradiction.

Suppose a>c; then a >= c+1.

2^c + 2^d < 2^c + 2^c = 2^(c+1).

so

2^c + 2^d <= 2^(c+1) - 1 < 2^(c+1) + 2^b <= 2^a + 2^b

a contradiction.