| TRWBW | Cale: yeah, that's waht i mean to say |
| Cale | This is the same as doing just one tensor contraction. |
| TRWBW | Cale: sure. Cale: i just wanted to make sure the term made the same thing, clearly it's identical usage. Cale: just like taking some function f(x,y):R^2->R and construction g:R->(R->R) such that (g(x))(y)=f(x,y) |
| Cale | right |
| _llll_ | c(A,C(B,C)) ~ c(BxA,C) :) oops, bad notation |
| Cale | Hom(A, Hom(B,C)) ~ Hom(AxB,C) |
| TRWBW | _llll_: actually i did some research in currying in non-functional languages, basic blocks, registers, pointers, etc. got a paper out of it. ;) |
| _llll_ | you should write Hom(B,C) different to the outer one. i like HOM(B,C) for the internal one or B^C, possibly |
| Cale | C^B |
| _llll_ | that too at some point i decided it should be BxA instead, but i forget why now something to do with bundles |
| Seeb | could anybody tell number of combinations for picking a straight flush poker hand (including ace on top)? I know it's 4*10, but that's not explanatory |
| TRWBW | Seeb: well break it down Seeb: first pick the suit Seeb: well you know one card, the ace Seeb: how many ways can you pick the ret Seeb: multiply Seeb: misread that, but that still holds Seeb: first pick suit, then pick top card |
| Seeb | TRWBW: yeah TRWBW: C(4,1)*... |
| TRWBW | Seeb: only special case you might be confusing is 5,4,3,2,ace Seeb: or ace,k,q,j 10 |
| Seeb | I don't get how you get 4*10 |
| TRWBW | Seeb: make sense? for this problem ace can be both 1 or 14, so it's like you had 14 cards |
| Seeb | (which is the answer) |
| TRWBW | Seeb: how many different top cards can you have? Seeb: that is how many cards can be the top card in a straight? (ignoring suit, let's assume they are all clubs) |
| Seeb | 5 to ace <=> 10 different top cards |
| TRWBW | k, so where you missing it? that is, yes, why aren't you happy with that? |
| Seeb | C(4,1)*10 = 4*10, but uhm, doesn't feel like I picked 5 cards |
| TRWBW | Seeb: you didn't. you picked one card actually |