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[2014-10-07. : 3:33 am] sigsaucy -- so you just end up solving multiple problems of the form Ax = b for vector x and b[2014-10-07. : 3:31 am] sigsaucy -- if we have more variables then equations we will have undetermined variables meaning multiple solutions[2014-10-07. : 3:30 am] sigsaucy -- if we have more equations then variables we can only hope for a solution if some of the equations are repeated[2014-10-07. : 3:29 am] sigsaucy -- MasterJohnnyMasterJohnny shouted: Suppose A is nxn and nonsingular and B is nxm how do I efficiently solve AX=B Johnny, i think it will either be non solvable or have infinite solutions[2014-10-07. : 3:22 am]  jjf28 -- http://www.cplusplus.com/reference/cstring/strcpy/ or http://www.cplusplus.com/reference/cstdio/sprintf/[2014-10-07. : 3:01 am]  Dem0n -- If I just do table->buckets[x].data.key = "fuck", it says they are incompatible types when assigning to type char[501] (501 is the max_str_length + 1) from type char[2014-10-07. : 3:00 am] Dem0n -- yo guys, if I have this shit, and I'm trying to access the key and value through a pointer to the hash_table, how can I modify those two strings?[2014-10-07. : 2:48 am] Devourer -- Though, if S2 really tried and wouldn't be so random at their balance decisions, HoN would have had a good chance[2014-10-07. : 2:47 am] Devourer -- Basically, every spell in LoL is limited to like, approximated, 300 pixels at most[2014-10-07. : 2:37 am] MasterJohnny -- Suppose A is nxn and nonsingular and B is nxm how do I efficiently solve AX=B[2014-10-07. : 2:30 am] Dem0n -- l)ark_ssj9kevinl)ark_ssj9kevin shouted: if you haven't made your OWN shortcut keys, you don't truly know all the shortcut keys. I have not. I'm still using shift+insert to paste D:[2014-10-07. : 1:57 am] l)ark_ssj9kevin -- if you haven't made your OWN shortcut keys, you don't truly know all the shortcut keys.[2014-10-07. : 1:36 am] trgk -- https://raw.githubusercontent.com/bwapi/bwapi/master/bwapi/BWAPI/Source/BW/CUnit.h[2014-10-07. : 1:30 am] Roy -- Or, more importantly, how you'd try to manage an array of groups of variables without using structs.[2014-10-07. : 1:08 am] jjf28 -- (was going to say "then think how ugly a function call with that would be" but that's not what you meant by not seeing a diff  )[2014-10-06. : 10:03 pm]  Moose -- Proof by contradiction. Assume x > 0 does not imply (1/x) > 0, then x > 0 must imply (1/x) <= 0. Since x > 0, you can multiply both sides by x without reversing the inequality, so (1/x) * x <= 0 * x. Since x != 0, that becomes 1 <= 0. Contradiction, GG. ∎[2014-10-06. : 9:29 pm] Moose -- ZoanZoan shouted: btw, how do you prove [ x > 0 ] => [ (1/x) > 0 ] Uh, did you try multiplying both sides by x?  |
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