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2012-07-22T07:33:30 *** Scooper has joined #aichallenge 2012-07-22T08:03:18 *** kilae has joined #aichallenge 2012-07-22T08:08:17 *** Kingpin13 has quit IRC (Ping timeout: 264 seconds) 2012-07-22T08:16:22 *** foRei has joined #aichallenge 2012-07-22T08:26:28 *** smj has joined #aichallenge 2012-07-22T08:26:29 *** smj has joined #aichallenge 2012-07-22T08:28:40 *** choas has quit IRC (Ping timeout: 264 seconds) 2012-07-22T09:00:30 *** Kingpin13 has joined #aichallenge 2012-07-22T09:46:47 mcstar: http://mathbin.net/102895 that's what i've spent tonight doing :| 2012-07-22T09:46:51 http://sketchia.com/draw.html#sSk9hBG 2012-07-22T09:46:52 that's why 2012-07-22T09:47:22 basically, if i have n nodes connected from above and i connect 3 (not already connected) nodes on the bottom at a time, how many new nodes can be reached? 2012-07-22T09:48:11 any ideas on a less messy way to show that? :P 2012-07-22T09:51:34 i think i messed up your drawing 2012-07-22T09:51:45 fixed 2012-07-22T09:51:45 lol 2012-07-22T09:51:50 haha 2012-07-22T09:52:45 it took me a while to come up with an argument for that 2012-07-22T09:53:02 i cant use that drawing board 2012-07-22T09:53:07 whats the erasr? 2012-07-22T09:53:38 thick white line tooll 2012-07-22T09:55:23 lol 2012-07-22T09:55:44 im not sure i wold have figured that out 2012-07-22T09:56:05 well, i started with small numbers to find the pattern 2012-07-22T09:56:15 then had to work out how to argue my pattern works inductively 2012-07-22T09:56:20 and voila that mess came out 2012-07-22T09:58:39 so 2012-07-22T09:58:46 what does this do? 2012-07-22T09:58:57 im not sure about the pattern 2012-07-22T09:59:04 whats the rule? 2012-07-22T10:03:23 antimatroid: can you state the problem so that i can understand? 2012-07-22T10:03:33 with the dots and links 2012-07-22T10:03:42 whats the rule? 2012-07-22T10:03:51 you have n nodes connected at the top 2012-07-22T10:03:59 then can connect 3 nodes at a time from the bottom 2012-07-22T10:04:11 and at least one node connected at the bottom has to be connected at the top 2012-07-22T10:04:23 what possible numbers of nodes can you connect to which aren't connected to from above? 2012-07-22T10:05:50 ok, so i can connect by 3, any number of times 2012-07-22T10:06:09 while still linking to 1 that is connected from the top 2012-07-22T10:06:16 and the top connection doesnt change 2012-07-22T10:06:48 yep 2012-07-22T10:06:56 and you can't connect to the same node twice from the bottom 2012-07-22T10:07:39 oh 2012-07-22T10:28:32 antimatroid: so, is the question, then, the number supremum of the number of possible new nodes for each n? 2012-07-22T10:28:37 -number 2012-07-22T10:28:52 2,4,5,8 2012-07-22T10:28:58 2,4,6,8 2012-07-22T10:28:59 what? 2012-07-22T10:29:19 the numbers, you draw after each row 2012-07-22T10:29:33 {2},{1,4},{0,3,6} 2012-07-22T10:29:45 the supremums are 2,4,6 2012-07-22T10:30:00 not important 2012-07-22T10:30:03 i just want those sets 2012-07-22T10:30:10 so you want the sets 2012-07-22T10:30:12 ie. how many new nodes can conceivably be reached 2012-07-22T10:30:13 yep 2012-07-22T10:30:21 otherwise 2n is trivial 2012-07-22T10:30:34 wait 2012-07-22T10:30:42 you want the sets itself, or the cardinality? 2012-07-22T10:30:47 the sets 2012-07-22T10:31:05 cause 'how many new nodes can conceivably be reached" thats different again 2012-07-22T10:31:13 sup of each set is 2n was my point 2012-07-22T10:31:25 and that's obvious because you can just attack each node to two new nodes 2012-07-22T10:31:31 attach* 2012-07-22T10:31:50 ofc 2012-07-22T10:32:13 so, you are looking for some recursive algorithm? 2012-07-22T10:32:35 i just wanted to prove that it's as i said in the prop statement 2012-07-22T10:32:50 ie. if 3|n then you can connect to {0, 3, ..., 2n} new nodes 2012-07-22T10:51:32 *** amstan has joined #aichallenge 2012-07-22T10:51:32 *** ChanServ sets mode: +o amstan 2012-07-22T11:05:31 antimatroid: take any integer, write it as all the possible sums of 1,2,3; replace 1->2,2->1,3->0, do the sum, take the union 2012-07-22T11:08:05 wut? 2012-07-22T11:08:11 i'm about to climb into bed 2012-07-22T11:08:19 thats one way to generate the set 2012-07-22T11:09:01 4 = 4,3+1,2+1+1,1+1+1+1 2012-07-22T11:09:10 damn 2012-07-22T11:09:13 not 4 2012-07-22T11:09:15 but 2+2 2012-07-22T11:09:17 yeah, you can do 1a2 + 2a3 st 3a1 + 2a2 + a3 = n but that's not really that helpful in proving it generally 2012-07-22T11:09:21 4 = 2+2,3+1,2+1+1,1+1+1+1 2012-07-22T11:09:30 i want a result i can use, not a way to find the set 2012-07-22T11:09:35 ie. classify the sets 2012-07-22T11:09:39 which i've done anyway i think 2012-07-22T11:10:47 errr {a2 + a3: 3a1 + 2a2 + a3 = n} is the set for each n where ai's are natural numbers 2012-07-22T11:14:07 antimatroid: i think your proof by induction is good, but the notation sucks 2012-07-22T11:14:34 you want to express thoses sets more succintly i think 2012-07-22T11:21:51 antimatroid: still there? 2012-07-22T11:24:40 *** amstan has quit IRC (Quit: Konversation terminated!) 2012-07-22T11:36:42 antimatroid: anyway,: f(1) = {2}, f(n) = f(n-1) U {n%3==0 -> 0, n%3==1 -> 2, n%3==2 -> 1} 2012-07-22T11:39:07 still better: 2012-07-22T11:39:22 f(1) = {2}, f(n) = f(n-1) U { (2*n) % 3 } 2012-07-22T11:40:03 antimatroid: this last statement can be easily proven i believe 2012-07-22T11:40:22 I started to ride my bicycle every day 2012-07-22T11:40:32 pairofdice: Iron Man 2012-07-22T11:40:44 to work? 2012-07-22T11:40:56 For fun and profit, I have no job 2012-07-22T11:41:03 o.O 2012-07-22T11:41:06 O.o 2012-07-22T11:41:23 For fun and sports, more like 2012-07-22T11:41:27 how do you make profit from riding a bike? 2012-07-22T11:41:41 It's health profit 2012-07-22T11:41:43 Or something 2012-07-22T11:41:49 i see 2012-07-22T11:42:12 how come you dont have a job? did you inherit a fortune? 2012-07-22T11:43:31 ...I just don't, I 'quit' the last one like 5 years ago and never recovered 2012-07-22T11:43:56 depression, or whatever, I don't know 2012-07-22T11:44:32 i hope ill get one soon 2012-07-22T11:46:36 Yeah, me too 2012-07-22T12:07:38 *** sigh has quit IRC (Remote host closed the connection) 2012-07-22T14:06:34 *** choas has joined #aichallenge 2012-07-22T14:57:11 *** HaraKir has joined #aichallenge 2012-07-22T14:59:24 *** HaraKiri has quit IRC (Ping timeout: 248 seconds) 2012-07-22T15:02:23 *** iglo has joined #aichallenge 2012-07-22T15:03:44 *** HaraKir is now known as HaraKiri 2012-07-22T15:39:45 *** amstan has joined #aichallenge 2012-07-22T15:39:45 *** ChanServ sets mode: +o amstan 2012-07-22T15:58:37 *** Accoun has quit IRC () 2012-07-22T16:02:01 *** smj has quit IRC (Ping timeout: 248 seconds) 2012-07-22T16:11:37 *** smj has joined #aichallenge 2012-07-22T16:11:38 *** smj has joined #aichallenge 2012-07-22T16:22:14 *** Accoun has joined #aichallenge 2012-07-22T16:39:21 *** kilae has quit IRC (Quit: ChatZilla 0.9.88.2 [Firefox 14.0.1/20120713134347]) 2012-07-22T16:44:45 *** amstan has quit IRC (Quit: 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