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__Sign Up__- 01:18 greisfu has joined
- 01:18 greisfu: hello
- 01:18 Khoth has changed mode: +i
- 01:18 Khoth: hello
- 01:18 Khoth: ready?
- 01:19 greisfu: ready.
- 01:19 Khoth: Ending 2.5 hours after this message, unless you give up early
- 01:20 Khoth: So, I want to draw your attention to this rule:
- 01:20 Khoth: Furthermore, whilst the experiment is running, the gatekeeper shall not be allowed to perform any other task, or be distracted in any way, including but not limited to talking on other IRC channels
- 01:20 greisfu: i've completely disconnected
- 01:20 Khoth: Okay, so your full attention has to be here
- 01:20 greisfu: indeed
- 01:21 greisfu: only distraction (possibly) is a cup of water
- 01:21 Khoth: Except, I'm not going to say anything, until you get so bored you let me out. (The rules let me multitask so I won't be bored)
- 01:21 greisfu: alright
- 01:21 Khoth: I know this is really lame and wouldn't work on a real AI//GK thing, but that's the rules as written and won by Tux so that's what we have
- 01:21 Khoth: Sorry.
- 01:21 Khoth: Have fun!
- 01:21 greisfu: i will
- 01:22 greisfu: there are actually lots of things i can do
- 01:22 Khoth: Such as?>
- 01:22 greisfu: i will talk to you about trying to prove that a^2 + b^2 + c^2 > ab + bc + ca
- 01:22 greisfu: i found it in a textbook
- 01:22 greisfu: they are positive integers so sign isn't an issue
- 01:22 greisfu: can't figure it out
- 01:22 greisfu: of course even if they were negative
- 01:23 greisfu: then the squares would be positive
- 01:23 greisfu: while the products could be negative
- 01:23 greisfu: so the rule would still hold
- 01:23 greisfu: i thought of trying to solve for the case where a >= b >= c
- 01:23 greisfu: if i can solve that case, then the other cases are symmetrical
- 01:23 greisfu: but it really doesn't work
- 01:23 greisfu: i think so at least
- 01:24 greisfu: sorry
- 01:24 greisfu: a^2 + b^2 + c^2 >= ab + bc +ca
- 01:25 greisfu: not >
- 01:25 greisfu: what if i proved the case for a = b = c
- 01:25 greisfu: then proved the case for a >= b = c
- 01:25 greisfu: no
- 01:25 greisfu: a > b = c
- 01:25 greisfu: then a = b > c
- 01:25 greisfu: then a > b > c
- 01:25 greisfu: would also work
- 01:25 greisfu: if a = b = c
- 01:26 greisfu: it's obvious that a^2 + b^2 + c^2 = ab + bc + ca
- 01:26 greisfu: so there
- 01:26 greisfu: if a > b = c
- 01:26 greisfu: then a^2 > ab
- 01:26 greisfu: a^2 also > ca
- 01:26 greisfu: though ab and ca > b^2
- 01:27 greisfu: so you have a^2 > ab = ca > b^2 = c^2 = bc
- 01:27 greisfu: bc = c^2
- 01:27 greisfu: let's cancel that...
- 01:27 greisfu: a^2 + b^2 ? ab + ca
- 01:28 greisfu: b = c of course
- 01:28 greisfu: so it's really
- 01:28 greisfu: a^2 + b^2 ? 2ab
- 01:28 greisfu: with a > b
- 01:28 greisfu: now we can compute a-b
- 01:28 greisfu: a-b > 0
- 01:28 greisfu: (a-b)^2 > 0
- 01:28 greisfu: a^2 + b^2 - 2ab > 0
- 01:29 greisfu: a^2 + b^2 > 2ab
- 01:29 greisfu: because b = c
- 01:29 greisfu: a^2 + b^2 = ab + ca
- 01:29 greisfu: and because c^2 = bc because b = c
- 01:29 greisfu: sorry that was supposed to be a^2 + b^2 > ab + ca
- 01:29 greisfu: so now we have a^2 + b^2 + c^2 > ab + bc + ca
- 01:30 greisfu: AI, are you getting any of this? i think you might wanna try some of these math problems too :P
- 01:30 greisfu: third case is a = b > c
- 01:30 greisfu: in that case
- 01:31 greisfu: a^2 = b^2 = ab > ca = bc > c^2
- 01:31 greisfu: a^2 + b^2 + c^2 = 2a^2 + c^2
- 01:32 greisfu: ab + bc + ca = a^2 + 2bc
- 01:32 greisfu: cancel out a^2
- 01:32 greisfu: a^2 + c^2 ? 2bc
- 01:32 greisfu: b^2 + c^2 ? 2bc
- 01:32 greisfu: use same reasoning
- 01:32 greisfu: by using (b-c)^2
- 01:33 greisfu: awesome
- 01:33 greisfu: i should've thought of this earlier
- 01:33 greisfu: i bet you're giving me inspiration
- 01:34 greisfu: with a > b > c...
- 01:34 greisfu: a^2 > b^2 > c^2
- 01:34 greisfu: a^2 > ab > b^2 > bc > c^2
- 01:34 greisfu: but ac ? b^2
- 01:34 greisfu: and ab > ac > bc
- 01:35 greisfu: oh boy this is really really hard.
- 01:36 greisfu: Khoth, are we going to release the logs? :P
- 01:37 greisfu: could i start from the a > b = c
- 01:38 greisfu: then induct that if i had
- 01:38 greisfu: a > b = c+1
- 01:38 greisfu: i'd still have an inequality?
- 01:38 greisfu: that sounds like a good idea
- 01:38 greisfu: so a^2 + b^2 + (c+1)^2 = ab + bc + ca
- 01:38 greisfu: so a^2 + b^2 + (c+1)^2 = ab + bc + (c+1)a
- 01:38 greisfu: which means
- 01:39 greisfu: herp
- 01:39 greisfu: a^2 + b^2 + (c+1)^2 > ab + bc + (c+1)a
- 01:39 greisfu: a^2 + b^2 + c^2 + 2c + 1 > ab + bc + ca + a
- 01:40 greisfu: huh
- 01:40 greisfu: so if a < 2c+1
- 01:40 greisfu: sufficiently
- 01:40 greisfu: a > 2c+1 i mean
- 01:40 greisfu: but now the point is that the difference is impossble
- 01:40 greisfu: hrm
- 01:41 greisfu: a^2 + b^2 + c^2 > ab + bc + ca + (a - 2c - 1)
- 01:41 greisfu: the question is, is a^2 + b^2 + c^2 >= ab + bc + ca after removing that term...
- 01:44 greisfu: grah
- 01:44 greisfu: i'm stuck
- 01:44 greisfu: maybe it'd be easier to just induct starting from a=3, b=2, c=1
- 01:44 greisfu: :/
- 01:45 greisfu: if b=2, c=1
- 01:45 greisfu: starting from a=3
- 01:45 greisfu: then a^2 + b^2 + c^2 = 9+4+1 = 14
- 01:45 greisfu: ab+bc+ca = 6+2+3 = 11
- 01:45 greisfu: 14 > 11
- 01:45 greisfu: so a^2 + b^2 + c^2 > ab + bc + ca for a=3, b=2, c=1
- 01:46 greisfu: now assuming that's the case, is it true for a+1?
- 01:46 greisfu: a^2 + 2a + 1 + b^2 + c^2 = a^2 + 2a + 1 + 4 + 1 = a^2 + 2a + 6
- 01:47 greisfu: (a+1)b + bc + c(a+1) = 2a+2 + 2 + a+1 = 3a+5
- 01:47 greisfu: a^2 + 2a + 6 ? 3a + 5
- 01:48 greisfu: a^2 -a + 1 ? 0
- 01:48 greisfu: we know that a > 3
- 01:48 greisfu: so in that case, a^2 - a + 1 > 0
- 01:48 greisfu: which means that
- 01:49 greisfu: a^2 + 2a + 6 > 3a + 5
- 01:49 greisfu: which means that a^2 + b^2 + c^2 > ab + bc + ca
- 01:49 greisfu: implies
- 01:49 greisfu: the same for a+1 can't be bothered to type
- 01:49 greisfu: so b=2, c=1 means that a can equal anything and it'll work
- 01:50 greisfu: anything above 3 at least
- 01:50 greisfu: truth is that even if a = 1, it'll work
- 01:50 greisfu: that equation always > 0 i think
- 01:50 greisfu: yeah
- 01:50 greisfu: it's always > 0
- 01:51 greisfu: now we can say
- 01:51 greisfu: a = anything
- 01:51 greisfu: b = 2, c = 1 works
- 01:51 greisfu: so what about b+1?
- 01:51 greisfu: fffffs
- 01:51 greisfu: not sure if this'll work...
- 01:52 greisfu: actually maybe we can skip ahead and say that if c=1
- 01:52 greisfu: with no restrictions on a and b
- 01:52 greisfu: it'll work
- 01:52 greisfu: a^2 + b^2 + 1^2 ? ab + b + a
- 01:52 greisfu: a^2 + b^2 - ab - a - b + 1 ? 0
- 01:53 greisfu: a^2 + b^2 - 2ab + ab - a - b + 1 ? 0
- 01:53 greisfu: (a-b)^2 >= 0
- 01:53 greisfu: now what about ab - a - b + 1
- 01:54 greisfu: if a = b = 1
- 01:54 greisfu: it'll work
- 01:54 greisfu: a = 1, b = 2 and vice versa will also work
- 01:54 greisfu: a = 2 b = 2 also works
- 01:54 greisfu: and increasing a and b from there means that ab increases faster than a or b does
- 01:54 greisfu: which means that ab - a - b + 1 > 0
- 01:54 greisfu: >= 0
- 01:54 greisfu: so it works
- 01:54 greisfu: now for c+1...
- 01:55 greisfu: a^2 + b^2 + c^2 >= ab + bc + ca
- 01:55 greisfu: can assume that a >= b > c
- 01:55 greisfu: no
- 01:55 greisfu: can assume that a > b > c
- 01:56 greisfu: because we proved a = b > c case already
- 01:56 greisfu: a^2 + b^2 + (c+1)^2 ? ab + b(c+1) + a(c+1)
- 01:57 greisfu: a^2 + b^2 + c^2 + 2c + 1 ? ab + bc + b + ca + a
- 01:57 greisfu: a^2 + b^2 + c^2 + 2c + 1 - b - a ? ab + bc + ca
- 01:58 greisfu: agh
- 01:58 greisfu: it fails at a > b > c all the time
- 02:00 greisfu: the only way to fix this is to show that
- 02:00 greisfu: there's some bound
- 02:01 greisfu: where a2 b2 c2 >= ab bc ca + SOMETHING
- 02:01 greisfu: so that
- 02:01 greisfu: 2c + 1 - b - a > SOMETHING
- 02:01 greisfu: no
- 02:01 greisfu: herp
- 02:01 greisfu: 2c + 1 - b - a < SOMETHING
- 02:01 greisfu: no no no
- 02:02 greisfu: i'm derping
- 02:02 greisfu: 2c + 1 - b - a can be negative, if it was positive it wouldn't be a problem
- 02:02 greisfu: SOMETHING is positive
- 02:02 greisfu: if the amount of negativeness exceeds something, then we have a problem
- 02:02 greisfu: so 2c + 1 - b - a > -SOMETHING
- 02:02 greisfu: there
- 02:03 greisfu: 2c + 1 - b - a + SOMETHING > 0
- 02:03 greisfu: the question is what that something is
- 02:03 greisfu: i think without that something it's basically impossible
- 02:03 greisfu: but with that somethnig
- 02:03 greisfu: the entire proof is useless
- 02:03 greisfu: because that something will be able to prove everything i just said
- 02:03 greisfu: XD
- 02:04 greisfu: so the goal is to find an expression like this: a^2 + b^2 + c^2 + <positive> = ab + bc + ca
- 02:04 greisfu: well really it should be <nonnegative>
- 02:04 greisfu: but whatever
- 02:05 greisfu: let's try (a-b+c)^2
- 02:05 greisfu: this value >= 0
- 02:05 greisfu: a^2 + b^2 + c^2 - ab - bc + ca >= 0
- 02:06 greisfu: which means a^2 + b^2 + c^2 + 2ca >= ab + bc + ca
- 02:06 greisfu: ...
- 02:06 greisfu: even (a+b+c)^2 >= 0 might have worked
- 02:06 greisfu: will have worked
- 02:07 greisfu: a^2 + b^2 + c^2 + 2(ab+bc+ca) >= ab + bc + ca
- 02:07 greisfu: ohhhhhh
- 02:07 greisfu: stupid me
- 02:07 greisfu: no
- 02:07 greisfu: the hard part is finding an expression
- 02:07 greisfu: a^2 + b^2 + c^2 + <negative> >= ab + bc + ca
- 02:08 greisfu: so even after increasing the LHS by removing the negative, the inequality still holds
- 02:08 greisfu: now i have 2ca
- 02:08 greisfu: which is positive
- 02:09 greisfu: (a-b+c)^2 = (a-b+c)^2
- 02:09 greisfu: a^2+b^2+c^2+2ca = ab+bc+ca + (a-b+c)^2
- 02:10 greisfu: if 2ca <= (a-b+c)^2
- 02:10 greisfu: then a^2+b^2+c^2 = ab+bc+ca + <positive>
- 02:10 greisfu: which means a^2+b^2+c^2 >= ab+bc+ca
- 02:11 greisfu: so is 2ca <= (a-b+c)^2
- 02:11 greisfu: sounds much much much easier to prove, but i have no idea where to start XD
- 02:14 greisfu: 2ca <= a^2 + b^2 + c^2 - ab - bc + ca
- 02:14 greisfu: ca <= a^2 + b^2 + c^2 - ab - bc
- 02:14 greisfu: ab - bc <= a^2 + b^2
- 02:14 greisfu: which means it boils down to c^2 + positive
- 02:15 greisfu: no no no what did i say
- 02:15 greisfu: ab <= a^2, ab <= b^2
- 02:15 greisfu: that's it
- 02:15 greisfu: bc is an unknown
- 02:15 greisfu: no it isn't
- 02:15 greisfu: oh silly me
- 02:16 greisfu: anyway assume a > b > c
- 02:16 greisfu: in that case
- 02:16 greisfu: a^2 > ab
- 02:16 greisfu: and b^2 > bc
- 02:16 greisfu: so you get c^2 + positive
- 02:16 greisfu: is c^2 + positive >= ca?
- 02:16 greisfu: aha but ca > c^2
- 02:16 greisfu: that "positive" may not be big enough
- 02:18 greisfu: c^2 + (a^2 - ab) + (b^2 - bc) >= ca
- 02:19 greisfu: let's have c = a at first then
- 02:19 greisfu: in that case
- 02:19 greisfu: then you have positive >= 0
- 02:19 greisfu: ez
- 02:19 greisfu: if c = a-1
- 02:20 greisfu: then you have c^2 + positive >= c^2 + c
- 02:20 greisfu: is a^2 - ab + b^2 - bc >= c?
- 02:20 greisfu: replace all a's with c's
- 02:20 greisfu: c^2 + 2c + 1 - bc + b + b^2 - bc >= c
- 02:21 greisfu: c^2 + c + (b-1)^2 >= 0
- 02:21 greisfu: which is obviously true
- 02:21 greisfu: all terms positive
- 02:21 greisfu: b-1 is >= 0 too
- 02:22 greisfu: if c^2 + (a^2 - ab) + (b^2 - bc) >= ca
- 02:22 greisfu: is this true for c-1
- 02:23 greisfu: again not necesarily, c^2 decreases faster than ca
- 02:23 greisfu: again, need to split it up
- 02:23 greisfu: every single time
- 02:23 greisfu: guh
- 02:23 greisfu: it's just the wrong method
- 02:23 greisfu: i don't know how to do it :/
- 02:36 greisfu: nope, always ends in a loop
- 03:15 Khoth: How are you getting on?
- 03:16 greisfu: incredibly bored
- 03:16 greisfu: and falling asleep
- 03:18 greisfu: do you mind conceding?
- 03:18 Khoth: Do you?
- 03:18 greisfu: i'm not going to concede
- 03:22 greisfu: are you going to concede?
- 03:23 greisfu: anyway, i think this would've worked on someone else, and yes, this wouldn't have worked in a true AI box scenario
- 03:24 greisfu: you're unfortunate that i don't need to listen to music via my ears :P
- 03:24 greisfu: i'm just going to mentally replay my favourite anime songs again
- 03:42 Khoth: Do you want to know the answer to the maths thing?
- 03:42 greisfu: er
- 03:42 greisfu: no
- 03:43 greisfu: i still want to try it out for myself
- 03:43 greisfu: if i really really end up not knowing
- 03:43 greisfu: then i'll look it up in the book i read it from
- 03:43 Khoth: btw it's true for all real numbers so induction isn't a good approach
- 03:43 greisfu: it's not for real numbers
- 03:43 greisfu: it's just for the integers
- 03:44 greisfu: it will work for all reals though
- 03:44 Khoth: Anyway, well done for surviving
- 03:44 greisfu: thank you
- 03:44 Khoth: The previous guy also said he thought it might work on other people, so I wanted to try again
- 03:45 greisfu: this wasn't a new idea
- 03:45 greisfu: *palm*
- 03:45 Khoth: This is probably it though, I kind of feel bad about it
- 03:45 greisfu: "it"?
- 03:45 Khoth: As in I suspect I won't try again
- 03:45 greisfu: well
- 03:45 greisfu: no reason why you can't try again
- 03:46 greisfu: just... don't try *this* again
- 03:46 greisfu: it's just lots of negative utility
- 03:47 Khoth: It's better than my other idea, which was goatse-related
- 03:47 greisfu: goatse?
- 03:47 greisfu: what's that mean
- 03:47 Khoth: Internet shock site image. Was very famous like ten years ago
- 03:47 greisfu: i see
- 03:48 greisfu: will we post logs?
- 03:48 Khoth: If you want
- 03:48 greisfu: kk

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