Math and Logic Puzzles: Redux

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Post Post #54 (isolation #0) » Tue Jun 12, 2018 7:37 pm

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Why was I not told this existed?
In post 39, Ircher wrote:Prove that the derivative of an odd function is an even function and that the derivative of an even function is an odd function.
Let f be a differentiable even function.
f'(x) = \lim_{h\to 0} [f(x+h)-f(x)]/h = \lim_{h\to 0} [f(-x-h) - f(-x)] / h = \lim_{h\to 0} [f(-x+-h) - f(- x)]/ - -h = \lim_{h\to 0} [f(-x+h) - f(- x)]/ -h = -f'(-x), thus f' is odd.
Let g be a differentiable odd function.
g'(x) = \lim_{h\to 0} [g(x+h)-g(x)]/h = \lim_{h\to 0} -[g(-x-h)-g(-x)]/h = \lim_{h\to 0} [g(-x-h)-g(-x)]/-h = \lim_{h\to 0} [g(-x+h)-g(-x)]/h = g'(-x), thus g' is even.
Last edited by Who on Tue Jun 12, 2018 7:44 pm, edited 1 time in total.
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Post Post #55 (isolation #1) » Tue Jun 12, 2018 7:43 pm

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Suppose you have 13 real numbers such that if you remove any one of them, you can divide the remaining 12 into two groups of 6 which each have equal sums. Prove that all 13 numbers are equal.
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Post Post #58 (isolation #2) » Tue Jun 12, 2018 8:14 pm

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Nice proofs, you got it.
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Post Post #64 (isolation #3) » Tue Jul 03, 2018 4:10 pm

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Spoiler: Above problem
Randomly generate 3 points, then randomly generate a 4th, then randomize which gets paired with which. If the 4th fits inside the triangle defined by the other three, no matter how you define the line segments they won't intersect. If it doesn't, then there is exactly one which it can be paired with to intersect the line defined by the other two. (Unless it lies on the line generated by two of them, which occurs with probability 0) The probability of it being paired with this one is 1/3. The probability that the fourth falls inside the area defined by the first three is the same as the expected value of the indicator variable, which is the same as the expected value of the area. Thus, the total probability is Expected area/3 = 11/(144*3).
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Post Post #67 (isolation #4) » Thu Jul 19, 2018 11:49 am

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On Implosion's triangle problem:
Spoiler: Why I was wrong
Even if one isn't in the triangle formed by the others, that doesn't mean you can't make a triangle out of other points such that the 4th is in the triangle formed by the other 3. If you tried doing the thing I said then the lines would intersect past the edge of one of the segments

Spoiler: Fixed
If and only if you have four points such that no matter which point you choose, it is not in the triangle formed by the other 3, then it is possible to assign the line segments such that they will intersect. Proof of this: Make a triangle, extend the line segments into lines, note that the areas where the fourth point can't be are inside the triangle, and in the areas which meet the triangle at a vertex (As opposed to meeting the triangle at a side). If it's inside the triangle we're done, if it's in one of the areas which ends at a vertex, then that vertex is inside the triangle formed by the other 3.

If A is in the triangle BCD then B is almost surely not in ACD. (And neither is C nor D). In any set of four points, at most one of them can be in the triangle formed by the other 4. Proof of this: Either A or B must be further from line CD, the triangle defined by the closer one clearly cannot contain the further one, since all points in the triangle are closer to CD than the third vertex.

For i=1-4, let x_i be the event "the ith point is in the triangle formed by the other 3". Note that each x_i has probability 11/144. Also, all events are mutually exclusive. Thus, the probability that none of them happen is 1-4*11/144 = 100/144.
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Post Post #80 (isolation #5) » Fri Jul 26, 2019 9:21 am

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In this specific case it doesn't have paradoxes because anything the super being could accomplish by predicting they could also accomplish by cheating (Have the box set up so that it automatically burns the money if you take both or some other similar thing). If we accept the super being's superness as fact, then the problem is the same as facing a cheating being, so obviously don't take the second box.

Other than the fact that it may be possible to use her to break quantum mechanics.
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Post Post #114 (isolation #6) » Mon Jan 20, 2020 3:14 pm

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(0!+0!+0!)! = 6
(1+1+1)! = 6
2+2+2 = 6
3!+ 3-3 = 6
(4-4/4)! = 6
5+5/5 = 6
6+6-6 = 6
7-7/7 = 6
(√(8 +8/8))! = 6
(√9)!+9-9 = 6
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Post Post #161 (isolation #7) » Wed Dec 09, 2020 12:22 pm

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I get the same result, also making some assumptions, but I feel reasonably confident in them.

How I get it:
Let A be the number of times a 1 is rolled minus the number of times a 4 is rolled, let B be the number of times a 3 is rolled minus the number of times a 6 is rolled, let C be the number of times a 2 is rolled minus the number of times a 5 is rolled. The event is now:
A+B>|C|
Also note that A, B, and C are identically distributed and approximately normal with mean 0 via CLT. Normalizing to make them variance 1 doesn't change the equation, so if we assume they are independent it's now
X>|Y| where X is normal with mean 0 variance 2, Y is normal with mean 0 variance 1. We then do basic integration (Set up the integral, note that it's the standard e^(-r^2/2) with r from 0 to infinity but over a sector with a weird angle) and end up with the answer scigatt gave. A,B, and C are obviously not independent, for example if A=N then that means that every roll was a 1 so B and C must both be 0. But they are [assumption] approximately independent because that is unlikely to happen.
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Post Post #171 (isolation #8) » Sat Oct 30, 2021 4:36 am

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If the field has odd characteristic:
For any nonzero x, x and -x are distinct and square to the same value. Thus there are (|F|-1)/2 nonzero squares so there must be (|F|-1)/2 nonzero nonsquares, so let c be a no square then x^2-c is irreducible and thus extending by the root of that is a quadratic extension.

If the field is characteristic 2:
Apply the same argument but instead of x and -x take x and x+1, and instead of both have the same square we have that both have the same x(x+1). Thus there’s some c which is not of the form x(x+1), and we have that x(x+1)+c is irreducible.
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Post Post #182 (isolation #9) » Mon Feb 14, 2022 4:23 pm

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10^27
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Post Post #197 (isolation #10) » Mon Feb 14, 2022 9:39 pm

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1/2^27

6/35
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Post Post #230 (isolation #11) » Mon Mar 20, 2023 9:21 am

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To clarify, the games are all being played sequentially, right? No game starts before the last one has ended. If so, are any of the following strategies allowed?
Use Stockfish against your crush
After beating her brother, before playing her, play another game against someone else you can beat, claim that you have won two games in a row and thus should be allowed to ask her out as per the exact words of the deal.
Beat her brother then forfeit against her before playing, then beat her brother again. Claim that because you forfeited before playing you didn't lose a game you played and thus you have won two games you played in a row and this one is just the last one but worse. Alternatively, lie and say you played her when you didn't.
What time control is there on each match, and are the games played immediately one after the other? Can you prolong a game against her brother to the point where she leaves/goes to sleep and then start the game against her and win on time when she doesn't show up?
Open with the bongcloud. She will assume that it is a meme game and proceed make meme moves, wait for her to screw up so badly that you can win starting a serious game there.
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Post Post #293 (isolation #12) » Tue Jan 09, 2024 9:59 am

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A puzzle which everyone here has probably already seen the base version of, but which has many variants which some people may not have seen:
There are 100 mathematicians in a prison. There's a room in the prison with a light switch, which starts switched off. Every day (They all have calendars and know what day it is) the warden chooses a (not necessarily new) mathematician uniformly at random and takes them into the room, and the mathematician sees the state of the light and may switch it, then they are returned to their room. Nobody other than the mathematicians can touch the light switch. If at any point, a mathematician correctly says "Every prisoner has been in this room at least once", they are all set free. If the mathematician is incorrect, they're all killed (So make sure that the probability of making a wrong declaration is exactly 0). In advance they're all allowed to agree on a strategy, give a strategy which will set them free. (Only important thing is that they end up free eventually with probability 1, how long it takes doesn't matter)

Variants (If you haven't solved the base variant don't look at them because they contain clues as to how to do it):
Spoiler: Easy variant
They have no information at all about what day it is the only information is how many times they've been to the room and what the state of the light has been each time), AND the starting state of the light is not known

Spoiler: More challenging variants

(All of these are separate)
The mathematicians now have families they want to get back to and want to minimize the amount of time they spend in prison. Find a solution which gets them out quickly on average/very probably. Also the same problem but N prisoners instead of 100, though I'm not sure how much of a difference that makes to the solution.

As soon as a prisoner correctly declares that everyone has been in the room, that prisoner (and only that prisoner) is set free. Taken out of the pool of prisoners who might visit the room in the future. Find a way to free everyone.

The warden cheats: If a prisoner exits the room with the light on he switches it off with some fixed probability p. The warden never switches an off light on. Also they can't count days like in the easy variant.

They can't count days like in the easy variant AND the mathematicians are all identical robots who must execute identical programs. The robots do have access to random number generators.

Same as the last one but instead of being taken at random the warden decides adversarially at the beginning which order to take them in, but this order must have each one must visit infinitely many times, and also the robots no longer have access to random number generators. I'm pretty sure this is impossible but can't prove it. (All the other variants are definitely possible)
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Post Post #296 (isolation #13) » Tue Jan 09, 2024 6:14 pm

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In post 294, tris wrote:
Spoiler:
assign a schedule of mathematicians, so they each get a day in each period of 100 days. if the mathematician enters the room on their day, they turn the light on. and the next one turns it off.
eventually
someone will have been after every other mathematician on their day. its going to take fucking forever tho
That works. I think it's the slowest strategy I know of, but speed was only required in one variant so it is 100% valid in the base variant. Testing it once it took about 3 million days, which is about 8,000 years. Testing it 100 times and taking an average is rather slow, but still seems to take on average around 3 million, but it seems to have a pretty high variability and I'd need more tests to be certain. I'm too lazy to calculate/estimate the expected value of number of days it will take by hand. There's a relatively simple modification you can make to make it take around 80,000-100,000 days, or about
30 years
(Edit: Missed a 0, it's actually 300 years), but even with the speedup it takes more than 1,000,000 days reasonably frequently. Even with the speedup it's still the slowest strategy I know of. (Not counting anything which is intentionally slow)

Since that was the first strategy you came up with the "easy" variant might not be so easy, I labelled it as "easy" expecting most people to come up with a different strategy first.
Last edited by Who on Wed Jan 10, 2024 8:41 pm, edited 1 time in total.
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Post Post #300 (isolation #14) » Wed Jan 10, 2024 8:51 pm

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In post 299, tris wrote:
Spoiler: easy variant
one person turns on the light every time they enter a dark room. everyone else turns off the light if it's on two times and then leaves it alone. once the one who turns on has seen the light go off 199 times, they know everyone has been in the room.
Correct!
Also, although not required by the variant, your new strategy is significantly faster. It takes on average about 20,000 visits, which at one visit per day is around 55 years. They might still be alive by the time they get out.
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