A well known puzzle among combinatorialists is the following: m men and n women, each with a different sexually transmitted disease, want to engage in all mn sexual encounters with no one catching anyone else's disease and with the minimal number of condoms being used. You are allowed to nest condoms and to turn them inside out, but once a surface becomes infected by a disease it stays infected for all time.
Well that's easy, you line them up and shoot them down.
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Anonymous2015-08-17 12:38
All dudes get 1 condom. Don't take it off until you're done. Superglue around the base.
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Anonymous2015-08-17 19:30
I have solved the general problem of the least number of condoms it takes for m men and n women to have all mxn heterosexual encounters. The answer is
m = n = 2 => 2 condoms
m = 2k+1, n = 1 => k+1 condoms
otherwise, it's the smallest integer greater or equal to m/2 + 2n/3. I assume that m >= n, otherwise interchange m and n.
The same method solves the problem for m homosexual men all having sex, and m bisexual men and n heterosexual women (posed by A. Orlitzky and L. Shepp).
how many diseases do you think ilan vardi caught before he figured it out?
What is the nesting threshold? Because wearing more than a couple condoms at a time decreases sensitivity to the point that it would be difficult to maintain erectile function.
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Anonymous2015-08-18 11:53
>>1 Is your own homeworkd broken? Just ask it to do it instead.
Thanks for perpetuating toxic masculinity in the tech world, asshole
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Anonymous2015-08-18 13:24
>>12 You forgot your "begone, troll, you lose", FFP-chan.
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Anonymous2015-08-18 13:29
>>9 when's the last time YOU called for increased funding for combinatorics? now they're going to breed and we'll have to find a meaner, scarier breed of mathematician to introduce in order to cull their population
Rob Pike solved the condom problem for homosexual men with a single condom. He now has an infinite number of sexually transmitted diseases.
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Anonymous2015-08-25 12:15
Prove or disprove.
One can acquire an infinite number of sexually transmitted diseases in a finite amount of time.
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Anonymous2015-08-25 12:30
/prog/ homeworkd 5025: The Infection Problem
A well known puzzle among combinatorialists is the following: m men and n women, each with a different sexually transmitted disease, want to exchange all of their diseases using the minimal amount of sexual encounters. At the end of the process, all participants will each be infected will all of their collective infections. You can assume the probability of infection from unprotected sex is 100% and all sex is unprotected.
/prog/ homeworkd 5026: The Infection Problem [Part 2] Generalize The Infection Problem to the case where there are ms straight men, mb bisexual men, mg gay men, fs straight women, fb bisexual women, and fg lesbians.
1. Two Chads, Chad_0 and Chad_1, are hanging out in a universe with an infinite number of women of cardinality aleph_0. Chad_0 brags that there is a bijection between the set of women he has slept with and the natural numbers, while Chad_1 claims that his set has a bijection with the real numbers. Chad_0 is not impressed, however, and claims that they're about even. Prove Chad_0 wrong using Cantor diagonalization.
2. An infinite number of Chads, of cardinality aleph_n, live in a universe with an infinite number of women, also of cardinality aleph_n. Each Chad wants to fuck an infinite number of women, of cardinality aleph_n, each without getting sloppy seconds. Prove to them that for that to happen they will need an infinite number of women of at least cardinality aleph_n+1.
3. Prove that Chad's "game" is in Nash-equilibrium with the set of betas and the set of women as players.
Both sides of the condom get infected upon sexual intercourse so they would need n*m condoms for n*m sexual encounters
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Anonymous2015-09-30 16:51
The solution is undefined unless m = n = 1, in which case no condom should be used at all. Any other proposed solution is an abhorrent in the eyes of The Lord.