The condom problem

Today I'm going to share a problem which I learned from Prof Vishnu Jejjela . As usual I won't spoil the solution for a few days so that people can mull over it. In the mean time I'll lay out the problem and a few special cases which I think are worth thinking about.

n men and m women want to have sex with each other. In particular they want all mn pairings to occur. The problem is that they all have a different STD and they all want to avoid getting anyone else's STD. For this they have condoms, these condoms have some fairly well defined properties.

1. They can be nested.
2. They can be used in either direction.
3. They never break.
4. Diseases never get through them.
5. On the other hand if a side of condom A is in contact with a side of condom B with disease x then disease x jumps to condom A.
6. If a person with disease y uses condom C then disease y appears on the side of condom C actually used.

The problem of course is to minimize the number of condoms used. The naive solution of course involves using mn condoms (1 for each encounter).

Two classically special interesting cases are
1. m=n=2.
2. n=1, m an arbitrary odd integer.

Happy New Year!!!

Happy new year guys (yes both of you). Anyway it's 2015 and I'm going to try to be a more active blogger this year (yes, yes easy to say at the start of the year when everything else going on is comparatively quite).

Anyway what better way to kick of the year than with the 2015 problem of the year . I've played with it a bit but I think I'll leave the ones I have out for a bit. There are a few nice tricks which I don't really want to spoil for anyone reading. I'll leave it for a commenter to spoil. :-)

Problem of the year 5775

OK so it's 5775 according to the Hebrew calendar. Which means of course that it's Problem of the year time (well ok it's a bit late but so what).

5775 turns out to be kinda tough (or I'm being silly).

(5+7)/(7+5)=1 is pretty easy.
getting 5-sqrt(7+7-5)=2 took me quite a while.

I have at the time of writing no way to get 3.

5-(7/7)^5=4
5*((7/7)^5)=5
5+(7/7)^5=6

all sort of fit together. 

7 like 3 I don't have yet, although I've spent less time on it.

5+sqrt(7+7-5)=8

5-(7/7)+5=9
5*(7/7)+5=10
 5+(7/7)+5=11

 again fit into a nice group.

This is about where I stopped thinking about these (because 3 and 7 where incomplete).

 So naturally the point of this post is to see what people come up with for the 3,7 and things bigger than 11.  Enjoy!

SATMO 2014

Last Saturday, the 3rd annual South African Tertiary Mathematics Olympiad was held. For US based readers think of an SA version of the Putnam contest.  Full results, questions and solutions,  can be found here.

Two of the questions (8 and 17) this year where submitted by me. Question 8 I first heard from Michael Lugo and question 17 I first heard from Pravesh Ranchod. 

A huge thank you for organizing the event goes out to Stephan Wagner of Stellenbosch Univesity, as does a huge congrats to the winners.

I've recently been involved in the formation of the Wits Undergrad Maths Society

This has involved going to a bunch of lectures and handing out sign up sheets. It's also involved a lot of talking to lecturers and other organizers to timetable this but I want to focus on the signup sheets.

Turns out that a fair number of students use there wits e-mail addresses and a fair number use something else (g-mail the most common). By inspection there seems to be a very very large correlation between what sheet you signed and which e-mail you put down. that is to say that some of the sheets are almost completely (or in some smaller cases completely) covered with students using there wits e-mail while others are almost devoid of student e-mails. 

Presumably students look at the first few people putting down a wits e-mail address and decide to copy or see the first few haven't and decide to use the non-wits address. Anyone have any thoughts on what causes this?

Hats

Dmytro Yeroshkin and myself have uploaded a paper to the archive here .
It's on a  hats game which is well described here by Tanya Khovanova, and originally proposed by Lionel Levine

There are a lot of opportunities for future work left. In particular the original problem with alot of wise men instead of the 2 player case described in our paper.

and the next line is...

An old puzzle today. this one is great for a blog format as it allows someone who has solved the puzzle to verify that they know the answer without giving away the answer to everyone
The problem is to find the next line. Enjoy and show the next line when you have it.

1
1 1
2 1
1 2 1 1
1 1 1 2 2 1
3 1 2 2 1 1