Finite Projective Plane

Posted by tpc at May 11th, 2010

Learned a cool trick today. The finite projective plane of order n has
n^2 + n + 1 points,
n^2 + n + 1 lines,
n + 1 points on each line,
n + 1 lines passing each point.

The best known example is that of a fano plane which is of order 2. Now for order 3, there are a total of 13 lines, 13 points and 4 points on each line. It turns out you can use a standard deck of cards to represent this. You can divide the deck into 13 lines of 4 cards each, with each face value (regardless of suit) representing the same point.
Now if
P = \{0, 1, \ldots 12 \} \mod 13,
each line is 0+i, 1+i, 3+i, 9+i

The fun thing is to distribute the each line of 4 cards to different persons. Then you can randomly call out two face value and be sure that exactly one person has both of these cards - two points line on a unique line. Moreover, any two line intersect at a unique point - means that any two persons should have exactly one card in common.

Posted in Combinatorics, Geometry/Topology| No Comments | 

A combinatorial identity

Posted by tpc at March 23rd, 2009

I came across the following identity today.
\displaystyle{1 \choose k} + {2 \choose k} + \ldots + {99 \choose k} = {100 \choose k+1}
It is not exactly difficult to establish if one uses the Pascal relation repeatedly.
\displaystyle{m \choose k+1} = {m-1 \choose k} + {m-1 \choose k+1}
\displaystyle = {m-1 \choose k} + {m-2 \choose k} + {m-2 \choose k+1} = \ldots
\displaystyle = \sum_{j=1}^{m-1} {m-j \choose k} + {0 \choose k+1}

An alternative proof is to extract the coefficient of x^k from the generating function identity:
\displaystyle \sum_{j=0}^{99} (x+1)^j = \frac{ (x+1)^{100} - 1}{ (x+1)-1}.

I’m thinking that this must have a nice combinatorial proof, but all my references are in my office.

Posted in Combinatorics| 2 Comments | 

A Calendar Puzzle

Posted by tpc at December 25th, 2005

A simple little puzzle. Suppose you are given two ordinary 6 sided dice. Is it possible to put the numbers 0-9 (with repetition) onto the faces of both dice, such that using both dice you can display all the days of the month i.e. 01 - 31 .

Posted in Combinatorics, Fun Stuff, Problems| 9 Comments | 

Web Sudoku

Posted by tpc at July 26th, 2005

I reached this link via Mathforge. What can I say, the game is addictive. My current record is 6 min for the easy game and I’ve yet to try the hard and 55 min for the evil level. Now, the very natural question is computing the total possible combinations offered by the Sudoku board and this has stumped me. The wiki page has links to work done by mathematicians on this calculation.

Posted in Combinatorics, Fun Stuff| No Comments | 

Cranks

Posted by tpc at March 23rd, 2005

in mathematics does not refer to eccentric old men (although there are many around) but refer to certain statistics related to partitions. The name probably came about because there is a related notion of ranks of partitions.

This article in New Scientist reports some sort of breakthrough. I’m not sure how significant it is, but it is significant enough that Mahlburg’s site is currently the first result if you google “cranks, partition”. I would like to take a look at the paper.

Posted in Combinatorics, Number Theory| 1 Comment | 

Binomial Coefficients

Posted by tpc at January 11th, 2005

{n \choose r} is the number of ways of picking r objects out of n possibilities without order. A simple but elegant identity is this
{n+1 \choose r }= {n \choose r} + {n \choose r-1}
which appears when you are proving the binomial theorem by induction.

Posted in Combinatorics| No Comments |