## ugly graph of the day

Posted in Books, Statistics with tags , , on February 25, 2012 by xi'an

Julien pointed out to me this terrible graph, where variations cannot be perceived and where the circles are meaningless! Two three-point curves would have been way more explicit!!! ## Back from Philly

Posted in R, Statistics, Travel, University life with tags , , , , , , , , , on December 21, 2010 by xi'an ## Le Monde puzzle [48: resolution]

Posted in R, Statistics with tags , , , on December 4, 2010 by xi'an

The solution to puzzle 48 given in Le Monde this weekend is rather direct (which makes me wonder why the solution for 6 colours is still unavailable..) Here is a quick version of the solution: Continue reading

## Le Monde puzzle 

Posted in R, Statistics with tags , , , , , , on December 2, 2010 by xi'an

This week(end), the Le Monde puzzle can be (re)written as follows (even though it is presented as a graph problem):

Given a square 327×327 symmetric matrix A, where each non-diagonal entry is in {1,2,3,4,5} and $\text{diag}(A)=0$, does there exist a triplet (i,j,k) such that $a_{ij} = a_{jk} = a_{ki}$

Solving this problem in R is very easy. Or appears to be. We can indeed create a random matrix A and check whether or not any of the five triple indicator matrices $(A==c)^3\,,\qquad c=1,\ldots,5,$

has a non-zero diagonal entry. Indeed, since $B=A^3$ satisfies $b_{uu} = \sum_{v,w} a_{uv}a_{vw}a_{wu}$

there is a non-zero entry iff there exists a triplet (u,v,w) such that the product $a_{uv}a_{vw}a_{wu}$ is different from zero. Here is the R code:

chec=1
for (mc in 1:10^3){

#random filled matrix
A=matrix(sample(1:5,(357)^2,rep=TRUE),357)
A=A*upper.tri(A)+t(A*upper.tri(A))

agr=0
for (t in 1:5){
B=(A==t)
agr=max(agr,max(diag(B%*%B%*%B)>0))
if (agr==1){ break()}
}

chec=min(chec,agr)
if (chec==0){ break()}
}


I did run the above code and did not find any case where no triplet was sharing the same number. Neither did I for 326, 325,… But Robin Ryder told me that this is a well-known problem in graph theory that goes under the name of Ramsey’s problem and that 327 is an upper bound on the number of nodes for the existence of the triplet with 5 colours/numbers. So this is another illustration of a case when crude simulation cannot exhibit limiting cases in order to void a general property, because of the number of possible cases. To improve the chances of uncovering the boudary value, we would need a simulation that dis-favours triplets with the same number. I am also curious to see Le Monde solution tomorrow, since finding Ramsey’s numbers seems to be a hard problem, no solution being provided for 6 colours/numbers.

Incidentally, I wonder if there is a faster way to produce a random symmetric matrix than the cumbersome

A=matrix(sample(1:5,(357)^2,rep=TRUE,357)
A=A*upper.tri(A)+t(A*upper.tri(A))


Using the alternative saving on the lower triangular part

A=matrix(sample(1:5,(357)^2,rep=TRUE,357)
A[upper.tri(A)]=sample(1:5,357*178,rep=TRUE)
A=A*upper.tri(A)+t(A*upper.tri(A))


certainly takes longer…

## Random graphs with fixed numbers of neighbours

Posted in Books, R, Statistics with tags , , , , , , , , , on November 25, 2010 by xi'an In connection with Le Monde puzzle #46, I eventually managed to write an R program that generates graphs with a given number n of nodes and a given number k of edges leaving each of those nodes. (My early attempt was simply too myopic to achieve any level of success when n was larger than 10!) Here is the core of the R code:

## Terrible graph of the weekend

Posted in Books, Statistics with tags , , , , , on May 2, 2010 by xi'an Every week, the weekend edition of Le Monde contains a tribune around some statistics and almost irremediably the illustrations are terrible. Witness the one above where the move from 940 millions to more than a billion is completely out of proportions, both for the disk areas and the number of people! Same thing for the split pie chart below: What’s wrong with a regular pie-chart?! And the increase in disabled planes below is not much better, moving from 5 to 8 plane silhouettes… The illustrators should read Tufte’s books instead of USA Today…! Ps-I seem to be always picking at Le Monde, but this is only because I have a subscribtion to its weekend edition!