Sorting rows and colums in a matrix (with some music, and some magic)

February 14, 2013
By

(This article was first published on Freakonometrics » R-english, and kindly contributed to R-bloggers)

This morning, I was working on some paper on inequality measures, and for computational reasons, I had to sort elements in a matrix. To make it simple, I had a rectangular matrix, like the one below,

> set.seed(1)
> u=sample(1:(nc*nl))
> (M1=matrix(u,nl,nc))
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    7    5   11   23    6   17
[2,]    9   18    1   21   24   15
[3,]   13   19    3    8   22    2
[4,]   20   12   14   16    4   10

I had to sort elements in this matrix, by row.

> (M2=t(apply(M1,1,sort)))
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    5    6    7   11   17   23
[2,]    1    9   15   18   21   24
[3,]    2    3    8   13   19   22
[4,]    4   10   12   14   16   20

Nice, elements are sorted by row. But for symmetric reasons, I also wanted to sort them by column. So from this sorted matrix, I decided to sort elements by column,

> (M3=apply(M2,2,sort))
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    3    7   11   16   20
[2,]    2    6    8   13   17   22
[3,]    4    9   12   14   19   23
[4,]    5   10   15   18   21   24

Nice, elements are sorted by column now. Wait… elements are also sorted by row. How comes ? Is it some coincidence ? Actually, no, you can try…

> library(scatterplot3d)
> nc=6; nl=5
> set.seed(1)
> u=sample(1:(nc*nl))
> (M1=matrix(u,nl,nc))
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    8   23    5   30   10   15
[2,]   11   27    4   29   21   28
[3,]   17   16   13   24   26   12
[4,]   25   14    7   20    1    3
[5,]    6    2   18    9   22   19
> M2=t(apply(M1,1,sort))
> M3=apply(M2,2,sort)
> M3
     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    3    7   14   19   22
[2,]    2    6    9   15   20   25
[3,]    4    8   10   17   23   26
[4,]    5   11   16   18   24   29
[5,]   12   13   21   27   28   30

or use the  following function is two random matrices are not enough,

> doublesort=function(seed=2,nl=4,nc=6){
+ set.seed(seed)
+ u=sample(1:(nc*nl))
+ (M1=matrix(u,nl,nc))
+ (M2=t(apply(M1,1,sort)))
+ return(apply(M2,2,sort))
+ }

Please, feel free to play with this function. Because this will always be the case. Of course, this is not a new result. Actually, it is mentioned in More Mathematical Morsels by Ross Honsberger, related to some story on marching band. The idea is simple: consider a marching band, a rectangular one. Here are my players

> library(scatterplot3d)
> scatterplot3d(rep(1:nl,nc),rep(1:nc,each=nl), as.vector(M1),
+ col.axis="blue",angle=40,
+ col.grid="lightblue", main="", xlab="", ylab="", zlab="",
+ pch=21, box=FALSE, cex.symbols=1,type="h",color="red",axis=FALSE)

Quite messy, isn’t it ? At least, this is what the leader of the band though, since some tall players were hiding shorter ones. So, he brought the shorter ones forward, and moved the taller ones in the back. But still on the same line,

> m=scatterplot3d(rep(1:nl,nc),rep(1:nc,each=nl), as.vector(M2),
> col.axis="blue",angle=40,
+ col.grid="lightblue", main="", xlab="", ylab="", zlab="",
+ pch=21, box=FALSE, cex.symbols=1,type="h",color="red",axis=FALSE)

From the leader’s perspective, everything was fine,

> M=M2
> for(i in 1:nl){
+ for(j in 1:(nc-1)){
+ pts=m$xyz.convert(x=c(i,i),y=c(j,j+1),z=c(M[i,j],M[i,j+1]))
+ segments(pts$x[1],pts$y[1],pts$x[2],pts$y[2])
+ }}

But someone in the public (on the right of this graph) did not have the same perspective.

> for(j in 1:nc){
+ for(i in 1:(nl-1)){
+ pts=m$xyz.convert(x=c(i,i+1),y=c(j,j),z=c(M[i,j],M[i+1,j]))
+ segments(pts$x[1],pts$y[1],pts$x[2],pts$y[2])
+ }}

So the person in the audience ask – one more time – players to move, but this time, to match with his perspective. Since I consider someone on the right, some minor adjustments should be made here

> sortrev=function(x) sort(x,decreasing=TRUE)
> M3b=apply(M2,2,sortrev)

This time, it is much bettter,

> m=scatterplot3d(rep(1:nl,nc),rep(1:nc,each=nl), as.vector(M3b),
+ col.axis="blue",angle=40,
+ col.grid="lightblue", main="", xlab="", ylab="", zlab="",
+ pch=21, box=FALSE, cex.symbols=1,type="h",color="red",axis=FALSE)

And not only from the public’ perspective,

> M=M3b
> for(j in 1:nc){
+ for(i in 1:(nl-1)){
+ pts=m$xyz.convert(x=c(i,i+1),y=c(j,j),z=c(M[i,j],M[i+1,j]))
+ segments(pts$x[1],pts$y[1],pts$x[2],pts$y[2])
+ }}

but also for the leader of the marching band

Nice, isn’t it ? And why is this property always valid ? Actually, it comes from the pigeonhole theorem (one more time), a nice explanation can be found in The Power of the Pigeonhole by Martin Gardner (a pdf version can also be found on http://www.ualberta.ca/~sgraves/..). As mentioned at the end of the paper, there is also an interpretation of that result that can be related to some magic trick, discussing – in picture – a few month ago on http://www.futilitycloset.com/… : deal cards into any rectangular array:

2012-01-26-ranks-and-files-1

Then put each row into numerical order:

ranks and files 2

Now put each column into numerical order:

ranks and files 3

That last step hasn’t disturbed the preceding one: rows are still in order. And that’s a direct result from  pigeonhole theorem. That’s awesome, isn’t it ?

Arthur Charpentier

Arthur Charpentier, professor in Montréal, in Actuarial Science. Former professor-assistant at ENSAE Paristech, associate professor at Ecole Polytechnique and assistant professor in Economics at Université de Rennes 1.  Graduated from ENSAE, Master in Mathematical Economics (Paris Dauphine), PhD in Mathematics (KU Leuven), and Fellow of the French Institute of Actuaries.

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