Find matrix dimension so as to contain numbers

Question

Suppose N is the number of elements that I want to store in a matrix. I don't know how much this matrix has to be big. I want to easily find the minimum required size for the matrix so as to make it contain the N numbers.

As a requirement, I would like the matrix to be as compact as possible, so that:

1 2 3
4 5 6
7

is preferred to:

1 2 3 4 5 6
7

Any suggestions?

Thank you.

What about `[1 2 3 4 5 6 7]`? How do you define "compact"? Minimize the number of wasted elements? Minimize the product of the dimensions? Minimize the sumo f the dimensions? — Dan, Oct 23 '13 at 10:09
Is `[1 2 3; 4 5 6; 7 NaN NaN]` also preferred to `[1 2 3 4; 6 7 8 NaN]`? And if so, why... can you give a rule that can be applied to all vectors? — Dennis Jaheruddin, Oct 23 '13 at 10:27

Luis Mendo · Accepted Answer · 2013-10-23T11:25:14.667

4

You should specify your optimization criterion, that is, how much weight you give to "squareness" versus "emptiness".

As an example of such criterion, the following code minimizes emptiness subject to the number of rows and columns being greater than 1; and picks the more square-like option if there are several minimizing sizes:

mm = ceil(sqrt(N)):-1:2; %//possible numbers of rows. Reverse order. Do not consider 1
nn = ceil(N./mm); %//corresponding numbers of columns
excess = mm.*nn-N; %//number of empty cells
[val ind] = min(excess);
m = mm(ind)
n = nn(ind)

Note that mm in the code is defined in reverse order so that min will find the last minimizing value (more square-like) if there are more than one.

For example, N=113 gives the solution m=6, n=19, resulting in 1 empty cell (=6*19-113). This solution is preferred to m=57, n=2 or m=38, n=3 (which also leave give 1 empty cell) because it is more square-like.

edited Oct 23 '13 at 11:25

answered Oct 23 '13 at 10:02

Luis Mendo

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It may happen that a huge part of the matrix remains empty, and this is undesired. I would prefer a rectangular matrix having a very small number of empty cells and fitting very well the input number. – Eleanore Oct 23 '13 at 10:07
1

@PhamTrung What do you mean? My solution does work for composite of for prime numbers. For example, for N=120 it correctly gives 12, 10 – Luis Mendo Oct 23 '13 at 10:42
@Luis Mendo: what about 55, sorry I am not familiar with the programming language :) – Pham Trung Oct 23 '13 at 10:47
@PhamTrung 55 would correctly give 5 and 11 (try it). I think I don't get your point – Luis Mendo Oct 23 '13 at 10:48
@Luis Mendo:sorry, I think I misunderstand your answer :) – Pham Trung Oct 23 '13 at 10:55
1

Here is a nice way to illustrate this solution: `a = NaN(m,n); a(1:N) = rand`. Especially if you put it in a for loop. – Dan Oct 23 '13 at 11:05

score 0 · Answer 2 · answered Oct 23 '13 at 11:15

From your example I would think that the 'minimum size' may mean the solution that minimizes max(size).

In this case the solution is very simple, assuming you have L elements:

In one dimension you have:

m = ceil(sqrt(L))

And in the other dimension you have:

n = ceil(L/m)

Now a matrix of size mxn or nxm will be the matrix of minimum size with (given that size) the least amount of empty elements.

Find matrix dimension so as to contain numbers

2 Answers2