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Asking this question on SO, I have been advised to post it here. I will be using Javascript to implement :

Please consider a row of size 12. On that row, I want to place some items that have 3 different width : 1, 2 and 3.

I would like to get all the possible combination of items on that row such that the sum of their width is equal to the row size.

For example, I could have :

[1,1,1,1,1,1,1,1,1,1,1,1] (12 items of size 1)

[2,2,2,2,2,2] (6 items of size 2)

[3,3,3,3] (4 items of size 3)

[3,3,3,1,1,1] (3 items of size 3 + 3 items of size 1)

[2,2,1,1,3,3] (2 items of size 2 + 2 items of size 1 + 2 items of size 3)

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Presumably, order matters - that is, $[3,3,3,2,1]$ and $[3,2,1,3,3]$ are considered different. –  Thomas Andrews Jun 20 '12 at 15:10
    
What are you going to do with all the arrangements, once you have them? There will be a lot of them, you know. Just using four 2s and four 1s, there will be 70 arrangements. Just using two 3s, two 2s, and two 1s, there will be another 90. All told, I'm sure there are over 1,000. –  Gerry Myerson Jun 21 '12 at 1:55
    
@GerryMyerson Good guesstimate! There are 927. –  Chris Taylor Jun 21 '12 at 8:57
    
@Gerr Myerson. I am trying to tile object of different width on rows on a screen. Similar to : images.google.com/… Except I set 3 ratio for the images ! –  500 Jun 21 '12 at 11:56

1 Answer 1

The following Haskell code generates all possible solutions where the row is of width $n$. It works on a simple principle - for the base cases ($n=0$, $n=1$ and $n=2$) we write out the answer explicitly. Otherwise (if $n>2$) we can put a block of width 1, 2 or 3 in the first position, and then we run the algorithm recursively on the remaining space.

Translate to Javascript at your leisure.

getCombinations :: Int -> [[Int]]
getCombinations 0 = [[]]
getCombinations 1 = [[1]]
getCombinations 2 = [[1,1], [2]]
getCombinations n = map (1:) (getCombinations (n-1)) ++
                    map (2:) (getCombinations (n-2)) ++
                    map (3:) (getCombinations (n-3))

Note that this is inefficient for large $n$. In fact, it is exponentially bad. For example, computing the answer for n = 12 computes the answer for n = 9, 10 and 11 in the process. Computing the answer for n = 11 also computes n = 8, 9 and 10, so we have duplication.

You could improve the efficiency by writing a memoizing algorithm, which stores the answer for each $n$ in a map when it is first computed, and looks up subsequent calls with the same n in the map rather than recomputing them.

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thank you, I don`t know Haskell at all, but might have a friend able to help :-) –  500 Jun 20 '12 at 15:28
    
Basically, you enumerate the cases for $n-3$ and add a $3$ to each of them, enumerate the $n-2$ cases and add a $2$ to them, and enumerate the $n-1$ cases and add a $1$ to them. @500 –  Thomas Andrews Jun 20 '12 at 15:36
    
@ChrisTaylor You could avoid specifying the cases for $n=1$ and $n=2$ if you could just define the function as the empty list $[]$ for negative $n$. –  Thomas Andrews Jun 20 '12 at 15:48
    
@ThomasAndrews Good point! –  Chris Taylor Jun 21 '12 at 8:56

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