The goal of this challenge is to produce a function of n which computes the number of ways to partition the n X 1 grid into triangles where all of the vertices of the triangles are on grid points.


For example, there are 14 ways to partition the 2 x 1 grid, so f(2) = 14 via the following partitions Partitions of 2 x 1 where the partitions have 2, 2, 2, 2, 4, and 2 distinct orientations respectively.


This is , so shortest code wins.

  • 10
    Some additional test cases would be beneficial, so we can verify our submissions are correct. – AdmBorkBork Nov 27 at 21:19
  • 8
    You may want to specify non-degenerate triangles. – Arnauld Nov 27 at 21:24
  • I've made edits to OEIS sequence A051708 to reflect this interpretation. – Peter Kagey Nov 30 at 18:32
up vote 1 down vote accepted

05AB1E, 13 bytes


Port of @Bubbler's Jelly answer.

Very slow due to the permutations builtin.

Try it online or verify the first four inputs.


·                # Double the (implicit) input
 L               # Create a list in the range [1, doubled_input]
  É              # Check for each if they're odd (1 if truthy, 0 is falsey)
                 # We now have a list of n 0s and n 1s (n being the input)
   œ             # Get all permutations of that list
    Ù            # Only leave the unique permutations
     ε     }     # Map each permutation to:
      Åγ         #  Run-length encode the current value (short for `γ€g`)
        o        #  Take 2 to the power for each
         ;       #  Halve each
          P      #  Take the product of the mapped permutation
            O    # Sum all mapped values together (and output implicitly)

Haskell, 60 55 54 52 bytes

After a drawing and programming a lot of examples, it occured to me that this is the same as the problem of the rooks:

On a \$(n+1) \times (n+1)\$ chessboard, how many ways are there for a rook to go from \$(0,0)\$ to \$(n,n)\$ by just moving right \$+(1,0)\$ or up \$+(0,1)\$?

Basically you have the top and the bottom line of the \$1 \times n\$ grid. Now you have to fill in the non-horizontal line. Each triangle must have two non-horizontal lines. Whether one of its sides is part of the top or the bottom line corresponds to the direction and length you'd go in the rooks problem. This is OEIS A051708. As an illustration of this correspondence consider following examples. Here the top line corresponds to up-moves, while the bottom line corresponds to right-moves.

Thanks @PeterTaylor for -6 bytes and @PostLeftGarfHunter for -2 bytes!

b 0=1
b 1=2
b n=div((10*n-6)*b(n-1)-9*(n-2)*b(n-2))n

Try it online!

  • I found the OEIS sequence by searching with the first few values. Nice explanation for why it matches. Do you want to edit it to add a comment about this alternative combinatorial interpretation? If not, I might. – Peter Taylor Nov 27 at 22:36
  • BTW you need to adjust the indexing, because the correct answer here is A051708(n+1). So I posted the first correct answer :-P – Peter Taylor Nov 27 at 22:39
  • I take it the rook moves map to triangles by making triangles with top and bottom edges correspond to up or right moves? – Neil Nov 27 at 22:40
  • @PeterTaylor Damn, thanks for pointing out my mistake :) – flawr Nov 27 at 22:41
  • 5
    @Neil I added a graphical explanation. – flawr Nov 27 at 23:10

Haskell, 42 bytes

f n=n?n

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A fairly direct implementation that recurses over 2 variables.

Here's how we can obtain this solution. Start with code implementing a direct recursive formula:

54 bytes

f n=n%n

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Using flawr's rook move interpretation ,a%b is the number of paths that get the rook from (a,b) to (0,0), using only moves the decrease a coordinate. The first move either decreases a or decreases b, keeping the other the same, hence the recursive formula.

49 bytes

f n=n%n

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We can avoid the repetition in map(a%)[0..b-1]++map(b%)[0..a-1] by noting that the two halves are the same with a and b swapped. The auxiliary call a?b counts the paths where the first move decreases a, and so b?a counts those where the first move decreases b. These are in general different, and they add to a%b.

The summation in a?b can also be written as a list comprehension a?b=sum[a%i|i<-[0..b-1]].

42 bytes

f n=n?n

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Finally, we get rid of % and just write the recursion in terms of ? by replacing a%i with a?i+i?a in the recursive call.

The new base case causes this ? to give outputs double that of the ? in the 49-byte version, since with 0?0=1, we would have 0%0=0?0+0?0=2. This lets use define f n=n?n without the halving that we'd other need to do.

  • Your 49-byte solution uses the same recursion as my answer, but I haven't yet figured out the 42-byte one. An explanation would be nice. – Peter Taylor Nov 28 at 8:57
  • I think I used the same approach in one of my earlier programs: The idea is generating (or counting) all partitions by generating the non-horizontal lines from right to left. You start out with the vertical line. Then you can recurse: Take one of the end nodes of the previous line and connect it to a node on the opposite horizontal line that is farther to the left of all previous nodes on this line. – flawr Nov 28 at 9:14
  • The operator a%b counts the number of partitions using the nodes 0,1,...,a on the top line, and the nods 0,1,..,b on the bottom line. The operator a?b counts the number of ways you can add a new line from the top node a if the bottom node b is already in use. (You can connect a to all of the nodes [0,1,...,b-1], but you then have to recurse for each of those.) – flawr Nov 28 at 9:14
  • @flawr, that's the 49-byte one, which I understand. It's the ? of the 42-byte one which I don't, and what's particularly curious is that it's not symmetric. – Peter Taylor Nov 28 at 11:12
  • @PeterTaylor Sorry for the confusion, I somehow mixed up the two solutions. I think we can quite easily transform the two solutions into eachother: In the first step we can replace map... with a list comprehension, in the second step we just plug in the definition of %: a?b=sum$map(a%)[0..b-1], a%b=a?b+b?a \$ \iff \$ a?b=sum[a%i|i<-[0..b-1]], a%b=a?b+b?a \$ \iff \$ a?b=sum[a?i+i?a|i<-[0..b-1]] – flawr Nov 28 at 19:24

CJam (24 bytes)


Online demo

This uses Bubbler's approach of summing over permutations of n 0s and n 1s.


{         e# Define a block
  2,*     e#   Given input n, create an array of n 0s and n 1s
  e!      e#   Generate all permutations of that array
  {       e#   Map:
    e`    e#     Run-length encode
    0f=:( e#     Extract just the lengths and decrement them
    1b    e#     Sum
    2\#   e#     Raise 2 to the power of that sum
  1b      e#  Sum the mapped values

Alternative approach (28 bytes)


Online demo


The triangles all have one horizontal edge and two edges which link the horizontal lines. Label the non-horizontal edges by a tuple of their two x-coords and sort lexicographically. Then the first edge is (0,0), the last edge is (n,n), and two consecutive edges differ in precisely one of the two positions. This makes for a simple recursion, which I've implemented using the memoised recursion operator j:

{            e# Define a block
  _          e#   Duplicate the argument to get n n
  1aa        e#   Base case for recursion: 0 0 => 1
  {          e#   Recursive body taking args a b
    _2$,f{j} e#     Recurse on 0 b up to a-1 b
    @@,f{j}  e#     Recurse on a 0 up to a b-1
    +1b      e#     Combine and sum
  }2j        e#   Memoised recursion with 2 args


This is not the first time I've wanted fj to be supported in CJam. Here it would bring the score down to 24 bytes also. Perhaps I should try to write a patch...

  • Yay, I beat you by 10 seconds, I don't think I was ever that close :) – flawr Nov 27 at 22:32
  • @flawr, I did consider posting before writing a dissection, but I thought I had time to knock it out quickly. Then I saw "New answer", so I deleted my part-written dissection, posted, and edited. – Peter Taylor Nov 27 at 22:34
  • 1
    Thanks for -5 bytes btw :D – flawr Nov 27 at 22:47

Jelly, 15 14 bytes


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-1 byte based on Peter Taylor's comment.

Uses flawr's illustration directly, instead of the resulting formula.

How it works

Ø.xŒ!QŒɠ€’§2*S    Main link (monad). Input: positive integer N.
Ø.x               Make an array containing N zeros and ones
   Œ!Q            All unique permutations
      Œɠ€         Run-length encode on each permutation
         ’§       Decrement and sum each
           2*S    Raise to power of 2 and sum

Take every possible route on a square grid. The number of ways to move L units in one direction as a rook is 2**(L-1). Apply this to every route and sum the number of ways to traverse each route.

  • Nice approach. When I ported it to CJam it was shorter to decrement the lengths, sum, and then raise 2 to the sum; rather than raising 2 to the length, halving, and then multiplying. Don't know whether it might save you a byte. – Peter Taylor Nov 28 at 11:27

Charcoal, 44 31 bytes

crossed out 44 is still regular 44


Try it online! Explanation: Works by calculating the number of ways to partition a trapezium of opposite side lengths m,n into triangles which all lie on integer offsets. This is simply a general case of the rectangle of size n in the question. The number of partitions is given recursively as the sums of the numbers of partitions for all sides m,0..n-1 and n,0..m-1. This is equivalent to generalised problem of the rooks, OEIS A035002. The code simply calculates the number of partitions working from 0,0 up to n,n using the previously calculated values as it goes.


Loop over the rows 0..n.


Start with an empty row.


Loop over the columns in the row 0..n.


Take the row so far and the values in the previous rows at the current column, and add the sum total to the current row. However, if there are no values at all, then substitute 1 in place of the sum.


Add the finished row to the list of rows so far.


Output the final value calculated.

JavaScript (ES6),  45 44  42 bytes

Uses the recursive formula found by Peter Taylor and flawr.


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Pari/GP, 43 bytes

According to OEIS, the generating function of this sequence is



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Python 3, 51 bytes

lambda n:-~n*(n<2)or(10-6/n)*f(n-1)-(9-18/n)*f(n-2)

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Port of flawr's answer

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