Tournament bracket placement algorithm

Question

Given a list of opponent seeds (for example seeds 1 to 16), I'm trying to write an algorithm that will result in the top seed playing the lowest seed in that round, the 2nd seed playing the 2nd-lowest seed, etc.

Grouping 1 and 16, 2 and 15, etc. into "matches" is fairly easy, but I also need to make sure that the higher seed will play the lower seed in subsequent rounds.

An example bracket with the correct placement:

1 vs 16             1 vs 8 8 vs 9                         1 vs 4 4 vs 13             4 vs 5 5 vs 12                                     1 vs 2 2 vs 15             2 vs 7 7 vs 10                         2 vs 3 3 vs 14             3 vs 6 6 vs 11

As you can see, seed 1 and 2 only meet up in the final.

Answer 1

This JavaScript returns an array where each even index plays the next odd index

function seeding(numPlayers){   var rounds = Math.log(numPlayers)/Math.log(2)-1;   var pls = [1,2];   for(var i=0;i<rounds;i++){     pls = nextLayer(pls);   }   return pls;   function nextLayer(pls){     var out=[];     var length = pls.length*2+1;     pls.forEach(function(d){       out.push(d);       out.push(length-d);     });     return out;   } }  > seeding(2) [1, 2] > seeding(4) [1, 4, 2, 3] > seeding(8) [1, 8, 4, 5, 2, 7, 3, 6] > seeding(16) [1, 16, 8, 9, 4, 13, 5, 12, 2, 15, 7, 10, 3, 14, 6, 11] 

Answer 2

With your assumptions, players 1 and 2 will play in the final, players 1-4 in the semifinals, players 1-8 in the quarterfinals and so on, so you can build the tournament recursively backwards from the final as AakashM proposed. Think of the tournament as a tree whose root is the final.

In the root node, your players are {1, 2}.

To expand the tree recursively to the next level, take all the nodes on the bottom layer in the tree, one by one, and create two children for them each, and place one of the players of the original node to each one of the child nodes created. Then add the next layer of players and map them to the game so that the worst newly added player plays against the best pre-existing player and so on.

Here first rounds of the algorithm:

 {1,2}  --- create next layer         {1, _}       /         --- now fill the empty slots  {1,2}       \{2, _}         {1, 4}   --- the slots filled in reverse order       /           {1,2}       \{2, 3}   --- create next layer again                /{1, _}        {1, 4}       /      \{4, _}  {1,2}                  --- again fill       \      /{2, _}        {2, 3}              \{3, _}               /{1, 8}        {1, 4}       /      \{4, 5}    --- ... and so on  {1,2}       \      /{2, 7}        {2, 3}              \{3, 6} 

As you can see, it produces the same tree you posted.