How to, given a predetermined set of keys, reorder the keys such that the minimum number of nodes are used when inserting into a B-Tree?

Question

So I have a problem which i'm pretty sure is solvable, but after many, many hours of thinking and discussion, only partial progress has been made.

The issue is as follows. I'm building a BTree of, potentially, a few million keys. When searching the BTree, it is paged on demand from disk into memory, and each page in operation is relatively expensive. This effectively means that we want to need to traverse as few nodes as possible (although after a node has been traversed to, the cost of traversing through that node, up to that node is 0). As a result, we don't want to waste space by having lots of nodes nearing minimum capacity. In theory, this should be preventable (within reason) as the structure of the tree is dependent on the order that the keys were inserted in.

So, the question is how to reorder the keys such that after the BTree is built the fewest number of nodes are used. Here's an example:

I did stumble on this question In what order should you insert a set of known keys into a B-Tree to get minimal height? which unfortunately asks a slightly different question. The answers, also don't seem to solve my problem. It is also worth adding that we want the mathematical guarantees that come from not building the tree manually, and only using the insert option. We don't want to build a tree manually, make a mistake, and then find it is unsearchable!

I've also stumbled upon 2 research papers which are so close to solving my question but aren't quite there! Time-and Space-Optimality in B-Trees and Optimal 2,3-Trees (where I took the above image from in fact) discuss and quantify the differences between space optimal and space pessimal BTrees, but don't go as far as to describe how to design an insert order as far as I can see.

Any help on this would be greatly, greatly appreciated.

Thanks

Research papers can be found at:

http://www.uqac.ca/rebaine/8INF805/Automne2007/Sujets2007Automne/p174-rosenberg.pdf

http://scholarship.claremont.edu/cgi/viewcontent.cgi?article=1143&context=hmc_fac_pub

EDIT:: I ended up filling a btree skeleton constructed as described in the above papers with the FILLORDER algorithm. As previously mentioned, I was hoping to avoid this, however I ended up implementing it before the 2 excellent answers were posted!

Answer 1

The algorithm below should work for B-Trees with minimum number of keys in node = d and maximum = 2*d I suppose it can be generalized for 2*d + 1 max keys if way of selecting median is known.

Algorithm below is designed to minimize the number of nodes not just height of the tree.

Method is based on idea of putting keys into any non-full leaf or if all leaves are full to put key under lowest non full node.

More precisely, tree generated by proposed algorithm meets following requirements: It has minimum possible height; It has no more then two nonfull nodes on each level. (It's always two most right nodes.)

Since we know that number of nodes on any level excepts root is strictly equal to sum of node number and total keys number on level above we can prove that there is no valid rearrangement of nodes between levels which decrease total number of nodes. For example increasing number of keys inserted above any certain level will lead to increase of nodes on that level and consequently increasing of total number of nodes. While any attempt to decrease number of keys above the certain level will lead to decrease of nodes count on that level and fail to fit all keys on that level without increasing tree height. It also obvious that arrangement of keys on any certain level is one of optimal ones. Using reasoning above also more formal proof through math induction may be constructed.

The idea is to hold list of counters (size of list no bigger than height of the tree) to track how much keys added on each level. Once I have d keys added to some level it means node filled in half created in that level and if there is enough keys to fill another half of this node we should skip this keys and add root for higher level. Through this way, root will be placed exactly between first half of previous subtree and first half of next subtree, it will cause split, when root will take it's place and two halfs of subtrees will become separated. Place for skipped keys will be safe while we go through bigger keys and can be filled later.

Here is nearly working (pseudo)code, array needs to be sorted:

PushArray(BTree bTree, int d, key[] Array)
{
    List<int> counters = new List<int>{0};
    //skip list will contain numbers of nodes to skip 
    //after filling node of some order in half
    List<int> skip  = new List<int>();
    List<Pair<int,int>> skipList = List<Pair<int,int>>();

    int i = -1;
    while(true)
    {     
       int order = 0;
       while(counters[order] == d) order += 1;
       for(int j = order - 1; j >= 0; j--) counters[j] = 0;

       if (counters.Lenght <= order + 1) counters.Add(0);
       counters[order] += 1;

       if (skip.Count <= order)
              skip.Add(i + 2);    
       if (order > 0)
           skipList.Add({i,order}); //list of skipped parts that will be needed later
       i += skip[order];

       if (i > N) break;

       bTree.Push(Array[i]);
    }

    //now we need to add all skipped keys in correct order
    foreach(Pair<int,int> p in skipList)
    {
        for(int i = p.2; i > 0; i--)
            PushArray(bTree, d, Array.SubArray(p.1 + skip[i - 1], skip[i] -1))
    }
}

Example:

Here is how numbers and corresponding counters keys should be arranged for d = 2 while first pass through array. I marked keys which pushed into the B-Tree during first pass (before loop with recursion) with 'o' and skipped with 'x'.

                                                              24
        4         9             14             19                            29 
0 1 2 3   5 6 7 8   10 11 12 13    15 16 17 18    20 21 22 23    25 26 27 28    30 ...
o o x x o o o x x o  o  o  x  x  x  x  x  x  x  x  x  x  x  x  o  o  o  x  x  o  o ...
1 2     0 1 2     0  1  2                                      0  1  2        0  1 ...
0 0     1 1 1     2  2  2                                      0  0  0        1  1 ...
0 0     0 0 0     0  0  0                                      1  1  1        1  1 ...
skip[0] = 1 
skip[1] = 3 
skip[2] = 13

Since we don't iterate through skipped keys we have O(n) time complexity without adding to B-Tree itself and for sorted array;

In this form it may be unclear how it works when there is not enough keys to fill second half of node after skipped block but we can also avoid skipping of all skip[order] keys if total length of array lesser than ~ i + 2 * skip[order] and skip for skip[order - 1] keys instead, such string after changing counters but before changing variable i might be added:

while(order > 0 && i + 2*skip[order] > N) --order;

it will be correct cause if total count of keys on current level is lesser or equal than 3*d they still are split correctly if add them in original order. Such will lead to slightly different rearrangement of keys between two last nodes on some levels, but will not break any described requirements, and may be it will make behavior more easy to understand.

May be it's reasonable to find some animation and watch how it works, here is the sequence which should be generated on 0..29 range: 0 1 4 5 6 9 10 11 24 25 26 29 /end of first pass/ 2 3 7 8 14 15 16 19 20 21 12 13 17 18 22 23 27 28