Given a fixed number of keys or values(stored either in array or in some data structure) and order of b-tree, can we determine the sequence of inserting keys that would gene
This is how I would add elements to b-tree.
Thanks to Steve314, for giving me the start with binary representation,
Given are n elements to add, in order. We have to add it to m-order b-tree. Take their indexes (1...n) and convert it to radix m. The main idea of this insertion is to insert number with highest m-radix bit currently and keep it above the lesser m-radix numbers added in the tree despite splitting of nodes.
1,2,3.. are indexes so you actually insert the numbers they point to.
For example, order-4 tree
4 8 12 highest radix bit numbers
1,2,3 5,6,7 9,10,11 13,14,15
Now depending on order median can be:
The choice of median (left/right) to be promoted will decide the order in which I should insert elements. This has to be fixed for the b-tree.
I add elements to trees in buckets. First I add bucket elements then on completion next bucket in order. Buckets can be easily created if median is known, bucket size is order m.
I take left median for promotion. Choosing bucket for insertion.
| 4 | 8 | 12 |
1,2,|3 5,6,|7 9,10,|11 13,14,|15
3 2 1 Order to insert buckets.
Example
Bucket median first
12,
Add elements to left
11,12,
Then after all elements inserted it looks like,
| 12 |
|11 13,14,|
Then I choose the bucket left to it. And repeat the same process.
Median
12
8,11 13,14,
Add elements to left first
12
7,8,11 13,14,
Adding rest
8 | 12
7 9,10,|11 13,14,
Similarly keep adding all the numbers,
4 | 8 | 12
3 5,6,|7 9,10,|11 13,14,
At the end add numbers left out from buckets.
| 4 | 8 | 12 |
1,2,|3 5,6,|7 9,10,|11 13,14,|15
For mid-median (even order b-trees) you simply insert the median and then all the numbers in the bucket.
For right-median I add buckets from the left. For elements within the bucket I first insert median then right elements and then left elements.
Here we are adding the highest m-radix numbers, and in the process I added numbers with immediate lesser m-radix bit, making sure the highest m-radix numbers stay at top. Here I have only two levels, for more levels I repeat the same process in descending order of radix bits.
Last case is when remaining elements are of same radix-bit and there is no numbers with lesser radix-bit, then simply insert them and finish the procedure.
I would give an example for 3 levels, but it is too long to show. So please try with other parameters and tell if it works.