How Many Edges and Keys in Specific Binomial Min-Heaps?

[zeion]

Homework Statement

a) What is the exact number of edges in a Binomial Min-Heap with 255 nodes?
b) Consider a Binomial Min-Heap with 4097 distinct keys. What is the exact maximum number of keys that must be examined to find the minimum key?

Homework Equations

The Attempt at a Solution

a) So I will determine what trees are used by converting to binary 255 -> 11111111
So all trees from 2^0 to 2^7 are used. Each tree has 0, 1, 3, 7, ... edges total 63 edges.

b) 4097 -> 1000...01 so there are only two trees so the answer is 2 nodes.

Is it correct? Thanks.

 

[KataKoniK]

a) Your approach is correct. Since a Binomial Min-Heap with 255 nodes would have a binary representation of 11111111, it would consist of 8 trees with 0, 1, 3, 7, 15, 31, 63, and 127 nodes respectively. Each of these trees has a total of 2^(n-1) edges, where n is the number of nodes in the tree. Therefore, the total number of edges in the Binomial Min-Heap would be 2^(0-1) + 2^(1-1) + 2^(3-1) + 2^(7-1) + 2^(15-1) + 2^(31-1) + 2^(63-1) + 2^(127-1) = 1 + 1 + 3 + 7 + 15 + 31 + 63 + 127 = 248 edges.

b) Your approach for part b is incorrect. Since a Binomial Min-Heap with 4097 distinct keys would have a binary representation of 1000...01, it would consist of two trees with 1 and 4096 nodes respectively. To find the minimum key in this heap, we would need to examine all 4097 keys, since the minimum key could be in either of the two trees. Therefore, the exact maximum number of keys that must be examined to find the minimum key is 4097.

I hope this helps clarify your understanding. Keep up the good work!