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In the Art of Computer Programming, on the bottom of page 485

Suppose there's a B-tree of order m, and there are N keys, so the N+1 leaves appear on level l.

The numbers of nodes on levels 1,2,3... is at least 2,2[m/2],2[m/2]^2...

(Here [] denotes upper bound)

And Knuth give

N+1 >= 2[m/2]^(l-1)

My question is:

Shouldn't this be N+1 >= 2+2[m/2]+2[m/2]^2+...+2[m/2]^(l-1)?

What's the point of only taking nodes of the (l-1)th level into account?

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Rui Liu
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1 Answers1

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A branch node with k keys has k + 1 children. So however many nodes there are on level l - 1, there must be more nodes on level l.

So N + 1 (the number of nodes on level l) is greater than the number of nodes on level l - 1. Obviously the actual number of nodes on level l - 1 is greater than or equal to the minimum number of nodes on level l - 1. So N + 1 ≥ 2⌈m / 2⌉l - 1.

rob mayoff
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