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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Size and depth of a balanced cluster tree

Statement

Let $I$ be a finite ordered set of size $n \geq 1$ , and let $b$ be an integer with $1 \leq b \leq n$ . In the balanced cluster tree of Finite cluster trees and geometric admissibility, every cluster at depth $h$ has size at most $⌈ n / 2^{h} ⌉$ . The tree has depth at most $⌈ \log_{2} (n / b) ⌉$ , at most $2 n / b$ leaves, and at most $4 n / b - 1$ nodes.

Facts & Assumptions

Given: The finite ordered index set $I$ , its size $n$ , the leaf threshold $b$ , and the balanced cluster tree.

[F1]

Finite cluster trees and geometric admissibility: each cluster larger than $b$ splits into two disjoint children of sizes $⌊ m / 2 ⌋$ and $⌈ m / 2 ⌉$ , and clusters of size at most $b$ are leaves.

Proof

1.1givenF1inductionalgebra

At depth zero the root has size $n = ⌈ n / 2^{0} ⌉$ . If a cluster at depth $h$ has size at most $⌈ n / 2^{h} ⌉$ , each child has size at most $⌈ ⌈ n / 2^{h} ⌉ / 2 ⌉ = ⌈ n / 2^{h + 1} ⌉$ , since the split sizes differ by at most one. Induction proves the size bound at every depth.

2.1givenF1step 1.1algebra

Put $H = ⌈ \log_{2} (n / b) ⌉$ . Then $2^{H} \geq n / b$ , so $⌈ n / 2^{H} ⌉ \leq b$ . By step 1.1 every cluster at depth $H$ has size at most $b$ and is a leaf; hence the tree depth is at most $H$ .

3.1givenF1algebra∎

If a leaf is not the root, its parent has an integer size $m > b$ , so the smaller child has size $⌊ m / 2 ⌋ \geq ⌊ (b + 1) / 2 ⌋ \geq b / 2$ . The same lower bound on leaf size holds when the root is a leaf because then there is one leaf and $n \geq b / 2$ . Since the leaves partition $I$ , their number $L$ satisfies $L b / 2 \leq n$ , hence $L \leq 2 n / b$ . A full binary tree with $L$ leaves has $L - 1$ internal nodes, so it has $2 L - 1 \leq 4 n / b - 1$ nodes.

Remarks

The bounds include $b = 1$ and $b = n$ . For $b = n$ the root alone is a leaf; for $b = 1$ every leaf is a singleton.

Depends on

Finite cluster trees and geometric admissibility

Used by

Balanced H and H² storage and matvec complexity Theorem

Dependency tree · one level

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Sources

Steffen Börm, Lars Grasedyck, and Wolfgang Hackbusch, Hierarchical Matrices, MiS Lecture Note 21 (2003; revised June 2006), §§2.1.1–2.1.2 (standard reference, not scraped)
Ralf Hiptmair, AdvNumCSE Lecture Document, Chapter 2, “Local Low-Rank Compression of Non-Local Operators” (ETH Zürich, AT’23), §2.3.2 (standard reference, not scraped)