Proof of the one-dimensional capacity theorem #
Fix a split U with |U| = a and process the blocks of a decomposition in order, keeping a
rectangle P × Q on which the current history is a function of the second side q alone.
- A one-dimensional block with
T ≥ 1changes isΨ_q(α(p))on the rectangle, where⟨w, glue p q⟩ = α(p) + β(q). The sorted-group lemma with2Tgroups keeps a group of⌊|P| / 2T⌋elements ofPand at least half ofQ. - A summary block with
Vvalues keeps the largest fiber ofσ, of⌊|P| / V⌋elements.
At the end the output is a function of q on P × Q; keep a majority value on at least half of
Q. Choosing a = ⌈log₂ (K Dₚ)⌉, where Dₚ is the product of the first-side factors, both sides
have at least K elements whenever 4 K² · cost ≤ 2ⁿ, contradicting two-sided
rectangle-freeness.
The weighted sum of a glued input splits into the two sides.
The factor by which a block shrinks the first side.
Equations
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The factor by which a block shrinks the second side.
Equations
- (Algebraic.Threshold.BlockKind.oneDim T).qcost = if T = 0 then 1 else 2
- (Algebraic.Threshold.BlockKind.summary V).qcost = 1
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After the first i blocks, some rectangle with sides of at least NP and NQ elements
has the history determined by the second side.
Equations
- One or more equations did not get rendered due to their size.
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Theorem U (one-dimensional capacity). If f is two-sided K-rectangle-free, every
block decomposition of f has 2ⁿ < 4 K² · cost.