Anchored bidirectional sequence gain #
In the general case of paper Step 4, the position carrying the smallest scale need not be the first position. The paper then sums "left and right separately" from that position. This file implements exactly that orientation:
- the left word starts at the anchor and reads the original word backwards;
- the right word starts at the anchor and reads the original word forwards;
- positions past either endpoint repeat the endpoint, hence contribute gain one;
- one global exception set is partitioned and reindexed into the two outward words.
For a fixed anchor, the original word injects into the pair consisting of
its padded left and right words. The two unrestricted finite sums can
therefore be bounded independently by the rectangular form of Lemma 5.13.
Finally the at most m anchor positions cost only another exponential
factor.
Words read outwards from an anchor #
Reindexing one global exception set #
Exceptions on original edges strictly left of the anchor, reflected.
Equations
- Anderson4D.anchoredLeftExceptions a E = Finset.image (Anderson4D.anchoredLeftExceptionIndex a) ({j ∈ E | ↑j < ↑a})
Instances For
Exceptions on original edges at or right of the anchor, translated.
Equations
- Anderson4D.anchoredRightExceptions a E = Finset.image (Anderson4D.anchoredRightExceptionIndex a) ({j ∈ E | ↑a ≤ ↑j})
Instances For
Dyadic gains and the two outward products #
At a genuine original left edge, the outward-left word gives the reverse
ratio: its current endpoint is j+1 and its next endpoint is j.
At a genuine original right edge, the outward-right ratio is forward.
Equations
- Anderson4D.anchoredLeftCodeWeight θ e E a w = ∏ j ∈ Finset.univ \ Anderson4D.anchoredLeftExceptions a E, Anderson4D.anchoredCodeEdgeGain θ e (Anderson4D.anchoredLeftWord a w) j
Instances For
Equations
- Anderson4D.anchoredRightCodeWeight θ e E a w = ∏ j ∈ Finset.univ \ Anderson4D.anchoredRightExceptions a E, Anderson4D.anchoredCodeEdgeGain θ e (Anderson4D.anchoredRightWord a w) j
Instances For
The bidirectional weight: both products start at the same anchor, the left one in reverse original order and the right one in original order.
Equations
- Anderson4D.anchoredBidirectionalCodeWeight θ e E a w = Anderson4D.anchoredLeftCodeWeight θ e E a w * Anderson4D.anchoredRightCodeWeight θ e E a w
Instances For
Fixed-anchor and all-anchor bounds #
For a fixed anchor, left and right are bounded separately by Lemma 5.13. The map from the original word to the pair of outward words is injective, so enlarging to all pairs is a valid nonnegative overcount.
Sum also over the choice of anchor position. Since there are n+1
positions and n+1 ≤ 2^(n+1), that final choice is absorbed into the
universal exponential base.
Reindexing to the active paper carrier #
One consecutive-pair gain on the actual active P carrier.
Instances For
Actual-carrier form of the two products read outwards from the anchor.
Equations
- One or more equations did not get rendered due to their size.
Instances For
Assembly-facing active-P form for an arbitrary positive exponent and
fixed global exception budget. The constant is outside the tree, marking,
multiplicity, length, exception, anchor, and word quantifiers.
The local inequalities initially produce the 1/8 gain. This specialization
is retained for the inner-estimate boundary.
Counting interface at the 1/16 exponent used by the separately proved
even/odd geometric-mean assembly in Step 4.
Counting interface at the paper's weakened 1/20 exponent, for use after
the separately proved outer/inner Hölder allocation.
The literal paper-length specialization. Here n = m - 1 is the number
of adjacent edges and the outer sum ranges over all m possible anchor
positions.
Paper specialization #
The generic theorem above handles an arbitrary anchor and one global
exception set of any fixed budget. The 1/8, budget-100 result is
stronger than the paper's intermediate 20/40-exception uses. In
particular, no pointwise comparison between a forward and a reverse gain
is used: the two outward finite sums are independent applications of
Lemma 5.13.