claimroutineinferred_dep
Descriptor processing fits within child volume
Descriptor handling at each child is absorbed in
shows descriptor cost adds nothing to the recurrence
Claim·§4.2 Transferring a finite circuit to address permutations·inside propPower-width chunk interchange in O(Ve^τ)
Statement (paper)
Even when the payload is one bit, descriptor processing at a child of depth in a power-width call of width and root volume takes time, where is the volume of that child; in particular, it fits within the additional term at each node. This is because and , so that .
In plain English
Even with a one-bit payload, processing the descriptor at a child of depth in a power-width call of width and root volume takes time, where is that child's volume. So it is absorbed by the additional term at each node.
Why it's here
This shows the descriptor handling left over from the depth-first schedule adds no new cost to the swap recurrence. The reason is that and , so , and a polylogarithm in is therefore bounded by .
Classified routine:
A direct bookkeeping calculation inside the proof. It bounds by using and , so the polylogarithmic descriptor cost is absorbed into the per-node term of the depth-first schedule (the previous claim).
Uses
Used by
Nothing in the paper
Sources
PaperOpen
- Section
- §4.2 Transferring a finite circuit to address permutations
- Page
- —
- Label
—- Line
sections/04-swap.tex:244
Leannot aligned
Phase 2 aligns this node to a Lean declaration.
Friction
- infoinferred_depoutline: "Descriptor processing absorbed by child volume" rests on claim-prop-power-interchange-5