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Powering amplifies a small unsatisfaction gap
Statement
Let , let be a finite alphabet with , and let be a constant. Then there are a constant depending only on , and an integer with the same dependence, such that for every , every binary constraint graph over whose underlying graph is -regular in the adjacency-slot convention with and normalized second eigenvalue bound , and every labeling of the powered graph of Constraint graph powering with local-view labels, In particular, if for a constant then , and for larger gaps the lower bound saturates at . No parity hypothesis on is needed in this parameterization: the powered slots have length and the analysed window is symmetric about the midpoint . The bound is uniform in the labeling .
Facts & Assumptions
Given: integers and ; a finite alphabet with ; a -regular binary constraint graph over with and ; a labeling of the powered graph .
For a labeling of , is one minus the fraction of ordinary edges satisfied, and ; for a labeling of , is the fraction of ordinary edges violated, equivalently the violation probability for a uniformly random edge oriented by its unique copy- incidence slot (Constraint graph and labeling value, Constraint graph powering with local-view labels).
The plurality decoding of satisfies for every , where is the law of the value claimed for by the view at the endpoint of a uniformly random lazy-walk pattern of length from (Plurality decoding of powered local views).
Draw a uniformly random powered edge oriented by its unique copy- incidence slot, equivalently a uniform start and a uniform lazy-walk pattern . With and , for every with and every slot of from to , the probability that both and , conditioned on the -th lazy step of being the option , is at least (Plurality opinions agree with local views in middle positions).
For any set of ordinary edges of with , with the event that the -th step of a uniformly random lazy-walk pattern of length from a uniform start vertex traverses an edge of , one has (Violated-edge positions have controlled collisions).
If finitely many events satisfy for some , then (Overlap control gives a union lower bound).
Proof
If then the right-hand side is zero and there is nothing to prove, so assume . Let be the plurality decoding of and put , so by [F1]; let be the set of edges violated by , of size . Choose a set as follows: if let be any subset of with , and otherwise let for some single violated edge . Writing and , in both cases , and . Put and . Since , this interval lies inside ; since , , and the integer positions give , so .
For in the window of step 1.1 and let be the event that the -th lazy step of the sampled representative walk traverses a slot of the edge carrying the relation in its specified endpoint order (for a loop, either of its two slots), and that both endpoint views report the decoded labels of the endpoints of . In the uniform powered-edge model of [F3] the position before the -th step is uniform, so each of the slots of is traversed by that step with probability , and combining this with the conditional bound of [F3] gives . Moreover implies that the sampled powered edge is violated: the tested pair at position is exactly the pair of decoded labels at the endpoints of , and this pair fails because violates .
Let be the event that the -th lazy step of the sampled representative walk traverses an edge of . If and then , a single step traversing one edge only; otherwise . The consecutive lazy steps in the central sub-window have the law of a length- lazy walk from a uniform start: the walk starts uniformly and the uniform distribution remains stationary at the beginning of this sub-window, while its step choices are independent. Thus [F4] applies to this block and this , and .
Summing the bound of step 2.1 over the positions and the edges gives , and the events and the collision sum are finite, so [F5] is applicable. Applying it with the ratio gives .
Two regimes. If then and hence , where . If then , so and with . Every implies that the sampled powered edge is violated, so , and because for while , both regimes give with and .
Remarks
- The two regimes are the two halves of the promise. For the union bound loses only the constant and delivers ; for the same computation yields a constant lower bound, which dominates because that quantity is at most . This is exactly Dinur's Lemma 6.1 with , and it is what makes the iteration of A complete uniform graph gap-amplification step terminate in rounds.
- The window is a sub-window of the central window. The positions used lie in and number , so every event is a violation of the powered slot relation and the sub-window still gives the gain. Over the full central window the middle-position lemma would lose a constant and the collision term would grow.
- Where each constant comes from. depends only on : through and the factor of the collision bound, and through in the same bound. No dependence on , on , on or on the labeling remains, and the proof chooses nothing beyond the subset and the slot of each edge, both of which are fixed deterministically.
- The statement above is uniform over labelings of , including non-liftable ones; that is the content of the decoding step 1.1, which replaces an arbitrary powered labeling by one base labeling at the cost of the factor in .
Depends on
Used by
Dependency tree · two levels
11 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Irit Dinur, The PCP theorem by gap amplification, §6 Lemma 6.1 and its proof (Lemmas 6.2-6.4), printed pp. 19-24. (standard reference, not scraped)
- Arora and Barak, Computational Complexity: A Modern Approach, §18.5.1 Lemma 18.29 part 3 with Claims 18.32-18.34, printed pp. 372-377. (standard reference, not scraped)