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Paper:

TR19-162 | 15th November 2019 20:26

The Random-Query Model and the Memory-Bounded Coupon Collector

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TR19-162
Authors: Ran Raz, Wei Zhan
Publication: 15th November 2019 20:33
Downloads: 120
Keywords: 


Abstract:

We study a new model of space-bounded computation, the {\it random-query} model. The model is based on a branching-program over input variables $x_1,\ldots,x_n$. In each time step, the branching program gets as an input a random index $i \in \{1,\ldots,n\}$, together with the input variable $x_i$ (rather than querying an input variable of its choice, as in the case of a standard (oblivious) branching program). We motivate the new model in various ways and study time-space tradeoff lower bounds in this model.

Our main technical result is a quadratic time-space lower bound for zero-error computations in the random-query model, for XOR, Majority and many other functions. More precisely, a zero-error computation is a computation that stops with high probability and such that conditioning on the event that the computation stopped, the output is correct with probability~1. We prove that for any Boolean function $f: \{0,1\}^n \rightarrow \{0,1\}$, with sensitivity $k$, any zero-error computation with time $T$ and space $S$, satisfies
$T\cdot (S+\log n) \geq \Omega(n \cdot k)$. We note that the best time-space lower bounds for standard oblivious branching programs are only slightly super linear and improving these bounds is an important long-standing open problem.

To prove our results, we study a memory-bounded variant of the coupon-collector problem that seems to us of independent interest and to the best of our knowledge has not been studied before. We consider a zero-error version of the coupon-collector problem. In this problem, the coupon-collector could explicitly choose to stop when he/she is sure with zero-error that
all coupons have already been collected. We prove that any zero-error coupon-collector that stops with high probability in time $T$, and uses space $S$, satisfies $T\cdot (S+\log n) \geq \Omega(n^2)$, where $n$ is the number of different coupons.



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