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Consider the 3SUM problem: given a set $S$ of $n$ numbers identify $x$,$y$,$z$, s.t $x + y = z$. It is believed that the simple $O(n^2)$ algorithm is the best possible; reductions from 3SUM have been used to prove lower bounds for other problems assuming 3SUM is hard [Patrascu 2010].

Is it possible to prove that a small amount $n^\delta$ ($\delta < 1$) of free cache does not help for 3SUM? (An algorithm can store information in the cache and access it for free while accesses to regular RAM cost 1.)

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I don't understand the question. Is the cache supposed to be read-only information chosen prior to the input (i.e., "non-uniform advice")? Or can one both read and write to these $n^{\delta}$ bits, at literally no cost? Using "almost-linear" hashing, you can reduce 3SUM on $n$ numbers to $O(t^2)$ instances of 3SUM on $O(n/t)$ numbers. So if you can really access this $n^{\delta}$ space for free, then by setting $t$ so that $(n/t)\log n \approx n^{\delta}$ you get a $n^{2-\varepsilon}$ 3SUM algorithm, as the $O(t^2)$ instances would take about $O(n/t)$ time each. –  Ryan Williams Dec 5 '12 at 10:19

1 Answer 1

Well, sure, it's possible. But since we basically don't know how to prove that anything is hard, it doesn't look easy.

In the mid 90s I proved an $\Omega(n^2)$ lower bound for 3SUM in a variant of the linear decision tree model; my proof was later simplified by Ailon and Chazelle. This model of computation has nothing corresponding to "memory"; the running time of an algorithm is defined simply as the worst-case number of (generalized) comparisons. So adding a cache won't break these lower bounds; to beat $O(n^2)$ time, you have to do some non-obvious computation with the data in the cache.

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