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    <p>The technical term for "no-feedback or blind" is "non-adaptive". O(e log n) calls still suffice, but the algorithm is rather more involved.</p> <p>Instead of testing the evenness of products, we're going to test the evenness of sums. Let E ≠ F be distinct subsets of {1, …, n}. If we have one array x<sub>1</sub>, …, x<sub>n</sub> with even numbers at positions E and another array y<sub>1</sub>, …, y<sub>n</sub> with even numbers at positions F, how many subsets J of {1, …, n} satisfy</p> <p>(∑<sub>i in J</sub> x<sub>i</sub>) mod 2 ≠ (∑<sub>i in J</sub> y<sub>i</sub>) mod 2?</p> <p>The answer is 2<sup>n-1</sup>. Let i be an index such that x<sub>i</sub> mod 2 ≠ y<sub>i</sub> mod 2. Let S be a subset of {1, …, i - 1, i + 1, … n}. Either J = S is a solution or J = S union {i} is a solution, but not both.</p> <p>For every possible outcome E, we need to make calls that eliminate every other possible outcome F. Suppose we make 2e log n calls at random. For each pair E ≠ F, the probability that we still cannot distinguish E from F is (2<sup>n-1</sup>/2<sup>n</sup>)<sup>2e log n</sup> = n<sup>-2e</sup>, because there are 2<sup>n</sup> possible calls and only 2<sup>n-1</sup> fail to distinguish. There are at most n<sup>e</sup> + 1 choices of E and thus at most (n<sup>e</sup> + 1)n<sup>e</sup>/2 pairs. By a union bound, the probability that there exists some indistinguishable pair is at most n<sup>-2e</sup>(n<sup>e</sup> + 1)n<sup>e</sup>/2 &lt; 1 (assuming we're looking at an interesting case where e ≥ 1 and n ≥ 2), so there exists a sequence of 2e log n calls that does the job.</p> <p>Note that, while I've used randomness to show that a good sequence of calls exists, the resulting algorithm is <em>deterministic</em> (and, of course, non-adaptive, because we chose that sequence without knowledge of the outcomes).</p>
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    1. POFinding even numbers in an array without using feedback
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    1. COThanks for the terminology. I'll think more about your solution and get back to you. Thanks for taking the time to help!
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    2. COThis is a lovely proof.
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    3. COI honestly didn't quite follow how to do this. Thanks for the answer. I voted for you but I'm going to accept the other answer.
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