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I am given n = 256 64-bit binary vectors ∈ GF(2)^64. I am also given a k and a target vector T, and I am required to choose exactly k out of the n vectors without repetition such that their sum (mod 2) is T. Such a solution is guaranteed to exist.

Obviously, a meet-in-the-middle attack would give a solution in approximately O(n^(k / 2)). However, I am wondering if the there exists faster solutions when k is really small compared to the number of dimension 64, say for k = 16. The reason I think so is that 16 vectors can only span a 16-dimensional space and so the existence of such solution is a strong information.

I have also thought of reducing the n to 64, since we can always choose 64 vectors that span GF(2)^64. However, I believe that would violate the "exactly k" part of the requirement. Or am I missing something obvious here?

Thank you for reading, any discussion would be helpful.

Aivean
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Gareth Ma
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  • It's a hard problem: https://madhu.seas.harvard.edu/papers/1999/dms-journ.pdf – David Eisenstat Nov 13 '21 at 15:28
  • Unrelated to the dimension but if k is small enough then a simple enumeration of all n choose k choices of k vectors will be faster than meet-in-the-middle. For your example of (n,k)=(256,16), 256 choose 16 is about 2^84, much less than 2^128. – President James K. Polk Nov 13 '21 at 19:19
  • @PresidentJamesK.Polk Won't a meet in the middle attack be 256^(k/2) = 256^8 = 2^64 instead? Thank you for your answer anyways. – Gareth Ma Nov 14 '21 at 01:20

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