tofael1104's blog

By tofael1104, 4 weeks ago, In English

We have numbers from $$$\ a_1, \ldots, a_n$$$, where each number is between $$$(0 \leq a_i < 2^k)$$$. We're trying to find a sequence of these numbers $$$(1 \leq i_1 < i_2 < \ldots < i_T \leq n)$$$ , where $$$T$$$ is $$$(T \geq 1)$$$ , and when we combine them using XOR $$$(a_{i_1} \oplus \ldots \oplus a_i)$$$, the result is $$$0$$$. This problem is well-known and can be solved using Gaussian Elimination technique($$$\mathcal{O}(\frac{nk^2}{w} + \frac{k^3}{w})$$$) in $$$(O(k^2))$$$ time.

But what if we want to find the smallest number of elements $$$T$$$ (still with $$$(T \geq 1.)$$$)? So far, I've only thought of a method that takes about $$$(O^*(n^k))$$$ time by trying all subsets with sizes up to $$$k$$$ and checking if their XOR sum is $$$0$$$.

Can we find a faster way, maybe something like $$$(O(\text{polylog}(n, k)))$$$? Any help would be appreciated, thank you.

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4 weeks ago, # |
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Ignore this. I misunderstood the problem.

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4 weeks ago, # |
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I remember a problem that was something like this. You can do the multiplication of (1 + x^a_i * y) mod y^(size of base + 1), where y^i * y^j = y^(i+j) and x^i * x^j = x^(i xor j). This instantly already gives us a O(2^k * k^2) solution I think, maybe k^something higher instead of k^2. The problem had a more specific setting, iirc it was finding the number of subsets of size multiple of M (with M <= 20) that had xor 0. Perhaps someone remembers the exact problem and is able to link it, I think it's a problem from a camp that tdas participated and asked about the problem to me during upsolving, which would mean some Petrozavodsk camp or Osijek from maybe 2-3 years ago.

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4 weeks ago, # |
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According to this paper, the problem is NP-hard, so if n and k are large you're out of luck. However, you can do a bit better ($$$\mathcal O\left(n 2^k\right)$$$) with a simple DP: for each $$$0 \le i \le n$$$, $$$0 \le a < 2^k$$$, let $$$dp_{i,a}$$$ be the minimum positive number of the first $$$i$$$ elements you need to get an XOR of a. $$$dp_{0,a} = \infty$$$, $$$dp_{i,a_i} = 1$$$, and otherwise $$$dp_{i,a} = \min \left ( dp_{i-1,a},1+dp_{i-1,a \oplus a_i} \right ) $$$. The answer is $$$dp_{n,0}$$$.