AAhaoxuan's blog

I have a problem
By AAhaoxuan, history, 4 months ago, In English

Given a sequence $$$( A )$$$ , where each element $$$( A_i )$$$ is a randomly chosen integer from $$$( 0 )$$$ to $$$( 2^k - 1 )$$$ , what is the probability that there exists a subsequence of $$$( A )$$$ whose XOR sum is equal to $$$0$$$ ?

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4 months ago, # |
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$$$\dfrac{1}{2^k}$$$

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    4 months ago, # ^ |
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    Can you please explain how?

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      4 months ago, # ^ |
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      Here's what I think: we notice that the each binary bit are independent, so we only need to calculate the probability of each bit.

      To make a bit's xor sum to $$$0$$$, you have to choose an even number of positions to be $$$0$$$ and the rest to be $$$1$$$, and the probability of such a number is

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      . Multiply the probabilities of all the bits together to give the answer: $$$\dfrac{1}{2^k}$$$
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    4 months ago, # ^ |
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    If $$$n=1$$$ ，It's right.

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4 months ago, # |
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Let $$$n$$$ be the length of $$$A$$$. Numbers from $$$0$$$ to $$$2^k-1$$$ with $$$xor$$$ is a $$$k$$$-dimensional vector space over $$$\mathbb{F}_2$$$, therefore if $$$n > k$$$ it is guaranteed that $$$A$$$ has $$$0$$$-sum subsequence, because $$$A$$$ is linearly dependent collection of vectors.

If $$$n\leq k$$$ we can calculate the probability by counting number of linearly independent collections of length $$$n$$$: there are $$$2^k - 1$$$ ways to choose the fisrt vector, $$$2^k-2$$$ ways for the second... $$$2^k-2^t$$$ for the $$$t+1$$$'th vector, so overall number is $$$\prod_{i=1}^{n}(2^k-2^{i-1})$$$.

Then the needed probability is $$$1- \frac{\prod_{i=1}^{n}(2^k-2^{i-1})}{2^{nk}}$$$.

Correct me if I am wrong.

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    4 months ago, # ^ |
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    Thank you! It help me prove the probability of the Problem 799F randomization algorithm being correct.

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    4 months ago, # ^ |
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    But the if $$$n\le k$$$，$$$2^k-2^{i-1}$$$ may be less than $$$0$$$.

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      4 months ago, # ^ |
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      Minimum of $$$2^k - 2^{i-1}$$$ reached when $$$i=n$$$ and $$$n = k$$$, so it's $$$2^k-2^{k-1} = 2^{k-1} > 0$$$.

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