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Prove this XOR property
By Red_Before_GTA6, history, 2 years ago, In English

Did you know that the minimum XOR pair of an array can be found by sorting the array, then checking the XOR of adjacent elements.

Can anyone prove why?

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2 years ago, hide # |
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Consider a greedy algorithm on a 0-1 Trie.

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2 years ago, hide # |
 
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this might help

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2 years ago, hide # |
 
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Let's look at all possible pairs of elements in the array. The minimum XOR is achieved in one of those pairs where the length of the common prefix in the bit representation is the longest. This is because it corresponds to the maximum prefix of zeros at the beginning of the XOR. Hence, the minimum XOR is achieved in one of such pairs. Now, notice that for any such pair, it is true that its elements are consecutive in the sorted array because the elements have the form $$$prefix0 \dots$$$ and $$$prefix1 \dots$$$ If there were at least one more element $$$a_i$$$ between them, the common prefix of $$$a_i$$$ with one of the elements of the pair would be greater than the prefix of the pair. $$$(a_i = prefix0\dots\ or\ prefix1\dots)$$$ But we initially considered pairs with the longest matching bit prefix. Contradiction!

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2 years ago, hide # |
 
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does anyone has list of problems like this where you can learn new XOR properties , such as state above.

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https://atcoder.jp/contests/abc308/tasks/abc308_g

you can go through the editorial of this problem

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2 years ago, hide # |
 
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I've understood it from the above comments now. Cheers

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Similar approach can be applied to find a maximal AND pair.

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    2 years ago, hide # ^ |
     
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    Does not work for maximal AND pair. Consider 1 4 9 Here (1 & 4) = 0, (4 & 9) = 0 but maximal AND (1 & 9) = 1

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Given some set of non-negative integers $$$S$$$, we want to show the following:

$$$a \oplus b = \min_{x, y \in S} x \oplus y \implies \not \exists c \in S, \text{such that } a \lt c \lt b$$$

We strive for a contradiction. Suppose there does exist such a $c \in S$. Let $$$i$$$ be the first (leftmost) index such that $$$c_i = 0$$$ and $$$b_i = 1$$$, where $$$x_i$$$ represents the $$$i$$$-the digit from the left in the binary representation of $$$x$$$. Now if $$$a_i = 0$$$, then $$$a \oplus c \lt a \oplus b$$$. If $$$a_i = 1$$$, then there must exist some $$$j \lt i$$$ such that $$$b_j = 1$$$ and $$$a_j = 0$$$ (since $$$a \lt c \lt b$$$). Here, $$$a_j \oplus b_j = 1$$$ but $$$b_j = c_j \implies b_j \oplus c_j = 0$$$. Thus $$$b \oplus c \lt a \oplus b$$$. In both cases, we have arrived at a contradiction. $$$\square$$$

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