I don't understand then.

You stated the condition for a match as ($A & $B) == 1 where $A and $B are (64-bit, unsigned) integers.
Let $As = $A >> 1 and $Bs = $B >> 1.
Isn't then the original condition equivalent to ($As & $Bs) == 0 ?

For a given $A (and $As), what other $B (and $Bs) does satisfy that condition than $Bs = ~$As?

In reply to Re^5: Need a faster way to find matches by kikuchiyo
in thread Need a faster way to find matches by remzak

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