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Use xor for common route #1

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Use xor for common route #1

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25 changes: 12 additions & 13 deletions src/tree.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -121,25 +121,24 @@ impl HashTree {
debug!("Current path is {:032b} ({})", self.path, self.path);
debug!("Target path is {:032b} ({})", path, path);

// We do (p1 AND p2) OR (!p1 AND !p2) to get a common route
// represented with 1s, the first 0 represents the first
// forking node
// We do (p1 XOR p2) to get a common route represented
// with 0s, the first 1 represents the first forking node
//
// Examples
// (000 AND 001) OR (111 AND 110) = 110
// (001 AND 101) OR (110 AND 010) = 011
// (101 AND 111) OR (010 AND 000) = 101
// (000 XOR 001) = 001
// (001 XOR 101) = 100
// (101 XOR 111) = 010
//
// Note: leading zeros of !p1 and !p2 will prepend our result
// with 1s, this is useful as we are looking for the first 0
let common = (self.path & path) | (!self.path & !path);
// Note: leading zeros of (p1 XOR p2) will prepend our result
// with 0s, this is useful as we are looking for the first 1
let common = self.path ^ path;
debug!("Common path is {:032b}", common);

let ones = common.leading_ones() as usize;
debug!("Reusable path is {} of {MAX_DEPTH}; useful = {}", ones, self.depth);
let zeros = common.leading_zeros() as usize;
debug!("Reusable path is {} of {MAX_DEPTH}; useful = {}", zeros, self.depth);

// We now just have to calculate first 0 index (first forking node)
let n = self.depth - (MAX_DEPTH - ones) as u8;
// We now just have to calculate first 1 index (first forking node)
let n = self.depth - (MAX_DEPTH - zeros) as u8;
debug!("Reusing {} nodes", n);

// Assigning new path
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