83 lines
2.5 KiB
Plaintext
83 lines
2.5 KiB
Plaintext
// @ts-check
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/**
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* We must remap all the old bits to new bits for each set variant
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* Only arbitrary variants are considered as those are the only
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* ones that need to be re-sorted at this time
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*
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* An iterated process that removes and sets individual bits simultaneously
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* will not work because we may have a new bit that is also a later old bit
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* This means that we would be removing a previously set bit which we don't
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* want to do
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*
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* For example (assume `bN` = `1<<N`)
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* Given the "total" mapping `[[b1, b3], [b2, b4], [b3, b1], [b4, b2]]`
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* The mapping is "total" because:
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* 1. Every input and output is accounted for
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* 2. All combinations are unique
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* 3. No one input maps to multiple outputs and vice versa
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* And, given an offset with all bits set:
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* V = b1 | b2 | b3 | b4
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*
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* Let's explore the issue with removing and setting bits simultaneously:
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* V & ~b1 | b3 = b2 | b3 | b4
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* V & ~b2 | b4 = b3 | b4
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* V & ~b3 | b1 = b1 | b4
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* V & ~b4 | b2 = b1 | b2
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*
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* As you can see, we end up with the wrong result.
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* This is because we're removing a bit that was previously set.
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* And, thus the final result is missing b3 and b4.
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*
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* Now, let's explore the issue with removing the bits first:
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* V & ~b1 = b2 | b3 | b4
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* V & ~b2 = b3 | b4
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* V & ~b3 = b4
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* V & ~b4 = 0
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*
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* And then setting the bits:
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* V | b3 = b3
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* V | b4 = b3 | b4
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* V | b1 = b1 | b3 | b4
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* V | b2 = b1 | b2 | b3 | b4
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*
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* We get the correct result because we're not removing any bits that were
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* previously set thus properly remapping the bits to the new order
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*
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* To collect this into a single operation that can be done simultaneously
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* we must first create a mask for the old bits that are set and a mask for
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* the new bits that are set. Then we can remove the old bits and set the new
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* bits simultaneously in a "single" operation like so:
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* OldMask = b1 | b2 | b3 | b4
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* NewMask = b3 | b4 | b1 | b2
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*
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* So this:
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* V & ~oldMask | newMask
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*
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* Expands to this:
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* V & ~b1 & ~b2 & ~b3 & ~b4 | b3 | b4 | b1 | b2
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*
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* Which becomes this:
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* b1 | b2 | b3 | b4
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*
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* Which is the correct result!
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*
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* @param {bigint} num
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* @param {[bigint, bigint][]} mapping
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*/
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export function remapBitfield(num, mapping) {
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// Create masks for the old and new bits that are set
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let oldMask = 0n
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let newMask = 0n
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for (let [oldBit, newBit] of mapping) {
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if (num & oldBit) {
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oldMask = oldMask | oldBit
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newMask = newMask | newBit
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}
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}
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// Remove all old bits
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// Set all new bits
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return (num & ~oldMask) | newMask
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}
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