5 ms·
Another fun trick I've discovered. `XOR[0...n] = 0 ^ 1 .... ^ n = [n, 1, n + 1, 0][n % 4]`
by st0le 1y ago
Another fun trick I've discovered.
`XOR[0...n] = 0 ^ 1 .... ^ n = [n, 1, n + 1, 0][n % 4]`
- tialaramex 1y agoRight, or in summary, no you don't need to all that extra work up front.
- nullc 1y agoTables yuck :P, maybe XOR[0...x] = (x&1^(x&2)>>1)+x*(~x&1)
- bsdz 1y ago~Is there a simple proof for this type of identity?~ Actually I found something through Gemini based on the table mod 4 idea in previous post. Thanks.