Another fun trick I've discovered.

`XOR[0...n] = 0 ^ 1 .... ^ n = [n, 1, n + 1, 0][n % 4]`

nullc 2 days ago | parent | next [-]

Tables yuck :P, maybe

XOR[0...x] = (x&1^(x&2)>>1)+x*(~x&1)

	▲	bsdz 2 days ago \| parent [-]
		~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.

tialaramex 2 days ago | parent | prev [-]

Right, or in summary, no you don't need to all that extra work up front.