Post #3020181

Alison's version uses powers of two, so you can recover the items by writing down the binary representation of the total. I realised you can use Zeckendorf arithmetic: if the scores are alternating terms of the Fibonacci sequence, each total can be decomposed uniquely. But the only property I need is \(f(n+2) \geq f(n+1) + f(n)\), so I add some random noise on to each score to make it extra-spooky.