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There's not a lot interesting about "facts about numbers" when you place such a low upper bound on your achievements. If he'd only done this up to 100, he'd nev
by Retra 10y ago
There's not a lot interesting about "facts about numbers" when you place such a low upper bound on your achievements. If he'd only done this up to 100, he'd never have had trouble finding a solution for 10958. But why 11111? Why not 111111111111?
To me this is just as interesting as any other arbitrary list of permutations of symbols.
- matthewbauer 10y agoI assume 111112 is unrepresentable as well?
- Mizza 10y agoTo me, the pattern of unrepresentable numbers (like the pattern of primes) would also be an "interesting" result.
- javajosh 10y agoIntuitively, this problem seems isomorphic to the famous NP-complete problem, 3SAT[1]. Instead of picking binary states, you have to pick symbols to insert, and where, that yield the desired outcome. If this intuition is true, then if you can write a program to generate these expressions (in polynomial time) then you'd have also proven P=NP. You could probably brute force it though as the difficulty scales with the length of your target number. 1. https://en.wikipedia.org/wiki/Boolean_satisfiability_problem https://en.wikipedia.org/wiki/Boolean_satisfiability_problem