3 ms·
it's not. it's np-hard. the easiest proof is that the best known algorithm is greater than O(2^N)
by adgjlsfhk1 3y ago
it's not. it's np-hard. the easiest proof is that the best known algorithm is greater than O(2^N)
- tylerhou 3y ago0/1 ILP is NP-hard and the trivial algorithm takes O(2^N), and it's also in NP.
- adgjlsfhk1 3y agoright, but tfa was about the general case where the fancy new algorithm is log(n)^n