| ▲ | Jun8 4 days ago |
| An interesting meta problem is to determine antagonistic set of denominations, like the [10,9,1] example given in the post, to maximize the number of coins selected by the gradient method. |
|
| ▲ | mgradowski 4 days ago | parent [-] |
| Isn't it trivially [1]? |
| |
| ▲ | zahlman 4 days ago | parent [-] | | Perhaps what is meant is "maximize the difference between the optimal result and the one calculated by the naive greedy algorithm". | | |
| ▲ | Jun8 3 days ago | parent [-] | | Thanks for clarifying my poorly worded description, that’s exactly what I meant. Like in the example given, the difference is 10-4=6, let’s call this the naive_greedy_miss_factor. Can we choose three other denominations so that NGMF is > 6? |
|
|
