4 ms·
I would be curious to see a more detailed writeup of this, as I haven't understood how this algorithm is applied. Question that I didn't understand the answer
by ronaldx 13y ago
I would be curious to see a more detailed writeup of this, as I haven't understood how this algorithm is applied.
Question that I didn't understand the answer to:
> There are implemented 3 cost functions:
> 1. sum of all tiles in the playing field
is this a useful cost function at all? Surely the sum of the tiles is not affected by strategy played, only by whether a 2 or 4 was randomly received.
> 2. number of all unassigned tiles in the playing field
> 3. average value of an occupied tile
Then, for similar reason, I would be surprised if these were not equivalent.
- felixN 13y agoYou are right, the first cost function doesn't make a lot of sense, but I didn't divide by the number of leafs - so it basically also was based on the number of unassigned tiles. So in the end you are right these approaches are more or less the same.