Okay, I'm seriously looking for help now, and instead of spamming bojster and Flummi with it in the chat, I'll post here:
In Minesweeper Advanced, I have a couple of factors...and from these factors a score should be calculated (we all know RND scores, right? the more score the better played).
The factors avalible are:
hex or octal mode (don't know which one is hardest - if anyone), mine playfield sizex, mine playfield sizey, amount of mines, time elapsed
As we all know with today's CE techniques, making a formula to RND CEs is not a very big problem.
In normal Windows Minesweeper, the 3 pre-programmed modes are:
9*9, 10 mines
16*16, 40 mines
16*30, 99 mines
Any ideas for a score calculation?
BTW: Minesweeper Advanced will be released soon...2 big things left - minesweeper-flags mode (multiplayer, get as many mines as you can) and score calculation...
Calculating score in Minesweeper Advanced
Moderators: Flumminator, Zomis
Perhaps the formula should be something like (Difficulty Bonus)+(Time Bonus). The hard part is ironing out what those bonuses should be, and how they should be weighted against each other... The difficulty bonus should probably be based on the percentage of squares that are mines. The time bonus would be based on the time taken, as compared how many squares it's estimated that the player had to push... sort of comparing the time to a "par time". However, I'm not sure how to calculate such a "par time".
Actually, it might be better to multiply the two bonuses instead of adding them... that might give better results for extreme cases like a huge board with only one mine, or a 4 square board with 3 mines.
As far as hex vs. octal, I'm as stumped as you... hex has less possiblities for a single square, but the lack of "corners" tends to make it harder to eliminate those possibilities.
Actually, it might be better to multiply the two bonuses instead of adding them... that might give better results for extreme cases like a huge board with only one mine, or a 4 square board with 3 mines.
As far as hex vs. octal, I'm as stumped as you... hex has less possiblities for a single square, but the lack of "corners" tends to make it harder to eliminate those possibilities.
Ah, yes. I've also thought about mines per square
So the difficulty bonus would then be mines/(x*y)
(which on expert level on normal Minesweeper results in difficulty 4.848484)
But then the question is, since I have access to time elapsed, how can I make it so that the less elapsed time, the more score I should get?
Should there be a maximum amount of time avalible so this time can be decreased with the time elapsed? (Since the aim is to have the highest score, and not the lowest).
ouch...just thought about something...
decimal division in RND!? BAH!!!
that's not implemented yet.
RND is a little bad with decimals and negative numbers..
So step 1: Implement decimal division (with decimals if possible)
Step 2: Score system
Step 3: Minesweeper Flags for Multiplayer
So the difficulty bonus would then be mines/(x*y)
(which on expert level on normal Minesweeper results in difficulty 4.848484)
But then the question is, since I have access to time elapsed, how can I make it so that the less elapsed time, the more score I should get?
Should there be a maximum amount of time avalible so this time can be decreased with the time elapsed? (Since the aim is to have the highest score, and not the lowest).
ouch...just thought about something...
decimal division in RND!? BAH!!!
that's not implemented yet.
RND is a little bad with decimals and negative numbers..
So step 1: Implement decimal division (with decimals if possible)
Step 2: Score system
Step 3: Minesweeper Flags for Multiplayer