Somewhat OT... Computer playing Minesweeper
Matteo Dell'Amico
della at toglimi.linux.it
Fri Jul 23 15:47:10 CEST 2004
Heiko Wundram wrote:
[...]
> Well, I had two thorough looks at what you're doing, but I can't seem to find
> an error (well, I'm no expert in probability theory). I can easily construct
> a consistent field which has six mines surrounding the numbers, the actual
> field has seven mines surrounding it, and I can also construct other fields
> with seven mines.
I think this example clearly demonstrates the error:
1 1 1
1 . 1
1 1 1
This creates an estimate of eight for the unknown point, which is
obviously wrong. The error lies in the fact that the point is a neighbor
of eight other points, and their "estimates" gets added.
Besides, I found your algorithm pretty good, but I think there are three
possible ways of making your algorithm better:
(1) Better efficience
(2) Better estimate of the probability of founding a mine
(3) Better choice for the square you're going to test
(1) should be accomplished quite simply by separating the calculations
for non-contiguous subfields. Of course, when some subfield gives a
probability of 0 to some adjacent square, short-circuit the algorithm
and immediatly discover it.
(2) could be done putting into account the undiscovered squares that are
non-adjacent to a number. Now you don't take it into account. Let's take
the case where there are 10 such squares, and one mine left: there are
10 consistent possibilities in this case, whereas, when there are two
mines left, the consistent possibilities are 10*9=90.
The general formula for n squares and x mines is n!/(x! * (n-x)!).
I'm almost sure I'm so contorted that you're not following me, so let's
do an example ('.'s are undiscovered squares):
1 . .
. . .
1 . .
Suppose in this case we have 2 bombs in all the field.
Your algorithm (if I understood it correctly) would ignore the three
rightmost squares, and find these three consistent states ('x' is a
bomb, '-' is a free square and '?' is a place where you don't know):
1 x ?
- - ?
1 x ?
1 - ?
- x ?
1 - ?
1 - ?
x - ?
1 - ?
You'd label all the three cases as equiprobable, so each square has
1/3rd of probability of having a mine.
This is not true, since - counting the whole field - there are 7
distinct possibilities: the one in the topmost state, with no bombs
where there are the questionmarks, and three for each one of the other
two, with a bomb in the place of one of the questionmarks.
So, (0,1) and (2, 1) have a probability of 1/7, while (1, 0) and (1, 1)
have a probability of 3/7.
By the way, you can infer the probability for '?' subtracting from the
total number of mines the probabilities that you calculate and dividing
evenly the outcome; in this case, (2 - 2 * 1/7 - 2 * 3/7) / 3 = 2/7.
This kind of improved accounting could be accomplished by giving a
weight to each consistent solution which is equal to s! / (m! * (s-m)!),
where s is the number of squares that I'd mark with a '?', and 'm' is
the number of remaining mines.
This kind of solution leaves you with some headaches if you want to
combine it with the optimization proposed in point (1). This is left as
an exercise to the reader. :-)
While (2) was a bit complicated, (3) looks to me as a nightmare ;-) :
the value you're really interested for each square is not the
probability of finding a mine when discovering it, but the probability
of finishing successfully the game if you "click" it. Exploring all
possible game outcomes looks pretty unfeasible, so you need an
heuristics for this. The probability of going "boom" in the next step is
of course a good idea :-), but you can consider, as someone else
suggested, taking into account also the closeness of a subfield for
which you have some information. Another idea could be the number of
mines you could immediately identify, or squares that you could
immediately reveal, after discovering that square.
By the way, something that could be revealed by experimentation: which
is the best square to click as a first move? We can simulate many games
starting by clicking some different squares, and see the percentages of
success starting from all different parts. :-)
I hope I have been clear (I'm not a native English speaker, so please
forgive me if not)... this problem is quite interesting, and we *have*
to beat KMines. :-)
--
Ciao,
Matteo
More information about the Python-list
mailing list