Hello all, I've been playing Nono for about 3 days now, loving it (not that I'm all that good... yet! :D)

Anyways, I've been working out various chances of certain things on the Beginner board:

The total number of possible beginner boards is 36C9 = 94143280
(for those unfamiliar with this notation, nCr means the number of ways of choosing r objects from a set of n. The formula is n! ÷ r!(n-r)!)

Therefore the chances of just choosing a particular board and getting it is 1 in 94143280.

You could expect a 50% probability that you will get this in the first 50% of 94143280 games, i.e. 47071640.

At 1 game every 1/2 second, you would need to play beginner for 23535820s = 272 days non-stop to reach this total. And then you have half a chance of getting the board you want :P

To make things a bit more likely, say you wanted a board with a blank row or column. In this case, there are 12 possible rows or columns the blanks could occupy, so the remaining 3 blanks have to be in the remaining 30 squares:
12 × 30C3 = 48720

Therefore the chances of getting a board with a blank row or column is 48720 / 94143280 which is about 1 in 1932 (attainable!)

Another good way to get a nice beginner level is to have lots of 1-1-2 or 1-2-1 or 2-1-1. Ideally, we want 3 of these and 3 of 1-4, 2-3, 3-2 or 4-1 along the rows (by my beginner tactics, and I think I saw that someone else used this as well).
The 1-1-2s could be in any of 6C3 places, and there are 3 options for the order. Then there are 4 options for each of the remaining 3 rows:
3 × 6C3 × 4³ = 3840
The chances of getting one of these boards is about 1 in 24516, and by the half-second half-chance timing estimate, you'd need to play for about 1h42m to get a 50% probability of getting one.

Happy sweeping!

Interesting calculations. Good to see someone else take an interest in that aspect of the game :D Oh, and don't get too discouraged if you aren't high on the boards yet: It's amazing how much people speed up when they've played for a while.

Hmm, I feel like figuring out some of this type of stuff myself:

A Corner beginner board like the fake board screenshot that was posted here (along with a fake 0s time) is 4 possibilities so 68 days of constant playing by ians working. However, any board with a 3x3 gap, gives 16 possibilities, so 17 days. It might be interesting to find out what % of boards exist that can be solved with a single click.

The current tournament (5x5, 9 mines), each board has a probability of 25c9 = 8,171,900. The chance, though, of a baord where the mines fill a single row plus a single column, since there are 25 boards which fit that pattern, is 326876, which at 2 boards a second, you have a 50% chance of finding in the first 23 hours. Okay, thats still a bit too much to brute force, but for anyone wondering how some people got the times they did in season 10, touney 5 (5x4, 4 mines), there are only 4845 possible boards, so a vertical row board, at 2 games a second, should have appeared every 8 minues on average.

On the other hand, there are 1,977,204,582,144,932,989,443,770,175 different intermediate boards, of which only 29,033,531,588,150,684,937,045 have a blank top row, so roughly 580 hexillion have a blank row or column, thats 1/3333 or so? I may have that wrong...

To get an example "perfect int board", of the type where either the bottom, top, left, or right 6 rows are all full, and 4 remaining mines placed randomly elsewhere, at 10 seconds a game average, should have a 50% chance to occur every 856 billion years of constant playing. So if anyone gets a board like that, you'd better not waste it!

Hehe, thanks Qetuth :D

As for beginner boards that can be solved in a single click, somehow we need to find a method of quickly counting all possible sets of 9 squares connected by edges. Hmm... ???

I guess this is essentially a Graph Theory problem, so take the squares as vertices and edges between squares as edges joining two vertices? Eek, this is hard. I think some of the critical problems could be if you try and build from a starting point, you could cross over yourself, or making sure you don't count things twice.

Maybe I'll just write a program to test every possibility...

Hmm... 1 in 1932 for beginner. Thought it would be much more likely than that. Could you perhaps work out the probability of an expert game with a blank row? Here's one right here.


http://www.planete-demineur.com/phpBB2/download.php?id=1879

Just saw another!

http://uselessfiles4.free.fr/uploads/wow2__143.jpg

Hey, i have this one:

Expert grid with empty Row & Column ???

http://hexx.250free.com/nonosweeper/x/xxxne64.png

I'd figure out the probability myself but I don't exactly know how to factor the number 180  without getting a serious head ache  ::)

I remember I got once also a board with empty row and column. I was pretty bad on it^^

A specific empty row and empty column on expert means you're basically putting 90 mines into a 14x11 area, or, 154c90 = 154c64 = 154! / 90!*64! = 1.6x10^44.

There are 180c90 boards for expert: 9.1x10^52

So, chance of a board with a specific row and column empty is 1 in 555 million.

But, there are 15 columns and 12 rows, so 180 possible choices of specific row and column, so if you don't care WHICH row and column, its a mere 1 in 3 million (That's not taking into account the rather negligible number of boards which have more than 2 blanks: only 2 per million boards with a blank row and column also have at least one other blank column. So even though we're counting those boards twice, my rounding has more effect than they do.)

Thought I'd make a start on the single click beginner boards:

If you have a blank row, the remaining 3 blocks can be put in in:
12c3=220 ways using up to one row on either side for the 3. 20 of these use only a single side.
12c2 x 2 = 132 ways placing 2 blocks as above, then one more on top of one of those. 30 of these use a single side.
8x3+4x2 = 32 ways of placing the three with only one connecting square. 16 on each side, of which 6 use 3 rows.

so, with the top row blank, can be 20+30+16=66 one-click games.
With second row blank, can be 220+66+16=302
With third row blank, can be 220+132+24=376

So, using symmetry of the board, there are ~~2976~~ boards solvable in a single click which have a blank row or column.

100 have 5 blanks in a row and 5 in a column, intersecting. Ignore those for a sec. Now, for boards with 5 blanks in a row, starting on the left, you have 4 mines left to place.

using one row on either side: 10c4+9c2+9c2+1=283
of these, 6c4-4=11 are one-sided

with one on r+2, rest on r+1: 5(4c2+1)+7=42
with two on r+2, rest on r+1: 4x7+6+4c2=40
with three on r+2, rest on r+1: 11
so, r+2, one sided, sum is 93

with one on r+2, rest on r+1/-1: 9+4(9c2+2)+10c2+1=207
with two on r+2, rest on r+1/-1: 4x17+10+4c2=84
sum of r+2 is 291

with one on r+2, one on r-2: 25, minus 5 already counted, 20.

blank on r+3, but no negatives:  5*8+1=41
blank on r+3, allow negatives: 5*13-1 = 64, minus 5 we've already counted elsewhere = 59

and anything with r+4 is counted in the 5 by 5s.

So, top row has 5 blank from left, but mine in top right corner: 11+93+41=145
second row: 283+291+59=633
third row: 283+291+291+20+59=944

now, multiple for left or right, top or bottom, row or column, then add the 100 with both, number of boards with 5 blanks in a row or column but not 6, solvable in one click, is ~~7652~~

So, 10628 single click boards with at least 5 blanks in a row. To be continued.....

@Q: EVERY board is solvable using only one click  ;D It´s a bit tricky though:
1) you can leave the grid to get into it at another place
2) you can move the cursor exactly at the line between cells - it doesn´t cause cells to behave as being clicked

OK, this was just for fun  ;), good job using all the math stuff  8) looking forward to see how many one click genius boards exist  ;D

Nikolaj I don't think you can go between cells   ???But maybe I think that because I am a spaz and can't control the mouse very well  >:(

<Edit> @nikolaj: gah!! Oh well, I'll keep going, I guess, since that's sort of cheating. I dont imagine a single click which goes out of the board is any faster than 2 clicks. Actually, many of these aren't actually fast to solve, but anyway.... </Edit>

we've covered everything with 5 in a row. So, category 3: boards within a 4x4 square:

You cant have 2 blank rows plus a blank column in that square. A blank row and column leaves two blanks remaining:
two sides give: 4x(5c2+4) = 56.
Side and middle: 8x(5c2+2) = 96
two middles: 4x(8c2+2) = 120
So subtotal for that section: ~272

Now, with a blank top row, you have 5 remaining, and we've counted anything that uses a blank column as well. So:
Blank top row, 5 in 2nd and 3rd rows: 56-12=44
Blank top row, one in bottom row, no complete columns: 4+8+8+4=24
Blank top row, two in bottom row: 4
total with blank row: 72

Blank 2nd row, 5 in 3rd and 4th: 44
blank 2nd row, 1 above, 4 below: 28+16+16+28=88
blank 2nd row, 2 above, 9+12+9+20=50
blank 2nd row, 3 above, 11x4=44
blank 2nd row, 4 above is counted
so, blank second row gives 226

total with a blank row or a column but not both is then 298x4= ~1192

So now we only need to find the possible sets of 9 with no blank row or column, in a 4x4 area.

With a 3x2 section (can go in 6 places), placing 3 more around it without completing a row or column, gives 4x4+2x13=42. so with a 2x3 section must also be 42. 12 of these have both 2x3 and 3x2 so we've counted them twice. 42+42-12=72

With a 3 row, then an adjacent but displaced 3 row, but without creating any of the above: 4 for top row, so 16 for top and bottom, mirrored.13 for middle row, so 26 for mirrored: 16+26=42

We can now try and list all the shapes which have none of the above formations.

With a ring: 8
With a part-ring (missing a side): 8+8=16
With a part ring (missing a corner): 8+40+24=72
ring missing side and corner: 4+24=28
Claw: 4
Cup: 8
b3t3b2b1: 4
b2b3t3t1: 2

Thats all I can find, so 142+42+72=~256

If that's right, 272+1192+256=1720 formations within a 4x4 square, of which there are 9 in a beginner board: ~~15480~~

So, our running total is 26108, with a fairly small but very hard to find set remaining to count.

@Nik: in fact EVERY board is solvable using 0 clicks ;)
look at  http://www.nonosweeper.com/cgi-bin/forums/YaBB.cgi?board=news;action=display;num=1105367487;start=150

I wrote a program to find single-click boards, and there are only apparently 17955... but of course I could be wrong :P

Doing it a case-by-case way, it's very easy to count ones twice.

wrote on Aug 9^th, 2006 at 10:28am:

Nikolaj I don't think you can go between cells   ???But maybe I think that because I am a spaz and can't control the mouse very well  >:(

I didn´t say that  ;D

http://sweb.cz/FarnikR/nonosweeper/oneclickbeg.avi

If I had left button pressed while F2ing, that board would have been solved in 0 clicks. Xvid encoding again ~160 kB. Enjoy  ;)

lol  :)

Heh, blank row in int.

:) :) :)

http://img78.imageshack.us/img78/8940/5065mt8.png