What would be your next deduction in this game of Minesweeper?

Question

Currently stuck in this game of Minesweeper. Can't seem to find any openings, so what would be the moves you would take and why?

Answer 1

Red n is a mine. The pattern, subtracting off known flags, is 2-2-1 (above n-o-p) against a wall. The middle 2 must be next to 2 mines, but only one mine can be on o and p, so the second mine must be n.

Red q is not a mine. If it were, then the 3 above red p would be satisfied, which leaves only one square for the 2 above red o.

Also, as Kevin Crussijen points out, i, j, k, and l can have only one mine between them, so at worst, if nothing else comes up, guess there.

Answer 2

Red q cannot be a mine, because if it were, the 2 above red o can only be surrounded by one mine on red n.

Answer 3

Short answer:

Click the red q first.

Longer answer, no spoiler tags:

To be systematic about it, first we can update the numbers in squares that are adjacent to both blue squares and flags. By "update the numbers" I mean decrease the numbers by however many flags are adjacent to the square. So, for example, the square above the $\color{red}{A}$ decreases from $2$ to $1$ because it's already adjacent to $1$ flag. Doing this results in the following picture:

Now all the numbers we updated (they're all in black) tell us how many mines are in the adjacent blue squares. So we can deduce the following:

1. Exactly one of $\color{red}{A}$ and $\color{red}{B}$ must be a mine.
2. Exactly two of $\color{red}{A},\color{red}{B},\color{red}{C},\color{red}{d},\color{red}{e}$ must be mines.
   And $\#1$ tells us that one of these mines must be $\color{red}{A}$ or $\color{red}{B}$. So we can conclude $\#3$ below.
3. Exactly one of $\color{red}{C},\color{red}{d},\color{red}{e}$ must be a mine.
4. Exactly one of $\color{red}{e}$ and $\color{red}{f}$ must be a mine.
5. Exactly one of $\color{red}{f}$ and $\color{red}{g}$ must be a mine.
6. Exactly two of $\color{red}{f},\color{red}{g},\color{red}{h},\color{red}{i}$ must be mines.
7. Exactly one of $\color{red}{h}$ and $\color{red}{i}$ must be a mine.
   (We can deduce this from the previous two or we can get this directly from the game board.)
8. Exactly one of $\color{red}{i}, \color{red}{j}, \color{red}{k}, \color{red}{l}$ must be a mine.
9. Exactly two of $\color{red}{m}, \color{red}{n},\color{red}{o}$ must be mines.
10. Exactly two of $\color{red}{n}, \color{red}{o},\color{red}{p}$ must be mines.
11. Exactly one of $\color{red}{o}, \color{red}{p}, \color{red}{q}$ must be a mine.

Since $11$ says we can't have a mine at both $\color{red}{o}$ and $\color{red}{p}$, then from $10$ we can deduce that $\color{red}{n}$ is definitely a mine. Therefore either $\color{red}{o}$ or $\color{red}{p}$ (but not both) is definitely a mine, which, combined with $\#11$ (again), tells us $\color{red}{q}$ is definitely not a mine. So let's update our board. I'll use a solid yellow square to indicate safety:

Now we can continue our list, but let's update some of the items first:

12. Exactly one of $\color{red}{m}$ and $\color{red}{o}$ must be a mine.
13. Exactly one of $\color{red}{o}$ and $\color{red}{p}$ must be a mine.
14. Exactly one of $\color{red}{s}$ and $\color{red}{t}$ must be a mine.
15. Exactly one of $\color{red}{t}$ and $\color{red}{x}$ must be a mine.
16. Exactly two of $\color{red}{t},\color{red}{u},\color{red}{v},\color{red}{w},\color{red}{x}$ must be mines.
17. Exactly one of $\color{red}{u},\color{red}{v},\color{red}{w}$ must be a mine. This is deduced from $\#15$ and $\#16$.
18. Exactly one of $\color{red}{x},\color{red}{y},\color{red}{z}$ must be a mine.
19. Exactly one of $\color{red}{y},\color{red}{z}, a$ must be a mine.
20. Exactly one of $\color{red}{z}, a, b$ must be a mine.
21. Exactly one of $a,b,c$ must be a mine.
22. Exactly two of $b,c,d,e$ must be mines.
23. At least one of $d$ and $e$ must be a mine. This follows from $\#21$ and $\#22$ together (since we can't have a mine at both $b$ and $c$ but we may have a mine at neither).
24. Exactly one of $e$ and $f$ must be a mine.
25. Exactly one of $f$ and $n$ must be a mine.
26. Exactly two of $n,m,o$ must be mines.
27. Exactly one of $o$ and $p$ must be a mine.
28. Exactly one of $p,q,r$ must be a mine.
29. Exactly two of $q,r,s,t,u$ must be mines.
30. Exactly one of $t$ and $u$ must be a mine.
31. Exactly one of $q,r,s$ must be a mine. This follows from $\#29$ and $\#30$ together.

At this point we can't make any other immediate solid conclusions (unless I missed something). What I would do next is update the board in this same manner using the number you get from clicking on $\color{red}{q}$. See if anything falls out of that. If not, then make some guesses using the list above and see if any contradictions fall out of it, thus allowing you to rule out other possibilities.

Answer 4

This answer, although it does address the question itself, is more of an interesting observation coming from analysing the situation by expressing the system in terms of simple linear equations (eg/ using capitals for the reds, you might say A+B=1 to say that exactly one of A and B can be a mine, or F+G+H+I=2 because there must be exactly two mines amongst those four). I don't think it's reasonable to create these sorts of equations for every step.

When you ask something like Maxima to solve it, you get a lot of equations that don't help much... but you also get, for example (substituting away the arbitrary introduced variables), N=Q+1. This can only be satisfied in one way if all variables can be only 0 or 1 - namely, N=1 and Q=0. So there's a mine at red N and no mine at red Q.

Similarly, you get (using lower-caps for black) o=1+q+r, which can only be satisfied if o=1, and q=r=0. So there's a mine at black o and no mine at black q and black r. Re-checking has shown it is o=q+r, which isn't as helpful. Substituting the above result doesn't determine any further values with any confidence, unfortunately (in contradiction with a previous version of this answer).

Some interesting other observations that can be made from this analysis...

At least one of black d and black e must be a mine, and if both of them are, then so are black a and red S and X (if only one is, then black a, red S and X are not mines).

Once you eliminate duplicates and the two certain values, there are 26 equations in 44 unknowns.

Answer 5

Same answer as greenturtle3141 except with longer explanation.

Answer

q

Explanation

Among n/o/p, there are two bombs. The two bombs can't be on both o and p because then the 3 above p would have four bombs adjacent. The two bombs can't be on neither o and p because then the 2 above o could not have two bombs. Therefore one of o/p has a bomb and the other doesn't. This means that we have accounted for all three bombs below the 3 above p. Therefore q is safe.

Answer 6

i would click on 'q'.

the 2 above the 'o' implies that two mine are between 'n', 'o', 'p'.

the 3 adjacent to it, means there is a mine on 'n', and another between 'o' and 'p'.

the 3 above 'p' has already two adjacent mines.

so, in conclusion, there should be no mine on 'q'.

Answer 7

Same answer as greenturtle3141, except with a picture.

Red q cannot be a mine, because if it were, 3 above p would be "full". Hence, red o or red p (next to the left of red q) cannot be a mine. However, 2 above red o would have only one mine, red n.

Answer 8

Note: Through this answer, red 'X' squares will be referred to as rX, and black 'X' squares as bX. For example, red 'A' is rA, and black 'A' is bA.

So far, there is one known mine:

rN: Starting from rP to the right, we have a 1-2-x pattern: There is one mine shared between rO and rP (it must be shared, because of the 2), which means that the other mine touching the 2 has to be rN.

There's also one known safe spot:

rQ: Thanks to rO and rP, we also have a 1-1-x pattern: There's one mine shared between those two, which means there can't be a mine touching rQ (because the 3 adjacent to rP reduces to a 1, thanks to having two exposed mines touching it).

If these don't allow further logic, then there are three optimal squares for guessing, which each have a 1/4 chance of being a mine.

rJ, rK, rL: Out of those three and rI, there can be only one mine, thanks to the 3 (reduced to 1). HOWEVER, because of the 2 (reduced to 1) adjacent to rI, there must be a mine shared between rH and rI, leaving rI with a 1/2 chance and the other three with 1/4.

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Considering everything we know so far, we can calculate the probability of each unknown square being a mine, and group them together with related squares:
[Fractions are unreduced, with the number of mines in a set over the number of squares.]

• rA: 1/2, shared between rA and rB: The 2 adjacent to rA reduces to 1.
  rB: 1/2, shared between rA and rB: The 2 adjacent to rA reduces to 1.

• rC: 1/3, shared between rC, rD, rE (rE is 1/2): The 3 reduces to a 2, where one mine is in rA/rB, and thus the other one must be shared between these squares.
  rD: 1/3, shared between rC, rD, rE (rE is 1/2): The 3 reduces to a 2, where one mine is in rA/rB, and thus the other one must be shared between these squares.

• rE: 1/2, shared between rE and rF (overrides rC/rD/rE): Both the 4 and the 2 reduce to 1, which gives a higher probability than rC/rD/rE. If this square has a mine, rG must also be mined.
  rF: 1/2, shared between rE and rF: Both the 4 and the 2 reduce to 1. Similarly, 1/2, shared between rF and rG: The 3 reduces to a 1.
  rG: 1/2, shared between rF and rG: The 3 reduces to a 1. If this square has a mine, rE must also be mined.

• rH: 1/2, shared between rH and rI: The 2 adjacent to rI reduces to a 1.
  rI: 1/2, shared between rH and rI (overrides rI/rJ/rK/rL): The 2 adjacent to rI reduces to a 1.

• rJ: 1/4, shared between rI, rJ, rK, and rL (rI is 1/2): The 3 reduces to a 1.
  rK: 1/4, shared between rI, rJ, rK, and rL (rI is 1/2): The 3 reduces to a 1.
  rL: 1/4, shared between rI, rJ, rK, and rL (rI is 1/2): The 3 reduces to a 1.

• rM: 1/2, shared between rM and rO: Considering rN, the 3 reduces to a 1. If this square is mined, rP must also be mined.
  rO: 1/2, shared between rM and rO: Considering rN, the 3 reduces to a 1. Similarly, 1/2, shared between rO and rP: Considering rN, the 2 reduces to a 1.
  rP: 1/2, shared between rO and rP: Considering rN, the 2 reduces to a 1. If this square is mined, rM must also be mined.

• rR: UNKNOWN. Depends on rQ.

• rS: 1/2, shared between rS and rT: The 3 and the 2 both reduce to 1.
  rT: 1/2, shared between rS and rT: The 3 and the 2 both reduce to 1. Similarly, 1/2, shared between rT and rX: The 3 touching 2 exposed mines reduces to 1.
  rX: 1/2, shared between rT and rX (overrides rX/rY/rZ): The 3 touching 2 exposed mines reduces to 1.

• rU: 1/3, shared between rU, rV, and rW: Considering rT/rX, the 3 jutting out reduces to 1.
  rV: 1/3, shared between rU, rV, and rW: Considering rT/rX, the 3 jutting out reduces to 1.
  rW: 1/3, shared between rU, rV, and rW: Considering rT/rX, the 3 jutting out reduces to 1.

• rY: 1/3, shared between rX, rY, and rZ (rX is 1/2): The 3 reduces to a 1. Similarly, 1/3, shared between rY, rZ, and bA: The 2 reduces to a 1.
  rZ: 1/3, shared between rX, rY, and rZ (rX is 1/2): The 3 reduces to a 1. Similarly, 1/3, shared between rY, rZ, and bA: The 2 reduces to a 1. Similarly, 1/3, shared between rZ, bA, and bB (bB is 2/4), due to the 1.
  bA: 1/3, shared between rY, rZ, and bA: The 2 reduces to a 1. Similarly, 1/3, shared between rZ, bA, and bB (bB is 2/4), due to the 1.
  [Note: If bA is safe, then rX is safe, rT is mined, and rS is safe.]

• bB: 2/4, shared between bB, bC, bD, and bE (overrides rZ/bA/bB; bD and bE are 4/6): The 3 reduces to a 2.
  bC: 2/4, shared between bB, bC, bD, and bE (bD and bE are 4/6): The 3 reduces to a 2.
  [Note: If bA is safe, then 2/4 set bB/bC/bD/bE breaks into 1/2 sets bB/bC and bD/bE.]

• bD: 4/6, shared between bD, bE, bF, bG, bH, and bI (overrides bB/bC/bD/bE): Can't reduce the 4.
  bE: 4/6, shared between bD, bE, bF, bG, bH, and bI (overrides bB/bC/bD/bE): Can't reduce the 4.
  bF: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4.
  bG: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4.
  bH: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4.
  bI: 4/6, shared between bD, bE, bF, bG, bH, and bI: Can't reduce the 4. Similarly, 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
  [Note: bE/bF is also 1/2. bF/bN is also 1/2 (2 reduces to 1). bF/bH/bI/bJ/bN is also 3/5.]

• bJ: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
  bK: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
  bL: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4.
  bM: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4. Similarly, 2/3, shared between bM, bN, and bO: The 3 reduces to a 2.
  bN: 4/6, shared between bI, bJ, bK, bL, bM, and bN: Can't reduce the 4. Similarly, 2/3, shared between bM, bN, and bO: The 3 reduces to a 2.
  bO: 2/3, shared between bM, bN, and bO: The 3 reduces to a 2.
  [Note: bF/bN is also 1/2 (2 reduces to 1). bF/bH/bI/bJ/bN is also 3/5.

• bP: 1/2, shared between bO and bP (bO is 2/3 & 4/6): Both the 2 and the 3 reduce to 1s.

• bQ: 1/3, shared between bP, bQ, and bR (bP is 1/2): The upper 2 reduces to a 1. Similarly, 1/3, shared between bQ, bR, and bS: Considering bT/bU, the lower 2 reduces to a 1.
  bR: 1/3, shared between bP, bQ, and bR (bP is 1/2): The upper 2 reduces to a 1. Similarly, 1/3, shared between bQ, bR, and bS: Considering bT/bU, the lower 2 reduces to a 1.
  bS: 1/3, shared between bQ, bR, and bS: Considering bT/bU, the lower 2 reduces to a 1.

• bT: 1/2, shared between bT and bU, due to the 1.
  bU: 1/2, shared between bT and bU, due to the 1.

Not sure if any other answers analysed the probabilities for each square. Looked at the question, did some checks, took a look at a couple answers, noticed that they only looked at one or two squares, and decided to make a more complete one.

Answer 9

Your mistake was not clicking 5 random squares in 4 quadrants and somewhere in the center. If you can hit all 5 and pass, you're much more likely to make a strong final run of it. If in doubt, manually write up the squares and perform math calculations for each square and determine the least probable fail square. If you're a minesweeper pro, you do that in your head on the go with each and every click. You don't use the question mark clicker or you'd make a better run of it. Red I,J,K, & L are obviously the next clicks. 1 in 4 seems to be your best odds at the moment. The 3/2/3 under Red NOP might have a solution but you may need to grid that out. Cheers.