Timeline for Can you create a Minesweeper puzzle with 'information crossover'?
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2015 at 9:45 | comment | added | JBSregath | @justhalf I think ABCD are just about on the edge of the board. But be my guest if you can optimize it even further, no problem. | |
| Mar 30, 2015 at 21:26 | comment | added | justhalf | Btw can we adapt this to make ABCD on the outer edge of the board to fulfill all the requirements? | |
| Mar 30, 2015 at 21:23 | comment | added | justhalf | Now it looks correct. +1 | |
| Mar 30, 2015 at 14:56 | comment | added | JBSregath | @justhalf Ah, yes indeed, I see what you mean. I have to place the ? correctly and put the right numbers in the blocks. I've forgotten a few. Look at this pdf explaining logic ports in minesweeper. The 'signal' from left to right is v -> r but I had the D on the v' (which is the opposite of the signal v). | |
| Mar 27, 2015 at 22:55 | history | edited | JBSregath | CC BY-SA 3.0 |
Corrected image
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| Mar 27, 2015 at 14:59 | comment | added | justhalf | I don't think this is a valid board. It can't have any solution at the four unknowns just four squares to the left of B. Also if D is mine, C will also need to be a mine following the diagonal from D to C. Then A will also be safe. | |
| Mar 26, 2015 at 8:33 | history | edited | JBSregath | CC BY-SA 3.0 |
Ok, a better explanation added.
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| Mar 25, 2015 at 13:42 | history | edited | JBSregath | CC BY-SA 3.0 |
Attempt #..oh well, I'm not counting anyway
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| Mar 25, 2015 at 8:56 | comment | added | Lopsy | I know a solution. But it was pretty dang hard for me to find one! And the solution I have is nowhere near optimal. | |
| Mar 25, 2015 at 8:52 | history | edited | JBSregath | CC BY-SA 3.0 |
Add explanation
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| Mar 25, 2015 at 8:42 | comment | added | JBSregath | @Lopsy I hope you have the solution to this problem and not just wondering if there is one. Because I do not see one valid solution that satisfies every combination of A=C and B=D simultaneously. But if you look at your comment "...all 4 mine-settings of A, B, C, D satisfying A=C and B=D extend to a solution", my puzzle is a possible solution. Therefor my question: do you know the answer? | |
| Mar 24, 2015 at 16:10 | comment | added | KSmarts | What I'm saying is that there is a valid solution where A is a mine and the dot next to C is a mine, so C is not a mine. So A$\neq$C. | |
| Mar 24, 2015 at 16:08 | comment | added | Ian MacDonald | No, @KSmarts is right. Rule #2's statement of "in any valid solution" does not say "there is a combination that satisfies A=C, B=D", it says "the only possible combinations satisfy A=C, B=D". | |
| Mar 24, 2015 at 16:04 | comment | added | JBSregath | If A is a mine, the dot next to it is not (because of the 1). And there is a solution in which C is a mine and the dot next to it is not. This is typically that moment of the 'last guess' (here) | |
| Mar 24, 2015 at 15:58 | comment | added | KSmarts | I don't think your new answer implies that A=C and B=D. For example, you could have A with a mine and the bottom-right dot has a mine, but C does not. | |
| Mar 24, 2015 at 15:55 | history | edited | JBSregath | CC BY-SA 3.0 |
Possible solution?
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| Mar 24, 2015 at 15:14 | history | edited | JBSregath | CC BY-SA 3.0 |
Work in progress, with discarded answers
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| Mar 24, 2015 at 15:10 | comment | added | JBSregath | @Lopsy Ok, I think it is clear now, let me (and others) work on a possible solution | |
| Mar 24, 2015 at 15:04 | comment | added | Lopsy | Maybe I made the rules unclear. Rule 2 says "In any valid solution, A=C and B=D." Your first option breaks this because there is a solution where A and D have mines and B and C don't. Your second option breaks this because anything is a solution. Is it clear how Rule 2 works now? | |
| Mar 24, 2015 at 15:03 | history | answered | JBSregath | CC BY-SA 3.0 |
