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# logical deduction – A Minesweeper Sudoku

Some normal observations:

• Each 3×3 field accommodates precisely three mines. They cannot be all in a single row, column, or diagonal, since then the $$1$$ in that field can be unplaceable. If the $$3$$ in a field is within the nook, then the $$1$$ should be within the reverse nook. The centre cell of a field can by no means be $$1$$ or $$2$$.
• Two different forbidden mixtures of mine placement are $${$$top-left, top-middle, middle-left$$}$$ and $${$$top-left, top-middle, bottom-left$$}$$ (additionally rotations and reflections of those patterns), since then the $$2$$ in that field can be unplaceable.
• Every mine cell should be at the least $$4$$, so in Killer Sudoku packing containers or row/column sums, two mines may give a sum from $$9$$ to $$17$$ inclusive, three mines from $$15$$ to $$24$$ inclusive, 4 mines from $$22$$ to $$30$$, 5 mines from $$30$$ to $$35$$, six mines should be $$39$$. Also, if some mine cells sum to $$9$$, then they should be both a single cell containing $$9$$ or two cells containing $$4,5$$, whereas if some mine cells sum to $$10$$, then they should be two cells containing both $$4,6$$ or $$5,5$$.

Of explicit curiosity on this puzzle:

Killer Sudoku Box 14 has two mines ($$5,9$$ or $$6,8$$). Column 28 has 4 mines ($$4,7,8,9$$ or $$5,6,8,9$$).

Step-by-step deductions

Start with the Killer Sudoku field summing to $$34$$.

It can comprise at most 5 mine cells (three from the top-middle field, two from the top-right field), however lower than 5 cannot sum to $$34$$, so it should be precisely 5. Therefore all the mines of the top-middle field should be in that Killer Sudoku field. Then there’s just one potential place for the mine adjoining to the already-placed $$1$$, and the $$3$$ should even be in that field. Using pink for mines and gray for not-mines, now we have this, with the numbers inside that 2×2 sq. being $$3$$ and three of $$9,8,7,6,4$$ (these being the 5 mine cells summing to $$34$$).

Also, the Killer Sudoku field summing to $$22$$.

It can comprise at most 4 mine cells (the decrease three of the 5 cannot all be mines), however lower than three cannot sum to $$22$$, so it should be three or 4. If it is three, they should be $$9,8,5$$ or $$9,7,6$$; if it is 4, they should be $$4,5,6,7$$.

And the Killer Sudoku field summing to $$27$$.

By the final observations above about what given numbers of mines can sum to, this field should comprise precisely 4 mines, $$4,7,8,9$$ or $$5,6,8,9$$. In the bottom-left and bottom-middle packing containers, that is both three (counting the centre cell) and one, or two and (each) two. Also the 2 cells within the $$9$$ diagonal cannot each be mines, as a result of if there are two mines on this diagonal they should be $$4$$ and $$5$$, which might’t each seem as mines within the $$27$$ field.

Now think about the row with $$32$$ sum.

Any row or column can comprise at most six mines. Here among the numbers $$4,5,6,7,8,9$$ should sum to $$32$$, so it could solely be the 5 numbers $$4,5,6,8,9$$. Considering the highest three 3×3 packing containers, two of them will need to have two mines within the center row, and the third will need to have precisely one mine within the center row.

This continues to be a really partial reply, however it’s a very laborious puzzle! Maybe another person can proceed from right here, utilizing a few of my methodology and deductions, or I’ll come again later to increase on this.

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