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Tired of solving Sudokus on paper? Look no further! The solution is, literally, under your nose.

Why use a 9x9 grid, when you can use your own QWERTY keyboard!


enter image description here


RULES:

  • Fill each row with the numbers from 1 to the number of cells in that row (1-10 for the top row, 1-9 for the middle row and 1-8 for the bottom row). No number can be repeated in the same row.

  • As a qwerty keyboard doesn't have columns nor boxes, there will be two 'diagonals' instead:

    enter preformatted text here enter image description here

    • The same number cannot be repeated in a diagonal.
    • Furthermore, no two numbers in a diagonal will be consecutive. E.g. 827 would not be valid, but 713 would.
    • (Note all cells are part of two diagonals except the 'Q' and the 'P' which are part of only 1.)
  • Finally, the diagonals of cells highlighted in red sum to the same amount. I.E the green diagonal and blue diagonal of a red cell, will add up to the same number.

    • All cells that fit this criteria are highlighted.

Please provide some explanation of key deductions/steps in your answer (preferably step by step)

Enjoy!!

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  • $\begingroup$ I know this has probably been twisted too much to label it as a sudoku, but I started it by trying to find different places a sudoku or similar could fit, and it still has some similar characteristics so I'll keep it tagged as such for now :) $\endgroup$ Sep 8 at 18:25
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    $\begingroup$ I think the sudoku tag fits. Most of the rules are inherited from sudoku or a sudoku varient. Great puzzle! $\endgroup$
    – Ankit
    Sep 8 at 19:03
  • $\begingroup$ @Ankit I wasn't going to add it until I read the description, which I think fits the puzzle, so I agree. And thanks! It took me a while to find a set of clues that I thought made it difficult enough but should be quite hard, but still fun now! $\endgroup$ Sep 8 at 19:08
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Completed grid:

Final

Logic:

Let's look at the bottom row first.

Consider the location of the 2 on the bottom row. It cannot be in entry 2 (same diagonal as 2 in the top row), position 3 or position 8 (same diagonal as a 1). If it were in position 5, then R1P5 would have to be 10, since the two diagonals on R2P5 have to have the same sum. [EDIT: The original logic (which follows) had an error in it...this actually shows that R1P4 is 6 greater than R3P3, not the other way around: but then the two diagonals on R2P4 would have to have the same sum, which implies R3P3 is 6 greater than R1P4, which is at least 3.] This implies that we must have R3P3 = 3 and R1P4 = 9: 1 and 2 are already used in the third row, and 10 is already used in the top row. But R2P2 must equal R3P3 to keep the diagonal sums on R1P2 equal, which forces R2P2 = 3, a contradiction given the 2 in its diagonal. So R3P7 must be 2.

Now let's look for the 3 in the bottom row. R3P2 cannot be 3, because it is on the same diagonal as the 2 in R1P2. From the red square in R1P3, we know that R2P2 and R3P3 have to be equal, so R3P3 cannot be 3 from the same diagonal. Note that R2P6 cannot be 7-9 from the 8 in R1P6, cannot be 5-6 from the 6 in R3P6, and cannot be 1 from the 1 in its row. So R2P6 must be one of 2, 3, or 4, which prevents R3P5 from being 3. Hence R3P8 must be 3.

Now look at the 3 in the second row. The 2 in R1P2 blocks R2P1-2. The 4 in R3P4 blocks R2P4-5, the 2 in R3P7 blocks R2P7 and the 3 in R3P8 blocks R2P9. So the only place this 3 can go is in R2P6. The grid thus far:

Grid 1

Continuing in the bottom row:

Let's try to place the 5. As before, R2P2 and R3P3 have to be the same, and neither can be 5 due to the given 5 in R2. In addition, this forces R3P3 to be either 7 or 8. Now suppose R3P5 were 5. Then R1P5 would have to be 7, since the diagonals through R2P5 have the same sum. This then forces R1P3 to be 8, and the same diagonal sums on R2P4 force R1P4 to be 11, a contradiction. Thus R3P2 must be 5.

Hoping this is the breakout:

Let's look at some possible values in the middle of the board. We have the obvious 7 and 8 in R3P3,5. The equal diagonal sums on R2P5 forces R1P5 to be 4 or 5. The equal diagonal sums on R2P4 force R1P4 to be either 7 or 9. Now in the red squares in R2, we have R2P4 is blocked from 1,3,5 in its row, 4 from R3P4, and it also cannot be either 7 or 8, as R3P3 in its diagonal must be one of these values. Hence it is 2, 6 or 9. Similar analysis shows R2P5 must be 2 or 6. The grid thus far:

Candidates

Picking a winner:

Suppose R1P5 is 5. Then neither R2P4 nor R2P5 can be 6, which forces them to be 9 and 2, respectively. But R2P4 being 9 forces both R1P4 and R3P3 to be 7, which contradicts the equal diagonal sum on R2P4. So R1P5 is 4, which resolves R1P4 = R3P3 = 7, which also gives R2P2 = 7, and R3P5=8. Some progress, but not the breakout I was hoping for...the grid thus far:

Progress

The road goes ever onward:

But there is continuing progress! Look at R2P4. From our analysis above, it can only be 2, 6 or 9, and 6 is eliminated by the 7s in its diagonals. Notice that it cannot be 2, due to the constraint that ALL squares with equal diagonal sums are red, and R2P4 being 2 would give equal diagonal sums to R1P4. Thus we must have R2P4 = 9, and the only place the 2 can then go in R2 is R2P5.

Let's finish this off. First note that R1P9 can be only 9 or 10, since 3, 5 and 6 are blocked by its diagonal to the lower left. In the upper left corner, we must have the two diagonals on R3P1 have the same sum, so we must have R1P1 in {3,5,6,9,10} + R2P1 in {4,6,8} add up to 7 + R1P3 in {3,5,9,10} (cannot be 6 due to the 7). Moreover, {R1P1,R1P3} cannot be {9,10}. If R1P3 is 9 or 10, then its diagonal sum is at least 16, which forces R1P1 to be 9 or 10. Hence R1P3 must be 3 or 5. But it cannot be 5, since there is no combination for R1P1,R2P1 that sum to 12. Hence R1P3 must be 3, and (R1P1,R1P2) must be (6,4). The rest of the grid fills in trivially. The final grid:

Final

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    $\begingroup$ @Ankit incorrect there is one solution, and your answer is incorrect. Check the 3 in row 2 in your solutions. In one it would be a red cell, however all reds are given. Therefore one solution. Jeremey will get the tick as he was the first to provide a correct solution with full working. $\endgroup$ Sep 8 at 21:35
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    $\begingroup$ @Ankit yes the H key, in the question it says all red cells are given. If Q is a 3, then U would be a 6, meaning that the H key, 3 should be a red cell. However as it is not, U cannot be a 6, and Q cannot be a 3. There is only 1 solution, Jeremy's answer literally proves that if you read it, which is why you are not correct. $\endgroup$ Sep 8 at 21:47
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    $\begingroup$ @Ankit Please see my comment on your answer. The assertion "diagonals of cells highlighted in red sum to the same amount" is a positive assertion which tells you nothing about cells that are not highlighted in red. However, the OP also states "All cells that fit this criteria are highlighted." This is the negative assertion: if a cell is not highlighted in red, then it DOES NOT have equal diagonal sums. $\endgroup$ Sep 8 at 22:18
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    $\begingroup$ In defense of the OP, I did not find that statement vague at all. $\endgroup$ Sep 8 at 22:28
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    $\begingroup$ The logic for why R3P7 must be 2 is incorrect. R2P4 needing its diagonals to be equal actually implies R1P4 is 6 greater than R3P3. You can find the correct contradiction in one more step - R1P3 requires R2P2 and R3P3 to be equal, so they would both need to be 3, but R2P2 can't be 3. $\endgroup$
    – Rob Watts
    Sep 8 at 22:48
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I have an answer. I have not proved uniqueness.

enter image description here

My start is very similar to Ankit's. Put in all the possible numbers. Start reducing at the red squares. However, what really helped was

"All cells that fit this criteria are highlighted"
This allowed me to start reducing at squares where the combinations would have otherwise matched.

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  • $\begingroup$ Answers to [grid-deduction] questions are generally expected to "show their work". Could you provide an outline of how you got here? $\endgroup$
    – bobble
    Sep 8 at 20:22
  • $\begingroup$ @bobble to be fair, I missed out that request in the question, added it now. But I can confirm this is the correct solution, however I can also confirm this is entirely solvable logically, and is a unique solution. If you could show how you got this, and work out all the logic I can accept, but not till then! $\endgroup$ Sep 8 at 20:24
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    $\begingroup$ Yes. My problem was actually that I had reduced to a certain point and tried a value at a square, then discovered that the entire grid fell into place. $\endgroup$ Sep 8 at 20:26
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    $\begingroup$ Yeah that can happen sometimes with grid-deductions. But like I said, there's no guessing required to solve so if you manage to backtrack and find the solution path feel free to edit it in and I can accept! $\endgroup$ Sep 8 at 20:29
  • $\begingroup$ The other answer appears to have done a more complete job of explaining the reasoning, and to me, it is much more fun to solve than to keep track of each step along the way, so I'm fine letting that one get complete and get the checkmark. $\endgroup$ Sep 9 at 13:29

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