Which Sudoku technique can be used to solve this? The screenshot is from the open source Sudoku app by SECUSO (Security Usability Society) for the Android platform.
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$\begingroup$ @Rajesh: Sorry for the nitpicking :x Could you please edit the attribution in the question? (Comments are supposed to be "deletable") $\endgroup$– Matthieu M.Commented Dec 5, 2023 at 13:14
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$\begingroup$ I'm new to this. Let me know step by step on what I need to do. $\endgroup$– RajeshCommented Dec 6, 2023 at 1:52
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3 Answers
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You can do sudoku to
whatever is in row 4, column 5.
That will eventually place the same digit in a spot where a three cannot go:
After that deduction, the grid will basically fill itself.
The technique that finds relationships like this is called "colouring the pair". When you have a lot of squares around the grid with the same two options only, it's a handy way to give an identity to the two options of the pair. Even if it hadn't conveniently placed a 1-or-3 in a spot that already sees a three, continuing to colour in the "squares that can only be 1 or 3, but cannot be green" in a different colour would have shown that the square in the bottom left corner (r6c1) sees both colours (both numbers of the 1-3 pair) and must therefore be a 2.
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1$\begingroup$ Thank you for the details on "coloring the pair" technique $\endgroup$– RajeshCommented Dec 5, 2023 at 2:18
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$\begingroup$ Thanks for the technique! For clarity of my understanding, you mean R6C1 sees both the green 1-3 R3C1 and the white 1-3 R6C5, right? $\endgroup$– justhalfCommented Dec 5, 2023 at 6:09
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$\begingroup$ @justhalf Exactly. Usually one would use two distinct colours like red and green, with white being "undecided", but here I happened to find a shortcut that resolved everything before I coloured in the red 1-3s, so my reds are still white, if that makes any sense :-) $\endgroup$– BassCommented Dec 5, 2023 at 8:38
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$\begingroup$ Yep, that's what prompted my clarification question, haha. No biggies. $\endgroup$– justhalfCommented Dec 5, 2023 at 9:44
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Using chess notation (rows 1 to 6 bottom to top and columns a to f, left to right)
.---------------------------------.
| 145 6 14 | 135 2 134 | 6
| 1245 3 124 | 15 6 14 | 5
|----------------+----------------|
|*13 5 6 | 2 4 *13 | 4
| 1234 124 1234 | 6 *13 5 | 3
|----------------+----------------|
| 6 14 134 | 13 5 2 | 2
| 2-13 12 5 | 4 *13 6 | 1
'---------------------------------'
a b c d e f
Double Kite (1 and 3) [* marked cells] => 1 and 3 are false at a1.
due to the linked cells e1,e3,f4,a4 we have a chain of implications: if 1 is false at e1, it is true at e3, false at f4 and true at a4, so either 1 is true at a4 or 1 is true at e1, and (same idea) either 3 is true at a4 a4 or 3 is true at e1.
Another way:
Remote Pair (13) [* marked cells] => 1 and 3 are false at a1. We have either a4=1,e1=3 or a4=3,e1=1 because of the sequence 1-3-1-3 or 3-1-3-1 at the marked cells.
With a1=2 the puzzle is easily solved.
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If the upper right hand square is a 3, you can follow around the 1,3 squares until you force the bottom left squares to be a 1,2 pair which forces row 5, column 4 to be a 1. This, of course forces a second 3 to be in the upper right box, which is not allowed. The upper right square becomes a 1,4. This immediately answers the entire 1,3 system, the upper right box and the makes the second row, third column a 2 (by uniqueness).
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$\begingroup$ I'm not following. If R1C6 is 3, then R3C6 is 1, R4C5 is 3, R6C5 is 1. Then R6C2 is 2 and R6C1 is 3. R5C2 and R5C3 are then 1,4 pair. No contradiction there yet, it seems? $\endgroup$– justhalfCommented Dec 5, 2023 at 11:07
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2$\begingroup$ The first step ("If the upper right hand square [1,3, or 4] is a 3") feels a lot like "try something and see if it turns out impossible". This is called bifurcation (which in its extreme form is basically just guessing), and even if you were a person that doesn't mind a method being somewhat distasteful as long as it produces results, the specific thought chain is unnecessarily long in both directions: You could just say: "if r3c6 is a 1, then both the bottom left squares are twos", and be done with it. $\endgroup$– BassCommented Dec 5, 2023 at 15:20