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6x6 Sudoku

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") $\endgroup$ Dec 5, 2023 at 13:14
  • $\begingroup$ I'm new to this. Let me know step by step on what I need to do. $\endgroup$
    – Rajesh
    Dec 6, 2023 at 1:52

3 Answers 3

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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:

enter image description here

After that deduction, the grid will basically fill itself.

enter image description here

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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    $\begingroup$ Thank you for the details on "coloring the pair" technique $\endgroup$
    – Rajesh
    Dec 5, 2023 at 2:18
  • $\begingroup$ 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? $\endgroup$
    – justhalf
    Dec 5, 2023 at 6:09
  • $\begingroup$ @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 :-) $\endgroup$
    – Bass
    Dec 5, 2023 at 8:38
  • $\begingroup$ Yep, that's what prompted my clarification question, haha. No biggies. $\endgroup$
    – justhalf
    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? $\endgroup$
    – justhalf
    Dec 5, 2023 at 11:07
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    $\begingroup$ 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. $\endgroup$
    – Bass
    Dec 5, 2023 at 15:20

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