2
$\begingroup$

enter image description here

I'm stuck on this sudoku. If anyone can solve this, then please share and explain the method that you used.

$\endgroup$
2
$\begingroup$

As Glorfindel has found, you can

eliminate the 9 in R1C6 and the 2's in R3C6 and R7C6

Then, there's a chain:

If R5C4 = 9, then R5C9 = 8, R7C9 = 9, R7C8 = 8, R1C8 = 9, R1C5 = 7, R6C5 = 2, R4C5 = 9, R5C4 is not 9.
This is a contradiction, so you can eliminate the 9 from R5C4.

Then another chain:

If R6C4 = 7, then R6C5 = 2, R4C5 = 9, R1C5 = 7, R1C6 = 8, R3C6 = 5, R3C4 = 2, R7C4 = 7, R6C4 is not 7.
This is a contradiction, so you can eliminate the 7 from R6C4.

The rest is trivial.
Not an easy puzzle. I could not find easier techniques to use.

| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ This is not too long, but it is still trial and error to me. "If ... (trial) then .... then contradiction (error)". $\endgroup$ – Florian F Jun 7 at 13:58
  • 1
    $\begingroup$ @FlorianF If it can be done without trial and error, I'd like to know. I couldn't find any easier ways to solve this one. $\endgroup$ – Dennis_E Jun 7 at 20:55
  • $\begingroup$ @Dennis_E When it comes to identifying chains with contradictions, is there a pattern or technique involved in doing so? For example, how did you know to start with R5C4 and R6C4. I get stuck on puzzles like this all the time. $\endgroup$ – Anita Taylor Jun 25 at 19:14
  • $\begingroup$ @AnitaTaylor I'm not an expert, so I don't know what the best way is. But I look at chains of cells that contain pairs, like the 89, 89, 89, 89, 79, 72, 29. R5C4 is connected to both the start and the end of that chain, so I try one of those numbers. (Yes, I suppose this is trial and error but all the basic techniques I know were exhausted) $\endgroup$ – Dennis_E Jun 26 at 8:40
  • $\begingroup$ Agree with @FlorianF. Logically the phrase "without trial and error" does not make sense. $\endgroup$ – WhatsUp Jul 23 at 12:42
4
$\begingroup$

A simple hint:

In column 6, you've correctly identified that rows 2 and 8 can only be 2 and 9. That means row 1 can't be a 9; row 3 can't be a 2; and row 7 can't be a 2 either.

This technique is called a naked pair; it seems you've used it already in row 7 to eliminate the 9's in row 4 and 6.

| improve this answer | |
$\endgroup$
1
$\begingroup$

We can solve this puzzle with two sudoko techniques.

First with Naked Pair (already @Glorfindel discussed). This one will fit with column $6^{th}$, where $2$ and $9$ will follow the naked pair, and remaining $2$ and $9$ from column $6^{th}$ would be eliminated.

Naked Pair

second with XY-Wing, where R6C5, R7C3 and R7C6 fit for the XY-wing technique. We will choose $7$ from R6C5, and this number will fit (other number $2$ from R6C5 will not fit here). Please see the below in the picture.

XY-Wing

Final solution looks like.

Final Solution

| improve this answer | |
$\endgroup$
  • $\begingroup$ I don't see how that hidden triple works. There should be a 4 pencilmark at R5C4 but it somehow went missing from the hidden triple picture. $\endgroup$ – Jaap Scherphuis Jul 23 at 11:40
  • $\begingroup$ @JaapScherphuis I have corrected my printing mistake. $2$, $3$, and $4$ of R5C2, R5C7 and R5C8 will follow the hidden triple technique. That why 2, 3 and 4 will be in either R5C2 or R5C7 or R5C8. Therefore we will remove 4 from R5C4, 8 and 9 from R5C7. $\endgroup$ – dtc348 Jul 23 at 11:55
  • $\begingroup$ But how can it be a hidden triple if the 4 pencilmark occurs more than 3 times in that row? If that was allowed you might as well say the 8 and 9 form a hidden pair in that row. $\endgroup$ – Jaap Scherphuis Jul 23 at 12:03
  • $\begingroup$ @JaapScherphuis I think, it is naked triple. Am I right? $\endgroup$ – dtc348 Jul 23 at 12:32
  • $\begingroup$ I don't think that row is anything. There are only 5 open cells. A naked triple would correspond to a hidden pair on the other unused digits in the row, and a hidden triple would correspond to a naked pair on the other unused digits. A priori it is perfectly possible for R5C4=4, as that leaves two naked pairs 2&3 in R5C2+R5C8 and 7&9 in R5C7+R5C9. Nothing in that row rules that out, although the rest of the grid may well make that impossible. $\endgroup$ – Jaap Scherphuis Jul 23 at 12:39
0
$\begingroup$

It can be solved with elementary techniques. Using my solver (CSP-Rules, available here: https://github.com/denis-berthier/CSP-Rules-V2.1):

naked-pairs-in-a-column: c6{r2 r8}{n2 n9} ==> r7c6 ≠ 2, r3c6 ≠ 2, r1c6 ≠ 9

biv-chain[2]: r2n9{c7 c6} - c5n9{r1 r4} ==> r4c7 ≠ 9

whip[1]: b6n9{r5c9 .} ==> r5c4 ≠ 9`

biv-chain-rc[4]: r3c4{n5 n2} - r7c4{n2 n7} - r7c6{n7 n4} - r9c6{n4 n5} ==> r3c6 ≠ 5, r9c4 ≠ 5

singles to the end

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.