## Stuck for 2 hours in the same sudoku :(

Post the puzzle or solving technique that's causing you trouble and someone will help

### Stuck for 2 hours in the same sudoku :(

Hi,

I already tried a few techniques that I learned (x-wing, xy-wing and swordfish) but I still find no way to solve this sudoku attached. I took around 15 minutes to get to this point where I am at but I spent the next 2 hours and couldnt finish it!

I am not looking for the solution, I am looking for a technique or logic that allows me to find the next number!
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batata004

Posts: 3
Joined: 26 December 2018

### Re: Stuck for 2 hours in the same sudoku :(

This might not be what you're looking for (depends on how much you know or want to know about AICs, ALSs, and the Eureka notation), but it just happened to be the first thing I found manually. It also happens to solve the puzzle without further ado (except singles):

Code: Select all
`.4..5....7..864....592.1.....6..9..198...6.......8....87.5...94..1.7.82.......1.7.-------------------.--------------.---------------------.| b13    4      8   |  9     5   7 | 23     b16      236 ||  7     13     2   |  8     6   4 | 359     15      39  ||  6     5      9   |  2     3   1 | 47      47      8   |:-------------------+--------------+---------------------:|  5     23     6   |  37    24  9 | 247     8       1   ||  9     8      47  |  1     24  6 | 23457   457     23  || c1(3)  123  c(47) | c(37)  8   5 | 2479  a(6)-47   269 |:-------------------+--------------+---------------------:|  8     7      3   |  5     1   2 | 6       9       4   ||  4     9      1   |  6     7   3 | 8       2       5   ||  2     6      5   |  4     9   8 | 1       3       7   |'-------------------'--------------'---------------------'`

(6)r6c8 = (61-3)r1c81 = (347)r6c134 => -47 r6c8; stte

original: Show
(6)r6c8 = (61)r1c81 - (1=347)r6c134 => -47 r6c8; stte

In other words, the short chain proves that either r6c8 is 6 or there's a triple 347 on row 6. Either way, r6c8 can't be 4 or 7, which leaves just one candidate left: 6.

Edit: made a stylistic change to the chain.
Last edited by SpAce on Sun Dec 30, 2018 2:11 pm, edited 1 time in total.
-SpAce-: Show
Code: Select all
`   *             |    |               |    |    *        *        |=()=|    /  _  \    |=()=|               *            *    |    |   |-=( )=-|   |    |      *     *                     \  ¯  /                   *    `

"If one is to understand the great mystery, one must study all its aspects, not just the dogmatic narrow view of the Jedi."

SpAce

Posts: 2558
Joined: 22 May 2017

### Re: Stuck for 2 hours in the same sudoku :(

Code: Select all
`*----------------------------------------*| 3-1 4    8  | 9  5  7 | 23    a16  236 || 7   13   2  | 8  6  4 | 359    15  39  || 6   5    9  | 2  3  1 | 47     47  8   ||-------------+---------+----------------|| 5   23   6  | 37 24 9 | 247    8   1   || 9   8    47 | 1  24 6 | 23457  457 23  ||b13  123 b47 |b37 8  5 | 2479  b467 269 ||-------------+---------+----------------|| 8   7    3  | 5  1  2 | 6      9   4   || 4   9    1  | 6  7  3 | 8      2   5   || 2   6    5  | 4  9  8 | 1      3   7   |*----------------------------------------*`

ALS XZ Rule: X = 6, Z = 1: (1=6) r1c8 - (6=1) r6c1348 => - 1 r1c1; stte

Leren

<edit> Perhaps I should back up the forum gobbldegook notation above, with a wordy description of this technique, since that is what you are after.

This move is an example of an ALS chain. ALS means Almost Locked Set, which is a group of N cells (in the same row, column or box) with N+1 digits. In this case there are 2 ALSs in the chain, the bi-value cell a and the 4 cells in Row 6.

The trick is to observe that if cell a is not 1 it must be 6. That would mean that r6c8 would not be 6. The 4 b cells would then have 4 digits 1347, which would be a locked quad, and since there is only one 1 in that set, r6c1 would be 1.

So, if r1c8 is not 1, r6c1 is 1. You can reverse this argument and assume that r6c1 is not 1 and by an analogous line of reasoning show that r1c8 is 1.

So this shows that at least one of r1c8 and r6c1 must be 1. Since r1c1 can see both of these cells, it can't be 1, so it's 3.

As it turns out this one extra solved cell is enough to completely solve the puzzle via a cascade of singles, which is what the stte term means (short for SinglesTo The End) the forum equivalent of Yee haaah, gotcha !

Leren
Last edited by Leren on Mon Dec 31, 2018 9:54 pm, edited 2 times in total.
Leren

Posts: 3927
Joined: 03 June 2012

### Re: Stuck for 2 hours in the same sudoku :(

Here are some more steps, you could make (but finally a short chain is still needed):
Code: Select all
` *---------------------------------------------------------* | #13   4     8    |  9    5    7  | @23      16    236   | |  7   #13    2    |  8    6    4  |  359     15    39    | |  6    5     9    |  2    3    1  |  47      47    8     | |------------------+---------------+----------------------| |  5   @23    6    |  37   24   9  |  47-2    8     1     | |  9    8     47   |  1    24   6  |  23457   457   23    | |  13   123   47   |  37   8    5  |  2479    467   269   | |------------------+---------------+----------------------| |  8    7     3    |  5    1    2  |  6       9     4     | |  4    9     1    |  6    7    3  |  8       2     5     | |  2    6     5    |  4    9    8  |  1       3     7     | *---------------------------------------------------------*`
There is a w-wing 23 with strong link for 3 in box 1 -> one of r1c7 and r4c2 must be 2 -> r4c7<>2, giving you the pair 47 in col. 7.
Code: Select all
` *---------------------------------------------------------* |  13   4     8    |  9    5    7  |  23      16    236   | |  7    13    2    |  8    6    4  |  359     15    39    | |  6    5     9    |  2    3    1  |  47     @47    8     | |------------------+---------------+----------------------| |  5    23    6    |  37   24   9  |  47      8     1     | |  9    8    @47   |  1    24   6  |  235     45-7  23    | |  13   123  #47   |  37   8    5  |  29     #467   269   | |------------------+---------------+----------------------| |  8    7     3    |  5    1    2  |  6       9     4     | |  4    9     1    |  6    7    3  |  8       2     5     | |  2    6     5    |  4    9    8  |  1       3     7     | *---------------------------------------------------------*`
With the strong link for 4 in row 6 you have another w-wing, eliminating 7 in r5c8.
Code: Select all
` *---------------------------------------------------------* |  13   4     8    |  9    5    7  | #23     a16   #23+6  | |  7    13    2    |  8    6    4  |  39-5   b15    39    | |  6    5     9    |  2    3    1  |  47      47    8     | |------------------+---------------+----------------------| |  5    23    6    |  37   24   9  |  47      8     1     | |  9    8     7    |  1    24   6  | #23+5    4-5  #23    | |  13   123   4    |  37   8    5  |  29      467   269   | |------------------+---------------+----------------------| |  8    7     3    |  5    1    2  |  6       9     4     | |  4    9     1    |  6    7    3  |  8       2     5     | |  2    6     5    |  4    9    8  |  1       3     7     | *---------------------------------------------------------*`
There is an xy-wing eliminating 9 from r2c7 and r6c9, but it does not solve the puzzle.
Instead you can look at the unique rectangle 23 in r15c79.
One of 5r5c7 or 6r1c9 must be true. Since 6r1c9 -> 1r1c8 -> 5r2c8, you can eliminate 5 from r5c8 and r2c7.
eleven

Posts: 2461
Joined: 10 February 2008

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