## Diabolical one has me stumped

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

### Diabolical one has me stumped

Ok, I've never been able to not figure out a sudoku, but I found this one on a plane ride in the American Way magazine and I have tried every/any trick on the internet and I must be overlooking something. I dont want to know the answer, I want to know the trick to getting to the next number in the puzzle. Any help is greatly appreciated.

------------------------------------
| 37 -8 -47 | 5- 1 -39 | 49- 2- 6 |
| 6 -5 -24 | 7- 29- 8 | 349- 134- 13 |
| 1 -9 -23 | 6 -23 -4 | 8 -5 -7 |
|-----------------------------------|
| 24 -7- 8 | 23 -5- 1 | 6 -34- 9 |
| 9 -36- 1 | 23 -47- 67 | 234 -8- 5 |
| 24 -36- 5 | 8 -49- 69 | 7 -134- 123 |
|-----------------------------------|
| 35 -1- 9 | 4 -6- 2 | 35 -7- 8 |
| 8 -2- 37 | 9 -37- 5 | 1 -6- 4 |
| 357 -4- 6 | 1 -8- 37 | 235 -9- 23 |
------------------------------------
packersfan

Posts: 3
Joined: 16 April 2008

Code: Select all
`+-----------------------------------------------------------+|  37    8     47   |  5     1     39   |  49    2     6    ||  6     5     24   |  7     29    8    |  349   134   13   ||  1     9     23   |  6     23    4    |  8     5     7    ||-------------------+-------------------+-------------------||  24    7     8    |  23    5     1    |  6     34    9    ||  9     36    1    |  23    47    67   |  234   8     5    ||  24    36    5    |  8     49    69   |  7     134   123  ||-------------------+-------------------+-------------------|| *35    1     9    |  4     6     2    | *35    7     8    ||  8     2     37   |  9     37    5    |  1     6     4    || *35+7  4     6    |  1     8     37   | *35+2  9     23   |+-----------------------------------------------------------+`

This is a Humpty-Dumpty PM. It's just waiting for a nudge to cause it to crack open and be solved. Besides numerous XY-Chains, there is a Unique Rectangle in [r79c17] for candidates 35 where [r9c1]=7 or [r9c7]=2 must be true. If you were to just guess [r9c1]=7, then the puzzle would solve easily and you would save yourself the hassle of a long chain to show that [r9c7]<>2.

An alternate choice is this chain/loop which also cracks the puzzle:

Code: Select all
`[r5c2]-3-[r5c4]-2-[r5c7]-4-[r4c8]-3-[r4c4]-2-[r5c4]-3-[r5c2]`

Note: Cells [r79c17] form a UR Type-4, but it doesn't seem to help once the 3s are eliminated from [r9c17].
Last edited by daj95376 on Wed Apr 16, 2008 4:52 pm, edited 1 time in total.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

So does the chain mean I can safely exclude the 3? I've never used the xy-chain method. I would rather not guess as I like to try to solve all puzzles without guessing. Thanks a lot for your help.
packersfan

Posts: 3
Joined: 16 April 2008

Also, r5c7 has 3 candidates. I looked up xy-chains and it says it has to be all sets of 2 candidates. Can you explain? Thanks in advance.
packersfan

Posts: 3
Joined: 16 April 2008

I'm sorry to have been confusing. The comment about XY-Chains was just that ... a comment.

The chain/loop that I listed was just your garden-variety chain/loop. No special type, other than it took advantage of eliminations along the way in row 5.

Bottom Line: I didn't find a simpler technique, but that doesn't mean one doesn't exist. I'm sure the advanced members will find something more elegant.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

daj95376 wrote:Bottom Line: I didn't find a simpler technique, but that doesn't mean one doesn't exist. I'm sure the advanced members will find something more elegant.

I think the following ALS-xz is more elegant:
Code: Select all
`*--------------------------------------------------*| 37   8    47   | 5    1    39   | 49   2    6    || 6    5    24   | 7    29   8    | 349  134  13   || 1    9    23   | 6    23   4    | 8    5    7    ||----------------+----------------+----------------||#24   7    8    | 23   5    1    | 6   #34   9    || 9    36   1    | 23   47   67   | 234  8    5    ||*24  *36   5    | 8   *49  *69   | 7   -134 -123  ||----------------+----------------+----------------|| 35   1    9    | 4    6    2    | 35   7    8    || 8    2    37   | 9    37   5    | 1    6    4    || 357  4    6    | 1    8    37   | 235  9    23   |*--------------------------------------------------*ALS-xz:ALS A = r4c18 = {234}ALS B = r6c1256 = {23469}restricted common x = 2 (r46c1)common z = 3 (r4c8+r6c2)Therefore r6c89, seeing both r4c8+r6c2, can't have 3The rest are singles.`

Another way to explain the logic of this move:

r46c1 can't be both 2, one of them must not be 2.

If r4c1<>2, r4c1=4 => r4c8=3
If r6c1<>2, r6c1=4 => r6c5=9 => r6c6=6 => r6c2=3

Either way, one or both of r4c8 or r6c2 must be 3, and r6c89 can't have 3.

udosuk

Posts: 2698
Joined: 17 July 2005

udosuk: non-elegant XY-Chain.

Code: Select all
`3-[r4c8]-4-[r4c1]-2-[r6c1]-4-[r6c5]-9-[r6c6]-6-[r6c2]-3; => [r6c89]<>3`
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

I don't consider ALS to be simple but it is quick.

Code: Select all
`.---------------.---------------.----------------.| 37   8    47  | 5    1    39  | 49    2    6   || 6    5    24  | 7    29   8   | 349   134  13  || 1    9    23  | 6    23   4   | 8     5    7   |:---------------+---------------+----------------:| 24   7    8   |#23   5    1   | 6    #34   9   || 9   *36   1   |*23  *47  *67  | 23-4  8    5   || 24   36   5   | 8    49   69  | 7     134  123 |:---------------+---------------+----------------:| 35   1    9   | 4    6    2   | 35    7    8   || 8    2    37  | 9    37   5   | 1     6    4   || 357  4    6   | 1    8    37  | 235   9    23  |'---------------'---------------'----------------'`

# sign als locks with *sign als on 2. Threes give nothing but all cells that see four cannot be 4.
wintder

Posts: 297
Joined: 24 April 2007

Code: Select all
` +-----------------------------------------------------+ |  37   8    47   |  5    1    39   |  49   2    6    | |  6    5    24   |  7    29   8    |  349  134  13   | |  1    9    23   |  6    23   4    |  8    5    7    | |-----------------+-----------------+-----------------| |  24   7    8    | c23   5    1    |  6   d34   9    | |  9    36   1    | b23   47   67   | a23-4 8    5    | |  24   36   5    |  8    49   69   |  7    134  123  | |-----------------+-----------------+-----------------| |  35   1    9    |  4    6    2    |  35   7    8    | |  8    2    37   |  9    37   5    |  1    6    4    | |  357  4    6    |  1    8    37   |  235  9    23   | +-----------------------------------------------------+`

Code: Select all
`Forcing Chain:    [r5c7]-2-                                             => [r5c7]<>4                  [r5c7]=2=[r5c4]=3=[r4c4]-3-[r4c8]-4-[r5c7];           => [r5c7]<>4Subset Counting:  [r4c48]={2,3,4}; [r5c7]=4 => [r5c4]=2 => [r4c48]={3}; => [r5c7]<>4daj95376:         4r5c7  2r5c4  [r4]~3`
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

Or one could leverage the bi-values presented by the 2's in box 6 for one very short forcing chain that reduces the puzzle to singles:

r5c7=2 r5c5=4 and r6c9=2 r6c1=4 force r6c5<>4

Knocks out a 4, not unlike wintder's ALS, but in a different location.

Cheers...

- drac
Draco

Posts: 143
Joined: 14 March 2008