The New Sudoku Players' Forum

[gurth]

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Gurth's 2+2 : A New Technique

Look at this grid :

  |---------------------|---------------------|---------------------|
  |12                   |                     |                     |
  |                     |                     |                     |
  |                     |                     |                     |
  |---------------------|---------------------|---------------------|
  |                     |                     |                     |
  |                     |                     |                     |
  |126                  |                     |                     |
  |---------------------|---------------------|---------------------|
  |                     |                     |                     |
  |                     |                     |                     |
  |1289                 |         489         |               89    |
  |---------------------|---------------------|---------------------|

Note the naked pair at a1 (the one "2"), the naked pair at k9 (the other "2"), and the naked quad at k1 (The "2+2"). Those make this technique very easy to spot. Call the naked pairs mn and uv, the quad mnuv. Note also the presence of the trips mnp and uvq at f1 and k5.

See if you can discover the answer for yourself. What can you deduce?

If I had to wait for a Sudoku to come along containing a 2+2, I might have to wait a long time. How long do you think, Ruud? (or any other walking encyclopedia). Can you trace any example?



[Carcul]

If we have a bivalue cell in f5 with candidates 4,6 (p,q) then:

- candidates 4,6 (p,q) can be eliminated from column 5 and row f;
- candidates 8,9 (u,v) can be eliminated from row k;
- candidates 1,2 (m,n) can be eliminated from column 1.

Carcul

[ronk]

So f5 cannot be the bivalue 'pq'

[ravel]

-4k5 > 6f1
-6f1 > 4k5
More ?

[Myth Jellies]

(6 = 1&2)[top ALS] - ((1or2) = (8or9))[quad cell] - (8&9 = 4)[bottom ALS]

Therefore there is a stong-only link between the 4 and the 6.

[r.e.s.]

(6 = 1&2)[top ALS] - ((1or2) = (8or9))[quad cell] - (8&9 = 4)[bottom ALS]

Therefore there is a stong-only link between the 4 and the 6.

Myth,
Wouldn't it be better to just write 6 = 8v9 - 1v2 = 4 ?
6 = 8v9 to avoid the impossible 12-12-12
8v9 - 1v2 obviously
1v2 = 4 to avoid the impossible 89-89-89

[udosuk]

The only direct result I could foresee is:

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Either r6c1=6 (6f1) or r9c5=4 (4k5)

It's one of those more obvious "OR" statements... The ones deduced by Carcul are often much much harder to see...

[Myth Jellies]

(6 = 1&2)[top ALS] - ((1or2) = (8or9))[quad cell] - (8&9 = 4)[bottom ALS]

Therefore there is a stong-only link between the 4 and the 6.

Myth,
Wouldn't it be better to just write 6 = 8v9 - 1v2 = 4 ?
6 = 8v9 to avoid the impossible 12-12-12
8v9 - 1v2 obviously
1v2 = 4 to avoid the impossible 89-89-89

Your analysis is equally valid and shorter, but...

(6 = 1&2)af1 - (12 = 89)k1 - (8&9 = 4)k59

actually shows all of the candidates and cells used in the deduction without need for further explanation; and, if you actually close the loop, say with

(6=1&2)af1 - (12=89)k1 - (8&9=4)k59 - (4=6)f5 -...

when you convert all the weak links into strong links because of the loop, it is fairly simple to see that you eliminate
all other 6's in row f
all other 4's in column 5
all other 1's and 2's in column 1 (either af1 = 1&2 or k1 = 12 which forms naked pair with a1)
and all other 8's and 9's in row k (for similar reasons)

[gurth]

Excellent replies! I could not have imagined better. Everybody gets full marks.

In my notation:

AIC: 6af1 = 12 - k1 = 89k19 - k5 = 4.