Luke451 wrote:Hmm, lemmee check that issue for something you could use...ah, here we go: Chelsea blanked Newcastle United 2-nil last Saturday. You'll thank me come November!

Mmm, believe it or not that result isn't as certain as say Obama winning the election or one of Angels/Dodgers becoming the champ. For the moment I'll say Chelsea 4 Newcastle 1. Let's come back by that time and see who will be more correct!

As for Draco's nice puzzle, I can offer 2 different solving paths:

After singles, locked candidates, plus a skyscraper on 5 (and another single):

- Code: Select all
`+-------------------+-------------------+-------------------+`

| 6 7 348 | 238 138 1238 | 34 9 5 |

| 1348 138 5 | 389 7 389 | 6 348 2 |

| 38 2 9 | 4 5 6 | 7 38 1 |

+-------------------+-------------------+-------------------+

| 34 6 234 | 1 349 239 | 8 5 7 |

| 7 9 248 | 258 6 258 | 24 1 3 |

| 5 38 1 | 238 348 7 | 249 46 69 |

+-------------------+-------------------+-------------------+

| 138 1358 38 | 7 2 4 | 359 36 69 |

| 2 35 6 | 3589 389 3589 | 1 7 4 |

| 9 4 7 | 6 13 135 | 35 2 8 |

+-------------------+-------------------+-------------------+

Ignoring a useless x-wing on 4, there are 2 ways to crack it:

1. The short but tricky way:

r1c45+r2c46 from {12389} must have 1 or 2

=> r6c245+r9c5 can't be [8231]

=> r6c5 can't be 3

-or-

ALS-xy-wing:

ALS A: r1c45+r2c46={12389}

ALS B: r6c24={238}

ALS C: r9c5={13}

restricted common between A,B: x=2

restricted common between A,C: y=1

common between B,C: z=3

Therefore r6c5, seeing r6c24+r9c5, can't be 3.

That opens up an xy-wing @ r4c1+r6c25 which solves the puzzle.

2. The long but easy (or easier) way:

r4c13+r5c3 from {2348} must have 3 or 8

=> r3c1+r17c3 can't be [383]

=> r1c3 can't be 8

-or-

ALS-xz:

ALS A: r34c1={348}

ALS B: r457c3={2348}

restricted common: x=4

common: z=8 @ r3c1+r57c3

Therefore r1c3, seeing r3c1+r57c3, can't be 8.

=> r1c37={34} (naked pair @ r1)

=> Hidden pair @ b2: r2c46={39} (naked pair @ r2)

Turbot fish (kite):

3 @ r1 locked @ r1c37

3 @ c8 locked @ r37c8

Now r1c7+r3c8, both @ b3, can't be both 3

=> r7c3, seeing r1c3+r7c8, can't be 3, must be 8

=> Hidden single @ b4: r6c2=8

r1c45 from {128} must have 1 or 2

=> r6c4+r9c5 can't be [21], must have 3

=> r6c5, seeing r6c4+r9c5, can't be 3, must be 4

-or-

ALS-xz:

ALS A: r16c4={238}

ALS B: r19c5={138}

restricted common: x=8

common: z=3 @ r6c4+r9c5

Therefore r6c5, seeing r6c4+r9c5, can't be 3, must be 4.

And then the puzzle can be solved via naked singles.