BMF
Sudoku Player
Registered: 05/19/10
Posts: 3
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| | 05/20/10 at 04:03 PM | Reply with quote | #1 |
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I think this one may be harder than the one I posted yesterday.
Can anyone figure out the next number? Help would be much appreciated.
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Rob
Sudoku Master
Registered: 05/01/08
Posts: 302
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| | 05/23/10 at 06:08 PM | Reply with quote | #2 |
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Another difficult one. Looking at it for myself I found a y-chain but then nothing after that.
I ended up plugging it into the solver at http://www.scanraid.com which came up with another couple of very esoteric moves.
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leeo
Sudoku Solver
Registered: 07/15/10
Posts: 8
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| | 07/23/10 at 08:37 PM | Reply with quote | #3 |
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...6..3...3..28769...9...1.319246578...781...867395421.8...2...12356......48.....
Code:
BMF Special
2479 479 128 | `6 157 `47 | ```3 `458 ``245
``45 ``3 `15 | 14 ``2 ``8 | ```7 ```6 ````9
2467 `47 268 | `9 357 347 | ``28 ```1 ``245
-------------+------------+-----------------
```3 ``1 ``9 | `2 ``4 ``6 | ```5 ```7 ````8
`245 `45 `25 | `7 ``8 ``1 | ``69 ``39 ```36
```8 ``6 ``7 | `3 ``9 ``5 | ```4 ```2 ````1
-------------+------------+-----------------
5679 ``8 `56 | 14 137 ``2 | `169 3459 34567
```1 ``2 ``3 | `5 ``6 479 | ``89 `489 ```47
5679 579 ``4 | `8 137 379 | 1269 `359 23567
I find two moves that unlock the puzzle. The first is a discontinuous "nice loop" that proves r8c9 != {4}
Code:
-4- r8c9 =7= r8c6 =4= r7c4 =1= r2c4 =4= r2c1 =5= r2c3 =1= r1c3 =8= r1c8 -8- r8c8 -4-
thus r8c9 = {7}.
Code:
Sudoku Grid
2479 479 128 | `6 157 `47 | ```3 `458 `245
``45 ``3 `15 | 14 ``2 ``8 | ```7 ```6 ```9
2467 `47 268 | `9 357 347 | ``28 ```1 `245
-------------+------------+----------------
```3 ``1 ``9 | `2 ``4 ``6 | ```5 ```7 ```8
`245 `45 `25 | `7 ``8 ``1 | ``69 ``39 ``36
```8 ``6 ``7 | `3 ``9 ``5 | ```4 ```2 ```1
-------------+------------+----------------
5679 ``8 `56 | 14 137 ``2 | `169 3459 3456
```1 ``2 ``3 | `5 ``6 `49 | ``89 `489 ```7
5679 579 ``4 | `8 137 379 | 1269 `359 2356
This permits the following ALS XZ move that shows r8c7 cannot be {9}
b8.16 -1- c7.57 {9} r8c7
b8.16 means the first and sixth positions in box 8 or r7c4 and r8c6. Here there are three candates in two positions.
c7.57 means the fifth and seventh position of column 7, or r5c7 and r7c7. Here there are also three candidates in two positions.
All the {1}'s of the first set contact all the {1}'s of the second set, meaning that {1} is a restricted common. the candidate {9} in r8c7 contacts all the candidate {9}'s in both sets, meaning that r8c7 cannot be a nine, otherwise both sets would turn into naked pairs, violating that the {1}'s must be in one or the other set.
Thus r8c7 must be {8}
From that point, hidden and naked singles are enough to solve. --leeo
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