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Jan 13th X-treme
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Para
Yokozuna
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PostPosted: Thu Jan 18, 2007 2:56 pm    Post subject: Jan 13th X-treme Reply with quote

Let's make it a thread a we all have reached the same point in the puzzle

Code:
.------------------------.------------------------.------------------------.
| 178     4678    14     | 3       5       9      | 248     4678    2678   |
| 57      3579    2      | 6       8       4      | 1       3579    379    |
| 4568    34568   39     | 7       2       1      | 358     345689  34689  |
:------------------------+------------------------+------------------------:
| 2459    2459    8      | 25      37      37     | 6       149     149    |
| 3       1       6      | 9       4       8      | 7       2       5      |
| 24579   24579   459    | 25      1       6      | 348     3489    3489   |
:------------------------+------------------------+------------------------:
| 245689  2345689 1359   | 148     3679    2357   | 358     345678  234678 |
| 24568   23568   7      | 148     36      235    | 9       138     123468 |
| 125689  2345689 345    | 148     3679    2357   | 23458   1345678 2378   |
'------------------------'------------------------'------------------------'


Let's see what we can make of this one.

Para
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rcbroughton
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PostPosted: Thu Jan 18, 2007 5:42 pm    Post subject: Reply with quote

5 cannot be in r4c1
r4c1=5 -> r6c4=5 -> r3c8=5 -> r2c2=5 -> r9c37 both forced to 5 - so r4c1 cannot be 5
5 cannot be in r9c1
r9c1=5 -> r4c4=5 -> r3c38 both forced to 5 - so r9c1 cannot be 5
5 cannot be in r2c2 (5)r2c2->(2)r4c4->(5)r6c4->(5)r3c8->
r2c2=5 -> r6c4=5 -> r3c8=5 - no valid placement for 5 in n4
5 cannot be in r7c1:
ALS [r1c1 r2c2 r3c3 r7c7 r8c8] and [r2c1] - common value 7 must be in one or the other. the other common candidate 5 at r2c1 and r7c7 eliminates 5 from r7c1

Ed has a couple of moves to follow on,
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sudokuEd
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PostPosted: Fri Jan 19, 2007 4:29 am    Post subject: Reply with quote

Finally caught up.

Can only add 1 good move to this newie. Looks like its going to be an awful puzzle - can feel some ugly moves coming on. Brick wall

Love that ALS move Richard - 2 candidates in one ALS but still restricted common. I think it can go one step further.

5. With that ALS move, when 7 is in r2c1 -> 5 in r7c7 -> 5 in r6c4
But this leaves no 5's for r2 (or 2 5's in r3 for n13 - take your pick)
5a. 7 cannot be in r2c1
5b. r2c1 = 5

Been trying to find a spot to try out my double-xychains-shortcut. None yet.

Should be here
Code:
.------------------------.------------------------.------------------------.
| 178     4678    14     | 3       5       9      | 248     4678    2678   |
| 5       379     2      | 6       8       4      | 1       379     379    |
| 468     3468    39     | 7       2       1      | 358     345689  34689  |
:------------------------+------------------------+------------------------:
| 249     2459    8      | 25      37      37     | 6       149     149    |
| 3       1       6      | 9       4       8      | 7       2       5      |
| 2479    24579   459    | 25      1       6      | 348     3489    3489   |
:------------------------+------------------------+------------------------:
| 24689   2345689 1359   | 148     3679    2357   | 358     345678  234678 |
| 2468    23568   7      | 148     36      235    | 9       138     123468 |
| 12689   2345689 345    | 148     3679    2357   | 23458   1345678 2378   |
'------------------------'------------------------'------------------------'
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rcbroughton
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PostPosted: Fri Jan 19, 2007 9:59 am    Post subject: Reply with quote

sudokuEd wrote:
But this leaves no 5's for r2 (or 2 5's in r3 for n13 - take your pick)
5a. 7 cannot be in r2c1
5b. r2c1 = 5

Finally saw how this elimination works on the train this morning. I was definitely not on the ball yesterday.

Saw another one this morning, though.
6. eliminate 3 from r8c2 from all possibilities {379} at r2c8 (ALS?)
6a. r2c8=3 -> no 3 in r8c2
6b. r2c8=7 -> r4c6=3 -> no 3 in r8c2
6c. r2c8=9 -> r3c3=9 -> 3 must be in r23c3 for n1 -> no 3 in r8c2

[edit]
Just realised the markup doesn't include another one of Ed's moves.

7. 2 cannot be in r9c2
7a. r9c2=2 -> 2 must be in r89c9 in n9
7b. r9c2=2 -> r4c4=2 D\ -> r1c9=2 D/ -> 2 cannot be in r89c9 in n9
so 2 cannot be in r9c2
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Para
Yokozuna
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PostPosted: Fri Jan 19, 2007 9:16 pm    Post subject: Reply with quote

k just started again caught up with you all got this one elimination so far.

8. No 3 in R9C9
8a. R9C9= 3 -->> R3C2=3 -->> R2C8=3 -->> No room for 3 in N7

Para

sorry was sleeping and mistyped lol.


Last edited by Para on Sun Jan 21, 2007 8:02 pm; edited 1 time in total
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rcbroughton
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PostPosted: Sat Jan 20, 2007 2:46 pm    Post subject: Reply with quote

This is getting tough !!

9. No 3 in r9c5
9. r9c5=3 -> r4c6=3 -> r7c2=3 -> r3c3=3 -> r2c9 and r8c9 =3

[edit]

10. 8 can't be in r9c7
10a r9c7=8 -> r1c1 & r9c2=8 -> r8c8=1 -> r8c4=4
10b. r9c7=8 -> r1c1=8 -> r1c3=1 ->r9c1=1 -> r9c4=4 contradiction

11. similarly 8 can't be in r1c7
11a. r1c7=8 -> r1c9=2 -> r6c4=5 -> r3c7=3 -> r3c3=9
11b. r1c7=8 -> r1c9=2 -> r6c4=5 -> r3c7=3 -> r6c7=4 -> r6c3=9 contradiction

Richard
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rcbroughton
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PostPosted: Sun Jan 21, 2007 10:51 am    Post subject: Reply with quote

A couple more for a Sunday morning:

12. 9 cannot be in r6c3
12a. r6c3=9 -> r3c3=3 ->r2c2=9 ->r9c1=9 -> r7c3=1 -> r1c3=4 -> r3c12={68}, r1c1=1
12b. r1c1=1 -> r8c8=8 -> r7c7=5 -> r3c7=8 -> no 8 in r3c12 contradiction

13. ALS [r6c3] and [r6c7 r3c7 r7c7] sharing 4&5 -> no 5 at r7c3

14. 2 can't be in r9c1
14a. r9c1=2 -> r6c4<>2 => r4c4=2 -> r4c4<>5 => r4c2=5 -> r6c3<>5 => r6c3=4 -> r1c3<>4 => r1c3=1 -> r1c1<>1 => r9c1=1 contradiction
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Para
Yokozuna
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PostPosted: Sun Jan 21, 2007 8:00 pm    Post subject: Reply with quote

This one is going number by number. How many eliminations do we have left d'oh! ?

15. No 2 in R8C9
15a. 2's on D/
15a. R1C9=2 -->> no 2 in R8C9
15b. R6C3=2 -->> R9C9=2 -->> no 2 in R8C9
15c. R8C2=2 -->> no 2 in R8C9

16. No 5 in R9C6
16a. 5's on R8
16b. R8C6=5 -->> no 5 in R9C6
16c. R8C2=5 -->> R6C3=5 -->> R4C4=5 -->> R3C8=5 -->> R9C7=5 -->> no 5 in R9C6


Para
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rcbroughton
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PostPosted: Mon Jan 22, 2007 2:25 pm    Post subject: Reply with quote

Para wrote:
This one is going number by number. How many eliminations do we have left d'oh! ?

Too many!! Crying or Very sad

17. 7 not in r9c9. (long implication chain condensed to the salient points ...)
17a. 7 forced to r1c2, r6c1, r2c8, r4c5, r7c6 -> r4c6=3 -> r8c6=2 -> r7c9=2
17b. -> r4c4=2 -> r6c2=2
17c. 17a. and 17b. mean all positions for no 2 in D/ blocked, so 7 can't be in r9c9

18. 7 now locked in D\ in n1

19. ALS using [r1c3 r1c2 r3c1 r3c2] and [r7c3 r2c8 r4c6] on 2
19a. Other common digit 3 removed from r3c7

Richard
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Para
Yokozuna
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PostPosted: Mon Jan 22, 2007 7:49 pm    Post subject: Reply with quote

rcbroughton wrote:

19. ALS using [r1c3 r1c2 r3c1 r3c2] and [r7c3 r2c8 r4c6] on 2
19a. Other common digit 3 removed from r3c7


on 1 i assume?

Para
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rcbroughton
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PostPosted: Mon Jan 22, 2007 7:59 pm    Post subject: Reply with quote

Para wrote:
on 1 i assume?

sorry! you're right. typo

I was getting too excited that I'd found more than one elimination in a row Embarassed
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Para
Yokozuna
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PostPosted: Tue Jan 23, 2007 6:44 pm    Post subject: Reply with quote

Ok we have another number in place through these steps.

Started with a bit of a step 32(of jan 12th)-like chain

20. No 2 or 5 in R6C2
20a. R1C3 = 4 -->> R6C3=5 -->>R6C4 =2 -->> R6C2 canít be 2 or 5
20b. R1C3 = 1 -->> Naked pair {39} in R37C3 -->> no 3 in R7C7 -->> naked triple {258} in D\ in R4C4, R7C7 and R9C9 -->> R1C1 = 7 -->> R6C2 = 7 -->> R6C2 canít be 2 or 5

21. R1C7 is not 4
21a. R1C7 = 4 -->> R1C9 = 2 -->> R7C9 = 2 -->> R4C4 = 2 -->> R6C1 = 2 -->> R7C2 = 2 -->> R8C6 =2
21b. R1C7 = 4 -->> R1C9 = 2 -->> R4C6 = 5 -->> R8C6 = 5
21c. R8C6 needs both 2 and 5 if 4 is in R1C7, so no 4 in R1C7

22. Naked single 2 in R1C7

Para

p.s. I think the chain in step 20 can be extended but i found this one already long enough.
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rcbroughton
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PostPosted: Tue Jan 23, 2007 7:46 pm    Post subject: Reply with quote

This puzzle is driving me mad! It is so tenacious.

23. Can't have 8 in r7c8
23a. r7c8=8 -> r9c9=2 -> r4c4=5 -> r7c7 =3 ->r8c8 =1, r3c3=9 -> r2c2=7, r8c3=1 -> r1c1=8 -> r1c3=4 -> r1c2 & r3c1 =6 contradiction

[edit]
had another look and almost thought I had it.

24. 2 cannot be in r7c8
24a. r7c8=2 -> r9c9=8 ->r4c4=2 ->r6c4=5 ->r6c3=4, r3c7=8 -> r1c3=1 -> r1c1=7 -> r8c8=1, r1c9=6
24b. r8c8=1 -> r4c9=1 -> r4c8=4 -> r1c8=6 contradiction

25. Hidden single 2 in column 9 at r9c9 -> r4c4=5 -> r6c4=2

26. Hidden single 5 in row 6 at r6c3

27. Naked pair {37} r49c6 for col 6

28. ALS [r1c3 r1c2 r1c8 r1c9]=1=[r7c3 r3c3 r7c7] -> No 8 in r1c1

29. 8 now locked in N9 for D\

30. ALS [r9c3]=4=[r9c7 r3c7 r7c7] -> No 3 in r9c8

31. r6c1 cannot be 4
31a. r6c1=7 -> r6c1<>4
31a. only other place r6c2=7->r6c1<>7=>r1c1=7=>r1c3=1=>r9c3=4->r9c7<>4=>r6c7=4->r6c1<>4

32. r8c1 cannot be 8
32a. r8c1=2 -> r8c1 <>8
32b. r8c1<>2=>r8c6=2->r8c6<>5=>r8c2=5->r3c7<>5=>r3c7=8->r7c7<>8=>r8c8=8 -> r8c1<>8

Thought it was going to unravel but ground to a halt again !
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Para
Yokozuna
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PostPosted: Tue Jan 23, 2007 10:41 pm    Post subject: Reply with quote

Nice moves

rcbroughton wrote:

24. 2 cannot be in r7c8
24a. r7c8=2 -> r9c9=8 ->r4c4=2 ->r6c4=5 ->r6c3=4, r3c7=8 -> r1c3=1 -> r1c1=7 -> r8c8=1, r1c9=6
24b. r8c8=1 -> r4c9=1 -> r4c8=4 -> r1c8=6 contradiction


R7C9 Wink

rcbroughton wrote:

I was getting too excited that I'd found more than one elimination in a row


still got the problem Wink
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Para
Yokozuna
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PostPosted: Tue Jan 23, 2007 11:16 pm    Post subject: Reply with quote

33. No 3 in R7C3
33a. R3C7 = 8 -->> R7C7 = 3 -->> no 3 in R7C3
33b. R3C7 = 5 -->> naked pair {43) in R9C37 -->> R9C6 =7 -->> R4C6 = 3 -->> no 3 in R7C3

34. No 3 in R2C2
34a. R7C3 = 9 -->> R3C3 = 3 -->> no 3 in R2C2
34b. R7C3 = 1 -->> R1C1 = 1 -->> R2C2 = 7 -->> no 3 in R2C2

35. 3 locked in N1 for R3. no 3 anywhere else in R3.



Para
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