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12 Jan X-treme - MOAX!
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sudokuEd
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PostPosted: Mon Jan 15, 2007 10:10 am    Post subject: Reply with quote

Para and Richard,
Wow - you've given so much to chew over. Taken a few days to catch up to you - so thanks for slowing down for me Wink

Richard those HUGE ALS are so hard to find. Congratulations. Any tips for a wanna be?

Para wrote:
I am resulting to raw moves nothing beautiful.
Step 14 is quite pretty actually Para - especially compared to the desperation moves later. Wink

15. is great - love those diagonals when they 'line' up like that - at least, they lined up after you pointed it out. Good one.

16. is very clever. This uses 2 7's in one ALS as restricted common. Never thought of that before - only ever look for ALS with a single isolated digit.

17-19. Chains. Finally get rid of some of those ubiquitous 6's. Couldn't work them out at first - but are very clever, and are pretty easy to see once you get the hang.

Question With these chains, why just stop at 9 links? When is an xy chain too long?

For example, using the same technique with both the 2's in n8 starting from the end of step 16, the xy chains for both just keep going and going. It seems to me that in this case, why not just keep them going? There is so much info they can tell you.

1. An xy chain may eventually reach a contradiction. Of course, that tells you that candidate cannot be in that first cell ( which does happen with one of these chains with the 2's)

But even if there is no contradiction, there is still so much they can tell you.
2.For example, with these 2's in n8, many of the same cells are filled for both chains -> only those two candidates can be in that cell, any others eliminated.

3. Some of the cells have identical placements for both chains - so obviously they are definites.

Any-way, just for the exercise I followed this through. Not right to the end, but with an equal number of links for both chains.

XY Chains with 2 in r8c4 end up with marks here
Code:
.---------------.---------------.---------------.
| 3    1    689 | 7    5    689 | 4    2    69  |
| 2    689  5   | 4    3    689 | 1    689  7   |
| 469  4689 7   | 689  2    1   | 3    689  5   |
:---------------+---------------+---------------:
| 69   5    3   | 689  1    2   | 69   7    4   |
| 1    7    69  | 69   4    3   | 2    5    8   |
| 8    2    4   | 5    7    69  | 69   3    1   |
:---------------+---------------+---------------:
| 469  4689 1   | 3    68   7   | 5    689  2   |
| 5    68   68  | 2    9    4   | 7    1    3   |
| 7    3    2   | 1    68   5   | 689  4    69  |
'---------------'---------------'---------------'

XY Chains with 2 in r7c6
Code:
.---------------.---------------.---------------.
| 3    1    69  | 8    5    79  | 67   4    2   |
| 2    8    56  | 4    137  137 | 167  9    1567|
| 49   459  7   | 6    2    19  | 3    8    15  |
:---------------+---------------+---------------:
| 169  59   3569| 2    137  8   | 167  567  4   |
| 16   7    356 | 9    4    13  | 2    56   8   |
| 8    2    4   | 5    17   6   | 9    3    17  |
:---------------+---------------+---------------:
| 49   49   1   | 3    8    2   | 5    67   67  |
| 5    6    2   | 7    9    4   | 8    1    3   |
| 7    3    8   | 1    6    5   | 4    2    9   |
'---------------'---------------'---------------'


Conclusion: r1c4 = {78}, r1c8 = {24}, r2c1 = 2, r4c6 = {28}, r6c2 = 2, r7c3 = 1, r7c6 = {27}, r8c4 = {27}, r8c7 = {78}, r8c8 = 1, r9c1 = 7, r9c3 = {28}, r9c8 = {24}

Which gets the puzzle to here. There are more chains in one of the marks pics, so more info could be had this way. And this is just with one of the potential chains.

Code:
.---------------------.---------------------.---------------------.
| 3      1      689   | 78     5      6789  | 46789  24     269   |
| 2      689    5689  | 4      13678  136789| 16789  689    15679 |
| 469    45689  7     | 689    2      1689  | 3      689    1569  |
:---------------------+---------------------+---------------------:
| 169    569    3569  | 2689   1367   28    | 1679   5679   4     |
| 169    7      3569  | 69     4      1369  | 2      569    8     |
| 8      2      4     | 5      167    69    | 1679   3      1679  |
:---------------------+---------------------+---------------------:
| 469    4689   1     | 3      68     27    | 5      26789  2679  |
| 5      68     268   | 27     9      4     | 78     1      3     |
| 7      3      28    | 1      68     5     | 4689   24     269   |
'---------------------'---------------------'---------------------'


I'm not proposing to use this approach - but in one sense, why not? If we have used it with 9 links, why not 25? If not with 25, why at all? Just musing.

OK, now getting back on track.

This is where I think we are at at the end of step 24
Code:
.---------------------.---------------------.---------------------.
| 3      1      2689  | 6789   5      6789  | 46789  246789 269   |
| 269    689    25689 | 4      13678  136789| 16789  689    125679|
| 469    45689  7     | 689    2      1689  | 3      689    1569  |
:---------------------+---------------------+---------------------:
| 1269   2569   23569 | 2689   1367   26789 | 1679   5679   4     |
| 169    7      3569  | 69     4      1369  | 2      569    8     |
| 8      269    4     | 5      17     269   | 1679   3      1679  |
:---------------------+---------------------+---------------------:
| 2479   2489   1     | 3      68     2678  | 5      26789  2679  |
| 5      68     268   | 27     9      4     | 78     1      3     |
| 279    3      289   | 1      678    5     | 46789  246789 269   |
'---------------------'---------------------'---------------------'


25. naked triple {689} in r2c28 + r8c2 with 9 locked for r2, so gets rid of lots of 9's in r2. Very Happy
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rcbroughton
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PostPosted: Mon Jan 15, 2007 4:19 pm    Post subject: Reply with quote

sudokuEd wrote:
25. naked triple {689} in r2c28 + r8c2 with 9 locked for r2, so gets rid of lots of 9's in r2. Very Happy

Sweet move!! I'm still not fully into the swing of using the diagonals. This is a great elimination.

Richard
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Para
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PostPosted: Mon Jan 15, 2007 4:40 pm    Post subject: Reply with quote

sudokuEd wrote:
Para wrote:
I am resulting to raw moves nothing beautiful.
Step 14 is quite pretty actually Para - especially compared to the desperation moves later. Wink


Yeah, with those Bowman's Bingo like moves, anything else starts to look pretty.

sudokuEd wrote:

15. is great - love those diagonals when they 'line' up like that - at least, they lined up after you pointed it out. Good one..

Knew you would Wink

sudokuEd wrote:

25. naked triple {689} in r2c28 + r8c2 with 9 locked for r2, so gets rid of lots of 9's in r2. Very Happy


Didn't know you could that over the diagonals as well. We are learning new things here.

Now for finding new steps.

Para

p.s. Last seconds on the internet for today, so i'll try and post something tomorrow. (if i can find anything Confused )
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rcbroughton
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PostPosted: Mon Jan 15, 2007 5:36 pm    Post subject: Reply with quote

Don't know if this is Bowman's Bingo or whatever. Not elegant either way.

26. Any choice of candidate in r4c4 {2689} leads to r9c1=7.
26a. r4c4=2 -> r4c6=8 -> r8c2=6 -> r2c8=9 -> r9c1=7
26b. r4c4=6 -> r5c4=9 -> r6c6=2 -> r4c6=8 -> r8c2=6 -> r2c8=9 -> r9c1=7
26c. r4c4=8 -> r8c4=2 -> r1c4 =7, r8c7=7 -> r2c9=7 -> r4c8=7 -> r6c5=7 -> r7c6=7 -> r9c1=7
26d. r4c4=9 -> r5c4=6 -> r6c6=8 -> r8c2=6 -> r2c8=9 -> r9c1=7

Then follows:

27. 7 now locked in col 5 in n5 - 7 nowhere else in the column

28. naked pair {68} in col 5 in n8 -> no 6/8 anywhere else in column or in n8

29. Hidden {137} in r4c5 r5c6 r6c5 n5 -> no 6/9 in r5c6
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sudokuEd
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PostPosted: Tue Jan 16, 2007 3:16 am    Post subject: Reply with quote

30.no 2 in r7c9
30a. 2 in r9c3->r8c4 -> D\r6c6 ->D/r1c9 -> no 2 r7c9
30b. 2 elsewhere r9 only in n9 -> no 2 r7c9

31. no 2 in r1c3
31a. 2 in r9c3 -> no 2 r1c3
31b. 2 in r9c8 -> 2 in D\ must be in n5 -> 2 in D/ must be in r1c9 -> no 2 in r1c3
31c. 2 in r9c9 -> 2 in r1c8 -> no 2 in r1c3
31d. -> no 2 in r2c9

32.no 2 in r7c1 or r2c3
32a. 2 in r8c4 -> r1c4 & r8c7 = 7
32b.-> naked quint {13689} r2c25678 -> r2c139 = [257] -> no 2 in r7c1 or r2c3
32c. 2 in r7c6 -> no 2 in r7c1
32d. 2 in r7c6 -> 7 in r8c4 -> 2 in r8c3 -> no 2 in r2c3.
32e. r2c1 = 2

How does step 32 seem? If its logical enough, then lets just cut to the chase and do the big short-cut as in my previous post. 'Cause that's all step 32 does (just the first tiny bit). But my previous post shows a much simpler/quicker way to do this type of step.

Why not make our lives easy??? Lets do the double xy-chain for 2's in n8, find the definites, find the cells with just 2 candidates, find the 2 that has the contradiction.

What do you guys think?


Code:
.---------------------.---------------------.---------------------.
| 3      1      689   | 6789   5      6789  | 46789  246789 269   |
| 2      689    568   | 4      13     13678 | 1678   689    1567  |
| 469    45689  7     | 689    2      1689  | 3      689    1569  |
:---------------------+---------------------+---------------------:
| 169    2569   23569 | 2689   137    2689  | 1679   5679   4     |
| 169    7      3569  | 69     4      13    | 2      569    8     |
| 8      269    4     | 5      17     269   | 1679   3      1679  |
:---------------------+---------------------+---------------------:
| 49     2489   1     | 3      68     27    | 5      26789  679   |
| 5      68     268   | 27     9      4     | 78     1      3     |
| 7      3      289   | 1      68     5     | 4689   24689  269   |
'---------------------'---------------------'---------------------'
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Para
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PostPosted: Tue Jan 16, 2007 7:55 am    Post subject: Reply with quote

Hi all

this one is beautiful. Really beautiful.

Code:

.---------------------.---------------------.---------------------.
| 3      1      689   | 6789   5      6789  | 46789  246789 269   |
| 2      689*   568   | 4      13     13678 | 1678   689*   1567  |
| 469    45689  7     | 689    2      1689  | 3      689    1569  |
:---------------------+---------------------+---------------------:
| 169    2569   23569 | 2689   137    2689  | 1679   5679-  4     |
| 169    7      3569  | 69     4      13    | 2      569-   8     |
| 8      269*   4     | 5      17     269*  | 1679*  3      1679* |
:---------------------+---------------------+---------------------:
| 49     2489   1     | 3      68     27    | 5      26789  679   |
| 5      68     268   | 27     9      4     | 78     1      3     |
| 7      3      289   | 1      68     5     | 4689   24689  269   |
'---------------------'---------------------'---------------------'

Step 33.
The squares marked with a star form an finned x-wing for number 9 with R6C79 being the fins. It doesn't look like a finned x-wing at first because R6C6 seems to ruin it, but of R2C2 and R6C26 only 1 number can be a 9, because R6C6 shares the diagonal with R2C2. which gives it the same properties as an x-wing that would only contain R26C6. Thus we can eliminate the 9 from R45C8.

Look it over a little if you don't get it but it fits perfectly. Smile

Para

p.s. by this logic step 15 is actually a variant on the x-wing as well.


Last edited by Para on Wed Jan 17, 2007 12:25 pm; edited 1 time in total
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rcbroughton
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PostPosted: Tue Jan 16, 2007 9:58 am    Post subject: Reply with quote

Following on from Ed's

34. 2 locked in row 2 in n1 -> remove 2 from r2c9

Following on from Para's

35. no 1 in r7c1.
XY-Chain (1)r5c1->r5c6->r3c6->r3c9->(5)r3c9->r2c9->r2c3->r3c2->(4)r3c2->r3c1->r7c1->(9)r7c1

36. no 6 in r4c3
XY chain (3)r4c3->r4c5->r2c5->r2c6->r5c6->(1)r5c6->r5c1->(6)r5c1
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Para
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PostPosted: Tue Jan 16, 2007 6:05 pm    Post subject: Reply with quote

rcbroughton wrote:

35. no 1 in r7c1.
XY-Chain (1)r5c1->r5c6->r3c6->r3c9->(5)r3c9->r2c9->r2c3->r3c2->(4)r3c2->r3c1->r7c1->(9)r7c1


Don't we already have a 1 in r7C3? I think this chain does eliminate the 9 in R5C1 after using step 32.

rcbroughton wrote:

36. no 6 in r4c3
XY chain (3)r4c3->r4c5->r2c5->r2c6->r5c6->(1)r5c6->r5c1->(6)r5c1


after eliminating the 9 in R5C1 this step works.

Para

I'll try and add now. Took me a while to see how these steps work.
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Para
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PostPosted: Tue Jan 16, 2007 6:15 pm    Post subject: Reply with quote

ok, next ones. Just playing with the 9. (i didn't think when i started i would finish the puzzle)

Step 37. no 9 in R6C6
Nishio on 9(at least i think this is Nishio)
R6C6 = 9 -->> R4C7 =9 -->> R5C3 =9 -->> R7C12=9 -->> R9C8=9 -->> R2C2 =9 which clashes with R6C6

Step 38. no 9 in R7C9
R9 9 either in R9C789(all in N9) or R9C3
R9C3 = 9 -->> R5C4=9 -->> R2C2=9 -->> R1C9=9
R7C9 seen by both chain R9C3 as R9C789.

Step 39
xy-wing in R6C6 and R7C69 eliminates 6 from R6C9 and R9C9

Step 40
xy-wing in R7C59 and R8C7 eliminates 8 from R7C8

Step 41
xy-wing in R7C9 and R8C27 eliminates 6 from R1C9

Step 42
Naked pair {29} in R19C9

Step 43
Naked pair {17} in R6C59

Step 44
xy-wing in R6C67 and R9C9 eliminates 9 from R9C7

Step 45
9 locked in N6 for C7. no 9 anywhere else in C7 (this kinda makes step 44 unnecessary but just the way i saw them Wink )

Step 46
xy-wing in R5C4 R6C6 and R9C9 eliminates 9 from R4C4

Step 47
Empty Rectangle for 9.
R2C8 = 9
R2C2 = 9 -->> R79C8 = 9
Either way no 9 in R13C8

Step 48
9 locked on D/ in N3.
No 9 anywhere else on D/
Hidden Single 9 in R5C4

Step 49
Naked pair {68} in R3C48

Step 50
Naked pair {49} in R37C1

Step 51
Naked Pair {16} R45C1

Step 52
Empty rectangle for digit 1
R6C5 = 1
R6C9 = 1 -->> R2C7=1
Either way R2C5 isn't 1

The rest is singles.

We are done. Was on a roll after those steps you came up with.

Para

p.s. number nine... number nine Wink
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rcbroughton
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PostPosted: Tue Jan 16, 2007 9:08 pm    Post subject: Reply with quote

Great work

I get a slightly different concusion after your step 37

38. remove 9 from r5c3 from ALS of [r6c2 r4c1 r5c1] and [r6c6 r4c5 r5c4 r5c6 r6c5] using common candidate 2

39. 9 now a hidden single at r5c4 in row 5

40. naked pair {68} in row 3
41. naked pair {49} in col 1
42. naked pair {16} in col 1

43. 6 rmeoved from r1c9
r1c9(6)-> r2c2=6 (using diagonals in n5) -> r8c2=6 which clashes with r1c9 through D/

44. now r2c8->r2c2->r8c2 on 6 contradiction means r2c8 can't be 6
45. now r9c90>r2c2->r8c2 on 6 contradiction means r9c9 can't be 6

46. naked pair {29} now in col 9 (9 now locked in n6 col 7)

48. now either r6c6 or r7c9 must be 6 (linked through r7c6) so r6c9 cannot be 6

but gets to the same conclusion eventually.

Guess I just worked the sixes, not the nines after your key move 37.
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Para
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PostPosted: Tue Jan 16, 2007 10:58 pm    Post subject: Reply with quote

yeah not really an ALS master. i work better with 1 number techniques(even starting to find a proper way to use Nishio) and xy-wings and xyz-wings(which are in theory mini ALS's). Sometimes i can squeeze out an xy-chain.

But i did the 6's too but they were in my xy-wings.

Glad you are there for those ALS's Wink

Para

p.s. did you guys try the January 13th X-file yet, cause i can't get true it.
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rcbroughton
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PostPosted: Wed Jan 17, 2007 9:36 am    Post subject: Reply with quote

Para wrote:

p.s. did you guys try the January 13th X-file yet, cause i can't get true it.

Just had a quick look this morning on the way in to work. I can place 5 numbers and a half dozen eliminations and then grind to a halt.

Code:
.-------------------------.----------------.-----------------------.
| 178     4678      14    | 3   5    9     | 248    4678    2678   |
| 57      3579      2     | 6   8    4     | 1      3579    379    |
| 4568    34568     359   | 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      1358    123468 |
| 125689  2345689   3459  | 148 3679 2357  | 23458  1345678 2378   |
'-------------------------.----------------.-----------------------'

I'll have a closer look later on today.

Richard
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