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A humbling puzzle

 
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Tue Apr 04, 2006 5:08 pm    Post subject: A humbling puzzle Reply with quote
Just when I think I'm getting proficient, a puzzle like this comes along to quickly disabuse me of that notion. I've solved exactly five cells, all in the initial cross-hatching. I feel figuratively constipated; I can't eliminate anything.

There was an early jellyfish, but the one candidate I could erase was of zero help.

Code:
----------------------------------------------------------
|6     1239  239   |2349  8     2349  |145   7     25    |
|5     279   2789  |1     69    24679 |48    3     268   |
|2378  1237  4     |237   5     2367  |18    9     268   |
----------------------------------------------------------
|23789 2379  6     |3589  4     389   |3589  258   1     |
|389   349   5     |6     2     1     |7     48    389   |
|1     2349  2389  |3589  7     389   |6     2458  3589  |
----------------------------------------------------------
|3479  8     1379  |479   169   4679  |2     156   3569  |
|2349  6     1239  |2489  19    5     |389   18    7     |
|279   5     1279  |2789  3     26789 |89    168   4     |
----------------------------------------------------------
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
PostPosted: Tue Apr 04, 2006 5:19 pm    Post subject: Not much help Reply with quote
Look at <6> in B3 and C9. There is an XY-wing in R8C8, but that is not much help.

Keith
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George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
PostPosted: Tue Apr 04, 2006 10:54 pm    Post subject: Reply with quote
No expert but the 6 in box9 col 9 can be removed
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ravel
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PostPosted: Wed Apr 05, 2006 9:12 am    Post subject: Reply with quote
Pretty hard puzzle. Like Keith said, you can eliminate a 6 and 3 9's.
You can also show, that r1c9=2:
if r1c9=5, triple 148 in r123c7, r8c7=3, r9c7=9
no number for r7c9 left

But then ... ??
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Wed Apr 05, 2006 4:56 pm    Post subject: Reply with quote
Thanks guys. I'll see where that takes me.

Quote:
You can also show, that r1c9=2:
if r1c9=5, triple 148 in r123c7, r8c7=3, r9c7=9
no number for r7c9 left


Ravel, I'm not sure why there's no number for r7c9. The way I follow it is that with the "5" in r1c9 and the "3" and "9" in r8c7 and r9c7, a "6" is forced into r7c9. That then leaves no "6" and a duplicate "2" in box 3. Of course, either way proves that a "5" can't dwell in r1c9.

Once again, thanks to all.
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Guest
PostPosted: Thu Apr 06, 2006 7:45 am    Post subject: Reply with quote
The 6 in r7c9 can be eliminated with simple box(3)/line(col 9) intersection.
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Thu Apr 06, 2006 5:00 pm    Post subject: Reply with quote
Anonymous wrote:
The 6 in r7c9 can be eliminated with simple box(3)/line(col 9) intersection.

Thanks, I was clearly asleep at the xx.
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ravel
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PostPosted: Thu Apr 06, 2006 5:24 pm    Post subject: Reply with quote
I managed to solve the puzzle in the meantime, but i needed 4 complicated and long steps (with more than 10 cells involved) to get 3 more numbers and 2 forcing chains after that.
I wonder, if there is an easier solution.
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Guest
PostPosted: Thu Apr 06, 2006 9:23 pm    Post subject: Reply with quote
I have written down my solution now:
After xy-wing and r1c9<:
Code:
 *--------------------------------------------------------------------*
 | 6      1      39     | 349    8      349    | 5      7      2      |
 | 5      279    2789   | 1      69*    2679   | 4      3      68*    |
 | 2378*  237    4      | 237    5      2367   | 1      9      68*    |
 |----------------------+----------------------+----------------------|
 | 23789  2379   6      | 3589   4      389    | 389    258    1      |
 | 389*   349    5      | 6      2      1      | 7      48*    39     |
 | 1      2349   2389   | 3589   7      389    | 6      2458   359    |
 |----------------------+----------------------+----------------------|
 | 3479   8      1379   | 479    169    4679   | 2      156    359    |
 | 2349   6      1239   | 2489   19*    5      | 38     18*    7      |
 | 279    5      1279   | 278    3      2678   | 89     168    4      |
 *--------------------------------------------------------------------*
If you look at the marked cells, you can see, that there are only 2 possibilities in this loop, where r3c1 and r5c1 are either 8 or not 8, the rest has to be one of the 2 candidates. From this follows, that
r469c8<>8 (either r5c8 or r8c8 must be 8)
r4c1<>8 (either r3c1 or r5c1 must be 8)
r2c6<>6 (either r2c5 or r2c9 must be 6)
r7c5<>9 (either r2c5 or r7c5 must be 9)

Next eliminate 1 from r9c8.
If r9c8=1, then r8c8=8 and
1. r8c5=1, r7c5=6, r7c8=5
2. r5c8=4, pair r46c8=25, r7c8<>5
=> r9c8=6

Eliminate 8 from r9c7:
If r9c7=8,
1. pair r9c45=27, r9c1=9
2. r5c8=8 and r7c9=9,r5c9=3 -> r5c1=9

Then we come here:
Code:
 *-----------------------------------------------------------*
 | 6     1     39    | 349   8     349   | 5     7     2     |
 | 5     279   2789  | 1     69    279   | 4     3     68    |
 | 2378  237   4     | 237   5     2367  | 1     9     68    |
 |-------------------+-------------------+-------------------|
 | 2379  2379  6     | 3589  4     389   | 38    25    1     |
 | 389   349   5     | 6     2     1     | 7     48    39    |
 | 1     2349  2389  | 3589  7     389   | 6     245   359   |
 |-------------------+-------------------+-------------------|
 | 3479  8     379   | 479   16    4679  | 2     15    35    |
 | 2349  6     239   | 249   19    5     | 38    18    7     |
 | 27    5     1     | 278   3     278   | 9     6     4     |
 *-----------------------------------------------------------*

From this point, susser found a better solution than mine was. I give it in my words:

I dont know, how this fish monster is called, but a 9 in r8c3 leaves an x-wing in r17c46, which kills all 9's in box. So r8c3<>9.

if r5c1=9:
r5c8=8
r5c9=9, r4c7=8
=>r5c1<>9

if r6c3=2:
r8c3=3
r5c1=8,r5c8=4,r8c8=1,r8c7=3
=> r6c3<>2

If r2c3=8:
r7c3=7
r8c3=2,r9c1=7
=> r2c3=8, r2c9=8

This finally solves the puzzle.
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ravel
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PostPosted: Thu Apr 06, 2006 9:26 pm    Post subject: Reply with quote
It was me Smile
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
PostPosted: Thu Apr 06, 2006 10:07 pm    Post subject: One (long) chain will do the trick. Reply with quote
Hi, Ravel!

Thanks for the explanation. Here's a slightly simpler way to solve this puzzle.

After incorporating all the excellent suggestions people had made in this forum, and making all the obvious moves that followed I arrived at this position.

Code:
  6    139   39    349    8    349    5     7     2
  5    279  2789    1    69   2679    4     3    68
2378   237    4    237    5   2367    1     9    68
23789 2379    6   3589    4    389   389   258    1
 389   349    5     6     2     1     7    48    39
  1   2349  2389  3589    7    389    6   2458  3589
3479    8   1379   479   169  4679    2    156   359
2349    6   1239  2489   19     5    38    18     7
 279    5   1279   278    3   2678   89    168    4


We can prove that r2c5 <> 9 as follows.

r2c5 = 9 ==> r8c5 = 1 ==> r7c5 = 6
r8c5 = 1 ==> r8c8 = 8 ==> r8c7 = 3 & r9c7 = 9
r8c8 = 8 ==> r5c8 = 4 ==> r5c1 = 8

With these changes applied the grid looks like this.

Code:
  6    139   39    34     8    34     5     7     2
  5    27    278    1     9    267    4     3    68
 237   237    4    237    5   2367    1     9    68
2379  2379    6   3589    4    389    8    25     1
  8    39     5     6     2     1     7     4    39
  1   2349   239  3589    7    389    6    25    39
3479    8   1379   479    6    479    2    15    59
 249    6    29   2489    1     5     3     8     7
 27     5    127   278    3    278    9    16     4


Now we can see the contradiction:

{3, 4} pair in r1c4&6 ==> r1c3 = 9
r9c7 = 9 ==> r7c9 = 5 ==> r7c8 = 1
r1c3 = 9 ==> r8c3 = 2 ==> r6c3 = 3 ==> r7c3 = 7

and there are now no candidates left for r9c1.

With r2c5 = 6 the puzzle is easily completed. dcb
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ravel
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PostPosted: Fri Apr 07, 2006 9:31 am    Post subject: Reply with quote
Hi David,

fine solution.

I had a similar proof to show that r3c1<>9, but after setting the 2 numbers.
So i am approved now, that this is a hard puzzle and that there is more than one way to skin a sudoku Smile
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