24 March, 2006

[David Bryant]

I've been taking private lessons from someone_somewhere.

Today's puzzle is extremely tough. I found a 9-star constellation ("alpha star" at r8c6), then a 5-star constellation (same alpha star), and finally a 7-star constellation (alpha star at r6c9) before this one finally crumbled.

Great puzzle, Ruud! dcb

[keith]

David,

Maybe I can take lessons from you!

I get to this point:

Code: Select all

+----------------------+----------------------+----------------------+
| 4      369    379    | 269    5      1      | 27     69     8      | 
| 69     1      79     | 269    8      4      | 3      5      27     | 
| 2      8      5      | 369    7      369    | 1      4      69     | 
+----------------------+----------------------+----------------------+
| 369    5      4      | 1      236    236789 | 6789   689    3679   | 
| 1      7      39     | 3469   346    3689   | 689    2      5      | 
| 8      369    2      | 3679   36     5      | 4      1      3679   | 
+----------------------+----------------------+----------------------+
| 7      4      8      | 5      9      26     | 26     3      1      | 
| 39     239    6      | 8      1      23     | 5      7      4      | 
| 5      23     1      | 3467   2346   2367   | 2689   689    269    | 
+----------------------+----------------------+----------------------+

There is one deduction that breaks the puzzle wide open:

R6C2 is not <9>. (And then you have a pair <36> in R6, etc)

How on earth does a normal person make this deduction??

Keith

[David Bryant]

Hi, Keith!

Yes, this is the spot where I found the first big "constellation" in this puzzle. When I posted my message earlier I called it a "9-star constellation," but looking at it again I see how it can be reduced to seven "stars." Anyway, here's how that works.

Code: Select all

+----------------------+----------------------+----------------------+
| 4      369    379    | 269    5      1      | 27     69     8      |
| 69     1      79     | 269    8      4      | 3      5      27x    |
| 2      8      5      | 369    7      369    | 1      4      69     |
+----------------------+----------------------+----------------------+
| 369    5      4      | 1      236    236789 | 6789   689    3679   |
| 1      7      39     | 3469   346    3689   | 689    2      5      |
| 8      369    2      | 3679   36     5      | 4      1      3679   |
+----------------------+----------------------+----------------------+
| 7      4      8      | 5      9      26     | 26     3      1      |
| 39     239    6      | 8      1      23*    | 5      7      4      |
| 5      23     1      | 3467   2346   2367   | 2689   689    269    |
+----------------------+----------------------+----------------------+ 

The "Alpha Star" is at r8c6. The "target" is r2c9.

A. r8c6 = 2 ==> r7c7 = 2 ==> r1c7 = 7 ==> r2c9 = 2.
B. r8c6 = 3 ==> r4c1 = 3 ==> r5c3 = 9 ==> r2c3 = 7 ==> r2c9 = 2.

So r2c9 has to be "2". If you follow this up and do some coloring on the digit "6" you'll soon reach a point where assuming that r8c6 = 3 causes a contradiction ... that's the second "constellation" I was talking about earlier.

Oh -- I should point out that some of the coloring on "6" is already possible in the grid you posted. There are only two ways to fit a "6" in column 2, and there are also just two ways to fit a "6" in the top right 3x3 box.

r1c2 = 6 ==> r3c9 = 6 ==> r6c9 <> 6
r1c2 <> 6 ==> r6c2 = 6 ==> r6c9 <> 6

You'll be able to extend this pattern after you put the "2" in at r2c9. dcb

PS I don't know how to prove that r6c2 <> 9. Not directly, anyway. And looking at my notes, I see that I still had "9" as a possibility at r6c2 even after I found the third "constellation," which cracked the puzzle wide open for me.

[Ruud]

Hi guys,

always interesting to see how people find different ways to solve these Nightmares.

The fish-style eliminations of digit 6 are a red herring. They accomplish nothing.

This was my intention:

Code: Select all

.---------------------.---------------------.---------------------.
| 4     *369    379   | 269    5      1     | 27   bc69     8     |
| 69     1      79    | 269    8      4     | 3      5      27    |
| 2      8      5     | 369    7     b369   | 1      4      69    |
:---------------------+---------------------+---------------------:
| 369    5      4     | 1      236    236789| 6789   689    3679  |
| 1      7      39    | 3469   346    3689  | 689    2      5     |
| 8     a369    2     |b3679   36     5     | 4      1     c3679  |
:---------------------+---------------------+---------------------:
| 7      4      8     | 5      9      26    | 26     3      1     |
| 39     239    6     | 8      1      23    | 5      7      4     |
| 5      23     1     | 3467   2346   2367  | 2689   689    269   |
'---------------------'---------------------'---------------------'

There are 3 candidates for digit 9 in row 6. I named them a, b, c. Check how each of these 3 candidates kills candidate 9 in *r1c2.

The next step is an XY wing rooted in #r2c1 with pincers in @r1c2 and @r8c1. (didn't we just cause that bivalue in r1c2?)

Code: Select all

.---------------------.---------------------.---------------------.
| 4     @36     379   | 269    5      1     | 27     69     8     |
|#69     1      79    | 269    8      4     | 3      5      27    |
| 2      8      5     | 369    7      369   | 1      4      69    |
:---------------------+---------------------+---------------------:
| 369    5      4     | 1      236    236789| 6789   689    3679  |
| 1      7      39    | 3469   346    3689  | 689    2      5     |
| 8      369    2     | 3679   36     5     | 4      1      3679  |
:---------------------+---------------------+---------------------:
| 7      4      8     | 5      9      26    | 26     3      1     |
|@39    -239    6     | 8      1      23    | 5      7      4     |
| 5     -23     1     | 3467   2346   2367  | 2689   689    269   |
'---------------------'---------------------'---------------------'

r8c2 and r9c2 are the victims here. The remainder of the puzzle can be solved with only a hidden pair.

Cheers,

Ruud.

[David Bryant]

Hi, Ruud!

The "template" (aka Nishio) eliminations are interesting. But any logical path to the solution is valid, so long as it works.

Beginning at this position:

Code: Select all

  4    369   379   269    5     1    27    69     8
 69     1    79    269    8     4     3     5    27
  2     8     5    369    7    369    1     4    69
 369    5     4     1    236 236789 6789   689  3679
  1     7    39   3469   346  3689   689    2     5
  8    369    2    379   36     5     4     1    379
  7     4     8     5     9    26    26     3     1
 39    239    6     8     1    23*    5     7     4
  5    23     1   3467  2346  2376  2689   689   269

We have a "7-star constellation" rooted in r8c6.

A. r8c6 = 2 ==> r7c7 = 2 ==> r7c1 = 7 ==> r2c9 = 2.
B. r8c6 = 3 ==> r4c1 = 3 ==> r5c3 = 9 ==> r2c3 = 7 ==> r2c9 = 2.

1. r2c9 = 2 (by DIC); r1c7 = 7 (sole candidate).
2. r1c4 = 2; r2c3 = 7 (unique horizontal).
3. Naked triplet {6, 8, 9} in center right 3x3 box.
4. Coloring eliminates "6" at r4c8, r5c4, r7c7, & r9c7 (it's not really a red herring!).
5. r7c7 = 2; r7c6 = 6 (sole candidate).
6. Coloring in rows 2 & 6 reveals that r1c2 <> 9.

Now a "5-star constellation" reveals that r8c6 <> 3:

Code: Select all

  4    36    39     2     5     1     7    69     8
 69     1     7    69     8     4     3     5     2
  2     8     5    369    7    39     1     4    69
 369    5     4     1    236  23789  689   89    37
  1     7    39    349   346   389   689    2     5
  8    369    2    379   36     5     4     1    37
  7     4     8     5     9     6     2     3     1
 39    239    6     8     1    23*    5     7     4
  5    23     1    347   234   237   89    689   69

r8c6 = 3 ==> r8c1 = 9 ==> r2c1 = 6 ==> r2c4 = 9 ==> r3c6 = 3.

We can't have two "3"s in column 6, therefore r8c6 = 2

7. r8c6 = 2 (sole candidate); r4c5 = 2; r9c2 = 2 (unique horizontal).

Here I found another "7-star constellation." The Alpha Star is r6c9.

Code: Select all

  4    36    39     2     5     1     7    69     8
 69     1     7    69     8     4     3     5     2
  2     8     5    369    7    39     1     4    69
 369    5     4     1     2   3789   689   89    37
  1     7    39    349   346   389   689    2     5
  8    369    2    379   36     5     4     1    37*
  7     4     8     5     9     6     2     3     1
 39    39     6     8     1     2     5     7     4
  5     2     1    347   34    37    89    689   69

A. r6c9 = 7 ==> r4c6 = 7 ==> r9c6 = 3 ==> r3c6 = 9.
B. r6c9 = 3 ==> r6c5 = 6 ==> {3, 9} pair in r5c3 & r6c2 ==> r4c1 = 6;
B1 r6c9 = 3 ==> r4c9 = 7 ==> r6c6 = 3 ==> r3c6 = 9.

And with r3c6 set equal to "9" this stubborn puzzle finally gives up the ghost! dcb

[Ron Moore]

Again, this puzzle is long forgotten, I'm sure, but while looking at this thread I saw the answer to the question Keith posed.

Code: Select all

+----------------------+----------------------+----------------------+
| 4      369    379    | 269    5      1      | 27     69     8      |
| 69     1      79     | 269    8      4      | 3      5      27     |
| 2      8      5      | 369    7      369    | 1      4      69     |
+----------------------+----------------------+----------------------+
| 369    5      4      | 1      236    236789 | 6789   689    3679   |
| 1      7      39     | 3469   346    3689   | 689    2      5      |
| 8      369    2      | 3679   36     5      | 4      1      3679   |
+----------------------+----------------------+----------------------+
| 7      4      8      | 5      9      26     | 26     3      1      |
| 39     239    6      | 8      1      23     | 5      7      4      |
| 5      23     1      | 3467   2346   2367   | 2689   689    269    |
+----------------------+----------------------+----------------------+

There is one deduction that breaks the puzzle wide open:

R6C2 is not <9>. (And then you have a pair <36> in R6, etc)

How on earth does a normal person make this deduction??

Keith

Maybe Ruud answered the question indirectly, but here's a more direct answer. Consider r1c2.

If r1c2=9, then immediately we have r6c2 <> 9.

If r1c2 is not 9, then r1c2 is 3 or 6, so that r1c2, r2c1, and r8c1 form the XY wing which Ruud mentioned. This eliminates 3 from r8c2 and r9c2. So r9c2=2 and then r8c2=9 and r6c2 <> 9.

Here's another way, which seems simpler to me for Keith's specific question. In column 2, r1c2 and r6c2 are a conjugate pair (strongly linked) for the digit 6. So

r6c2=9 --> r1c2=6 --> r2c1=9

These two placements of 9's would eliminate both 9 candidates in box 7.

Ron