Feb 1 Nightmare

[lac]

Anybody got a way to solve it that doesnt involve uniqueness?

Laura

[David Bryant]

Hi, Laura!

I hadn't planned to do this, but since you asked I went ahead and gave it a try.
After doing the obvious stuff the grid looks like this.

Code: Select all

 24     5    147   126   79     8    369  23469  36
  3    278    6    25     4    79    58    29     1
 248    9    148  1256    3    16     7    246   58
  1   4678   478    9    568    2   3568   367  3568
5689   678    3    14    568   14     2    679   568
25689  268   89     7    568    3   15689  169    4
 469  1346    5    246   279  4679   136    8    27
  7    36     2     8     1     5     4    36     9
4689  1468  1489    3    279  4679   16     5    27

It's apparent that there aren't many ways to place the "9"s in this puzzle. And
there are only two ways to put a "9" in the top center 3x3 box. So let's see what
happens if we set r1c5 = 9. This immediately forces the rest of the "9"s, plus a "7"
at r2c6 and an "8" at r6c3.

After these hypothetical moves the grid looks like this.

Code: Select all

 24     5    147   126    9     8    36    24    36
  3    28     6    25     4     7    58     9     1
 248    9    14   1256    3    16     7    24    58
  1    467   47     9    568    2   3568   367  3568
  9    67     3    14    568   14     2    67    568
 256   26     8     7    568    3     9    16     4
 46   1346    5    46    27     9    136    8    27
  7    36     2     8     1     5     4    36     9
 468  1468    9     3    27    46    16     5    27

Notice that setting r1c5 = 9 creates a naked pair {3, 6} in the top right 3x3 box,
and that allows us to identify another pair {2, 4} in column 8. The same move
(r1c5 = 9) also creates the {2, 7} pair in r7c5 & r9c5, and uncovers {4, 6} in
r7c4 & r9c6.

Now the path to a contradiction is fairly simple.

-- We can eliminate the "2" and the "6" at r1c4 because of the pairs in row 1. So r1c4 = 1.

-- r1c4 = 1 ==> r5c5 = 4 ==> r7c4 = 6 ==> r7c1 = 4.

And now it's clear that we can't complete the top left 3x3 box, because there are
only two values -- "2" and "8" -- that can possibly fit in the three squares r1c1,
r2c2, & r3c1. Therefore r1c5 <> 9 and we must have r2c6 = 9. dcb

[lac]

We can eliminate the "2" and the "6" at r1c4 because of the pairs in row 1. So r1c4 = 1.

Fantastic. I missed that altogether. thank you.

Laura