Just to let you know,
todays Daily Nightmare is not from my own collection. It's a scrambled version of one presented by somebody on a forum far far away...
It lacks symmetry, but makes up by letting you choose to solve it the very hard way, or perform an ingenious technique that I will explain tomorrow (unless the Nightmare gang already discovers it)
I'm confident you can do it.
Enjoy, Ruud.
April 19, an outsider
April 19, an outsider
“If the human brain were so simple that we could understand it, we would be so simple that we couldn't.” - Emerson M Pugh
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- Location: Denver, Colorado
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Look at column 1
Hi, Ruud!
I'm not sure if this is the "easy way" or not, but I noticed a naked triplet in column 1, which made the rest of the puzzle relatively straightforward. Some coloring on the "5"s and "7"s finished it off. dcb
I'm not sure if this is the "easy way" or not, but I noticed a naked triplet in column 1, which made the rest of the puzzle relatively straightforward. Some coloring on the "5"s and "7"s finished it off. dcb
Defeated?
I get to a stage where all the possibilities are pairs, except for four squares which each have three possibilities. If I can eliminate <7> from three of these squares, the fourth solves the puzzle with a BUG.
Except, I cannot find a reasonable way to resolve the <7>'s.
Keith
Except, I cannot find a reasonable way to resolve the <7>'s.
Code: Select all
+-------------+-------------+-------------+
| 89 19 3 | 18 57 4 | 2 6 57 |
| 157 6 4 | 3 17 2 | 9 57 8 |
| 2 78 57 | 9 578 6 | 1 4 3 |
+-------------+-------------+-------------+
| 17 3 9 | 4 2 8 | 57 157 6 |
| 68 5 16 | 7 9 13 | 38 2 4 |
| 4 78 2 | 6 13 5 | 38 17 9 |
+-------------+-------------+-------------+
| 369 19 16 | 5 38 7 | 4 89 2 |
| 39 2 57 | 18 4 13 | 6 89 57 |
| 57 4 8 | 2 6 9 | 57 3 1 |
+-------------+-------------+-------------+
Hi guys,
I knew I could count on you 2 to give it a try.
Here is my solution:
Starting position:
After basics,
a naked pair in column 7... more basics...
a naked pair in box 9,
the naked triple in column 1,
a hidden pair in row 8,
a naked pair in column 3... more basics...
brings us here:
At this moment, you can find a swordfish for digit 5 in rows 1,3,8... which takes you on a path with more subsets, another swordfish, some coloring, an XYZ-wing culminating in a N-star constellation.
...but wait...there is an interesting development:
Look at the cells I marked with a star. Normally, you do not care about cells that are solved, but take another look.
What if R6C2 contains a 7? Then there would be an isolated group of digits 5 and 7 in box 458, rows 567 and columns 246. This isolates the remaining candidates for 5 and 7 with no given value. As a result, there are 2 ways to place the remaining candidates 5 and 7. There would be at least 2 solutions to the puzzle.
Not when we assume a unique solution. So we can eliminate R6C2 digit 7 as a candidate, which allows us to solve the remainder of the puzzle with singles only.
This is a reversed uniqueness situation, a.k.a. reversed BUG. I like it because it is so easy to see, it uses completed cells, and it is so easy to build into the next version of my solver
In this puzzle, it significantly shortens the solving path.
David, have you been solving the same puzzle?
Ruud.
I knew I could count on you 2 to give it a try.
Here is my solution:
Starting position:
Code: Select all
. . 3|. . .|. 6 .
. 6 4|. . .|9 . 8
2 . .|. . .|1 . .
-----+-----+-----
. 3 9|4 2 8|. . 6
. . .|. 9 .|. . .
4 . 2|. . 5|. . 9
-----+-----+-----
. . .|5 . 7|4 . 2
. . .|. . .|. . .
. . 8|2 . .|. 3 1
a naked pair in column 7... more basics...
a naked pair in box 9,
the naked triple in column 1,
a hidden pair in row 8,
a naked pair in column 3... more basics...
brings us here:
Code: Select all
.------------------.------------------.------------------.
| 89 1789 3 | 189 14578 14 | 2 6 57 |
| 157 6 4 | 13 1357 2 | 9 57 8 |
| 2 789 57 | 89 578 6 | 1 4 3 |
:------------------+------------------+------------------:
| 17 3 9 | 4 2 8 | 57 157 6 |
| 68 5 16 | 7 9 13 | 38 2 4 |
| 4 178 2 | 6 13 5 | 38 17 9 |
:------------------+------------------+------------------:
| 369 19 16 | 5 38 7 | 4 89 2 |
| 39 2 57 | 138 1348 134 | 6 89 57 |
| 57 4 8 | 2 6 9 | 57 3 1 |
'------------------'------------------'------------------'
...but wait...there is an interesting development:
Code: Select all
.------------------.------------------.------------------.
| 89 1789 3 | 189 14578 14 | 2 6 57 |
| 157 6 4 | 13 1357 2 | 9 57 8 |
| 2 789 57 | 89 578 6 | 1 4 3 |
:------------------+------------------+------------------:
| 17 3 9 | 4 2 8 | 57 157 6 |
| 68 *5 16 |*7 9 13 | 38 2 4 |
| 4 *178 2 | 6 13 *5 | 38 17 9 |
:------------------+------------------+------------------:
| 369 19 16 |*5 38 *7 | 4 89 2 |
| 39 2 57 | 138 1348 134 | 6 89 57 |
| 57 4 8 | 2 6 9 | 57 3 1 |
'------------------'------------------'------------------'
What if R6C2 contains a 7? Then there would be an isolated group of digits 5 and 7 in box 458, rows 567 and columns 246. This isolates the remaining candidates for 5 and 7 with no given value. As a result, there are 2 ways to place the remaining candidates 5 and 7. There would be at least 2 solutions to the puzzle.
Not when we assume a unique solution. So we can eliminate R6C2 digit 7 as a candidate, which allows us to solve the remainder of the puzzle with singles only.
This is a reversed uniqueness situation, a.k.a. reversed BUG. I like it because it is so easy to see, it uses completed cells, and it is so easy to build into the next version of my solver
In this puzzle, it significantly shortens the solving path.
David, have you been solving the same puzzle?
Ruud.
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- Posts: 86
- Joined: Fri Jan 20, 2006 6:21 pm
- Location: Denver, Colorado
- Contact:
Oops! I goofed.
Yes, I was working the same puzzle. But I made a lucky(?) mistake, so my "solution" wasn't really valid.Ruud wrote:David, have you been solving the same puzzle?
At the point you illustrated I started chasing the chain of "5"s and "7"s, and I must have miscolored a link somewhere, because I thought I had proved that r4c8 = 1. Looking at it again I see that coloring doesn't allow that deduction.
The thing is, r4c8 = 1 is enough to solve the puzzle, so I didn't notice my logic error until I went back and re-examined the steps I had taken. My bad. dcb
PS I like your explanation, Ruud. It's really just a variation on the "non-unique rectangle", but this one's a "non-unique hexagon." Neat!
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- Joined: Tue Apr 04, 2006 7:07 am
I commented on the other site, so I didn't want to wreck your surprise. Nice explanation by the way. Remember that your starred cells forming your non-unique deadly pattern (aka unavoidable set) have to at least contain all of the initially given clues for those digits. I think you can have other solved clues outside.