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Paladin
Joined: 10 Feb 2006
Posts: 15
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 Posted: Mon Oct 16, 2006 9:44 pm Post subject: "Coloring" |
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| Code: |
+----------------------+------------------------------+---------------------------+
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| 6 8 2* | 1,3,9 1,3,9 4* | 5* 1,9 7* |
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| 4 3 5 | 1,6,9 1,6,9 7* | 2* 1,9 8 |
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| 1 9* 7 | 5 2 8* | 3,6 4* 3,6 |
+----------------------+------------------------------+---------------------------+
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| 8* 6 3* | 1,9 1,5,9 2 | 4 7* 1,5 |
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| 5 2* 4* | 8 7 1,3 | 9* 6* 1,3 |
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| 7 1* 9 | 3,6 4 3,5,6 | 8* 3,5 2* |
+----------------------+------------------------------+---------------------------+
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| 2 4* 1 | 7* 3,5,6 3,5,6,9 | 3,6 8* 5,9 |
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| 3 5 8* | 4* 1,6 1,6,9 | 7 2 6,9 |
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| 9* 7 6* | 2* 8 3,5 | 1* 3,5 4 |
+----------------------+------------------------------+---------------------------+
This is a "Super Hard" puzzle that appeared on June 6, 2006 (Starred numbers signify original grid).
After setting 28 cells with normal methods, it appeared as if I was at an impasse(I did not notice the X-wings of 5's at r47c59 and r69c68).
I then attempted to "color" on 3's, as follows:
r1c4 (+) --> r1c5 --> r7c5 --> r9c6 --> r9c8 --> r7c7 --> r3c7 --> r3c9 --> r5c9 (+);
r1c4 (+) --> r6c4 (-).
If r5c9 is (+) and r6c4 is (-), then r5c6 cannot equal 3; r5c6 must equal 1.
Is this a correct application of "coloring"?
Are there any other techniques that will solve this puzzle?
Thank you for your responses.
Paladin
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David Bryant
Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado
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| Posted: Tue Oct 17, 2006 12:56 am Post subject: Well, ... |
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Hi, Paladin! Haven't heard from you in quite a while.
You've reached a valid conclusion, but I can't quite follow your logic. Here's the problem.
r1c4 (+) --> r1c5 --> r7c5 --> r9c6 --> r9c8 --> r7c7 --> r3c7 --> r3c9 --> r5c9 (+);
This isn't a valid "coloring" chain. The problem comes at the move from r7c5 --> r9c6. This isn't valid for "simple coloring", because r7c6 might also be a "3". So there isn't a binary chain leading through the bottom center 3x3 box.
There's a much shorter chain that makes the same exclusion, though.
| Code: | *--------------------------------------------------*
| 6 8 2 | 139 139 4 | 5 19 7 |
| 4 3 5 | 169 169 7 | 2 19 8 |
| 1 9 7 | 5 2 8 | 36 4 36 |
|----------------+----------------+----------------|
| 8 6 3 | 19 159 2 | 4 7 15 |
| 5 2 4 | 8 7 13 | 9 6 13- |
| 7 1 9 | 36 4 356 | 8 35+ 2 |
|----------------+----------------+----------------|
| 2 4 1 | 7 356 369 | 36 8 59 |
| 3 5 8 | 4 16 169 | 7 2 69 |
| 9 7 6 | 2 8 35+ | 1 35- 4 |
*--------------------------------------------------* |
There is a binary chain r9c6 --> r9c8 --> r6c8 --> r5c9 that does force the conclusion r5c5 <> 3. dcb |
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TKiel
Joined: 22 Feb 2006
Posts: 292
Location: Kalamazoo, MI
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| Posted: Tue Oct 17, 2006 1:43 am Post subject: |
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Paladin,
| David Bryant wrote: | | There is a binary chain r9c6 --> r9c8 --> r6c8 --> r5c9 that does force the conclusion r5c5 <> 3. |
Not that it's needed for solving the puzzle, but the chain can be extended to include r5c6 (as a +) at which point row 6 has two +'s , so - must be where the 3's are. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Tue Oct 17, 2006 12:54 pm Post subject: Re: "Coloring" |
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| Paladin wrote: | | Are there any other techniques that will solve this puzzle? |
| Code: | *--------------------------------------------------*
| 6 8 2 | 139 139 4 | 5 19 7 |
| 4 3 5 | 169 169 7 | 2 19 8 |
| 1 9 7 | 5 2 8 | 36 4 36 |
|----------------+----------------+----------------|
| 8 6 3 | 19 159 2 | 4 7 15 |
| 5 2 4 | 8 7 -13 | 9 6 13 |
| 7 1 9 |#36 4 #356 | 8 35 2 |
|----------------+----------------+----------------|
| 2 4 1 | 7 356 3569 | 36 8 59 |
| 3 5 8 | 4 16 169 | 7 2 69 |
| 9 7 6 | 2 8 #35 | 1 35 4 |
*--------------------------------------------------* |
There is an xyz-wing in the marked cells (either r9c6=3 or r9c6=5 => r6c46=36), that leads to the same elimination. |
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Paladin
Joined: 10 Feb 2006
Posts: 15
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| Posted: Tue Oct 17, 2006 5:22 pm Post subject: "Coloring" |
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Thank you all very much for your comments. I am very grateful for your assistance, clarifications and comments.
Paladin |
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Myth Jellies
Joined: 27 Jun 2006
Posts: 64
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| Posted: Thu Oct 19, 2006 3:51 am Post subject: |
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| Paladin wrote: | | Are there any other techniques that will solve this puzzle? |
Here are a couple of unusual techniques based on the uniqueness premise. The first is called a MUG (Multivalued Universal Grave), which is related to a BUG-Lite (Bivalued Universal Grave).
It turns out that another pattern which results in multiple solutions is the following...
| Code: | *--------------------------------------------------*
| . . . | abcd abcd . | . abcd . |
| . . . | abcd abcd . | . abcd . |
| . . . | . . . | . . . |
|----------------+----------------+----------------|
| . . . | abcd abcd . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
|----------------+----------------+----------------|
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
*--------------------------------------------------* |
| Code: | *--------------------------------------------------*
| 6 8 2 |*139 *139 4 | 5 *19 7 |
| 4 3 5 |*169 *169 7 | 2 *19 8 |
| 1 9 7 | 5 2 8 | 36 4 36 |
|----------------+----------------+----------------|
| 8 6 3 |*19 *19+5 2 | 4 7 15 |
| 5 2 4 | 8 7 13 | 9 6 13 |
| 7 1 9 | 36 4 356 | 8 35 2 |
|----------------+----------------+----------------|
| 2 4 1 | 7 356 369 | 36 8 59 |
| 3 5 8 | 4 16 169 | 7 2 69 |
| 9 7 6 | 2 8 35 | 1 35 4 |
*--------------------------------------------------* |
...in this grid, the digits 1, 3, 6, and 9 can substitute in for abcd (not all digits need to be represented in every cell). The only way to avoid this deadly MUG pattern is for r4c5 = 5.
Another related option is the Extended Uniqueness argument. Consider the following grid....
| Code: | *--------------------------------------------------*
| . . . | ab* ab* . | . ab* . |
| . . . | ab* ab* . | . ab* . |
| . . . | . . . | . . . |
|----------------+----------------+----------------|
| . . . | ab* ab* . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
|----------------+----------------+----------------|
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
*--------------------------------------------------* |
If digits 'a' and 'b' are actually forced to occupy the cells shown above, then there is no way that you can possibly arrange the a's and b's to avoid a uniqueness rectangle or other BUG-Lite formation
Now look at our puzzle grid again...
| Code: | *--------------------------------------------------*
| 6 8 2 |*139 *139 4 | 5 *19 7 |
| 4 3 5 |*169 *169 7 | 2 *19 8 |
| 1 9 7 | 5 2 8 | 36 4 36 |
|----------------+----------------+----------------|
| 8 6 3 |*19 *159 2 | 4 7 15 |
| 5 2 4 | 8 7 13 | 9 6 13 |
| 7 1 9 | 36 4 356 | 8 35 2 |
|----------------+----------------+----------------|
| 2 4 1 | 7 356 369 | 36 8 59 |
| 3 5 8 | 4 16 169 | 7 2 69 |
| 9 7 6 | 2 8 35 | 1 35 4 |
*--------------------------------------------------* |
Consider the 1's and the 9's as our 'a' & 'b', and the starred cells as our extended uniqueness pattern. In box 3 there is no way for either a one or a nine to escape the deadly cells. The same is true for box 2. In box 5, the only way for a one or a nine to escape the deadly pattern is for r5c6 = 1. Thus we can make that placement and solve the puzzle that way as well. Note that one could have looked for escape candidates in the combo box 2, col 8, row 4; or the combo row 1, 2, & 4; or the combo columns 4, 5, & 8 as well--any combo that covers the extended pattern. |
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