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Bonus VH for 100217

 
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immpy



Joined: 06 May 2017
Posts: 252

PostPosted: Mon Oct 02, 2017 11:51 pm    Post subject: Bonus VH for 100217 Reply with quote

A compliment to arkietech's fine daily puzzle. Enjoy...

Code:

+-------+-------+-------+
| . . . | . . 2 | 8 . . |
| . . . | 5 . . | . 2 3 |
| . . . | . 3 8 | . . 7 |
+-------+-------+-------+
| . 9 . | 4 . . | . . 2 |
| . . 4 | . . 1 | . 7 . |
| 2 . 1 | . 9 . | . . 4 |
+-------+-------+-------+
| 9 . . | . . . | 2 . 6 |
| . 1 . | . 6 . | . . . |
| . 2 3 | 8 . 7 | 9 . 5 |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Oct 03, 2017 2:19 am    Post subject: Reply with quote

I wasted too much time on this to no avail.
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arkietech



Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA

PostPosted: Tue Oct 03, 2017 5:58 am    Post subject: Reply with quote

Code:
 *-----------------------------------------------------------*
 |c357  b3456  67    | 69   a47    2     | 8     569   1     |
 | 17    8     9     | 5     17    6-4   | 46    2     3     |
 | 15    456   2     | 169   3     8     | 456   569   7     |
 |-------------------+-------------------+-------------------|
 |d357   9     67    | 4     8    e356   | 1     56    2     |
 | 8     356   4     | 36    2     1     | 356   7     9     |
 | 2     356   1     | 7     9     356   | 356   8     4     |
 |-------------------+-------------------+-------------------|
 | 9     7     8     | 13    5    f34    | 2     14    6     |
 | 4     1     5     | 2     6     9     | 7     3     8     |
 | 6     2     3     | 8     14    7     | 9     14    5     |
 *-----------------------------------------------------------*
[4r1c5=(4-3)r1c2=r1c1-r4c1=r4c6-(3=4)r7c6]-4r2c6; ste
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Aj Dimonios



Joined: 01 May 2017
Posts: 339
Location: Sassari Italy

PostPosted: Tue Oct 03, 2017 8:19 am    Post subject: Reply with quote

Hi everyome


Code:

+-------------+------------+-----------+
| 357 3456 67 | 69  47 2   | 8   569 1 |
| 17  8    9  | 5   17 46  | 46  2   3 |
| 15  456  2  | 169 3  8   | 456 569 7 |
+-------------+------------+-----------+
| 357 9    67 | 4   8  356 | 1   56  2 |
| 8   356  4  | 36  2  1   | 356 7   9 |
| 2   356  1  | 7   9  356 | 356 8   4 |
+-------------+------------+-----------+
| 9   7    8  | 13  5  34  | 2   14  6 |
| 4   1    5  | 2   6  9   | 7   3   8 |
| 6   2    3  | 8   14 7   | 9   14  5 |
+-------------+------------+-----------+

Play this puzzle online at the Daily Sudoku site


[(4)R1C2-(4)R3C2=(4)R3C7-(4=6)R2C7-(6)R2C6 and (4-3)R1C2=(3)R1C1-(3)R4C1=(3)R4C6-(3=4)R7C6-(4)R2C6]=>contradiction R2C6==>-4R1C2=>solution stte.



Ciao a Tutti

Paolo
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immpy



Joined: 06 May 2017
Posts: 252

PostPosted: Tue Oct 03, 2017 9:37 am    Post subject: Reply with quote

Hello all...

My first step was to use the UR 69 r13c48 with a Unique Side (5) in c8. r3c7<>5; r4c8<>5(=6). Applying a rinse cycle then took the puzzle to the next stage here...

Code:

+----------+-----------+---------+
| 37 34  6 | 9  47 2   | 8  5  1 |
| 17 8   9 | 5  17 46  | 46 2  3 |
| 15 45  2 | 16 3  8   | 46 9  7 |
+----------+-----------+---------+
| 35 9   7 | 4  8  35  | 1  6  2 |
| 8  356 4 | 36 2  1   | 35 7  9 |
| 2  356 1 | 7  9  356 | 35 8  4 |
+----------+-----------+---------+
| 9  7   8 | 13 5  34  | 2  14 6 |
| 4  1   5 | 2  6  9   | 7  3  8 |
| 6  2   3 | 8  14 7   | 9  14 5 |
+----------+-----------+---------+

Play this puzzle online at the Daily Sudoku site

At this point you could use two separate W-Wings of 35 in b46 to eradicate the 3s in r56c2, or you could use a finned XY-Wing 356 with pivot at r6c6 (which also contains the fin (3)). In this case the XY-Wing was false, resulting in a contradiction; therefore the fin (3) is true and r6c6=3. stte.

Thanks for your input and interest.


Last edited by immpy on Tue Oct 03, 2017 7:22 pm; edited 1 time in total
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Oct 03, 2017 2:10 pm    Post subject: Reply with quote

Quote:
Code:
 *-----------------------------------------------------------*
 |c357  b3456  67    | 69   a47    2     | 8     569   1     |
 | 17    8     9     | 5     17    6-4   | 46    2     3     |
 | 15    456   2     | 169   3     8     | 456   569   7     |
 |-------------------+-------------------+-------------------|
 |d357   9     67    | 4     8    e356   | 1     56    2     |
 | 8     356   4     | 36    2     1     | 356   7     9     |
 | 2     356   1     | 7     9     356   | 356   8     4     |
 |-------------------+-------------------+-------------------|
 | 9     7     8     | 13    5    f34    | 2     14    6     |
 | 4     1     5     | 2     6     9     | 7     3     8     |
 | 6     2     3     | 8     14    7     | 9     14    5     |
 *-----------------------------------------------------------*
Code:


Quote:
My first step was to use the UR 69 r13c48 with a Unique Side (5) in c8. r3c7<>5; r4c8<>5(=6). Applying a rinse cycle then took the puzzle to the next stage here...


Immpy, I'm not understanding the UR. Yes, r13c8 contain 5s, each of which would break the deadly pattern, but how does the 1 in r3c4 get accounted for?
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immpy



Joined: 06 May 2017
Posts: 252

PostPosted: Tue Oct 03, 2017 6:46 pm    Post subject: Reply with quote

Hello all...

Marty, I don't have an answer for you as far as accounting for the 1 in r3c4. But my experience has been that a unique side has worked every time I come across it. Many times the extra candidate is also a part of one of the unique side squares, and I just leave it for the time being; but surely one of those two squares that contain the 5 will be 5. I simply don't concern myself with any extra candidates. Maybe someday I will get burned by my ignorance, but it hasn't happened so far.

There are still a few facets to the six different types of URs that I have not completely mastered, like opposite corners with extra candidate(s). My key is to just recognize the UR, and in this instance the 69s immediately caught my eye. Sometimes things don't appear as neatly as a classic #1UR which has just one extra candidate in one of four corners.

Hoping this sheds some light on the UR technique. Surely it sheds some light on my ignorance....LOL.

Thanks for the input.
immpy
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immpy



Joined: 06 May 2017
Posts: 252

PostPosted: Tue Oct 03, 2017 7:48 pm    Post subject: Reply with quote

After revisiting this puzzle several times in the last day or two, I noticed that a W-Wing of 14 in b89 connected by 4 would have solved it in one step.
r7c4<>1(=3); r9c8<>1(=4) stte.
I can't see the forest because of the trees sometimes, or maybe in this case I can't pick out every tree in a forest.....LOL

immpy
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Tue Oct 03, 2017 7:56 pm    Post subject: Reply with quote

Quote:
But my experience has been that a unique side has worked every time I come across it. Many times the extra candidate is also a part of one of the unique side squares,.


"Unique side " is a new term to me.

Quote:
but surely one of those two squares that contain the 5 will be 5


I don't see it. For each of those two cells that contain a 5, there are three possibilities for 5 in row, column and box. What I learned is that either the 1 in r3c4 or a 5 in r1c8 or r3c8 is needed to kill the deadly pattern. Then you see what happens with a 5 and the 1 and look for a common outcome. I did not find a common outcome, but I didn't look thoroughly. If r4c8=6 solves the puzzle, I'd have to call it a lucky guess or lucky mistake.
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immpy



Joined: 06 May 2017
Posts: 252

PostPosted: Thu Oct 05, 2017 12:04 am    Post subject: Reply with quote

Hello all...

Marty, I really don't have all of the UR techniques solidly in my grasp yet. But I have enjoyed 100% success with a common candidate on one side of the UR. That would be the 5 here in r13c8. "Unique side" is just my personal term for it. I guess it really is a subset of sorts in the UR, but with just one common candidate, in this case the two 5s present on one side of the rectangle. I make eliminations based on what both of the squares that contain the 5 can see. If this is just luck on my part, then I have been 100% lucky in numerous puzzles where this pattern occurs.

I may be entirely wrong in my perception of this concept, but it is hard to buck the success rate. I also may have to go back and study the various forms of the UR so as to better understand and apply them.

Thanks for taking the time to respond to my posts.

immpy
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Thu Oct 05, 2017 1:34 am    Post subject: Reply with quote

If the 1 were not present in r3c4 and the four corners were 69-69-695-695 it would be a Type 2 and an elimination would be made from cells that see both of the 5s, i.e., r3c7 and r4c8.
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immpy



Joined: 06 May 2017
Posts: 252

PostPosted: Thu Oct 05, 2017 4:19 pm    Post subject: Reply with quote

Marty, you are absolutely correct! I shall have to be more mindful of how I use the UR in the future. Thanks again.
immpy
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