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immpy
Joined: 06 May 2017 Posts: 571
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Posted: Wed Oct 31, 2018 1:35 am Post subject: Advanced 103018 |
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Hello everyone, enjoy the puzzle...
Code: |
+-------+-------+-------+
| . 3 2 | . 4 5 | . . . |
| 4 . 6 | 3 . . | . . . |
| 7 . . | . 1 9 | . . . |
+-------+-------+-------+
| . 2 4 | . . . | . 3 . |
| . . . | 2 . 8 | . . . |
| . 6 . | . . . | 1 5 . |
+-------+-------+-------+
| . . . | 5 8 . | . . 7 |
| . . . | . . 6 | 5 . 3 |
| . . . | 7 3 . | 6 8 . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
cheers...immp |
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Clement
Joined: 24 Apr 2006 Posts: 1111 Location: Dar es Salaam Tanzania
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Posted: Tue Nov 06, 2018 10:14 am Post subject: Advanced 103018 |
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I have to settle with two steps Code: |
+--------------------+-----------------+---------------------+
| 19 3 2 | 8 4 5 | 79 e1679 d169 |
| 4 19 6 | 3 27 27 | 8 19 5 |
| 7 m58 58 | 6 1 9 | 3 2 4 |
+--------------------+-----------------+---------------------+
| 5 2 4 | 19 6 h17 | g79 3 8 |
| 3 179 179 | 2 5 8 | 4 f679 c69 |
| 89 6 j789 | 4 i79 3 | 1 5 2 |
+--------------------+-----------------+---------------------+
| 6 o14-9 3 | 5 8 14 | 2 149 7 |
| 128 L1478 k178 | 19 29 6 | 5 14 3 |
|a129 na1459 a159 | 7 3 124 | 6 8 b19 |
+--------------------+-----------------+---------------------+
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1): (9)r9c123 = r9c9 - (9=6)r5c9 - r1c9 = (6-7)r1c8 = r1c7 - r4c7 = r4c6 - r6c5 = r6c3 - r8c3 = (7-4*8)r8c2 = (8-5)r3c2 = (5-4*)r9c2 = (4)r7c2 => - 9r7c2
Code: |
+--------------------+------------------+------------------+
| 1 3 2 | 8 4 5 | 79 67 69 |
| 4 9 6 | 3 27 27 | 8 1 5 |
| 7 58 58 | 6 1 9 | 3 2 4 |
+--------------------+------------------+------------------+
| 5 2 4 | d19 6 17 | 79 3 8 |
| 3 17 179 | 2 5 8 | 4 67 69 |
| 8-9 6 789 | 4 e79 3 | 1 5 2 |
+--------------------+------------------+------------------+
| 6 14 3 | 5 8 b14 | 2 9 7 |
| 28 178 178 | c19 29 6 | 5 4 3 |
| a29 45 59 | 7 3 a24 | 6 8 1 |
+--------------------+------------------+------------------+
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2): (9=24)r9c16 - (4=1)r7c6 - (1=9)r8c4 - r4c4 = (9)r6c5 => - 9r6c1; stte |
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immpy
Joined: 06 May 2017 Posts: 571
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Posted: Tue Nov 06, 2018 2:51 pm Post subject: |
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Very elegant Clement, my friend! This one was tricksy, indeed.
I opted for a (non) Uniqueness Test on the 19 DP (deadly pattern) in r27c28. I only call it that because the pattern exists in four separate boxes, instead of the normal two boxes. Under normal circumstances, i.e. the two boxes, any cell that sees the extra candidate (4) common to one side of the rectangle can have the 4 eliminated from candidacy. The cell in question becomes r7c6. It is a bi-value cell with 1 and 4 as possibilities. Can we eliminate the 4?? I have seen this often enough to know we cannot, because of the four-box variety of the Uniqueness Test. That means we MUST place the 4 in r7c6, by my crude thinking....but some have cautioned me in the past that I have no logic to back up my doing this, and they are most probably correct in that regard.
Sooo.....let us assume we don't know nuttin' from nuttin".....we can still attack this puzzle from that focal point cell right there at r7c6, a la Gianni. And it doesn't take too long to see that if r7c6=1, a contradiction awaits us down the road. I am not writing this out in eureka, others can work this out on their own, I am sure. You will have to trust me on this one.
So if r7c6 = 4 the puzzle is solved by stte. As Gianni would say, a solution in one step......and.......Ciao a Tutti......... |
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immpy
Joined: 06 May 2017 Posts: 571
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Posted: Tue Nov 06, 2018 3:09 pm Post subject: |
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My apologies to Paolo and Gianni for confusing them. It is Paolo who likes to find the one-step solutions by identifying the contradiction of a possible cell value. Thank you Paolo and Gianni for all of your astute observations on this great forum!!
cheers....immp |
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Clement
Joined: 24 Apr 2006 Posts: 1111 Location: Dar es Salaam Tanzania
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Posted: Tue Nov 06, 2018 5:02 pm Post subject: |
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immpy wrote: | Very elegant Clement, my friend! This one was tricksy, indeed.
I opted for a (non) Uniqueness Test on the 19 DP (deadly pattern) in r27c28. I only call it that because the pattern exists in four separate boxes, instead of the normal two boxes. Under normal circumstances, i.e. the two boxes, any cell that sees the extra candidate (4) common to one side of the rectangle can have the 4 eliminated from candidacy. The cell in question becomes r7c6. It is a bi-value cell with 1 and 4 as possibilities. Can we eliminate the 4?? I have seen this often enough to know we cannot, because of the four-box variety of the Uniqueness Test. That means we MUST place the 4 in r7c6, by my crude thinking....but some have cautioned me in the past that I have no logic to back up my doing this, and they are most probably correct in that regard.
Sooo.....let us assume we don't know nuttin' from nuttin".....we can still attack this puzzle from that focal point cell right there at r7c6, a la Gianni. And it doesn't take too long to see that if r7c6=1, a contradiction awaits us down the road. I am not writing this out in eureka, others can work this out on their own, I am sure. You will have to trust me on this one.
So if r7c6 = 4 the puzzle is solved by stte. As Gianni would say, a solution in one step......and.......Ciao a Tutti......... | Doesn't that amount to Brute Force?(which is not really a technique). Place a digit in a cell and look, whether you get a solution or not. Some one must find a logical path leading to the elimination or contradiction of 1 in r7c6. |
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immpy
Joined: 06 May 2017 Posts: 571
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Posted: Wed Nov 07, 2018 8:18 am Post subject: |
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I 'm not sure. Does it amount to Brute Force? Paolo solves puzzles in single steps like this quite frequently. Maybe he (Paolo) could add some insight to this if he is so inclined.
I never understood how, or even IF, Paolo would know where to begin the hunt for a contradiction. In this very puzzle, one could begin in many places, almost anywhere, really. My logic, if it can be termed logic, is that r7c6 is THE ideal place to start, based on my past experience with 4-box DPs. It is juicy indeed when one adds the slight lean in favor of the candidate 4 in this instance.
Surely, and logically, the original rectangle contains an X-Wing on 9. After eliminations one can use other skills, i.e. Kites, Chains, Wings, etc. to arrive at a viable solution. I just relished the opportunity to speed the process along with my earlier pattern observance.
And lastly, Brute Force is currently listed (albeit at the very bottom) as a technique that can be used to solve a puzzle by my program of puzzle generation. And isn't a simple XY-Chain a SEMI form of Brute Force when we begin by assuming "if cell ?? is not X, then cell ?? is Y; and so on...etc."?
Just some thoughts rolling around in my otherwise empty skull. I continue to learn and be amazed by these sometimes confounding things called Sudoku Puzzles.
cheers...immp |
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Clement
Joined: 24 Apr 2006 Posts: 1111 Location: Dar es Salaam Tanzania
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Posted: Wed Nov 07, 2018 11:12 am Post subject: |
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immpy wrote: | I 'm not sure. Does it amount to Brute Force? Paolo solves puzzles in single steps like this quite frequently. Maybe he (Paolo) could add some insight to this if he is so inclined.
I never understood how, or even IF, Paolo would know where to begin the hunt for a contradiction. In this very puzzle, one could begin in many places, almost anywhere, really. My logic, if it can be termed logic, is that r7c6 is THE ideal place to start, based on my past experience with 4-box DPs. It is juicy indeed when one adds the slight lean in favor of the candidate 4 in this instance.
Surely, and logically, the original rectangle contains an X-Wing on 9. After eliminations one can use other skills, i.e. Kites, Chains, Wings, etc. to arrive at a viable solution. I just relished the opportunity to speed the process along with my earlier pattern observance.
And lastly, Brute Force is currently listed (albeit at the very bottom) as a technique that can be used to solve a puzzle by my program of puzzle generation. And isn't a simple XY-Chain a SEMI form of Brute Force when we begin by assuming "if cell ?? is not X, then cell ?? is Y; and so on...etc."?
Just some thoughts rolling around in my otherwise empty skull. I continue to learn and be amazed by these sometimes confounding things called Sudoku Puzzles.
cheers...immp | I agree with you. You have to look for an ideal place to start, and I usually do that and I think every one does that. I did the same in my above solution. What I do not prescribe is to place a digit in a cell, and when by Mercy of Providence it solves the puzzle you say If cell A is x or not y it is solved by stte, then you start working backwards for a chain or any other technique. I don't know if I have made myself clear. |
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immpy
Joined: 06 May 2017 Posts: 571
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Posted: Wed Nov 07, 2018 7:05 pm Post subject: |
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You have made yourself quite clear, my friend. Clement, I am thinking you are, by now, familiar with my personal, and more frequently used, ways and means of puzzle solving, which I employ 99 percent of the time. I am not one to arbitrarily pick a cell and make a hypothesis (which is what this was, simply)...EXCEPT...in those rare instances (less than 1 percent, actually) where a hypothesis seems to coincide with what I am observing in the pattern of candidates. That was the case in this particular instance, for me personally. I will continue to use it as a solving technique in these rare instances.
The fact still remains that if R7C6=1=>contradiction =>-1R7C6=>solution. Therefore R7C6=4=>stte. |
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immpy
Joined: 06 May 2017 Posts: 571
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Posted: Fri Nov 09, 2018 10:41 pm Post subject: |
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As this puzzle still rolls around in my head, I began to try and form an explanation of the pattern I am seeing and its implications.
The DP on 19 is a UR that occurs in 4 boxes instead of the customary 2 boxes. There is an extra candidate (4) in two of the cells which make up one side of the rectangle. There is a single cell (a fin of a sort?) which sees BOTH cells that contain the extra candidate (4). This is a bi-value cell (14). It does not contain a possibility for (9). No other extra candidates exist in the corner cells to clutter the pattern. From past experience, EITHER these four cells consist of a perpetual Remote Pair of 19 (or call it a double X-Wing of 19) in which ALL 1s and 9s can be removed from both the rows AND the columns affected; OR one of the 4s will be true in one of the corners....because of the presence of the bi-value fin cell.
One of these two possibilities MUST be true in this instance. And I used this to cut my solving time by quite a bit. I completely understand all who feel this is a form of guessing and that it should not be employed at all whatsoever. To each his/her own I feel. I quite like finned XY-Wings as well, which many equate to guessing also, as they work in a very similar manner. I rather like these situations where only two possibilities exist, AND, very important I think, the puzzle becomes solved by their implementation. It does no good to use these techniques when too many cells will still remain to be solved.
Hope I shed a little light....immp |
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