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Free Press May 15, 2015

 
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keith



Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA

PostPosted: Mon Jun 01, 2015 8:49 am    Post subject: Free Press May 15, 2015 Reply with quote

Code:
Puzzle: FP051515
+-------+-------+-------+
| 9 . 1 | . . . | . . . |
| 2 8 . | . . 6 | . 7 . |
| . 4 . | . . . | 9 . . |
+-------+-------+-------+
| 4 6 . | 1 . 7 | . 2 . |
| 3 . . | . . . | . . 7 |
| . 5 . | 2 . 4 | . . 9 |
+-------+-------+-------+
| . . 6 | . . . | . 8 . |
| . 7 . | 8 . . | . 6 1 |
| . . . | . . . | 3 . 5 |
+-------+-------+-------+

Keith
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Marty R.



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

PostPosted: Tue Jun 02, 2015 1:41 am    Post subject: Reply with quote

Code:

+-------+---------------+------------+
| 9 3 1 | 47  47 28   | 28 5  268 |
| 2 8 5 | 39  19   6    | 14  7  34  |
| 6 4 7 | 35  1258 2358 | 9   13 28  |
+-------+---------------+------------+
| 4 6 9 | 1   358  7    | 58  2  38  |
| 3 1 2 | 569 5689 589  | 568 4  7   |
| 7 5 8 | 2   36   4    | 16  13 9   |
+-------+---------------+------------+
| 1 9 6 | 345 245  235  | 7   8  24  |
| 5 7 3 | 8   249  29   | 24  6  1   |
| 8 2 4 | 67  67   1    | 3   9  5   |
+-------+---------------+------------+

Play this puzzle online at the Daily Sudoku site

(7=6)r9c5-(6=35947)r23571c4=>r1c5<>7
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Tue Aug 04, 2015 3:40 pm    Post subject: Reply with quote

Hi Marty! I don't see how you concluded that r1c5<>7! Could you explain to me in English (my Eureka must not be so good) how the remaining 5 numbers for the 5 rows listed leads to the conclusion. I don't see it! What am I missing? Thanks, Don.
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Marty R.



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

PostPosted: Tue Aug 04, 2015 5:04 pm    Post subject: Reply with quote

Marty R. wrote:
Code:

+-------+---------------+------------+
| 9 3 1 | 47  47 28   | 28 5  268 |
| 2 8 5 | 39  19   6    | 14  7  34  |
| 6 4 7 | 35  1258 2358 | 9   13 28  |
+-------+---------------+------------+
| 4 6 9 | 1   358  7    | 58  2  38  |
| 3 1 2 | 569 5689 589  | 568 4  7   |
| 7 5 8 | 2   36   4    | 16  13 9   |
+-------+---------------+------------+
| 1 9 6 | 345 245  235  | 7   8  24  |
| 5 7 3 | 8   249  29   | 24  6  1   |
| 8 2 4 | 67  67   1    | 3   9  5   |
+-------+---------------+------------+

Play this puzzle online at the Daily Sudoku site

(7=6)r9c5-(6=35947)r23571c4=>r1c5<>7


Quote:

Hi Marty! I don't see how you concluded that r1c5<>7! Could you explain to me in English (my Eureka must not be so good) how the remaining 5 numbers for the 5 rows listed leads to the conclusion. I don't see it! What am I missing? Thanks, Don.


Don, I don't know if I can answer and I don't know if my solution is correct.

This term (6=35947) is an ALS (Almost Locked Set). An ALS is when there are N candidates for N-1 cells. Naked pairs, triples, quads, etc are locked sets with an equal number of candidates and cells. This ALS has six candidates, 345679, for five cells. If it's not 6, then the other numbers remain and the only 7 is in r1c4 which is a pincer along with r9c5.

As I look at the puzzle, it doesn't seem possible because if r9c5 is not 7 then r9c4 is =7 and r1c4 can't be 7. Every time I use an ALS it's an adventure and I can now see that this one's wrong. The number that leads into the ALS, 6 in this case, must see all the 6's in the ALS and the 6 in r9c5 doesn't see the 6 in r5c4. ALS is a very powerful technique that almost feels like cheating in that you can see a number you want that is not available and use it if there's an ALS there.

As you may know, Dan (arkietech) posts a puzzle every day on the other forum (he used to post it here until the site went down for a while.

Virtually every puzzle has an ALS in one of the posted solutions. Here's one from the other day in which my ALS is OK. The 3 leads into the ALS which is 3=615 in box 7.

Code:


Postby Marty R. » Sun Aug 02, 2015 10:29 pm




Postby Marty R. » Sun Aug 02, 2015 10:29 pm

Code: Select all
    +-------------+-------------+----------+
    | 67  5    2  | 3   67  1   | 9  8  4  |
    | 146 146  9  | 46  8   2   | 7  5  3  |
    | 8   347  37 | 5   47  9   | 2  6  1  |
    +-------------+-------------+----------+
    | 3   47   57 | 469 2   567 | 8  1  69 |
    | 457 8    1  | 469 456 567 | 3  2  69 |
    | 2   9    6  | 1   3   8   | 5  4  7  |
    +-------------+-------------+----------+
    | 167 1367 8  | 2   9   4   | 16 37 5  |
    | 9   1367 4  | 8   156 356 | 16 37 2  |
    | 156 2    35 | 7   16  36  | 4  9  8  |
    +-------------+-------------+----------+[/code


Play this puzzle online at the Daily Sudoku site
Same eliminations as Leren, less elegance

5r4c6=r4c3-(5=3)r9c3-(3=615)b8p985=> -5r5c5,r8c6
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Tue Aug 04, 2015 5:40 pm    Post subject: Reply with quote

Wow, mind-boggling stuff! I've got to look those over and read about them and see if it sinks in! Thanks a lot Marty! Don.

You know what? When I saw your solution (i.e. r1c5<>7), I checked to make sure that it only left single steps (it did) so I tried to find a chain by assuming that r1c5=7 that led to a contradiction. All I came up with was a long-winded two-branch contradiction. I sure would like to see a 'nice' chain that solves this in one step.
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Tue Aug 04, 2015 8:12 pm    Post subject: Reply with quote

dongrave wrote:
... I tried to find a chain by assuming that r1c5=7 that led to a contradiction. All I came up with was a long-winded two-branch contradiction. I sure would like to see a 'nice' chain that solves this in one step.

The 2 solutions of R26 + B39 + C9 -> +6r9c4; ste

Proof :
Code:
+---------+--------------------+-----------------+
| 9  3  1 | 47      47    28   | 28   5     6    |
| 2  8  5 | (39)    19    6    | 14   7     4(3) |
| 6  4  7 | 35      1258  2358 | 9    1(3)  28   |
+---------+--------------------+-----------------+
| 4  6  9 | 1       358   7    | 58   2     38   |
| 3  1  2 | 5-6(9)  5689  589  | 568  4     7    |
| 7  5  8 | 2       (36)  4    | 16   1(3)  9    |
+---------+--------------------+-----------------+
| 1  9  6 | 345     245   235  | 7    8     24   |
| 5  7  3 | 8       249   29   | 24   6     1    |
| 8  2  4 | 67      67    1    | 3    9     5    |
+---------+--------------------+-----------------+
[(6=3)r6c5-3r6c8=3r3c8-3r2c9=(3-9)r2c4=9r5c4] - 6r5c4=6r9c4; ste
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Wed Aug 05, 2015 2:07 pm    Post subject: Reply with quote

Hey! What happened to Bat's solution? I was trying to reason it out last night and couldn't. Did you delete it because it wasn't valid Bat?

Thanks for the solution JC! But you can't fool me - that chain is a one of your famous (A or B; A implies C; B implies C; therefore C) solutions cleverly disguised as a chain! Laughing If I deciphered your Eureka expression correctly (which might not be the case), the English-like 'Van Hay' solution (as I like to call them) would be expressed as follows:
r6c5 is 3 or 6; if it's 6 then r5c4<>6 so r9c4=6; if it's 3 then r6c8<>3 so r3c8=3 so r2c9<>3 so r2c4=3 (not 9) so r5c4=9 (not 6) so r9c4=6; therefore r9c4=6. I never would have found that one! Good one!
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Marty R.



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

PostPosted: Wed Aug 05, 2015 3:37 pm    Post subject: Reply with quote

Quote:
Hey! What happened to Bat's solution? I was trying to reason it out last night and couldn't. Did you delete it because it wasn't valid Bat?

Thanks for the solution JC! But you can't fool me - that chain is a one of your famous (A or B; A implies C; B implies C; therefore C) solutions cleverly disguised as a chain! Laughing If I deciphered your Eureka expression correctly (which might not be the case), the English-like 'Van Hay' solution (as I like to call them) would be expressed as follows:
r6c5 is 3 or 6; if it's 6 then r5c4<>6 so r9c4=6; if it's 3 then r6c8<>3 so r3c8=3 so r2c9<>3 so r2c4=3 (not 9) so r5c4=9 (not 6) so r9c4=6; therefore r9c4=6. I never would have found that one! Good one!


Don, change your opening statement. Before I started to learn notation (and some will say I didn't learn it very well) I had to learn how to read a term such as (X=Y). That says that the cell in question must be X or Y and if it's not X,then it's =Y. In other words, left of the = is false and right true. JC's solution flows logically and easy to understand. But I always have at least one question for JC and it's here

(3-9)r2c4=9r5c4] - 6r5c4=6r9c4. I would've been able to follow it better if it was written (3-9)r2c4=(9-6)r5c4=6r9c4

I don't understand why he wrote it the way he did.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Wed Aug 05, 2015 4:57 pm    Post subject: Reply with quote

Hi Marty! The thing that stuck out to me was the sub-expression enclosed within the square brackets - so what I think he was trying to point out is that the sub-expression in the brackets takes us from the chain beginning with the assumption of r6c5=3 but that the remainder of the formula outside the brackets is also necessary (which assumes the other starting value in r6c5) in order to show that they both result with r9c4=6. I might be completely wrong here but that's why I think it was necessary for him to split it up with the brackets - because I don't think your replacement ending leads to the conclusion by itself (or at least I don't see it if it does). Let me know what you think! Or maybe even JC can tell us what he was thinking! Thanks, Don.
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Marty R.



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

PostPosted: Wed Aug 05, 2015 5:26 pm    Post subject: Reply with quote

Don, I don't understand the brackets. This is a run-of the-mill chain that mostly is written in one string without brackets. I see a lot of puzzle solutions, most of which are chains, that are written from start to finish without brackets or other stuff that segments things. Take away the brackets and it says if r6c5 is not 6 then r9c4 is a 6.

As to my replacement chain, (3-9)r2c4=(9-6)r5c4=6r9c4. R2c4 is 3 and if not 9, then r5c4 is 9 and if not 6, then r9c4 is 6. I combined JC's last term in the bracket plus the next term into one term.

JC has forgotten more about Sudoku than I'll ever know, so keep that in mind when you read my comments. JC is not wrong, but he does do some things differently than some other folks.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Wed Aug 05, 2015 6:30 pm    Post subject: Reply with quote

Hi again Marty! I must be missing something here. When I look at the run of the mill chain, I don't see how r6c5 not being a 6 and r9c4 being a 6 concludes anything - unless the other sub-expression is also included showing that r6c5 being a 6 also leads to r9c4 being a 6. Am I missing something here? Thanks a lot, Don.
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bat999



Joined: 09 Jul 2015
Posts: 55
Location: UK

PostPosted: Wed Aug 05, 2015 7:22 pm    Post subject: Reply with quote

dongrave wrote:
Hey! What happened to Bat's solution? I was trying to reason it out last night and couldn't. Did you delete it because it wasn't valid Bat?

No, you had asked why r1c5<>7.
"Hi Marty! I don't see how you concluded that r1c5<>7!"

The other guy showed how r1c5<>7 (because r9c5<>6, it is 7 so r1c5<>7).

If r6c5=6 then r9c5<>6 (obviously).
If r6c5<>6 then through the chain (a,b,c,d,e,f,g) r9c4=6 (which kills the 6 in r9c5).

[(6=3)r6c5-3r6c8=3r3c8-3r2c9=(3-9)r2c4=9r5c4] - 6r5c4=6r9c4; ste

The expression is OK without those square brackets imho:
(6=3)r6c5 - (3)r6c8 = (3)r3c8 - (3)r2c9 = (3-9)r2c4 = (9-6)r5c4 = (6)r9c4 => -6 r9c5; stte
Code:
.---------.-------------------.--------------.
| 9  3  1 |  47    47    28   | 28   5   6   |
| 2  8  5 | e39    19    6    | 14   7  d34  |
| 6  4  7 |  35    1258  2358 | 9   c13  28  |
:---------+-------------------+--------------:
| 4  6  9 |  1     358   7    | 58    2   38 |
| 3  1  2 | f569   5689  589  | 568   4   7  |
| 7  5  8 |  2    a36    4    | 16   b13  9  |
:---------+-------------------+--------------:
| 1  9  6 |  345   245   235  | 7     8   24 |
| 5  7  3 |  8     249   29   | 24    6   1  |
| 8  2  4 | g67    7-6    1   | 3     9   5  |
'---------'-------------------'--------------'


The solution I posted and deleted before was different.
It solved the puzzle but didn't answer your question. Razz
Code:
.---------.------------------.---------------.
| 9  3  1 | 47   b47    28   |  28   5   6   |
| 2  8  5 | 39    19    6    | e14   7  d34  |
| 6  4  7 | 35    1258  2358 |  9   c13  28  |
:---------+------------------+---------------:
| 4  6  9 | 1     358   7    |  58    2   38 |
| 3  1  2 | 569   5689  589  |  568   4   7  |
| 7  5  8 | 2     3-6    4   | f16   b13  9  |
:---------+------------------+---------------:
| 1  9  6 | 345   245   235  |  7     8   24 |
| 5  7  3 | 8    c249   29   | d24    6   1  |
| 8  2  4 | 67   a67    1    |  3     9   5  |
'---------'------------------'---------------'
(6=7)r9c5 - (7=4)r1c5 - (4)r8c5 = (4)r8c7 - (4=1)r2c7 - (1=6)r6c7 => -6 r6c5; stte
Cool
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Marty R.



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

PostPosted: Wed Aug 05, 2015 8:02 pm    Post subject: Reply with quote

dongrave wrote:
Hi again Marty! I must be missing something here. When I look at the run of the mill chain, I don't see how r6c5 not being a 6 and r9c4 being a 6 concludes anything - unless the other sub-expression is also included showing that r6c5 being a 6 also leads to r9c4 being a 6. Am I missing something here? Thanks a lot, Don.


I'm still totally confused by "sub-expressions". If r6c5<>6 we have shown that that forces r9c4=6. Now let's turn it around and do the opposite of the first premise. The opposite of r6c5 not 6 is r6c5=6. This is by definition a classic pincer situation; Either r6c5 or r9c4 must be =6. R5c4 sees both cells and cannot be =6. It's just like starting with a bivalue cell, say 38. If the 3 proves an 8 somewhere, fine, if the bivalue is not 3 then it's 8 and no 8 can survive in a cell seeing both the original bivalue cell and the cell in which an 8 is proven by the opening premise of bivalue=3 (same as <>8). because one of those two cells must be =8.
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dongrave



Joined: 06 Mar 2014
Posts: 568

PostPosted: Wed Aug 05, 2015 8:55 pm    Post subject: Reply with quote

OH! NOW I SEE IT! Bat 'completing' the chain by including r9c5 - and then Marty's explanation of the chain actually beginning and ending with pincers and r5c4 being the target FINALLY clicked! How obvious! Sorry about being so blind! Thanks for the explanation!
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Marty R.



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

PostPosted: Wed Aug 05, 2015 9:51 pm    Post subject: Reply with quote

dongrave wrote:
OH! NOW I SEE IT! Bat 'completing' the chain by including r9c5 - and then Marty's explanation of the chain actually beginning and ending with pincers and r5c4 being the target FINALLY clicked! How obvious! Sorry about being so blind! Thanks for the explanation!


No problem I've been there and done that many times.
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Thu Aug 06, 2015 7:22 pm    Post subject: Reply with quote

Code:
+---------+-----------------+-------------+
| 9  3  1 | 47   47    28   | 28   5   6  |
|         |      +          | aA          |
|         |                 |             |
| 2  8  5 |*39   19    6    | 14   7  *34 |
|         | aA   Aa         | aA       Aa |
|         |                 |             |
| 6  4  7 | 35   1258  2358 | 9   *13  28 |
|         |                 |      Aa  Aa |
+---------+-----------------+-------------+
| 4  6  9 | 1    358   7    | 58   2   38 |
|         |                 |          aA |
|         |                 |             |
| 3  1  2 |*569  5689  589  | 568  4   7  |
|         |  -a             |             |
|         |                 |             |
| 7  5  8 | 2   *36    4    | 16  *13  9  |
|         |      aA         | Aa   aA     |
+---------+-----------------+-------------+
| 1  9  6 | 345  245   235  | 7    8   24 |
|         |                 |          aA |
|         |                 |             |
| 5  7  3 | 8    249   29   | 24   6   1  |
|         |                 | Aa          |
|         |                 |             |
| 8  2  4 | 67   67    1    | 3    9   5  |
|         | +     +         |             |
+---------+-----------------+-------------+
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