February 26, 2017

Post puzzles for others to solve here.

Re: February 26, 2017

Postby SteveG48 » Sun Feb 26, 2017 3:43 pm

eleven wrote:Another trouble is that relatively easy things look that complicated.
When you look at Steve's solution:
Code: Select all
 *-----------------------------------------------------------*
 | 39    78    5     | 1379  6     4     | 138   1389  2     |
 | 39    6    c2-8   | 1239 c35    1359  |d138   7     4     |
 | 1     47  bc24    | 2379  8     379   | 5     39    6     |
 *-------------------+-------------------+-------------------|
 | 58    158 af18    | 6     9     2     | 7     4     3     |
 | 6     2     7     | 134   34    13    | 9     58    58    |
 | 4     9     3     | 5     7     8     | 6     2     1     |
 *-------------------+-------------------+-------------------|
 | 258   458   9     | 34    1     6     | 238   358   7     |
 | 258   3   af148   | 479 bc45    579   |e128   6     589   |
 | 7     15    6     | 8     2     359   | 4     135   59    |
 *-----------------------------------------------------------*

it is easy to spot:
(8=4)r48c3, then 5r8c5 (last digit) and 1r8c7 (last 1 in row) - (5|1=8)r2c57 => -8r2c3,
but hard to formulate.


So very true. My solution would actually be easier to follow as Kraken cell [138]r2c7 => -8 r2c3, but my personal goal is to write all solutions as proper bi-directional AICs whenever I can. That doesn't always result in a solution that's easy to follow.
Steve
User avatar
SteveG48
2016 Supporter
 
Posts: 1735
Joined: 08 November 2013
Location: Orlando, Florida

Re: February 26, 2017

Postby bat999 » Sun Feb 26, 2017 3:53 pm

eleven wrote:... hard to formulate.

Code: Select all
.------------------.-------------------.------------------.
|  39   78    5    |  1379   6    4    |  138  1389   2   |
|  39   6     2-8  |  1239  c35   1359 | c138  7      4   |
|  1    47    24   |  2379   8    379  |  5    39     6   |
:------------------+-------------------+------------------:
|  58   158  a18   |  6      9    2    |  7    4      3   |
|  6    2     7    |  134    34   13   |  9    58     58  |
|  4    9     3    |  5      7    8    |  6    2      1   |
:------------------+-------------------+------------------:
|  258  458   9    |  34     1    6    |  238  358    7   |
| b258  3    a148  | b479   b45  b579  | b128  6     b589 |
|  7    15    6    |  8      2    359  |  4    135    59  |
'------------------'-------------------'------------------'

(8=4)r48c3 - (4=15)r8c145679 - (1|5=8)r2c57 => -8 r2c3; stte

With a 4 in r8c3 row 8 becomes 2 3 4 79 5 79 1 6 8.
But we are now Monday morning quarterbacks, I think. :lol:
8-)
User avatar
bat999
2017 Supporter
 
Posts: 624
Joined: 15 September 2014
Location: UK

Re: February 26, 2017

Postby Cenoman » Sun Feb 26, 2017 6:44 pm

SteveG48 wrote:
I think the highlighted 4 is a typo.

Right, Steve. Message edited and typo corrected.
Thank you for reading the path and spotting the typo.

Best regards, Cenoman.
Cenoman
 
Posts: 135
Joined: 21 November 2016
Location: Paris, France

Re: February 26, 2017

Postby JC Van Hay » Sun Feb 26, 2017 7:32 pm

eleven wrote:
bat999 wrote:Yes, I had a lot of trouble with the Eureka notation.

Another trouble is that relatively easy things look that complicated.
When you look at Steve's solution:
Code: Select all
 *-----------------------------------------------------------*
 | 39    78    5     | 1379  6     4     | 138   1389  2     |
 | 39    6    c2-8   | 1239 c35    1359  |d138   7     4     |
 | 1     47  bc24    | 2379  8     379   | 5     39    6     |
 *-------------------+-------------------+-------------------|
 | 58    158 af18    | 6     9     2     | 7     4     3     |
 | 6     2     7     | 134   34    13    | 9     58    58    |
 | 4     9     3     | 5     7     8     | 6     2     1     |
 *-------------------+-------------------+-------------------|
 | 258   458   9     | 34    1     6     | 238   358   7     |
 | 258   3   af148   | 479 bc45    579   |e128   6     589   |
 | 7     15    6     | 8     2     359   | 4     135   59    |
 *-----------------------------------------------------------*

it is easy to spot:
(8=4)r48c3, then 5r8c5 (last digit) and 1r8c7 (last 1 in row) - (5|1=8)r2c57 => -8r2c3,
but hard to formulate.
Yes, easy to spot or as the simplest key of this puzzle is the 2 solutions of C5 :
Code: Select all
+-----------------+------------------+------------------+
| 39   78   5     | 1379  6     4    | 138    1389  2   |
| 39   6    2-8   | 1239  (35)  1359 | (138)  7     4   |
| 1    47   24    | 2379  8     379  | 5      39    6   |
+-----------------+------------------+------------------+
| 58   158  (18)  | 6     9     2    | 7      4     3   |
| 6    2    7     | 134   34    13   | 9      58    58  |
| 4    9    3     | 5     7     8    | 6      2     1   |
+-----------------+------------------+------------------+
| 258  458  9     | 34    1     6    | 238    358   7   |
| 258  3    (148) | 479   (45)  579  | 28(1)  6     589 |
| 7    15   6     | 8     2     359  | 4      135   59  |
+-----------------+------------------+------------------+
3r2c5 -> WWing(81)r2c7.r8c3,1r8c35
||
5r2c5 -> r8c5=4, NP(18)r48c3
Conclusion : -8r2c3; stte

In Eureka notation : Kraken Cell (138)r2c7 : AAIC[NP(81)r48c3=4r8c3-(4=53)r82c5-3r2c7=*WWing(81)r2c7.r8c3,1r8c37]-8r2c3 because [8r48c3==8r2c7]; stte

The exclusion of 8r2c3 can also be explained by Steve's chain which is nothing else as a way of writing an nrczt-chain, an oriented chain [IOW, not a bi-directional AIC].
However, the best way to comprehensively represent such a chain is to write down an exclusion matrix (never used by DB, to say the least) :
Code: Select all
8r2c3
8r4c3 1r4c3
      1r8c3 1r8c7
8r2c7       1r2c7 3r2c7
                  3r2c5 5r2c5
                        5r8c5 4r8c5
8r8c3 1r8c3                   4r8c3

An exclusion matrix makes it possible to enumerate all the solutions of a sub-puzzle, so the exclusion of 8r2c3 is determined by the solutions of an ordered set P of 6 constraints :
P= {(81)r4c3, 1r8c37, (183)r2c7, (35)r2c5, (54)r8c5, (418)r8c3}; Note : 8r2c7 is a z-candidate and 1r8c3 is a t-candidate

Reading the exclusion matrix from the last line, one gets

Kraken Cell (148)r8c3 -> [8r4c3==8r2c7==8r8c3]-8r2c3; stte
1r8c3-(1=8)r4c3
||
4r8c3-(4=53)r82c5-3r2c7=*WWing[(8=*1)r2c7-1r8c7=1r8c3-(1=8)r4c3]
||
8r8c3

or from the 4th line, as above :

[NP(81)r48c3=4r8c3-(4=53)r82c5-3r2c7=*WWing(81)r2c7.r8c3,1r8c37]-8r2c3; stte

Last, but not least, -8r2c3 is equivalent to -4r8c3. This allows to write down another 6 constraints chain equivalent to Phil's one :
Code: Select all
+------------------+------------------+------------------+
| 39   78   5      | 1379  6     4    | 138    1389  2   |
| 39   6    (28)   | 1239  (35)  1359 | (138)  7     4   |
| 1    47   (24)   | 2379  8     379  | 5      39    6   |
+------------------+------------------+------------------+
| 58   158  18     | 6     9     2    | 7      4     3   |
| 6    2    7      | 134   34    13   | 9      58    58  |
| 4    9    3      | 5     7     8    | 6      2     1   |
+------------------+------------------+------------------+
| 258  458  9      | 34    1     6    | 238    358   7   |
| 258  3    8-4(1) | 479   (45)  579  | 28(1)  6     589 |
| 7    15   6      | 8     2     359  | 4      135   59  |
+------------------+------------------+------------------+
Kraken Cell (138)r2c7 : AAIC : 1r8c3=1r8c7-1r2c7=*XY-Chain[(4=28)r32c3-(8=*3)r2c7-(3=54)r28c5] -> [1r8c3==4r3c2==4r8c5]-4r8c3
JC Van Hay
 
Posts: 604
Joined: 22 May 2010

Re: February 26, 2017

Postby SteveG48 » Sun Feb 26, 2017 7:50 pm

JC Van Hay wrote:The exclusion of 8r2c3 can also be explained by Steve's chain which is nothing else as a way of writing an nrczt-chain, an oriented chain [IOW, not a bi-directional AIC].


JC, I'm pretty sure that my chain is a bi-directional AIC. That was my intent in any case. Otherwise, I'd have posted it as a Kraken cell.
Steve
User avatar
SteveG48
2016 Supporter
 
Posts: 1735
Joined: 08 November 2013
Location: Orlando, Florida

Re: February 26, 2017

Postby eleven » Sun Feb 26, 2017 8:43 pm

I like the w-wing chain
w-wing=3r2c7-(3=4)r28c5-(4=8)r48c3,
[edit: sorry Steve, where was my head ?]
but exclusion matrices and nrczt chains are a horror for me.
I can't imagine, that manual solvers could do such things.
eleven
 
Posts: 1381
Joined: 10 February 2008

Re: February 26, 2017

Postby eleven » Sun Feb 26, 2017 11:53 pm

Sorry, big mistake from my side (corrected above). Steve's chain IS something, i think manual solvers would do.
eleven
 
Posts: 1381
Joined: 10 February 2008

Re: February 26, 2017

Postby JC Van Hay » Sun Feb 26, 2017 11:58 pm

SteveG48 wrote:
JC Van Hay wrote:The exclusion of 8r2c3 can also be explained by Steve's chain which is nothing else as a way of writing an nrczt-chain, an oriented chain [IOW, not a bi-directional AIC].


JC, I'm pretty sure that my chain is a bi-directional AIC. That was my intent in any case. Otherwise, I'd have posted it as a Kraken cell.

Maybe I don't understand the symbols | and & !
However, here is your chain : (8=14)r48c3 - (4r3c3)|(4r8c5) = ((35)r28c5)&((28)r23c3) - (3|8=1)r2c7 - r8c7 = (18)r48c3 => -8 r2c3 ; stte

In order to understand it, I wrote the associated exclusion matrix :
Code: Select all
8r2c3
8r4c3 1r4c3
8r8c3 1r8c3 4r8c3
            4r3c3 2r3c3
                  2r2c3 8r2c3
            4r8c5             5r8c5
                              5r2c5 3r2c5
                        8r2c7       3r2c7 1r2c7
                                          1r8c7 1r8c3
8r4c3                                           1r4c3
Strangely, your chain and the exclusion matrix contain the target 8r2c3 ! Straightfowardly, this means that r2c3=2, r3c3=4 and r8c3≠4. That's why I simplified your chain by removing (28)r2c3 and (24)r3c3 to avoid cannibalism.

Nevertheless, reading the exclusion matrix top to bottom and bottom to top, i only can write the 2 following "different" chains :
(81=4)r48c3-(4r3c3)&(4r8c5)=((28)r32c3)&((53)r82c5)-(8&3=1)r2c7-1r8c7=(18)r48c3 -> -8r2c3; stte
(81)r4c3=1r8c7-(1=3|8)r2c7-(3r2c5|8r2c3)=((54)r28c5)|((24)r23c3)-(4=18)r84c3 -> -8r2c3; stte

What am I missing ?
JC Van Hay
 
Posts: 604
Joined: 22 May 2010

Re: February 26, 2017

Postby SteveG48 » Mon Feb 27, 2017 12:54 am

Thanks for the reply, JC, but honestly I don't understand your discussion about the exclusion matrix.

My symbols | and & are the standard Boolean symbols for OR and AND respectively. In writing my chain, I've followed the rules posted by David P Bird for bi-directional AICs:

Nevertheless for an alternating inference chain to be bi-directional the same basic rules must always be followed.
1) Each term in the chain must be a Boolean which will be interpreted as either being true or false
2) The link types must alternate between weak and strong
3) Each link must be stand-alone and independent of the truth states of candidates elsewhere in the chain.


I think if you look at the chain again you'll see that it reads correctly from right to left or left to right indicating that 8 must be true in r48c3.
Steve
User avatar
SteveG48
2016 Supporter
 
Posts: 1735
Joined: 08 November 2013
Location: Orlando, Florida

Re: February 26, 2017

Postby Cenoman » Mon Feb 27, 2017 3:02 pm

Grid after initial lcls eliminations (for memory)
Code: Select all
 +-------------------+--------------------+--------------------+
 | 39    78    5     | 1379   6    4      | 138   1389   2     |
 | 39    6     28    | 1239   35   1359   | 138   7      4     |
 | 1     47    24    | 2379   8    379    | 5     39     6     |
 +-------------------+--------------------+--------------------+
 | 58    158   18    | 6      9    2      | 7     4      3     |
 | 6     2     7     | 134    34   13     | 9     58     58    |
 | 4     9     3     | 5      7    8      | 6     2      1     |
 +-------------------+--------------------+--------------------+
 | 258   458   9     | 34     1    6      | 238   358    7     |
 | 258   3     148   | 479    45   579    | 128   6      589   |
 | 7     15    6     | 8      2    359    | 4     135    59    |
 +-------------------+--------------------+--------------------+

As I have proved to be expert at logics on February 23, I'm going to throw my own pinch of salt in the debate...
(@JC "je vais mettre mon grain de sel...")

Steve's chain, for memory: (8=14)r48c3 - (4r3c3)|(4r8c5) = ((35)r28c5)&((28)r23c3) - (3|8=1)r2c7 - r8c7 = (18)r48c3 => -8 r2c3 ; stte
I write it another way:
Code: Select all
               /|4r3c3=28r23c3-8r2c7|\\
(18)r48c3=4r8c3-|4r8c5=35r28c5-3r2c7|=1r2c7-r8c7=(18)r48c3

Note a slight variant in the head ALS (148)r48c3; that could be written as well (8=4)r48c3 as in use on this forum.
The core of Steve's demonstration is the sub-chain:
Code: Select all
     /|4r3c3=28r23c3-8r2c7|\\
4r8c3-|4r8c5=35r28c5-3r2c7|=1r2c7

Read from left to right this demonstrates that if 4r8c3 equals TRUE, then 1r2c7 equals TRUE.
Read from right to left the sub-chain demonstrates that if 1r2c7 equals FALSE, then 4r8c3 equals FALSE.
This is exactly what is needed to read the chain from right to left:
whether rightest (18)r48c3 equals TRUE, or 1r8c7 equals TRUE, 1r2c7 equals FALSE, 4r8c3 equals FALSE, leftest (18)r48c3 equals TRUE; Q.E.D. !

For me, Steve's chain, either with his own writting, or with my "net" writing above, works bidirectional. I can't see any memory effect hidden inside.

A minor remark to Steve: using (18)r48c3 is a bit ambiguous. Out of the context, such a member can be interpreted as 1r4c3&8r8c3|1r8c3&8r4c3. Here only 1r8c3&8r4c3 is possible although 18 is present in both cells, and personally, I would have written your chain:
(8=4)r48c3 - (4r3c3)|(4r8c5) = ((35)r28c5)&((28)r23c3) - (3|8=1)r2c7 - r8c7 = r8c3 - (1=8)r4c3 => -8 r2c3
but I am entering a debate about tastes and colours...

And now, the icing on the cake. Why not, with the same core logic:
Code: Select all
Kraken cell (138)r2c7 => -4 r8c3; stte
(1)r2c7 - r8c7 = r8c3 - 4r8c3
(3)r2c7 - (3=4)r28c5 - 4r8c3
(8)r2c7 - (8=4)r23c3 - 4r8c3

...my preference, but you could also write it:
(4r3c3)|(4r8c5) = ((35)r28c5)&((28)r23c3) - (3|8=1)r2c7 - r8c7 = (1)r8c3 => -4 r8c3; stte
It sounds familiar, does it not ?

Cenoman
Cenoman
 
Posts: 135
Joined: 21 November 2016
Location: Paris, France

Re: February 26, 2017

Postby SteveG48 » Mon Feb 27, 2017 3:51 pm

Cenoman wrote:A minor remark to Steve: using (18)r48c3 is a bit ambiguous. Out of the context, such a member can be interpreted as 1r4c3&8r8c3|1r8c3&8r4c3. Here only 1r8c3&8r4c3 is possible although 18 is present in both cells, and personally, I would have written your chain:
(8=4)r48c3 - (4r3c3)|(4r8c5) = ((35)r28c5)&((28)r23c3) - (3|8=1)r2c7 - r8c7 = r8c3 - (1=8)r4c3 => -8 r2c3
but I am entering a debate about tastes and colours...


I agree that your way is nicer.

And now, the icing on the cake. Why not, with the same core logic:
Code: Select all
Kraken cell (138)r2c7 => -4 r8c3; stte
(1)r2c7 - r8c7 = r8c3 - 4r8c3
(3)r2c7 - (3=4)r28c5 - 4r8c3
(8)r2c7 - (8=4)r23c3 - 4r8c3



Yes. As I indicated a few posts up, I could have presented it as this Kraken cell solution. I didn't because I'm trying to hone my skill at creating proper bi-directional AICs whenever possible. Sometimes it results in solutions that are a bit complex, but it's good practice for me.
Steve
User avatar
SteveG48
2016 Supporter
 
Posts: 1735
Joined: 08 November 2013
Location: Orlando, Florida

Re: February 26, 2017

Postby StrmCkr » Tue Feb 28, 2017 12:01 am

Code: Select all
.---------------.----------------.----------------.
| 39   78   5   | 1379  6   4    | 138  1389  2   |
| 39   6    28  | 1239  35  1359 | 138  7     4   |
| 1    47   24  | 2379  8   379  | 5    39    6   |
:---------------+----------------+----------------:
| 58   158  18  | 6     9   2    | 7    4     3   |
| 6    2    7   | 134   34  13   | 9    58    58  |
| 4    9    3   | 5     7   8    | 6    2     1   |
:---------------+----------------+----------------:
| 258* 458* 9   | 34*   1   6    | 238  358   7   |
| 258* 3    148 | 479   45  579  | 128  6     589 |
| 7    1-5  6   | 8     2   359* | 4    135   59*  |
'---------------'----------------'----------------'



Almost Disjointed Distributed Subset:

Cells 55,56,58,64,78,81
digit Set 34589
Row 7: 4
Row 9: 9
box 8: 3
box7: 28

unrestricted cell candidate 5
=>> R9C2 <> 5

singles to the end.

xsudo logic copy:

Code: Select all
Sue  [41,107] 16 Candidates
     9 Truths = {4R7 9R9 7N124 8N1 9N69 3B8}
     4 Links = {5r9 258b7}
     1 Elimination --> r9c2<>5 
Some do, some teach, the rest look it up.
User avatar
StrmCkr
 
Posts: 597
Joined: 05 September 2006

Re: February 26, 2017

Postby Alex_Popov_92 » Tue Feb 28, 2017 4:17 am

Code: Select all
 39   78   5    | 1379 6    4    | 138  1389 2   
 39   6    28   | 1239 35   1359 | 138  7    4   
 1    47   24   | 2379 8    379  | 5    39   6   
----------------+----------------+----------------
 58   158  18   | 6    9    2    | 7    4    3   
 6    2    7    | 134  34   13   | 9    58   58 
 4    9    3    | 5    7    8    | 6    2    1   
----------------+----------------+----------------
 258  d458  9   | e34   1    6   | a28-3  358  7   
 258  3    c148 | 479  45   579  | b128  6    589
 7    15   6    | 8    2    359  | 4    135  59

(3-2)r7c7 = (2-1)r8c7 = (1-4)r8c3 =4r7c2 - (4=3)r7c4 => -3 r7c7
stte
Alex
Alex_Popov_92
 
Posts: 22
Joined: 22 February 2017

Re: February 26, 2017

Postby bat999 » Tue Feb 28, 2017 12:02 pm

Alex_Popov_92 wrote: => -3 r7c7
stte

Hi Alex.
=> -3 r7c7 doesn't solve the puzzle with stte (singles to the end).
It still needs some locked sets to finish it with lclste (locked cells and sets to end).

So your solution is like this...
Code: Select all
.-----------------.-----------------.------------------.
| 39    78    5   | 1379  6    4    |  138   1389  2   |
| 39    6     28  | 1239  35   1359 |  138   7     4   |
| 1     47    24  | 2379  8    379  |  5     39    6   |
:-----------------+-----------------+------------------:
| 58    158   18  | 6     9    2    |  7     4     3   |
| 6     2     7   | 134   34   13   |  9     58    58  |
| 4     9     3   | 5     7    8    |  6     2     1   |
:-----------------+-----------------+------------------:
| 258  d458   9   | e34    1   6    | a28-3  358   7   |
| 258   3    c148 | 479    45  579  | b128   6     589 |
| 7     15    6   | 8      2   359  |  4     135   59  |
'-----------------'-----------------'------------------'
(3-2)r7c7 = (2-1)r8c7 = (1-4)r8c3 = 4r7c2 - (4=3)r7c4 => -3 r7c7; lcste

Some people think that lcste solutions are not as good as stte solutions.
But it's your call, submit your solution for "peer review" and see if you get any complaints. :D
8-)
User avatar
bat999
2017 Supporter
 
Posts: 624
Joined: 15 September 2014
Location: UK

Re: February 26, 2017

Postby JC Van Hay » Wed Mar 01, 2017 1:41 pm

Thanks, Steve, eleven and Cenoman for your comments even though they didn't answser why Steve's chain was an AIC.
I had to analyse each link to be convinced of the AIC nature of the chain. By the way, it reminds me of a very long discussion here !
Furthermore, as a consequence, I learned at least one way to translate an nrzct-chain into an AIC by adding extra constraints !
Because of the above given link, I will however refrain to comment on my favorite way of writing the wysiwyg proof of the exclusion of a candidate, the "antique" Andrei Zelevinsky's Exclusion [or Forbidding] Matrices.
JC Van Hay
 
Posts: 604
Joined: 22 May 2010

PreviousNext

Return to Puzzles