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My mutli-digit coloring test

 
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Sun Dec 09, 2007 9:10 pm    Post subject: My mutli-digit coloring test Reply with quote

This puzzle was posted at the player's forum some days ago:
Code:
 *-----------*
 |..3|25.|.9.|
 |8..|9..|64.|
 |...|..7|3..|
 |---+---+---|
 |6..|7.1|...|
 |.84|5..|...|
 |...|4..|...|
 |---+---+---|
 |..5|.8.|..4|
 |...|...|.3.|
 |.3.|...|.71|
 *-----------*
 *-----------*
 |..3|25.|.9.|
 |8..|913|64.|
 |...|8.7|3..|
 |---+---+---|
 |6..|7.1|48.|
 |.84|5..|...|
 |...|4.8|...|
 |---+---+---|
 |..5|38.|.64|
 |...|17.|.3.|
 |.3.|6..|.71|
 *-----------*
 *-----------------------------------------------------------------------------*
 | 147     1467    3       | 2       5       46      | 178     9       78      |
 | 8       257     #2b7a   | 9       1       3       | 6       4      #5a7b    |
 | 1459    1469    169     | 8       46      7       | 3      #1a5b    2       |
 |-------------------------+-------------------------+-------------------------|
 | 6       259     29      | 7       239     1       | 4       8       359     |
 | 12379   8       4       | 5       2369    269     | 179     12      3679    |
 | 123579  1279   #1a27b9  | 4       2369    8       | 79     #1b25    3679    |
 |-------------------------+-------------------------+-------------------------|
 | 17      17      5       | 3       8       29      | 29      6       4       |
 | 249     2469    2689    | 1       7       2459    | 2589    3       89      |
 | 249     3       289     | 6       249     2459    | 2589    7       1       |
 *-----------------------------------------------------------------------------*

The guys there must be more dead than me, because until today there was no solution posted Wink

I studied it for hours, but couldn't find a nice solution either, only a first step. The marked cells form a cycle, with ab being the only possibiltities. This gives
r2c2<>7, r3c12<>1, r6c3<>29.

An xyz-wing, triple and ER follows, but then stuck again. I couldn't find anything, that would give me a number.
Code:
 *-----------------------------------------------------------*
 | 147   1467  3     | 2     5     46    | 178   9     78    |
 | 8     25    27    | 9     1     3     | 6     4     57    |
 | 459   469   169   | 8     46    7     | 3     15    2     |
 |-------------------+-------------------+-------------------|
 | 6     259   29    | 7     239   1     | 4     8     359   |
 | 2379  8     4     | 5     2369  269   | 179   12    3679  |
 | 235   179   17    | 4     236   8     | 79    25    36    |
 |-------------------+-------------------+-------------------|
 | 17    17    5     | 3     8     29    | 29    6     4     |
 | 249   246   2689  | 1     7     2459  | 2589  3     89    |
 | 249   3     289   | 6     249   2459  | 2589  7     1     |
 *-----------------------------------------------------------*
So i just continued the multidigit coloring (did not get far in a), until i got a contradiction in b. At this time only a few cells were not colored b.
It was a relatively simple method to proceed with this awkward puzzle (another cycle solved it then). I did not really like it, but it would have been worse, if the color had solved it at the end.

So multidigit coloring for me is the last thing to resort.
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Dec 10, 2007 12:41 am    Post subject: Reply with quote

ravel,

In your first grid, I follow everything except that I don't see the "b" marking for the <1> in r6c8. Instead, I see that the <1> in r3c3 should be marked "b", consistent with the elimination of <1>s in r3c12, with which I agree. As you've marked it, the <1>s in r6c12 would be eliminated. However, I don't agree with that. Am I not understanding something?

I suspect that not everyone will see how this coloring was done (or how I did it, at least). It involves locked candidates in the left "tower" (B147) based on the coloring in B3 and the {17} locked pair in r7c12. If "a" is true, there is an X-Wing on <1> in r17c12 leaving the <1> in r6c3 as the only "a" <1> in c3 (because of the "a" <1> in r3c8). And, since this "a" <1> displaces one of the only 2 <7>s in c3, the other <7> in r2c3 must be "a". (A similar approach could be used for the "b" case with a <7> X-Wing, etc.)
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Dec 10, 2007 1:18 am    Post subject: Reply with quote

I, too, next extended the "b" implication and found a contradiction after involving almost the entire grid. (Awkward, indeed!) This allowed placement of all "a" values. But, I only needed an XYZ Wing after that to finish. Here's the grid I got:
Code:
+--------------+------------+-------------+
| 14  146 3    | 2 5   46   | 78   9 78   |
| 8   2   7    | 9 1   3    | 6    4 5    |
| 5   469 69   | 8 46  7    | 3    1 2    |
+--------------+------------+-------------+
| 6   5   29   | 7 239 1    | 4    8 39   |
| 379 8   4    | 5 369 69   | 1    2 3679 |
| 23  79  1    | 4 236 8    | 79   5 36   |
+--------------+------------+-------------+
| 17  17  5    | 3 8   29   | 29   6 4    |
| 249 46  2689 | 1 7   2459 | 2589 3 89   |
| 249 3   289  | 6 249 2459 | 2589 7 1    |
+--------------+------------+-------------+

The Pivot is r1c2.
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Mon Dec 10, 2007 9:15 am    Post subject: Reply with quote

Asellus wrote:
In your first grid, I follow everything except that I don't see the "b" marking for the <1> in r6c8. Instead, I see that the <1> in r3c3 should be marked "b", consistent with the elimination of <1>s in r3c12, with which I agree.
Oops, this wasn't my day, this puzzle has confused me after all.

When i tried to reconstruct this step, i had the right eliminations in mind, but somehow saw a strong link for 1 in row 6 instead of column 3. So i mixed the cells r3c3 and r8c8.
So this is the right cycle now, using the strong links for 1 and 7 in column 3:
r2c3=7 => r2c9=5 => r3c8=1 => r3c3<>1 => r6c3=1 => r2c3=7
And the opposite direction.
r2c3=2 => r6c3=7 => r3c3=1 => r3c8=5 => r2c9=7 => r2c3=2
Code:
 *-----------------------------------------------------------------------------*
 | 147     1467    3       | 2       5       46      | 178     9       78      |
 | 8       257    #2b7a    | 9       1       3       | 6       4      #5a7b    |
 | 1459    1469   #1b69    | 8       46      7       | 3      #1a5b    2       |
 |-------------------------+-------------------------+-------------------------|
 | 6       259     29      | 7       239     1       | 4       8       359     |
 | 12379   8       4       | 5       2369    269     | 179     12      3679    |
 | 123579  1279   #1a27b9  | 4       2369    8       | 79      125     3679    |
 |-------------------------+-------------------------+-------------------------|
 | 17      17      5       | 3       8       29      | 29      6       4       |
 | 249     2469    2689    | 1       7       2459    | 2589    3       89      |
 | 249     3       289     | 6       249     2459    | 2589    7       1       |
 *-----------------------------------------------------------------------------*
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Mon Dec 10, 2007 7:02 pm    Post subject: Reply with quote

This is my notation exercise (whats its name ?)
(2=7)r2c3-(7=5)r2c9-(5=1)r3c8-(1)r3c3=(1-7)r6c3=(7-2)r2c3
Then either r2c3=7 or r2c9=7 etc.
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Mon Dec 10, 2007 9:07 pm    Post subject: Reply with quote

ravel wrote:
This is my notation exercise (whats its name ?)
(2=7)r2c3-(7=5)r2c9-(5=1)r3c8-(1)r3c3=(1-7)r6c3=(7-2)r2c3
Then either r2c3=7 or r2c9=7 etc.



The Eureka notation looks just fine to me.
"It"s name is continuous nice loop. Continuous means there is always weak link, strong link, weak, strong ... and so on, and loop means that the chain is closed. What it does is, it effectively makes ALL links into strong links and that allows the elimination of all candidates that see any two candidates in the loop separated by an odd number of links, i.e. r2c2<>7, r3c12<>1, r6c3<>29 like you said in your first post.

7 in r2c2 sees 7 in r2c39,
1 in r3c12 sees 1 in r3c38,
2 and 9 in r6c3 see 1 and 7 in r6c3

____

I am not sure, but in my view the loop does not involve 2 in r2c3, just 7 in that cell
loop: (7)r2c3-(7=5)r2c9-(5=1)r3c8-(1)r3c3=(1-7)r6c3=(7)r2c3


Last edited by nataraj on Mon Dec 10, 2007 10:25 pm; edited 1 time in total
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Mon Dec 10, 2007 10:25 pm    Post subject: Reply with quote

Ok, thanks again, nataraj,

i think i understood enough now to be able to read it (also noticed that - like with an ALS with 2 extra candidates - its not an exclusive either/or). But - like with (classical) nice loops and others - i do not intend to use it myself. Such a closed loop is exactly the only sample, where i see an advantage for this notation.

The problems i have with all these notations (instead of basic implication chains, marked grids and everyday speech) are:
- they dont reflect the way i spot the things (this includes ALS)
- you need an explanation to be able to read them (and e.g. the nice loop explanations are deterrently difficult)
- there are always (relatively) simple deductions, that cannot be expressed at all or only in a very difficult way
- i cant see any need for them (for humans)

Whats the Eureka notation for this simple elimination ?
Code:
12345 12 . | 123
 .    .  . |  .   
14    .  . | 34       
r1c1=1 => r1c2=2 => r1c4=3 => r3c4=4 => r3c1=1
=> r1c1<>1
(such chains are the basic trick of Denis Berthier's chains - and notation)
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nataraj



Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria

PostPosted: Mon Dec 10, 2007 10:54 pm    Post subject: Reply with quote

ravel,
I had to edit my previous post - the continuous loop only allows eliminations that use an odd number of links in the loop. Which goes to show it is a terribly tricky subject and I am no expert at all.

This said, I think the value of the concept of AICs or nice loops for me lies in their simple and elegant basic structure. It does not make the patterns easier to find. It helps me to be sure, though. Write down the links. Make sure there is an alternating sequence. Use the ends for elimination.

I do not think it is the best formal language for all situations. Maybe I will, given some more time. Which still leaves the question: is it effective? As you pointed out, it only works when the message is understood by the recipient. So the answer is probably: depends. Depends on the situation (more complex situations tend to need more formal structuring, I think) and depends on the partners in the conversation. Most of the time, simple sentences and basic implications like r1c2=3 => r1c6=7 are certainly more effective means of communication.
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Asellus



Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

PostPosted: Mon Dec 10, 2007 11:33 pm    Post subject: Reply with quote

ravel,

The Eureka notation for your simple elimination (a short ALS Chain) is:

(1)r1c1-({12}=3)r1c24-(3=4)r3c4-(4=1)r3c1-(1)r1c1; r1c1<>1

While it looks mysterious to those who don't know it, it is really not difficult to understand. (I don't attempt to explain the notation here, though.) Every ALS contains a strong link. This is obvious in bivalue cells. In the 2-cell {123} ALS of r1c24, the <3> is strongly linked ("=") to the {12} locked pair. If the <1> in r1c1 is true, then the {12} pair is false (as a locked pair, not as two digits), the <3> true, etc. In any case, to me, this notation is no more complicated than the way you wrote out the implication. It has the virtue of showing that one is exploiting an ALS sequence in a consistent way rather than just forcing values true.

Obviously, such notation isn't necessary for eliminations that are simple to see and describe in ordinary language. I believe the notation has value in (at least) two situations: (1) communicating accurately and concisely to others; (2) keeping track of more complex AICs when trying to solve a difficult puzzle on paper.

I started learning it mostly so that I could understand it when others used it. After I did, I gradually found that it increased my solving ability and altered how I looked at puzzles. The discipline of the notation does make one see things differently, in my experience. But then, that may have just been my route to such insights. The nice loops lingo isn't necessary for constructing AICs or using Eureka. For instance, I didn't know that what I was doing was a "discontinuous loop" until nataraj started using the term. But, as with all such things, it no doubt adds some more value.
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Tue Dec 11, 2007 12:02 am    Post subject: Reply with quote

I knew, that it can be expressed as xyz-ALS, but this is an unneccessary complicated thing here. One simple step more and you need an ALS-chain.

Just can say that i did not make the experience. i really never saw a nice loop, which i wouldnt have found easier without thinking in the notation.

Worse, i noticed that people were not willing or able to understand solutions, which could have been described in simple words to them.

Another sample:
I dont even know, how to write this chain in Eureka notation (suppose there is a good way, but just to demonstrate, that there is a lot to learn to express - and probably understand - simple things):
Code:
1239 . . | 29
.    . . | .
.    . . | .
-------------
3456 . . | 23

strong link for 3 in c1
r1c1=9 => r4c1=3 => r4c4=2 => r1c4=9
=> r1c1<>9
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Asellus



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PostPosted: Tue Dec 11, 2007 3:41 am    Post subject: Reply with quote

The Eureka for the second example is
(9-3)r1c1=(3)r1c4-(3=2)r4c4-(2=9)r4c4-(9)r1c1; r1c1<>9

It's exactly the same as what you wrote but just looks different.

I agree that it is unnecessarily complicated (not helpful) for casual communication and simple cases, and myself only post Eureka for complex situations intended for more advanced readers or in addition to a "plain language" explanation. I wouldn't have used it for either of your simple examples, for instance; so, I'm with you there. I would still encourage folks to learn it if they have any interest.
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ravel



Joined: 21 Apr 2006
Posts: 536

PostPosted: Tue Dec 11, 2007 9:43 am    Post subject: Reply with quote

(9-3)r1c1=(3)r4c1-(3=2)r4c4-(2=9)r1c4-(9)r1c1; r1c1<>9

What was confusing me is, that this way you dont have the alternating links starting and ending with a strong one. So (without the weakly linked ends) i just see directly, that one of r1c1=3 and r1c4=9 must be true - which of course both imply r1c1<>9.

On the other hand the contradiction above is obviously at the first glance.
But i must admit, that when exploring strong links i would not start with r1c1=9, but use a similar, but a bit more general way as looking for xy-chains, that is better reflected by this Eureka chain:
Can i find a "way" from r1c1=3 to one of 456 in a cell that sees r4c1 or one from r4c1=3 to one of 129 in a cell that sees r1c1 ?
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Asellus



Joined: 05 Jun 2007
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PostPosted: Tue Dec 11, 2007 12:55 pm    Post subject: Reply with quote

I agree: we actually tend to work from the "pincers" (i.e., the strongly linked ends that kill the victim). The convention in Eureka notation, however, is to place the node of the discontinuity (i.e., the victim(s)) at the two ends of the chain. The weak links indicate that it is false (an elimination). One quickly learns to look past the two ends.

It is possible to have a discontinuity node that is strongly linked on both ends. In that case, the node is proved true.

In either case, the affected node is usually one or more instances of some candidate digit. However, it could be anything... a locked set, for instance or a wing of some sort, I suppose.

Another quick point: focussing on the weak link starting point enables one to see what wouldn't typically be considered a pincer, such as the companion <3> sharing the cell with the victim <9> in this example. There is something nicely unifying about the AIC way of seeing things.

Quote:
But i must admit, that when exploring strong links i would not start with r1c1=9, but use a similar, but a bit more general way as looking for xy-chains, that is better reflected by this Eureka chain:
Can i find a "way" from r1c1=3 to one of 456 in a cell that sees r4c1 or one from r4c1=3 to one of 129 in a cell that sees r1c1 ?

If I understand your comment correctly, I would respond that you seem to be looking from "outside" the strong link rather than seeing it as a strong link option along a chain. A strict ALS chain alternates strong links within an ALS with weak links between each ALS. But, one can mix and match. A strong link between two candidate digits in peer cells can replace an ALS along the chain with no problem. The two digits needn't even be peers if the strong link results from a pincer relationship, for example... or from a UR, say. In any case, however creative one gets in discovering such mixes and matches, the notation helps keep track of it.
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