January 5, 2015

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Re: January 5, 2015

Postby David P Bird » Wed Jan 07, 2015 1:46 pm

Don, my view is that Steve's solution is very clever but contains concealed branching in an unrecorded pattern.

What needs to be shown is (9)r1c45 = (57)r12c4 to take (7)r1c5 out of consideration. This then needs something like
(9)r1c45 = (9-5)r2c6 = (5)r2c4
&
(9=7)r1c4

The last thing I want to do is to discourage innovative solutions like this because that's how we expand our collective talent, but as ever I have two things on my mind:
1) How can that be explained to a newcomer
2) If that's considered acceptable what precedent would be set for finding other one-off patterns on the fly which could become very elaborate.

It's a recurring problem for me that in Sudoku it's virtually impossible to find well defined demarcating lines for what is acceptable and what isn't. As soon as a line is set, it will be found that either some quite reasonable method has been outlawed, or the door has been opened to one which is distasteful.

One way out would be to describe the pattern and prove its inferences, but how long will it be before we find another like it?

The thing is that the puzzle is very simple and it's only because we're trying to find single step solutions that this device has emerged. It's therefore very unlikely that it would ever provide an essential opening for a tougher puzzle.

Nowadays there's no such thing as consensus, but for these reasons I won't include this in my personal repertoire, but then I'm not trying to solve single steppers!
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Re: January 5, 2015

Postby SteveG48 » Wed Jan 07, 2015 3:50 pm

David P Bird wrote:Don, my view is that Steve's solution is very clever but contains concealed branching in an unrecorded pattern.

What needs to be shown is (9)r1c45 = (57)r12c4 to take (7)r1c5 out of consideration. This then needs something like
(9)r1c45 = (9-5)r2c6 = (5)r2c4
&
(9=7)r1c4

The last thing I want to do is to discourage innovative solutions like this because that's how we expand our collective talent, but as ever I have two things on my mind:
1) How can that be explained to a newcomer
2) If that's considered acceptable what precedent would be set for finding other one-off patterns on the fly which could become very elaborate.


David, interesting discussion. I don't disagree with anything that you've said. I would like to address your second question (and perhaps the first). This particular pattern may be a one-off, but in even the brief time that I've been working these puzzles I've seen a similar situation arise a number of times. I'd describe it this way:

1. We have an ALS that becomes a locked set when one or more candidates are removed.
2. Once the set is locked, the position of one or more of the remaining candidates in the set becomes obvious.
3. The information in (2) is critical to forging the next link in the chain.

The question, then, is whether it's acceptable to assume that the reader will look at the locked set on the grid and see how the pattern is used, or is it necessary for the solver to describe it somehow. It seems to me that leaving it to the reader is reasonable, as long as we confine ourselves to a locked set in one house. That seems simple enough to describe to a newcomer- after all, I am one! I also don't see how the precedent would be expanded to the distasteful. How complicated could a locked set in one house become? Anyway, those are my thoughts. I'm interested in what others think about it.
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Re: January 5, 2015

Postby eleven » Wed Jan 07, 2015 4:59 pm

To be able to understand any ALS-chains (in AIC or whatever notation), a newcomer must first understand the principle of an ALS. If he knows, that if one of n+1 candidates in n cells in a house is false, then all the other digits must be true (wherever they can be in these cells), this should be enough also to understand this link.
If 9 is false, 5r3c4 must be true, because it is the only one in the ALS cells. Obviously 7r1c4 must be true without the 9. Thats all. No special notation needed.
To learn to understand the basic AIC notation is a lot harder for a newcomer.
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Re: January 5, 2015

Postby David P Bird » Wed Jan 07, 2015 9:39 pm

Steve, as I hope I made clear, I considered your use of forced placements was on the acceptability borderline. As the current emphasis for in this puzzles section is a quest for clever one-step solutions I guess most participants would be happy with it. If so, to give readers with limited knowledge the chance to understand how the heck the chain works, one suggestion would be to use an &So notation to lead them onto the right path:
(9)r1c45 = (4567)r13c45 &So (57)r13c4 – (57=9)r5c4 .....
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