The New Sudoku Players' Forum

[andre43]

Code: Select all

 *-----------------------------------------------------------*
 | 24     x      x   |    x     x      x | 257     x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |-------------------+-------------------+-------------------|
 |  x     x      x   |    x     x      x | 28      x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |-------------------+-------------------+-------------------|
 | 127   17      x   |    x     x      x |  x      x       x |
 | 1278  1478    x   |    x     x      x |  x      x       x |
 |  x     x     48   |    x     x      x | 248     x       x |
 *-----------------------------------------------------------*

Hi, I’d like to show you a situation, similar to one I confronted today solving an “extreme” puzzle. Well, near the end of the solving process, and after using all my “advanced” weapons (mainly through a network), I spotted an ALS chain,that is (see above): -[r1c7]=2=[r1c1]-2-[ALS r7c12|r8c12]-4-[r9c3]=4=[r9c7]-8-[r9c7]=8=[r4c7]-2-[r4c7]=2=[r1c7], that is r1c7=1. But then I saw that, if -[r1c7]=2=[r1c1]-2-[ALS r7c12|r8c12]-8-[r9c3]=8=[r9c7]-8-[r4c7]=2=[r4c7]-2-[r1c7], that is we reach in a “continuous nice loop”!!! The reason, of course, is that the ALS can not be valid, since in that case the r9c3 will be empty the ALS contains both digits). My question is this: may both ways (cont. nice loop AND nice loop, rule 1)be valid in our route to r1c7? Or we cannot make the assumption (and consequently all the nice loop process) of that ALS, since the latter is invalid a priori? I’d greatly appreciate your help. Thanks! (and sorry for failing to send the right image of the grid-I just didn't understand the way to do it yet!!)
Grid corrected using [ code ] tags - Ruud

[re'born]

I spotted an ALS chain,that is (see above): -[r1c7]=2=[r1c1]-2-[ALS r7c12|r8c12]-4-[r9c3]=4=[r9c7]-8-[r9c7]=8=[r4c7]-2-[r4c7]=2=[r1c7], that is r1c7=1.

The first thing I see is that your chain isn't well formed. You've effectively written:
r9c3<>4 => r9c7=4 and then r9c7=8 =>r9c7<>8 => r4c7=8 and then r4c7=2 => r4c7<>2.

But why even leave c123? As you noted everything you need is already there.

Edit: To make this last point more explicit, the 5 given cells in box 7 form a naked quint. Hence, the 2's in column 1 are locked in box 7 and hence, r1c1<>2.

[andre43]

re'born, you are right...my example, as I said, is a similar one to that of the puzzle I solved, but I have done the mistake in r9c6. Please, see it again without the digit 2 in that cell. Thanks.

[andre43]

A new mistake! I mean you must take out the digit 2 from r9c7 (and not in r9c6).

[re'born]

Andre,

So that I understand what you're asking, let me try to rephrase your question, and you can correct me if I've missed something. Given the grid:

Code: Select all

 *-----------------------------------------------------------*
 | 24     x      x   |    x     x      x | 257     x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |-------------------+-------------------+-------------------|
 |  x     x      x   |    x     x      x | 28      x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |  x     x      x   |    x     x      x |  x      x       x |
 |-------------------+-------------------+-------------------|
 | 127   17      x   |    x     x      x |  x      x       x |
 | 1278  1478    x   |    x     x      x |  x      x       x |
 |  x     x     48   |    x     x      x | 48      x       x |
 *-----------------------------------------------------------*

where we assume that the 2's in row 1 and column 7 are conjugate and the 8's in column 7 are conjugate. Forgetting about the naked quint in box 7, then you're question is whether the following two deductions are both valid:

Code: Select all

[r1c7]=2=[r1c1]-2-[ALS: r78c12]-4-[r9c3]-8-[r9c7]=8=[r4c7]=2=[r1c7], => r1c7=2

[r1c7]=2=[r1c1]-2-[ALS: r78c12]-8-[r9c3]-4-[r9c7]-8-[r4c7]-2-[r1c7], => r1c7=2


The first I believe, no problem. The second, I don't since usually deductions in a continuous loop occur outside of the pattern. In this case, I would think all we can conclude is that r234569c1<>2, r78c3,r9c12<>8, r9c1245689<>4, r235678c7<>2,8. However, all of these deductions follow either from our assumptions on the grid, or because of naked pairs and locked candidates.

[andre43]

Thanks, re'born. At first I think that by oversight you give => r1c7=2 in the second case, while in that case we have => r1c7<>2. I think you have catched what I meant, but I must confess that I am a little confused by your hints. As I see it, you say that the correct way is the first one, fixing the 2 in r1c7 (and this is also the solution of that puzzle I have solved!). But I didn’t understand your reasoning for the second case (of the continuous loop): finally, may we or may not make all these deductions you refer, by the way of the cont. loop alone? (By the way, the naked quint and the naked pair 48 are occurred accidentally of course, since my example was clearly impromptu…). In conclusion, my one direct and essential question is: reaching at the ALS ([r1c7]=2=[r1c1]-2-[ALS: r78c12]), which of the two ways I should follow? May I follow both ways, making first all the deductions produced by a continuous loop AND THEN fixing the 2 in r1c7, using the other rule of the discontinuous nice loops? That’s all!! Many thanks for your analysis!

[re'born]

At first I think that by oversight you give => r1c7=2 in the second case, while in that case we have => r1c7<>2.

The abridged English translations of the nice loops above are:
1. If r1c7<>2 then somehow we get r1c7 = 2. Hence, r1c7=2.

2. If r1c7<>2, then somehow we get r1c7<>2. Hence r1c7=2.

The first is okay. The second, I say implies r1c7=2 and you say it should be r1c7<>2. In fact, we are both wrong. The English translation makes it clear that we get no information about r1c7 from the second loop. However, both loops you propose are valid, they just make different exclusions.

But I didn’t understand your reasoning for the second case (of the continuous loop): finally, may we or may not make all these deductions you refer, by the way of the cont. loop alone?

Yes we may make all of the exclusions I suggested, but to emphasize, this list does not include r1c7.

In conclusion, my one direct and essential question is: reaching at the ALS ([r1c7]=2=[r1c1]-2-[ALS: r78c12]), which of the two ways I should follow? May I follow both ways, making first all the deductions produced by a continuous loop AND THEN fixing the 2 in r1c7, using the other rule of the discontinuous nice loops?

Probably you should use the discontinuous nice loop first. Placing the 2 in r1c7 (generally speaking) will almost certainly lead to easy eliminations of all of the candidates you would eliminate using the continuous loop.

That’s all!! Many thanks for your analysis!

You're welcome. I apologize for not understanding what you were getting at sooner. Thanks for bearing with me.

[andre43]

That’s it re'born! You have made them all clear to me by now..Thanks again.