The New Sudoku Players' Forum

[arsoncupid]

I'm working on the hardest puzzle I've found, by hand. It's a deeply troubling puzzle

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..1 .8. 6.4
.37 6.. ...
5.. ... ...

... ..5 ...
..6 .1. 8..
... 4.. ...

... ... ..3
... ..7 52.
8.2 .9. 7..

Through logic, I reach this state in the middle "row of groups":

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12379 124789 349 | 2789 2367 5 | 12349 134679 12679
2379  2479 6 | 279 1 29 | 8 34579 2579
12379 12789 5 | 4 2367 2689 | 1239 13679 12679

Notice in the middle row, the 3's and 4's. There are a pair of each, and the pairs intersect at the 8'th column. Late one night I assumed this meant I could exclude 5,7,9 from the 8th column because either a 3 or a 4 must go in that square. But now I see no way of saying that without excluding the other options.

I did prove that this must be the case, in fact the eighth column's number must be a 3. But I had to do it by assuming a 5,7,9 in that possition, then proving it leads to a contradiction. This is a rather large branch of logic and my work on this puzzle, still largely unfinished, is 16 pages!

...
Another sample, one I've not proved but is the same sort of generalization, is with this puzzle:

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.96 ... .1.
.5. 6.. 7..
..1 8.. ...

5.. .94 .68
.6. ... .4.
97. 16. ..5

... ..1 3..
..5 ..2 .7.
.1. ... 59.

Through basic steps, the bottom right group (3x3) is:

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3 28 246
468 7 1
5 9 24

The same reduction applied here would drop the 4 from 4,6,8.

Is there a logical way to say the following?
In any row, column, or group, where two pairs intersect, you can exclude the other numbers from the intersecting square.

[ronk]

Is there a logical way to say the following?
In any row, column, or group, where two pairs intersect, you can exclude the other numbers from the intersecting square.

I'd be very surprised if that was true. But you really think it is, go ahead and assume it's true. If it's a valid assumption, it will work for you every time. If it's not valid, you will soon run into a situation where it doesn't work. Then you'll know.

If your group is for rows 1, 2, and 3 ...

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12379 124789 349 | 2789 2367   5    | 12349 134679 12679
2379  2479   6   | 279  1      29   | 8     34579  2579
12379 12789  5   | 4    2367   2689 | 1239  13679  12679

... the best one can say IMO is ... if r2c1<>3, then r2c8=3, r2c8<>4, and r2c2=4 ... and other implications from those.

And yes, it's a very difficult puzzle to do by hand.

[bennys]

It’s obvious that it’s not true.
According to that argument if I have 12, 13, and 23 I can eliminate all the candidates.

[arsoncupid]

It’s obvious that it’s not true.
According to that argument if I have 12, 13, and 23 I can eliminate all the candidates.

No no! You've misread it, I'm sure.
With that example:
12, 13, 23
*These numbers occur in a row, column or box. No other 2 or 3 occur in that row, column or box.

My line of reasoning says we can eliminate any other numbers from the 23 cel. So:
124, 1346, 236
We can drop the 6 from 236. Which resolves:
124, 1246, 23

If you literally mean only 12, 13, 23. My line of reasoning says only that either a 2 or 3 must reside in the 23 cel. Which is rather obvious.

[Shazbot]

If you have

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12379 124789 349 | 2789 2367 5    | 12349 134679 12679
2379  2479   6   | 279  1    29   | 8     34579  2579
12379 12789  5   | 4    2367 2689 | 1239 13679  12679

your second row is 2379 2479 6 | 279 1 29 | 8 34579 2579.

This is not an example of a hidden pair - a hidden pair is where two candidates are limited to TWO cells, but your two are limited to THREE cells.

Without going through the whole puzzle, it's possible you could end up with r2c1=3, r2c2=4 and r2c8=5 7 or 9. Perhaps you were just lucky in this case, but I don't think you can apply it as a rule to all puzzles.

You're saying "In any row, column, or group, where two pairs intersect, you can exclude the other numbers from the intersecting square." I don't follow your terminology - you're looking at a single candidate in column 1, a single candidate in column 2 (not 2 pairs), which happen to be candidates in the same cell in column 8. You're implying that where two candidates in one cell occur separately in two other cells in intersecting groups, that one of those candidates in the two other cells must be incorrect - I don't see how that can be stated as a rule.

If you literally mean only 12, 13, 23. My line of reasoning says only that either a 2 or 3 must reside in the 23 cel. Which is rather obvious.

This is true, but if you had 12, 13, 2346, you couldn't assume that the third cell MUST be either 2 or 3 and therefore eliminate the other two numbers UNLESS none of the remaining cells in the row/column/box contained candidates 123, in which case these three cells become a hidden triple of 123 - THEN you could reduce it down to 12, 13, 23.

[bennys]

In 13 123 32 you cant eliminate anything

[arsoncupid]

Thank you ronk, shazbot.

This is I suppose a case of wonderful coincidence. It is a particularly tempting idea, because it's shown up true in two complicated puzzles (I've since solved the second problem).

It's easy to illustrate how to drop the 4 in the second puzzle I originally posted.

The starting position:

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.96 ... .1.
.5. 6.. 7..
..1 8.. ...

5.. .94 .68
.6. ... .4.
97. 16. ..5

... ..1 3..
..5 ..2 .7.
.1. ... 59.

After finding the givens:

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23478 9   6  | 234 2345 357 | 48  1  234
2348  5   32 | 6   1    39  | 7   28 2349
2347  234 1  | 8   234  379 | 46  5  23469

5     23  23 | 7   9    4   | 1   6  8
1     6   8  | 23  235  35  | 9   4  7
9     7   4  | 1   6    8   | 2   3  5

246   248 9  | 5   7    1   | 3   28 246
346   348 5  | 9   34   2   | 468 7  1
234   1   7  | 34  8    6   | 5   9  24

Notice these co-conspiring facts:
the 1st column pair of 8's at r1c1, r2c1.
the 3rd 3x3 box pair of 8's at r1c7, r2c8.
the 9th 3x3 box pair of 8's at r7c8, r8c7.
the 2nd column pair of 8's at r7c2, r8c2.

the 1st column pair of 6's at r7c1, r8c1.
the 3rd 3x3 box of 6's at r3c7, r3c9.
the 9th 3x3 box pair of 6's at r7c9, r8c7.

These form a large, connected loop that intersects on r8c7. This is the square my guess-of-a-rule says cannot contain a 4.

Set r7c1=6, and go through the remaining cels, setting them to what they must become. r8c7=6.
The other option, set r8c1=6. Simplify: r8c7=8.

So the 4 can be dropped.

But when I try to show the general case I find that it is logically possible to set the two candidate numbers throughout the entire puzzle wihtout conflict.


I guess I still find it interesting because it is close to another, simple rule. In this case:

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123 1234567 1234567 | 128 129 1236 | 789 68 1236

The cells with 4's and 5's, being the only cells with those numbers, can be reduced:

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123 45 45 | 128 129 1236 | 789 68 1236

Which, in that example, nicely uncovers further.

If it was logically equivalent, to split one of those cels into two cels, like this:

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A      B          C
1234 123567 1234567 | ...

The original 4,5 pair has been split into two cels. One cel will contain the resolution of the original 4,5 cel -- it will contain the number that would have been there.
'C' still loses other numbers, because it remains the same. And, whichever of 'A', 'B' contains the resolved number from the original 4,5 pair, will lose all it's numbers.

I know none of that is meaningful. Just illustrating the point.