The New Sudoku Players' Forum

[fermat]

I have two questions on this puzzle, which can be solved with xys, I know.

Code: Select all

.------------.------------.------------.
| 6   1   3  | 4   57  9  | 8   57  2  |
| 8   7   4  | 36  2   56 | 1   9   35 |
| 2   5   9  | 13  17  8  | 34  6   47 |
:------------+------------+------------:
| 37  24  1  | 5   9   24 | 6   37  8  |
| 39  6   5  | 8   14  7  | 2   13  49 |
| 79  24  8  | 16  3   26 | 45  157 479|
:------------+------------+------------:
| 1   3   7  | 2   45  45 | 9   8   6  |
| 5   8   2  | 9   6   3  | 7   4   1  |
| 4   9   6  | 7   8   1  | 35  2   35 |
'------------'------------'------------'


Question 1 is: can I assume, using uniqueness, that r6c6 is 6?

Question 2: using Bug can I see that r6c8=7 ?

[udosuk]

For question 1:

Yes, there is an UR type 1 in r46c26 which prohibit r6c6 from being 2.


I'm still learning BUG and would love if somebody could demonstrate it for question 2...

[ronk]

I have two questions on this puzzle, which can be solved with xys, I know.

Code: Select all

.------------.------------.--------------.
| 6   1   3  | 4   57  9  | 8   57   2   |
| 8   7   4  | 36  2   56 | 1   9    35  |
| 2   5   9  | 13  17  8  | 34  6    47  |
:------------+------------+--------------:
| 37  24  1  | 5   9   24 | 6   37   8   |
| 39  6   5  | 8   14  7  | 2   13   49  |
| 79  24  8  | 16  3   26 | 45  15+7 79+4|
:------------+------------+--------------:
| 1   3   7  | 2   45  45 | 9   8    6   |
| 5   8   2  | 9   6   3  | 7   4    1   |
| 4   9   6  | 7   8   1  | 35  2    35  |
'------------'------------'--------------'


Question 1 is: can I assume, using uniqueness, that r6c6 is 6?

Question 2: using Bug can I see that r6c8=7 ?

Answer 1: Yes, nice observation. There is no placement of a 4 in the row, column, and box of r6c6. Therefore, a candidate 4 may be "freely invented" there ... resulting in UR(24) Type 1 in r46c26 for r6c6=6.

Answer 2: I've added the "+"s to your BUG+2 grid. From these we know that at least one of r6c8=7 and r6c9=4 must be true ... but we don't know which. So r6c8=7 is not directly a valid deduction that I can see.

However, you can make deductions common to both. For example ...

r6c9=4 => r6c9<>9 and
r6c8=7 => r6c1=9 => r6c9<>9

... together implying that r6c9<>9.

[RW]

I don't know how you got rid of the '4' in r6c6, but even if it was there there's a simple BUG+3 solution.

Code: Select all

 *--------------------------------------------------*
 | 6    1    3    | 4    57   9    | 8    57   2    |
 | 8    7    4    | 36   2    56   | 1    9    35   |
 | 2    5    9    | 13   17   8    | 34   6    47   |
 |----------------+----------------+----------------|
 | 37   24   1    | 5    9    24   | 6    37   8    |
 | 39   6    5    | 8    14   7    | 2    13   49   |
 | 79   24   8    | 16   3    26+4 | 45   15+7 79+4 |
 |----------------+----------------+----------------|
 | 1    3    7    | 2    45   45   | 9    8    6    |
 | 5    8    2    | 9    6    3    | 7    4    1    |
 | 4    9    6    | 7    8    1    | 35   2    35   |
 *--------------------------------------------------*

Extra candidates:
r6c6=4, r6c8=7 and r6c9=4, one of these must be true! And here's the key to the solution:

r6c8=5 (r6c8<>7) => r6c7=4 (r6c69<>4)
=> r6c8<>5.

[Edit: I forgot to answer your question. Question 2: Yes, if you make these eliminations:

r6c8=5 (r6c8<>7) => r6c7=4 (r6c9<>4) =>r6c8<>5
r6c8=1 (r6c8<>7) => r6c4=6 => r6c6=2 =>r6c2=4 (r6c9<>4) => r6c8<>1

But your not likely to use that solution as either one of the mentioned chains alone solves the puzzle.]

RW

[fermat]

I don't know how you got rid of the '4' in r6c6, but even if it was there there's a simple BUG+3 solution.

I was at this point (roughly) and used uniquness to eliminate the 4.

Code: Select all

.---------------.---------------.---------------.
| 6    1    3   | 4    57   9   | 8    57   2   |
| 8    7    4   | 1356 2    56  | 135  9    35  |
| 2    5    9   | 13   17   8   | 134  6    347 |
:---------------+---------------+---------------:
| 37   24   1   | 25*  9    245*| 6    357  8   |
| 39   6    5   | 8    14   7   | 2    13   349 |
| 79   24   8   | 1256 3    2456| 145  157  4579|
:---------------+---------------+---------------:
| 1    3    7   | 25*  45   245*| 9    8    6   |
| 5    8    2   | 9    6    3   | 7    4    1   |
| 4    9    6   | 7    8    1   | 35   2    35  |
'---------------'---------------'---------------'

Code: Select all

 *--------------------------------------------------*
 | 6    1    3    | 4    57   9    | 8    57   2    |
 | 8    7    4    | 36   2    56   | 1    9    35   |
 | 2    5    9    | 13   17   8    | 34   6    47   |
 |----------------+----------------+----------------|
 | 37   24   1    | 5    9    24   | 6    37   8    |
 | 39   6    5    | 8    14   7    | 2    13   49   |
 | 79   24   8    | 16   3    26+4 | 45   15+7 79+4 |
 |----------------+----------------+----------------|
 | 1    3    7    | 2    45   45   | 9    8    6    |
 | 5    8    2    | 9    6    3    | 7    4    1    |
 | 4    9    6    | 7    8    1    | 35   2    35   |
 *--------------------------------------------------*

I noticed that both bug candidates are correct (r6c8, r6c9)when the four (r6c6) was eliminated and that made me hope there was an easy rule involved. Is it significant that the two that work are in the same box, or just serendipity? (luck)

With the BUG +3 the 4 in r6c6 is the "+" that is not true, thus damping my hope for something easy.

[udosuk]

Thanks for the BUG demonstration guys...

Answer 1: Yes, nice observation. There is no placement of a 4 in the row, column, and box of r6c6. Therefore, a candidate 4 may be "freely invented" there ... resulting in UR(24) Type 1 in r46c26 for r6c6=6.

I don't see any need to "freely invent" a 4 in r6c6... Since r4c26=r6c2={24}, it should automatically follows that r6c6 cannot have either 2 or 4... It could have been {246}. {26}, {46} or {123456789}... 2 and 4 simply must be eliminated...

And I don't see any need to check that if there is any placement of 4 in the row, column, and box of r6c6... For if there is such a placement then either r4c6 or r6c2 could not have been {24} and must have been determined as 2 instead...

[ronk]

I don't see any need to "freely invent" a 4 in r6c6...

In general, I agree with your comment ... but the classic definition of a UR has both UR digits in all four corners of the rectangle.

And I don't see any need to check that if there is any placement of 4 in the row, column, and box of r6c6... For if there is such a placement then either r4c6 or r6c2 could not have been {24} and must have been determined as 2 instead...

You may be right. Rules on "freely inventing" candidates don't exist AFAIK, so I tend to be overly cautious about doing so. Also everyone doesn't adjust pencilmarks correctly after making a placement. I certainly miss removing candidates from time to time.

[fermat]

Also everyone doesn't adjust pencilmarks correctly after making a placement. I certainly miss removing candidates from time to time.

LOL, I neglect to include some from time to time, those hurt worse!