The New Sudoku Players' Forum

[RW]

Eh, some confusion about the BUGs I can see...

A Bivalue Universal Grave (BUG) is any grid in which all the unsolved cells have two candidates, and if a candidate exists in a row, column, or box, it shows up exactly twice.

Removing all the +n candidates from a BUG+n should leave a BUG according to the definition above. Wapati, your BUG+3 reduction doesn't. Neither does this BUG+2 you suggested to use after my BUG+3 move:

Code: Select all

 *------------------------------------------*
 | 15   6    9 | 2    3    4 | 7    8    15 |
 | 12   17+4 3 | 5    79   8 | 6    29   14 |
 | 25+4 47   8 | 6    79   1 | 29   3    45 |
 |-------------+-------------+--------------|
 | 3    5    7 | 1    2    6 | 8    4    9  |
 | 6    9    1 | 8    4    7 | 3    5    2  |
 | 8    2    4 | 9    5    3 | 1    6    7  |
 |-------------+-------------+--------------|
 | 7    14   5 | 3    6    9 | 24   12   8  |
 | 14   3    2 | 7    8    5 | 49   19   6  |
 | 9    8    6 | 4    1    2 | 5    7    3  |
 *------------------------------------------*

This is not a BUG+2 grid, remove the +n candidates and you have three 1s and only one 4 in row 2, column 1 and box 1. If the elimination that you falsely assumed to be a BUG+2 elimination leads to a solution, it is plain luck.

Corollary 4 of the BUG principle ... ... allows us to set r2c1=2 which solves. I'm not sure if that ties to your r2c1=2+14 idea or not.

Same misunderstanding here. The grid left after setting r2c1=2 is definitely not a BUG+1 grid.

Some form of colouring would achive the same result.

So it does, and I'm afraid colouring isn't Extreme enough for this thread...

RW

[ravel]

But is r2c1<>4 a valid BUG+3 move then ?
A 4 in r2c1 would not only remove 4 from the other 2 BUG cells, but also the second 2 in b1, c1 and r2.

Code: Select all

 *------------------------------------------*
 | 15   6    9 | 2    3    4 | 7    8    15 |
 | 124  147  3 | 5    79   8 | 6    29   14 |
 | 245  47   8 | 6    79   1 | 29   3    45 |
 |-------------+-------------+--------------|
 | 3    5    7 | 1    2    6 | 8    4    9  |
 | 6    9    1 | 8    4    7 | 3    5    2  |
 | 8    2    4 | 9    5    3 | 1    6    7  |
 |-------------+-------------+--------------|
 | 7    14   5 | 3    6    9 | 24   12   8  |
 | 14   3    2 | 7    8    5 | 49   19   6  |
 | 9    8    6 | 4    1    2 | 5    7    3  |
 *------------------------------------------*


[ab]

I note you have finned swordfish puzzles. How about a couple of finned jellyfish puzzles?

Code: Select all

 7 . 4 | 1 . 2 | . . 8
 . . 6 | . . 9 | . . .
 . . . | . . 4 | . . 3
 ------+-------+------
 . 6 8 | . . . | 4 . 1
 . . . | . 6 . | . . .
 9 . 7 | . . . | 3 2 .
 ------+-------+------
 1 . . | 9 . . | . . .
 . . . | 4 . . | 8 . .
 6 . . | 7 . 8 | 9 . 2

 . 1 . | . 9 . | . . .
 . . . | 4 . 3 | . . 7
 . . 5 | 1 . 8 | 2 . .
 ------+-------+------
 . 2 7 | . . . | 9 3 .
 3 . . | . . . | . . 8
 . 8 9 | . . . | 4 1 .
 ------+-------+------
 . . 6 | 7 . 9 | 8 . .
 8 . . | 3 . 4 | . . .
 . . . | . 2 . | . 7 .


[ronk]

Corollary 4 of the BUG principle ... ... allows us to set r2c1=2 which solves. I'm not sure if that ties to your r2c1=2+14 idea or not.

Same misunderstanding here. The grid left after setting r2c1=2 is definitely not a BUG+1 grid.

Darn! Thought I'd finally spotted one of those Corollary 4 thingies on my own.

But is r2c1<>4 a valid BUG+3 move then ?

From the BUG+3 we know that at least one of r2c1=1, r2c2=4 and r3c1=4 must be true. Each individually true implies r2c1<>4.

[RW]

But is r2c1<>4 a valid BUG+3 move then ?
A 4 in r2c1 would not only remove 4 from the other 2 BUG cells, but also the second 2 in b1, c1 and r2.

The idea in a BUG+n grid is that at least one of the '+n' candidates must be true. So if you can divide all unsolved cells in the grid into A+n where A is a bivalue, and if a candidate exists in A it comes up exactly twice in the sets A of that row, column and box, then at least one of the +n candidates must be true.

In my elimination I had divided the grid into A+n cells, the only way possible for the particular grid. Therefore I know that one of the +n cells must be true, and a 4 in r2c1 would eliminate them all, therefore r2c1<>4. This doesn't imply that setting r2c1=4 would force a BUG grid. It only implies that if the solution has r2c1=4, then it would be possible to eliminate candidate 1 from r2c1 and candidate 4 from r2c2 and r3c1 with some forcing chains or similar, and thus end up with a BUG grid which must have multiple solutions. If it in fact would be possible to make a placement in a grid that would leave a BUG grid, then the puzzle would have multiple solutions.

I underlined the word if earlier, because it is not alway possible to divide the puzzle into a BUG+n grid. After my BUG+3 elimination you cannot divide it into a valid BUG+2 grid. I also noticed that when n gets really big it might be impossible to make the division.

RW

[RW]

I note you have finned swordfish puzzles. How about a couple of finned jellyfish puzzles?

Finned jellyfish are of course welcome, as long as it's not possible to make the same elimination with multiple colors.

Your first puzzle solved with multiple colors and a UR. In the second I can find a finned jellyfish/jellyfish and a finned swordfish/squirmbag, but none of them solves the puzzle alone. Could you show your intended one step solution?

RW

[udosuk]

I note you have finned swordfish puzzles. How about a couple of finned jellyfish puzzles?

Code: Select all

 7 . 4 | 1 . 2 | . . 8
 . . 6 | . . 9 | . . .
 . . . | . . 4 | . . 3
 ------+-------+------
 . 6 8 | . . . | 4 . 1
 . . . | . 6 . | . . .
 9 . 7 | . . . | 3 2 .
 ------+-------+------
 1 . . | 9 . . | . . .
 . . . | 4 . . | 8 . .
 6 . . | 7 . 8 | 9 . 2

 . 1 . | . 9 . | . . .
 . . . | 4 . 3 | . . 7
 . . 5 | 1 . 8 | 2 . .
 ------+-------+------
 . 2 7 | . . . | 9 3 .
 3 . . | . . . | . . 8
 . 8 9 | . . . | 4 1 .
 ------+-------+------
 . . 6 | 7 . 9 | 8 . .
 8 . . | 3 . 4 | . . .
 . . . | . 2 . | . 7 .


I thought we were supposed to spot some finned jellyfish from the following patterns:

Code: Select all

 . 5 . | . 5 . | 5 5 .
 5 . . | 5 . . | . 5 .
 5 5 . | . . . | . 5 .
-------+-------+-------
 5 . . | 5 . 5 | . 5 .
 . 5 . | . . 5 | 5 . .
 . 5 . | . . 5 | . . .
-------+-------+-------
 . . 5 | . 5 5 | . . .
 . . . | . . . | . . .
 . . 5 | . 5 . | . . .



 . . . | 5 . 5 | 5 . 5
 . . . | . 5 . | . 5 .
 . . . | . . . | . . .
-------+-------+-------
 . . . | 5 . . | . . 5
 . 5 . | . . 5 | . 5 .
 5 . . | 5 . . | . . 5
-------+-------+-------
 5 . . | . 5 . | . . .
 . . . | . 5 . | 5 5 5
 5 5 . | . . 5 | 5 . 5



 . . . | 6 . 6 | 6 . 6
 . 6 . | . 6 . | . 6 .
 6 . . | . . . | . 6 6
-------+-------+-------
 . . . | 6 . . | . . 6
 . 6 . | . . 6 | . 6 .
 6 . . | 6 . . | . . 6
-------+-------+-------
 . . . | . . . | . . .
 . . . | . 6 . | 6 6 6
 . . . | . . 6 | 6 . 6

The first puzzle, after SSTS, comes to the following state:

Code: Select all

 *-----------------------------------------------------------*
 | 7    *359   4     | 1     35    2     | 56   *569   8     |
 | 2358  123   6     | 35    78    9     | 12    457   47    |
 | 58   *59    12    | 6     78    4     | 12   *579   3     |
 |-------------------+-------------------+-------------------|
 | 235   6     8     | 235   9     357   | 4     57    1     |
 | 4     1235  123   | 23    6     1357  | 57    8     9     |
 | 9     15    7     | 8     4     15    | 3     2     6     |
 |-------------------+-------------------+-------------------|
 | 1     8     235   | 9     235   356   | 67    3467  47    |
 | 23    7     9     | 4     123   36    | 8     136   5     |
 | 6     4     35    | 7     135   8     | 9     13    2     |
 *-----------------------------------------------------------*

Here, the 9 is strongly linked on r13c28.
If r1c8=5, a deadly pattern of {59} would be formed in r13c28.
Therefore r1c8<>5 and a swordfish of 5 would follow in r234c148.

I think this is the UR RW talked about...

And I guess here is RW's finned swordfish/squirmbag for the 2nd puzzle:

Code: Select all

 .  .  . | 6  . @6 |@6  . -6
 . @6  . | . @6  . | . @6  .
#6  .  . | .  .  . | . *6 #6
---------+---------+---------
 .  .  . |#6  .  . | .  . #6
 . @6  . | .  . @6 | . @6  .
#6  .  . |#6  .  . | .  . #6
---------+---------+---------
 .  .  . | .  .  . | .  .  .
 .  .  . | . @6  . |@6 @6  6
 .  .  . | .  . @6 |@6  .  6

#+*: (row) finned swordfish r346c149 + r3c8 => r1c9<>6
@+*: (col) finned squirmbag r12589c25678 + r3c8 => r1c9<>6

... And the finned jellyfish for the 2nd puzzle:

Code: Select all

 .  .  . | 5  . #5 |#5  . -5
 .  .  . | . @5  . | . *5  .
 .  .  . | .  .  . | .  .  .
---------+---------+---------
 .  .  . |@5  .  . | .  . @5
 . #5  . | .  . #5 | . #5  .
@5  .  . |@5  .  . | .  . @5
---------+---------+---------
@5  .  . | . @5  . | .  .  .
 .  .  . | .  5  . |#5 #5  5
 5 #5  . | .  . #5 |#5  .  5

#+*: (col) finned jellyfish r1589c2678 + r2c8 => r1c9<>5
@+*: (row) finned jellyfish r2467c1459 + r2c8 => r1c9<>5
(Alternatively, r2c8=5  => r1c9<>5
                r2c8<>5 => r2c5=5 => r7c1=5 => x-wing in r46c49 => r1c9<>5)

Combined, they force r1c9=4 and the puzzle is solved...

[ab]

I note you have finned swordfish puzzles. How about a couple of finned jellyfish puzzles?

Finned jellyfish are of course welcome, as long as it's not possible to make the same elimination with multiple colors.

Your first puzzle solved with multiple colors and a UR. In the second I can find a finned jellyfish/jellyfish and a finned swordfish/squirmbag, but none of them solves the puzzle alone. Could you show your intended one step solution?

RW

I think i misunderstood the rules. I'll go back to the drawing board. I was just pleased to show of the finned fish my generator can gullet!

[udosuk]

ab, I give up... This toughie finned jellyfish has eluded me...

Please tell me how you could spot one from this pattern:

Code: Select all

 .  5  . | .  5  . | 5  5  .
 5  .  . | 5  .  . | .  5  .
 5  5  . | .  .  . | .  5  .
---------+---------+---------
 5  .  . | 5  .  5 | .  5  .
 .  5  . | .  .  5 | 5  .  .
 .  5  . | .  .  5 | .  .  .
---------+---------+---------
 .  .  5 | .  5  5 | .  .  .
 .  .  . | .  .  . | .  .  .
 .  .  5 | .  5  . | .  .  .


[ravel]

From the BUG+3 we know that at least one of r2c1=1, r2c2=4 and r3c1=4 must be true. Each individually true implies r2c1<>4.

The idea in a BUG+n grid is that at least one of the '+n' candidates must be true. So if you can divide all unsolved cells in the grid into A+n where A is a bivalue, and if a candidate exists in A it comes up exactly twice in the sets A of that row, column and box, then at least one of the +n candidates must be true.

Thats clear, when you have the division, but you didnt explain, how you find (and validate) it. Is there a simpler way like this?

Code: Select all

*------------------------------------------*
 | 15   6    9 | 2    3    4 | 7    8    15 |
 | 124  147  3 | 5    79   8 | 6    29   14 |
 | 245  47   8 | 6    79   1 | 29   3    45 |
 |-------------+-------------+--------------|
 | 3    5    7 | 1    2    6 | 8    4    9  |
 | 6    9    1 | 8    4    7 | 3    5    2  |
 | 8    2    4 | 9    5    3 | 1    6    7  |
 |-------------+-------------+--------------|
 | 7    14   5 | 3    6    9 | 24   12   8  |
 | 14   3    2 | 7    8    5 | 49   19   6  |
 | 9    8    6 | 4    1    2 | 5    7    3  |
 *------------------------------------------*


- For a BUG+1 you need a candidate, which you have two more times in the same row, column, box. So only 4 remains for r2c2 and r3c1 and 14 for r2c1.
- With 4 in r2c1 the other 2 cells become bivalue cells, and only one 2 is left for box/column 1 - no BUG.
- r2c1<>1 and r2c2<>4 leave a BUG+1 with 4 in r3c1
- r2c1<>1 and r3c1<>4 leave a BUG+1 with 4 in r2c2
- r2c2<>4 and r3c1<>4 leave a BUG+1 with 1 in r2c1

[Ocean]

A 19-clues puzzle (double diagonal reflectional symmetry) that can be solved with a Nice Loop:

Code: Select all

 *-----------------*
 |. . . . . 1 . 2 .|
 |. 1 . . . . . . 3|
 |. . 4 . 5 . . . .|
 |. . . 6 . . . . 7|
 |. . 8 . 4 . 9 . .|
 |5 . . . . 3 . . .|
 |. . . . 9 . 4 . .|
 |3 . . . . . . 5 .|
 |. 7 . 4 . . . . .|
 *-----------------*
#
 *-----------------------------------------------------------*
 | 9     5     3     |#78    67    1     | 68    2     4     |
 |#678   1     67    |#28    26    4     | 5     9     3     |
 | 268   28    4     | 3     5     9     | 168   7     168   |
 |-------------------+-------------------+-------------------|
 | 4     9     2     | 6     18    5     | 3     18    7     |
 |#17    3     8     | 1-7   4     2     | 9     6     5     |
 | 5     6     17    | 9     178   3     | 128   4     128   |
 |-------------------+-------------------+-------------------|
 | 1268  28    16    | 5     9     7     | 4     3     1268  |
 | 3     4     9     | 12    12    68    | 7     5     68    |
 | 1268  7     5     | 4     3     68    | 1268  18    9     |
 *-----------------------------------------------------------*
#
-7-[r1c4]-8-[r2c4]=8=[r2c1]=7=[r5c1] -> r5c4<>7: Either r1c4 is 7, or r5c1 is 7, therefore '7' can be eliminated from those cells that see both of those.

#
Edit: Modified the puzzle, so it's no longer solvable with UR.
#

[ronk]

Thats clear, when you have the division, but you didnt explain, how you find (and validate) it. Is there a simpler way like this?

Code: Select all

 *------------------------------------------*
 | 15   6    9 | 2    3    4 | 7    8    15 |
 | 124  147  3 | 5    79   8 | 6    29   14 |
 | 245  47   8 | 6    79   1 | 29   3    45 |
 |-------------+-------------+--------------|
 | 3    5    7 | 1    2    6 | 8    4    9  |
 | 6    9    1 | 8    4    7 | 3    5    2  |
 | 8    2    4 | 9    5    3 | 1    6    7  |
 |-------------+-------------+--------------|
 | 7    14   5 | 3    6    9 | 24   12   8  |
 | 14   3    2 | 7    8    5 | 49   19   6  |
 | 9    8    6 | 4    1    2 | 5    7    3  |
 *------------------------------------------*


- For a BUG+1 you need a candidate, which you have two more times in the same row, column, box. So only 4 remains for r2c2 and r3c1 and 14 for r2c1.

Close. Do the Bug+Nonbug candidate split first for all rows, columns, and boxes that have only one cell with more than two candidates. For this puzzle that's r2 and c2 and the B+N split gets us to ...

Code: Select all

 *------------------------------------------*
 | 15   6    9 | 2    3    4 | 7    8    15 |
 | 124  17+4 3 | 5    79   8 | 6    29   14 |
 | 25+4 47   8 | 6    79   1 | 29   3    45 |
 |-------------+-------------+--------------|
 | 3    5    7 | 1    2    6 | 8    4    9  |
 | 6    9    1 | 8    4    7 | 3    5    2  |
 | 8    2    4 | 9    5    3 | 1    6    7  |
 |-------------+-------------+--------------|
 | 7    14   5 | 3    6    9 | 24   12   8  |
 | 14   3    2 | 7    8    5 | 49   19   6  |
 | 9    8    6 | 4    1    2 | 5    7    3  |
 *------------------------------------------*

Then repeat the step pretending the already-split-cells are the bivalues previously determined. For this puzzle, we are left with only r2c1 with more than two candidates. Now ignoring the 4s in r2c2 and r3c1, we note there are three 1s in r2, c1, and b1. Therefore, the correct B+N split for r2c1 is "24+1".

How to validate? Mentally discard all the Nonbug candidates and examine the grid ...

Code: Select all

 *------------------------------------------*
 | 15   6    9 | 2    3    4 | 7    8    15 |
 | 24   17   3 | 5    79   8 | 6    29   14 |
 | 25   47   8 | 6    79   1 | 29   3    45 |
 |-------------+-------------+--------------|
 | 3    5    7 | 1    2    6 | 8    4    9  |
 | 6    9    1 | 8    4    7 | 3    5    2  |
 | 8    2    4 | 9    5    3 | 1    6    7  |
 |-------------+-------------+--------------|
 | 7    14   5 | 3    6    9 | 24   12   8  |
 | 14   3    2 | 7    8    5 | 49   19   6  |
 | 9    8    6 | 4    1    2 | 5    7    3  |
 *------------------------------------------*

... to check that each candidate appears exactly twice in each row, column and box. That's the Bivalue Universal Graveyard.

[Ocean]

This might qualify as an "Effortless Extreme". Can be solved with a single "Nice Loop".

Code: Select all

 *-----------*
 |..6|...|5..|
 |.7.|.3.|.9.|
 |4..|...|..6|
 |---+---+---|
 |1..|...|..2|
 |.2.|...|.3.|
 |..4|.5.|6..|
 |---+---+---|
 |...|1.7|...|
 |.8.|.9.|.7.|
 |...|4.5|...|
 *-----------*

 *-----------------------------------------------------------------------*
 | 2389    139    6      | 2789    12478  12489  | 5       1248   13478  |
 | 258     7      1258   | 2568    3      12468  | 1248    9      148    |
 | 4       1359   123589 | 25789   1278   1289   | 12378   128    6      |
 |-----------------------+-----------------------+-----------------------|
 | 1       3569   35789  | 36789   4678   34689  | 4789    458    2      |
 | 56789   2      5789   | 6789    14678  14689  | 14789   3      145789 |
 | 3789    39     4      | 23789   5      12389  | 6       18     1789   |
 |-----------------------+-----------------------+-----------------------|
 | 23569   34569  2359   | 1       268    7      | 23489   24568  34589  |
 | 2356    8      1235   | 236     9      236    | 1234    7      1345   |
 | 23679   1369   12379  | 4       268    5      | 12389   1268   1389   |
 *-----------------------------------------------------------------------*

 After basic reductions:
 *-----------------------------------------------------------------------*
 | 2389    139    6      | 2789    1247   12489  | 5       1248   37     |
 | 258     7      1258   | 2568    3      12468  | 1248    9      148    |
 | 4       1359   123589 | 25789   127    1289   | 37      128    6      |
 |-----------------------+-----------------------+-----------------------|
 | 1      #3569   35789  | 36789   467    34689  | 4789   #458    2      |
 | 56789   2      5789   | 6789    1467   14689  | 14789   3      145789 |
 | 3789    39     4      | 23789   5      12389  | 6       18     1789   |
 |-----------------------+-----------------------+-----------------------|
 | 23569   4      2359   | 1       268    7      | 2389   #2568   3589   |
 | 256     8      125    | 236     9      236    | 124     7      145    |
 | 23679  #1369   12379  | 4       268    5      | 12389  #1268   1389   |
 *-----------------------------------------------------------------------*

[r4c2]=6=[r9c2]-6-[r9c8]=6=[r7c8]=5=[r4c8]-5-[r4c2]
=> r4c2<>5, which solves the puzzle.

Edit: Candidate grid corrected - original grid substituted with candidates after basic reductions. Especially, the placement r7c2=4 (hidden single) must be done before the nice loop can be applied. Thanks to ronk for pointing this out!

[RW]

Nice puzzles Ocean! Both your latest nice loops are added to the list.

RW

[Ocean]

Locked Candidates plus one Bidirectional Cycle (5-ring) is enough to solve this puzzle:

Code: Select all

001002000030040050500600700008000006010000090600000800006007004090010030000500200
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 *--------------------------------------------------------------------*
 |#479    6      1      | 3789   5      2      |-349   #48     389    |
 | 279    3      279    | 1789   4      189    | 6      5      1289   |
 | 5      8      249    | 6      39     139    | 7      124    1239   |
 |----------------------+----------------------+----------------------|
 | 39     24     8      | 12349  2379   5      | 134    1247   6      |
 | 37     1      357    | 2348   6      348    | 34     9      235    |
 | 6      24     359    | 12349  2379   1349   | 8      1247   1235   |
 |----------------------+----------------------+----------------------|
 | 123    5      6      | 239    2389   7      |-19    #18     4      |
 | 8      9      24     | 24     1      6      | 5      3      7      |
 |#134    7      34     | 5      389    349    | 2      6     #189    |
 *--------------------------------------------------------------------*
#
# [r1c1]=4=[r9c1]=1=[r9c9]-1-[r7c8]=8=[r1c8]-4-[r1c1] => r1c7<>4, r7c7<>1, r9c1<>3.
#

(Came across the puzzle while searching for hard ones.)
Edit: Overlooked the elimination r9c1<>3 (which is not necessary for solving the puzzle though).