The New Sudoku Players' Forum

[RW]

BUG+1 seems to be a widely used technique by now, BUG+2 and BUG+3 are also quite common, but why is it that most people seem to stop there? I noticed that bigger Bugs often can provide quite simple solutions to tougher puzzles when you have it reduced to mostly bivalue cells. In these cases there usually is some XY-chain also that can solve the puzzle, but I think it's easier to find the BUG solution. Mainly because when looking for the BUG solution I know exactly where to look for the move, around the non-bivalued cells. A XY-chain could start and end anywhere in the grid, and I think it takes a lot more time to find the links in such a chain, than to separate the bug candidates from the +candidates and find an elimination based on that.

Here's a BUG+7 example:

Code: Select all

.6..9..2.2..6.3..1..5...6...9.831.4.1..4.2..9.8.569.3...6...9..8..9.6..3.4..5..1.
 *--------------------------------------------------*
 | 4    6    18   | 17   9    5    | 3    2    78   |
 | 2    7    89   | 6    48   3    | 45   59+8 1    |
 | 39   13   5    | 27+1 18+2 47   | 6    89   47+8 |
 |----------------+----------------+----------------|
 | 6    9    2    | 8    3    1    | 7    4    5    |
 | 1    5    3    | 4    7    2    | 8    6    9    |
 | 7    8    4    | 5    6    9    | 1    3    2    |
 |----------------+----------------+----------------|
 | 5    23+1 6    | 13+2 12   47   | 9    78   48   |
 | 8    12   17   | 9    24+1 6    | 45   57   3    |
 | 39   4    79   | 37   5    8    | 2    1    6    |
 *--------------------------------------------------*

Here it's easy to see that a 1 in r7c4 would immediately kill four of the extra candidates (r7c4<>2, r3c4 & r7c2 & r8c5<>1). Without too much effort it's also easy to spot that

Code: Select all

if r7c4=1 => r7c5=2 (r3c5<>2)
          => r1c4=7 => r1c9=8 (r2c8 & r3c9<>8)

That killed all the remaining extra candidates so we know that r7c4<>1. That advances the puzzle to a BUG+5:

Code: Select all

 *--------------------------------------------------*
 | 4    6    18   | 17   9    5    | 3    2    78   |
 | 2    7    89   | 6    48   3    | 45   59+8 1    |
 | 39   13   5    | 27+1 28   47   | 6    89   47+8 |
 |----------------+----------------+----------------|
 | 6    9    2    | 8    3    1    | 7    4    5    |
 | 1    5    3    | 4    7    2    | 8    6    9    |
 | 7    8    4    | 5    6    9    | 1    3    2    |
 |----------------+----------------+----------------|
 | 5    23+1 6    | 23   12   47   | 9    78   48   |
 | 8    12   17   | 9    24+1 6    | 45   57   3    |
 | 39   4    79   | 37   5    8    | 2    1    6    |
 *--------------------------------------------------*

I can find several different eliminations here, but I think the easiest is:

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if r3c8=8 (r2c8 & r3c9<>8)
          => r3c5=2 => r7c5=1 (r7c2 & r8c5<>1)
          => r1c9=7 => r1c4=1 (r3c4<>1

That would also kill all the exra candidates => r3c8<>8

Anybody else also using these? I wonder what's the biggest n for BUG+n that could give a useful (readable) solution.

RW

[Carcul]

Before your BUG+7:

Code: Select all

 *---------------------------------------------*
 | 4    6    18 | 17   9    5  | 3    2    78  |
 | 2    7    89 | 6    48   3  | 45   589  1   |
 | 39   13   5  | 127  128  47 | 6    789  478 |
 |--------------+--------------+---------------|
 | 6    9    2  | 8    3    1  | 7    4    5   |
 | 1    5    3  | 4    7    2  | 8    6    9   |
 | 7    8    4  | 5    6    9  | 1    3    2   |
 |--------------+--------------+---------------|
 | 5    123  6  | 123  12   47 | 9    78   478 |
 | 8    12   17 | 9    124  6  | 45   57   3   |
 | 39   4    79 | 37   5    8  | 2    1    6   |
 *---------------------------------------------*


Note the AUR in cells r78c25:

[r7c2]=3|4=[r8c5]=2=[r8c2](-2-[r7c2])=1=[r8c3]-1-[r7c2],

and so r7c2=3 which solves the puzzle.
In my opinion, this is much more easier and direct than using colors/BUG+7/BUG+5.

Carcul

[re'born]

Before your BUG+7:

Code: Select all

 *---------------------------------------------*
 | 4    6    18 | 17   9    5  | 3    2    78  |
 | 2    7    89 | 6    48   3  | 45   589  1   |
 | 39   13   5  | 127  128  47 | 6    789  478 |
 |--------------+--------------+---------------|
 | 6    9    2  | 8    3    1  | 7    4    5   |
 | 1    5    3  | 4    7    2  | 8    6    9   |
 | 7    8    4  | 5    6    9  | 1    3    2   |
 |--------------+--------------+---------------|
 | 5    123  6  | 123  12   47 | 9    78   478 |
 | 8    12   17 | 9    124  6  | 45   57   3   |
 | 39   4    79 | 37   5    8  | 2    1    6   |
 *---------------------------------------------*


Note the AUR in cells r78c25:

[r7c2]=3|4=[r8c5]=2=[r8c2](-2-[r7c2])=1=[r8c3]-1-[r7c2],

and so r7c2=3 which solves the puzzle.

Here is a different way to see the same exclusion:

(9,1)3 ( > (7,2)!3) > (9,4)7 > (7,6)4 > (8,5)!4
This creates a deadly pattern in ([78],[25])<12>, a contradiction. Therefore, (9,1)9 (<=> (7,2)3).

In my opinion, this is much more easier and direct than using colors/BUG+7/BUG+5.

Carcul

In a BUG+7 grid, it seems that one should expect to find AUR's often and then it becomes a matter of taste. To me, the AUR's are easier to spot and make exclusions with, but I like RW's approach and would love to see an example that didn't have an AUR alternative.

[ronk]

Code: Select all

 4    6   *18 |*17   9    5  | 3    2    78 
 2    7   *89 | 6   *48   3  | 45   589  1   
 39   13   5  | 127  128 *47 | 6    789  478

r2c3-8-r1c3-1-r1c4-7-r3c6-4-r1c5-8-r2c3

... implies r2c3<>8 which solves the puzzle
... without multiple inferences, URs and overlaps

It's a small ALS xz-rule deduction too. But big BUGs have an appeal, so I'll be looking for them.

[RW]

Note the AUR in cells r78c25:

[r7c2]=3|4=[r8c5]=2=[r8c2](-2-[r7c2])=1=[r8c3]-1-[r7c2],

and so r7c2=3 which solves the puzzle.
In my opinion, this is much more easier and direct than using colors/BUG+7/BUG+5.

I agree that in this case it's probably easier to spot. But as always in manual solving, there is no definite technique hierarchy, I usually look for interesting connections in the grid and when I find something I follow it up. In this case I saw the possibility of a big BUG so I went with that and never looked for URs, becaue I don't get to do big BUG eliminations as often as UR.

In a BUG+7 grid, it seems that one should expect to find AUR's often and then it becomes a matter of taste.

That's also probably true, a BUG+1 cannot have a smaller uniqueness pattern, but any bigger BUG can. Should be some big BUGs out there without AURs or BUG-lites though.

r2c3-8-r1c3-1-r1c4-7-r3c6-4-r1c5-8-r2c3

... implies r2c3<>8 which solves the puzzle
... without multiple inferences, URs and overlaps

As I said, with mostly bivalue cells there's usually some XY-chain around also. I find it unlikely that any puzzle unsolvable by other widely used techniques would be solvable with a big BUG, but they're fun to play around with.

RW

[Finlip]

Sorry. I'm off topic.

RW, I'm surprised that you are using pencilmarks to solve sudokus. I thought you solved even the toughest ones without using pencilmarks.

[RW]

RW, I'm surprised that you are using pencilmarks to solve sudokus. I thought you solved even the toughest ones without using pencilmarks.

Well, before I joined this forum I had never used any pencilmarks, and to this day I have still never written down a pencilmark on a paper sudoku I've been solving. However, I've had to check loads of puzzles for the Effortless thread for techniques that do require pencilmarks, so I've done it in SS with pencilmarks on. As I got used to the SS pm-grid I also sometimes just generate tough puzzles and skip them to the "no hint available"-stage in SS and look for possible new cool techniques with pms on, that's how I found this example. However, I see this more as looking for new techniques than solving. When I solve I still do it on paper without pencilmarks (and then I would of course never use a BUG+7). I also skipped the pencil lately and changed to ink pen as this forces me to consider every move more carefully when there's no "undo" available.

RW

[claudiarabia]

Isn't this a nice big BUG?

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*--------------------------------------------------*
 | 79   47   5    | 1    3    29   | 8    24   6    |
 | 39   8    6    | 25   4    259  | 1    7    23   |
 | 14   2    13   | 8    6    7    | 35   9    45   |
 |----------------+----------------+----------------|
 | 24   9    8    | 3    7    1    | 25   6    45   |
 | 12   34   13   | 9    5    6    | 7    24   8    |
 | 6    5    7    | 4    2    8    | 9    3    1    |
 |----------------+----------------+----------------|
 | 8    37   4    | 27   1    23   | 6    5    9    |
 | 35   1    2    | 6    9    35   | 4    8    7    |
 | 57   6    9    | 57   8    4    | 23   1    23   |
 *--------------------------------------------------*


Claudia