The New Sudoku Players' Forum

[Mathimagics]

My solver reports a backdoor value 1r2c3.

Confirmed, give that man a cigar!

[SpAce]

My solver reports a backdoor value 1r2c3.

Confirmed, give that man a cigar!

Hodoku actually reports quite a few more: 1r2c3, 1r7c2, 1r8c4, 6r2c8, 6r3c1 for singles, and 7r2c2 and 7r6c3 for basics. However, I don't think they're very easy to hit accidentally (like when coloring) or even by deliberate guessing (or so I assume, never really tried that with any puzzle). Such well-hidden backdoors don't really ruin a puzzle for a manual solver.

[Mathimagics]

Such well-hidden backdoors don't really ruin a puzzle for a manual solver.

A curious observation! In what way does a backdoor "ruin" a puzzle for P&P solvers?

[m_b_metcalf]

Hodoku actually reports quite a few more: 1r2c3, 1r7c2, 1r8c4, 6r2c8, 6r3c1 for singles

I confirm that. My program doesn't normally search beyond the first one found.

[SpAce]

Such well-hidden backdoors don't really ruin a puzzle for a manual solver.

A curious observation! In what way does a backdoor "ruin" a puzzle for P&P solvers?

We had a discussion about that earlier in this thread, starting here. Of course not everyone may find it a problem, but I think it's disappointing if a solution is found in minutes for a supposedly hardish puzzle. After all, the fun is in finding the logical solving steps, not the solution. Stumbling on a backdoor kind of ruined my very first 9.0 experience too. Such a solution is also impossible to document because it's indistinguishable from a lucky guess (even if found via a logical solving technique like coloring).

[Mathimagics]

Fair enough. Personally I like it when my educated guess for a cell hits the jackpot! Cigar time!

[SpAce]

Fair enough. Personally I like it when my educated guess for a cell hits the jackpot! Cigar time!

I should try that some time! But what exactly is an educated guess? Would you have guessed any of the backdoors in this puzzle? I need education on this topic!

Seems to me that a solving strategy based on guessing could be a very inefficient process, unless you can work out the testing in your head. Otherwise you may end up testing lots of candidates that look promising if true but are ineffective as eliminations (which is the more likely result). And what if the puzzle happens to be a beast without any single-cell backdoors?

For those reasons I would imagine that the normal logical solving is probably more efficient in general. It can still find backdoors as well, but in the mean time it can find more useful eliminations that advance the puzzle (which makes guessing easier too, if that's desired). Of course things are different in a simple puzzle where almost every reasonable guess has a 50% chance of succeeding, but then it's just as simple to solve normally anyway.

[Mathimagics]

Seems to me that a solving strategy based on guessing could be a very inefficient process

Ok, let me say firstly that my P&P solving strategy is not exactly one that is based on guessing, but uses guessing as a last resort, when I can't make any further progress through deductive logic. I don't use any "AST's" per se, but do use methods such as exploitation of the uniqueness property, as these are, to me at least, instinctive.

But what exactly is an educated guess? Would you have guessed any of the backdoors in this puzzle?

Let me answer the second question first - no! I have formed the habit of only considering bi-values and bi-locals as guess targets, and all of the identified backdoor cells still had 3 or 4 possible values when I got stuck.

I just took Mike's suggestion (1r2c3) and applied it, and hey presto ...

But I have learned something from all this - my idea of an educated guess is to consider the various cells and look for guesses that have at least some obvious initial chain of implications that might make it worth pursuing. Bi-values and bi-locals are the low-hanging fruit here, and I have not had to consider higher branches so far.

I want the guess to turn out to be a backdoor (for the current state) or lead to a contradiction. In the magazine puzzles I favour, I usually guess correctly the first time (so my eraser doesn't get deployed), or else I get a contradiction that itself leads to a solution.

IDBA's puzzle above is actually the very first one that I have had genuine trouble with - I was stuck without the backdoor hints - and what I've learned from that is, just because a cell has more than 2 pencilmarks, that does not mean that it can't be considered as a guess candidate - it's the value of the guess that matters (ie: does it have an obvious chain of consequences?).

[SpAce]

Hi Mathimagics,

Sorry about the late reply.

Ok, let me say firstly that my P&P solving strategy is not exactly one that is based on guessing, but uses guessing as a last resort, when I can't make any further progress through deductive logic.

Fair enough. I didn't think it was all you did anyway

I don't use any "AST's" per se, but do use methods such as exploitation of the uniqueness property, as these are, to me at least, instinctive.

What's "AST"?

But what exactly is an educated guess? Would you have guessed any of the backdoors in this puzzle?

Let me answer the second question first - no! I have formed the habit of only considering bi-values and bi-locals as guess targets, and all of the identified backdoor cells still had 3 or 4 possible values when I got stuck.

Indeed. That's why I said they didn't ruin the puzzle Bivalues and bilocations form the backbone of chaining and netting, so these weren't likely to be used as major chain components revealing their special nature. Thus no accidental stumbling on them via binary coloring either. One would need to specifically try them as placements to see that they lead to the solution, and I don't really see why anyone would do that here. In much more difficult puzzles that might be more likely, when binary coloring methods get harder to use but basic contradiction networks are still possible. I would really hate to solve that way, though.

But I have learned something from all this - my idea of an educated guess is to consider the various cells and look for guesses that have at least some obvious initial chain of implications that might make it worth pursuing. Bi-values and bi-locals are the low-hanging fruit here, and I have not had to consider higher branches so far.

I don't really think you should either. At least learn some better solving methods first! All I used to solve this was bivalues and bilocals as starting points for binary chains/nets. What makes that method powerful is that you can simultaneously track the results of both options at the same time, which finds traps and loops (with multiple eliminations), contradictions (proving the other half of the the binary option true), and also backdoors (which is normally an unwanted result). If you test a single candidate placement, all you can find is either a contradiction or a backdoor or a dead end, which is both inefficient and tedious.

I want the guess to turn out to be a backdoor (for the current state) or lead to a contradiction. In the magazine puzzles I favour, I usually guess correctly the first time (so my eraser doesn't get deployed), or else I get a contradiction that itself leads to a solution.

Are you sure of that ratio? How can you guess correctly more than 50% of the time (implied by "usually") even if you're fairly sure that a binary option contains a backdoor? If you can, it's no longer a pure guess, because you must have seen something that makes one or the other option more likely to succeed or to produce a contradiction.

Like I said before, in magazine puzzles and even in most of Dan's one-trick-ponies close to 50% is quite achievable, because it's usually easy to see which binary options are the most likely to contain backdoors. That's why I almost always find a one-step-solution for Dan's puzzles on the first go without knowing the backdoors beforehand (there are some rare exceptions when the backdoors are few and less obvious). However, that involves no guessing on which way the two options are false and true. That part is revealed through actual solving, i.e. looking for a pattern or a chain that tells the answer without doubt. If need be, I use coloring to find it.

IDBA's puzzle above is actually the very first one that I have had genuine trouble with - I was stuck without the backdoor hints

Because your method seems limited to finding contradictions for a single starting point.

and what I've learned from that is, just because a cell has more than 2 pencilmarks, that does not mean that it can't be considered as a guess candidate - it's the value of the guess that matters (ie: does it have an obvious chain of consequences?).

Yes, an obvious chain of consequences is needed, of course. However, that doesn't guarantee anything. It can just as well (or in fact more probably) lead to a contradiction or a dead end, instead of the solution. A contradiction is obviously better than a dead end, because you get an elimination, but if it's in the middle of nowhere it doesn't necessarily advance the puzzle at all. That's why I said this:

Otherwise you may end up testing lots of candidates that look promising if true but are ineffective as eliminations (which is the more likely result).

So... If you're interested in solving puzzles more efficiently, I would recommend studying some better tools. Personally I think coloring is the best friend of a manual solver. It worked here quite nicely, too. Like I said, I didn't take detailed notes but below you'll find what I did write down.

my solution steps: Show

Starting grid after basics:

Code: Select all

.----------------------.--------------------.--------------------.
| 4789*  1478*  1379-8 | 6       47*   2    | 5     1378   78    |
| 5678   15678  1578   | 57      3     9    | 1267  12678  4     |
| 4567*  2      357    | 8       457*  1    | 9     367    67    |
:----------------------+--------------------+--------------------:
| 56789  3      579-8  | 579     2     4    | 67    678    1     |
| 26789  678    4      | 1379    1679  368  | 267   5      23678 |
| 1      5678   2578   | 357     567   3568 | 4     9      23678 |
:----------------------+--------------------+--------------------:
| 257    157    6      | 1235    15    35   | 8     4      9     |
| 245*   9      125    | 124-5*  8     56   | 3     1267   2567  |
| 3      1458*  1258   | 12459*  1569  7    | 126   126    256   |
'----------------------'--------------------'--------------------'


Step 1. Coloring 4s => -5 r8c4, -8 r14c3 (traps)

Code: Select all

.---------------------.-------------------.----------------------.
| 4789    1478   1379 | 6      47    2    | 5     1378*   8-7*   |
| 5678    15678  1578 | 57     3     9    | 1267  12678*  4      |
| 4567    2      357  | 8      457   1    | 9     367     67     |
:---------------------+-------------------+----------------------:
| 56789*  3      579  | 579    2     4    | 67    678*    1      |
| 26789   678    4    | 1379   1679  368  | 267   5       23678* |
| 1       5678   2578 | 357    567   3568 | 4     9       23678* |
:---------------------+-------------------+----------------------:
| 257     157    6    | 1235   15    35   | 8     4       9      |
| 245     9      125  | 124    8     56   | 3     1267    2567   |
| 3       1458   1258 | 12459  1569  7    | 126   126     256    |
'---------------------'-------------------'----------------------'

Step 2. Coloring (78)r1c9 / 8s -> contradiction in r2c4 => -7 r1c9

Code: Select all

.---------------------.-------------------.------------------.
| 479    147    1379  | 6      47    2    | 5     137   8    |
| 5678   15678  157-8 | 57     3     9    | 1267  1267  4    |
| 4567   2      357   | 8      457   1    | 9     367   67   |
:---------------------+-------------------+------------------:
| 5679   3      579   | 579    2     4    | 67    8     1    |
| 26789  678    4     | 1379   1679  368  | 267   5     2367 |
| 1      5678   2578  | 357    567   3568 | 4     9     2367 |
:---------------------+-------------------+------------------:
| 257    157    6     | 1235   15    35   | 8     4     9    |
| 245    9      125   | 124    8     56   | 3     1267  2567 |
| 3      458-1  28-15 | 12459  1569  7    | 126   126   256  |
'---------------------'-------------------'------------------'

Step 3. Coloring 4s (again) => -1 r9c23, -5 r9c3, -8 r2c3 (traps)

Code: Select all

.--------------------.-------------------.------------------.
| 479    147    1379 | 6      47    2    | 5     137   8    |
| 5678   15678  157  | 57     3     9    | 1267  1267  4    |
| 4567   2      357  | 8      457   1    | 9     367   67   |
:--------------------+-------------------+------------------:
| 5679   3      579  | 579    2     4    | 67    8     1    |
| 26789  678    4    | 1379   1679  368  | 267   5     2367 |
| 1      5678   2578 | 357    567   3568 | 4     9    ^2367 |
:--------------------+-------------------+------------------:
| 257    157    6    | 1235   15    35   | 8     4     9    |
| 245    9      125  | 124    8     6-5* | 3     1267  2567 |
| 3      458    28   | 12459  1569  7    | 126   126   256  |
'--------------------'-------------------'------------------'

Step 4. Coloring (56)r8c6 -> contradiction in r6c9 => -5 r8c6

(In this step I used some URs to advance the coloring, so it was a bit more complicated. Can't remember which URs, though.)

Code: Select all

.--------------------.-------------------.------------------.
| 479    147    1379 | 6      47    2    | 5     137   8    |
| 5678   15678  157  | 57     3     9    | 1267  1267  4    |
| 4567   2      357  | 8      457   1    | 9     367   67   |
:--------------------+-------------------+------------------:
| 5679   3     ^579  | 579    2     4    | 67    8     1    |
| 26789  678    4    | 1379   1679  38   | 267   5     2367 |
| 1      5678   2578 | 357    567   358  | 4     9     2367 |
:--------------------+-------------------+------------------:
| 257    157    6    | 1235   15    3-5* | 8     4     9    |
| 245    9      125  | 124    8     6    | 3     127   257  |
| 3      458    28   | 12459  159   7    | 126   126   256  |
'--------------------'-------------------'------------------'

Step 5. Coloring (35)r7c6 -> contradiction in r4c3 -> -5 r7c6

Code: Select all

.-------------------.----------------.------------------.
| 479   147    1379 | 6      47    2 | 5     137   8    |
| 8     167-5  157  | 57     3     9 | 1267  1267  4    |
| 4567  2      357  | 8      457   1 | 9     367   67   |
:-------------------+----------------+------------------:
| 5679  3      579  | 79     2     4 | 67    8     1    |
| 2679  67     4    | 1379   1679  8 | 267   5     2367 |
| 1     678    278  | 37     67    5 | 4     9     2367 |
:-------------------+----------------+------------------:
| 257   157    6    | 125    15*   3 | 8     4     9    |
| 245   9      125  | 124    8     6 | 3     127   257  |
| 3     458    28   | 12459  159   7 | 126   126   256  |
'-------------------'----------------'------------------'

Step 6. Coloring (15)r7c5 =>

(5)r79c2 = r2c2 - r2c4 = r3c5 - (5=1)r7c5 - r8c4 = r8c3 - (1=75)r2c34 => -5 r2c2

(This was the only chain I actually wrote down, because it was so simple and obvious.)

Code: Select all

.-------------------.----------------.-----------------.
| 479   14-7  139-7 | 6     47     2 | 5     13   8    |
| 8     167   157   | 57    3      9 | 1267  126  4    |
| 4567  2     357   | 8     45-7   1 | 9     36   67   |
:-------------------+----------------+-----------------:
| 567-9  3    579   | 79    2      4 | 67    8    1    |
| 2679   67   4     | 1379  16-79  8 | 267   5    2367 |
| 1      8    27    | 37    67     5 | 4     9    2367 |
:-------------------+----------------+-----------------:
| 27     157  6     | 125   15*    3 | 8     4    9    |
| 24     9    12    | 124   8      6 | 3     7    5    |
| 3      45   8     | 45-9  59     7 | 126   126  26   |
'-------------------'----------------'-----------------'

Step 7. Coloring (15)r7c5 (again) => -79 r5c5, -7 r1c23, -7 r3c5, -9 r4c1,r9c4 (loops)

Code: Select all

.----------------.--------------.-----------------.
| 9-47  14   139 | 6     47   2 | 5     13   8    |
| 8     167  157 | 57    3    9 | 1267  126  4    |
| 4567  2    357 | 8     45   1 | 9     36   67   |
:----------------+--------------+-----------------:
| 567   3    579 | 79    2    4 | 67    8    1    |
| 2679  67   4   | 1379  16   8 | 267   5    2367 |
| 1     8    27  | 37    67   5 | 4     9    2367 |
:----------------+--------------+-----------------:
| 27    157  6   | 12-5  15*  3 | 8     4    9    |
| 24    9    12  | 124   8    6 | 3     7    5    |
| 3     45   8   | 45    9    7 | 126   126  26   |
'----------------'--------------'-----------------'

Step 8. Coloring (15)r7c5 (once more) => -47 r1c1, -5 r7c4 (traps)

Code: Select all

.----------------.----------.------------.
| 9     4     3  | 6   7  2 | 5   1   8  |
| 8     16+7  17 | 5   3  9 | 27  26  4  |
| 6-7   2     5  | 8   4  1 | 9   3   67 |
:----------------+----------+------------:
| 5     3     9  | 7   2  4 | 6   8   1  |
| 26+7  6-7   4  | 9   1  8 | 27  5   3  |
| 1     8     27 | 3   6  5 | 4   9   27 |
:----------------+----------+------------:
| 27    17    6  | 12  5  3 | 8   4   9  |
| 4     9     12 | 12  8  6 | 3   7   5  |
| 3     5     8  | 4   9  7 | 1   26  26 |
'----------------'----------'------------'

Step 9. BUG+2 => -7 r3c1,r5c2; stte

[Mathimagics]

AST = "Advanced Solving Technique"

If you can, it's no longer a pure guess, because you must have seen something that makes one or the other option more likely to succeed or to produce a contradiction.

Indeed, and that's where I derive my P&P pleasure, from making "educated" and/or intuitive guesses ...

If you're interested in solving puzzles more efficiently, I would recommend studying some better tools.

Ah, there's the rub. You see, I actually prefer the process I currently use!

It does have the advantage that I can do puzzles in the bath ...

I am aware of its limitations, that there are some puzzles I might not be able to solve - but these are exceptional cases generally.

One example looks to be this "Easter Monster" puzzle that Mauriès Robert used in a recent post:

Code: Select all

1.......2.9.4...5...6...7...5.9.3.......7.......85..4.7.....6...3...9.8...2.....1

On quick inspection this looks like a puzzle that initially offers NO singles, NO bi-values and NO bi-locals. That's a challenge ...

[SpAce]

AST = "Advanced Solving Technique"

Ok, thanks

Indeed, and that's where I derive my P&P pleasure, from making "educated" and/or intuitive guesses ...
...
Ah, there's the rub. You see, I actually prefer the process I currently use!
...
It does have the advantage that I can do puzzles in the bath ...

Fair enough! I have no intention of ruining your fun!

I am aware of its limitations, that there are some puzzles I might not be able to solve - but these are exceptional cases generally. One example looks to be this "Easter Monster" puzzle that Mauriès Robert used in a recent post

Well, any normal techniques fail with puzzles of that level. That's why things like JExocets and multi-digit fish patterns (including SK-Loops and MSLS) had to be discovered, to reduce such puzzles to something more manageable. The Easter Monster is famous because it gave birth to the SK-Loop, named after its genius discoverer SteveK. Here's how he did it. Around last Easter we had another kind of challenge with that puzzle.

On quick inspection this looks like a puzzle that initially offers NO singles, NO bi-values and NO bi-locals. That's a challenge ...

What makes you think it doesn't have bi-locals? I see ten if I counted correctly (all box-based, though), and more with group links. Only 3s and 8s have none. If I remember correctly, no one has found a puzzle that doesn't have any of the above.

[Mathimagics]

Oops, my goof, my gaffe!

There are certainly a number of box bi-locals in there ... but it's a brute of a puzzle nevertheless

[International_DBA]

I have been concentrating on other things over the last few weeks so I missed this ongoing conversation.
It's good to see that I am giving you something to talk about.
I just published this one.
See what you think of it:
https://andrews-sudoku.blogspot.com/201 ... treme.html

[SpAce]

Hi International_DBA,

I just published this one.
See what you think of it:
https://andrews-sudoku.blogspot.com/201 ... treme.html

Code: Select all

.1.58.73..9.........5.76...2...548.....8..37.9..3..6..........854....2....7....4.

Could you post your future puzzles with a string like that? That can be pasted into software solvers directly.

First what the different graders said:

• SE: 8.5
• Hodoku: 6484
• SudokuWiki: 1024

It has quite a few backdoors, some of which are probably possible to guess or to hit with coloring, though nothing blatantly obvious.

So, clearly easier than the last one, though not trivial. I found a pretty easy (though tedious and inelegant) one-stepper with the first pattern that caught my eye:

my solution: Show

Code: Select all

.-----------------.------------------------.----------------------.
| 46   1    246   | 5       8       29     | 7     3         269  |
| 7    9    268   | 124     1234    123    | 145   1258      1256 |
| 38   238  5     | 1249    7       6      | 149  *19+[2]8  *19+2 |
:-----------------+------------------------+----------------------:
| 2    367  36    | 67      5       4      | 8    *19+      *19+  |
| 146  56   146   | 8       1269    129    | 3     7         245  |
| 9    578  148   | 3       12      127    | 6     25        245  |
:-----------------+------------------------+----------------------:
| 136  236  12369 | 124679  123469  123579 | 159   1569      8    |
| 5    4    13689 | 169     1369    1389   | 2     169       7    |
| 168  268  7     | 1269    1269    12589  | 159   4         3    |
'-----------------'------------------------'----------------------'

UR(19)r34c89 using +internals

(2)r3c8
||
(8)r3c8 -[singles]-> contradiction
||
(2)r3c9 -[basics]-> contradiction

=> +2 r3c8; stte

--
Some might consider that a two-stepper, but it could be written as a complicated net resulting in a single-step verity:

(2)r3c8 == (2)r6c5 => -2 r6c8; stte

The contradiction version is easier to understand and to verify, though.

[Mauriès Robert]

Hi SpAce,

Here is my resolution by TDP which does not take into consideration the UR r34c89.
I remind you that P=track, P' = anti-track

1) P(2r5c56)={6r4c4, ...} and P'(2r5c56)={2r6c56, 5r6c8, 5r79c7, 5r2c9, 6r1c9, 9r1c6, 9r5c5, 6r4c4,...} => r4c4=6 + 5 validation and 1 elimination by basic techniques.

2) P(1r9c1) and P(2r9c2) are conjugated, because P'({1r9c1,2r9c2})={6r8c5, 3r8c3, 8r9c6,... contradiction} => P'({1r9c1,2r9c2}) invalid.
P(1r9c1)={1r9c1, 46r15c9, 3r7c1, 8r3c1, 8r2c8, ...} and P(2r9c2)={2r9c2, 38r3c12, 8r2c8, ...} => r2c8=8.

3) Like you, to make it quick, P(19r3c8) invalid, P(2r3c8) solution.

You can look more elegant, but it's a little longer.

Sincerely
Robert