The New Sudoku Players' Forum

[813554]

https://imgur.com/a/teAkOYX I've shared another one a couple of days ago and I was shocked how blind I was. But now, I've been trying to solve this for over 1.5 hours but nothing, nothing's there. No hiddens, no nakeds no pointings nothing. Actually I am really curious about what I'm missing out this time.

[SpAce]

You're right that you have exhausted all basic techniques. The easiest next step for this puzzle would be a Skyscraper or a 2-String Kite. More detailed instructions below (hidden, in case you want to first try to find one yourself).

Hidden Text: Show

Code: Select all

.-----------------.----------------.--------------------.
| 7    59    13   | 13  6      59  | 248   248    48    |
| 2    15    8    | 7   15     4   | 9     3      6     |
| 4    39    6    | 8   2      39  | 5     1      7     |
:-----------------+----------------+--------------------:
| 5    13    13   | 9   4      8   | 7     6      2     |
| 6    48   *4(9) | 2   57     57  | 3     8-9    1     |
| 8-9  2     7    | 6   13     13  | 48    5     *48(9) |
:-----------------+----------------+--------------------:
| 389  4678  2    | 13  13789  137 | 1468  4789   5     |
| 1    678   5    | 4   789    2   | 68    789    3     |
| 389  478  *4(9) | 5   13789  6   | 1248  24789 *48(9) |
'-----------------'----------------'--------------------'

Skyscraper: (9)r5c3 = r9c3 - r9c9 = (9)r6c9 => -9 r6c1, r5c8; stte

or:

Code: Select all

.-----------------.----------------.--------------------.
| 7    59    13   | 13  6      59  | 248   248    48    |
| 2    15    8    | 7   15     4   | 9     3      6     |
| 4    39    6    | 8   2      39  | 5     1      7     |
:-----------------+----------------+--------------------:
| 5    13    13   | 9   4      8   | 7     6      2     |
| 6    48   *4(9) | 2   57     57  | 3    *8(9)   1     |
| 89   2     7    | 6   13     13  | 48    5     *48(9) |
:-----------------+----------------+--------------------:
| 389  4678  2    | 13  13789  137 | 1468  4789   5     |
| 1    678   5    | 4   789    2   | 68    789    3     |
| 389  478   4-9  | 5   13789  6   | 1248  24789 *48(9) |
'-----------------'----------------'--------------------'

2-String Kite: (9)r5c3 = r5c8 - r6c9 = (9)r9c9 => -9 r9c3; stte

[813554]

It always surprises me how easy it is... Thank you. I don't understand how you people can see it this easy, in fact you make it seem too easy so I don't understand how I couldn't see it. Thank you again.

[SpAce]

It always surprises me how easy it is... Thank you. I don't understand how you people can see it this easy, in fact you make it seem too easy so I don't understand how I couldn't see it. Thank you again.

It gets easier with practice. In fact, this is a good puzzle for learning about the easiest single-digit chains, because most variants (Skyscrapers, 2-String Kites, Empty Rectangles, and other unnamed Turbot Fishes or generic X-Chains) can be found on the same cluster of 9s. Some of them can also be seen as Finned or Sashimi X-Wings, which is another point of view of the same eliminations. I recommend that you save this puzzle and try to find as many variants of those single-digit patterns as you can.

At first it's probably hard to see the patterns directly, but you can cheat by using Simple Coloring, when you notice a certain digit has lots of bilocation strong links (i.e. just two instances in a box, row, or column). For example, here I've "colored" the interesting cluster of 9s with ' and " to depict the two opposing parities:

Code: Select all

.-------------------.----------------.---------------------.
| 7     59    13    | 13  6      59  | 248   248    48     |
| 2     15    8     | 7   15     4   | 9     3      6      |
| 4     39    6     | 8   2      39  | 5     1      7      |
:-------------------+----------------+---------------------:
| 5     13    13    | 9   4      8   | 7     6      2      |
| 6     48    4(9") | 2   57     57  | 3     8(9')  1      |
| 8(9') 2     7     | 6   13     13  | 48    5      48(9") |
:-------------------+----------------+---------------------:
| 389   4678  2     | 13  13789  137 | 1468  4789   5      |
| 1     678   5     | 4   789    2   | 68    789    3      |
| 389   478   4(9') | 5   13789  6   | 1248  24789  48(9') |
'-------------------'----------------'---------------------'

Notice the bottom row which has two 9s with parity '. That's a contradiction ("color wrap"), which means we can eliminate all 9s of that parity (r6c1, r5c8, r9c3, r9c9) and place all 9s of the opposite parity (r5c3 and r6c9). We can also see individual elimination possibilities using "color traps", i.e. when a candidate sees both parities it can't be true. That applies to r9c3 and r9c9 here, but those eliminations are already covered by the more powerful color wrap so they're redundant.

Now that you know what can be eliminated, you can (and should) try to see how each of those eliminations could be spotted and described as individual patterns or chains. That's how you eventually learn to see them directly as well, and can choose surgical strikes (patterns, chains) instead of weapons of mass destruction (coloring).

Notice that Simple Coloring is the weakest of coloring techniques and only shows pretty obvious eliminations (i.e. they become obvious once you get more experience). To see more, you need more powerful variants such as X-Coloring, 3D-Medusa, or GEM.

[Yogi]

7...6....2.87.49364.682.5175..9487626..2..3.1.276...5...2.....51.54.2..3...5.6...
If that is the correct code for this puzzle, why could r1c3 not be a 9 ?

[SpAce]

7...6....2.87.49364.682.5175..9487626..2..3.1.276...5...2.....51.54.2..3...5.6...
If that is the correct code for this puzzle, why could r1c3 not be a 9 ?

Hi Yogi. Basic techniques (some locked candidates + naked/hidden pair) clear the 9 from r1c3. Did you possibly mean some other cell? In any case, these are the basic moves left if we start from scratch with your string:

LC-1: (9)b2 => -9 r7c6
LC-1: (4)b4 => -4 r5c8
LC-1: (3)b5 => -3 r6c1
LC-2: (3)c1 => -3 b7p289
NP(49)/HP(13)c3 => -9 r1c3
LC-1: (9)b1 => -9 r4789c2
NP(13)/HP(59)r1 => -13 r1c26

Added:

LC-1 : locked candidates type 1 = pointing pair/triple
LC-2 : locked candidates type 2 = claiming / box-line reduction
NP : naked pair
HP : hidden pair

[Yogi]

I didn’t understand that, but thanx anyway. Another solver showed that identifying r1234c2 as a quad confines candidate 9 in Box1 to the middle column, taking it out of r1c3.
That makes my intended comment more applicable. For many of us the next step would be to look for the short-chain eliminations such as Kites & Skyscrapers, also X Wings and ERs (or Hinges) but no-one has mentioned the simple box-analysis technique that shows which candidates could be open to such eliminations, in this case only 3, 8 or 9.
It’s a lot easier than trying to find conjugate pairs in all candidates throughout the whole puzzle.

[SpAce]

I didn’t understand that, but thanx anyway.

I added a translation for the acronyms used. Did that help?

Another solver showed that identifying r1234c2 as a quad confines candidate 9 in Box1 to the middle column, taking it out of r1c3.

Sure you can do it that way, if you actually look for quads before pointing pairs, claiming, and naked or hidden pairs (which are usually considered simpler). It's indeed more efficient in this case, because then all you need is:

Naked Quad (1359)r1234c2: -3 r79c2, -9 r1c3, r789c2

and then one of the several Kites or Skyscrapers on 9s to get a singles-only state. That way you can skip all those other basic steps.

That makes my intended comment more applicable. For many of us the next step would be to look for the short-chain eliminations such as Kites & Skyscrapers, also X Wings and ERs (or Hinges) but no-one has mentioned the simple box-analysis technique that shows which candidates could be open to such eliminations, in this case only 3, 8 or 9.
It’s a lot easier than trying to find conjugate pairs in all candidates throughout the whole puzzle.

I don't really know what that box-analysis technique is. If it only helps with said types of eliminations, I'd probably opt for something more generic that helps with not only them but others as well. I don't think it's very difficult to mark all conjugate pairs if you're using candidates in the first place (speaking of pencil&paper solving here). In fact, I think marking candidates without marking their conjugacy relationships is a waste of time. Once done, finding any kinds of AICs is much easier.

[Pat]

7...6....2.87.49364.682.5175..9487626..2..3.1.276...5...2.....51.54.2..3...5.6...

why could r1c3 not be a 9 ?

what SpAce showed,
i would describe thus --

(3) r4\b4
==> (3) c1\b7
==> c3 {1,3}

-- a sequence of 3 moves

Another solver showed that
identifying r1234c2 as a quad
confines candidate 9 in Box1 to the middle column,
taking it out of r1c3.

yes,

c2 {4,6,7,8} ==> (9) c2\b1

[SpAce]

Yogi, is this what you call the "box-analysis technique"? I guess it could be somewhat useful, especially if solving without pencil marks. I'll try to remember to test it some time. Then again, I think the basic single-digit chains are among the easiest patterns to spot anyway, almost regardless of the tools available. The grouped variants have a slightly better chance to fly under the radar, however.

[Yogi]

That’s it, suggested by Keith and adapted in this way to suit me. It really is a very simple idea. Here’s how it works:
Candidate Box Analysis for this puzzle 7...6....2.87.49364.682.5175..9487626..2..3.1.276...5...2.....51.54.2..3...5.6...

Draw a grid representing the 9 boxes and note the unsolved candidates for each box which are NOT confined to a single column or row.
Then identify those that appear in at least four boxes which are arranged in a rectangle. In this example you get 3, 8 and 9.
This is actually made easier for me by my having already noted, scribbled outside the puzzle, those candidates which ARE confined to a single row or column or are known to be part of a locked set. And I find, as Keith noted, that when I go into this box analysis I often discover other moves or constraints which I had not spotted before.
The next step is to search for and note any conjugate pairs in these candidates.
Remember to discard or ignore any cases that include a cell which is in a box which you have not marked with that candidate. In this case we find these CPs:

813554.png (4 KiB) Viewed 831 times

The two conjugate pairs in candidate 3 form a 2-string Kite which eliminates 3 from r7c2, but this is not new information as the 3 has already been excluded from there by the quad above it in that column. The 8CP in row 5 on its own comes to nothing as there is no ER which can see into either end of it.
However, the five 9CPs (that I found) offer a number of eliminations which can solve the puzzle. For example, the Skyscraper {9C39} => 9r5c3r6c9.
The value of this technique is more than just finding short-chain eliminations which may or may not reduce the puzzle to solving by singles. If it produces no useful eliminations it will usually have generated a number of conjugate pairs that can provide a good startpoint for longer or more complicated chains or loops. It seems like a natural progression to me.

[SpAce]

That’s it, suggested by Keith and adapted in this way to suit me. It really is a very simple idea. Here’s how it works:

Thanks, Yogi! I do think it's probably more useful than I initially thought. It seems like a very quick way to check if and where those kinds of short X-Chains are possible (and where not), so one can narrow the search easily. I'd say drawing that box is quite optional, though, as it's quite easy to check each digit on the grid directly (especially with pencil marks).

Then identify those that appear in at least four boxes which are arranged in a rectangle. In this example you get 3, 8 and 9.
...
The two conjugate pairs in candidate 3 form a 2-string Kite which eliminates 3 from r7c2, but this is not new information as the 3 has already been excluded from there by the quad above it in that column.

Actually, if you'd performed all basic eliminations (that I listed earlier) before using this technique, the 3s wouldn't even appear in your box. There are only two 3s left in box 7 and they're lined up, so you don't have a rectangle. In general, I highly recommend that you don't skip basic steps before moving on to non-basics like this (or look for quads before much easier basic steps, for that matter -- even if one did shortcut the solution in this particular case). Of course you can use harder moves first if you happen to spot them, but it's probably not efficient to actively look for them prematurely, unless you're really trying to optimize your solve path.

The value of this technique is more than just finding short-chain eliminations which may or may not reduce the puzzle to solving by singles. If it produces no useful eliminations it will usually have generated a number of conjugate pairs that can provide a good startpoint for longer or more complicated chains or loops. It seems like a natural progression to me.

You mean that the process has made you notice some already existing conjugate pairs -- not generated them (which can only happen through eliminations)? I guess it's a side-effect which may or may not be relevant depending on one's solving style. For me that wouldn't be a relevant benefit of this technique at all, because I usually identify all conjugate pairs during basic solving (how else do you find line-based hidden pairs, for example?) and thus before even considering this. I see conjugate pairs as such a fundamental part of solving any non-trivial puzzles that I really don't see value in finding just a few of them separately. Why not find them all at once and be done with it? It's not hard. That way you have the ability and freedom to use any chains available and not just the ones those few may or may not yield.

I find the potential usefulness of this technique specifically in its ability to shortcut the process of finding those simple X-Chains, but of course it might have additional benefits for others. Then again, I haven't yet test-driven it in practice so I'll have to see how (or if) it fits into my solving process. I suspect it's less beneficial for my solving exactly because I like to be aware of those conjugates early on (with or without pencil marks), and they can also be used directly to find the same chains quite easily. But, I'll try it and we'll see.