The New Sudoku Players' Forum

[Yogi]

.3..9.167..96.1342..12.3985..3..951....31..9.19.56.7233.6...27..17...83..2..3..5.
Here, once you spot the locked pair 1=9 at r7c49 you can then note that this forces the 1 in Row9 to be diagonally opposite the 1 in Row7,
forming an XWing sort of pattern in 1s over the two boxes. After identifying a few other locked pairs and triples, AS’s Solver then said
Unique Rectangle: Uniqueness Type 4B (Row) removing 9 from roof because of G4,G9,J4,J9 (or r79c49)
I do understand about avoidable URs which would result in multiple solutions, but I don’t get this one at all. What is meant by ‘Roof?’ Either r7c4 or r7c9 must be 9.
Are there any particular techniques which can be brought to bear in this case?

[keith]

Let me give you the Sudoku Susser explanation:

Code: Select all

+----------------------+----------------------+----------------------+
| 2458   3      25     | 48     9      458    | 1      6      7      |
| 578    578    9      | 6      578    1      | 3      4      2      |
| 467    467    1      | 2      47     3      | 9      8      5      |
+----------------------+----------------------+----------------------+
| 24678  4678   3      | 478    248    9      | 5      1      468    |
| 245678 45678  25     | 3      1      27     | 46     9      468    |
| 1      9      48     | 5      6      48     | 7      2      3      |
+----------------------+----------------------+----------------------+
| 3      458    6      | 19     458    458    | 2      7      19     |
| 459    1      7      | 49     245    26     | 8      3      469    |
| 489    2      48     | 14789  3      67     | 46     5      1469   |
+----------------------+----------------------+----------------------+

* Hint: A Type-3 Unique Rectangle will set you on the path to progress.

* Squares [x=4,y=7], [x=9,y=7], [x=4,y=9] and [x=9,y=9] form a Type-3B Unique Rectangle on <19>.  Because they share two rows, two columns, and two blocks, if they all had possibilities <19> then the puzzle would have two solutions; you could simply exchange the <1>s with the <9>s in the squares to get the other solution, and their common rows, columns and blocks would still contain one of each value.  Since a valid Sudoku can have only one solution, either [x=4,y=9] or [x=9,y=9] must contain one of the values <4678>; if neither does, then the squares form a two-solution configuration.  While we don't know which of the squares contains one of the values <4678>, the fact that one of them must have one of those values lets us look for naked sets in row 9; we can treat [x=4,y=9] and [x=9,y=9] as if they were a single square with possibilities <4678>.  Upon close inspection, it is clear that:

([x=4,y=9]|[x=9,y=9])<4678>, [x=7,y=9]<46>, [x=6,y=9]<67> and [x=3,y=9]<48> form a naked quad on <4678> in row 9.  No other squares in the row can contain these possibilities.

   [x=1,y=9] - removing <48> from <489> leaving <9>.

+----------------------+----------------------+----------------------+
| 2458   3      25     | 48     9      458    | 1      6      7      |
| 578    578    9      | 6      578    1      | 3      4      2      |
| 467    467    1      | 2      47     3      | 9      8      5      |
+----------------------+----------------------+----------------------+
| 24678  4678   3      | 478    248    9      | 5      1      468    |
| 245678 45678  25     | 3      1      27     | 46     9      468    |
| 1      9      48     | 5      6      48     | 7      2      3      |
+----------------------+----------------------+----------------------+
| 3      458    6      | 19     458    458    | 2      7      19     |
| 45     1      7      | 49     245    26     | 8      3      469    |
| 9      2      48     | 1478   3      67     | 46     5      146    |
+----------------------+----------------------+----------------------+


Keith

[Leren]

Code: Select all

*---------------------------------------------------------------------------------*
| 2458    3       25       |  48      9       458      | 1       6       7        |
| 578     578     9        |  6       578     1        | 3       4       2        |
| 467     467     1        |  2       47      3        | 9       8       5        |
|--------------------------+---------------------------+--------------------------|
| 24678   4678    3        |  478     248     9        | 5       1       468      |
| 245678  45678   25       |  3       1       27       | 46      9       468      |
| 1       9       48       |  5       6       48       | 7       2       3        |
|--------------------------+---------------------------+--------------------------|
| 3       458     6        | F19      458     458      | 2       7      F19       |
| 459     1       7        |  49      245     26       | 8       3       469      |
| 9-48    2      *48       |*R19+478   3     *67       |*46      5     *R19+46    |
*---------------------------------------------------------------------------------*

Here's my explanation. There is a possible UR in digits 19 in cells r79c49.

The two Floor cells contain only the UR digits. The two Roof cells contain at least one UR digit (and if one each they must be different), plus extra digits that prevent the Deadly 19 pattern from being fully exposed.

For this puzzle, what this means is that at least one of 4678 must be true in r9c49. You can think of this as a pseudo single cell containing only those digits.

Now notice that there are 3 other cells containing 4678 in Row 9. Together with the pseudo cell there are 4 virtual cells that must contain 4678 in Row 9. In other words 4678 must be distributed amongst the 5 cells marked * in Row 9.

So you can eliminate 4678 from all cells in Row 9 but not in the 5 cells marked *, which leads to - 48 in r9c1 in this puzzle.

This uniqueness pattern is referred to as Type 3 (N guard digits in the Roof cells plus an N-1 cell ALS containing the guard digits that the Roof cells can see). Obviously N must be at least 2.

Compare this description with the Hodoku example/explanation here or Andrew Stuart's example/explanation here.

Leren

[keith]

Yogi,

Paste this puzzle (the string you posted) into Helmut's Sudoku Formatter,

http://sudoku.saueregger.at/HSH/HSF.htm

and then go to his Sudoku Helper (HSH). This puzzle can be solved by a long sequence of not very advanced moves.

I think HSH will help immensely with many of the questions you have been asking.

Keith

[David P Bird]

A much simpler logic can be used in this case:
We must stop (19)r9c49 also being locked pair as together with (19)r7c49 this would make the UR.
(1) is already locked in these cells so (9) must be true somewhere else in the row and the only available cell is r9c1.

When a potential UR exists we know that two things must be true:
1) 0ne of the two 'internal' digits (1) or (9) must be true somewhere else in one of the containing houses
2) An 'external' digit, one of (4678), must be true in one of the cells.

Once a computer solver has found a unique option for one of these it may stop looking for a simpler unique option for the other.

DPB

[Yogi]

That's what we want - Simple!

Thank you David.

[Leren]

I'm all for keeping things simple, but David has spotted a specific simplified pattern within the Type 3 UR case.

However, this isn't always the case, in fact it's the first such example I've been made aware of. In the more common case you have to apply the full logic I outlined in my previous post.

To illustrate, here is an example Type 3 UR puzzle : 3.......7...98...3.5.....2...1.....5.968.12......237....4.....9...71..5.......36.

After some singles you get to here :

Code: Select all

*--------------------------------------------------------------------------------*
| 3       1468-2 *89+2     | 125     46      2456     |*89      14      7        |
| 146     146-27 *27       | 9       8       2467     | 5       14      3        |
| 1489    5      *89+7     | 13      347     47       |*89      2       6        |
|--------------------------+--------------------------+--------------------------|
| 2       3       1        | 4       79      79       | 6       8       5        |
| 7       9       6        | 8       5       1        | 2       3       4        |
| 458     48      58       | 6       2       3        | 7       9       1        |
|--------------------------+--------------------------+--------------------------|
| 568     268     4        | 235     36      2568     | 1       7       9        |
| 689     268     3        | 7       1       2689     | 4       5       28       |
| 1589    1278    589-27   | 25      49      24589    | 3       6       28       |
*--------------------------------------------------------------------------------*

As you can see, there is a type 3 UR (89+27) in the cells marked *, but there are more than two 8's and 9's in the Roof Column 3, so you can't place them anywhere, but you can make non UR digit eliminations in the Roof column and box as shown.

Leren

[David P Bird]

However, this isn't always the case, in fact it's the first such example I've been made aware of.

Methinks you haven't been looking very hard!!

The alternative (simpler?) view for your example is that to disrupt the UR cell r58c3s must contain either 58 or 59 which excludes 27 from r9c3 locking these digits in r789c2 leading to the box/line eliminations in r12c2.

As I said, the route taken by most computer solvers is the first one tried that produces a reduction which is not necessarily the simplest. For example, Hodoku always uses paths through large Almost Naked Sets rather than those through their complementary Almost Hidden Sets that would involve fewer digits and cells.

Solver/helper programs that considered both paths and presented the simpler one would be friendlier.

David