The New Sudoku Players' Forum

[Yogi]

7249561381684235979357186245..3..81..4..8175..81.7.24..13....72...1...85.5...7.61

ER.png (20.17 KiB) Viewed 1488 times

This puzzle is an exercise in Empty Rectanges, which makes it all new to me. Yahoo!
How to go about it? As I understand it, the only Boxes here that could have ERs in them would be Boxes 4578, and looking at each of these, I think they could only have ERs for candidates 269 (also 4 in Box8). However, the options sharply reduce when you look for the outside Rows or Columns they would need to work with. For Candidate 2 there are Columns 4&6 and Row4, and for 9 there is only c2 and r5. So we end up with a short-list of 5 possible arrangements for ERs, only for candidates 2 and 9 and only working with these units.
The 2ER in Box4 has it’s possible 2s in r4c3 and r5c13, and the intersecting point of the two lines these 2-candidate cells are in is r4c3. This cell can see one of the two instances of candidate 2 in c4. This eliminates 2 from r9c3, but then what?
Yeah, I missed it. Eliminating 2 from r9c3 solves that cell as a 9 and that solves the whole puzzle. There were then only three clues in Box7 and I did not realise that r9c3 was bi-valued.
I went looking through all the other possibles in my short-list, none of which came to anything that would solve the puzzle. So I will have to be more careful. Bring on the ERs!

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 7     2     4      | 9     5     6      | 1     3     8      |
| 1     6     8      | 4     2     3      | 5     9     7      |
| 9     3     5      | 7     1     8      | 6     2     4      |
|--------------------+--------------------+--------------------|
| 5     79   *2679   | 3     469  *249    | 8     1     69     |
| 236   4     269    | 26    8     1      | 7     5     369    |
| 36    8     1      | 56    7     59     | 2     4     369    |
|--------------------+--------------------+--------------------|
| 468   1     3      | 568   469   459    | 49    7     2      |
| 246   79    2679   | 1     3469 *249    | 349   8     5      |
| 248   5     9-2    |*28    349   7      | 349   6     1      |
*--------------------------------------------------------------*

Empty Rectangle in 2's: Base Cells r4c3 r4c6 + Empty Rectangle Box 8 + Target Cell r9c3 => - 2 r9c3; stte

Personally, I've never liked the way ER moves are usually explained. In the ER Box there are two ER's in 2, r78c45 and r79c56. So when is the ER box suitable for this move and when isn't it suitable ?

What's important to realize, is that the ER box is suitable for the move when and only when the ER digit candidates (in the ER box) can be covered by exactly 1 horizontal line and one vertical line (Row 9 and Column 6 in this puzzle).

The number of ER digits in the ER box can be 2, 3, 4 or 5 and yes, there is at least one Empty Rectangle of the ER digit in the ER box when the 1 horizontal line and one vertical line pattern exists.

Personally I think this is a simpler way of looking at this move, although no doubt others may differ in this opinion.

Leren

PS: You can find Andrew Stuart's explanation of Empty Rectangles here. If you read this you'll notice that he does mention the two lines but he does not cover the important (and quite common) case where there are just two ER candidates in the ER box (and two Empty Rectangles). You could be forgiven for incorrectly concluding that this case would not be suitable if you believe Andrew is a Guru and has given a full explanation of all possible variations of the move.

[David P Bird]

From a manual solver's perspective it's easy to see the box inferences for a digit when it only occupies a few cells, but they can be missed when there are 4 or 5 instances of it that are confined to single rows and columns.

The Empty Rectangle name signifies that within a box if a rectangle of cells can't contain a focus digit - ie are 'empty' – the ones that are occupied will provide two useful group nodes, one restricted to a single row and the other restricted to a single column.

Code: Select all

*-------*---
| 5 \ \ | .      \ = Empty with (5) absent
| 5 \ \ | .       
| 5 5 5 | 5      . = Any digit 
*-------*---
| 5 . . | .

In the diagram digit (5) can't be false both in row3box1 and column1box1 making it impossible for (5) to be true outside the box in both in r3 and c1.
As a chain: (5)r4c1 – (5)r123c1 = (5)r3c123 – (5)r3c4

Which group node should include cell r3c1 is optional and above it has been shown as a member of both of them as this emphasises that this is a 'strong-only' inference.
Strong-only inferences are those that can't alternatively be used as weak. As shown the two sides can't both be false but could both be true when (5)r3c1 is true.

Allocating r3c1 to just one of the group nodes wouldn't make any difference
(5)r4c1 = (5)r123c1 – (5)r3c23 = ???
??? – (5)r12c1 – (5)r3c123 = (5)r3c4
On one side or another (5)r3c1 will prevent a strong link being made to a further (5) outside the box.

So when the common cell between the row and column is empty either type of link can be used, but when it is occupied only a strong link can be used.

TAGdpbEmptyRectangles

[Yogi]

Yes Andrew’s explanation is a little confusing and he does not describe the situation where there are only two possibilities of the elimination candidate in a box. The fact that this method is called Empty Rectangle seems to focus on the cells that are not possible for the candidate you are interested in, when the way they work is actually about the candidate cells and the shape they are arranged in. It is helpful to understand that this technique can work for 2 up to 5 possible candidate cells, as long as they are all included in two lines, one of which must be a row and the other must be a column. I note that in the case of there being 5 candidate cells, they can only work if they form a Star (four corners plus the centre cell) a T (in some orientation) or what you might call a Frame, being the cells along two neighbouring edges of a box.
All cases of 3 to 5 candidate cells will point to a specific Intersecting Cell for the two lines (which may or may not be a candidate cell.) More interesting are the cases where there are only two candidate cells in a box, but not in the came row or column. Here you can choose two ways of drawing the lines to produce two different ICs, one or both of which may work to produce an elimination elsewhere.
This came up in a puzz I did recently: 47.....2.93.42...78125.734.364.72.8.129845673587...4.2741.532.82587...3.6932..75.
By the look of it, boxes 235689 could all have had possible ERs for a number of candidates, so to cut down on the work I looked for the rows & columns that have two cells only with particular candidates. This shortened the list to 135689, and 1358 were not productive.
Candidate 6 had two-cell options in c8 and r3. Could there be an ER to work with these? Not in Box2 (wrong shape) and the 6ER in Box8 has an IC which cannot see into a 6cell in either of these. Box9 has the option of drawing the IC for the 6ER at r8c8 or r7c9. Unfortunately, neither of these can see a cell that can produce an elimination.
The puzzle was finally solved by identifying the 9ER in Box5 which worked with the 9s in Row7 to show that r6c8 could not be 9.
I agree with the comment about the language used to describe ERs and what they do. RCB said:
1. Empty Rectangle on candidate 9 at n5(r6->c4) - r7c4=r7c8
1a. Removed candidate 9 from r6c8
I’d suggest just 9ER in Box5 + r7 => r6c8<>9

[Leren]

Yogi wrote :The fact that this method is called Empty Rectangle seems to focus on the cells that are not possible for the candidate you are interested in, when the way they work is actually about the candidate cells and the shape they are arranged in.

Yogi, you and I are definitely on the same wavelength about the apparently confusing explanations of the so-called Empty Rectangle move. It used to be called a Hinge, which reflected the real issue of the focus digit (call it k) being confined to exactly one row and one column in the Hinge box. In fact, if there are two or three k's in the box, and they are all in one row or one column, there are 3 empty rectangles of K's in the box, but it is not suitable for the Hinge move. How ridiculous is that ?

Here is a method for finding suitable hinge boxes that is suitable for both solver programmers and pattern matching solvers (which I think you are).

1. Count the number of focus digits in the box. It must be 2, 3, 4 or 5.

2. Look for a mini-row in the box (intersection of the box and a row) that has at least one focus digit but not all of the focus digits.

3. Look for a mini-column in the box that has at least one focus digit but not all of the focus digits.

4. Add up the focus digits found in Steps 3 and 4.

5. Check whether the intersection of the mini-row and mini-column has a k or not.

6. If the intersection has a k, and the total k's in the mini-row and mini-column is one more than the number of k's in the box, then the box is a suitable hinge box.

If the intersection has no k, and the total k's in the mini-row and mini-column is equal to the number of k's in the box, then the box is a suitable hinge box.

This method works every time, with no exceptions. And guess what ? There is no mention of empty rectangles at all. They are not necessary. They are merely a distraction.

Well, I guess I've vented my spleen enough for one post. I wish you happy ER hunting, or I should say happy Hinge hunting.

Leren

[JasonLion]

The advantage of Andrew's description is that explains how to find Empty Rectangles in one sentence that is both clear and easy to apply.

An Empty Rectangle occurs within a box where four cells form a rectangle which does NOT contain a certain candidate.

If you focus just on that sentence, it also makes the two candidates case abundantly clear. There are two different ways to use empty rectangles with two candidate cells in the box because there are two different empty rectangles. Of course, some people don't count the two candidate case as an empty rectangle at all, since it is covered by a simpler technique (conjugate pair).

I agree that this puts the emphasis in a potentially misleading place, but it does serve to dramatically simplify the description.

Another spin on things is to note that empty rectangles are just a special case of grouped links, and to call them grouped links instead of empty rectangles. This kind of goes along with treating as many things as possible as AIC, which is kind of implied by the common notation.

It is almost never a good idea to make up new notation for things that are already well understood. It almost always creates confusion and also has the potential to alienate others.

[David P Bird]

In fact, if there are two or three k's in the box, and they are all in one row or one column, there are 3 empty rectangles of k's in the box, but it is not suitable for the Hinge move. How ridiculous is that ?

The three 'Hinge' or 'ER' derived inferences are not ridulous and still apply: cells outside the box in the same row and one of the three columns and can only contain one true k at most (they are weakly linked).

The fact is that, as k is locked in the box in the row, the external instances of it in that row must be false anyway.
This isn't any different to other mechanically found derived inferences that don't yield any new eliminations.