The New Sudoku Players' Forum

[Luke]

Back to the rectangles for a minute.

This puzzle had what to me was an interesting one.

Ruud's Nightmare Wed Nov 21 2007

Code: Select all

 085007000000030040040102600800000501000000000207000008004608030090070000000300790
 *-----------------------------------------------------------*
 | 69    8     5     | 49    46    7     | 23    1     23    |
 | 167   16    2     |*589   3    *569   | 89    4     57    |
 | 79    4     3     | 1     58    2     | 6     58    79    |
 |-------------------+-------------------+-------------------|
 | 8     36    9     | 47    246   346   | 5     27    1     |
 | 4     156   16    | 78    28    156   | 39    27    39    |
 | 2     135   7     |*59    15   *1359  | 4     6     8     |
 |-------------------+-------------------+-------------------|
 | 157   7     4     | 6     9     8     | 12    3     25    |
 | 3     9     16    | 2     7     145   | 18    58    46    |
 | 156   2     8     | 3     145   145   | 7     9     46    |
 *-----------------------------------------------------------*


This starts out as a UR on 59 but I'm trying to figure out if there's something a little more here. This pattern is one of those "strong elbows" on one base candidate and an additional strong link on the other.

Thumbnail


Whenever this pattern exists, the base candidate not involved with the elbow cannot be in the elbow cell. Thus r6c6 <> 5. This is pretty straight forward, altho I've never seen this particular pattern identified outside of this forum.

Let's say you didn't notice this. Instead, you decided to try 5 and 9 in the rectangle in cells where they're not involved with the strong links. For example, if 5 was in r2c6, then the strong links would force this result to avoid the Deadly Pattern:

Code: Select all

8|5
5|9

Well, glory be, there's already a 58 in r3c5, so that's three cells for two candidates. I'm lousy at math, but this seems to add up to r2c6<>5.

Now let's say you didn't do that either. Here's a third scenario. If 9 was in r2c4 the strong links would force this result to avoid the DP:

Code: Select all

9|6
5|9

This time, r1c45 form a triple on 469 and with the 9|6 there's four cells for three candidates. r2c4<>9, right?

My question is, can I make all three eliminations legitimately? My cake and eat it too? The latter two depend on the threat of a deadly pattern, which evaporates if scenario one is exercised first.

[udosuk]

I'm no expert in URs, but I can achieve your 3 deductions with some "branching analysis". Perhaps it's not convincingly logical or elegant but should be easy to understand at the very least.

In this pattern you have a potential deadly pattern {59} in r26c46:

Code: Select all

+-------------------+-------------------+-------------------+
| 69    8     5     | 49    46    7     | 23    1     23    |
| 167   16    2     |*589   3    *569   | 89    4     57    |
| 79    4     3     | 1     58    2     | 6     58    79    |
+-------------------+-------------------+-------------------+
| 8     36    9     | 47    246   346   | 5     27    1     |
| 4     156   16    | 78    28    156   | 39    27    39    |
| 2     135   7     |*59    15   *1359  | 4     6     8     |
+-------------------+-------------------+-------------------+
| 157   7     4     | 6     9     8     | 12    3     25    |
| 3     9     16    | 2     7     145   | 18    58    46    |
| 156   2     8     | 3     145   145   | 7     9     46    |
+-------------------+-------------------+-------------------+

To avoid this DP, at least one of the following 3 premises must be true:

r2c4=8 or r2c6=6 or r6c6=1|3

So let's work them out one by one:


Task 1: to prove r6c6<>5

If r2c4=8 => r5c4=7 => r4c4=4 => r1c4=9 => r6c4=5 => r6c6<>5
If r2c6=6 => r6c6=9 (strong link of 9 @ c6) => r6c6<>5
If r6c6=1|3 => obviously r6c6<>5

So in all 3 cases r6c6<>5, therefore it must be true.


Task 2: to prove r2c6<>5

If r2c4=8 => r3c5=5 => r2c6<>5
If r2c6=6 => obviously r2c6<>5
If r6c6=1|3 => r2c6=9 (strong link of 9 @ c6) => r2c6<>5

So in all 3 cases r2c6<>5, therefore it must be true.


Task 3: to prove r2c4<>9

If r2c4=8 => obviously r2c4<>9
If r2c6=6 => r1c5=4 => r1c4=9 => r2c4<>9
If r6c6=1|3 => r2c6=9 (strong link of 9 @ c6) => r2c4<>9

So in all 3 cases r2c4<>9, therefore it must be true.


All 3 proofs make use of the deadly pattern as well as a few bivalue cells/strong links outside it. I don't know the correct terminology of this logic, perhaps "UR inference chains"?

[daj95376]

This starts out as a UR on 59 but I'm trying to figure out if there's something a little more here. This pattern is one of those "strong elbows" on one base candidate and an additional strong link on the other.

Whenever this pattern exists, the base candidate not involved with the elbow cannot be in the elbow cell. Thus r6c6 <> 5. This is pretty straight forward, altho I've never seen this particular pattern identified outside of this forum.

Let's say you didn't notice this. Instead, you decided to try 5 and 9 in the rectangle in cells where they're not involved with the strong links. For example, if 5 was in r2c6, then the strong links would force this result to avoid the Deadly Pattern:

Code: Select all

8|5
5|9

Well, glory be, there's already a 58 in r3c5, so that's three cells for two candidates. I'm lousy at math, but this seems to add up to r2c6<>5.

Now let's say you didn't do that either. Here's a third scenario. If 9 was in r2c4 the strong links would force this result to avoid the DP:

Code: Select all

9|6
5|9

This time, r1c45 form a triple on 469 and with the 9|6 there's four cells for three candidates. r2c4<>9, right?

My question is, can I make all three eliminations legitimately? My cake and eat it too? The latter two depend on the threat of a deadly pattern, which evaporates if scenario one is exercised first.

Just because a UR exists doesn't always mean that it needs to be used.

Your [r2c6]<>5 elimination is half of two eliminations that can be obtained through a Skyscraper. A Triple in [box 2] for {469} results and [r2c4]<>9 follows. In addition, [r2c6]<>5 lets Colors eliminate [r56c6]<>5, and this leads to [r5c2]=5. Another Skyscraper in 6 and Singles follow.