The New Sudoku Players' Forum

[Yogi]

Or Remote Crash? Original 17C puzzle:
........1.....2..3.45..........45....6...7...8......2....8..7...7....5..9..1..... but now at
6..7.4.517..562.43.45.18.67....456785642871398.76.14254.685.7.2.7842.5.6952176384

Code: Select all

+-------------------------+-------------------------+-------------------------+
| 6       28      39      | 7       39      4       | 28      5       1       |
| 7       189     19      | 5       6       2       | 89      4       3       |
| 23      4       5       | 39      1       8       | 29      6       7       |
+-------------------------+-------------------------+-------------------------+
| 123     1239    139     | 39      4       5       | 6       7       8       |
| 5       6       4       | 2       8       7       | 1       3       9       |
| 8       39      7       | 6       39      1       | 4       2       5       |
+-------------------------+-------------------------+-------------------------+
| 4       13      6       | 8       5       39      | 7       19      2       |
| 13      7       8       | 4       2       39      | 5       19      6       |
| 9       5       2       | 1       7       6       | 3       8       4       |
+-------------------------+-------------------------+-------------------------+


There may be other paths, but I was solving by P&P and was interested in what was happening with the (39) Remote Pair chain r6c25 – r43c4 – r1c53
The Rule (as I understand it) is that if you label the cells along the chain alternately A – B – A –B etc. (and it can branch) then any other cell which can see an A and a B from the connected network cannot be either of the paired candidates in the chain. The same rule can be used in Simple Colouring.
What I found interesting was that if you work right through the chain as I’ve written it, you find that r6c2 is an A and r1c3 is a B, which would indicate that r4c3 can be solved for 1. However, r4c3 also sees r4c4 which is an A in the chain. I wondered at the time if this would counter the RP deletions, but decided it didn’t, and 1r4c3 certainly solved the puzzle in singles. It seems the Remote Pairs and Simple Colouring methods are immune to non-eliminating connections when there are effective eliminating connections.
The other point is that this RP solution of r4c3 to 1 could not occur earlier before I had reduced r1c3 to (39) only.

[Leren]

Code: Select all

*---------------------------------------*
| 6    28   a39B  | 7 b39A 4  | 28 5  1 |
| 7    18-9  19   | 5   6  2  | 89 4  3 |
| 23   4     5    | 39B 1  8  | 29 6  7 |
|-----------------+-----------+---------|
| 123  1239  1-39 | 39A 4  5  | 6  7  8 |
| 5    6     4    | 2   8  7  | 1  3  9 |
| 8   d39A   7    | 6 c39B 1  | 4  2  5 |
|-----------------+-----------+---------|
| 4    13    6    | 8   5  39 | 7  19 2 |
| 13   7     8    | 4   2  39 | 5  19 6 |
| 9    5     2    | 1   7  6  | 3  8  4 |
*---------------------------------------*

(39) r1c3 = r1c5 - r5c5 = (39) r5c2 => -39 r4c3, - 9 r2c2; stte

The Skyscraper on 3's alone is enough to solve, but Hodoku played it as Remote Pairs (or 2 skyscrapers on 3 & 9 in the same cells) as depicted. Why did it do that ? Maybe the Hodoku experts can explain.

Leren

<edit> Added parity tags AB. The answer to Yogi's question is simple. The elimination cell r4c3 can see two A's but only one B, and although the elimination appears to have been "twice" it relies on the establishment of the single B cell that it can see.

[Maxito_Bahiense]

Hi Yogi, Leren!

There may be other paths, but I was solving by P&P and was interested in what was happening with the (39) Remote Pair chain r6c25 – r43c4 – r1c53
The Rule (as I understand it) is that if you label the cells along the chain alternately A – B – A –B etc. (and it can branch) then any other cell which can see an A and a B from the connected network cannot be either of the paired candidates in the chain. The same rule can be used in Simple Colouring.
What I found interesting was that if you work right through the chain as I’ve written it, you find that r6c2 is an A and r1c3 is a B, which would indicate that r4c3 can be solved for 1. However, r4c3 also sees r4c4 which is an A in the chain. I wondered at the time if this would counter the RP deletions, but decided it didn’t, and 1r4c3 certainly solved the puzzle in singles. It seems the Remote Pairs and Simple Colouring methods are immune to non-eliminating connections when there are effective eliminating connections.

Not sure to understand you correctly. There are two pairs tagged as "A", r6c2 and r4c4. Is that what you mean? You could use either of them with "B" r1c3 to eliminate (3,9) from r4c3. Indeed, if you see this as a remote pair 'chain', then the extremes of the chain are what we normally see to eliminate candidates. In this case, the extremes are "A" r6c2/ "B" r1c3. But noticing that you have the "A" r4c4, you could also have taken the (short) chain r43c4 r1c53, that also wipes (3,9) in r4c3. So, multiple chains are possible here. As long as you take an "A" pair with a "B" as extremes, you're safe. Notice that the latter chain does not kill c9 in r2c2, which is eliminated with your RP chain, since that 9 sees both nines in pairs (3,9) of the extremes. Notice also that you could have taken the direct chain r6c25 r1c53.

The Skyscraper on 3's alone is enough to solve, but Hodoku played it as Remote Pairs (or 2 skyscrapers on 3 & 9 in the same cells) as depicted. Why did it do that ? Maybe the Hodoku experts can explain.

Leren

That's because you have remote pairs up in the solver scale, coming before skycrapers. If you consider skycrapers easier than remote pairs (I do) you could move the order in Edit/Preferences Solver tab.

[Maxito_Bahiense]

Indeed, if you see this as a remote pair 'chain', then the extremes of the chain are what we normally see to eliminate candidates. In this case, the extremes are "A" r6c2/ "B" r1c3. But noticing that you have the "A" r4c4, you could also have taken the (short) chain r43c4 r1c53, that also wipes (3,9) in r4c3. So, multiple chains are possible here. As long as you take an "A" pair with a "B" as extremes, you're safe. Notice that the latter chain does not kill c9 in r2c2, which is eliminated with your RP chain, since that 9 sees both nines in pairs (3,9) of the extremes. Notice also that you could have taken the direct chain r6c25 r1c53.

To expand and clarify my answer, let us take your example as one for Medusa colouring, instead of remote pairs. Starting with seed r6c2 3 blue (9 red), you could make the following Medusa cluster:

Code: Select all

.-------------------.------------.-----------.
| 6    2B8R  3R9B   | 7   3B9R 4 | 2B8R 5  1 |
| 7    18B9! 19     | 5    6  2  | 8R9B 4  3 |
| 2R3B  4    5      | 3R9B 1  8  | 2B9R 6  7 |
:-------------------+------------+-----------:
| 12B3 12R39 13!9!  | 3B9R 4  5  | 6   7   8 |
| 5    6     4      | 2   8   7  | 1   3   9 |
| 8    3B9R  7      | 6  3R9B 1  | 4   2   5 |
:-------------------+------------+-----------:
| 4    13    6      | 8   5   39 | 7   19  2 |
| 13   7     8      | 4   2   39 | 5   19  6 |
| 9    5     2      | 1   7   6  | 3   8   4 |
'-------------------'------------'-----------'


As you are using a colouring scheme, not a chain, any pair of blue/red coloured candidates for a digit are enough to eliminate any candidate seeing both. So, you could use any of 3/9 pairs coloured 3 blue, 9 red with r1c3 inversely coloured to eliminate candidates 3/9 in r4c3. Notice that candidate 9 in r2c2 is also targeted.
However, I'd like to point out that instead of eliminating 3/9 directly from r4c3, you could notice that candidate 3 in r4c3 can be coloured blue from r1c3 3 red, since there is a strong link between 3 r1c3=r4c3. In this way, you get a collision of two blue 3's in r4c34, which proves all red candidates true.

[Yogi]

Thanks guys. I hadn’t even seen the double skyscraper r6c25 – r1c53, which is an almost instant solve, as I was looking at the longer RP chain (which does work) and thinking more about the non-eliminating spanner in the works which appeared to argue against it.
It’s the sudoku equivalent of what I used to call chess blindness. You get a bright idea and focus on it so totally that you fail to see something else that is more obvious and possibly more effective.
Cheers!