The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |97.|...|...|
 |18.|.29|64.|
 |..2|3..|...|
 |---+---+---|
 |2.8|4..|..3|
 |.1.|.8.|.6.|
 |4..|..6|1.8|
 |---+---+---|
 |...|..5|3..|
 |.36|17.|.94|
 |...|...|.56|
 *-----------*


Play/Print this puzzle online

[SpAce]

Code: Select all

.-------------------.-------------------------.------------------.
| 9     7      45  |    568    156      48    | c258    3   125  |
| 1     8      3   |  ea(5)7*  2        9     |  6      4   57   |
| 6   ea4(5)*  2   |    3      1-5  edba47#8* | c5789  d18  1579 |
:------------------+--------------------------+------------------:
| 2     6      8   |    4      59       1     |  59     7   3    |
| 3     1      579 |    2579   8        27    |  4      6   59   |
| 4     59     579 |    579    3        6     |  1      2   8    |
:------------------+--------------------------+------------------:
| 78    249    49  |    2689   469      5     |  3      18  127  |
| 5     3      6   |    1      7       b28    | c28     9   4    |
| 78    249    1   |    289    49       3     |  278    5   6    |
'------------------'--------------------------'------------------'

Almost XY-Wing:[54-47#8-75]r3c26,r2c4

XYW:[54-47-75] = 8r(3-8)c6 = r(8-13)c7 = 8r3c(8-6) = XYW:[54-47-75]

=> -5 r3c5, (-8 r3c6); stte

[SteveG48]

Code: Select all

 *-----------------------------------------------------------*
 | 9     7    b45    |b568   156 ab48    |c258   3     125   |
 | 1     8     3     | 57    2     9     | 6     4     57    |
 | 6     45    2     | 3     15   a478   | 5789  18    1579  |
 *-------------------+-------------------+-------------------|
 | 2     6     8     | 4     59    1     | 59    7     3     |
 | 3     1     579   | 2579  8     27    | 4     6     59    |
 | 4     59    579   | 579   3     6     | 1     2     8     |
 *-------------------+-------------------+-------------------|
 | 78    249   49    | 2689  469   5     | 3     18    127   |
 | 5     3     6     | 1     7     2-8   |d28    9     4     |
 | 78    249   1     | 289   49    3     | 278   5     6     |
 *-----------------------------------------------------------*


8r13c6 = (458)r1c346 - (5|8=2)r1c7 - (2=8)r8c7 => -8 r8c6 ; stte

[Ngisa]

Code: Select all

+------------------+--------------------+---------------------+
| 9     7      45  | 568     156    48  | 258     3     c125  |
| 1     8      3   |i7-5      2      9  | 6       4      57   |
| 6     45     2   | 3      a15    h478 | 5789   b18    b1579 |
+------------------+--------------------+---------------------+
| 2     6      8   | 4       59     1   | 59      7      3    |
| 3     1      579 | 2579    8     g27  | 4       6      59   |
| 4     59     579 | 579     3      6   | 1       2      8    |
+------------------+--------------------+---------------------+
| 78    249    49  | 2689    469    5   | 3       18    d127  |
| 5     3      6   | 1       7     f28  | e28     9      4    |
| 78    249    1   | 289     49     3   | 278     5      6    |
+------------------+--------------------+---------------------+


(5=1)r3c5 - r3c89 = (1-2)r1c9 = r7c9 - (2=8)r8c7 - (8=2)r8c6 - (2=7)r5c6 - r3c6 = (7)r2c4 =>
- 5r2c4; stte

Clement

[eleven]

8r13c6 = (458)r1c346 - (5|8=2)r1c7 - (2=8)r8c7 => -8 r8c6 ; stte

Can't see that link. Why not 6 in r1c4 ?

[SteveG48]

8r13c6 = (458)r1c346 - (5|8=2)r1c7 - (2=8)r8c7 => -8 r8c6 ; stte

Can't see that link. Why not 6 in r1c4 ?

The only 8 remaining in box 2 is r1c4. Perhaps I should have written (HT458)r1c346 .

[SpAce]

8r13c6 = (458)r1c346 - (5|8=2)r1c7 - (2=8)r8c7 => -8 r8c6 ; stte

Can't see that link. Why not 6 in r1c4 ?

The only 8 remaining in box 2 is r1c4. Perhaps I should have written (HT458)r1c346 .

Hi Steve! Thanks for the clarification. I have to admit that I fell into the same trap as eleven, but I now see how it works. The formed triple was hard to see because it was born out of multiple mothers so to speak (i.e. the different digits got locked for different reasons). I don't think adding the 'HT' helps at all in that regard, and I don't think it's even correct. Weak links should be inside the triple's cells in that case, but yours are external, which makes it act more like a naked triple. It's a combo of a NP(45, row) and a HS(8, box) which together form a row-based triple, but I don't think it can be classified as either HT or NT. If anything, NT is a closer match based on the weak links. In general, I think those HP/HT etc. markings are totally useless clutter anyway (but that's just my opinion). I don't really know if there's a good way to make this clearer, except writing it more explicitly:

(8)r13c6 = (45)r1c63&(8)r1c4 - (5|8=2)r1c7...

...but then again, (45)r1c63&(8)r1c4 <-> (458)r1c634, so your original was correct anyway (just a bit hard to decipher). I probably would have written it your way (and I have, in similar situations), so I can't complain about anything but my own blindness here

Edit: changed "eliminations" -> "weak links".

[blue]

The only 8 remaining in box 2 is r1c4. Perhaps I should have written (HT458)r1c346 .

I think the term "pseudo naked triple" has been used in this kind of situation .... "PNT".

Side note: This is a situation where reversing the chain, would have made it more readable

(8=2)r8c7 - (2=458)r1c367 - 8r1c4 = 8r13c6

... or the condensed version:

(8=2)r8c7 - (2=8)r1c367 - r1c4 = 8r13c6

[SpAce]

The only 8 remaining in box 2 is r1c4. Perhaps I should have written (HT458)r1c346 .

I think the term "pseudo naked triple" has been used in this kind of situation .... "PNT".

PNT -- that makes perfect sense! Thanks for that, blue!

Side note: This is a situation where reversing the chain, would have made it more readable

(8=2)r8c7 - (2=458)r1c367 - 8r1c4 = 8r13c6

Yes, that's great! Here's another (less readable) variant:

(8=2)r8c7 - r1c7 = (21-6)r1c95 = (6'48)r1c4,r13c6

... or the condensed version:

(8=2)r8c7 - (2=8)r1c367 - r1c4 = 8r13c6

I'm not a fan of the condensed ALS form. I think your previous chain was the most readable one. I'd be tempted to write it like this, though, but it's just me (and I don't expect much sympathy ):

(8=2)r8c7 - (2=45'8)r1c734 - 8.b2p(1=39)

[SteveG48]

Thanks, Blue. I try to check for increased clarity by reversal, but I wasn't thinking this time.

[SpAce]

Thanks, Blue. I try to check for increased clarity by reversal, but I wasn't thinking this time.

Actually, the significant bit was not the reversal of the chain. Blue's chain works just as well the other way around:

(8)r13c6 = r1c4 - (8=45'2)r1c637 - (2=8)r8c7

The improved clarity comes mostly from a different choice of nodes, not the reversal.

[blue]

Yes, that's great! Here's another (less readable) variant:

(8=2)r8c7 - r1c7 = (21-6)r1c95 = (6'48)r1c4,r13c6

I would avoid this one in favor of something else, because

1) The 4 in the final node, is superfluous.
2) It isn't clear (without looking at the PM) that in final node, the 8 would be forced to c6 (in sight of the target: 8r8c6)

Either of these, seem easier to read:

(8=2)r8c7 - r1c7 = (216-8)r1c459 = 8r13c6
(8=2)r8c7 - r1c7 = (HT: <216>r1c459) - 8r1c4 = 8r13c6

---

Actually, the significant bit was not the reversal of the chain. Blue's chain works just as well the other way around:

(8)r13c6 = r1c4 - (8=45'2)r1c637 - (2=8)r8c7

It does. I'm pleasently surprised
Nice catch

[SpAce]

Yes, that's great! Here's another (less readable) variant:

(8=2)r8c7 - r1c7 = (21-6)r1c95 = (6'48)r1c4,r13c6

I would avoid this one in favor of something else, because

1) The 4 in the final node, is superfluous.

True. I actually saw and wrote it without the 4, but then changed it because it seemed to contradict what I said next about hating the condensed form (in this case two digits in three cells) I know it's fully correct that way, but I thought it might be a bit less confusing as a complete triple.

2) It isn't clear (without looking at the PM) that in final node, the 8 would be forced to c6 (in sight of the target: 8r8c6)

True. However, it's also true about many if not most of our chains using ALS nodes. I'm the first to admit that it's annoying, and I used to complain about it when I was more of a newbie and had actual difficulty following such chains. Since then I've given in and considered seeing the PM as a required part of following any chains. I think it's a built-in weakness of Eureka, but I'm not really sure how it could be fixed as the nodes are considered sets instead of ordered tuples (and many times there's no known ordering anyway).

On my part I try to improve clarity by separating the linking digits with a ' and keeping the digits and cells in as logical an order as possible, so that most of the links could be followed without seeing the PM (not always possible), but from someone else's POV it probably has the opposite effect. I also use a comma to imply an ordered tuple when absolutely needed, but that seems to confuse some readers too.

Either of these, seem easier to read:

(8=2)r8c7 - r1c7 = (216-8)r1c459 = 8r13c6

Much better! I don't know why I didn't see it.

(8=2)r8c7 - r1c7 = (HT: <216>r1c459) - 8r1c4 = 8r13c6

Not to my liking. As I said before, I think those 'HT'-prefixes are unnecessary clutter. I hated them when I was (more of a) beginner, and I hate them now even more. For a manual solver such as myself, a small AHS is usually easier to see than a huge ANS, and it often results in a simpler chain notation too. Adding unnecessary prefixes totally defeats the latter benefit. I really don't see why they should be treated as some special cases. (I'm pretty sure my views here are in line with David's.)

Nice catch

Thanks!

[eleven]

A note about the condensed form:

When you see an almost locked set (n cells, n+1 digits), you can look at it in different ways.
One is "if one digit is false, all the others must be true"
Another "there is a strong link between each 2 digits"

The second view corresponds to the condensed form.

[SpAce]

A note about the condensed form:

When you see an almost locked set (n cells, n+1 digits), you can look at it in different ways.
One is "if one digit is false, all the others must be true"
Another "there is a strong link between each 2 digits"

The second view corresponds to the condensed form.

Yes, these days I understand that. I have no problem with the logic or the correctness of that notation per se. It's just that the condensed form leaves more to check for the reader if one wants to validate the ALS. It's much easier if all the digits are listed already. The other thing is that with loops those bystander digits are actually important because they get locked and have their own eliminations. If the bystanders are not listed but their eliminations are, then things get really confusing (or worse, those eliminations may be missed). I'm fine with anything else, but I think they should always be listed with ALS loops. It also serves an educational purpose for newcomers, as I don't think those extra ALS eliminations in loops are self-evident for everyone (I know it wasn't for me).