- Code: Select all
*-----------*
|...|..3|.2.|
|..9|2..|.8.|
|.2.|..1|9.6|
|---+---+---|
|...|..4|3..|
|5..|9.7|..8|
|..1|8..|...|
|---+---+---|
|2.5|3..|.6.|
|.3.|..2|1..|
|.4.|7..|...|
*-----------*
Play/Print this puzzle online
*-----------*
|...|..3|.2.|
|..9|2..|.8.|
|.2.|..1|9.6|
|---+---+---|
|...|..4|3..|
|5..|9.7|..8|
|..1|8..|...|
|---+---+---|
|2.5|3..|.6.|
|.3.|..2|1..|
|.4.|7..|...|
*-----------*
.----------------------.-------------.-------------------.
| 6 1578 478 | 45 9 3 | f457 2 1457 |
| a[34]1 b15 9 | 2 7 6 | c45 8 1345 |
| h(7)-34 2 fg347 | 45 8 1 | 9 g3457 6 |
:----------------------+-------------+-------------------:
| 789 789 2 | 1 6 4 | 3 579 579 |
| 5 6 e34 | 9 23 7 | d24 1 8 |
| a[34] 79 1 | 8 23 5 | 6 479 2479 |
:----------------------+-------------+-------------------:
| 2 179 5 | 3 14 89 | e478 6 479 |
| 789 3 78 | 6 45 2 | 1 4579 4579 |
| b19 4 6 | 7 15 c89 | d258 359 2359 |
'----------------------'-------------'-------------------'
SpAce wrote:Too hot weather for solving, I guess.
eleven wrote:I think, your last link needs to remember the 3 in r3c3.
*--------------------------------------------------------------*
| 6 g1578 g478 | 45 9 3 | af457 2 1457 |
|cg134 g15 9 | 2 7 6 | b5-4 8 1345 |
| 347 2 347 | 45 8 1 | 9 3457 6 |
*--------------------+-------------------+---------------------|
| 789 789 2 | 1 6 4 | 3 579 579 |
| 5 6 bi34 | 9 23 7 | aj24 1 8 |
|ch34 79 1 | 8 23 5 | 6 479 2479 |
*--------------------+-------------------+---------------------|
| 2 179 5 | 3 14 89 | af478 6 479 |
| 789 3 78 | 6 45 2 | 1 4579 4579 |
| c19 4 6 | 7 15 d89 | e258 359 2359 |
*--------------------------------------------------------------*
SteveG48 wrote:Yuck.
4r157c7 = (4*)r5c3&(4#)r2c7 - (4=139)r269c1 - (9=8)r9c6 - r9c7 = (78)r17c7 - (7|*4|#4=1358)b1p2345 - (3=4)r6c1 - r5c3 = 4r5c7 => -4 r2c7 ; stte
.---------------------.-------------.--------------------.
| 6 c1578 c478 | 45 9 3 | d457 2 1457 |
| f13(4) c15 9 | 2 7 6 | 5-4 8 1345 |
| 347 2 347 | 45 8 1 | 9 3457 6 |
:---------------------+-------------+--------------------:
| 789 789 2 | 1 6 4 | 3 579 579 |
| 5 6 bh34 | 9 23 7 | ah2[4] 1 8 |
| g34 79 1 | 8 23 5 | 6 479 2479 |
:---------------------+-------------+--------------------:
| 2 179 5 | 3 14 89 | d478 6 479 |
| 789 3 78 | 6 45 2 | 1 4579 4579 |
| e19 4 6 | 7 15 e89 | d258 359 2359 |
'---------------------'-------------'--------------------'
Double Kraken AALS (58+147)b1p235 & AALS (134)r2c1
(1)r12c2 --------------------------------
|| \
(7)r1c23 - r1c7 = (78)r79c7 - (8=91)r9c61 - (1)r2c1
|| ||
|| (3)r2c1 - r6c1 = (3-4)r5c3 = (4)r5c7
|| ||
|| (4)r2c1
(4)r1c3 - r5c3 = (4)r5c7
=> -4 r2c7; stte
*-------------------------------------------------------------*
| 6 e1578 e478 | 45 9 3 |af457 2 1457 |
|cd134 e15 9 | 2 7 6 | b5-4 8 1345 |
| d347 2 347 | 45 8 1 | 9 3457 6 |
*--------------------+-------------------+--------------------|
| 789 789 2 | 1 6 4 | 3 579 579 |
| 5 6 b34 | 9 23 7 |af24 1 8 |
|cd34 79 1 | 8 23 5 | 6 479 2479 |
*--------------------+-------------------+--------------------|
| 2 179 5 | 3 14 89 | a478 6 479 |
| 789 3 78 | 6 45 2 | 1 4579 4579 |
| e19 4 6 | 7 15 e89 | f258 359 2359 |
*-------------------------------------------------------------*
SteveG48 wrote:I like this variant on mine better than my original:
4r157c7 = r2c7&r5c3 - 4r26c1 = 4r3c1&(13)r26c1 - (1|4=578)b1p235&(89)r9c26 - (7|8=245)r159c7 = -4 r2c7 ; stte
SteveG48 wrote:I personally prefer to just list the candidates in numerical order, but I don't think it's important one way or the other. I think my preference came from a discussion with David P Bird, but I'm not sure of that.
David P Bird wrote:As you have discovered, tying to read memory chains in the reverse direction requires fortune foretelling abilities!
SpAce wrote:Me too, but you can't write the colored part like that (and I'm not talking about the minor typo -> r9c16). A valid counter-argument is that you just did, but as previously discussed (and I think you agreed), it's ambiguous. Should be:
... - (1|4)b1p235,r9c1 = (578)b1p235&(89)r9c16
It would be interesting to know what David's argument for that style might have been. To me it's the hardest to read of all possibilities (including omitted bystanders, though I don't like that either).
While not critical (as in making anything incorrect), I don't quite agree that it's totally unimportant. Having the linking digits on the proper sides keeps the chain flowing more naturally both ways, imho. That being said, you're of course most welcome to stick to your own preference. I won't stop reading your chains because of it
SteveG48 wrote:I wonder more and more why I don't prefer the omitted bystander option.
Sometimes including the bystanders is downright confusing. In those cases, I'll omit them.
The 7/28 puzzle is an example if I recall from earlier today.
SteveG48, July 28 wrote:(3=567)r289c1 - 5r4c1 = (35)b5p1238 => -3 r3c5 ; stte
I wrote:(3=675)r389c1 - r4c1 = (53)b5p1238 => -3 r3c5; stte
Cenoman tends to put some bystanders on the left side of the link in situations where I'd put them on the right side. I can't argue for either as being preferable.
The argument that I've seen for including bystanders is based on clarity. It's presumed that including them makes things easier to understand, particularly for newcomers. I just don't know if that's true. I really don't know if leaving them out would have made things easier or harder for me when I was first learning the notation.
And there's an argument for just leaving the bystanders out. The chain certainly flows naturally when only the relevant things are there.
SpAce wrote:SteveG48 wrote:I wonder more and more why I don't prefer the omitted bystander option.
Well, that wasn't on my wish list It's the second worst option, as far as I'm concerned. Or maybe the worst. Depends on the day. It's also much farther from your current style than mine is. Why do you hate "my" way (also the standard Eureka convention, and probable preference of most) so much that you'd rather consider such a drastic move?
I'd really love to hear your reasons for disliking the standard in this case. I'm all for kicking standards and conventions in the face when they don't make sense, but I strongly think this one does. I've still missed seeing your opposing argument. (Not that you need one -- I accept if you simply state that it's your preference, period. I'm just extremely confused if you now start speaking for the omitted-bystanders side. That makes no sense.)
SteveG48 wrote:I really don't dislike the standard. Obviously, it's what I use now and have used for a long time.
Sudopedia wrote:When an embedded ALS is part of the chain, the digit linked to the previous node is isolated from the remaining digits with a strong link symbol. The remaining digits are placed in such an order that the digit linked to the next node is the last one.
(1)r5c4-(1=264)r5c789-(4)r4c8
I just question my own preferences from time to time.
Leaving the bystanders out clearly makes for a shorter notation. People here were calling it "abbreviated" notation for awhile. The reaction against it was based on clarity.
The clarity argument is important to me, but I began wondering whether the standard was really clearer to most people than the abbreviated. Apparently it is to you, and that's a good data point for me. I don't anticipate changing.
SpAce wrote:The standard style is the ONLY one that makes sense with loops because both bystanders and linking digits have their own eliminations, and it's practically impossible to depict that with any other digit ordering (and definitely impossible by omitting the bystanders). Can you deny that? It's funny that I've made that loop argument many times but don't remember receiving any response from anyone.
...
I'm especially curious why my loop argument has been ignored by everyone thus far. I really don't see how you'd write an understandable ALS loop (with bystander eliminations) with either your style or with omitted bystanders.
SpAce wrote:My real question was about your current style and why you prefer it over the "standard" digit ordering. I'm still missing your reasons.
SpAce wrote:The standard style is the ONLY one that makes sense with loops because both bystanders and linking digits have their own eliminations, and it's practically impossible to depict that with any other digit ordering (and definitely impossible by omitting the bystanders). Can you deny that? It's funny that I've made that loop argument many times but don't remember receiving any response from anyone.