The New Sudoku Players' Forum

[Marty R.]

I have some questions and comments on the subject. Whether or not there is a standard method of notating, I don't know, but people do write them differently.

Let's use as an example an ALS 13579 in r1c2468.

Consider this partial string written two different ways:

1) (6=5)r7c2-(1359=7)r1c2468
2) (6=5)r7c2-(5=1379)r1c2684

As a relative newcomer to notation and a reader trying to figure out what others are doing, my preference is:

(6=5)r7c2-(5139=7)r1c2684

I prefer the multiple numbers to the left of the equal sign so the reader can immediately see that the single digit to the right of the sign is the ending number. We know the starting number is 5 based on the previous term, but the 5139 is out of sequence just to quickly show that the 5 is the end of the previous term and the beginning of the next. This, however, is not critical since even if it's written sequentially as 1359 we know it has to start with 5.

Note that the cells of the ALS are not the sequential r1c2468, but the non-sequential r1c2684. This is so the ending number of the ALS is located in the last cell in the string, i.e., r1c4. It's a little harder to figure out if it's written as r1c2468.

Finally, some people like to shorten the ALS to (5=7). I realize that shortening is a goal, but this saves only three characters and makes it more difficult for this particular reader to follow along.

I'd welcome the comments of others.

[ArkieTech]

I have some questions and comments on the subject. Whether or not there is a standard method of notating, I don't know, but people do write them differently.

Let's use as an example an ALS 13579 in r1c2468.

Consider this partial string written two different ways:

1) (6=5)r7c2-(1359=7)r1c2468
2) (6=5)r7c2-(5=1379)r1c2684

As a relative newcomer to notation and a reader trying to figure out what others are doing, my preference is:

(6=5)r7c2-(5139=7)r1c2684

I prefer the multiple numbers to the left of the equal sign so the reader can immediately see that the single digit to the right of the sign is the ending number. We know the starting number is 5 based on the previous term, but the 5139 is out of sequence just to quickly show that the 5 is the end of the previous term and the beginning of the next. This, however, is not critical since even if it's written sequentially as 1359 we know it has to start with 5.

Note that the cells of the ALS are not the sequential r1c2468, but the non-sequential r1c2684. This is so the ending number of the ALS is located in the last cell in the string, i.e., r1c4. It's a little harder to figure out if it's written as r1c2468.

Finally, some people like to shorten the ALS to (5=7). I realize that shortening is a goal, but this saves only three characters and makes it more difficult for this particular reader to follow along.

I'd welcome the comments of others.

Let's use as an example an ALS 13579 in r1c2468

You have 5 digits in 4 cells. if one digit is removed -(5 then you have a four digit locked set. 4 digits in 4 cells. -(5=1379)r1c2468

since the set is locked -(5=7) would also be correct as would -(5=1), -(5=3), or -(5=9) This seems simpler to me.

[Leren]

Marty R wrote : Note that the cells of the ALS are not the sequential r1c2468, but the non-sequential r1c2684. This is so the ending number of the ALS is located in the last cell in the string, i.e., r1c4. It's a little harder to figure out if it's written as r1c2468.

One disadvantage of this is that (in the most general case) the linking numbers in each ALS (in an ALS chain) may be in more than one cell of an ALS. That's why I prefer to use (6=5) r7c2 - (5=7) r1c2468. I then always read this as (for example, for the second ALS) : "If all of the 5's in r1c2468 are false (there may be more than one 5 in the ALS) then exactly one of the 7's in r1c2468 must be true (there may be more than one 7 in the ALS)."

Leren

[sultan vinegar]

Finally, some people like to shorten the ALS to (5=7). I realize that shortening is a goal, but this saves only three characters and makes it more difficult for this particular reader to follow along.

I like to shorten it like that because if you end up making a nice loop, all the strong links become conjugate, so exactly one of (57) is true and thus (139) in your example are all locked, and it's easier to see the ensuing eliminations I find. Plus the other fact is that often you'll continue the chain with a weak link off the (7), so it's easier to follow if you don't have the (139) getting in the way.

[daj95376]

To tell the truth, I don't recall all of the details of my solver's ALS notation. The unit(s) containing the ALS and eliminations factor into it, though.

Here is an example where the ALS is in the same unit as the resulting "traditional" naked subset, and the eliminations are restricted to [r2].

Code: Select all

 (5)r6c2 - ALS(5=123)r2c258 - <123>[r2]
 +-----------------------------------+
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  ~ 125 ~  |  ~  23 ~  |  ~  13 ~  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  . =5  .  |  .  .  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 |  .  .  .  |  .  .  .  |  .  .  .  |
 +-----------------------------------+


Here is an example where the ALS has a non-traditional naked subset and possible eliminations outside the ALS unit.

Code: Select all

 (5)r6c2 - ALS(5=2)r2c285 - (2)[b2+c5+r2]
 +-----------------------------------+
 |  .  .  .  |  ~  ~  ~  |  .  .  .  |
 |  ~ 135 ~  |  ~ 123 ~  |  ~  13 ~  |
 |  .  .  .  |  ~  ~  ~  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  ~  .  |  .  .  .  |
 |  .  .  .  |  .  ~  .  |  .  .  .  |
 |  . =5  .  |  .  ~  .  |  .  .  .  |
 |-----------+-----------+-----------|
 |  .  .  .  |  .  ~  .  |  .  .  .  |
 |  .  .  .  |  .  ~  .  |  .  .  .  |
 |  .  .  .  |  .  ~  .  |  .  .  .  |
 +-----------------------------------+


When the previous ALS is in a loop, then I manually alter my solver's output to read ALS(513=132)r2c285. This is akin to the notation for an ERI in a loop ... and justifies eliminations for <13> in [r2].

There are additional scenarios, but I think these two will help you in most cases.

Bottom Line: everyone has their preferred way of notating an ALS.