View previous topic :: View next topic |
Author |
Message |
Victor
Joined: 29 Sep 2005 Posts: 207 Location: NI
|
Posted: Mon Dec 10, 2007 9:57 am Post subject: ALS? |
|
|
Menneske 794832.
Code: |
+-------+-------+-------+
| . 4 . | . . . | . 5 . |
| 7 . . | . . . | 4 . 1 |
| 9 . 6 | 1 . . | . . 2 |
+-------+-------+-------+
| 4 5 . | . . . | 7 . . |
| . 7 . | . 3 . | . 9 . |
| . . 9 | . . . | . . 4 |
+-------+-------+-------+
| . 2 . | 5 8 . | . 4 . |
| . . . | . . . | . . 8 |
| . . . | . . 9 | 1 . . |
+-------+-------+-------+
|
Play this puzzle online at the Daily Sudoku site
Took me ages, until I spotted an ALS, which I'm not good at. (It's the simple 2-set kind, called xz rule.) Any other methods?
PS. Why's it xz? - rather than say xy. |
|
Back to top |
|
|
re'born
Joined: 28 Oct 2007 Posts: 80
|
Posted: Mon Dec 10, 2007 10:46 am Post subject: |
|
|
Code: | .---------------------.---------------------.---------------------.
| 12 4 12 | 3678 67 3678 | 3689 5 369 |
| 7 38 5 | 23689 269 2368 | 4 368 1 |
| 9 38 6 | 1 45 45 | 38 7 2 |
:---------------------+---------------------+---------------------:
| 4 5 238 | 2689 269 268 | 7 1 36 |
| 268 7 28 | 4 3 1 | 2568 9 56 |
| 2368 1 9 | 2678 2567 25678 | 2368 2368 4 |
:---------------------+---------------------+---------------------:
| 13 2 137 | 5 8 367 | 369 4 3679 |
| 35 9 347 | 2367 1 23467 | 2356 236 8 |
| 358 6 3478 | 237 247 9 | 1 23 357 |
'---------------------'---------------------'---------------------' |
(6)r5c1 = (6-3)r6c1 = (3)r4c3 - (3=6)r4c9, => r5c79<>6, solving the puzzle. |
|
Back to top |
|
|
ravel
Joined: 21 Apr 2006 Posts: 536
|
Posted: Mon Dec 10, 2007 4:00 pm Post subject: |
|
|
Suppose, this is the same as Victor's ALS with A {2368} in r4c3, r5c13, B {36} in r4c9, x=3, z=6, r5c79<>6.
I dont know, why Benny chose x and z to denote that and also cannot see a good alternative for this puzzle.
In other words:
r5c79=6 => r4c9=3
r5c79=6 => r5c13=28 => r4c3=3
re'born, i still dont understand the notation, maybe you can explain it with this line. |
|
Back to top |
|
|
Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England
|
Posted: Mon Dec 10, 2007 4:46 pm Post subject: |
|
|
In his original paper on ALSs Benny S considered pairs A and B with a restricted common candidate x and also triples A, B, C with restricted common candidates x for (A, B) and y for (A, C). In both cases z was used for the candidate to be eliminated. Custom has followed this in calling the first rule xz and the second xyz. xz is convenient, I think, in that there is no risk of a beginner inferring a connection with an xy-chain.
The traditional way of writing re’born’s chain is
r5c1 ≠ 6 => r6c1 = 6 => r6c1 ≠ 3 => r4c3 = 3 => r4c9 = 6
Steve |
|
Back to top |
|
|
ravel
Joined: 21 Apr 2006 Posts: 536
|
Posted: Mon Dec 10, 2007 5:40 pm Post subject: |
|
|
Thanks, i saw that here r6c1 was used instead of r4c3 (which was easier to spot for me than the 2 strong links).
But it is still confusing for me, that - reading it from left to right -
(6)r5c1 means r5c1<>6 and (3)r4c3 then means r4c3 = 3.
So i cannot find out, what the symbols stand for.
Is there a good link, where it is explained ? |
|
Back to top |
|
|
Steve R
Joined: 24 Oct 2005 Posts: 289 Location: Birmingham, England
|
Posted: Mon Dec 10, 2007 5:51 pm Post subject: |
|
|
I don’t know of a link. Perhaps wrongly, I read these things like nice loops: it’s only an alternative notation after all.
Thus I would see:
(3)r4c3 as meaning r4c3 ≠ 3
- (3)r4c3 as meaning r4c3 ≠ 3
=(3)r4c3 as meaning r4c3 = 3
Steve |
|
Back to top |
|
|
nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
|
Posted: Mon Dec 10, 2007 6:48 pm Post subject: |
|
|
It is an AIC (alternating inference chain).
Quote: | (6)r5c1 = (6-3)r6c1 = (3)r4c3 - (3=6)r4c9, => r5c79<>6 |
shows alternating strong and weak links writen as "=" and "-" resp.
(6)r5c1 means the statement r5c1=6
(it is confusing that the same character "=" is used both for equal and strong link, that is why the notation reserves = for strong link and puts the candidates in parentheses)
r5c1=6 strong link r6c1=6 weak link r6c1=3 strong ... etc.
strong link means that at least one of the two statements must be true
(alternatively think "if not one, then the other"), weak links means they cannot be both true.
the chain must begin and end with a strong link and have alternating strong and weak links throughout. Then any candidate that "sees" both the beginning and the end of the chain can be eliminated.
Read as a sentence, the chain means: if r5c1<>6 then r6c1=6 then r6c1<>3 then r4c3=3 then r4c9<>3 then r4c9=6
(the chain can also be read from the other end, by the way)
Together it works just like the pincers in your old xy wing (if a is not 6 then b is 6) and it means any cell that sees the two ends cannot contain 6, in our case r5c79
link for the notation: http://www.sudopedia.org/wiki/Eureka
edited for meaning of AIC ("inference" chain) and the sudopedia link
Last edited by nataraj on Mon Dec 10, 2007 8:28 pm; edited 2 times in total |
|
Back to top |
|
|
ravel
Joined: 21 Apr 2006 Posts: 536
|
Posted: Mon Dec 10, 2007 6:51 pm Post subject: |
|
|
Thanks nataraj, at the same time i found it by myself
Ah, then maybe i got it. It says a bit more than a nice loop (though the result is the same).
So '=' means strong links either between cells or within a cell and '-' weak links.
Then like in x-chains (and AIC's ?) the 1st, 3rd, ...., last link must be strong (if not left side then right side and vice versa). Then everything left of '=' is equivalent and an either or to everything right of '='. |
|
Back to top |
|
|
nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
|
Posted: Mon Dec 10, 2007 7:04 pm Post subject: |
|
|
ravel wrote: | . Then everything left of '=' is equivalent and an either or to everything right of '='. |
not quite.
everything: no. only the statement left/right of =
equivalent: no. The = means strong link and has nothing to do with equivalence. There is an assertion to the left that can be true or false.
either/or,no. Mostly it will be like you said.
But strong link really leaves open the possibility that both are true.
Not in the simple case of a bi-value cell but there are more exotic strong links.
For example, the chain we discussed a minute ago constitutes one single strong link between first and last cell (if not a then b). And it is quite possible that both ends of a chain may end up with the same candidate. |
|
Back to top |
|
|
nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
|
Posted: Mon Dec 10, 2007 7:15 pm Post subject: |
|
|
nataraj wrote: | And it is quite possible that both ends of a chain may end up with the same candidate. |
oops I did it again.
both ends of the chain are statements, not cells.
The sentence should read: ...possible that both statements at the end of a chain end up as being true.
Like for example:
(3)r1c4=....=(5)r1c9;r1c4<>5
which in plain English reads: if r1c4 is not 3, then r1c9=5, either way 5 in r1c4 is toast. But (since the chain is symmetrical) it might be better to say "either r1c4=3 or r1c9=5 or both" |
|
Back to top |
|
|
Victor
Joined: 29 Sep 2005 Posts: 207 Location: NI
|
Posted: Wed Jan 02, 2008 9:11 pm Post subject: |
|
|
Thanks re'born & others. I do get all that, tho' whether I'll be able to spot what I suppose is a sort of non-specific AIC (i.e. not a named structure) is another story!
And thanks Steve, for the x / z explanation. |
|
Back to top |
|
|
Asellus
Joined: 05 Jun 2007 Posts: 865 Location: Sonoma County, CA, USA
|
Posted: Thu Jan 03, 2008 12:52 am Post subject: |
|
|
While this thread is a little old, no one wrote out the Eureka notation for the ALS solution described above. (Re'born's chain, while having the same result, is different since it involves r6c1.) Here is how the ALS can be expressed using Eureka:
(6)r5c79-(6={238})r5c13|r4c3-(3=6)r4c9-(6)r5c79; r5c79<>6
The "|" character means "and" and is used to concatenate cells, etc. The {2368} ALS in r5c13|r4c3 (Set "A", above) is considered as a <6> strongly linked with a {238} locked set. The {238} locked set is, in turn, weakly linked to the <3> at r4c9. (The link is conjugate, which is both strong and weak for inference purposes.) So, it can be expressed in ordinary language as:
If r5c7 or r5c9 is <6>, then the r5c13|r4c3 ALS becomes a {238} locked set (since the <6> is false), so r4c9 is not <3> so must be <6>, so r5c7 or r5c9 cannot be <6>. This contradiction means that the initial assumption is false, so r5c79<>6.
The Eureka notation make it apparent that the shared exclusive <3> is weakly linked between the two ALS's and the shared common <6> is strongly linked within the two ALS's to the complementary locked sets. |
|
Back to top |
|
|
storm_norm
Joined: 18 Oct 2007 Posts: 1741
|
Posted: Thu Jan 03, 2008 6:46 am Post subject: |
|
|
asellus wrote:
Quote: | The "|" character means "and" and is used to concatenate cells, etc
|
looked up "concatenate" in the old dictionary... ehem...
that word was made for sudoku, hands down. don't care what others say. |
|
Back to top |
|
|
|