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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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 Posted: Thu Jun 19, 2008 2:30 am Post subject: ending proves challenging |
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| Code: | 3 . .|. 8 .|. 7 .
2 6 .|7 . .|. . 4
. . 9|. . 6|. . .
-----+-----+-----
9 2 .|. 1 .|. . .
. 7 .|. . .|. 9 .
. . .|. 2 .|. 4 8
-----+-----+-----
. . .|5 . .|1 . .
1 . .|. . 8|. 6 9
. 5 .|. 9 .|. . 3 |
| Code: | .---------------------.---------------------.---------------------.
| 3 14 145 | 12 8 125 | 9 7 6 |
| 2 6 15 | 7 3 9 | 8 15 4 |
| 7 8 9 | 14 45 6 | 3 15 2 |
:---------------------+---------------------+---------------------:
| 9 2 468 | 468 1 47 | 67 3 5 |
| 4568 7 3468 | 3468 456 345 | 2 9 1 |
| 56 13 136 | 9 2 357 | 67 4 8 |
:---------------------+---------------------+---------------------:
| 468 9 23468 | 5 46 234 | 1 28 7 |
| 1 34 234 | 234 7 8 | 5 6 9 |
| 68 5 7 | 126 9 12 | 4 28 3 |
'---------------------'---------------------'---------------------' |
nothing nice to look at here. some coloring, wing, but the ending proves challenging. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Jun 19, 2008 6:17 am Post subject: |
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coloring - yes I did that (skyscraper 3, something on 6 can't remember exactly)
wing - yes, did that, too (xyz-wing 4,6,8 )
challenge - indeed:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 3 14 145 | 12 8 125 | 9 7 6 |
| 2 6 15 | 7 3 9 | 8 15 4 |
| 7 8 9 | 14 45 6 | 3 15 2 |
+--------------------------+--------------------------+--------------------------+
| 9 2 68 | 68 1 4 | 7 3 5 |
| 48 7 468 | 368 56 35 | 2 9 1 |
| 5 13 13 | 9 2 7 | 6 4 8 |
+--------------------------+--------------------------+--------------------------+
| 468 9 2348 | 5 46 23 | 1 28 7 |
| 1 34 234 | 234 7 8 | 5 6 9 |
| 68 5 7 | 126 9 12 | 4 28 3 |
+--------------------------+--------------------------+--------------------------+
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My friend Medusa told me this:
starting at r1c2=1, we can look at row 1 box 2 and find: r1c2=1 => r3c4=1
But we can also look at col 2 and find r1c2=1 => r8c2=4 => r8c4<>4 r3c4=4.
A contradiction (Medusa trap), thus r1c2=4.
Hm, there should be a better way, let's see what we can do without Medusa:
There is a UR type ? (maybe type 6?), anyway:
potential DP 12 in r19c46 and strong link on 1 in r9, thus r1c6<>1
(if r1c6=1 then r1c4 and r9c6=2 and r9c4=1 which is the DP)
After that, some naked singles and then:
| Code: |
+--------------------------+--------------------------+--------------------------+
| 3 14 145 | 12 8 25 | 9 7 6 |
| 2 6 15 | 7 3 9 | 8 15 4 |
| 7 8 9 | 14 45 6 | 3 15 2 |
+--------------------------+--------------------------+--------------------------+
| 9 2 68 | 68 1 4 | 7 3 5 |
| 48 7 468 | 368 56 35 | 2 9 1 |
| 5 13 13 | 9 2 7 | 6 4 8 |
+--------------------------+--------------------------+--------------------------+
| 468 9 2348 | 5 46 23 | 1 28 7 |
| 1 34 234 | 234 7 8 | 5 6 9 |
| 68 5 7 | 26 9 1 | 4 28 3 |
+--------------------------+--------------------------+--------------------------+
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From here, a number of short xy-chains finish the puzzle.
I used another UR instead: to avoid the potential DP 68 in r45c34, either r5c3=4 or r5c4=3. Both cases very quickly lead to r4c3=6. There is one more xy-wing (36-26-23 in boxes 5 and 8 ) and we are done. |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Thu Jun 19, 2008 8:49 am Post subject: |
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| nataraj wrote: | There is a UR type ? (maybe type 6?), anyway:
potential DP 12 in r19c46 and strong link on 1 in r9, thus r1c6<>1 |
Not Type 6, but akin to that or Type 4 because it exploits one or more internal strong links. By exactly the same logic, due to the strong link on 2 in r1, r9c4<>2. And, with both of these eliminations, the puzzle is solved!
However, if we ignore that very helpful 12 UR, then this puzzle provides good examples of how strong (inference) links induced by DPs can be exploited in AICs.
For instance, that 68 UR in r45c34: Rather than testing the results of the extra <3> and <4>, we note that they are strongly linked (one or both must be true) due to the UR:
(3)r5c4=(4)r5c3
and incorporate this link into what is otherwise an XY Chain:
(6=5)r5c5 - (5=3)r5c6 - UR[(3)r5c4=(4)r5c3] - (4=8)r5c1 - (8=6)r4c3;
r5c3|r4c4<>6
For me, that is definitely more satisfying that testing those "extra" digits.
After those <6> eliminations, we arrive here:
| Code: | +---------------+--------------+----------+
| 3 14 145 | 12 8 125 | 9 7 6 |
| 2 6 15 | 7 3 9 | 8 15 4 |
| 7 8 9 | 14 45 6 | 3 15 2 |
+---------------+--------------+----------+
| 9 2 6 | 8 1 4 | 7 3 5 |
| 48 7 48 | 36 56 35 | 2 9 1 |
| 5 13 13 | 9 2 7 | 6 4 8 |
+---------------+--------------+----------+
| 468 9 2348 | 5 46 23 | 1 28 7 |
| 1 34 234 | 234 7 8 | 5 6 9 |
| 68 5 7 | 126 9 12 | 4 28 3 |
+---------------+--------------+----------+ |
The 48 UR in r57c13 induces the following strong inference:
(6)r7c1=({23})r7c3
If we look to the right in r7, we see that there is a 238 ALS in r7c68. As with any ALS (such as the familiar bivalue), there is an internal strong inference. Here, I want to consider ({23}=8) in the r7c68 ALS, which I'll notate as:
ALS[({23})r7c68=(8)r7c8]
I can weakly link the matching grouped digits, {23}, provided they can all see each other, which they can in this case since they all share r7. This gives me a short and, otherwise, simple AIC:
(8=6)r9c1 - UR[(6)r7c1=({23})r7c3] - ALS[({23})r7c68=(8)r7c8];
r9c8<>8
This, too, solves the puzzle.
Note: If you're trying to follow this and the weakly linked grouped digits aren't making sense to you, you might think of it this way... either the <2>s are weakly linked or the <3>s are weakly linked, we just don't know which it is and don't need to know for the inferences to hold.
In other words, if you follow the inferences in the chain, then if r7c1 is not <6>, then r7c3 must be either <2> or <3>. Whichever it is, that same digit cannot also be true in the r7c68 ALS and, thus, it will contain <8> no matter what. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Thu Jun 19, 2008 10:37 am Post subject: |
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There are several ways to use this DP 48.
Looking at the strong links (4 in c1, 8 in c3) you immediately can eliminate 4 from r7c1 and 8 from r7c3. But it does not get you far.
The same result you get, if you look, where 48 can be columns 13 else:
r9c1=8 or r18c3=4 (=> r5c3=8)
An alternative to see Asellus' elimination is to look, where 48 else can be rows 57:
r7c8=8 or r7c5=4 (=> r9c4=6 => r9c1=8)
I also looked at the 14/13/34 DP in columns 23, but only found, that r8c3<>3. |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Thu Jun 19, 2008 2:42 pm Post subject: Final |
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xyz-wing (468) in boxes 7&8.
UR (12) in boxes 2&8, where either solution has <1> in R9C6.
xy-chain with <6> pincers at R9C4 and R5C5 finishes the puzzle.
Earl |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Jun 19, 2008 4:16 pm Post subject: |
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| Asellus wrote: | By exactly the same logic, due to the strong link on 2 in r1, r9c4<>2. And, with both of these eliminations, the puzzle is solved!
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Quite! And isn't this another good example of how UR eliminations can be destroyed by solving a few cells and how important it is to look at ALL possible eliminations before proceeding.
Thanks.
The use of the other DP as a strong link is indeed much more satisfying. I never thought of it that way. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Jun 19, 2008 6:02 pm Post subject: |
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| nataraj wrote: | | Asellus wrote: | By exactly the same logic, due to the strong link on 2 in r1, r9c4<>2. And, with both of these eliminations, the puzzle is solved!
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Quite! And isn't this another good example of how UR eliminations can be destroyed by solving a few cells and how important it is to look at ALL possible eliminations before proceeding.
Thanks.
The use of the other DP as a strong link is indeed much more satisfying. I never thought of it that way. |
that will hopefully be the next rung of my learning ladder. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Jun 20, 2008 6:07 am Post subject: |
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| ravel wrote: | There are several ways to use this DP 48.
Looking at the strong links (4 in c1, 8 in c3) you immediately can eliminate 4 from r7c1 and 8 from r7c3. |
I must admit that for me it was not at all immediately that I understood what ravel is talking about here. At first I went like, ok this is just another one of these not quite type sixes, but then it occured to me that the two 48 cells are not diagonal but aligned on a row, just like in a plain old type 1 or 2.
I had to go through several iterations of incorrect logic until I arrived at this explanation:
If r7c3=8 then r5c3=4 then r5c1=8 then (because of the strong link 4 in col 1) r7c1=4 which is the deadly pattern.
So it seems that if there is a strong link in one of the two candidates of a UR situation, we can always find additional possibilities to remove one of the two UR candidates
| Code: |
case 1:
ab ab
--------
a-bx a-by ... strong link a in row
case 2:
ab ab
--------
abx a-by
.
. strong link a in column
.
case 3:
abx ab
--------
ab -aby
.
. strong link a in column
.
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case 1: ab,ab cells on same row, strong link in row
because of the strong link, one of the candidates (a) MUST appear in the two bottom cells of the rectangle. If if the other candidate appeared as well, we'd have a deadly pattern. It is possible to remove b from BOTH cells in the bottom row of the UR.
case 2: ab,ab cells on same row, strong link in column 1
We start with the cell in the column opposite the one with the strong link and choose the cell with more than 2 candidates.
If this (bottom right) cell were b (the OTHER candidate, not the one we have a strong link on!), then the top right cell must be a, then the top left cell must be b (both cells only have 2 possibilities).
Now the strong link on a in the left column comes into play: if the top left cell is NOT a, then the bottom left cell must be a (this is one of the definitions of a strong link).
Again this is the deadly pattern and we can therefore remove b from the bottom right cell.
case 3: ab,ab cells are diagonally opposite each other. Strong link in column 1
Again we start with the column opposite the strong link and choose the cell with more than one candidate (bottom left cell). This time we choose the SAME candidate (a) that we have the strong link in.
If bottom right cell is a, then bottom left cell is b (as is top right).
Same strong link logic as before: if bottom left is NOT a, then top left is a.
This would be the deadly pattern and we can thus remove a from the bottom right cell.
----
I am sure many of you guys are already very familiar with this and my thanks go to ravel, Asellus and keith for my continuing education on URs. I had to to go through the logic step by step and write it down even if it is immediately obvious to some ...
Hope I got it right
_____
"The rain in Spain falls mainly in the plains" - "I think she's GOT IT!"
"You teach best what you most need to learn" Richard Bach, "Illusions - The Adventures of a Reluctant Messiah" |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Jun 20, 2008 8:35 pm Post subject: |
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Some might find it interesting to see how the UR induced strong links in nataraj's three UR cases work in AIC terms for these eliminations.
I will suppose the URs are located at r12c34. In all cases we have an induced x=y strong link. x and/or y may be grouped digits (such as {257}), it doesn't matter.
Case 1: (This is just Type 4)
AIC Loop: (a-x)r2c3=(y-a)r2c4=(a-x)r2c3, etc.; r2c34<>b
All of the AIC links in this tiny loop become conjugate, squeezing the b's out of the two cells.
Case 2:
(b-y)r2c4=(x-a)r2c3=(a)r1c3-(a=b)r1c4-(b)r2c4; r2c4<>b
Case 3: ("Type 6A")
(a)r2c4-(a)r2c3=(a-x)r1c3=(y-a)r2c4; r2c4<>a
This may or may not help some folks see the eliminations. But, it is interesting nonetheless. |
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