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ravel
Joined: 21 Apr 2006
Posts: 536
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 Posted: Thu Dec 13, 2007 5:11 pm Post subject: Next from Pattern Game |
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By g.r.emlin
| Code: | +-------+-------+-------+
| . 9 . | . 2 . | 5 . . |
| . . 4 | . . 5 | . 1 . |
| . 6 . | . . . | . . 3 |
+-------+-------+-------+
| . . 1 | 8 . . | . 6 . |
| . . . | 9 . . | . . 2 |
| . 8 . | . 7 2 | . . . |
+-------+-------+-------+
| 5 . . | . . 1 | . 7 . |
| . . . | 3 . . | 9 . 1 |
| . . 3 | . . . | . . . |
+-------+-------+-------+ |
After coloring in 9 and a harder step showing that r1c6 or r6c1 must be 3 i came to this nice grid.
| Code: | *----------------------------------------------*
| 17 9 8 | 167 2 3 | 5 4 67 |
| 237 23 4 | 67 89 5 | 268 1 679 |
| 127 6 5 | 4 189 79 | 28 89 3 |
|--------------+---------------+---------------|
| 239 235 1 | 8 35 4 | 7 6 59 |
| 34 35 7 | 9 135 6 | 148 38 2 |
| 349 8 6 | 15 7 2 | 14 39 459 |
|--------------+---------------+---------------|
| 5 4 9 | 2 6 1 | 3 7 8 |
| 6 7 2 | 3 4 8 | 9 5 1 |
| 8 1 3 | 57 59 79 | 46 2 46 |
*----------------------------------------------* |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Dec 13, 2007 6:37 pm Post subject: |
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| One technique, applied three times, but in two different ways, does the trick. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Dec 13, 2007 8:35 pm Post subject: |
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| Quote: | | ... i came to this nice grid. |
Not so hard from here. 2 xy-chains sharing some cells:
(3)r4c5 35-59-79-79-89-38 (3)r5c8 ; r5c5<>3
(5)r5c2 35-38-89-79-79-59 (5)r9c5 ; r5c5<>5
r5c5=1.
BUT ... getting there ?
From the initial puzzle,
after some coloring removed
5 from r6c8,
9 from r6c3, r7c5, r9c1
and 6 from r2c5,
this is it. Stuck right here:
| Code: | +--------------------------+--------------------------+--------------------------+
| 137 9 8 | 167 2 367 | 5 4 67 |
| 237 23 4 | 67 389 5 | 2678 1 6789 |
| 127 6 5 | 147 1489 4789 | 278 289 3 |
+--------------------------+--------------------------+--------------------------+
| 2349 235 1 | 8 345 34 | 47 6 4579 |
| 34 35 7 | 9 13456 346 | 148 358 2 |
| 349 8 6 | 145 7 2 | 14 39 459 |
+--------------------------+--------------------------+--------------------------+
| 5 4 9 | 2 68 1 | 3 7 68 |
| 68 7 2 | 3 4568 468 | 9 58 1 |
| 68 1 3 | 567 5689 6789 | 2468 258 4568 |
+--------------------------+--------------------------+--------------------------+
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Some bad voodoo involving multiple URs ? Or plain old shortsighted me ...
Please shed some light. Thx. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Dec 13, 2007 9:32 pm Post subject: |
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| Code: | +--------------------------+--------------------------+--------------------------+
| 137 9 8 | 167 2 367 | 5 4 67 |
| 237 23 4 | 67 389 5 | 2678 1 6789 |
| 127 6 5 | 147 1489 4789 | 278 289 3 |
+--------------------------+--------------------------+--------------------------+
| 2349 235 1 | 8 345 34# | 47 6 4579 |
| 34 35 7 | 9 13456 346# | 148 358 2 |
| 349 8 6 | 145 7 2 | 14 39 459 |
+--------------------------+--------------------------+--------------------------+
| 5 4 9 | 2 68# 1 | 3 7 68 |
| 68 7 2 | 3 4568 468A | 9 58 1 |
| 68 1 3 | 567 5689 6789B | 2468 258 4568 |
+--------------------------+--------------------------+--------------------------+ |
nataraj,
I marked the cells that I want you to concentrate on
especially r89c6
lets make:
r8c6 ( A ) | r9c6 ( B )
there is a couple combination of numbers that won't work in these two cells and we can eliminate them thus eliminating a number from r9c6
like this:
lets make A=8 and B=6 : no good, would make r7c5 =nothing
make A=4 | B= 6 : no good, then r45c6 would both equal 3
and A and B can't both equal 6
this eliminates 6 as a candidate from r9c6 because it can't be paired with any number in r8c6 |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Thu Dec 13, 2007 9:36 pm Post subject: |
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From ravel's posted grid there are some potential DP's which can be used in several ways to solve the puzzle, using the potential [67] DP* in R12C49, either R1C4=1 or R2C9=9 or both for avoiding the DP.
When R2C9=9 => R3C8=8 => R3C7=2 => R2C7=6 => R2C4=7
=> R4C9=5 => R6C9=4 => R6C7=1 => R6C4=5 => R9C4=7
Now we have a contradiction in C4,two <7>'s in the same column, this means that R1C4 must be <1>
| Code: |
*----------------------------------------------*
| 17 9 8 |*167 2 3 | 5 4 *67 |
| 237 23 4 |*67 89 5 | 268 1 *679 |
| 127 6 5 | 4 189 79 | 28 89 3 |
|--------------+---------------+---------------|
| 239 235 1 | 8 35 4 | 7 6 59 |
| 34 35 7 | 9 135 6 | 148 38 2 |
| 349 8 6 | 15 7 2 | 14 39 459 |
|--------------+---------------+---------------|
| 5 4 9 | 2 6 1 | 3 7 8 |
| 6 7 2 | 3 4 8 | 9 5 1 |
| 8 1 3 | 57 59 79 | 46 2 46 |
*----------------------------------------------*
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Dec 13, 2007 9:41 pm Post subject: |
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wow!
that's nifty. Thanks Norm!
The method sounds very similar to APE (aligned pair exclusion) just not only with pairs. I never thought I'd run into one of these babies in normal sudokus. 
edit - now the 6 is gone from r9c6, but I did not make any significant progress. I'll let it rest until tomorrow ...
Last edited by nataraj on Thu Dec 13, 2007 9:50 pm; edited 1 time in total |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Thu Dec 13, 2007 9:45 pm Post subject: |
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xy-wing chains come to mind too. they can help in looking for these pair exclusions.
it gets harder when you are looking for a multitude of candidates, I not like |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Thu Dec 13, 2007 10:00 pm Post subject: |
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I think I've got it. Some sort of wing with coloring to bridge the gap between the pincers:
(4=3)r4c6-(3)r1c6=(3)r1c1-(3=4)r5c1;r4c1,r5c56<>4
let's see what it does to the puzzle ... |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Fri Dec 14, 2007 4:33 am Post subject: |
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I agree with nataraj that one should really begin from the start, especially since ravel doesn't explain the bit with those <3>s. The initial grid after basics is rather a mess. That <9> elimination in r6c3 can be seen in two ways, marked below.
| Code: | +------------------+--------------------+-------------------+
| 1378 9 78 |e1467 2 e34678 | 5 48 4678 |
| 2378 23 4 |e67 3689g e5 | 2678 1 6789r |
| 1278 6 5 | 147 1489 4789 | 2478 *2489g 3 |
+------------------+--------------------+-------------------+
| 23479 235 1 | 8 345 34 | 47 6 4579 |
| 3467 35 67 | 9 13456 346 | 1478 3458 2 |
| 3469 8 6-9 | 1456 7 2 | 14 *3459r 459 |
+------------------+--------------------+-------------------+
|e5 e4 2689G |e26 689R e1 | 3 7 68 |
|e68 e7 268 |e3 4568 e468 | 9 2458 1 |
| 689 1 3 | 24567 45689 46789 | 2468 2458 4568 |
+------------------+--------------------+-------------------+ |
One way is to use the strong link in c8, marked *, and the 3 ERs in Box 2, then Box 8, then Box 7. r6c3 sees r8c6 directly and r3c8 via those ERs.
Or, this is a Color Wing (Multi-Coloring) with a weak link R-g "Bridge" in c5 that eliminates using r-G, which r6c3 sees from r6c8 and r7c3.
At some point, this puzzle seems to demand a complex step. That might as well be with the <3>s. I don't know how ravel did it, but here's a way:
| Code: | +------------------+--------------------+-------------------+
|F13f7 9 8 | 167 2 367 | 5 4 67 |
|F23f7 G23t 4 | 67 A3t689 5 | 2678 1 B6789t |
| 127 6 5 | 147 1489 4789 | 278 C289f 3 |
+------------------+--------------------+-------------------+
| 2349 235 1 | 8 345 34 | 47 6 4579 |
| 34 35 7 | 9 13456 346 | 148 358 2 |
|E3t49 8 6 | 145 7 2 | 14 D359t 459 |
+------------------+--------------------+-------------------+
| 5 4 9 | 2 68 1 | 3 7 68 |
| 68 7 2 | 3 4568 468 | 9 58 1 |
| 68 1 3 | 567 5689 6789 | 2468 258 4568 |
+------------------+--------------------+-------------------+ |
First, in Eureka:
(3-9)r2c5=(9)r2c9-(9)r3c8=(9-3)r6c8=(3)r6c1-(3)r12c1=(3)r2c2-(3)r2c5; r2c5<>3
If that is meaningless to you, then follow along with the marked grid starting with cell A and proceeding alphabetically. I have used "t" for true and "f" for false in the grid:
If A=3 then it isn't 9 => B=9 => C is not 9 => D=9 so isn't 3 => E=3
=> the two cells F are not 3 => G=3
But, we can't have two <3>s in r2. So, cell A (r2c5) cannot be <3>.
This gets us to ravel's first grid.
From that point, I used
1. {59} W-Wing r4c9|r9c5 removes <5> from r4c5
2. {67} UR r12c49: r1c4<>7 and r2c9<>6 (This is akin to Type 4 with both <6> and <7> strong links.)
3. XY Chain of 8 cells [rc sequence: 67-64-94-24-14-11-19-99] removes <4> from r9c7 and solves the puzzle.
BTW... nataraj's "wing with coloring to bridge the gap" that removes those <4>s is, of course, a W-Wing.
[Edit to fix cell address error as noted below.]
Last edited by Asellus on Fri Dec 14, 2007 11:22 pm; edited 1 time in total |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Fri Dec 14, 2007 4:53 am Post subject: |
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| Quote: | This gets us to ravel's first grid.
From that point, I used
1. {59} W-Wing r4c9|r9c5 removes <5> from r4c5
2. {67} UR r12c49: r1c4<>7 and r2c9<>6 (This is akin to Type 4 with both <6> and <7> strong links.)
3. XY Chain of 8 cells [rc sequence: 67-64-94-24-14-11-19-99] removes <4> from r9c7 and solves the puzzle. |
Alternatively:
Break up DP on 67 UR
Break up DP on 35 UR
Use Type 4 UR on 23 to finish it off. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Fri Dec 14, 2007 6:33 am Post subject: |
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Asellus,
There's a typo in your AIC at the very end:
That should read
-(3)r2c5; r2c5<>3
Had me puzzled there for a while, but then I read the plain text and it became clear. I had in fact tried many ways to remove that 3 in r1c1 but not via r2c5 yet. Very elegant. |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Fri Dec 14, 2007 10:40 am Post subject: |
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Phew, so many answers !
Let me start with the original puzzle.
| Code: | *--------------------------------------------------------------*
| 1378 9 78 | 1467 2 34678 | 5 48 4678 |
| 2378 23 4 | 67 3689 5 | 2678 1 6789 |
| 1278 6 5 | 147 1489 B4789 | 2478 C2489 3 |
|-------------------+----------------------+-------------------|
| 23479 235 1 | 8 345 34 | 47 6 4579 |
| 3467 35 67 | 9 13456 346 | 1478 3458 2 |
| 3469 8 6-9 | 1456 7 2 | 14 C349 459 |
|-------------------+----------------------+-------------------|
| 5 4 A2689 | 26 A689 1 | 3 7 68 |
| 68 7 268 | 3 4568 468 | 9 2458 1 |
| 689 1 3 | 24567 45689 B46789 | 2468 2458 4568 |
*--------------------------------------------------------------*
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For the 9-coloring i used the 3 strong links A,B,C.
(9)r7c3=(9)r7c5-(9)r9c6=(9)r3c6-(9)r3c8=(9)r6c8
| Code: | *------------------------------------------------------*
| 137 9 8 | 167 2 3a67 | 5 4 67 |
| 237 23 4 | 67 3b89 5 | 2678 1 6789b |
| 127 6 5 | 147 1489 4789 | 278 289a 3 |
|---------------+-------------------+------------------|
| 2349 235 1 | 8 345 34 | 47 6 4579 |
| 34 35 7 | 9 13456 346 | 148 358 2 |
| 3b49 8 6 | 145 7 2 | 14 3a9b 459 |
|---------------+-------------------+------------------|
| 5 4 9 | 2 68 1 | 3 7 68 |
| 68 7 2 | 3 4568 468 | 9 58 1 |
| 68 1 3 | 567 5689 6789 | 2468 258 4568 |
*------------------------------------------------------*
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My chain for the elimination of 3 in r1c1 is part of Asellus' beside of the first link:
(3)r1c6=(3-9)r2c5=(9)r2c9-(9)r3c8=(9-3)r6c8=(3)r6c1; r1c1<>3
or
r1c6<>3 => r2c5=3 => r2c9=9 => r3c8<>9 => r6c8=9 => r6c1=3
so one of r1c6 and r6c1 must be 3.
Now this grid:
| Code: | *----------------------------------------------*
| 17 9 8 |#167 2 3 | 5 4 #67 |
|@237 @23 4 |#67 89 5 | 268 1 #679 |
| 127 6 5 | 4 189 79 | 28 89 3 |
|--------------+---------------+---------------|
|@239 @235 1 | 8 35 4 | 7 6 59 |
| 34 35 7 | 9 135 6 | 148 38 2 |
| 349 8 6 | 15 7 2 | 14 39 459 |
|--------------+---------------+---------------|
| 5 4 9 | 2 6 1 | 3 7 8 |
| 6 7 2 | 3 4 8 | 9 5 1 |
| 8 1 3 | 57 59 79 | 46 2 46 |
*----------------------------------------------*
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Probably the simplest way to solve it is over the UR 67. As Johan pointed out, one of the extra candidates r1c4=1 and r2c9=9 must be true.
r1c4=1 => r6c4=5
r2c9=9 => r4c9=5
So one of r6c4 and r4c9 must be 5, r6c9<>5 and the puzzle is solved.
The first way i tried it was to use the pair 67 in this UR with the strong links for 6 in row 1 and 7 in column 9. It gives either r2c4=7 or r2c9=7 (and one of r12c4=6) => r2c1<>7.
This makes the UR 23 a type 4 (with the strong link for 2 in row 4) and eliminates 3 fron r4c12 => r4c5=3.
Asellus came to the same result easier with the W-wing 59.
But it does not solve the puzzle.
Lets look at the other UR's. As Asellus pointed out, with the strong links in the UR67 you can eliminate r1c4<>7 and r2c9<>6.
Now the UR 35 in r45c25, which Marty spotted: The strong links for 3 in c5 and 5 in c2 and r5 give r4c2<>35, r5c5<>5, [added:]r5c2<>3. (btw note, that in this case the strong links dont make the 35's a remote naked pair, because they dont have a common 35-cell)
Together with r4c5<>5 this solves the puzzle.
Note that these deductions show something special for UR's. The order of applying them influences strongly, how easy it is to see other possibilities. If you first make the eliminations in the UR 67, you probably would not spot the possible transport of the extra candidates. If you apply the eliminations of the UR 35 first, you probably would not see the elimination of the type 4 UR 23 anymore. If you eliminate 5 in r4c5 first you probably miss the UR 35. |
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