| View previous topic :: View next topic |
| Author |
Message |
Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
|
 Posted: Tue Jan 15, 2008 9:53 pm Post subject: Not so easy |
 |
|
M6368062 (95)
| Code: |
+-------+-------+-------+
| 2 6 . | 8 . 1 | . . 5 |
| . . 3 | . . . | 9 . . |
| 1 9 . | . . . | . 6 . |
+-------+-------+-------+
| 3 . . | . 5 9 | . . 7 |
| . . . | 7 . 8 | . . . |
| 9 . . | 1 2 . | . . 8 |
+-------+-------+-------+
| . 2 . | . . . | . 9 6 |
| . . 6 | . . . | 2 . . |
| 8 . . | 2 . 6 | . 5 4 |
+-------+-------+-------+
|
Play this puzzle online at the Daily Sudoku site
I've got part way into this one, but am currently bogged down. Some straightforward techniques, but then ..... ? |
|
| Back to top |
|
 |
storm_norm
Joined: 18 Oct 2007
Posts: 1741
|
| Posted: Wed Jan 16, 2008 1:49 am Post subject: |
|
|
| Code: | 2 6 47 | 8 9 1 | 3 47 5
457 4578 3 | 45 6 2 | 9 478 1
1 9 458 | 345 347 3457 | 48 6 2
---------------------------------------------------------------------
3 148 1248 | 6 5 9 | 14 124 7
6 14 124 | 7 34 8 | 5 1234 9
9 57 57 | 1 2 34 | 6 34 8
---------------------------------------------------------------------
457 2 1457 | 345 13478 3457 | 178 9 6
457 1457 6 | 9 1487 457 | 2 18 3
8 3 9 | 2 17 6 | 17 5 4
|
|
|
| Back to top |
|
|
storm_norm
Joined: 18 Oct 2007
Posts: 1741
|
| Posted: Wed Jan 16, 2008 2:54 am Post subject: |
|
|
| Code: | 2 6 47 | 8 9 1 | 3 47 5
457 4578 3 | 45 6 2 | 9 478 1
1 9 458 | 345 347 3-(4)57 | R48 6 2
---------------------------------------------------------------------
3 148 1248 | 6 5 9 | 1G4 124 7
6 14 124 | 7 3G4 8 | 5 1234 9
9 57 57 | 1 2 3R4 | 6 3G4 8
---------------------------------------------------------------------
457 2 1457 | 345 13478 3457 | 178 9 6
457 1457 6 | 9 1487 457 | 2 18 3
8 3 9 | 2 17 6 | 17 5 4
|
one coloring move on 4 first, the 4 in r3c6 is toast.
then, re-polarizing the colors...
| Code: | 2 6 R47 | 8 9 1 | 3 G47 5
457 4578 3 | 45 6 2 | 9 478 1
1 9 458 | 345 347 357 | G48 6 2
---------------------------------------------------------------------
3 148 12-(4)8 | 6 5 9 | 1R4 124 7
6 14 124 | 7 34 8 | 5 1234 9
9 57 57 | 1 2 34 | 6 34 8
---------------------------------------------------------------------
457 2 1457 | 345 13478 3457 | 178 9 6
457 1457 6 | 9 1487 457 | 2 18 3
8 3 9 | 2 17 6 | 17 5 4 |
the coloring on 4 eliminates the 4 in r4c3 |
|
| Back to top |
|
|
storm_norm
Joined: 18 Oct 2007
Posts: 1741
|
| Posted: Wed Jan 16, 2008 3:03 am Post subject: |
|
|
xyz- wing on {4,5,7} in boxes 1 and 2, takes out the 4 in r2c2
doesn't solve it |
|
| Back to top |
|
|
storm_norm
Joined: 18 Oct 2007
Posts: 1741
|
| Posted: Wed Jan 16, 2008 3:32 am Post subject: |
|
|
| Code: | 2 6 47 | 8 9 1 | 3 47 5
457 4578 3 | 45 6 2 | 9 478 1
1 9 458 | 345 347 357 | 48 6 2
---------------------------------------------------------------------
3 148 128 | 6 5 9 | 14 124 7
6 14 124 | 7 34 8 | 5 1234 9
9 57 57 | 1 2 34 | 6 34 8
---------------------------------------------------------------------
457 2 1457 | 345 *13478 3457 |*178 9 6
457 1457 6 | 9 1487 457 | 2 18 3
8 3 9 | 2 *17 6 |*17 5 4 |
the UR on {1,7} the 1's can only go in col 7 so you can remove them from r9c5 and r7c5... is that right? that should solve the puzzle. |
|
| Back to top |
|
|
ravel
Joined: 21 Apr 2006
Posts: 536
|
| Posted: Wed Jan 16, 2008 8:44 am Post subject: |
|
|
| storm_norm wrote: | | the UR on {1,7} the 1's can only go in col 7 so you can remove them from r9c5 and r7c5... is that right? |
Cant see why. |
|
| Back to top |
|
|
ravel
Joined: 21 Apr 2006
Posts: 536
|
| Posted: Wed Jan 16, 2008 12:47 pm Post subject: |
|
|
There are exactly 2 ways, how the missing 5 8's can be placed. So i started to color them red and green. Then i first followed the red numbers - ah this pair then this and that single and so on - until the puzzle was solved.
Good luck or bad luck ? |
|
| Back to top |
|
|
Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
|
| Posted: Wed Jan 16, 2008 1:01 pm Post subject: |
|
|
Storm_norm, you can remove a 7 from R5C7 because of the UR, but I think that's it. (In this type, when there's a strong link in one number at right-angles to the naked pairs, you can remove the other number from the other corner -mentally put in the 7 and run it round clockwise.)
Ravel: is that what's known as Medusa colouring, which I haven't yet attempted?
(Guess I was thinking that somebody would come up with a sort of AIC, which I'm not good at seeing.) |
|
| Back to top |
|
|
nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
|
| Posted: Wed Jan 16, 2008 1:18 pm Post subject: |
|
|
Medusa. I did exactly what ravel proposed - i.e. to look at those "8"s
any candidate that sees both red and green can be eliminated, which gets rid of 4 in r4c2 and 1 in r4c3.
now with the "4" gone, the cell r4c2 becomes bi-value and the 1 in it can be colored green.
Again, there is a candidate that sees red and green: 1 in r4c8.
Another: 1 in r8c2.
This almost solves the puzzle.
I had to use an w-wing later on:
with 1 gone from r4c8 there is only {2,4} so a simple chain
24-47-47-24 ending in r5c3 (at that time r5c3 has lost the "1", too) makes r5c8=1 and we're done.
It turns out that 1 is indeed in r79c7 but I could not see the reasoning behind Norm's UR argument. But I am notoriously weak in URs ... |
|
| Back to top |
|
|
nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
|
| Posted: Wed Jan 16, 2008 1:25 pm Post subject: |
|
|
I should have added that the nice thing with (Medusa-) Coloring is that ANY combination of red and green can be seen as a strong link to complete a nice loop.
So - the first two eliminations could indeed be written as follows:
-(4-8)r4c2=[red to green]=(4)r4c7-; r4c2<>4
-(1-8)r4c3=[green to red]=(1)r4c7-; r4c3<>1
Those are 2 examples of chains that connect different numbers in the same house (see Victor's post in the solving techniques forum
http://www.dailysudoku.com/sudoku/forums/viewtopic.php?t=2289 |
|
| Back to top |
|
|
Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
|
| Posted: Wed Jan 16, 2008 2:38 pm Post subject: |
|
|
Storm_norm, being careless again, I wrote that you can kill the 7 in R7C5. That's wrong: you can kill the 1 in this cell, because there's a strong link in 7s in C7.
Nataraj, thanks. I do get your last post, tho' reading Eureka is a bit of a struggle for me. I guess everybody finds some techniques 'comfortable', others less so. I certainly prefer your chain-type explanation to Medusa colouring, even tho' they're equivalent. |
|
| Back to top |
|
|
ravel
Joined: 21 Apr 2006
Posts: 536
|
| Posted: Wed Jan 16, 2008 4:52 pm Post subject: |
|
|
| Victor wrote: | | Ravel: is that what's known as Medusa colouring, which I haven't yet attempted? |
No, what i did is "extended Medusa" in Asellus' sense and i did it in an extreme way - i just followed one color of the initial Medusa coloring (Medusa coloring only uses strong links - bivalue/bilocation - like nataraj showed in his sample) and - accidently - reached the solution. But this is the same as "trying" a number and look where it brings you (note that in harder puzzles this will bring you nowhere).
| Code: | *------------------------------------------------------*
| 2 6 47 | 8 9 1 | 3 47 5 |
| 457 578 3 | 45 6 2 | 9 478 1 |
| 1 9 458 | 345 347 357 | 48 6 2 |
|------------------+-------------------+---------------|
| 3 148 128 | 6 5 9 | 14 124 7 |
| 6 14 124 | 7 34 8 | 5 1234 9 |
| 9 57 57 | 1 2 34 | 6 34 8 |
|------------------+-------------------+---------------|
| 457 2 1457 | 345 13478 3457 | 178 9 6 |
| 457 1457 6 | 9 1478 457 | 2 18 3 |
| 8 3 9 | 2 17 6 | 17 5 4 |
*------------------------------------------------------*
|
I hoped to find a more satisfying solution, but as the title says - its not easy. Here is a contradiction net:
We know, that one 8 lets fall the others into place, if one of r4c2, r3c3, r2c8, r7c7, r8c5 is 8, then all.
The second chain i need is r7c7=8 => r3c7=4 => r1c8=7 => r1c3=4
Then we have
r7c7=8 => (r3c3=8 and r1c3=4) => r34c3=12 => r8c2=1 => r8c8=8 |
|
| Back to top |
|
|
nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
|
| Posted: Wed Jan 16, 2008 7:14 pm Post subject: |
|
|
| Victor wrote: | | Nataraj,.. I certainly prefer your chain-type explanation to Medusa colouring, even tho' they're equivalent. |
Same here. I much prefer writing down the chain because it is self-contained and shows ... hm, the "essence" of the argument. But for me the question is really HOW to find those chains.
Now we're getting into the vast topic of patterns. For me, to plot single candidates (a grid of positions for "1", a grid of positions for "2" etc.) gives me a superb visual clue and then finding those xwing,kite,turbot even ERs is fairly easy.
Same for xy-wing: to look only at bi-value cells and having found a pair of let's say 45-57 I start looking for "47". Nothing complicated, a simple search, but the whole process is standardized and guaranteed to yield the desired result (if it exists at all, that is, but even in case I did not find any I can be very confident there IS no xy-wing)
Not quite so easy with xy-chains. Very tough with AICs in general.
That's why I love the (extended) Medusa or Graded Equivalence Marks methods:
Start out with a promising single number (like the 8s here) that has a lot of connected strong links. Add all the strong links to other numbers. In case of GEM or extended Medusa add appropriate weak-strong combinations to extend the web. This yields a visual picture of possible strong link shortcuts. No need to follow through all the possible chains, but it is possible to concentrate on the ends of such chains - and it is always the ends that make the elimination...
Once I find a possible elimination in this way, I usually write down the whole chain in order to verify its validity (and to be able to post it  )
Or, I draw silly lines and circles on a screenshot.  |
|
| Back to top |
|
|
Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
|
| Posted: Wed Jan 16, 2008 10:23 pm Post subject: |
|
|
nataraj,
Why did you stop Medusa coloring after those few eliminations? Continuing where you left off, the grid has become:
| Code: |
+----------------------+---------------------+-------------------+
| 2 6 47 | 8 9 1 | 3 47 5 |
| 457 @4578G 3 | 45 6 2 | 9 @478R 1 |
| 1 9 @458R | 345 347 357 |@4R8G 6 2 |
+----------------------+---------------------+-------------------+
| 3 @1G8R @2R8G | 6 5 9 |@1R4G @2G4R 7 |
| 6 @1R4G @2G4R | 7 34 8 | 5 @1G2R-3-4 9 |
| 9 57 57 | 1 2 34 | 6 34 8 |
+----------------------+---------------------+-------------------+
| 457 2 1 | 345 @3478G 3457 |@7G8R 9 6 |
| 457 457 6 | 9 @1G8R 457 | 2 @1R8G 3 |
| 8 3 9 | 2 @1R7G 6 |@1G7R 5 4 |
+----------------------+---------------------+-------------------+
|
I've continued the conventional Medusa and used "@" to indicate all cells with color marks. The eliminations in r5c8 lead us here:
| Code: |
+----------------------+---------------------+----------------+
| 2 6 47 | 8 9 1 | 3 47 5 |
| 457 @4578G 3 | 45 6 2 | 9 @478R 1 |
| 1 9 @458R | 345 47 357 |@4R8G 6 2 |
+----------------------+---------------------+----------------+
| 3 @1G8R @2R8G | 6 5 9 |@1R4G @2G4R 7 |
| 6 @1R4G @2G4R | 7 3 8 | 5 @1G2R 9 |
| 9 57 57 | 1 2 4 | 6 3 8 |
+----------------------+---------------------+----------------+
| 457 2 1 | 345 @478G 357 |@7G8R 9 6 |
| 457 457 6 | 9 @1G8R 57 | 2 @1R8G 3 |
| 8 3 9 | 2 @1R7G 6 |@1G7R 5 4 |
+----------------------+---------------------+----------------+
|
This is the end of the conventional Medusa road. However, only a tiny bit of "extended" Medusa is needed to solve this thing. If red is true, then r1c8 is <7> and r1c3 is <4>. But, this is a Wrap: 2 red <4>s in c3. So, all Gs are true and all Rs are false. |
|
| Back to top |
|
|
Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
|
| Posted: Thu Jan 17, 2008 12:06 am Post subject: |
|
|
I, for one, sure find some of the recent discussions to be interesting!
Regarding the comments about "extending" Medusa coloring, I recommend that conventional Medusa coloring be carried as far as possible before using the "extension" approach. In the puzzle above, conventional Medusa was able to proceed quite far.
Also, I recommend that when the "extension" approach is used it be used for both colors when possible, not just a single color.
ravel's comment about extending only a single color from a limited Medusa start piqued my curiosity. So, I began with storm_norm's {17} UR grid and colored only digit <8> (a highly limited form of Medusa!). Then, I extended only one of the colors. I was still able to solve the puzzle using only this approach... though it was slightly trickier than following my recommended approach.
| ravel wrote: | | But this is the same as "trying" a number and look where it brings you (note that in harder puzzles this will bring you nowhere). |
Even in this more extreme approach, I don't agree that this is the same as trial and error. It is a bifurcation technique, and as valid in that sense as any other bifurcation technique it seems to me.
As for where it might bring you in harder puzzles, I have been using the extended Medusa on at least a couple dozen "extreme" or "fiendish" or "nightmare" puzzles in the past couple of weeks and it has been able to solve every single one of them. So far, the only puzzle I've attempted that I haven't managed to solve in this way is the "AI Escargot," though even there I made progress. I'm sure there are many other puzzles that can't be solved in this way, but I just haven't been finding them.
I agree with nataraj that, while AICs can be interesting and elegant (and they have good learning value), they are still tricky to uncover. Usually, I have uncovered those clever AICs by using a coloring approach. In fact, it was this process that led me to clarify my "extended Medusa" thinking and then realize that it alone is so powerful that going back to individual AICs seemed to be an unnecessary diversion.
Finally...
| nataraj wrote: | | In case of GEM or extended Medusa add appropriate weak-strong combinations to extend the web. |
While one can dissect "weak-strong combination" links in the "extended Medusa" by analyzing its logical underpinnings, I make no such explicit determinations in performing the technique. One can perform extended Medusa coloring without thinking about a weak link. |
|
| Back to top |
|
|
storm_norm
Joined: 18 Oct 2007
Posts: 1741
|
| Posted: Thu Jan 17, 2008 12:26 am Post subject: |
|
|
| Quote: | | the UR on {1,7} the 1's can only go in col 7 so you can remove them from r9c5 and r7c5... is that right? that should solve the puzzle. |
I meant to ask about the 7's in col 7 not the 1's.
sorry for the bad error on my part.
norm |
|
| Back to top |
|
|
nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
|
| Posted: Thu Jan 17, 2008 6:52 am Post subject: |
|
|
| storm_norm wrote: | | Quote: | | the UR on {1,7} the 1's can only go in col 7 so you can remove them from r9c5 and r7c5... is that right? that should solve the puzzle. |
I meant to ask about the 7's in col 7 not the 1's.
sorry for the bad error on my part.
norm |
Norm,
no need to be sorry at all.
Looking at that sentence again, I don't quite understand exactly what you are trying to say. Even when substituting "7" for "1" . The 's' in "the 1's ... in col 7" makes no sense. There can only be ONE 7 (or 1) in col 7.
There MUST be 7 in one of r79c7, but that still leaves another 7 to go into col 5 (or col 6) in box 8.
And, as I said, UR logic is one of my weak points so there might well be a valid elimination... |
|
| Back to top |
|
|
Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
|
| Posted: Mon Jan 21, 2008 9:31 pm Post subject: |
|
|
Thanks guys, I found this very helpful & revealing. Maybe I'll try to overcome my (quite unreasonable) prejudice against Medusa - I do take your point Nataraj about using Medusa to help to find chains.
(I pulled out my sheet and had another look at why the 4 gets killed in R4C2. Seems easy now - why didn't I spot it? !)
| Code: |
8z 48
C-------------D
+-------+-------+----|---+
|
48x 8y 4z|
A----B E
|
In fact, the links 8A=8B=8C=8D=4D=4E are all conjugate though the links from B to C & within D don't have to be. Anyway, one easy way to see it is to that if cell A does not contain 8, then cell E contains 4. And of course if A does contain 8 then it can't contain 4. Maybe I'll spot the next one that turns up.
I'm going to post another not-too-easy. Interested to see how you do it. |
|
| Back to top |
|
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
