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Postby re'born » Sat Apr 22, 2006 10:18 am

RW,

Those are some hard fought deductions. You might want to look through them again as there are a few typos (occasionally your row is off) and possibly some omitted steps (or maybe I just don't see something). All can be fixed with ease.


RW wrote:
rep'nA wrote:If one doesn't want to assume Mac's reductions (which I presume were based on trial & error), we can still make this "simple" reduction (using RW's first 2 deductions which do not depend on Mac's work).


I think the second of my reductions would need some extra steps to remove 2, 4 and 8 from r2c7, otherwise it works fine:

If r1c4=4 => r1c23=(26) => r1c8=1 => r3c4=1 => r2c3=4 => r2c4=5 => r2c6=8 => r2c9=2 => r2c7=7 => r3c8=4 => UR in r13c48 => r1c4<>4


Oops, I didn't read your first post closely enough. I just took your deduction and proved it for myself without using Mac's reductions. But you are correct, your first attempt did use them.
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Postby RW » Sat Apr 22, 2006 10:42 am

rep'nA wrote:You might want to look through them again as there are a few typos (occasionally your row is off) and possibly some omitted steps


Oops, you're right. The BUG-lite trail is missing r9c7=9 and the row is of in several steps of my last trail. Should be:

If r9c7=9 => r9c9=2 => r9c5=1 => r4c7=8 => r6c9=3 => r5c78=(24) => r5c45=(68) => r5c1=3 => r4c4=7 => r4c2=2 => r1c2=6 => r1c1=5 => r3c1=7 => r9c1=6 => r9c4= empty cell => r9c7<>9

EmilyS wrote:I cannot possibly understand how you could "smell" the BUG-lite in your reduction.


I saw this pattern:

Code: Select all
6  26 . |.  .  . |.  26 2
.  .  26|.  .  . |.  26 2
.  .  . |.  .  . |.  .  .
-------------------------
.  2  . |        |
6  .  . |        |
6  26 26|        |
-------------------------
        |        |
        |        |
        |        |2=====2

In boxes 1, 3 and 4 there is only three 2s and three 6s outside the BUG-lite pattern. Because of the strong link in row 9 I could immediately see that a 2 in r4c7 would eliminate all of the extra 2s, leaving only the 6s that all are in the same row. Now I knew that if r4c7=2 would also give me r8c1=6 or r9c1=6, then the BUG-lite would be completed, so I read on to see if that would happen. As rep'nA showed that the finned x-wing would remove candidate 3 from r9c1, we could actually leave out the step r5c1=3 from my trail, and thus the correct trail should be:

If r4c7=2 => r9c7=9 => r9c9=2 => r8c8=3 => r5c8=4 => r5c7=8 => r5c4=6 => r9c4=7 => r9c1=6 => BUG-lite in r1c28, r2c38, r6c23 => r4c7<>2

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Postby GreenLantern » Sat Apr 22, 2006 11:51 am

Here are the deductions I made to solve the puzzle. I'd be interested to hear
from the nice loop experts if my deductions are valid.

After some basic steps, we have the following starting grid:
Code: Select all
+----------------------+----------------------+----------------------+
| 56     26     246    | 1458   7      9      | 3      1246   1248   |
| 1      9      2467   | 458    3      48     | 24578  2467   248    |
| 357    37     8      | 145    2      6      | 457    147    9      |
+----------------------+----------------------+----------------------+
| 4      2378   1      | 78     5      2378   | 289    239    6      |
| 368    5      9      | 468    1468   12348  | 248    234    7      |
| 3678   23678  2367   | 46789  4689   23478  | 1      5      2348   |
+----------------------+----------------------+----------------------+
| 2      178    5      | 3      1489   1478   | 6      1479   14     |
| 36789  13678  367    | 2      14689  1478   | 479    13479  5      |
| 3679   4      367    | 679    169    5      | 279    8      123    |
+----------------------+----------------------+----------------------+

1) [r1c4]=1=[r1c89]=8=[r2c79]-8-[r2c6]-4-[r1c4] => r1c4<>4
2) =[r1c4]=5=[r1c1]=6=[r1c23|r2c3]=7=[r3c12]-7-[r3c78]=7=[r2c78]=5=[r2c4]-5-[r1c4]= => r3c4<>5
3) [r1c3]=4=[r2c3]=7=[r2c78]=6=[r1c8]-6-[r1c3] => r1c3<>6
4) =[r1c4]=8=[r1c9]-8-[r2c79]=8=[r2c46]=5=[r1c4]= => r1c4<>1
5) [r2c3]=7=[r2c78]=6=[r1c8]-6-[r1c2]-2-[r2c3] => r2c3<>2
6) =[r2c8]=2=[r2c79]=8=[r2c46]-8-[r1c4]-5-[r1c1]-6-[r1c8]=6=[r2c8]= => r2c8<>7
7) [r7c9]=4=[r7c8|r8c78]=9=[r9c7]=2=[r9c9]-2-[r2c9]-8-[r1c9]-1-[r7c9] -> r7c9<>1 => r7c9=4

And now the puzzle solves in straightforward fashion.
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Postby EmilyS » Sat Apr 22, 2006 2:43 pm

Thanks GreenLantern for that. I've read the posts with explanations, but there is some things I don't understand yet. Your first loop:

1) [r1c4]=1=[r1c89]=8=[r2c79]-8-[r2c6]-4-[r1c4] => r1c4<>4

I can understand that if r1c4=1, then r1c4<>4. I can also understand that if r1c9=1 then r2c7 or r2c9=8 and r2c6=4, ->r1c4<>4. But how does this loop explain that if r1c8=1 then r1c4<>4?

Can somebody please explain..?

-E
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Postby Carcul » Sat Apr 22, 2006 3:03 pm

Hi GreenLantern.

GreenLantern wrote:I'd be interested to hear from the nice loop experts if my deductions are valid.


You should revise your knowledge of nice loops, because all the ones you posted are wrong: you are using incorrectly the "grouped link" concept. For example, "=1=[r1c89]=8=[r2c79]" has no meaning. As I have already written elsewhere, a grouped node can only be between two strong links if the labels of both strong links are equal. And this one, "=4=[r7c8|r8c78]=9=[r9c7]=" which is not even a grouped strong link. GreenLantern, you should check the logic behind your deductions, instead of posting them blindily.

Regards, Carcul
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Postby GreenLantern » Sat Apr 22, 2006 3:10 pm

Thanks for the clarification, Carcul. Back to the drawing board...
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Postby RW » Sat Apr 22, 2006 3:23 pm

In my last post I advanced the puzzle to this point:

Code: Select all
 *--------------------------------------------------------------------*
 | 56     26     4      | 158    7      9      | 3      126    128    |
 | 1      9      267    | 458    3      48     | 57     2467   248    |
 | 357    37     8      | 145    2      6      | 457    147    9      |
 |----------------------+----------------------+----------------------|
 | 4      2378   1      | 78     5      2378   | 89     239    6      |
 | 368    5      9      | 468    1468   12348  | 48     234    7      |
 | 3678   23678  2367   | 46789  4689   23478  | 1      5      234    |
 |----------------------+----------------------+----------------------|
 | 2      178    5      | 3      1489   1478   | 6      1479   14     |
 | 36789  13678  367    | 2      14689  1478   | 479    13479  5      |
 | 679    4      367    | 679    169    5      | 2      8      13     |
 *--------------------------------------------------------------------*


After this, no more 10+ step trails were needed, but quite a lot of shorter ones. I'll list all deductions I made, even though they're not all neccessary, just to show you what kind of fun stuff you can find in a puzzle like this:

If r5c8=3 => r9c9=3 => r9c5=1 => r5c6=1 => nowhere to place 2 in row 5 => r5c8<>3
If r4c7=8 => r4c4=7 => r8c7=9 => r4c8=9 => r6c9=3 => r9c9=1 => r9c45=(69) => nowhere to place 9 in box 7 => r4c7<>8

=> r4c7=9 => r5c7=8

Forcing chain:
If r5c5=1 => r9c9=1 => r8c8=3 => r4c8=2 => r5c8=4 => r5c4=6 => r5c1=3
If r5c6=1 => r5c45=(46) => r5c1=3

Also gives us r3c2=3

If r7c2=7 => r9c4=7 => r4c4=8 => r6c45=(49) => r5c4=6 => r5c5=1 => r9c9=1 => r8c8=3 => r4c8=2 => r4c2= empty cell => r7c2<>7

If r7c8=7 => r7c5=9 => r8c7=4 => r7c9=1 => r9c5=1 => r27c6=(48) => r8c6=7 => r9c4=6 => r5c4=4 => r6c5=8 => r8c5= empty cell => r7c8<>7

=> r7c6=7

(I later checked in Simple Sudoku and it turned out that from this point on the puzzle can be solved with multiple colors and one xy-wing, but I'll tell you what I did anyway)

This reveals a possible uniqueness rectangle in r46c24 => r6c2<>8

If r4c2=8 => r8c1=8 => r9c1=9 => r9c4=6 => r9c5=1 => r8c6=4 => r5c4=4 => nowhere to place 4 in box2 => r4c2<>8

=> r6c1=8

uniqueness rectangle in r78c25 => r78c5<>1

xy-wing in r5c4, r9c4, r6c5 => r6c4 and r789c5<>9

=> r6c5=9 => r9c4=9 => r5c4=6 => r8c2=9 => r7c8=9

If r5c6=4 => r5c5=1 => r8c6=1 => r8c2=8 => r9c5=6 => r8c5=4 => UR in r58c56 => r5c6<>4

If r5c5=4 => r6c9=4 => r8c6=4 => nowhere to place 4 in box 9 => r5c5<>4

=> r5c5=1 => Puzzle solved

I'm not quite satisfied with the solution though, because of Macs T&E reductions that I used as a starting point. I did so because my original intention was to show Mac that he gave up just when he was about to break the puzzle, but it turned out that it wasn't that easy... So I suppose other solutions are still welcome.

Regards RW
Last edited by RW on Sun Apr 23, 2006 4:15 am, edited 1 time in total.
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Postby ravel » Sat Apr 22, 2006 11:50 pm

Very nice deductions:)

So for an overall solution it is only missing, that someone writes down 3 (hopefully not very long) chains to eliminate 2,4 and 8 from r2c7 at the beginning (not hard, but maybe cumbersome).
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Postby re'born » Sun Apr 23, 2006 2:17 am

ravel wrote:Very nice deductions:)

So for an overall solution it is only missing, that someone writes down 3 (hopefully not very long) chains to eliminate 2,4 and 8 from r2c7 at the beginning (not hard, but maybe cumbersome).




Here is a nice way to rule out the 4 from this position:

Code: Select all
  *--------------------------------------------------------------------*
 | 56     26     246    | 1458   7      9      | 3      1246   1248   |
 | 1      9      2467   | 458    3      48     | 24578  2467   248    |
 | 357    37     8      | 145    2      6      | 457    147    9      |
 |----------------------+----------------------+----------------------|
 | 4      2378   1      | 78     5      2378   | 289    239    6      |
 | 368    5      9      | 468    1468   12348  | 248    234    7      |
 | 3678   23678  2367   | 46789  4689   23478  | 1      5      2348   |
 |----------------------+----------------------+----------------------|
 | 2      178    5      | 3      1489   1478   | 6      1479   14     |
 | 36789  13678  367    | 2      14689  1478   | 479    13479  5      |
 | 3679   4      367    | 679    169    5      | 279    8      123    |
 *--------------------------------------------------------------------*


Notice that the 5's in row 2 and column 7 are strongly linked.

(2,7)4 ( > (2,4)5 ) > (3,7)5 > (3,4)4 and so we get a deadly pattern in ([23],[47])<45>. Therefore, (2,7)!4.
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Postby RW » Sun Apr 23, 2006 8:23 am

rep'nA wrote:Notice that the 5's in row 2 and column 7 are strongly linked.

(2,7)4 ( > (2,4)5 ) > (3,7)5 > (3,4)4 and so we get a deadly pattern in ([23],[47])<45>. Therefore, (2,7)!4.


Nice one rep'nA. But I'd add r1c3=4 to the chain to explain your last step r3c4=4. I suppose you assumed that one was only solved, as we did earlier, but you haven't done so in your pencilmarkgrid. Two numbers left to get rid of, that shouldn't be too hard.

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Postby RW » Sun Apr 23, 2006 3:05 pm

I found a short trail to reduce candidate 2 from r2c7, but couldn't find any shorter than 13 steps for candidate 8. So here's the complete solution, only the neccessary steps included:

r2c2=9
r4c5=5
r7c3=5

1. Finned x-wing in columns 3 and 9 => r9c1<>3

2. If r2c7=4 => r2c4=5 => r3c4=4 => r3c7=5 => UR in r23c47 => r2c7<>4

3. If r2c7=2 => r6c3=2 => r9c9=2 => r8c8=3 => r9c5=1 => r5c6=1 => r4c6=2 => r4c2=3 => nowhere to place 3 in row 5 => r2c7<>2

4. If r2c7=8 => r1c4=8 => r4c4=7 => r2c6=4 => r2c9=2 => r1c9=1 => r7c9=4 => r9c5=1 => r9c7=2 => r5c7=4 => r5c4=6 => r9c4=9 => r7c5=8 => r5c5=empty cell => r2c7<>8

5. If r9c7=7 => r2c7=5 => r3c7=4 => r9c9=2 => r2c9=8 => r2c6=4 => r2c4= empty cell => r9c7<>7

6. If r4c7=2 => r9c9=2 => r8c8=3 => r5c8=4 => r5c7=8 => r5c4=6 => r9c7=9 => r9c4=7 => r9c1=6 => BUG-lite in r1c28, r2c38, r6c23 => r4c7<>2

7. If r9c7=9 => r9c9=2 => r9c5=1 => r4c7=8 => r6c9=3 => r5c78=(24) => r5c45=(68) => r5c1=3 => r4c4=7 => r4c2=2 => r1c2=6 => r1c1=5 => r3c1=7 => r9c1=6 => r9c4= empty cell => r9c7<>9

=> r9c7=2

8. If r5c8=3 => r9c9=3 => r9c5=1 => r5c6=1 => nowhere to place 2 in row 5 => r5c8<>3

9. If r4c7=8 => r4c4=7 => r8c7=9 => r4c8=9 => r6c9=3 => r9c9=1 => r9c45=(69) => nowhere to place 9 in box 7 => r4c7<>8

=> r4c7=9
=> r5c7=8

10. Forcing chain:
If r5c5=1 => r9c9=1 => r8c8=3 => r4c8=2 => r5c8=4 => r5c4=6 => r5c1=3
If r5c6=1 => r5c45=(46) => r5c1=3

=> r3c2=3

11. If r7c2=7 => r9c4=7 => r4c4=8 => r6c45=(49) => r5c4=6 => r5c5=1 => r9c9=1 => r8c8=3 => r4c8=2 => r4c2= empty cell => r7c2<>7

12. If r7c8=7 => r7c5=9 => r8c7=4 => r7c9=1 => r9c5=1 => r27c6=(48) => r8c6=7 => r9c4=6 => r5c4=4 => r6c5=8 => r8c5= empty cell => r7c8<>7

=> r7c6=7

13. Multiple colors on 8 => r6c5<>8 (If r6c5=8 => r8c6=8 => r7c2=8 => nowhere to place 8 in box4)

14. xy-wing in r5c4, r9c4, r6c5 => r6c4 and r789c5<>9

=> r6c5=9
=> r9c4=9
=> r5c4=6
=> r8c1=9
=> r7c8=9
=> r6c1=8

15. Multiple colors on 4 => r5c5<>4 (If r5c5=4 => r6c9=4 => r8c6=4 => nowhere to place 4 in box 9)

=> r5c5=1

Interesting to note that I found a solution with only 3 trails for the "ancient toughest" - this one required a lot more. Doesn't prove that this one is harder, I'm sure there is a shorter solution than this one, but I'd at least say it's right up there in the same league. I checked the puzzle for "magic cells" and there's only one (r6c6=4) that lets us solve the rest with basic techniques and a few that leaves us with some coloring reductions. Most cells don't really help us at all (as I noticed while solving...)

So thanks Emily for the puzzle. If your friend makes any more of these, you can tell him to post them directly here, I think there's always people around here who like great challenges:)

RW
Last edited by RW on Mon Apr 24, 2006 6:28 am, edited 2 times in total.
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Postby re'born » Sun Apr 23, 2006 3:40 pm

RW wrote:
rep'nA wrote:Notice that the 5's in row 2 and column 7 are strongly linked.

(2,7)4 ( > (2,4)5 ) > (3,7)5 > (3,4)4 and so we get a deadly pattern in ([23],[47])<45>. Therefore, (2,7)!4.


Nice one rep'nA. But I'd add r1c3=4 to the chain to explain your last step r3c4=4. I suppose you assumed that one was only solved, as we did earlier, but you haven't done so in your pencilmarkgrid. Two numbers left to get rid of, that shouldn't be too hard.

RW


RW,

Actually, you don't need r1c3=4 to make the deduction r3c4=4. Once you set r2c7=4, r3c4 is pinned to 4 (its the only 4 left in row 3). To make things clearer in my deduction, maybe I should add that r2c7 = 4 => r3c8 <> 4.

I can't wait to go through the rest of your solution. Excellent work RW.
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Postby RW » Sun Apr 23, 2006 3:49 pm

rep'nA wrote:Actually, you don't need r1c3=4 to make the deduction r3c4=4. Once you set r2c7=4, r3c4 is pinned to 4 (its the only 4 left in row 3).


Sorry, it was too early in the morning so I actually missed that.:) Your deduction was of course clear enough. Within my trails I usually write down the information needed to solve all included cells as singles, either naked or hidden. I think adding lot's of direct reductions caused by each step only would make it more confusing.

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Postby EmilyS » Sun Apr 23, 2006 6:57 pm

Thanks RW for the solution. I sure couldn't have done all of that on my own. My friend was quite amazed when I showed your solution to him, even he didn't think the puzzle was that hard. He'll send his puzzles here in the future if he makes any similar ones.
:)

-E
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Postby keith » Sun Apr 23, 2006 11:11 pm

Emily,

No need to send more puzzles, but please send your friend's solution to this one that is "not that hard". (Not the answer, but how to solve it.)


Best wishes,

Keith
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