Throwing the kitchen sink at MM-14

Advanced methods and approaches for solving Sudoku puzzles

Throwing the kitchen sink at MM-14

Postby Myth Jellies » Thu Jun 15, 2006 9:27 am

AIC's, AUR's, ALS's with subsets, and even a MUG to solve M. Mepham's bifurcation puzzle #14. I'm showing all of my steps. I think they should all qualify as theoretical, i.e. providing an action at a distance. I haven't taken the time to weed out the unnecessary steps.

You might want to bone up on AIC's and also Advanced ALS's in AIC's before tackling this post.

Here is the original puzzle...
Code: Select all

 *-----------*
 |.4.|...|.2.|
 |9..|...|1.3|
 |..2|3.7|9..|
 |---+---+---|
 |...|1.4|...|
 |8..|...|..4|
 |...|7.5|.9.|
 |---+---+---|
 |..8|5.2|6..|
 |1.9|...|..8|
 |.6.|...|.7.|
 *-----------*

...and this is as far as you can get with basic methods.
Code: Select all
 
 *-----------------------------------------------------------------------------*
 | 356     4       1356    | 689     1569   D1689    | 578     2       567     |
 | 9      A578     567     |-2468    2456    68      | 1       4568    3       |
 | 56     B18      2       | 3      C14      7       | 9       48      56      |
 |-------------------------+-------------------------+-------------------------|
 | 23567   23579   3567    | 1       23689   4       | 23578   3568    2567    |
 | 8       123579  13567   | 269     2369    369     | 2357    1356    4       |
 | 2346    123     1346    | 7       2368    5       | 238     9       126     |
 |-------------------------+-------------------------+-------------------------|
 | 347    *37      8       | 5      *13479   2       | 6       13      19      |
 | 1      *2357    9       | 46     *3467    36      | 2345    35      8       |
 | 2345    6       345     |F489     1349   E1389    | 2345    7       1259    |
 *-----------------------------------------------------------------------------*

Step 1:
The starred cells in r78c25 locate a 37 uniqueness rectangle with 2 strong links in the 7's. This implies r8c5 <> 3.

Step 2:
The first AIC. (strong inference: '=', weak inference: '-')
A8 = B8 - B1 = C1 - D1 = E1 - E8 = F8 => r2c4 <> 8
The problem with this notation is that you need a picture to figure out where A, B, C, etc. is for every chain, and that is just not practical with this puzzle. So I am going to adopt the following shorthand. I start with the digit premise on one end of the chain along with its location. I include all strong and weak inferences. Whenever either the digit premise or the location changes, I note that. So the above AIC would be rewritten as follows...
8[r2c2]=[r3c2]-1=[r3c5]-[r1c6]=[r9c6]-8=[r9c4] => r2c4 <> 8
Hopefully, that is compact, readable, and easy enough to understand and duplicate.

Step 3:
3[r1c1]=[r1c3]-1=[r3c2]-[r3c5]=4-[r7c5]=[r7c1] => r7c1 <> 3

Step 4:
3[r1c1]=[r1c3]-1=[r3c2]-[r3c5]=4-[r7c5]=[r7c1]-7=[r4c1] => r4c1 <> 3

Step 5:
An AIC using ALS subsets. I'll use a picture for the first one...
Code: Select all
 *-----------------------------------------------------------------------------*
 | 356     4       1356    |A689'    1569   A1'689   | 578     2       567     |
 | 9      D578     567     | 246     2456   A68'     | 1       4568    3       |
 | 56     C18      2       | 3      B14      7       | 9       48      56      |
 |-------------------------+-------------------------+-------------------------|
 | 2567    23579   3567    | 1       23689   4       | 23578   3568    2567    |
 | 8       123579  13567   | 269     2369    369     | 2357    1356    4       |
 | 2346    123     1346    | 7       2368    5       | 238     9       126     |
 |-------------------------+-------------------------+-------------------------|
 | 47      37      8       | 5       13479   2       | 6       13      19      |
 | 1       2357    9       | 46      467     36      | 2345    35      8       |
 | 2345    6       345     | 489     1349    1389    | 2345    7       1259    |
 *-----------------------------------------------------------------------------*

Looking at ALS set A, note that if you exclude the 6's from the set, you end up with the primed digits remaining. Thus we have...
A6=A(1'&8'&9')-B1=C1-C8=D8-A(1'&8'&9')=A6 => A has to contain a 6, therefore r1c5 & r2c45 <> 6. Using only the primes we need in the new notation, this becomes...
6[r1c46,r2c6]=(1&8'&9)-1[r3c5]=[r3c2]-8=[r2c2] -(1&8'&9)[r1c46,r2c6]=6 => r1c5 & r2c45 <> 6

Step 6:
3[r1258c6]=(1&6&8'&9)-1[r3c5]=[r3c2]-8=[r2c2] -(1&6&8'&9)[r1258c6]=3 => r9c6 <> 3

Step 7:
9[r1258c6]=(1&3&6&8')-1[r3c5]=[r3c2]-8=[r2c2] -(1&3&6&8')[r1258c6]=9 => r9c6 <> 9

Step 8:
No primes in this one, so this is just a regular ALS in a chain
(2&4&6)[r258c4]=9-[r5c6]=[r1c6]-1=[r9c6]-8=[r9c4] => r9c4 <> 4

Step 9:
9[r5c4]=(8&9)[r19c4]-6[r1c4]=[r12c6]-[r58c6]=(3&9)-9[r5c4] loop => r5c25 & r4c5 <> 9, r4c2 = 9

Step 10:
When you have a UR with two strong links forming a corner on one of the base digits (b in this case)...
Code: Select all
 Wab---baX
   |
---|-------
   |
 Yab   abZ

...there is a hidden weak UR link between the 'a' in the corner Wab-cell and the 'ab' in the opposing abZ-cell. Cells r15c46 contain a 69-UR with two strong links and an exploitable weak link...
6[r128c6]=[r5c6]-UR-(6or9)[r1c4]=8-[r2c6]=6 => r5c6 <> 6

Step 11:
4[r9c7]=[r8c7]-[r8c4]=[r2c4]-2=[r5c4]-9=[r5c6]-3=[r8c6]- [r8c8]=5 => r9c7 <> 5

Step 12:
1[r1c3]=[r3c2]-[r3c5]=4-[r2c4]=[r8c4]-[r8c7]=[r9c7]-[r9c3]=[r6c3] => r6c3 <> 1

Step 13:
Really just coloring here...
1[r9c6]=[r1c6]-[r1c3]=[r5c3]-[r5c8]=[r7c8] => r7c5 & r9c9 <> 1

Step 14:
At this point I was stuck for awhile. Here is the grid...
Code: Select all
 *--------------------------------------------------------------------*
 | 356    4     #1356   |*689   *159   *1689   |#578    2      567    |
 | 9      578    567    | 24     245    68     | 1      4568   3      |
 | 56     18     2      | 3      14     7      | 9      48     56     |
 |----------------------+----------------------+----------------------|
 | 2567   9      3567   | 1      2368   4      | 23578  3568   2567   |
 | 8      12357  13567  | 269    236    39     | 2357   1356   4      |
 | 2346   123    346    | 7      2368   5      | 238    9      126    |
 |----------------------+----------------------+----------------------|
 | 47     37     8      | 5      3479   2      | 6      13     19     |
 | 1      2357   9      | 46     467    36     | 2345   35     8      |
 | 2345   6      345    |*89    *1349  *18     | 234    7     #259    |
 *--------------------------------------------------------------------*

I noticed a potential MUG with the digits 189 in the starred cells (r19c456). There seemed to be too many extra digits to do much with it, but then I noted that to avoid the deadly MUG pattern, you had to have (r1c3 = 1) or (r1c7 = 8) or (r9c9 = 9) be true. Now (r1c7 = 8 or r9c9 = 9 is true) is just another way of saying that there is a strong inference between those two candidates. Look what happens when we start creating a chain with that inference
8[r1c7]=9[r9c9]-[r7c9]=1-[r6c9]=[r6c2]-[r3c2]=[r1c3]
|
8[r3c8]=[r3c2]-1=[r1c3]
Amazing, but true! The second and third MUG avoidance candidates, when OR'ed together actually force the first MUG avoidance candidate. (When you have a big molecular method enzyme sitting in your face for days, this is actually fairly easy to spot.) Therefore, we can safely assume that the first MUG avoidance candidate is true. r1c3 = 1.

This unlocks a bit of basic solving, until you get stuck here...
Code: Select all
 *--------------------------------------------------------------------*
 | 3      4      1      | 69     5      689    | 78     2      67     |
 | 9      57     567    | 24     24     68     | 1      568    3      |
 | 56     8      2      | 3      1      7      | 9      4      56     |
 |----------------------+----------------------+----------------------|
 | 2567   9      3567   | 1      2368   4      | 23578  3568   2567   |
 | 8      12357  3567   | 269    236    39     | 2357   1356   4      |
 | 246    123    346    | 7      2368   5      | 238    9      126    |
 |----------------------+----------------------+----------------------|
 | 47     37     8      | 5      3479   2      | 6      13     19     |
 | 1      2357   9      | 46     467    36     | 2345   35     8      |
 | 245    6      345    | 8      349    1      | 234    7      259    |
 *--------------------------------------------------------------------*


Step 15:
(5&7)[r27c2]=3-[r7c8]=1-[r5c8]=[r5c2] => r5c2 <> 5 or 7

Step 16:
5[r3c9]=[r3c1]-[r2c2]=[r8c2]-[r9c13]=[r9c9] => r4c9 <> 5

Step 17:
5[r2c8]=[r3c9]-[r9c9]=[r9c13]-[r8c2]=[r2c2] => r2c3 <> 5

Step 18:
7[r5c7]=[r5c3]-[r2c3]=6-[r2c6]=8-[r2c8]=[r1c7] => r1c7 <> 7

You get to solve a few more things and get to here
Code: Select all
 *-----------------------------------------------------------*
 | 3     4     1     | 69    5     69    | 8     2     7     |
 | 9     57    67    | 24    24    8     | 1     56    3     |
 | 56    8     2     | 3     1     7     | 9     4     56    |
 |-------------------+-------------------+-------------------|
 | 2567  9     3567  | 1     236   4     | 2357  8     26    |
 | 8     123   3567  | 269   236   39    | 2357  1356  4     |
 | 246   123   346   | 7     8     5     | 23    9     126   |
 |-------------------+-------------------+-------------------|
 | 47    37    8     | 5     3479  2     | 6     13    19    |
 | 1     2357  9     | 46    467   36    | 2345  35    8     |
 | 245   6     345   | 8     349   1     | 234   7     259   |
 *-----------------------------------------------------------*

Step 19:
6[r3c1]=[r3c9]-[r46c9]=[r5c8]-[r5c45]=[r4c5] => r4c1 <> 6

Step 20:
Finned swordfish in 3's r469c357, fin in r6c2 => r5c3 <> 3

Step 21:
Even though a 6 is missing, you can reconstitute it and create a potential 69 deadly UR pattern in r15c46. Only the 2 or the 3 in those cells will avoid the deadly pattern. Since one of those digits must be true, a strong inference exists between them...
2[r5c4]=UR=3[r5c6]-[r8c6]=6-[r8c4]=4-[r2c4]=2 loop => r8c5 <> 6, and then a box-box interaction => r5c4 <> 6

Step 22:
2[r8c7]=[r8c2]-5=[r2c2]-7=[r2c3]-[r5c3]=[r5c7] => r5c7 <> 2

Step 23:
3[r6c7]=2-[r8c7]=[r8c2]-5=[r2c2]-7=[r2c3]-[r5c3]=[r5c7] => r5c7 <> 3

Step 24:
3[r56c2]=(1&2)-2[r8c2]=[r8c7]-[r6c7]=3 => r6c3 <> 3

Step 25:
3[r6c7]=[r6c2]-[r4c3]=[r9c3] => r9c7 <> 3

Step 26:
7[r7c2]=3-[r9c3]=[r9c5]-9=[r7c5] => r7c5 <> 7, r8c5 = 7, and a little clean up gives...

Code: Select all
 *-----------------------------------------------------------*
 | 3     4     1     | 69    5     69    | 8     2     7     |
 | 9     57    67    | 24    24    8     | 1     56    3     |
 | 56    8     2     | 3     1     7     | 9     4     56    |
 |-------------------+-------------------+-------------------|
 | 257   9     3567  | 1     236   4     | 2357  8     26    |
 | 8     123   567   | 29    236   39    | 57    1356  4     |
 | 246   123   46    | 7     8     5     | 23    9     126   |
 |-------------------+-------------------+-------------------|
 | 47    37    8     | 5     349   2     | 6     13    19    |
 | 1     235   9     | 46    7     36    | 2345  35    8     |
 | 245   6     345   | 8     349   1     | 24    7     259   |
 *-----------------------------------------------------------*

Step 27:
5[r3c9]=[r9c9]-9=[r9c5]-3=[r9c3]-[r7c2]=7-[r2c2]=5 => r3c1 & r2c8 <> 5 which cracks the puzzle.

[All done by hand using the molecular method to bring out AIC's]
Myth Jellies
 
Posts: 593
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Postby daj95376 » Thu Jun 15, 2006 2:57 pm

I'm behind the learning curve on documenting chains, so you get to fill in your own favorites below. My solver pointed out the weak areas in the puzzle and I finished refining it by hand. (The solution is for the original puzzle.)

Code: Select all
r3      -  56    Naked  Pair
    b3  -  7     Locked Candidate (1)
    b3  -  4     Locked Candidate (1)
    b5  -  8     Locked Candidate (1)
r8c6    =  3     [r8c6]=6 => Invalid Puzzle (complex chains of naked/hidden singles)
r8c8    =  5     Naked  Single
r7c8    =  3     [r7c8]=1 => Invalid Puzzle (complex chains of naked/hidden singles)
                 Naked/Hidden Singles remain
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Postby RW » Thu Jun 15, 2006 6:54 pm

Nice solution MJ! I like the fact that all steps are fairly short, no monsterchains. True that the puzzle can be cracked in two steps as daj95376 showed, but I don't see any greater value in solving puzzles with solver programs. I think the challenge in solving superhard puzzles is to manually find a solution where each step is as simple as possible. If a computer finds you 2 complicated steps, it tells you that the puzzle can be solved that way, but doesn't say that you have solved the puzzle. IMO it's closer to an analysis than a solution.

MJs solution on the other hand is very comprehensive with short chains, and also some very interesting steps, like the MUG. I had a little problem reading your notation there at first, personally I would have preferred a forcing chain:

r1c7=8 => r3c2=8 => r1c3=1
r9c9=9 => r7c9=1 => r6c2=1 => r1c3=1

but that's just because I'm not used to your "OR'ed" notation. Anyway, I like it a lot as it shows not only knowledge of techniques, but also the ability to apply the techniques to complex situations.

Couldn't help noticing this nice reduction in the starting grid:

Code: Select all
 *-----------*
 |.4.|...|.2.|
 |9..|...|1.3|
 |..2|3.7|9..|
 |---+---+---|
 |...|1.4|...|
 |8..|...|..4|
 |...|7.5|.9.|
 |---+---+---|
 |..8|5.2|6..|
 |1.9|...|..8|
 |.6.|...|.7.|
 *-----------*

If r7c5=7 => r8c2=7 => r8c78=25 => r9c456=189
=> r9c9=empty cell => r7c5<>7

=> r8c5=7


Unfortunately it doesn't really help in solving the puzzle, but it's a nice elimintaion that I hope you can see even without pms.

RW
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Postby Myth Jellies » Fri Jun 16, 2006 3:39 am

RW wrote:Nice solution MJ!

Thanks. My family thinks I am truly obsessed now, though.:)

RW wrote:I had a little problem reading your notation there [at the MUG step] at first, personally I would have preferred a forcing chain:

r1c7=8 => r3c2=8 => r1c3=1
r9c9=9 => r7c9=1 => r6c2=1 => r1c3=1

but that's just because I'm not used to your "OR'ed" notation.

The notation reflects my personal idiosyncrasy. I dislike it when otherwise perfectly good chains are presented as "if you assume cell A = n then...." I'd much rather see them presented as an extended "or".
Code: Select all
A = B means A or B must be true.
A = B -...- C = D means A or D must be true.

I think you're less likely to be confused with brute force methods by using the or-ed approach.

Actually when I was trying to figure out that MUG step I thought more than once, "What would RW do here--probably solve the whole thing in one fell swoop with one of his reverse uniqueness tricks.":)
RW wrote:Couldn't help noticing this nice reduction in the starting grid:

Code: Select all
 *-----------*
 |.4.|...|.2.|
 |9..|...|1.3|
 |..2|3.7|9..|
 |---+---+---|
 |...|1.4|...|
 |8..|...|..4|
 |...|7.5|.9.|
 |---+---+---|
 |..8|5.2|6..|
 |1.9|...|..8|
 |.6.|...|.7.|
 *-----------*

If r7c5=7 => r8c2=7 => r8c78=25 => r9c456=189
=> r9c9=empty cell => r7c5<>7

=> r8c5=7



You aren't the first one to come up with that as one of their initial steps. Someone in another forum used some complex braid analysis to make the same deduction. It may just be a mental block, but I haven't found a simple way of making the equivalent deduction without assuming r7c5 = 7.
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Re: Throwing the kitchen sink at MM-14

Postby ronk » Fri Jun 16, 2006 4:31 am

MJ, congratulations on your solution. I'm scratching my head about the MUG step though.

Myth Jellies wrote:Step 14:
At this point I was stuck for awhile. Here is the grid...
Code: Select all
 *-----------------------------------------------------------*
 | 356   4    #1356  |*689  *159  *1689  |#578   2     567   |
 | 9     578   567   | 24    245   68    | 1     4568  3     |
 | 56    18    2     | 3     14    7     | 9     48    56    |
 |-------------------+-------------------+-------------------|
 | 2567  9     3567  | 1     2368  4     | 23578 3568  2567  |
 | 8     12357 13567 | 269   236   39    | 2357  1356  4     |
 | 2346  123   346   | 7     2368  5     | 238   9     126   |
 |-------------------+-------------------+-------------------|
 | 47    37    8     | 5     3479  2     | 6     13    19    |
 | 1     2357  9     | 46    467   36    | 2345  35    8     |
 | 2345  6     345   |*89   *1349 *18    | 234   7    #259   |
 *-----------------------------------------------------------*

I noticed a potential MUG with the digits 189 in the starred cells (r19c456). There seemed to be too many extra digits to do much with it, but then I noted that to avoid the deadly MUG pattern, you had to have (r1c3 = 1) or (r1c7 = 8) or (r9c9 = 9) be true.

Aren't (r2c6=8) or (r3c5=1) or (r7c5=9) or (r5c4=9) also ways to avoid the deadly MUG pattern?
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Re: Throwing the kitchen sink at MM-14

Postby Myth Jellies » Fri Jun 16, 2006 5:50 am

ronk wrote:...Aren't (r2c6=8) or (r3c5=1) or (r7c5=9) or (r5c4=9) also ways to avoid the deadly MUG pattern?


Yes, the placements in other locations in boxes 2 and 8 avoid the MUG pattern, but they complement, rather than nullify, the conditions I used. The 8 outside the MUG in row 1 requires an 8 outside the MUG in box 2. Thus, you still need a 1, 8, or 9 in either row 1 or 9 which is outside of columns 456 to avoid the MUG.
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Re: Throwing the kitchen sink at MM-14

Postby ronk » Fri Jun 16, 2006 10:40 am

Myth Jellies wrote:
ronk wrote:...Aren't (r2c6=8) or (r3c5=1) or (r7c5=9) or (r5c4=9) also ways to avoid the deadly MUG pattern?

Yes, the placements in other locations in boxes 2 and 8 avoid the MUG pattern, but they complement, rather than nullify, the conditions I used. The 8 outside the MUG in row 1 requires an 8 outside the MUG in box 2.

I see that now for (r2c6=8) or (r3c5=1) or (r7c5=9) ... but what about for (r5c4=9)?

BTW this looks like a BUG-Lite rather than a MUG.
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Postby Myth Jellies » Fri Jun 16, 2006 2:46 pm

The MUG can have more than two avoided digits per cell (in this case three) as opposed to the BUG-lite which has up to two avoided digits per cell.
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Postby ronk » Fri Jun 16, 2006 3:41 pm

Myth Jellies wrote:The MUG can have more than two avoided digits per cell ...

OK, but what about my question about r5c4=9 being true to avoid the deadly MUG? Based only on the MUG and the 4 cells we've discussed, the following are possible placements ...
Code: Select all
 . . x | 6 1 8 | x . .
 . . . | . . . | . . .
 . . . | . . . | . . .
-------+-------+-------
 . . . | . . . | . . .
 . . . | 9 . . | . . .
 . . . | . . . | . . .
-------+-------+-------
 . . . | . . . | . . .
 . . . | . . . | . . .
 . . . | 8 9 1 | . . x

... which forces avoidance of the deadly pattern ... with r1c3=1, r1c7=8, and r9c9=9 all false.
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Postby r.e.s. » Fri Jun 16, 2006 4:54 pm

Myth Jellies wrote:
RW wrote:Couldn't help noticing this nice reduction in the starting grid [...]

If r7c5=7 => r8c2=7 => r8c78=25 => r9c456=189
=> r9c9=empty cell => r7c5<>7

=> r8c5=7[/code]

I haven't found a simple way of making the equivalent deduction without assuming r7c5 = 7.

Here's an AIC:
Code: Select all
7[r8c5]=(3&4&6)[r8c456]-(3v4v6)[r9c456]=(1&8&9)[r9c456]-(1v9)[r9c9]=
(2v5)[r9c9]-(2&5)[r8c78]=(3v4v6)[r8c78]-(3&4&6)[r8c456]=7[r8c5]

=> 7[r8c5] is True
where '&', 'v' stand for AND, OR respectively.
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Postby RW » Fri Jun 16, 2006 5:49 pm

ronlk wrote:OK, but what about my question about r5c4=9 being true to avoid the deadly MUG? Based only on the MUG and the 4 cells we've discussed, the following are possible placements ...

Code: Select all
 . . x | 6 1 8 | x . .
 . . . | . . . | . . .
 . . . | . . . | . . .
-------+-------+-------
 . . . | . . . | . . .
 . . . | 9 . . | . . .
 . . . | . . . | . . .
-------+-------+-------
 . . . | . . . | . . .
 . . . | . . . | . . .
 . . . | 8 9 1 | . . x


... which forces avoidance of the deadly pattern ... with r1c3=1, r1c7=8, and r9c9=9 all false.


Code: Select all
 *-----------------------------------------------------------*
 | 356   4    #1356  |*689  *159  *1689  |#578   2     567   |


Comparing to MJs pencilmarks, your diagram leaves no possible cell for '9' in row 1, that's why it doesn't work. If there would be a 9 outside the pattern in row 1, then r5c4=9 would function as a complement to that cell.

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Postby ronk » Fri Jun 16, 2006 6:46 pm

RW wrote:
Code: Select all
 *-----------------------------------------------------------*
 | 356   4    #1356  |*689  *159  *1689  |#578   2     567   |


Comparing to MJs pencilmarks, your diagram leaves no possible cell for '9' in row 1, that's why it doesn't work.

Precisely, and that's why I qualified my placements with "based only on the MUG and the 4 cells we've discussed".

RW wrote:If there would be a 9 outside the pattern in row 1, then r5c4=9 would function as a complement to that cell.

So one needs to consider ...
  • only those candidates outside the uniquenss pattern which destroy the deadly pattern, and
  • of those, only the candidates which have complements, and
  • of those, only one of each complementary pair (pair?) of candidates.
Is that correct? If so, that's a technique that seems prone to error. IOW it seems too easy to not consider a candidate that should be considered.
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Question to the initial posting

Postby absolute beginner » Fri Jun 16, 2006 9:20 pm

Myth Jellies uses at first:

Step 1:
The starred cells in r78c25 locate a 37 uniqueness rectangle with 2 strong links in the 7's. This implies r8c5 <> 3.

Why can't I use the same argument for the
1-9-rectangle in r79c59.
There are also two strong links in the 9's which
would imply, that r7c9<>9.
This would solve the puzzle

What made I wrong?
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Postby r.e.s. » Fri Jun 16, 2006 10:40 pm

absolute beginner wrote:Myth Jellies uses at first:

Step 1:
The starred cells in r78c25 locate a 37 uniqueness rectangle with 2 strong links in the 7's. This implies r8c5 <> 3.

Why can't I use the same argument for the
1-9-rectangle in r79c59.
There are also two strong links in the 9's which
would imply, that r7c9<>9.

In MJ's UR (r78c25), the strong links on 7 imply that the rectangle reduces to either
Code: Select all
37  | 1349      or        3 | 7
235 | A7                  7 | A346

In the case on the left, A is not 3 (it's 7); and in the case on the right, A is not 3 (because otherwise a uniqueness-destroying 3-7-3-7 rectangle would have entered the puzzle). Therefore, A is not 3.

Now in your rectangle (r79c59), the strong links on 9 imply that the rectangle reduces to either
Code: Select all
9    | 1        or     1347 | 9
1348 | B9             13489 | B125

which does not imply that B is not 9.
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UR's are different?

Postby keith » Fri Jun 16, 2006 10:53 pm

Absolute,

A good question! Let's look at MJ's reduction:
Code: Select all
37a   | 13479c
2357b | 3467d

and there are strong links on <7> in the right column and bottom row.

Now, if a is <7>, the strong links force d to be <7> and not <3>.
If a is <3>, d cannot be <3>, because that would force <7> at b, c, which is the deadly pattern.

You will see that the strong links do not involve the cell with only two candidates. If you look at the other UR you have noted, the situation is different.

Keith
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