Down Under Upside Down - a Sudoku puzzle

Post puzzles for others to solve here.

Postby gsf » Mon Nov 24, 2008 7:59 pm

Mauricio wrote:Have fun with the following puzzles:
Each one has 6 distinct automorphisms (one of them being the trivial one). See if you can find them all.

my solver -f%#aC gets these automorphism counts, respectively
Code: Select all
6 18 6 54 54 18 54 6 6 6 108
gsf
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Location: NJ USA

Postby eleven » Tue Nov 25, 2008 5:19 am

Mauricio wrote:Have fun with the following puzzles:

Thanks for the puzzles.

As to the first one (only boxes in band 1 might map on one another)
Code: Select all
 +-------+-------+-------+
 | . . . | . . 1 | . 2 . |
 | . 1 . | . . . | . . 3 |
 | . . 2 | . 3 . | . . . |
 +-------+-------+-------+
 | . . 4 | . 5 . | 6 . . |
 | 5 . . | . . 7 | . 8 . |
 | . 6 . | 8 . . | . . 9 |
 +-------+-------+-------+
 | . . 7 | . 8 . | 9 . . |
 | 4 . . | . . 9 | . 6 . |
 | . 5 . | 7 . . | . . 4 |
 +-------+-------+-------+

2 cyclic column changes (123,456) and one for rows 123 lead to
Code: Select all
 +-------+-------+-------+
 | . . 1 | . . . | . . 3 |
 | 2 . . | 3 . . | . . . |
 | . . . | . 1 . | . 2 . |
 +-------+-------+-------+
 | 4 . . | 5 . . | 6 . . |
 | . 5 . | . 7 . | . 8 . |
 | . . 6 | . . 8 | . . 9 |
 +-------+-------+-------+
 | 7 . . | 8 . . | 9 . . |
 | . 4 . | . 9 . | . 6 . |
 | . . 5 | . . 7 | . . 4 |
 +-------+-------+-------+

which is a well known symmetry with number cycles (132)(479)(586), the mapping is given by cyclic stack changes and row changes in all bands.

I dont know now, where the other 3 automorphisms are, but its already easy to solve.
[Added: ah, band 2 maps to band 3 with (457896)]
eleven
 
Posts: 1580
Joined: 10 February 2008

Postby Mauricio » Tue Nov 25, 2008 9:53 am

gsf wrote:
Mauricio wrote:Have fun with the following puzzles:
Each one has 6 distinct automorphisms (one of them being the trivial one). See if you can find them all.

my solver -f%#aC gets these automorphism counts, respectively
Code: Select all
6 18 6 54 54 18 54 6 6 6 108

I guess that counts the number of automorphisms of the solutions, not of the puzzles.
Mauricio
 
Posts: 1174
Joined: 22 March 2006

Postby eleven » Tue Nov 25, 2008 10:12 pm

Btw, i found [edit:] 2 automorphisms of the puzzle above as described (starting from the equivalent puzzle shown, A1 is " bands right, rows down in all bands, number cycling", A2 is change bands 23, number cycling [edit 2: when i retried it now , i needed band/stack/row/column switches and number cycles (13)(2)(46)(59)(78)]).
You can apply A1 2 times and you can mix them, e.g. A1A2A1 to get 6 different automorphic mappings. But should we count that ?
We also dont count the identity more times, though we can get it with A2A2 or A1A1A1.


[Added:]
The 3rd puzzle was quite interesting.
Code: Select all
 +-------+-------+-------+
 | . . . | . . 1 | . . 2 |
 | . . 3 | . . . | . 4 . |
 | 5 . . | . 6 . | 7 . . |
 +-------+-------+-------+
 | . . 6 | . . . | . . 5 |
 | . 1 . | . . 7 | . . . |
 | 8 . . | 4 . . | . 9 . |
 +-------+-------+-------+
 | . . 4 | . . 9 | . . . |
 | . . . | . 5 . | . . 8 |
 | . 2 . | 3 . . | 1 . . |
 +-------+-------+-------+

With bands down (cyclically), stacks right and number cycles (154)(269)(378) you have this familiar symmetry of cyclic boxes mapping 159|267|348
But this does not easily solve it.

If you move the bands up however you have a diagonal symmetry, and its solved with singles.
Code: Select all
 +-------+-------+-------+
 | . . 6 | . . . | . . 5 |
 | . 1 . | . . 7 | . . . |
 | 8 . . | 4 . . | . 9 . |
 +-------+-------+-------+
 | . . 4 | . . 9 | . . . |
 | . . . | . 5 . | . . 8 |
 | . 2 . | 3 . . | 1 . . |
 +-------+-------+-------+
 | . . . | . . 1 | . . 2 |
 | . . 3 | . . . | . 4 . |
 | 5 . . | . 6 . | 7 . . |
 +-------+-------+-------+
eleven
 
Posts: 1580
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Postby udosuk » Wed Nov 26, 2008 4:52 am

I think these are the 6 automorphisms of the 1st puzzle:

Code: Select all
.....1.2.
.1......3
..2.3....
..4.5.6..
5....7.8.
.6.8....9
..7.8.9..
4....9.6.
.5.7....4

Automorphism 1 (trivial)
r123456789
c123456789
d123456789 -> d123456789

Automorphism 2
r231564897
c456789123
d123456789 -> d231968457

Automorphism 3
r312645978
c789123456
d123456789 -> d312785964

Automorphism 4
r132987654
c132798465
d123456789 -> d213547698

Automorphism 5
r321879546
c798465132
d123456789 -> d321694875

Automorphism 6
r213798465
c465132798
d123456789 -> d132879546

:idea:
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Wed Nov 26, 2008 6:00 am

Thanks, i see, what is meant.

I cannot check that in detail now, but i suppose, your automorphisms 2 and 4 are equivalent to the two i found.
And you get number 3 by applying 2 two times, and number 5 by applying 2 and 4 and number 6 is 3+4.

What i wanted to say is, that there are only 2 relevant automorphisms, one with cycle 3 (to get the identity again) and the other with cycle 2. The rest are combinations of them. From a solvers point of view, those will not give you any more useful information.

So i am thinking in symmetries, not automorphisms.
eleven
 
Posts: 1580
Joined: 10 February 2008

Postby udosuk » Wed Nov 26, 2008 7:06 am

Yep, there are only 2 simultaneous symmetry classes for each puzzle, one with cycle-2 & one with cycle-3. Together they form 2x3=6 automorphisms.

I take it you haven't noticed what the cycle-2 symmetry really was?:)

Try reordering the rows/columns as:

Code: Select all
r456213789
c564213789

And you can do a similar thing for puzzle 4...:idea:
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Thu Nov 27, 2008 2:12 am

Not sure, what you mean. I tried nr 4 now. One symmetry is obvious in the given puzzle:
Code: Select all
 +-------+-------+-------+
 | . . . | . . 1 | . . 2 |
 | . 1 . | 3 . . | 4 . . |
 | 2 . . | . 4 . | . 5 . |
 +-------+-------+-------+
 | . . 6 | . . . | . . 5 |
 | 7 . . | . 5 . | 2 . . |
 | . 8 . | 6 . . | . 7 . |
 +-------+-------+-------+
 | . . 8 | . . 3 | . . . |
 | 6 . . | 9 . . | . 8 . |
 | . 9 . | . 1 . | 3 . . |
 +-------+-------+-------+

A1: bands down, stacks right, numbers (158)(263)(479)
This also can be used for solving relatively easy, boxes 149|267|348 map cyclically.

Since we have 3 boxes with 2 givens, they either can map cyclically or by mirroring in the diagonal or by exchanging two of them.
Diagonal symmetry is not possible here, because there are no candidates for the diagonal: All numbers appear at least twice in a band or stack.

So at last you can try to swap two of the 2 number boxes. It does not matter which ones here, because if one mapping works, you can get the others with the first symmetry.
Exchanging 2 boxes in the diagonal can be done by swapping the 2 bands and stacks. (Maybe you meant this, but i dont know, why this mapping also swaps rc12 and cycles c456). Then (if the numbers cant match) with row/col changes you can try to get other pattern matches. In this case with
A2: switch bands 12, stacks 12, rows 23,56,89, cols 12,45,78
you get
Code: Select all
 +-------+-------+-------+
 | . . . | . . 6 | . . 5 |
 | . 6 . | 8 . . | 7 . . |
 | 5 . . | . 7 . | . 2 . |
 +-------+-------+-------+
 | . . 1 | . . . | . . 2 |
 | 4 . . | . 2 . | 5 . . |
 | . 3 . | 1 . . | . 4 . |
 +-------+-------+-------+
 | . . 3 | . . 8 | . . . |
 | 1 . . | 9 . . | . 3 . |
 | . 9 . | . 6 . | 8 . . |
 +-------+-------+-------+

and the number cycles (16)(25)(38)(47)(9) map the original puzzle here.

This symmetry is hard to use for solving, because there is no easy way (i know) to see, where each cell maps to.

[Added:] I just noticed, that this second symmetry leaves r7c9 unchanged (so it must be 9). But it cant be equivalent to the 180 degree symmetry.
[Added 2:] No wait, think it is. (I really need a better tool than an editor.)
eleven
 
Posts: 1580
Joined: 10 February 2008

Postby udosuk » Thu Nov 27, 2008 8:15 pm

Mauricio wrote:Have fun with the following puzzles:
Code: Select all
.....1.2..1......3..2.3......4.5.6..5....7.8..6.8....9..7.8.9..4....9.6..5.7....4
.....1.2..1......3..2.3......45...6..5...78..6...8...9..97...8..6...49..4...5...7
.....1..2..3....4.5...6.7....6.....5.1...7...8..4...9...4..9.......5...8.2.3..1..
.....1..2.1.3..4..2...4..5...6.....57...5.2...8.6...7...8..3...6..9...8..9..1.3..
.....1..2.1.3..4..2...5..6...7.....65...6.2...8.7...9...8..3...7..9...8..4..1.3..
.....1..2.3.2..4..5...6..1...7.....48...5.7...4.9...2...8..6...6..1...9..7..8.3..
.....1.2...32....4.5..6.1...3......7..6..83..7...1..9...5.8....8....9..2.4.5...7.
..1..2..3.2..3..4.5..1..6....5..7..6.1..6..5.8..9..7....8..9..4.9..7..8.4..2..3..
..1..2.3..3.4....25...6.1...4...7..8..8.4.9..7..2...1...5.9...63....5.9..7.6..8..
.....1..2..3.2..4..5.6..7....7.....1.4...5.7.8...9.6....1..8....8..4...96..2...3.
.....1..2.1.3..4..2...5..6...7.....64...6.2...8.7...5...8..3...7..4...8..5..1.3..

The following row/column rearrangement will expose all symmetry classes associated with the corresponding puzzle:
Code: Select all
r978123456
c213546879

r897123456
c213546879

r456789123
c123456789

r213546879
c132465798

r213546879
c132465798

r213546879
c132465798

r213546879
c213546879

r231564897
c123456789

r123456789
c123456789

r456123789
c123456789

r213546879
c132465798

Some of them are dead easy and some a little bit more tricky.

For #4 I still can't find any easy path to crack it, even applying symmetrical tricks on both classes.:(
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby udosuk » Sun Nov 30, 2008 2:15 am

Okay, let's get things into perspective regarding the 11 doubly-symmetrical puzzles posted by Mauricio.

Two of them (1,2) have band/stack-permutation symmetry plus 180-degree-rotation symmetry
Four of them (3,8,9,10) have block-(diagonal-)permutation symmetry plus diagonal-reflection symmetry
Five of them (4,5,6,7,11) have block-(diagonal-)permutation symmetry plus 180-degree-rotation symmetry

The first 2 groups are all very easy to solve, as in the following summary:

Puzzle 1 - solvable by gludo+naked pairs
Puzzle 2 - solvable by gludo+naked pairs
Puzzle 3 - solvable by diagonal placements+singles
Puzzle 8 - solvable by triples or diagonal placements+naked singles
Puzzle 9 - solvable by diagonal placements+naked pairs
Puzzle 10 - solvable by naked pairs or diagonal placements+singles

However, the 3rd group (of 5 puzzles) are all quite tricky, as we have not (yet) developed any powerful technique for "block symmetry" that's comparable to gludo. Here I'll list out my solution paths for each of them, in ascending order of their difficulties (as perceived by me):

Code: Select all
Puzzle 5:

r213546879
c132465798
+----------------------+----------------------+----------------------+
| 689    569    1      | 3      26789  2789   | 4      5789   57     |
| 34689  34569  35679  | 468    1      4789   | 5789   2      357    |
| 2      349    379    | 48     4789   5      | 1789   13789  6      |
+----------------------+----------------------+----------------------+
| 5      1349   39     | 148    489    6      | 2      13478  1347   |
| 1349   7      239    | 12458  24589  23489  | 158    6      1345   |
| 1346   12346  8      | 7      245    234    | 15     1345   9      |
+----------------------+----------------------+----------------------+
| 7      12356  2356   | 9      2456   24     | 156    145    8      |
| 169    8      2569   | 2456   3      247    | 15679  14579  12457  |
| 69     2569   4      | 2568   25678  1      | 3      579    257    |
+----------------------+----------------------+----------------------+

GSP (Gurth's Symmetrical Placement): r5c5=4
BkS (Block Symmetry): r8c8=5, r2c2=9
Now r1c2 from {56} & r1c9 from {57} must have 6 or 7
=> r1c1+r9c9 can't be [67], must be [82]
BkS: r3c3+r4c4+r6c6+r7c7=[7136]
Naked Pair @ c2: r34c2={34}
All singles from here.

Code: Select all
Puzzle 6:

r213546879
c132465798
+----------------------+----------------------+----------------------+
| 179    169    3      | 2      5789   79     | 4      56789  5678   |
| 479    469    689    | 34578  1      3479   | 5689   2      35678  |
| 5      249    289    | 3478   34789  6      | 89     3789   1      |
+----------------------+----------------------+----------------------+
| 8      12369  1269   | 346    234    5      | 7      1369   36     |
| 1239   7      12569  | 368    238    123    | 15689  4      3568   |
| 13     1356   4      | 9      378    137    | 1568   13568  2      |
+----------------------+----------------------+----------------------+
| 6      2345   25     | 1      23457  2347   | 258    578    9      |
| 12349  8      1259   | 3457   6      23479  | 125    157    457    |
| 1249   12459  7      | 45     2459   8      | 3      156    456    |
+----------------------+----------------------+----------------------+

GSP: r5c5=3
BkS: r8c8=5, r2c2=9
SPP (Symmetrical Pointing Pair): r3c3+r7c7={28} => r3c7=9
BkS: r6c1=3, r9c4=5
Hidden Single @ b2: r1c5=5
SPP: r2c1+r8c9={47} => r8c1<>4
BkS: r2c4<>8, r5c7<>1
Intersection: 8 @ r3,b2 locked @ r3c45
All singles from here.

Code: Select all
Puzzle 11:

r213546879
c132465798
+----------------------+----------------------+----------------------+
| 5689   569    1      | 3      26789  2789   | 4      5789   79     |
| 35689  34569  34679  | 689    1      4789   | 5789   2      379    |
| 2      349    3479   | 89     4789   5      | 1789   13789  6      |
+----------------------+----------------------+----------------------+
| 4      1359   39     | 1589   589    6      | 2      13789  1379   |
| 1359   7      239    | 12589  24589  23489  | 189    6      1349   |
| 1369   12369  8      | 7      249    2349   | 19     1349   5      |
+----------------------+----------------------+----------------------+
| 7      12369  2369   | 4      2569   29     | 1569   159    8      |
| 169    8      2469   | 2569   3      279    | 15679  14579  12479  |
| 69     2469   5      | 2689   26789  1      | 3      479    2479   |
+----------------------+----------------------+----------------------+

GSP: r5c5=9
BkS: r8c8=9, r2c2=9
BkS: r2c4+r5c7+r8c1={168} => 1 locked @ r5c7+r8c1
Also 1 @ c9 locked @ r458c9
Grouped Turbot Fish: r8c19 can't be both 1
=> At least one of r45c9+r5c7 must be 1
=> r4c8+r7c78, seeing r45c9+r5c7, can't have 1
Intersection: 1 @ r6,b4 locked @ r6c12
Naked Pair @ b4: r4c2+r5c1={35} => r5c3=2
BkS: r8c6=7, r2c9=3
All singles from here.

Code: Select all
Puzzle 7:

r213546879
c213546879
+----------------------+----------------------+----------------------+
| 16789  169    3      | 579    2      57     | 568    56789  4      |
| 6789   469    4789   | 34579  34789  1      | 2      56789  35689  |
| 5      249    24789  | 6      34789  347    | 38     1      389    |
+----------------------+----------------------+----------------------+
| 129    12459  6      | 24579  479    8      | 145    3      15     |
| 3      12459  12489  | 2459   469    2456   | 14568  24568  7      |
| 28     7      248    | 1      346    23456  | 9      24568  568    |
+----------------------+----------------------+----------------------+
| 167    8      17     | 347    13467  9      | 13456  456    2      |
| 12679  12369  5      | 8      13467  23467  | 1346   469    1369   |
| 4      12369  129    | 23     5      236    | 7      689    13689  |
+----------------------+----------------------+----------------------+

GSP: r5c5=4
BkS: r8c8=6, r2c2=9
BkS: r3c7+r6c1+r9c4={238} => 2 locked @ r6c1+r9c4
Also 2 @ c8,b6 locked @ r56c8
Turbot Fish: r6c18 can't be both 2
=> At least one of r5c8+r9c4 must be 2
=> r5c4, seeing r5c8+r9c4, can't be 2
XYZ-wing: r1c4 from {579}, r1c6 from {57}, r5c4 from {59}
=> r1c6+r25c4 can't be [759]
=> r2c4, seeing r1c46+r5c4, can't be 5
Intersection: 5 @ r1,b2 locked @ r1c46
Now 5 @ r2,b3 locked @ r2c89
Also 5 @ c2,b4 locked @ r45c2
Turbot Fish: r2c8+r5c2 can't be both 5 (BkS)
=> At least one of r2c9+r4c2 must be 5
=> r4c9, seeing r2c9+r4c2, can't be 5, must be 1
Intersection: 1 @ r9,b7 locked @ r9c23
All singles from here.

Code: Select all
Puzzle 4:

r213546879
c132465798
+-------------------------+-------------------------+-------------------------+
| 589     579     1       | 3       256789  26789   | 4       6789    69      |
| 34589   34579   34567   | 578     1       6789    | 6789    2       369     |
| 2       379     367     | 78      6789    4       | 16789   136789  5       |
+-------------------------+-------------------------+-------------------------+
| 7       1349    34      | 148     489     5       | 2       134689  13469   |
| 1349    6       234     | 12478   24789   23789   | 189     5       1349    |
| 13459   123459  8       | 6       249     239     | 19      1349    7       |
+-------------------------+-------------------------+-------------------------+
| 6       123457  23457   | 9       2457    27      | 157     147     8       |
| 145     8       2457    | 2457    3       267     | 15679   14679   12469   |
| 45      2457    9       | 24578   245678  1       | 3       467     246     |
+-------------------------+-------------------------+-------------------------+

(Quick and dirty path)
Grouped Turbot Chain: 1 @ c9 locked @ r458c9
=> r6c78 & r8c1 can't both have 1
Now 1 @ r6 locked @ r6c1278 & 1 @ b7 locked @ r7c2+r8c1
=> At least one of r6c12+r7c2 must have 1
=> r4c2, seeing r6c12+r7c2, can't be 1
180 Symmetry: r6c8 can't be 3
BkS: r9c2 can't be 2
Hidden Pair @ c2: r67c2={12}
All singles from here.

For #4 there ought to be a more elegant way to crack it (I hate anything associated with the word "chain"), but I don't have time to dig deep now. Good luck to whoever trying it (eleven? Glyn? Mauricio?).

PS: Wholehearted regards for Glyn's mum who has encountered a little mishap recently. Fingers crossed for a great recovery!

Added later:

Here is an alternative path for #4:

GSP: r5c5=7
BkS: r8c8=9, r2c2=4
Now 2 @ c4 locked @ r589c4
=> r5c3+r7c6 can't both be 2
BkS: r5c3+r78c6 can't be [226]
=> r78c6 can't be [26], must have 7
=> 7 @ c6,b8 locked @ r78c6
X-wing: 7 @ c28 locked @ r19c28
Again, 2 @ c4 locked @ r589c4
=> r5c3+r7c5 can't both be 2
BkS: r4c23+r5c3+r7c5 can't be [3422]
=> r4c2 can't be 3
Now r3c2 from {39}, r4c2 from {19}
HEUR (Half Emerald Unlocking Rectangle): r37c2 can't be [31]
=> r347c2 can't be [391]
=> r7c2 can't be 1
Hidden Single @ b7: r8c1=1
Intersection: 1 @ c9,b6 locked @ r45c9
All singles from here.

It is not necessarily more elegant than the "quick and dirty path", but for me the critical moves are somehow easier to spot.

:idea:
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby Mauricio » Fri Dec 05, 2008 4:50 am

Directly from Red Ed's class 10
Code: Select all
+-------+-------+-------+
| . . . | . . 1 | . . 2 |
| . . . | 3 . . | 4 . . |
| . . . | . 5 . | . 6 . |
+-------+-------+-------+
| . . 1 | . . 7 | . 8 . |
| 3 . . | 9 . . | . . 7 |
| . 5 . | . 8 . | 9 . . |
+-------+-------+-------+
| . . 4 | 7 . . | . . 1 |
| 6 . . | . 9 . | 3 . . |
| . 2 . | . . 8 | . 5 . |
+-------+-------+-------+
Code: Select all
+-------+-------+-------+
| . . 1 | . . 2 | . . 3 |
| 2 . . | 4 . . | 5 . . |
| . 4 . | . 1 . | . 6 . |
+-------+-------+-------+
| . . 3 | . . 7 | . 2 . |
| 5 . . | 8 . . | . . 4 |
| . 6 . | . 9 . | 1 . . |
+-------+-------+-------+
| . . 4 | . . 9 | . . 8 |
| 1 . . | 7 . . | 9 . . |
| . 2 . | . 8 . | . 7 . |
+-------+-------+-------+
Mauricio
 
Posts: 1174
Joined: 22 March 2006

Postby udosuk » Sun Dec 07, 2008 4:41 am

Mauricio wrote:
Code: Select all
+-------+-------+-------+
| . . 1 | . . 2 | . . 3 |
| 2 . . | 4 . . | 5 . . |
| . 4 . | . 1 . | . 6 . |
+-------+-------+-------+
| . . 3 | . . 7 | . 2 . |
| 5 . . | 8 . . | . . 4 |
| . 6 . | . 9 . | 1 . . |
+-------+-------+-------+
| . . 4 | . . 9 | . . 8 |
| 1 . . | 7 . . | 9 . . |
| . 2 . | . 8 . | . 7 . |
+-------+-------+-------+

This (2nd) one is very easy. Just move the top band to the bottom (i.e. rearrange the rows into r456789123), then you have reflective symmetry about the non-leading diagonal. Very easy to solve after that diagonal is filled in.

However, the 1st one is not an automorphic puzzle I'm afraid. The solution might be highly automorphic, but the puzzle doesn't match any automorphism. I've tried the 6 main types (180, 90, diagonal, sticks, bands, blocks), none of them works. Perhaps a mistake from Mauricio?:(
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Sun Dec 07, 2008 6:59 am

The symmetry both puzzles have, is new for me. Just move the rows in the bands down cyclically and the columns in the stacks right, i.e.
r123456789 -> r231564897
c123456789 -> c231564897
The number cycles are (134)(246)(798) and (124)(356)(789) resp.

But for the first puzzle i only found one elimination using the symmetry (r1c7<>7, because r1c7=7 places all 7's and 9's, but with a contradiction in box 1).
eleven
 
Posts: 1580
Joined: 10 February 2008

Postby udosuk » Sun Dec 07, 2008 8:05 pm

Thanks eleven! I totally missed that obvious automorphism. Was too keen on morphing for "easy to use" symmetries.

I guess that makes the 7th main type of automorphism. Perhaps we could call it "mini-diagonal" (although it also includes "mini-broken-diagonals":) ).

eleven wrote:But for the first puzzle i only found one elimination using the symmetry (r1c7<>7, because r1c7=7 places all 7's and 9's, but with a contradiction in box 1).

I don't see any obvious contradiction. Could you elaborate more please?

From what I see placing 7 @ r1c7 enables one to place all {789} with 2 choices on c123. However the eventual contradictions all need to bring {135} or {246} into consideration, and occur after a whole lot of other placements, that it's very hard not to be considered as "major T&E".

On the other hand, I have this approach (which I haven't confirmed to work yet):
Code: Select all
Rearrange the rows to r231564897:

+-------+-------+-------+
| . . . | 3 . . | 4 . . |
| . . . | . 5 . | . 6 . |
| . . . | . . 1 | . . 2 |
+-------+-------+-------+
| 3 . . | 9 . . | . . 7 |
| . 5 . | . 8 . | 9 . . |
| . . 1 | . . 7 | . 8 . |
+-------+-------+-------+
| 6 . . | . 9 . | 3 . . |
| . 2 . | . . 8 | . 5 . |
| . . 4 | 7 . . | . . 1 |
+-------+-------+-------+

You can see the clue pattern has "diagonal reflective symmetry" now. However, that doesn't make it a puzzle with "diagonal reflective automorphism". Anyway, if we reflect the puzzle along the leading diagonal, only 6 clues are changed, which are all from {246}. That prompts me to think the key to crack this puzzle might be on the digits {246}. Anyway, after peeking at the solution (yes I know this is cheating) I found that the 3 missing digits on the leading diagonal are indeed {246}. If only we can develop a sound logical argument that it is a direct consequence of the "pattern symmetry" above...:idea:
udosuk
 
Posts: 2698
Joined: 17 July 2005

Postby eleven » Mon Dec 08, 2008 12:47 am

udosuk wrote:
eleven wrote:But for the first puzzle i only found one elimination using the symmetry (r1c7<>7, because r1c7=7 places all 7's and 9's, but with a contradiction in box 1).

I don't see any obvious contradiction. Could you elaborate more please?

Sorry, no, i managed to make the same mistake twice:(
On the other hand, I have this approach (which I haven't confirmed to work yet)...

Interesting, but from my feeling i cant believe, that it works. I guess, that partial symmetry does not say anything about the rest of the grid.
eleven
 
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Joined: 10 February 2008

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