## Down Under Upside Down - a Sudoku puzzle

Post puzzles for others to solve here.
eleven wrote:
Code: Select all
+----------------------+----------------------+----------------------+
| 1     234569  23456  | 3579  34567   8      | 267   2456    24567  |
| 348   3456    34568  | 2     134567  13456  | 1678  14568   9      |
| 2489  24569   7      | 159   1456    14569  | 3     124568  12456  |
+----------------------+----------------------+----------------------+
| 6     245     2458   | 1589  158     7      | 129   3       124    |
| 237   1       235    | 4     356     3569   | 2679  269     8      |
| 3478  347     9      | 138   2       136    | 5     146     1467   |
+----------------------+----------------------+----------------------+
| 379   8       36     | 1357  1357    2      | 4     1569    156    |
| 5     2467    246    | 178   9       14     | 1268  1268    3      |
| 2349  2349    1      | 6     3458    345    | 289   7       25     |
+----------------------+----------------------+----------------------+
Only a small forcing net is needed:
r7c3=3 -> (symm) r8c6=1 -> r7c45=57 -> r8c4=8 -> (symm) r7c1=7 - contradiction (2 7's in row 7)

This gives r7c3=6, r8c6=4, r9c9=5 and the rest is easy.

I'll present the move the following way:

r7c45 from {1357} must have at least one of {137}
=> r7c13+r8c46 can't be [7381|9371]
=> r7c3 can't be 3

Then it still takes turbot fish & xy-wing to finish the puzzle. But very nice solution, compared to the hellish forcing chains it would take without using symmetry.

udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:Then it still takes turbot fish & xy-wing to finish the puzzle.
The 48-xy-wing in r2c1 was easy to spot (r3c1<>9).
This puzzle is very tough also when using symmetry moves. After 3 long chains, that only brought me 2 (i.e. 6) numbers, i gave up.
eleven

Posts: 1710
Joined: 10 February 2008

on page 3, udosuk wrote:
on page 2, eleven wrote:
Code: Select all
. 9 6 | . . 5 | . 1 .
. . . | 3 . . | . . .
2 . . | . . 6 | 5 . 9
-------+-------+------
. . 5 | . . . | 6 . .
9 4 2 | . . . | 3 . .
. . 3 | . . . | 1 9 4
-------+-------+------
. 2 . | . . 4 | . . 1
5 . . | 8 . 2 | 4 . 7
. 8 4 | 7 . 1 | . 6 .

I decided to perform the following operations on the puzzle,
which transformed it back to the original puzzle --

Code: Select all
Swap r13
Swap r56
Swap r89

Permute c123 to c312
Permute c789 to c897
Swap left stack with right stack

swap digits 1,2
swap digits 5,6
swap digits 7,8

[ here i skipped much of udosuk's explanation,
mostly because i'd phrase it differently
]

because the "axis of symmetry" for this puzzle is r247c456,
an immediate first move is all 9 cells of r247c456 must be from {3,4,9}.

hey udosuk,

the transformation (which brought us back to the original puzzle)
left 3 digits stationary, also 3 stationary columns and 3 stationary rows;
assuming the puzzle has just one answer,
the stationary cells must take values from the stationary digits!
which means 3 "naked" trios in columns, and 3 "naked" trios in rows
-- as you said, r247c456 must be from {3,4,9}
is it possible to have a different quantity?

Pat

Posts: 3506
Joined: 18 July 2005

on page 3, eleven wrote:Lets take this one to demonstrate, how a box symmetry can be used to solve an otherwise extremely hard puzzle --
Code: Select all
Version 2

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

ER=10.4

Hint: Observe boxes 1-2-3,4-5-6,7-8-9 and digits 1-2-3,4-5-6,7-8-9.

It has a threefold symmetry,
each minirow maps to the (cyclically) next one in the neighbour box.
E.g if r7c3=3, you have r8c6=1 and r9c9=2.

hi eleven,

so i see this "symmetry" in the clues
and i see the same "symmetry" in the answer (for this specific example)
but i'm dense --
how does this "symmetry" in the clues prove the same "symmetry" in the answer?

Pat

Posts: 3506
Joined: 18 July 2005

Pat wrote:hey udosuk,

the transformation (which brought us back to the original puzzle)
left 3 digits stationary, also 3 stationary columns and 3 stationary rows;
assuming the puzzle has just one answer,
the stationary cells must take values from the stationary digits!
which means 3 "naked" trios in columns, and 3 "naked" trios in rows
-- as you said, r247c456 must be from {3,4,9}
is it possible to have a different quantity?

On page 3, I wrote:I have a conjecture that in all kinds of symmetry the "axis" must only be of 1 cell (e.g. rotational) or 9 cells spanning 3 blocks (e.g. diagonal reflection, or this puzzle). For 1-cell-axis only 1 digit maps to itself while for 9-cell-axis we have 3 self-mapping digits.

This applies to all 4 symmetries with a fixed "axis" (180 rotation, 90 rotation, diagonal reflection, sticks). So I believe there are either 1 or 3 stationary digits, and no other quantity. You can try to prove me wrong by creating a counter-example. I don't think I can!

Pat wrote:
Code: Select all
Version 2

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

ER=10.4

Hint: Observe boxes 1-2-3,4-5-6,7-8-9 and digits 1-2-3,4-5-6,7-8-9.

but i'm dense --
how does this "symmetry" in the clues prove the same "symmetry" in the answer?

Do the following transformation:

r123456789 -> r312645978
c123456789 -> c789123456
digit [123456789] -> digit [231564897]

You can transform the puzzle back to itself.

As a result, you can establish "symmetrical relationships" among all cell values in the solution grid.
udosuk

Posts: 2698
Joined: 17 July 2005

Aha, you are talking about the discovered and rediscovered and rerediscovered nice properties of automorphic sudokus:

http://forum.enjoysudoku.com/viewtopic.php?p=46233#p46233
http://forum.enjoysudoku.com/viewtopic.php?p=32833#p32833
Pat wrote:
on page 3, eleven wrote:Lets take this one to demonstrate, how a box symmetry can be used to solve an otherwise extremely hard puzzle --
Code: Select all
Version 2

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

ER=10.4

Hint: Observe boxes 1-2-3,4-5-6,7-8-9 and digits 1-2-3,4-5-6,7-8-9.

It has a threefold symmetry,
each minirow maps to the (cyclically) next one in the neighbour box.
E.g if r7c3=3, you have r8c6=1 and r9c9=2.

hi eleven,

so i see this "symmetry" in the clues
and i see the same "symmetry" in the answer (for this specific example)
but i'm dense --
how does this "symmetry" in the clues prove the same "symmetry" in the answer?

And BTW, that puzzle was first posted by me in http://forum.enjoysudoku.com/viewtopic.php?p=40236#p40236 in January 2007, exposing the properties here discussed.
Mauricio

Posts: 1174
Joined: 22 March 2006

Hi Mauricio,

The new thing was, that symmetry techniques also can be used to solve puzzles with an automorphism, where no cell is mapped to itself (but probably not, when each number maps to itself, i needed the number cycles for additional information).

If you still have a collection of such puzzles, it would be nice, if you could post them in your thread.
eleven

Posts: 1710
Joined: 10 February 2008

eleven wrote:Lets take this one to demonstrate, how a box symmetry can be used to solve an otherwise extremely hard puzzle:
Code: Select all
Version 2
1 . .|. . 8|. . .
. . .|2 . .|. . 9
. . 7|. . .|3 . .
-----+-----+-----
6 . .|. . 7|. 3 .
. 1 .|4 . .|. . 8
. . 9|. 2 .|5 . .
-----+-----+-----
. 8 .|. . 2|4 . .
5 . .|. 9 .|. . 3
. . 1|6 . .|. 7 . ER=10.4
Hint:Observe boxes 1-2-3,4-5-6,7-8-9 and digits 1-2-3,4-5-6,7-8-9.

It has a threefold symmetry, each minirow maps to the (cyclically) next one in the neighbour box. E.g if r7c3=3, you have r8c6=1 and r9c9=2.

Code: Select all
*--------------------------------------------------------------------*
| 1     234569  23456  | 3579  34567   8      | 267   2456    24567  |
| 348   3456    34568  | 2     134567  13456  | 1678  14568   9      |
| 2489  24569   7      | 159   1456    14569  | 3     124568  12456  |
|----------------------+----------------------+----------------------|
| 6     245     2458   | 1589  158     7      | 129   3       124    |
| 237   1       235    | 4     356     3569   | 2679  269     8      |
| 3478  347     9      | 138   2       136    | 5     146     1467   |
|----------------------+----------------------+----------------------|
| 379   8       36     | 1357  1357    2      | 4     1569    156    |
| 5     2467    246    | 178   9       14     | 1268  1268    3      |
| 2349  2349    1      | 6     3458    345    | 289   7       25     |
*--------------------------------------------------------------------*
Only a small forcing net is needed:
r7c3=3 -> (symm) r8c6=1 -> r7c45=57 -> r8c4=8 -> (symm) r7c1=7 - contradiction (2 7's in row 7)

This gives r7c3=6, r8c6=4, r9c9=5 and the rest is easy.

Here is my effort
If r7c1=7 => r8c4=8,r9c7=9 => r7c1=9 contradiction
If r6c1=7 => r4c4=8 => r6c1=8 contradiction
Leaving only r5c1=7 with immediate r6c4=8,r4c7=9.
Singles to follow
Glyn

Posts: 357
Joined: 26 April 2007

Nice solution, even simpler than mine (suppose it would be rated about ER=7.2 !)

Pat wrote:how does this "symmetry" in the clues prove the same "symmetry" in the answer?
The point is, that it must be guaranteed, that the puzzle is unique. If so, and you can transform it to itself with operations like udosuk noted above (those "symmetric" puzzles are defined to have such an automorphism), also the solution must have this "symmetry" (the original and the transformed puzzles are identical and have the same solution).
eleven

Posts: 1710
Joined: 10 February 2008

eleven wrote:After a second thought i cant believe, that other symmetries would not be useful for a solution. E.g. in this Puzzle from this post
http://forum.enjoysudoku.com/viewtopic.php?p=40236#p40236
Code: Select all
Version 1
1 . .|. . 4|. 8 .
. 9 .|1 . .|. . 5
. . 6|. 7 .|1 . .
-----+-----+-----
. 3 .|. . 2|4 . .
5 . .|. 3 .|. . 2
. . 2|6 . .|. 3 .
-----+-----+-----
. 8 .|3 . .|. . 1
. . 1|. 9 .|3 . .
3 . .|. . 1|. 7 . ER=??(Not tested)
Hint: Observe boxes 1-2-3,4-5-6,7-8-9 and digits 1-1,2-2,3-3,4-5-6,7-8-9.
*--------------------------------------------------------------------*
| 1      257    357    | 259    256    4      | 2679   8      3679   |
| 2478   9      3478   | 1      268    368    | 267    246    5      |
| 248    245    6      | 2589   7      3589   | 1      249    349    |
|----------------------+----------------------+----------------------|
| 6789   3      789    | 5789   158    2      | 4      1569   6789   |
| 5      1467   4789   | 4789   3      789    | 6789   169    2      |
| 4789   147    2      | 6      1458   5789   | 5789   3      789    |
|----------------------+----------------------+----------------------|
| 24679  8      4579   | 3      2456   567    | 2569   24569  1      |
| 2467   24567  1      | 24578  9      5678   | 3      2456   468    |
| 3      2456   459    | 2458   24568  1      | 25689  7      4689   |
*--------------------------------------------------------------------*

we only know from the symmetry, that a 4 in r2c13 implies a 5 in r3c46 and a 6 in r1c79, thus a 2 in r2c8 and so on.
I dont have the time now (and probably the next days) to look for eliminations from that, but the puzzle must be easier to solve with this additional informations.

If we made an additional constraint based on an observation that the givens 1,2,3 do not simultaneously share a row and box and apply that to the pencilmarks we get

Code: Select all
.------------------.------------------.------------------.
| 1     57    57   | 259   256   4    | 2679  8     3679 |
| 2478  9     3478 | 1     68    68   | 267   246   5    |
| 248   245   6    | 2589  7     3589 | 1     49    49   |
:------------------+------------------+------------------:
| 6789  3     789  | 5789  58    2    | 4     1569  6789 |
| 5     1467  4789 | 4789  3     789  | 6789  69    2    |
| 4789  47    2    | 6     1458  5789 | 5789  3     789  |
:------------------+------------------+------------------:
| 24679 8     4579 | 3     456   567  | 569   4569  1    |
| 467   4567  1    | 24578 9     5678 | 3     456   468  |
| 3     456   459  | 458   4568  1    | 25689 7     4689 |
'------------------'------------------'------------------'

solvable by singles. However, is there a good reasons to lock down these special candidates? Apart from that I've only been able to pick away at it slowly.
Glyn

Posts: 357
Joined: 26 April 2007

Glyn wrote:If we made an additional constraint based on an observation that the givens 1,2,3 do not simultaneously share a row and box and apply that to the pencilmarks we get

Code: Select all
.------------------.------------------.------------------.
| 1     57    57   | 259   256   4    | 2679  8     3679 |
| 2478  9     3478 | 1     68    68   | 267   246   5    |
| 248   245   6    | 2589  7     3589 | 1     49    49   |
:------------------+------------------+------------------:
| 6789  3     789  | 5789  58    2    | 4     1569  6789 |
| 5     1467  4789 | 4789  3     789  | 6789  69    2    |
| 4789  47    2    | 6     1458  5789 | 5789  3     789  |
:------------------+------------------+------------------:
| 24679 8     4579 | 3     456   567  | 569   4569  1    |
| 467   4567  1    | 24578 9     5678 | 3     456   468  |
| 3     456   459  | 458   4568  1    | 25689 7     4689 |
'------------------'------------------'------------------'

solvable by singles. However, is there a good reasons to lock down these special candidates? Apart from that I've only been able to pick away at it slowly.

Great discovery! However you haven't got the trick 100% accurate. Let me try to clarify things for you...

For this very puzzle, with the digit mapping [123456789]->[123564897], each min-row must either have the whole of {123}|{456}|{789}, or one each digit from these 3 sets (e.g. 147, 158, 349).

Proof:

Firstly we need to see that the 3 digits {456} must either exists as a whole mini-row or be distributed on 3 different mini-rows within a block.

Suppose this is not the case, i.e. two digits out of {456} are on the same mini-row and the remaining one is on a different mini-row. Then this pattern must also be duplicated in the other 2 blocks of the same band. So within each block we have one mini-row containing two of {456} (i.e. being {45},{56},{46} respectively), one mini-row containing one of {456} (i.e. 6,4,5 respectively), one mini-row without any of {456}.

Now, each (whole) row of this band must also consist of 3 mini-rows, and these must also have two, one, none of {456} respectively. However, the mini-row with one of {456} will always clash with the mini-row with two of {456} (i.e. that sole value will always be the same as one of the two values). To see this fact, just look at the following example:
Code: Select all
456 . 456 | 456 .  .  |  .  .  .
.  .  .  | 456 . 456 | 456 .  .
456 .  .  |  .  .  .  | 456 . 456

No matter how you assign values you'll always run into a contradiction.

Therefore, within a block, each of {456} must either occupy the same mini-row or occupy 3 different mini-rows.

Using the same reasoning, each of {789} must also behave the same.

As a result, each of {123} must also behave in the same manner!

(Explicit proof: Suppose {456} occupy the same mini-row. Then {789} can't occupy 3 different mini-rows, so must also occupy another single mini-row. Then {123} are forced to occupy the remaining single mini-row.

Suppose {456} occupy 3 different mini-rows. Then {789} can't occupy a single mini-row, so must also occupy 3 different mini-rows. Then {123}, with only one cell remaining available on each mini-row, are forced to also occupy 3 different mini-rows.)

(End of proof.)

With this trick, it's very easy to crack this puzzle. (Initially the top band can be both ways but in the middle and bottom band you must have each mini-row containing exactly one from {123},{456},{789} respectively. After a series of singles you can also confirm this property also holds true for the top band. And then you can complete the puzzles using singles only. )

(And also for "version 2" this trick is also applicable: first notice r7c1 must be 3 because r7c123 can't be {789}. So r8c2=7 (hidden single). Now seeing r6c123 can't be {789}, r6c1 must be 4, hence r4c4=5 (symmetry). And the rest are all naked singles. Way cool! )
udosuk

Posts: 2698
Joined: 17 July 2005

On Pg 3. eleven wrote:After a second thought i cant believe, that other symmetries would not be useful for a solution. E.g. in this Puzzle from this post
http://forum.enjoysudoku.com/viewtopic.php?p=40236#p40236
Code: Select all
Version 1
1 . .|. . 4|. 8 .
. 9 .|1 . .|. . 5
. . 6|. 7 .|1 . .
-----+-----+-----
. 3 .|. . 2|4 . .
5 . .|. 3 .|. . 2
. . 2|6 . .|. 3 .
-----+-----+-----
. 8 .|3 . .|. . 1
. . 1|. 9 .|3 . .
3 . .|. . 1|. 7 . ER=??(Not tested)

Because I was curious and had other appointments (and could let my computer grind away in my absence), I turned SE loose and came back after a couple of hours. ER=10.6.

Here's the analysis:
Triple click to see what SE wrote:
Analysis results
Difficulty rating: 10.6
This Sudoku can be solved using the following logical methods: 51 x Hidden Single
2 x Naked Single
10 x Pointing
2 x Claiming
5 x Naked Pair
1 x X-Wing
1 x Hidden Pair
1 x Naked Triplet
1 x XY-Wing
1 x Forcing X-Chain
4 x Forcing Chain
2 x Nishio Forcing Chains
2 x Cell Forcing Chains
4 x Region Forcing Chains
6 x Dynamic Cell Forcing Chains
10 x Dynamic Contradiction Forcing Chains
6 x Dynamic Region Forcing Chains
1 x Dynamic Contradiction Forcing Chains (+)
1 x Dynamic Region Forcing Chains (+ Forcing Chains)
1 x Dynamic Cell Forcing Chains (+ Forcing Chains)
13 x Dynamic Contradiction Forcing Chains (+ Forcing Chains)

Cheers...

- drac
Draco

Posts: 143
Joined: 14 March 2008

Great new technique, Glyn and udosuk (added: i call it Gludo )

So these symmetrical puzzles in the moment are the last resort to find essentially new techniques for solving classical sudokus (though only applicable for a special symmetry).

Hope the puzzle creators will jump on the train and bring fresh food.
eleven

Posts: 1710
Joined: 10 February 2008

udosuk Very neat proof of those properties cracks both very easily.
Glyn

Posts: 357
Joined: 26 April 2007

There are still 3 puzzles left in this post with threefold diagonal box symmetry
http://forum.enjoysudoku.com/viewtopic.php?p=40236#p40236

Code: Select all
Version 1
9 . .|2 . .|. . .
. 6 .|. . 8|. 3 .
. . 8|. 5 .|1 . .
-----+-----+-----
. 1 .|9 . .|. . 5
. . .|. 7 .|. 2 .
. . 3|. . 6|9 . .
-----+-----+-----
. 7 .|3 . .|6 . .
5 . .|. . 1|. 7 .
. . 2|. . .|. . 8 ER=9.6
Hint:Observe boxes 1-5-9,2-6-7,3-4-8; and digits 1-1,2-2,3-3,4-4,5-5,6-6,7-8-9.

The minirows in the box cycles map to the next one shifted right, e.g. r1c123 -> r5c564 -> r9c978.

This one is easy to solve with symmetrical simple coloring:
r9c1<>1 -> r9c8=1 -> r1c3=1, r2c1/r6c5/r7c9, r7c3/r6c8/r2c4<>1
r6c2<>2 -> r5c2=2 -> r4c6=2, r4c1/r8c5/r3c9, r6c5/r7c9/r2c1<>2
r1c5<>3 -> r4c5=3 -> r3c1=3, r1c2/r5c6/r9c7, r3c6/r4c7/r8c2<>3
r2c7<>5 -> r2c3=5 -> r7c8=5, r1c8/r5c3/r9c4, r9c7/r1c2/r5c6<>5

The rest are singles.

Code: Select all
Version 2
. . 9|. 3 .|. 2 .
. . .|4 . .|6 . .
. . .|. . 1|. . 4
-----+-----+-----
5 . .|. . .|1 . .
. . 2|7 . .|. . 3
. 4 .|. . .|. 5 .
-----+-----+-----
. . 6|. . 5|. . .
. 1 .|. 6 .|. . .
3 . .|2 . .|. 8 . ER=10.3
Hint: Observe boxes 1-5-9,2-6-7,3-4-8 and digits 1-1,2-2,3-3,4-5-6,7-8-9.

I made some easy moves, but dont have the time now to look deeper.

r2c8<>8 -> r2c2=3 -> r7c7=3, r3c7/r5c2/r8c6, r8c8/r3c2/r4c6<>3
r3c5<>2 -> r3c12=2, r4c56=2, r6c5/r7c9/r2c2<>2

r2c1<>7 -> r2c1=8 -> r7c9=7, r7c1<>7, r2c5<>8, r6c9<>9

r6c1=8 -> r7c5=9 -> r6c5=8, r6c1<>8, r7c5<>9, r2c9<>7

Leaves me here.
Code: Select all
*-----------------------------------------------------------------------------*
| 4       5678    9       | 568     3       678     | 578     2       1       |
| 78      23578   1       | 4       279     2789    | 6       379     589     |
| 2678    235678  578     | 5689    2789    1       | 5789    379     4       |
|-------------------------+-------------------------+-------------------------|
| 5       6789    378     | 689     2489    234689  | 1       4679    2789    |
| 1       689     2       | 7       5       4689    | 489     469     3       |
| 679     4       378     | 1       89      23689   | 2789    5       278     |
|-------------------------+-------------------------+-------------------------|
| 289     2789    6       | 389     478     5       | 23479   1       79      |
| 2789    1       4578    | 389     6       4789    | 234579  479     2579    |
| 3       579     457     | 2       1       479     | 4579    8       6       |
*-----------------------------------------------------------------------------*

This is the third puzzle:

Code: Select all
Version 3
6 . .|. . 9|. 5 .
. . .|. 3 .|. . 2
. 2 .|8 . .|9 . .
-----+-----+-----
. 7 .|. . 3|. 9 .
. . 6|. 4 .|7 . .
3 . .|. . .|. . 1
-----+-----+-----
2 . .|. 1 .|. . .
. . 7|. . 8|1 . .
. 8 .|4 . .|. . 5 ER=10.6
Hint: Observe boxes 1-5-9,2-6-7,3-4-8 and digits 1-2-3,4-5-6,7-8-9.
eleven

Posts: 1710
Joined: 10 February 2008

PreviousNext