## Down Under Upside Down - a Sudoku puzzle

Post puzzles for others to solve here.
What i really like here is, that no one posts program solutions

I finished the Version 2 puzzle now, but needed a rather complicated step.
There was one more easy elimination:
[Edit: This was a mistake, you first have to do the step below]
r2c1<>7 -> r2c1=8 -> r7c9=7 -> r4c8=7, r2c8<>7, r6c3<>8, r7c4<>9
Code: Select all
` *-----------------------------------------------------------------------------* | 4      *5678    9       | 568     3       678     | 578     2       1       | | 78     X23578   1       | 4       279     2789    | 6       379    *589     | | 2678   X235678  578     | 5689    2789    1       | 5789    379     4       | |-------------------------+-------------------------+-------------------------| | 5       6789    378     | 689     2489   X234689  | 1       4679    2789    | | 1       689     2       | 7       5       4689    | 489     469     3       | | 679     4       378     | 1       89     X23689   | 2789    5       278     | |-------------------------+-------------------------+-------------------------| | 289     2789    6       | 389     478     5       |X23479   1       79      | | 2789    1      *4578    | 389     6       4789    |X234579  479     2579    | | 3       579     457     | 2       1       479     |*4579    8       6       | *-----------------------------------------------------------------------------* `

Now i looked, where the 23's can go to.
If both are in the same minicolumn in r23c2, the cells marked X are occupied by 23.
Then also r2c9=5 -> r9c7=5 -> r8c3=5 -> r1c2=5.
But this is not possible, because r9c7=5 implies r1c2=6 from the symmetry.

So r3c1 must be 2 (which places all 2's) and a kite solves the puzzle.
Last edited by eleven on Tue Oct 07, 2008 2:46 am, edited 1 time in total.
eleven

Posts: 2377
Joined: 10 February 2008

eleven wrote:There was one more easy elimination:

r2c1<>7 -> r2c1=8 -> r7c9=7 -> r4c8=7, r2c8<>7, r6c3<>8, r7c4<>9

I don't understand here. How does r7c9=7 imply r4c8=7?

I also solved this one, but with much more complex logic. I'm still tidying it up, and is quite busy at the moment so probably won't post the result for another week or so.

Note I need to do some row/column transformations to make the blocks direct copies of each other. The mini-row/mini-column shifting is driving me crazy. As a result for me the first 2 puzzles are actually:
Code: Select all
`9....2....6..8...3..85...1....7....2..3.6..9..1...95....2...8...7...3.6.5...1...7.....14...9.3...2.....4...65.......1.2..7.3....4....5...16.....3...2..8..6...5...`

udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:
eleven wrote:r2c1<>7 -> r2c1=8 -> r7c9=7 -> r4c8=7, r2c8<>7, r6c3<>8, r7c4<>9

I don't understand here. How does r7c9=7 imply r4c8=7?
Thanks, this was a mistake. Originally i made this elimination after the 23-step (which places 2 in r6c7), but then thought, i had missed it before
Note I need to do some row/column transformations to make the blocks direct copies of each other.
Good idea, i needed some time to quickly see, where the numbers go to.

I also solved the third one, but with 2 long forcing nets, which i dont bother to write down (eliminations r5c1<>8 and r4c9<>6). Dont have an idea for a new technique for this symmetry either.
eleven

Posts: 2377
Joined: 10 February 2008

Version 1

r123456789 -> r123564978
c123456789 -> c123564978
Code: Select all
`+----------------------+----------------------+----------------------+| 9     -345    1457   |*1346   347    2      | 467    4578   4568   ||-1247   6      1457   |*149    8     *147    | 2479   2457   3      || 2347   234    8      | 5      3479   467    | 24679  1      469    |+----------------------+----------------------+----------------------+| 468   #4589  #4569   | 7     -345    1458   | 1346   348    2      || 2478  #2458   3      |-1248   6      1458   | 147    9      148    || 24678  1      467    | 2348   234    9      | 5      3478   468    |+----------------------+----------------------+----------------------+| 1346   349    2      | 469    4579   4567   | 8     -345    1459   || 148    7      149    | 2489   2459   3      |-1249   6      1459   || 5      3489   469    | 24689  1      468    | 2349   234    7      |+----------------------+----------------------+----------------------+`

1 @ b2 locked @ r12c4+r2c6 => r2c1+r5c4+r8c7 can't have 1

5 @ b4 locked @ r4c23+r5c2 => r1c2+r4c5+r7c8 can't have 5

Hidden pair @ b1: r12c3={15} (naked pair @ c3)

Hidden singles: r6c3=7 => r5c7=7 => r4c7=1 => r9c7=3

Symmetry: r3c1=r6c4=3

All naked singles from here.

Version 2

r123456789 -> r231456897
c123456789 -> c132546987
(After singles)
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 2678   -3578    235678  |*2789    5689    1       | 4       379     35789   || 4       9       5678    | 3       568     678     | 1       2       578     ||-278     1       23578   |*2789    4      *2789    | 5789    379     6       |+-------------------------+-------------------------+-------------------------+| 5      #378    #36789   | 2489   -3689    234689  | 2789    4679    1       || 1       2       689     | 5       7       4689    | 3       469     489     || 6789   #378     4       |-289     1       23689   | 2789    5       2789    |+-------------------------+-------------------------+-------------------------+| 2789    4578    1       | 6       389     34789   | 2579   -3479    234579  || 3       457     579     | 1       2       479     | 6       8       4579    || 2789    6       2789    | 4789    389     5       |-279     1       23479   |+-------------------------+-------------------------+-------------------------+`

2 of b2 locked @ r13c4+r3c6 => r3c1+r6c4+r9c7 can't have 2

3 of b4 locked @ r4c23+r6c2 => r1c2+r4c5+r7c8 can't have 3

Locked candidates: r1c9+r4c3+r7c6 can't have 3
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 2678    578     235678  | 2789    5689    1       | 4       379     5789    || 4       9       5678    | 3       568     678     | 1       2       578     || 78      1       23578   | 2789    4       2789    | 5789    379     6       |+-------------------------+-------------------------+-------------------------+| 5       378     6789    |#2489   *689     234689  | 2789    4679    1       || 1       2       689     | 5       7      *4689    | 3       469     489     || 6789    378     4       |*89      1      #23689   | 2789    5       2789    |+-------------------------+-------------------------+-------------------------+| 2789    4578    1       | 6       389     4789    | 2579    479     234579  || 3       457     579     | 1       2       479     | 6       8       4579    || 2789    6       2789    |#4789    389     5       | 79      1      #23479   |+-------------------------+-------------------------+-------------------------+`

4 @ c4 locked @ r49c4
4 @ r9 locked @ r9c49
But r4c5+r5c6+r6c4 from {4689} must have 4|6
=> r4c4+r6c6 can't be [46]
Symmetry: r4c4+r9c9 can't be [44]
=> r9c4 must be 4
Symmetry: r3c7=5, r6c1=6

Now r2c9 from {78}, r5c3 from {89}, r8c6 from {79}
Symmetry: r2c9+r5c3+r8c6=[789|897]
=> r2c9+r5c3 must have 8, r2c3+r5c9 can't have 8
=> r5c3+r8c6 must have 9, r5c6+r8c3 can't have 9
=> r2c9+r8c6 must have 7, r2c6+r8c9 can't have 7
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 278     578     235678  | 2789    5689    1       | 4       379     789     || 4       9      #567     | 3       568     68      | 1       2      #78      || 78      1       2378    | 2789    4       2789    | 5       379     6       ||-------------------------+-------------------------+-------------------------|| 5       378    *789     | 289     689     234689  | 2789    4679    1       || 1       2      *89      | 5       7       468     | 3       469     49      || 6       378     4       | 89      1       2389    | 2789    5       2789    ||-------------------------+-------------------------+-------------------------|| 2789    4578    1       | 6       389     789     | 279     479     234579  || 3       457     57      | 1       2       79      | 6       8       459     || 2789    6       2789    | 4       389     5       | 79      1       2379    |*-----------------------------------------------------------------------------*`

7 @ r2 locked @ r2c39
=> either r45c3={89} or r2c9+r5c3=[78]
=> 8 @ c3,b4 locked @ r45c3
(Alternatively: r45c3<>8 => r45c3=[79] => no 7 @ r2 => contradiction)
Symmetry: 7 @ c9,b3 locked @ r12c9, 9 @ c6,b8 locked @ r78c6

r13c8={39} (naked pair @ c8,b3)
r46c2={37} (naked pair @ c2,b4)
r79c5={38} (naked pair @ c5,b8)
r12c9={78} (naked pair @ c9)
r45c3={89} (naked pair @ c3)
r78c6={79} (naked pair @ c6)
Code: Select all
`+----------------------+----------------------+----------------------+|-278   -58     23567  | 2789   569    1      | 4      39    *78     || 4      9     #567    | 3      56     68     | 1      2     #78     ||*78     1      237    | 2789   4      28     | 5      39     6      |+----------------------+----------------------+----------------------+| 5      37     89     | 289    69     23468  | 2789   467    1      || 1      2      89     | 5      7      468    | 3      46     49     || 6      37     4      | 89     1      238    | 2789   5      29     |+----------------------+----------------------+----------------------+| 2789   458    1      | 6      38     79     | 279    47     23459  || 3      45     57     | 1      2      79     | 6      8      459    || 2789   6      27     | 4      38     5      | 79     1      239    |+----------------------+----------------------+----------------------+`

W-wing: 7 @ r2 locked @ r2c39, r1c9+r3c1 from {78}
=> r1c12, seeing r1c9+r3c1, can't have 8

Hidden single @ b1: r3c1=8
Symmetry: r6c4=9, r9c7=7

All naked singles from here.

Version 3 should also be crackable, but it'll take some more time...
udosuk

Posts: 2698
Joined: 17 July 2005

The 46 move is nice. With the strong links for 4 in row 9 and column 4 and the symmetry mapping 4-5-6 you get r4c4=4 <-> r6c6=6, thus leaving three 89 cells in block 5.
udosuk wrote:Version 3 should also be crackable, but it'll take some more time...

Good luck, if i did not miss something, it is as disgusting as a "normal" sudoku needing long chains.
eleven

Posts: 2377
Joined: 10 February 2008

Finally, I found a satisfying solving path for version 3:

Version 3

r123456789 -> r312465879
c123456789 -> c213654798
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 2       1457    1345    | 14567   567     8       | 9       3467   *13467   ||-1459    145789  14589   | 14567   3       1567    | 468     2       14678   || 134     6       1348    | 9       27     #127     | 348     3478    5       |+-------------------------+-------------------------+-------------------------+| 7       1458    12458   | 3       2568    1256    | 24568   468     9       || 459     3       24589   | 2567    256789  25679   | 24568   1       2468    || 159     1589    6       | 125     4       1259    | 7       38     *238     |+-------------------------+-------------------------+-------------------------+| 34569   459     7       | 8       2569    23569   | 1       3469   *2346    || 34569   2       3459    | 567     1       35679   |#3468    346789 *34678   || 8       19      139     | 267     2679    4       | 236     5      *2367    |+-------------------------+-------------------------+-------------------------+`

3 @ c9 locked @ r16789c9
=> either r1c9=3 or r6c9=3 or r789c9 have 3
=> either r1c9=3 or r6c9<>2 or r8c7<>3
Symmetry: r1c9+r2c1+r3c6 can't be [111]
=> r2c1 can't be 1

Now 1 @ c1 locked @ r36c1
Symmetry: r6c4 can't be 1
Symmetry: r9c7 can't be 2, r3c1 can't be 3

1 @ c4,b2 locked @ r12c4
2 @ c7,b6 locked @ r45c7
3 @ c1,b7 locked @ r78c1

r3c56={27} (NP @ r3,b2)
r6c89={38} (NP @ r6,b6)
r9c23={19} (NP @ r9,b7)

1 @ r3,b1 locked @ r3c13
r7c2+r8c3={45} (NP @ b7)
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 2      -457     345     | 1456   *56      8       | 9      -3467    13467   || 459     45789   4589    | 1456    3       56      | 468     2       14678   || 14      6       1348    | 9       27      27      | 348     348     5       |+-------------------------+-------------------------+-------------------------+| 7       1458    12458   | 3      -2568    1256    | 2456   *46      9       || 459     3       24589   | 2567    256789  25679   | 2456    1       46      || 159     159     6       | 25      4       1259    | 7       38      38      |+-------------------------+-------------------------+-------------------------+| 36     *45      7       | 8      -2569    23569   | 1       3469    2346    || 36      2       45      | 567     1       35679   | 3468    346789  34678   || 8       19      19      | 267     267     4       | 36      5       2367    |+-------------------------+-------------------------+-------------------------+`

Now r1c5 from {56}, r4c8 from {46}, r7c2 from {45}
Symmetry: r1c5+r4c8+r7c2=[564|645]
r1c5+r4c8 must have 6 => r1c8+r4c5 can't have 6
r1c5+r7c2 must have 5 => r1c2+r7c5 can't have 5
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 2      #47      345     | 1456    56      8       | 9      *347     13467   ||#459     45789   4589    | 1456    3       56      |*468     2       14678   ||#14      6       1348    | 9       27      27      |*348    *348     5       |+-------------------------+-------------------------+-------------------------+| 7       1458    12458   | 3       258     1256    | 2456    46      9       || 459     3       24589   | 2567    256789  25679   | 2456    1       46      || 159     159     6       | 25      4       1259    | 7       38      38      |+-------------------------+-------------------------+-------------------------+| 36      45      7       | 8       269     23569   | 1       3469    2346    || 36      2       45      | 567     1       35679   |#3468    346789  34678   || 8       19      19      | 267     267     4       |#36      5       2367    |+-------------------------+-------------------------+-------------------------+`

Now r1c8+r2c7+r3c78 from {34678} must have 6|7
=> either r1c8=7 or r2c7=6
=> r1c2+r8c7 & r1c2+r9c7 both can't be [76]
Symmetry: r1c2+r2c1 & r1c2+r3c1 both can't be [74]
=> r23c1 can't have 4

Hidden single @ c1: r5c1=4
Symmetry: r8c4=5, r2c7=6

All naked singles from here!

The first move feels a little bit forcing but it's the best I could do at that position. After that the rest is of the same level of elegancy as my path for version 2.
udosuk

Posts: 2698
Joined: 17 July 2005

Code: Select all
`+-------------------------+-------------------------+-------------------------+| 2       1457    1345    | 14567   567     8       | 9       3467   *13467   ||-1459    145789  14589   | 14567   3       1567    | 468     2       14678   || 134     6       1348    | 9       27     #127     | 348     3478    5       |+-------------------------+-------------------------+-------------------------+| 7       1458    12458   | 3       2568    1256    | 24568   468     9       || 459     3       24589   | 2567    256789  25679   | 24568   1       2468    || 159     1589    6       | 125     4       1259    | 7       38     *238     |+-------------------------+-------------------------+-------------------------+| 34569   459     7       | 8       2569    23569   | 1       3469   *2346    || 34569   2       3459    | 567     1       35679   |#3468    346789 *34678   || 8       19      139     | 267     2679    4       | 236     5      *2367    |+-------------------------+-------------------------+-------------------------+`
Interesting solution by udosuk for version 3.
I see the first step as "a 1 in in r2c1 kills all 3's in column 9":
r2c1=1 -> r8c7=3 and r1c9=1 and (r3c6=1 ->) r6c9=2.

The same holds for r2c2=1.

Following this tactic, i immediately found 2 more eliminations:
r1c2=1 kills all 3's in row 9: r1c2=1 -> r9c3=1 and r7c8=3
r1c9=7 kills both 9's in column 8: r1c9=7 -> r8c8=7 and r7c6=9
eleven

Posts: 2377
Joined: 10 February 2008

Great spotting, eleven!

eleven wrote:The same holds for r2c2=1.

As a matter of fact, at the very beginning you can apply a grouped turbot chain to eliminate 1 from r12c23:

r12c23 have 1 => r3c6=1 (r3) & r6c1=1 (c1) => no 1 @ r4 => contradiction

But with my initial elimination r2c1<>1 I can get the same eliminations through a box-line intersection a few steps later. Therefore I didn't bothered with that particular elimination in r2c2.
udosuk

Posts: 2698
Joined: 17 July 2005

Some side issues related to Version 3 and some 'spurious?' symmetries may be found moved to StrmCkr observations on Version 3 puzzle. Hopefully both threads are a little clearer now they have been split.
Glyn

Posts: 357
Joined: 26 April 2007

New puzzles with symmetries I had missed:
Code: Select all
`+-------+-------+-------+| 1 . . | 2 . . | 3 . . || . 2 . | . 3 . | . 1 . || . . 4 | . . 5 | . . 6 |+-------+-------+-------+| 5 . . | 6 . . | 4 . . || . 7 . | . 8 . | . 9 . || . . 1 | . . 2 | . . 3 |+-------+-------+-------+| . . 6 | 9 . . | . . . || . . . | . . 4 | 7 . . || 8 . . | . . . | . . 5 |+-------+-------+-------+`

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

Solutions?
Mauricio

Posts: 1174
Joined: 22 March 2006

Thanks for creating new puzzles

This is a new game now, find the symmetry first to solve it (with singles here).

[Added:] Hm, no reaction - "with singles" is not quite true.
In the first puzzle i started after the triple/locked candidates. For both puzzles i needed a short symmetry chain (4 and 2 cells resp. to a contradiction). And the second one needed a hidden pair after that.

For those, who have trouble to see the symmetry:
Puzzle 1: Cycle (cyclically change) the rows in band 3, cycle the stacks and cycle the numbers 123,456,789 to get the same puzzle again.
Puzzle 2: Cycle the stacks, cycle the rows in band 23, again cycle numbers 123,456,789.
eleven

Posts: 2377
Joined: 10 February 2008

Both Mauricio's puzzles above can be solved extremely easily using "Gludo". But firstly you must use "POCKET" to correctly identify and verify the symmetry/automorphism:

Puzzle 1:
Code: Select all
`1..2..3...2..3..1...4..5..65..6..4...7..8..9...1..2..3..69..........47..8.......5`

With the transformation:

r123456789 -> r123456897
c123456789 -> c456789123
d123456789 -> d312645978

You get back the original puzzle. Therefore there is a "whole stack permutation" symmetry/automorphism existing in the puzzle/solution grid. Particularly, you can apply "Gludo" in r789, the only band with "minirow shifting". Therefore each minirow in r789 must be either {123},{456},{789} or contain 1 digit each from these 3 sets.
Code: Select all
`+-------------------+-------------------+-------------------+| 1     5689  5789  | 2     4679  6789  | 3     4578  4789  || 679   2     5789  | 478   3     6789  | 589   1     4789  || 379   389   4     | 178   179   5     | 289   278   6     |+-------------------+-------------------+-------------------+| 5     389   2389  | 6     179   1379  | 4     278   1278  || 2346  7     23    | 1345  8     13    | 1256  9     12    || 469   4689  1     | 457   4579  2     | 568   5678  3     |+-------------------+-------------------+-------------------+| 2347  1345  6     | 9     1257  1378  | 128   2348  1248  || 239   1359  2359  | 1358  1256  4     | 7     2368  1289  || 8     1349  2379  | 137   1267  1367  | 1269  2346  5     |+-------------------+-------------------+-------------------+`

Hidden triple @ r5: r5c147={456}
Gludo: r7c123 with r7c3=6 and r7c2 from {1345}
=> r7c1 can't be from {23}, must be from {47}
Hidden single @ c1: r8c1=2
Symmetry: r9c4=3, r7c7=1
Gludo: r9c123 with r9c13 from {789} must be [897]

All naked singles from here.

Puzzle 2:
Code: Select all
`..........4..5..6...7..8..9...1....5..6...2..3....4.....47...3..1...58..9...2...6`

With the transformation:

r123456789 -> r123564897
c123456789 -> c456789123
d123456789 -> d312645978

You get back the original puzzle. Therefore there is a "whole stack permutation" symmetry/automorphism existing in the puzzle/solution grid. Particularly, you can apply "Gludo" in r456789, the two bands with "minirow shifting". Therefore each minirow in r456789 must be either {123},{456},{789} or contain 1 digit each from these 3 sets.
Code: Select all
`+-------------------------+-------------------------+-------------------------+| 12568   235689  123589  | 23469   134679  123679  | 13457   124578  123478  || 128     4       12389   | 239     5       12379   | 137     6       12378   || 1256    2356    7       | 2346    1346    8       | 1345    1245    9       |+-------------------------+-------------------------+-------------------------+| 2478    2789    289     | 1       36789   23679   | 34679   4789    5       || 14578   5789    6       | 3589    3789    379     | 2       14789   13478   || 3       25789   12589   | 25689   6789    4       | 1679    1789    178     |+-------------------------+-------------------------+-------------------------+| 2568    2568    4       | 7       1689    169     | 159     3       12      || 267     1       23      | 3469    3469    5       | 8       2479    247     || 9       3578    358     | 348     2       13      | 1457    1457    6       |+-------------------------+-------------------------+-------------------------+`

Gludo: r8c123 with r8c23 from {123} must be [213]
Symmetry: r9c456=[321], r7c789=[132]
Gludo: r9c123 from {5789} must be [978]
Symmetry: r7c456=[789], r8c789=[897]
Hidden pair @ c1: r45c1={47}
Hidden single @ b4: r6c3=1

All naked singles from here.

udosuk

Posts: 2698
Joined: 17 July 2005

udosuk wrote:d123456789 -> d312645978
I would say d123456789 -> d231564897.

The Gludo solutions are nice. I did not even look for them, because the puzzles were solved quickly also without, especially the second one.

There is an easy to spot box pattern for bands with "minirow shifting" and a 3 number cycle.
If you have a number in only 2 minirows, and only in one cell of a minirow, you can eliminate one of the other 2 numbers of the cycle there.
Code: Select all
`+-------------------------+| 2478    2789    289     || 14-578  5789    6       || 3       25789   1-2589  |+-------------------------+| 2568    2568    4       || 267     1      -23      || 9       357-8   358     |+-------------------------+`

The 5 in minirow 2 maps to a 4 in minirow 1, and thus would kill the second 4 in this box.
The 2 in minirow 3 maps to a 1 in minirow 2, and thus would kill the second 1 in this box.
The 2 in minirow 2 maps to a 3 in minirow 3, and thus would kill the other 3's in this box.
The 8 in minirow 3 maps to a 7 in minirow 2, and thus would kill the second 7 in this box.

Also in the first puzzle there are 2 of those eliminations, but dont help that much.

Anyway - these highrated puzzles were easier to solve than a newspaper puzzle needing a naked triple
eleven

Posts: 2377
Joined: 10 February 2008

eleven wrote:
udosuk wrote:d123456789 -> d312645978
I would say d123456789 -> d231564897.

I don't think so.

If you use the digit mapping: d123456789 -> d231564897,
the original puzzle (#2) becomes this:
Code: Select all
`..........5..6..4...8..9..7...2....6..4...3..1....5.....58...1..2...69..7...3...4`

Then you need the following row/column permutations to transform it back to the original:

r123645978
c789123456

If you use my row/column permutations (r123564897+c456789123), you'll need my digit mapping (d123456789 -> d312645978).

eleven wrote:There is an easy to spot box pattern for bands with "minirow shifting" and a 3 number cycle.
If you have a number in only 2 minirows, and only in one cell of a minirow, you can eliminate one of the other 2 numbers of the cycle there.
Code: Select all
`+-------------------------+| 2478    2789    289     || 14-578  5789    6       || 3       25789   1-2589  |+-------------------------+| 2568    2568    4       || 267     1      -23      || 9       357-8   358     |+-------------------------+`

The 5 in minirow 2 maps to a 4 in minirow 1, and thus would kill the second 4 in this box.
The 2 in minirow 3 maps to a 1 in minirow 2, and thus would kill the second 1 in this box.
The 2 in minirow 2 maps to a 3 in minirow 3, and thus would kill the other 3's in this box.
The 8 in minirow 3 maps to a 7 in minirow 2, and thus would kill the second 7 in this box.

Perhaps a better way to describe this move is:
If there is minirow shifting in the band (e.g. r7c123 corresponds to r8c456), and there is a cyclic triple of digits {xyz} (x->y->z->x) then x can't appear on a minirow directly above a minirow containing y. (Otherwise we will have two y's on the same row.)

Ditto for y,z and z,x.

Code: Select all
`+-------------------------+| 2478    2789    289     || 14-578  5789    6       || 3       25789   1-2589  |+-------------------------+| 2568    2568    4       || 267     1      -23      || 9       357-8   358     |+-------------------------+`

4 @ b4 locked @ r45c1, which can't be [45] => r5c1<>5
1 @ b4 locked @ r5c1+r6c3, which can't be [12] => r6c3<>2
3 @ b7 locked @ r8c3+r9c23, which can't be [23?|2?3] => r8c3<>2
7 @ b7 locked @ r8c1+r9c2, which can't be [78] => r9c2<>8

udosuk

Posts: 2698
Joined: 17 July 2005

Ok.

Its your notation, so you can define it as weird as you want. This is, how i interpreted it. Look at the 9 in r9c1:
Code: Select all
`..........4..5..6...7..8..9...1....5..6...2..3....4.....47...3..1...58..9...2...6`

r123456789 -> r123564897
row 9 goes to row 7, 9r9c1 -> 9r7c1
c123456789 -> c456789123
column 1 becomes column 4, 9r7c1 -> 9r7c4
d123456789 -> d312645978
mumber 9 becomes number 8, 9r7c4 -> 8r7c4

But the original puzzle has a 7 in r7c4.

You also are free to reformulate my eliminations. Personally i keep my own way to find and use them
eleven

Posts: 2377
Joined: 10 February 2008

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