## this puzzle is weird to me..

Post the puzzle or solving technique that's causing you trouble and someone will help

### this puzzle is weird to me..

Hi i play sudoku in my cellphone. sometimes i have some puzzles here that i cant solve using SSTS. Here is one
0030090010051302000800004070000410500085000700000000040000000090000906003506000002
i solved this using bifurcation at last.
Can u help?
sarker306

Posts: 16
Joined: 06 March 2009

This is not a valid puzzle.

What you have posted has 82 givens.
Leaving off the 2 at the end doesn't help as there remain more than one solution.
wintder

Posts: 297
Joined: 24 April 2007

030090010051302000800004070000410500085000700000000040000000090000906003506000002
it is corrected. perhaps itr is valid..
sarker306

Posts: 16
Joined: 06 March 2009

Here it is after SSTS :
Code: Select all
`.3..9..1..513.2...8....4.7....41.5...85...7.........4........9....9.6..35.6.....2     *--------------------------------------------------------------------* | 67     3      4      |*568    9     -578    | 2      1     *58     | | 79     5      1      | 3      78     2      | 89     6      4      | | 8      26     29     | 1     *56     4      | 3      7      59     | |----------------------+----------------------+----------------------| | 2369   267    2379   | 4      1      378    | 5      23     689    | | 4      8      5      | 26     236    9      | 7      23     1      | | 12369  1267   237    | 258    23578  3578   | 89     4      689    | |----------------------+----------------------+----------------------| | 123    124    238    | 258    2458   158    | 6      9      7      | | 12     1247   278    | 9      248    6      | 14     5      3      | | 5      9      6      | 7      34     13     | 14     8      2      | *--------------------------------------------------------------------*`
If you like them, this will fall with little more than XY chains. First, there's this xyz wing that will open up an x-wing.

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

### Another one that troubles me

Luke's reply helped a lot, thanks. Here is another one i need a detailed help until i can go to basic techniques.
..2......89..4.......5...1...5...7......8..3......9...........8.....2.49..67.....
It is called hardest 17 clued.
sarker306

Posts: 16
Joined: 06 March 2009

sarker306, there are a lot of much harder 17 clues.
This one can be solved with coloring on 8 (3 strong links) and 9 (3 strong links):
Code: Select all
`+----------------------------+----------------------------+----------------------------+| 134567   134567   2        |#13689    13679    13678    | 3489     6-89     3467     || 8        9        137      | 1236     4        1367     | 235      256      23567    || 3467     3467     347      | 5        23679    3678     | 23489    1        23467    |+----------------------------+----------------------------+----------------------------+|A123469  #123468   5        | 12346    1236     1346     | 7       @2689     1246     || 124679   12467    1479     | 1246     8        57       | 12459    3        12456    || 123467   1234678  13478    | 12346    57       9        | 12458    2568     12456    |+----------------------------+----------------------------+----------------------------+| 123459   12345    1349     |B13469    13569    13456    | 12356    7        8        || 1357     13578    1378     |@1368     1356     2        | 1356     4        9        ||B123459  @123458   6        | 7       A1359    #13458    | 1235     25       1235     |+----------------------------+----------------------------+----------------------------+`

Either r1c4=8 or
r8c4=8 -> r9c6<>8 -> r9c2=8 -> r4c2<>8 -> r4c8=8
So r1c8 cannot be 8.

Either r1c4=9 or
r7c4=9 -> r9c5<>9 -> r9c1=9 -> r4c1<>9 -> r4c8=9
So r1c8 cannot be 9.

[Edit: corrected r8c4=9 to r7c4=9]
Last edited by eleven on Fri Mar 06, 2009 6:34 pm, edited 1 time in total.
eleven

Posts: 1907
Joined: 10 February 2008

### Re: Another one that troubles me

eleven, not to take away any credit from you, but I have found a 1-move solution for that 17-clue puzzle 3 hours ago but was watching a movie on TV so delayed the posting. Here is my move:

Code: Select all
`+----------------------------+----------------------------+----------------------------+| 134567   134567   2        | 13689    13679    13678    | 345689  #56789    34567    || 8        9        137      | 1236     4        1367     | 2356     2567     23567    || 3467     3467     347      | 5       *23679    3678     |*234689   1        23467    |+----------------------------+----------------------------+----------------------------+|-912346   123468   5        | 12346    1236     1346     | 7       #2689     1246     || 124679   12467    1479     | 1246     8        14567    | 124569   3        12456    || 123467   1234678  13478    | 12346    123567   9        | 124568   2568     12456    |+----------------------------+----------------------------+----------------------------+| 1234579  123457   13479    | 13469    13569    13456    | 12356    2567     8        || 1357     13578    1378     | 1368     1356     2        | 1356     4        9        ||@123459   123458   6        | 7       @1359     13458    | 1235     25       1235     |+----------------------------+----------------------------+----------------------------+`

(*): 9 @ r3 locked @ r3c57
(#): 9 @ c8 locked @ r14c8
(@): 9 @ r9 locked @ r9c15

Either r3c5 or r3c7 must be 9 (*)
=> r9c5 & r1c8 can't be both 9
=> At least one of r9c1 & r4c8 must be 9 (@,#)
=> r4c1, seeing r9c1+r4c8, can't be 9

I guess this move could be called a "franken swordfish" or "simple x-chain" or "turbot chain with 3 strong links" but I decide to ignore all the nomenclature rubbish. It's a simple move involving 7 cells and 1 digit and that is it.
udosuk

Posts: 2698
Joined: 17 July 2005

eleven wrote:sarker306, there are a lot of much harder 17 clues.
This one can be solved with coloring on 8 (3 strong links) and 9 (3 strong links):
Code: Select all
`+----------------------------+----------------------------+----------------------------+| 134567   134567   2        |#13689    13679    13678    | 3489     6-89     3467     || 8        9        137      | 1236     4        1367     | 235      256      23567    || 3467     3467     347      | 5        23679    3678     | 23489    1        23467    |+----------------------------+----------------------------+----------------------------+|A123469  #123468   5        | 12346    1236     1346     | 7       @2689     1246     || 124679   12467    1479     | 1246     8        57       | 12459    3        12456    || 123467   1234678  13478    | 12346    57       9        | 12458    2568     12456    |+----------------------------+----------------------------+----------------------------+| 123459   12345    1349     | 13469    13569    13456    | 12356    7        8        || 1357     13578    1378     |@1368     1356     2        | 1356     4        9        ||B123459  @123458   6        | 7       A1359    #13458    | 1235     25       1235     |+----------------------------+----------------------------+----------------------------+`

Either r1c4=8 or
r8c4=8 -> r9c6<>8 -> r9c2=8 -> r4c2<>8 -> r4c8=8
So r1c8 cannot be 8.

Either r1c4=9 or
r8c4=9 -> r9c5<>9 -> r9c1=9 -> r4c1<>9 -> r4c8=9
So r1c8 cannot be 9.

Sarker306
To reach the position that Eleven begins from, note that :
- r7c8 must be 7 : no other spot available in box 9.
- 5 can only appear in box 1 in r1c12.
Those steps therefore eliminate 5 and 7 from r1c8, which is reduced to 689.
Deriving a contradiction from supposing that something is true (Eleven's method) is not "universally admired" in Sudoku circles, although it is of course rock-solid logically.
You can however reach those same conclusions here with another method :
examine r4c8.
Notice that two simple chains show that :
- if r4c8 is not 8 then r3c7 must be 8
- if r4c8 is not 9 then r3c7 must be 9.
There are three deductions which follow, of which only two are useful in the circumstances, though the third is the most interesting :
1. at least one of r4c8 and r3c7 must be 8 : hence r1c8 cannot be 8
2. at least one of r4c8 and r3c7 must be 9 : hence r1c8 cannot be 9
As a result r1c8 is reduced to 6 which is enough to solve the puzzle straight away.
3. the third deduction is that r4c8 must be 8 or 9, so anything else in the cell ie 26, can be removed.
Because if r4c8 were neither 8 nor 9, then r3c7 would be both 8 and 9...which under the current rules of Sudoku is impossible.
aran

Posts: 334
Joined: 02 March 2007

eleven wrote:Either r1c4=8 or
r8c4=8 -> r9c6<>8 -> r9c2=8 -> r4c2<>8 -> r4c8=8
So r1c8 cannot be 8.

Either r1c4=9 or
r7c4=9 -> r9c5<>9 -> r9c1=9 -> r4c1<>9 -> r4c8=9
So r1c8 cannot be 9.

Since both chains go through r1c4 and r4c8 there exists the continuous loop: r4c8 =8= ... r1c4 =9= ... r4c8.

Code: Select all
` 134567  134567   2     |*89-136 13679  13678 | 3489   689   3467 8       9        137   | 1236   4      1367  | 235    256   23567 3467    3467     347   | 5      23679  3678  | 23489  1     23467------------------------+---------------------+--------------------*123469 *123468   5     | 12346  1236   1346  | 7     *89-26 1246 12467-9 12467    1479  | 1246   8      57    | 12459  3     12456 123467  123467-8 13478 | 12346  57     9     | 12458  2568  12456------------------------+---------------------+-------------------- 12345-9 12345    1349  |*13469  1356-9 13456 | 12356  7     8 1357    1357-8   1378  |*1368   1356   2     | 1356   4     9*123459 *123458   6     | 7     *1359  *13458 | 1235   25    1235r4c8 =8= r4c2 -8- r9c2 =8= r9c6 -8- r8c4 =8= r1c4 =9= r7c4 -9- r9c5 =9= r9c1 -9- r4c1 =9= r4c8 =8= continuous loop==> r68c2<>8, r1c4<>136, r7c5<>9, r57c1<>9 and r4c8<>26 (10 eliminations)`

[edit: 1) added terse continuous loop; 2) included eleven's edit in quote]
Last edited by ronk on Fri Mar 06, 2009 7:21 pm, edited 2 times in total.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

Just realised my move is almost the same as the 2nd move listed by eleven, just eliminates the 9 in different cells. So your 1st move (elimination of 8) is totally redundant.

aran wrote:examine r4c8.
Notice that two simple chains show that :
- if r4c8 is not 8 then r3c7 must be 8
- if r4c8 is not 9 then r3c7 must be 9.

Also r4c8<>9 => r1c8=9 => r3c7<>9.
So r4c8 must be 9.
udosuk

Posts: 2698
Joined: 17 July 2005

There was a mistake in my post, the strong link for 9 in column 4 is between r1c4 and r7c4.

aran wrote:Deriving a contradiction from supposing that something is true (Eleven's method) is not "universally admired" in Sudoku circles, ...
Though i dont have a problem with it, especially when this method is shorter/easier to spot/better to describe etc, in this case i did not use a contradiction, but just (AIC like) showed that one of 2 cells have to be 8 (9).

ronk showed all the additional eliminations you get by combining the 2 chains. We had some similar examples some time ago in the thread Contrary "17" Puzzles. It turned out (for me), that those patterns can be found in harder 17 clue puzzles more often than in random ones with similar difficulty.
eleven

Posts: 1907
Joined: 10 February 2008

eleven wrote:There was a mistake in my post, the strong link for 9 in column 4 is between r1c4 and r7c4.

aran wrote:Deriving a contradiction from supposing that something is true (Eleven's method) is not "universally admired" in Sudoku circles, ...
Though i dont have a problem with it, especially when this method is shorter/easier to spot/better to describe etc, in this case i did not use a contradiction, but just (AIC like) showed that one of 2 cells have to be 8 (9).

You're right, apologies, misread you.
aran

Posts: 334
Joined: 02 March 2007

Code: Select all
` 134567  134567   2     |*89-136 13679  13678 | 3489   689   3467 8       9        137   | 1236   4      1367  | 235    256   23567 3467    3467     347   | 5      23679  3678  | 23489  1     23467------------------------+---------------------+--------------------*123469 *123468   5     | 12346  1236   1346  | 7     *89-26 1246 12467-9 12467    1479  | 1246   8      57    | 12459  3     12456 123467  123467-8 13478 | 12346  57     9     | 12458  2568  12456------------------------+---------------------+-------------------- 12345-9 12345    1349  |*13469  1356-9 13456 | 12356  7     8 1357    1357-8   1378  |*1368   1356   2     | 1356   4     9*123459 *123458   6     | 7     *1359  *13458 | 1235   25    1235r4c8 =8= r4c2 -8- r9c2 =8= r9c6 -8- r8c4 =8= r1c4 =9= r7c4 -9- r9c5 =9= r9c1 -9- r4c1 =9= r4c8 =8= continuous loop==> r68c2<>8, r1c4<>136, r7c5<>9, r57c1<>9 and r4c8<>26 (10 eliminations)`

The close relationship between 8 and 9 also demonstrated by the fact that a continuous loop on 9 starting from r4c8 following an identical or parallel path produces the same eliminations.
aran

Posts: 334
Joined: 02 March 2007

aran wrote:
ronk wrote:r4c8 =8= r4c2 -8- r9c2 =8= r9c6 -8- r8c4 =8= r1c4 =9= r7c4 -9- r9c5 =9= r9c1 -9- r4c1 =9= r4c8 =8= continuous loop
==> r68c2<>8, r1c4<>136, r7c5<>9, r57c1<>9 and r4c8<>26 (10 eliminations)

The close relationship between 8 and 9 also demonstrated by the fact that a continuous loop on 9 starting from r4c8 following an identical or parallel path produces the same eliminations.

Nice loops are meant to be read both left-to-right and right-to-left. Your "point" is the right-to-left read.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

ronk wrote:
aran wrote:
ronk wrote:r4c8 =8= r4c2 -8- r9c2 =8= r9c6 -8- r8c4 =8= r1c4 =9= r7c4 -9- r9c5 =9= r9c1 -9- r4c1 =9= r4c8 =8= continuous loop
==> r68c2<>8, r1c4<>136, r7c5<>9, r57c1<>9 and r4c8<>26 (10 eliminations)

The close relationship between 8 and 9 also demonstrated by the fact that a continuous loop on 9 starting from r4c8 following an identical or parallel path produces the same eliminations.

Nice loops are meant to be read both left-to-right and right-to-left. Your "point" is the right-to-left read.

Yes indeed.
aran

Posts: 334
Joined: 02 March 2007

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