## Challenge: New set of 11 'Unsolvables'

Advanced methods and approaches for solving Sudoku puzzles
. 1 .|. . .|. 5 .
7 . 8|. . .|. . .
. . 2|6 . 8|3 . .
-----+-----+-----
9 . .|3 . 5|. . 7
. . .|. . .|. . .
8 . .|4 . 2|. . 6
-----+-----+-----
. . 9|7 . 1|2 . .
. . 3|. . .|4 . .
. 8 .|. . .|. 9 .

puzzel #33

I have a strange method for solving a bunch of the squares that leads to singles.... ?

A4 = (45) - (8+7+6+5+4+3+1)
A4 = (11)

11 | (9+2), (8+3), (8+2+1), (7+4), (7+3+1), (6+5), (6+4+1), (6+3+2), (5+4+2), (5+3+2+1)

remove known numbers

11| (9+2), (5+4+2), (5+3+2+1)
check logic on (paired unknowns)
(9+2)
9 = (8+1), (7+2), (6+3), (5+4),(5+3+1) (remove the known number solutions
9 = (7+2) is the only known number in possible summations is the 2 we can apply logic to this
since in the original equation 2 was not seen by the cell A4, 9 was not directly revealed. thus eliminated the 9 as the solution.

there fore the only valid solution is A4(2)

J(4) sets to 5 (single)
(from here my solver i use can acutally solve the puzzle) and just copy paste in the results.

i can use the method above to sove the entire puzzel check out this article for a bit more ideas on how it functions.
http://forum.enjoysudoku.com/viewtopic.php?p=34634#p34634

POINTING PAIR: 5s at G9/H9 points to E9, removing 5

LBR: 5 exists only in box 4 and col 3, can remove from E1
LBR: 5 exists only in box 4 and col 3, can remove from E2
LBR: 5 exists only in box 4 and col 3, can remove from F2

MULTI-COLOURING (TYPE 2): 7 taken off E2 - using E6 H2

AIC Rule 2, on 7 (Alternating Inference Chain, length 6):
1[F8]=3[F8]-3[F2]=7[F2]-7[H2]=7[H8]-
- Chain ends F8 cannot be 7 and H8 cannot be 1

AIC Rule 2, on 8 (Alternating Inference Chain, length 10):
7[A7]=7[J7]-7[J3]=7[H2]-7[F2]=3[F2]-3[F8]=1[F8]-1[D7]=8[D7]-
- Chain ends A7 cannot be 8 and D7 cannot be 7

SINGLE: A9 set to 8, unique in Row

NAKED TRIPLE (Col): G9/H9/J9 removes 1 from B9
NAKED TRIPLE (Col): G9/H9/J9 removes 1 from C9
NAKED TRIPLE (Col): G9/H9/J9 removes 1/3 from E9
NAKED TRIPLE (Box): G9/H9/J9 removes 3 from G8
NAKED TRIPLE (Box): G9/H9/J9 removes 1 from J7

NAKED TRIPLE (Row): C1/C2/C9 removes 4/5/9 from C5
NAKED TRIPLE (Row): C1/C2/C9 removes 4 from C8

SINGLE: B5 set to 5, unique in Column

POINTING PAIR: 8s at G8/H8 points to D8, removing 8
POINTING PAIR: 8s at G8/H8 points to E8, removing 8

SINGLES CHAIN (Type 1): Removing 1 from E8

ALS/ALS: [A7|J7] and [B4|B7], 6 is restricted common, other common candidate 9 can be removed from B9

AIC Rule 1, 1 taken off B8 - chain ends: C8 F8
AIC Rule 1, 1 taken off D8 - chain ends: C8 F8
AIC on 1 (Alternating Inference Chain, length 8):
1[C8]=7[C8]-7[H8]=7[H2]-7[F2]=3[F2]-3[F8]=1[F8]-

ALS/ALS: [A7|J7] and [B8|B9|C9], 9 is restricted common, other common candidate 6 can be removed from B7

NAKED PAIR (Row): B4/B7 removes 9 from B2
NAKED PAIR (Row): B4/B7 removes 9 from B6
NAKED TRIPLE (Box): A1/A3/B2 removes 4 from C1
NAKED TRIPLE (Box): A1/A3/B2 removes 4 from C2

This sets a bunch of single squares leading to

SINGLE: A6 set to 9, unique in Column
SINGLE: D7 set to 8, unique in Row
SINGLE: D7 set to 8, unique in Column
SINGLE: F5 set to 9, unique in Row
SINGLE: H5 set to 8, unique in Row
SINGLE: H5 set to 8, unique in Column
SINGLE: J3 set to 7, unique in Row
SINGLE: J3 set to 7, unique in Column
SINGLE: J5 set to 2, unique in Column

POINTING PAIR: 1s at H1/J1 points to E1, removing 1

SINGLES CHAIN (Type 1): Removing 4 from A1
SINGLES CHAIN (Type 1): Removing 4 from G2

POINTING PAIR: 4s at G1/J1 points to E1, removing 4
Y-WING 4 taken off G5 - using A5 A1 G1
Then the puzzle becomes all singles solutions.

wow... just noticed this i can solve this puzzle in 4 steps - rest become single elimination.

solve with my set of rules found on the link and u get this.

a4 (2) sets J4(5)

use breakdown limitations on H2 to solve
7 is limited by 5 which is limited by a 2 this shows me the lowest break is actually the 5 (first number you cannot reduce fully)

H2 = 5

h6 is solved using break limits+ repeating limiation laws h6 = 6

and the rest of the puzzle comes down to naked and hidden singles.
Last edited by StrmCkr on Thu Sep 14, 2006 2:00 am, edited 1 time in total.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 732
Joined: 05 September 2006

#32 is longer than the last couple of solutions again using grouped strong links with overlap in the start and end nodes as key. This also has a grouped nice loop where two adjacent ALS share a common cell which doesn't contain the linking label.
1) Naked Single (46)
2) Locked Line/Box (12)
3) Hidden Single (10)
4) Grouped Nice Loops with 3 GSL/ALS (5)
5) Grouped Nice Loops with 4 GSL/ALS (5)
6) Nice Loops with 3 Strong Links/BV Cells (4)
7) Finned X-wing (3)
8) UR+2(X,D,B)/1SL (Type 4,...) (3)
9) Nice Loops with 4 Strong Links/BV Cells (2)
10) Strong Nice Loops with 4 GSL/BV Cells (2)
11) Naked Triple (1)
12) Locked Box/Box (1)
13) Hidden Pair (1)
14) Mutant Swordfish (1)
15) Generalized WXYZ-wing (1)
16) 4-node XY-ring (1)
17) UR+3(X,C,N,U,E)/2SL (1)
21) Nice Loops with 5 Strong Links/BV Cells (1)
22) A=2 cell ALS-xz rule (1)
23) B=1 cell ALS-xy rule (1)
Code: Select all
`Locked Row Line/Box: r3c46 => r3c9<>9Locked Row Line/Box: r9c46 => r9c23<>5Locked Row Line/Box: r5c78 => r5c26<>1Locked Row Line/Box: r5c78 => r5c456<>3Locked Column Line/Box Pair: r46c9 => r2c9<>5,r8c9<>6Locked Column Line/Box: r89c8 => r5c8<>6Naked Row Triple: r1c467 => r1c2<>68,r1c3<>2,r1c8<>8Locked Row Line/Box: r2c12 => r2c5<>63-element Nice Loop: r4c6=1=r4c1-1-r6c1-6-r6c9=6=r4c9~6~r4c6 => r4c6<>63-element Nice Loop: r8c2=5=r8c3-5-r2c3-2-r9c3~7~r8c2 => r8c2<>74-element Nice Loop: r8c2=5=r8c3-5-r2c3-2-r2c9-9-r7c9=9=r7c2~9~r8c2 => r8c2<>94-element Strong Nice Loop: r7c1=1=r7c2=9=r7c9-9-r2c9-2-r2c13=2=r3c1~2~r7c1 => r7c1<>25-element Advanced Colouring: r1c3=4=r5c3=9=r8c3-9-r8c7=9=r2c7=1=r2c8=5=r1c8~5~r1c3 => r1c3<>5,r1c8<>4Locked Row Line/Box: r3c89 => r3c2<>45-element Nice Loop: r2c3=5=r8c3=9=r7c2=1=r6c2-1-r6c1-6-r2c1=6=r2c2~5~r2c3 => r2c2<>5B=1 cell ALS xy-rule: r1c7-2-r2c9-9-r7c4689 => r9c7<>8,r23c8<>84-element Nice Loop: r8c2-5-r1c2=5=r1c8-5-r2c8-1-r5c8~3~ => r8c8<>3UR+3C/2SL (4,7): r38c89 => r8c9<>7A=2 cell ALS xz-rule: r7c46-7-r89c7|r7c9 => r7c8<>33-element Grouped Nice Loop: r6c6=4=r6c2-4-ALS:r12389c2-6-r5c2=6=r5c456~6~r6c6 => r6c6<>6+--------------------+-----------------------+------------------+|    9   457*b   47  |   268     1       26  |   28    57    3  || 2368   368*b   25  |     4   238        7  | 1289    15   29  || 2378   378*b    1  |  2389     5      239  |    6    47  247  |+--------------------+-----------------------+------------------+|  167      2     3  |  5678   678       15  |    4     9   56  ||    5   4679*  479  | 2679*c 267*c   2469*c |   13    13    8  ||   16   1469*    8  |  3569    36  13459-6* |    7     2   56  |+--------------------+-----------------------+------------------+| 1378  13789     6  |   237     4       23  |    5    78  279  ||  237    35*b 2579  |     1  2367        8  |  239   467  249  ||    4   378*b   27  | 23567     9     2356  |   23  3678    1  |+--------------------+-----------------------+------------------+Common 3-element Grouped Nice Loop: r9c7-2-ALS:r159c3-9-ALS:r89c3|r8c12~3~ => r9c2<>3,r8c7<>3+---------------------+---------------------+------------------+|    9    457    47*b |   268     1     26  |   28    57    3  || 2368    368     25  |     4   238      7  | 1289    15   29  || 2378    378      1  |  2389     5    239  |    6    47  247  |+---------------------+---------------------+------------------+|  167      2      3  |  5678   678     15  |    4     9   56  ||    5   4679   479*b |  2679   267   2469  |   13    13    8  ||   16   1469      8  |  3569    36  13459  |    7     2   56  |+---------------------+---------------------+------------------+| 1378  13789      6  |   237     4     23  |    5    78  279  || 237*c   35*c 2579*c |     1  2367      8  | 29-3   467  249  ||    4   78-3   27*bc | 23567     9   2356  |   23* 3678    1  |+---------------------+---------------------+------------------+Locked Row Line/Box: r9c78 => r9c46<>3UR+2D/1SL (7,8): r79c28 => r7c2<>8UR+2D/1SL (2,9): r28c79 => r8c9<>94-element Strong Nice Loop: r7c2=1=r6c2-1-r6c1-6-r6c5-3-r8c5=3=r7c46~3~r7c2 => r7c2<>33-element Grouped Nice Loop: ALS:r1c23-5-ALS:r89c2|r9c3|r78c1-1-ALS:r46c1~7~ => r3c1<>7+--------------------+--------------------+------------------+|     9   457*   47* |  268     1     26  |   28    57    3  ||  2368   368    25  |    4   238      7  | 1289    15   29  || 238-7   378     1  | 2389     5    239  |    6    47  247  |+--------------------+--------------------+------------------+|  167*c    2     3  | 5678   678     15  |    4     9   56  ||     5  4679   479  | 2679   267   2469  |   13    13    8  ||   16*c 1469     8  | 3569    36  13459  |    7     2   56  |+--------------------+--------------------+------------------+| 1378*b  179     6  |  237     4     23  |    5    78  279  ||  237*b  35*b 2579  |    1  2367      8  |   29   467   24  ||     4   78*b  27*b | 2567     9    256  |   23  3678    1  |+--------------------+--------------------+------------------+Overlap 3-element Grouped Nice Loop: ALS:r7c1468-1-ALS:r46c1|r5c23-4-ALS:r1357c6~2~ => r9c6<>2+--------------------+--------------------+------------------+|    9    457    47  |  268     1    26*c |   28    57    3  || 2368    368    25  |    4   238      7  | 1289    15   29  ||  238    378     1  | 2389     5   239*c |    6    47  247  |+--------------------+--------------------+------------------+| 167*b     2     3  | 5678   678     15  |    4     9   56  ||    5  4679*b 479*b | 2679   267  2469*c |   13    13    8  ||  16*b  1469     8  | 3569    36  13459  |    7     2   56  |+--------------------+--------------------+------------------+| 1378*   179     6  |  237*    4    23*c |    5    78* 279  ||  237     35  2579  |    1  2367      8  |   29   467   24  ||    4     78    27  | 2567     9   56-2  |   23  3678    1  |+--------------------+--------------------+------------------+3-element Grouped Nice Loop: ALS:r78c1|r9c23-1-ALS:r6c15-3-r2c5=3=r2c12~3~ => r3c1<>3+--------------------+--------------------+------------------+|     9   457    47  |  268     1     26  |   28    57    3  || 2368*c 368*c   25  |    4   238*     7  | 1289    15   29  ||  28-3   378     1  | 2389     5    239  |    6    47  247  |+--------------------+--------------------+------------------+|   167     2     3  | 5678   678     15  |    4     9   56  ||     5  4679   479  | 2679   267   2469  |   13    13    8  ||   16*b 1469     8  | 3569   36*b 13459  |    7     2   56  |+--------------------+--------------------+------------------+|  1378*  179     6  |  237     4     23  |    5    78  279  ||   237*   35  2579  |    1  2367      8  |   29   467   24  ||     4    78*   27* | 2567     9     56  |   23  3678    1  |+--------------------+--------------------+------------------+Overlap 4-element Grouped Nice Loop: r2c9-9-ALS:r7c14689-1-ALS:r6c15-3-ALS:r2c35789~2~ => r2c1<>2+--------------------+--------------------+--------------------+|     9   457    47  |  268     1     26  |    28    57     3  || 368-2   368   25*d |    4  238*d     7  | 1289*d  15*d  29*d ||    28   378     1  | 2389     5    239  |     6    47   247  |+--------------------+--------------------+--------------------+|   167     2     3  | 5678   678     15  |     4     9    56  ||     5  4679   479  | 2679   267   2469  |    13    13     8  ||   16*c 1469     8  | 3569   36*c 13459  |     7     2    56  |+--------------------+--------------------+--------------------+| 1378*b  179     6  | 237*b    4    23*b |     5   78*b 279*b ||   237    35  2579  |    1  2367      8  |    29   467    24  ||     4    78    27  | 2567     9     56  |    23  3678     1  |+--------------------+--------------------+--------------------+Mutant Swordfish (r19c1/c7b27, fins=r9c4|r3c1): r1c467|r9c347|r38c1 => r3c4<>2+-------------------+---------------------+------------------+|    9   457    47  |   268*    1     26* |   28*   57    3  ||  368   368    25  |     4   238      7  | 1289    15   29  ||   28#  378     1  | 389-2     5    239  |    6    47  247  |+-------------------+---------------------+------------------+|  167     2     3  |  5678   678     15  |    4     9   56  ||    5  4679   479  |  2679   267   2469  |   13    13    8  ||   16  1469     8  |  3569    36  13459  |    7     2   56  |+-------------------+---------------------+------------------+| 1378   179     6  |   237     4     23  |    5    78  279  ||  237*   35  2579  |     1  2367      8  |   29   467   24  ||    4    78    27* |  2567#    9     56  |   23* 3678    1  |+-------------------+---------------------+------------------+5-element Grouped Advanced Colouring: r3c1=2=r2c3=5=r2c8=1=r2c7=8=r1c7=2=r1c46~2~ => r3c6<>24-element Grouped Nice Loop: r6c5-6-r5c456=6=r5c2-6-ALS:r12389c2-4-r6c2=4=r6c6~3~r6c5 => r6c6<>3+-------------------+----------------------+------------------+|    9  457*c   47  |   268     1      26  |   28    57    3  ||  368  368*c   25  |     4   238       7  | 1289    15   29  ||   28  378*c    1  |   389     5      39  |    6    47  247  |+-------------------+----------------------+------------------+|  167     2     3  |  5678   678      15  |    4     9   56  ||    5  4679*  479  | 2679*b 267*b  2469*b |   13    13    8  ||   16  1469*    8  |  3569    36* 1459-3* |    7     2   56  |+-------------------+----------------------+------------------+| 1378   179     6  |   237     4      23  |    5    78  279  ||  237   35*c 2579  |     1  2367       8  |   29   467   24  ||    4   78*c   27  |  2567     9      56  |   23  3678    1  |+-------------------+----------------------+------------------+4-element Grouped Nice Loop: ALS:r39c2-3-r2c12=3=r2c5-3-r6c5-6-ALS:r5c3456~7~ => r5c2<>7+--------------------+---------------------+------------------+|    9    457    47  |   268     1     26  |   28    57    3  || 368*b  368*b   25  |     4   238*     7  | 1289    15   29  ||   28    378*    1  |   389     5     39  |    6    47  247  |+--------------------+---------------------+------------------+|  167      2     3  |  5678   678     15  |    4     9   56  ||    5  469-7  479*c | 2679*c 267*c 2469*c |   13    13    8  ||   16   1469     8  |  3569    36*  1459  |    7     2   56  |+--------------------+---------------------+------------------+| 1378    179     6  |   237     4     23  |    5    78  279  ||  237     35  2579  |     1  2367      8  |   29   467   24  ||    4     78*   27  |  2567     9     56  |   23  3678    1  |+--------------------+---------------------+------------------+4-element Grouped Nice Loop: ALS:r39c2-3-r2c12=3=r2c5-3-ALS:r6c15-1-ALS:r7c1468~7~ => r7c2<>7+--------------------+-------------------+------------------+|     9   457    47  |  268     1    26  |   28    57    3  ||  368*b 368*b   25  |    4   238*    7  | 1289    15   29  ||    28   378*    1  |  389     5    39  |    6    47  247  |+--------------------+-------------------+------------------+|   167     2     3  | 5678   678    15  |    4     9   56  ||     5   469   479  | 2679   267  2469  |   13    13    8  ||   16*c 1469     8  | 3569   36*c 1459  |    7     2   56  |+--------------------+-------------------+------------------+| 1378*d 19-7     6  | 237*d    4   23*d |    5   78*d 279  ||   237    35  2579  |    1  2367     8  |   29   467   24  ||     4    78*   27  | 2567     9    56  |   23  3678    1  |+--------------------+-------------------+------------------+4-element Grouped Nice Loop: ALS:r2c35789-3-r6c5-6-r5c456=6=r5c2-6-ALS:r34678c1~8~ => r2c1<>8+--------------------+---------------------+------------------+|     9   457    47  |   268     1     26  |   28    57    3  ||  36-8   368    25* |     4   238*     7  | 1289*   15*  29* ||   28*c  378     1  |   389     5     39  |    6    47  247  |+--------------------+---------------------+------------------+|  167*c    2     3  |  5678   678     15  |    4     9   56  ||     5   469*  479  | 2679*b 267*b 2469*b |   13    13    8  ||   16*c 1469     8  |  3569    36*  1459  |    7     2   56  |+--------------------+---------------------+------------------+| 1378*c   19     6  |   237     4     23  |    5    78  279  ||  237*c   35  2579  |     1  2367      8  |   29   467   24  ||     4    78    27  |  2567     9     56  |   23  3678    1  |+--------------------+---------------------+------------------+3-element Advanced Colouring: r7c9=7=r3c9=2=r3c1=8=r7c1~7~r7c9 => r7c1<>7Column Finned X-Wing: r48c1|r458c5 => r4c4<>73-element Nice Loop: r8c1=2=r3c1=8=r7c1-8-r9c2~7~r8c1 => r8c1<>7UR+2B/1SL (5,6): r46c49 => r6c4<>63-element Nice Loop: r6c2=1=r7c2=9=r8c3-9-r5c3~4~r6c2 => r6c2<>4WXYZ-wing: r2c12|r3c2, r9c2 => r1c2<>7Column Finned X-Wing: r39c2|r37c9 => r9c8<>74-node XY-ring (r1c3-4-r1c2-5-r2c3-2-r9c3-7-r1c3) => r8c3<>2,r8c3<>7Locked Row Line/Box: r9c23 => r9c4<>7Hidden Row Pair: r8c58 => r8c5=67,r8c8=67Locked Row Line/Box: r7c46 => r7c1<>3Row Finned X-Wing: r3c19|r8c17 => r7c9<>2Locked Column Line/Box: r89c7 => r12c7<>2Locked Row Box/Box: r2c39|r3c19 => r2c5<>2`
Last edited by Mike Barker on Sat Jan 26, 2008 1:01 pm, edited 2 times in total.
Mike Barker

Posts: 458
Joined: 22 January 2006

still wondering on the 9 = 9+0 (whered the 0 come from when we only look at numbers posible?)

i got a score of 11 as the value missing what makes up that number.

well a 9 + 2 makes the 11 score. as well as other options but what numbers result in a score of 9 and not in the orginal information.. if its not known then the resulting score would be limited by that information ie fallible. if it was already known then the puzzle would have tested that 9 was singled out. from this information i can derive the score of 11 is equal to some other set off numbers not impeded by that limitation. all other set of numbers that can equate to that answer all point out the 2 is the only unknown from all selecatble solutions.

to me all the numbers have to be identifiable uniques meaning all other numbers that could result in that numbers must be used or seen befor that number can acuatly be the resulting answer.

i can't figure out how to accuratly discribe what im doing... so plz read and see if some one could actually help re write this for calrity.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 732
Joined: 05 September 2006

Code: Select all
`.1.|...|.5.7.8|...|.....2|6.8|3..---+---+---9..|3.5|A.7...|...|...8..|4.2|..6---+---+---..9|7.1|2....3|...|4...8.|...|.9.`

Consider cell A (r4c7).

A=(45)-(2+3+4+5+6+7+9)=9
=1+8|2+7|3+6|4+5|1+2+6|1+3+5|2+3+4
Only 1+8|1+2+6|1+3+5 contain "unseen" candidates by A

1+8 are both "unseen" by A, therefore 8 must be broken up to other numbers:
8=1+7|2+6|3+5|1+2+5|1+3+4
Only 1+7|1+2+5|1+3+4 contain "unseen" candidates by A

But A couldn't be 1+(1+7)|1+(1+2+5)|1+(1+3+4) (duplication of 1)

So A must be 1!

But we know from the final solution, r4c7=8. Therefore your logic has given us a wrong conclusion.
Case closed.
udosuk

Posts: 2698
Joined: 17 July 2005

Code:
.1.|...|.5.
7.8|...|...
..2|6.8|3..
---+---+---
9..|3.5|A.7
...|...|...
8..|4.2|..6
---+---+---
..9|7.1|2..
..3|...|4..
.8.|...|.9.

Consider cell A (r4c7).

A=(45)-(2+3+4+5+6+7+9)=9
=1+8|2+7|3+6|4+5|1+2+6|1+3+5|2+3+4
Only 1+8|1+2+6|1+3+5 contain "unseen" candidates by A

1+8 are both "unseen" by A, therefore 8 must be broken up to other numbers:
8=1+7|2+6|3+5|1+2+5|1+3+4
Only 1+7|1+2+5|1+3+4 contain "unseen" candidates by A

But A couldn't be 1+(1+7)|1+(1+2+5)|1+(1+3+4) (duplication of 1) up to

So A must be 1!

the last line is where you go wrong with your logic string. because of the duplication of (1) and it has no break downs we know that A must be 8 must be the real solution as no other possible combination can yeild a score of 9 - this is a {self inclusive numbers} which eliminates the 1 as solutions.

i really need a clear way of writing what im trying to explain fully..

pic some other locations and i'll try explaining it again..

R4C5 is also selficlusive proving that 6 is the valid solution.
Last edited by StrmCkr on Thu Sep 14, 2006 4:35 am, edited 1 time in total.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 732
Joined: 05 September 2006

Here is what you wrote above:
StrmCkr wrote:A4 = (45) - (8+7+6+5+4+3+1)
A4 = (11)

11 | (9+2), (8+3), (8+2+1), (7+4), (7+3+1), (6+5), (6+4+1), (6+3+2), (5+4+2), (5+3+2+1)

remove known numbers

11| (9+2), (5+4+2), (5+3+2+1)
check logic on (paired unknowns)
(9+2)
9 = (8+1), (7+2), (6+3), (5+4),(5+3+1) (remove the known number solutions
9 = (7+2) is the only known number in possible summations is the 2 we can apply logic to this
since in the original equation 2 was not seen by the cell A4, 9 was not directly revealed. thus eliminated the 9 as the solution.

there fore the only valid solution is A4(2)

From that example, 2 was the number with no "break downs", and you decided to keep it and eliminate 9. Here you decide to keep 8 and eliminate 1... So in fact there is no logic... Just whatever you like...

I saw and still see all these stuffs as a hoax... Inconsistent "logic" really throws me off...
udosuk

Posts: 2698
Joined: 17 July 2005

in this case we have

9 = (8+1) - could use
9 = (7+2) - not
9 = (6+3) - not
9 = (6+2+1)- valid
9 = (5+4) - not
9 = (5+3+1 ) valid

so in reality there is 3 solutions that could be used y isn't 1 of them vaild ..

becasue 8 breaks into a group of knowns and 1 solution created from unkowns (7+1)

from here we know that 1 influces the choice of "8" so it casues solutions with single "1" to be no longer vaild as it limits the answers. leaving "8" + 1 as the only way to solve the solution. y is that if u plug the only other unknown sum of 8 into the inital equation it creates an error to get around the err of limit you have to use the larger number proving that "8" is the only solution.

look i can compute a missing score derived from the known puzzle this gives me a score

the way a score can be created is deifined and influencd by limitations of this puzzle and the limiations of summation of each individual number.

But each number individual is also inturn limited by the puzzles known number forceing it to be either a single unique or has a limited number possible sums creating limitions on equations.

as the above y is the 9 either a (8,1) how can it be soley the 8 is the fact all other numbers that create the 8 are singled out except when adding with a 1. this constricts the creation of the "8" portion of the equation.

so y isn't the solution a 1 for the same reason above the score of "9" is created with the aid of the 1 in sum of 8.

thus to satisfy this limit of the puzzel this solution must contain the "1' twice with out repeating, so it arrives in the form of "8"

becasue the "1" is a limit of 8 we can eliminated solutions containing singled out "1"s. but this restriction of "1" contained in the 8 shows that the (8+1) is still a vaild solution so we cannot eliminated it, but the limitation of the 1 does intern remove the 1 in that equation proving to me that the "8" is correct solution.

11| (9+2), (5+4+2), (5+3+2+1)
check logic on (paired unknowns)
(9+2)
9 = (8+1), (7+2), (6+3), (5+4),(5+3+1) (remove the known number solutions
9 = (7+2) is the only known number in possible summations is the 2 we can apply logic to this
since in the original equation 2 was not seen by the cell A4, 9 was not directly revealed. thus eliminated the 9 as the solution.

i should really reword that to show im removing the score solution of (9+2) proving y it becomes a false solution was the above quote this leaves the other solutions all with "2" as the valid answer.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 732
Joined: 05 September 2006

Dang, Mike, what techniques don't you have in that thing? (I don't want to waste my time looking for something the computer has already scoured. )

Just to make sure I understand these "Almost Swordfish" steps.

Is it true that...

The rule is that a potential swordfish and the group of candidates that could break it up in one dimension (rows, columns) are strongly linked (one or the other must be true)...

...and, the potential swordfish and any candidates that it would eliminate are weakly linked (one or the other must be false).

Of course this rule applies to all seafood...the swordfish in the rule is just a generic NxN swordfish.

Taking your example from puzzle 23...

Code: Select all
` *-----------------------------------------------------------------------------* | 5       3479    479     | 8       679     2       | 1       346     3469    | |X379    X6       2       | 1       4      x379     | 39      5      X8       | | 1       3489    489     | 56      569     359     | 3469    7       2       | |-------------------------+-------------------------+-------------------------| | 348     2348    5       | 9       1       6       | 7       234     34      | | 3467    23479   4679    | 257     8      A57      | 234569  1       34569   | |X679    X279     1       | 3      x257     4       | 8       26     X569     | |-------------------------+-------------------------+-------------------------| | 2       5       46789   | 67      679     789     | 346     3468    1       | | 4678    1       4678    | 2567    3       578     |D2456    9      B467     | |X679    X789     3       | 4       2569    1       |D256     268   CX567     | *-----------------------------------------------------------------------------*`

Almost Row Swordfish (r2c6|r6c5-7-r5c6-5-, r9c9-7-r9c9=5=r56c9): r2c16|r6c125|r9c129 => r5c7<>5
which, if I wanted to make an AIC would be
5[A]=7[A]-7[x]=7swordfish[X]-7[B]=7[C]-5[C]=5[D] => r5c6 <> 5

Just trying to clarify this, because the concept is fairly new, especially to the other sites.
Myth Jellies

Posts: 593
Joined: 19 September 2005

I've no experience with AIC's however I think it should look something like
5[A]=7[A]-7[x]=7swordfish[X]-7[B]-5[B]=5[C]-5[A] => r5c7 <> 5
Code: Select all
`Almost Row Swordfish (r2c6|r6c5-7-r5c6-5-, r9c9-7-r9c9=5=r56c9): r2c16|r6c125|r9c129 => r5c7<>5+---------------------+------------------+----------------------+|     5   3479    479 |     8   679    2 |       1   346   3469 ||  X379      6      2 |     1     4 x379 |      39     5      8 ||     1   3489    489 |    56   569  359 |    3469     7      2 |+---------------------+------------------+----------------------+|   348   2348      5 |     9     1    6 |       7   234     34 ||  3467  23479   4679 |   257     8  A57 | -234569     1 C34569 ||  X679   X279      1 |     3  x257    4 |       8    26   C569 |+---------------------+------------------+----------------------+|     2      5  46789 |    67   679  789 |     346  3468      1 ||  4678      1   4678 |  2567     3  578 |    2456     9    467 ||  X679   X789      3 |     4  2569    1 |     256   268   B567 |+---------------------+------------------+----------------------+`

As you can tell by my nomenclature I tend to think of the links as radiating out from the fish in a nice loop fashion and the elimination based on a discontinuous nice loop. There can actually be more than two of these radiating links and the links can be chains. Because my solver currently limits these links to a single ALS or GSL (grouped strong link), I haven't worried about continuous nice loop possibilities. The write up for all this is here. This technique should not be a surprise to the UK crowd since I first saw it over there in Anne's post for Ruud's #5000, though its been extended to include GSL.
Mike Barker

Posts: 458
Joined: 22 January 2006

.1.|...|.5.
7.8|...|...
..2|6.8|3..
---+---+---
...|...|...
8..|4.2|..6
---+---+---
..9|7.1|2..
..3|...|4..
.8.|...|.9.

If you look at r4c8

B = 45 - (9+7+6+5+3)
B = 15

8+6+1
(6+2)+6+1
(5+1)+2)+6+1 = 6 is the break limit of 8 meaning it is the true solveing number for this square is "6".
b= 6 (also confirmed when the twin DE is calculated in (removes the 1)

then from here i can look at R4C2 - C
leaving all other numbers unknown

C = 45 - (9+8+5+3+1)
C = 12

(9+3)
(8+1)+3
(7+2)+3 - first noted limitation
(6+3)+3
(6+2+1) + 3 - this is the real equation that limits satisfying the above score. shows the first forked path.

solutions for solving 9 are impeeded on the number 8 becasue it is impeeded by sub additons when (6+2) are expressed, sub numbers for the number 6 are impeeded by the addition of (4+2) this is a chain of events showing that 4+2 combination is a forked path of restrictions. meaning that 6 is not the solution as it has 2 limiting factors in creation of its self. this removes the 6, (confirmed by location b, and the twin DE removes the 4 - leaving 2 as the solution )

9 is also prevented with (5+4) - which transfers into (4+3+2) showing the fork path when further explored shows number permuations are all influeced by number "2" combined with the sub limit of 4 (3+1) which when used to subdived 3 creates a bi forked limit of the same answer of "2" twice. since math show 2+2 = 4 (and cant be used thanks to the puzzles restrictions we know that 4 is not a valid answer when limited by 2.

from this we can also conlude that 3+1 limit has no definable affect excpet makeing the 4 errorous when checking. this is "how my rule of break limits applies", since 4 is errorous on the check it is intern removed from the plasuable answers.

this removes all scores with a "4" in the selection of answers (varified by the twin locations of D&E - which when calculated to include the twin 1,4 show only a "2" as possible solution)

and what you have left is a "2" as the only valid answer.
c=2

i'll be back to finish this and explain how the D E are twins of eacher and cant be solved at this state (but can be used to solve other locations where the twin intercepts)

alright * edit back from lunch...

if we where to explore the limitations of D-(R4C3) and/or E-(R4C8) befor exploring the limits of C or B or A

we would find out they are identical bifucated permutations which cannot succesfully removed either the 4,1 solutions identifing its self as a twin location. they inital reduction equations of limits can remove all other unknown numbers except the (4,1 paried number) all of this is intern validated by the solution above).

D = 45 - (9+7+6+5+3)
D = 15

(9+6)
(8+1)+6 - bi limit of 8,1
(7+2)+6 - noted limit of 2
(6+3)+6
(5+4)+6 - limited by 4
(4+3+2)+6 - bi limited (4,2)
(3+1)+3)+2+6 (sub set of 4 showing 4 is limited by the 1(becasuse 2+2 subdivision can't occur with a 4)

so the number 8 sub set is tri-limited bye (4,2,1) we can eliminate it from the possible calculation (A shows this cant be an 8)

that leaves (4,2,1) as possiblities (C confirms the removal of the "2")
4 is directly limited by the 1 (3+1) and not the 2 as that breaks the rules of subdivision) this leaves 4,1 as the only solutions we cannot remove, why is that? it becasue these numbers are the root of each subdivision stoped by the creation of the number 4 - since the 2 cant manipulate the creation of a 4 it has a reduced limitation on aspects of sub permutations.(has less influence over the corase of subdvision compared to the other numbers). thus the 4,1 become the locked set of numbers used to achieve a singular anwser. Becasue they are locked as valid solutions they can't be removed with out addition information. Thus a posible twin location is discoverd.

with the (4,1) remaing in location D we can conclude that it is in fact a twin as location at E only has a 4,1 as an identical solution forcing both locations to be locked in sequence (both solutions are valid). this inturn removes the choice of 4,1 from all other equations as a result in that row.

At this point here is where i come to realize what udosuk was attemting to understand, how do i know that this process cannot remove the corret answer and not a fake number ?

which in the case of the twin i cannot without other information. proving my idea false so there has to be away to check for the twins...

so how do you discover that this location is not a twin of a number in the same coloumn or row???

well thats the easy part the inital score is the true unique number, if the score is duplicated in another cell along the same axsis and all the same seen numbers exist in both equations you have found its twin. This creates 2 location locked to those numbers and no method without other information can solve it.
Some do, some teach, the rest look it up.

StrmCkr

Posts: 732
Joined: 05 September 2006

Contratulations Mike, I didn't think these would be felled so quickly. My hit rate for the 1465 is 850, so well done on both counts. I'll have to digest your solutions before putting up the credit but first pass I can see some very nice extended strategies. Super stuff.
AndrewStuart

Posts: 21
Joined: 28 December 2005

### Unsolvable #33

u33

PROLOGUE

This is supposed to be the hardest of the Unsolvables, so I am not going to waste time explaining techniques contained in SSTS (the Simple Sudoku Technique Set). When I use a forcing net, starting with a candidate x such as 5e4 and proceeding by SSTS to a contradiction, I am not going to list the SSTS steps, I am merely going to state " ? 5e4, (SSTS), ?? -5e4." Anyone interested in the details of those steps merely has to use SS to trace them.

As usual, I concentrate on developing the technique of Forcing Nets. Here I have one novel technique to offer !

GURTH'S (MULTIPLE) ELIMINATION TECHNIQUE

When "bifurcating" to start a Forcing Net, it is customary to *place* a candidate in some cell and then prove a contradiction, thus allowing elimination of the placed candidate.

What is *not* customarily done (nor ever, as far as I have seen!) is to *eliminate* a candidate in some cell and then prove a contradiction, thus allowing *placement* of the eliminated candidate !!!

Why is it not customarily done? For no good reason. Simply because nobody has thought of doing it.

I have already described this idea at [url]http://forum.enjoysudoku.com/viewtopic.php?t=4830[\url]. This is Gurth's Elimination Technique. (GET).

Furthermore, one can go further : instead of trying a single elimination in a cell, one can try either :

(1) A multiple elimination of candidates in one cell. (A good reason for this would be to create a locked set.)

(2) A multiple elimination of candidates x in a row, column or box.

In both cases (1) and (2), proving a contradiction will allow elimination of the non-eliminated candidates !!! This is Gurth's Multiple Elimination Technique. (GMET).

SOLUTION OF U33

(1) (AIC: 7h2=k3 - k7=h8 - c8=a7 - a6=e6) -7e2.

(2) ? -9c2, (SSTS) ?? 9c2. (An example of GET).

(3) ? 9a5, (SSTS) ?? -9a5.

(4) ? 9b5, (SSTS) ?? -9b5.

(5) ? 3a5, (SSTS) ?? -3a5.

(6) ? 3a6 (==3b2==3e1), (SSTS) ?? -3a6, -3b2, -3e1, 3a1.

(7) ? 3e9, (SSTS) ?? -3e9, (SSTS) -3h8.

(8) ? 7a5, (SSTS), ( ? 5e7, (SSTS) ?? -5e7), (SSTS) ?? -7a5.
This is a net within a net. I could maybe get rid of this, but I don't see any reason to : this is a solution for advanced players, and I like this example of *real* bifurcation instead of misnamed bifurcation.

(9) ? 2e1, (SSTS), ?? -2e1, -2h2.

(10) ? 8h9, (SSTS) ?? -8h9, (SSTS) -5h1.

(11) ? 8e9, (SSTS) ?? -8e9.

(12) ? 8h8, (SSTS) ?? -8h8, -8f5.

(13) ? 8e7, (SSTS) ?? -8e7.

(14) ? -12k1, (SSTS) ?? 12k1. (-456k1 !!!) (GMET).

(15) ? 1e9, (SSTS) ?? -1e9.

(16) ? 1e7, (SSTS) ?? -1e7.

(17) ? 1f8, (SSTS) ?? -1f8, 3f8, 3e2, 2d2.

(18) ? 3g9, (SSTS) ?? -3g9, 3k9, 3g5, 3b6, -4k3.

(19) ? 7a6, (SSTS) ?? -7a6. It's all singles after that.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

### Solutions to Unsolvables #26, #25, #24.

Unsolvable #26 solved by Forcing Nets

(1) SSTS.

(2) ? 9e6, SSTS, ?? -9e6, 9k6.

(3) ? 7e5, SSTS, ?? -7e5, 9e5, SSTS.

(4) ? 1c2, SSTS, ?? -1c2, SSTS.

(5) ? 5e4, SSTS, ?? -5e4, 7e4, SSTS.

Unsolvable #25 solved by Forcing Nets

(1) SSTS.

(2) ? 1f3, SSTS, ?? -1f3, SSTS.

(3) ? 1f9, SSTS, ?? -1f9.

(4) ? 1f8, SSTS, ?? -1f8, 1d9, SSTS.

(5) ? 2g9, SSTS, ?? -2g9, 6g9, all naked singles after that.

Unsolvable #24 : Letting it all hang out.

As I always maintain that the main interest is not the solution but How the solution came about, hiding nothing, I have decided to show all incorrect and unnecessary tries made, just as they happened.

(1) SSTS. (2) ? 3c3, SSTS, ?? -3c3, 3b3, SSTS

(3) ? 3f2 led nowhere. ?9f8, SSTS, ?? -9f8, 5f8.

(4) ? 5h2, SSTS, ?? -5h2, SSTS.

(5) ? 5c3 led nowhere. ? 9f1, SSTS, -9f1.

(6) ? 4f6, rapid SSTS, ?? -4f6.

(7) ? 4a6, SSTS, ?? -4a6.

(8) ? 4c6, SSTS, ?? -4c6, 4k6.

(9) ? 1d5, SSTS, ?? -1d5.

(10) ? 1d4 led nowhere. ? 1h5, SSTS, ?? -1h5, SSTS.

(11) ? -1e5, SSTS, ?? 1e5, SSTS.

(12) ? 2c1, SSTS, ?? -2c1, SSTS to end.
gurth

Posts: 358
Joined: 11 February 2006
Location: Cape Town, South Africa

### Re: Challenge: New set of 11 'Unsolvables'

AndrewStuart wrote:A new set of eleven Michael Mepham 'unsolvable' sudokus has been released on http://www.sudoku.org.uk/bifurcation.htm.

Whats wrong here? I can only see 22 puzzles on this page.
ravel

Posts: 998
Joined: 21 February 2006

they where published on a sub link... i dont see it either... werid..
Some do, some teach, the rest look it up.

StrmCkr

Posts: 732
Joined: 05 September 2006

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