Strmckr's 11.4 puzzle partial solution using sk loop

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Strmckr's 11.4 puzzle partial solution using sk loop

Postby StrmCkr » Thu Mar 06, 2008 9:43 am

SK loop: no clue how to word it, but

i found a virus pattern loop in my puzzle that involves the hidden quad intersection on the pattern 3,5,8,9

this is exactly identical to the pattern first pointed out and
used to solve Easter Monster

listed on this site.
http://sudoku.com.au/sudokutips.aspx?Go=T3-1-1992&category=sudoku

if i read and understand how it works correctly.

which removes 17
candidates at once.

inital puzzle.
Code: Select all
 *-----------*
 |5..|...|..9|
 |.2.|1..|.7.|
 |..8|...|3..|
 |---+---+---|
 |.4.|..2|...|
 |...|.5.|...|
 |...|7.6|.1.|
 |---+---+---|
 |..3|...|8..|
 |.6.|..4|.2.|
 |9..|...|..5|
 *-----------*

Code: Select all
 candiates
 *-----------------------------------------------------------------------------*
 | 5       137A     1467    | 23468   234678  378     | 1246    468C     9       |
 | 346A     2       469D     | 1       34689   3589    | 456B     7       468C     |
 | 1467    179D     8       | 24569   24679   579     | 3       456B     1246    |
 |-------------------------+-------------------------+-------------------------|
 | 13678   4       15679   | 389     1389    2       | 5679    35689   3678    |
 | 123678  13789   12679   | 3489    5       1389    | 24679   34689   234678  |
 | 238     3589    259     | 7       3489    6       | 2459    1       2348    |
 |-------------------------+-------------------------+-------------------------|
 | 1247    157B     3       | 2569    12679   1579    | 8       469D     1467    |
 | 178C     6       157B     | 3589    13789   4       | 179D     2       137A     |
 | 9       178C     1247    | 2368    123678  1378    | 1467    346A     5       |
 *-----------------------------------------------------------------------------*

loop is:
A = 3
B = 5
C = 8
D = 9

intersections removed and reduce the grid to this placement

Code: Select all
 *-----------------------------------------------------------------------------*
 | 5       137     1467    | 23468   234678  378     | 12      468     9       |
 | 346     2       469     | 1       389     3589    | 456     7       468     |
 | 1467    179     8       | 24569   24679   579     | 3       456     12      |
 |-------------------------+-------------------------+-------------------------|
 | 13678   4       15679   | 389     1389    2       | 5679    3589    3678    |
 | 123678  389      12679   | 3489    5       1389    | 24679   389     234678  |
 | 238     3589    259     | 7       3489    6       | 2459    1       2348    |
 |-------------------------+-------------------------+-------------------------|
 | 24      157     3       | 2569    12679   1579    | 8       469     1467    |
 | 178     6       157     | 3589    389     4       | 179     2       137     |
 | 9       178     24      | 2368    123678  1378    | 1467    346     5       |
 *-----------------------------------------------------------------------------*


is this correct so far?
Last edited by StrmCkr on Thu Mar 06, 2008 5:16 pm, edited 2 times in total.
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Postby StrmCkr » Thu Mar 06, 2008 10:40 am

my next move is intresting.

i use this row + column contain 389 pairing to induce a 4 cells to only hold 3 candidates in a column and also in a row. via box line reduction.

target als is based off an implicit chain.

als cells.

r2c5 (389)
r5c2 (389)
r8c5 (389)
r5c8 (389)

linked to
pivoting cells in box 5 are

R4C5 (1389)
R6C5 (4389)
R5C4 (4389)
R5C6 ( 1389)

i target a chain of singles off of Row 5 and column 5
appling box line reduction in the piviot box: forcing Box 5 to from an Als Chain inducing a visible error.

if 3 is held in R5c1 or R5C9

R5c2(89)+R5c8(89) = r5c4(4) = R5c6 (1) > R4c5(389) + R6C5 (389) +R2C5(389) + R8c5(389) Error 4 cells in a Column only 3 candidates each cell.

next
i can use the same pattern to elliminate 8 in R5c1 or R5c9

if 8 is in R5c1 or R5c9 than

R5c2(39)+R5c8(39) = r5c4(4) = R5c6 (1) > R4c5(389) + R6C5 (389) +R2C5(389) + R8c5(389) Error 4 cells in a Column only 3 candidates each cell.

next move is to use the same pattern again on colum 5 to eliminate

3+8 from

if R1C5 or R9C5 = 3 than

R2C5 (89) + R8C5 (89) = R4C5 (4)= R6C5 (1) = R5C2 (389) + R5C4(389)+ R5C6 (389) + R5C8 (389)Error 4 cells in a Row only 3 candidates each cell.

If
R1c5 or R9C5
= 8 than

R2C5 (39) + R8C5 (39) = R4C5 (4)= R6C5 (1) = R5C2 (389) + R5C4(389)+ R5C6 (389) + R5C8 (389)Error 4 cells in a Row only 3 candidates each cell.

which implies
R1c5, R9c5, R5c1 , R5c9 Cannot = 3 or 8

next i utilize the same pattern again inducing the Same chain above. except boxline swaps to 3/8 pairing.

which
removes 9 From R3C5,R5C3,R5c7,R7C5

proof.

if R7c5 or R3C5 = 9 than

R2C5 (38) + R8C5 (38) = R4C5 (4)= R6C5 (1) = R5C2 (389) + R5C4(389)+ R5C6 (389) + R5C8 (389)Error 4 cells in a Row only 3 candidates each cell.

iF R5C3 or R5C7 = 9 than

R2C5 (38) + R8C5 (38) = R4C5 (4)= R6C5 (1) = R5C2 (389) + R5C4(389)+ R5C6 (389) + R5C8 (389)Error 4 cells in a Row only 3 candidates each cell.
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