Conjugated pair chains

Advanced methods and approaches for solving Sudoku puzzles

Conjugated pair chains

Postby StrmCkr » Wed Nov 29, 2006 12:37 am

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Last edited by StrmCkr on Sat Dec 13, 2014 6:00 am, edited 2 times in total.
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Postby StrmCkr » Wed Nov 29, 2006 7:46 am

inital form of the puzzle
[code ]
| 0 1 0 | 0 0 0 | 0 5 0 |
| 7 0 8 | 0 0 0 | 0 0 0 |
| 0 0 2 | 6 0 8 | 3 0 0 |
----------------------
| 9 0 0 | 3 0 5 | 0 0 7 |
| 0 0 0 | 0 0 0 | 0 0 0 |
| 8 0 0 | 4 02 | 0 0 6 |
----------------------
| 0 0 9 | 7 0 1 | 2 0 0 |
| 0 0 3 | 0 0 0 | 4 0 0 |
| 0 8 0 | 0 0 0 | 0 9 0 | [/code

lets look at unsolvable #33 Conjugated pairs. (#'s where they can only be placed in 1/2 cells)


Code: Select all
 
x's are cells that are solved (0's are none conjugated cells)

 | 3  x  0 | 2 0  |78 0 28 |
 | x 39 x | 0 0 0 | 0 0 0 |
 | 0  9 x | x 7 x | x 7 0 |
-----------------------------
 | x 2  0 | x 0 x | 0 2 x |
 | 0 0 0 | 0 0 0 | 0 0 0 |
 | x 3 0 | x 0 x | 0 3 x |
--------------------------
 | 0 0 x | x 0 x | x 0 0 |
 | 0 7 x | 0 0 0  | x 7 x |
 | 0 x 7 | 2 0 0 | 7 x 0 |



lets look at unsolvable #33 Conjugated pairs. (#'s where they can only be placed in 1/2 cells)

heres the noted ALS chain ( formed over 3+squares of mutiple choices )
not choices that when A= x: B is not x

A1 = 3
B2 = 39
D2 = 9

also to note:
B2 is also indirectly linked to a twin chain of 3 held in F2, F8

(not needed as its simply a A forces b = not # in A (just poiting this one out.

Another noteable ALS chain

A4 =2
A9= 2/8
A8 = 8/7

also noted
that A8 indirectly linked to a long twin chain H8, h2 , J3, J7
and that J4 is directly linked via selction of 2

As A4 =2 : J4 = not 2. (this forms a sub chain) that acts similar to a twin, but weakly linked, both commons are not found in the same cell.

since these are expressed paths in the puzzle these chain must remain intact to solve the puzzle. Chains to stay intact must be linked with a common number held in all linked Cells.

which implies

A1 = 3+n link B2 = 39+n link to C2 = 9+n

this issolates a n number to be exclusive to the chain (its the common number found in both parts of the chain)

in quad 1 this happens to be 4,

the chain stays intact. and 4 is removed all from other cells in the same quad (as in this case where the als is in the quad only).

A3 cannot = 4
C1 cannot = 4

As 4 can only be found inthe AlS to prevent the ALS from breaking apart (which leads to errors)

In chain 2 the restricted common is {2}

Which implies and sets the cells as follows

A4 = 2/9 - strongly linked to B2 - 2 is removed from B2 as 2 is restricted to chain
A7 = 278
A9 = 289

also noted effect from the twin chain
when A9= 2
A6 = 7,c8 = not 7, C4 = 7
now the other plasuable.
A9 =8, A6 = 7 C8 = not 7, C4 = 7

other effects of a chain.

Issolation of numbers where overlapps exists.

now this issolates the 7 to being exlcusive in the chain.

and from all permutatons of that chain the 7 is either assigned to row A in quad 3 or is assigned to C5 this removes the 7 from cells A5 and A6


i need a way to show a color map of it!!!

i'll finish this tommarow mayhaps i can think of a way to clearly ilistrate where this type of ALS removes candidates.

strmckr
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