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I am continuing my education of how to solve "very hard" puzzles and can even see an Xwing in this puzzle yet that seems to take me nowhere....Any suggestions most welcome...

**9 *** 26*
8*1 **2 59*
625 4*9 3*8

497 2** 6**
*** 9** 8**
*86 **7 4*9

9** 8** 1*5
**8 19* 7*6
**4 *** 98*

(Taken from Sudoku programme - "very hard"

I believe there is an Xwing of 2's but cannot see where this takes me...cannot see another means of solving this b****y puzzle!!

It has more then just an X-wing, it also has a Swordfish (in the same number)...

Not that the swordfish helps you but it's nice to see one

The key to solving it: box 5 + column 6 + box 7 + row 9

Both the xwing and swordfish are optional though. But it does require several subsets to eliminate candidates.

Code: Select all

row 5: unique subset 46 in cells (5,5)(5,6) - removing other candidates for these cells
column 6: unique subset 18 in cells (1,6)(4,6) - removing other candidates for these cells
blocks 2 and 5 must contain 3 in columns 4 and 5 - removing 3 from candidates for cell(s) (7,5)(9,4)(9,5)
blocks 2 and 5 must contain 5 in columns 4 and 5 - removing 5 from candidates for cell(s) (9,4)(9,5)
block 7: disjoint subset 235 in cells (7,3)(8,1)(8,2) - updating candidates for cell(s) (7,2)(9,1)(9,2)

plain sailing after that.

Sorry the original post was mine - forgot to log in...

Animator, I'm not following - i can see a pair of 4,6's in box 5 but cannot see how to take the puzzle forward....

Ok, that's what you need to know about box 5.

Then take a look at column 6... you will see that there is 'pair' of four numbers...
The crucial thing is to remove those numbers from box 8 (except ofcourse in column 6)

Then look at box 7, you will see a 'pair' of three numbers, remove them too

Now, when you are done with that then you should be able to do something with row 9

..trying to follow you...column 6 has a pair of 1,8 at r1c6 and r4c6..this, together wiht 6,4 pair in box 5 means cells r7c6 r8c6 and r9c6 must have no's 3,5.

in box 7 numbers 2,3,5 must be in cells r8c1,r8c2 and r7c3...so 5 is eliminated from r8c6 and hence 5 goes in r9c6...the rest follows...as 3 must go in r8c6....thanks!

What unlocked it for me was the three cell 3,4,6 grouping in col 6. That forced the 3 into one of two cells in the bottom-centre box, and once the three was eliminated as a possibility from the other cells in the bottom-centre box it all fell together (as created another 1,6,7 grouping on the bottom row)