## "FIENDISH" Is there a way without guessing?

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

### "FIENDISH" Is there a way without guessing?

I find iTunes incredibly annoying, so not too long ago I put rockbox on my iPod (rockbox is an alternative operating system for mobile media devices). Within the basic package for rockbox there is a sudoku puzzle generator! The generator rates its puzzles as very easy, easy, medium and hard. I'm at the point where I can do 'hard' in about 10 minutes and 'medium' or less is not even worth doing.

Well, my sudoku generator surprised me with a FIENDISH level of difficulty (I didn't know he had it in him!!!). It kind of sucks now that I know it can make 'em this hard (there is no option for difficulty and Fiendish is ultra-rare I want more!!!)...

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`..7|.1.|.8....|..9|...1..|7.8|.3.-------------71.|4..|6...4.|...|.2...8|..6|.51-------------.9.|1.5|..3...|6..|....2.|.4.|8..`

I got this far in less than a minute,
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`..7|.14|.8..8.|..9|...1..|7.8|.3.-------------71.|4..|698.4.|8.1|327..8|.76|451-------------.9.|1.5|..3.7.|6..|....2.|.47|8..`

but I had a ton of triple splits and one double, so I started there.
Hidden Text: Show
I eventually determined that the 23 option on c6 needed to put the 2 on r8, otherwise r3c9 would have to be occupied by both 2 and 4. I THEN had a similar issue with 25 on r4, but figured that 5 needed to be on c5.
I had to do trial by force to figure two separate double splits out, but I'm not really satisfied with that technique (it feels an awfully lot like guessing...).

Is there an algorithm for a situation like this? Or at least a way to do this without force-trials? Did I miss something?
spinsane

Posts: 1
Joined: 19 July 2010

### Re: "FIENDISH" Is there a way without guessing?

Check to see which eliminations you don't have and you'll know what you missed.

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`   c2b1  Locked Pair                     <> 56   r12c1,r23c3   c5b5  Locked Candidate 1              <> 5    r23c5 r2  b3  Hidden Pair                     =  17   r2c78   c9b3  Locked Candidate 1              <> 4    r8c9   c9b3  Locked Candidate 1              <> 6    r9c9 <29+3>  XY-Wing  r6c4/r4c6+r9c4         <> 3    r8c6 <39+2>  XY-Wing  r9c4/r6c4+r8c6         <> 2    r4c6   c7b3  Locked Pair                     <> 59   r123c9   c35   X-Wing                          <> 3    r2c1 <39+5>  XY-Wing  r8c5/r5c5+r8c3         <> 5    r5c3 W-Wing: (2=6)r1c9 - (6)r1c2 = (6)r3c2 - (6=2)r3c5  =>  r1c4,r3c9<>2`
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006