in some order. r2c9,r3c7,8 and 9 are 6789. To make 14 in r2c1 and r3c1, r3c1 has to be ...

Sorry, too abrupt. Additionally,r2c9 has to be the same as r3c4.

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Bigtone53 wrote:in some order. r2c9,r3c7,8 and 9 are 6789. To make 14 in r2c1 and r3c1, r3c1 has to be ...

Why could it not be

- Code: Select all
`.........`

5.......9

9.....678

given that we don't know anything else yet ?

- afjt
**Posts:**82**Joined:**07 September 2005

Ok, if we know that r2c9 = r3c4, then this does fix r3c1, but where does the r2c9 = r3c4 come from ?

As PaulIQ said at the top of the chain, this puzzle is bulging with 15s, either 96 or 87. Having got the 1 in r3c6, whichever r3c7/8 is (ie 96 or 87), r2/c3-r3c3 is the other one. So is r3c4-r4c4. To make this happen, r3c4 has to be the 'opposite' (9 from 6, 8 from 7) from r3c9. Between them in some order r4c4,7,8,9 contain 9876. QED

We probably do do things differently since I am averse to writing in every possibility for every square, as some do. I prefer to stare and then write in the numbers (OK, I sometimes note locked pairs) I appreciate that this makes doing Superior puzzles that much harder but I hopefully get there!

- Bigtone53
**Posts:**413**Joined:**19 September 2005

PaulIQ164 wrote:Well I for one certainly did.

Having happily slain every 'Killer' so far including all the 'Deadly' ones published, I was appalled to be beaten on this one. All the above may help but I'm still confused. I wonder if some kind person would start at PaulIQ164's original position and post the next few steps from there. That (except that I had filled in the 7 at r6c6) was as far as I got too.

Thanks.

- Bernard Stay
**Posts:**94**Joined:**22 March 2005

It was definitely more challenging than the deadly from the other day, but I didn't think it was as challenging as some of the deadlys in the killer book. I can't remember exactly how I did it (as it's more than 24 hours ago now) but I do know that the 15s were the key (they gave some quadruples). I also used normal sudoku techniques as well, more than I usually would in a killer. I hope they keep up the standard today (won't see the paper till lunchtime at the earliest)

- dalek
**Posts:**17**Joined:**28 September 2005

afjt wrote:Bernard Stay wrote:That (except that I had filled in the 7 at r6c6) was as far as I got too.

Do you mean r6c5 ?

Yes, sorry! And thanks BigTone53. it was OK after that.

- Bernard Stay
**Posts:**94**Joined:**22 March 2005

Bigtone53 wrote:Having got the 1 in r3c6, whichever r3c7/8 is (ie 96 or 87), r2/c3-r3c3 is the other one. So is r3c4-r4c4. To make this happen, r3c4 has to be the 'opposite' (9 from 6, 8 from 7) from r3c9. Between them in some order r4c4,7,8,9 contain 9876.

r2/c3-r3c3 should read r2c9-r3c9 and r4c4,7,8,9 should read r3c4,7,8,9

Anyway, I now understand Bigtone53's logic.

If anyone wants more hints about how to continue (either following my logic, or Bigtone53's) then let us know.

- afjt
**Posts:**82**Joined:**07 September 2005

Hmmm, I solved this in a slightly different way. Below is my method.

r1c6 and r2c6 must be an 8 and 9 one way or another. This means that r4c4 and r6c4 must also be 8 and 9 one way or another.

If r4c4 is a 9 then r3c4 is 6. It also follows that the 14 twosome in box 2 must be 5 and 9 (otherwise you repeat a 6 in box 2) and the 12 twosome must therefore be 4 and 8. This forces the 15 twosome in box one to be formed of 9 and 6. This is incompatible with the 14 twosome also in box 1 because that can only be formed of 8 and 6 or 9 and 5 (and both those options lead to a repeated number in box 1).

Therefore, r4c4 cannot be a 9 and must be an 8. This then gives r6c4 and with only a little bit of work r3c9 and r2c9. The rest of the puzzle then opens up pretty easily.

r1c6 and r2c6 must be an 8 and 9 one way or another. This means that r4c4 and r6c4 must also be 8 and 9 one way or another.

If r4c4 is a 9 then r3c4 is 6. It also follows that the 14 twosome in box 2 must be 5 and 9 (otherwise you repeat a 6 in box 2) and the 12 twosome must therefore be 4 and 8. This forces the 15 twosome in box one to be formed of 9 and 6. This is incompatible with the 14 twosome also in box 1 because that can only be formed of 8 and 6 or 9 and 5 (and both those options lead to a repeated number in box 1).

Therefore, r4c4 cannot be a 9 and must be an 8. This then gives r6c4 and with only a little bit of work r3c9 and r2c9. The rest of the puzzle then opens up pretty easily.

- mattlondon
**Posts:**3**Joined:**29 September 2005

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