## Killer 14/11 - Stumped!

For fans of Killer Sudoku, Samurai Sudoku and other variants
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.
Bigtone53

Posts: 413
Joined: 19 September 2005

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.......99.....678`

given that we don't know anything else yet ?
afjt

Posts: 82
Joined: 07 September 2005

Bigtone53 wrote:Additionally,r2c9 has to be the same as r3c4.

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

We obviously solved it in completely different ways !
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

Anyone else think that today's Killer in the Times was one of the most difficult puzzles for some while? I had to start again and it took me nearly 2 hours to finish it. I'd never have finished the puzzle in the alloted 40 mins.

zebedee

Posts: 26
Joined: 27 March 2005

Well I for one certainly did.
PaulIQ164

Posts: 533
Joined: 16 July 2005

Definitely - I had to make second attempt too but got there in the end
CathyW

Posts: 316
Joined: 20 June 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

I wonder if some kind person would start at PaulIQ164's original position and post the next few steps from there

I don't have it in front of me but the purpose of finding the 9876 in row 3 means that r3c1 can only be a 5 so r2c1 is a 9 so ...
Bigtone53

Posts: 413
Joined: 19 September 2005

I've been having trouble with it because I can't read the numbers in my downloaded copy! - Makes it a bit more difficult.
stuartdm

Posts: 2
Joined: 15 November 2005

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 ?
afjt

Posts: 82
Joined: 07 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.

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.
mattlondon

Posts: 3
Joined: 29 September 2005

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