The New Sudoku Players' Forum

[Crazy Girl]

I'm stuck on today's Killer, here's how far I've got.

Code: Select all

 
(95) (95) 8| - - - | - - -
(712) (712) (34) | - (12) - | - - -
6 (12) (34) | - (12) (98) | (98) - -
8 - - | - - 6 | - - -
- - -| - - - | - - -
- - 9 | 8 - - | - - -
- - -| - - - | - - -
- - -| - - - | - - -
- - -| - - - | - - -


Notation
6[3] means 3 cells totaling 6.

Here's how I got there:
Column 1 and 2 total 90, so 8 in R1C3. then R1C1 / R1C2 = 9 and 5.
this forces 14[2] in Column 1 to be 6 & 8.
the total of box 1 must be 45, so 45 - ( 22 + 10 + 6) = 7
so the part of 20[4] in box 1 totals 7 - only option 3&4.

I can't see the next step, so any help would be appreciated.

[CathyW]

Can't help yet - I went wrong so am starting over. Will post again later when (if?) I've done it.

The Times Killers have certainly been more challenging this week

[PaulIQ164]

The next step I used was the dreaded 180-rule!

That is, rows 6, 7, 8 and 9 have to total 180, and you can use that to work out r5c1, which sticks out.

Happy counting!

[Crazy Girl]

Have managed this week's without too much of a problem by came a cropper on this one.

thanks Paul for the tip.

[Bigtone53]

I can't see the next step, so any help would be appreciated.

My way was ro realise that the 9 in box 5 had to be in row 4, with only one possibility

[CathyW]

The next step I used was the dreaded 180-rule!

That was the one that cracked it for me too though I confess I used a calculator to check my adding up

[SammyJ]

I've been doing the Killers since they started coming out and this is the hardest I've seen.

I used that 180 rule as well, and it still took a while to come out. That was actually the first digit I got.

Not done yet but only a few columns on the right to go!

Not sure about the 9 that Bigtone got. He/she must have had some more info than where I was at, or I'm missing something.

Sam

[blankhalo]

Hmm... good spot that 180-rule. I used a slightly different attack from Crazy Girl's position. I focussed on row 2. We know 1,2,7 have already gone from the list (in the 10 in square 1 and the 3 in square 2) although we don't know in what order. There is an 11 in row 2 therefore its combinations can only be (83) or (65) as the 2 and 7 can't be used. There is also a 14 in row and since either 11 combination would prohibit it being (86) it must be (95). Therefore this means the 11 must be (83). Whew. This gives you r2c3, r3c3 and with a bit more logic r2c4 and r4c3.

BH

[Bigtone53]

Not sure about the 9 that Bigtone got. He/she must have had some more info than where I was at, or I'm missing something.

Sam,

I was commenting on CrazyGirl's 'so far' at the top of the chain

Looking at box 5, the 9 cannot be in row 6 as there is one there already It cannot be in row 5 as these make up 7's and 8's. So it is row 4. One of them is occupied and another would give a 6 in a row where a 6 already is. So it is the other one.

I am not sure that I am ready for dreaded 180 rules yet. I was pleased when I spotted a 90 last week.

Bigtone (bloke)