Today's Times Killer, No. 755; 22 Feb 2008

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

Today's Times Killer, No. 755; 22 Feb 2008

Postby dkybird » Fri Feb 22, 2008 6:41 pm

Please can someone offer some tips as to how to get started on this one. Having completed every other puzzle in this week's Times, this killer has been stumped.
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Today's Times Killer, No. 755; 22 Feb 2008

Postby Cec » Sat Feb 23, 2008 2:12 pm

Hi dkybird,

"Killer" puzzles are probably out of my solving capabilities but to get help (?) I suggest you post both the original puzzle and how far you advanced. See Here for procedure to post puzzles.

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Re: The Times (2008.Feb.22)

Postby Cec » Sun Feb 24, 2008 2:24 pm

Oops!... Sorry dkybird for my "useless" posting instructions in overlooking this was a "Killer" puzzle as distinct from those which can be posted.

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Postby dkybird » Sun Feb 24, 2008 7:20 pm

Don't worry Cec. You'll see the puzzle now. Give it a go.
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"Killer" No. 755

Postby Cec » Mon Feb 25, 2008 3:50 am

dkybird wrote:Don't worry Cec. You'll see the puzzle now. Give it a go."

Don't worry? When I saw the puzzle that's when I started to worry.. then when I saw the time limit of one hour that made me worry more:) I take my hat off to people who can solve these "creatures".

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Postby Bigtone53 » Mon Feb 25, 2008 12:47 pm

Dkybird

It is not easy to describe but

1. Insert the singles 4,4,2,7
2. Consider where the 9 is in column 3, given that the 1 in column 3 has to be in the 14
3. This gives two choices for the remaining 13 in the 17 in column 3
4. You know the two cells in box 7 where the 9 can be.
5. Playing around with the alternatives in step 3 and trying to complete boxes 4 and 7 shows that one of the alternatives does not work.
6. Pencil in the other one and you can firm up some cells in Box 4
7. Meanwhile, there are two alternatives for cells r8c9 /r9c9 totalling 15. Considering where the 8 and 9 in column 9 can fit cracks this and hence the rest of the puzzle.
Friday's Killers are always Deadly, having built up in difficulty over the week.
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"Killer" No 755

Postby Cec » Mon Feb 25, 2008 1:56 pm

Bigtone53 wrote:"...Insert the singles 4,4,2,7..."

I still don't understand these "Killer" puzzles? I can see three shaded cells within the grid but I don't understand why singles 4,4,2,7 are specifically chosen or what cells they are placed in. Help appreciated.

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Postby ronk » Mon Feb 25, 2008 3:20 pm

I think discussions about Killer Sudoku, including help with particular Killer puzzles, belong on ...

Sudoku variants -- For fans of Killer Sudoku, Samurai Sudoku and other variants
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Postby Bigtone53 » Mon Feb 25, 2008 3:22 pm

I still don't understand these "Killer" puzzles? I can see three shaded cells within the grid but I don't understand why singles 4,4,2,7 are specifically chosen or what cells they are placed in. Help appreciated.




Because every row, column and box consists of 1-9, it has to add up to 45.

Look at box 1. r3c3 has to be a 4. Consider boxes 1,4,7, adding up to 135. r9c3 has to be a 2, so r9c4 has to be a 7. Consider rows 6,7,8.9. These 4 rows add up to 180 so r5c9 is a 4

and so on. The shaded squares are a shorthand way of demonstrating that you have solved the puzzle, not anything to do with the solving itself.
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Postby Bigtone53 » Mon Feb 25, 2008 4:15 pm

ronk said
I think discussions about Killer Sudoku, including help with particular Killer puzzles, belong on ...

Sudoku variants -- For fans of Killer Sudoku, Samurai Sudoku and other variants


You are right of course, but how do us grunts move stuff around? I hope that our practice is not that first-time enquirers should be ignored if they happen to post in the wrong forum:(
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Killer No. 755

Postby Cec » Tue Feb 26, 2008 4:31 am

Bigtone53 wrote:"...Look at box 1. r3c3 has to be a 4. Consider boxes 1,4,7, adding up to 135. r9c3 has to be a 2, so r9c4 has to be a 7. Consider rows 6,7,8.9. These 4 rows add up to 180 so r5c9 is a 4
and so on...."


I'm sure I am missing something - yes, I know what you're thinking when I say that:) - but I still don't get this.

I can see why r9c3 has to be a 2 but I don't understand why r3c3 has to be a 4. As the numbers to be placed in each of the three cells in column 3 (r345c3) must total 17 then why is cell r3c3 restricted to 4? With a 2 later placed in r9c3 following Bigtone's above comments then there are alternative choices for different numbers between 1 to 9 (excluding 2 of course) which could be placed in r345c3 and still give a total of 17.

On the matter of posting this puzzle in the wrong forum, I too agree with Ronk's above post that the puzzle should belong in the Sudoku Variants Forum. That said, Bigtone's comments are also relevant as to how this "problem" can be avoided.

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Postby Bigtone53 » Tue Feb 26, 2008 11:22 am

I have responded to Cec in a new thread in the right place, Sudoku Variants.
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Postby Pat » Tue Feb 26, 2008 12:43 pm

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