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 Post subject: Re: CCC
PostPosted: Thu Nov 18, 2010 4:10 am 
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Grand Master
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Posts: 1893
Location: Lethbridge, Alberta, Canada
If you want some practice in findings CCCs, Assassin 196 is a good one to try. They are available from the starting position of the puzzle; there's a code string in the A196 thread if you need it.

Image

See how many CCCs you can find. My walkthrough contains 8 of them.

Thanks Ed for pointing out that these eliminations can also be obtained using IOUs. I've also posted this position in the IOU thread.


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 Post subject: Re: CCC
PostPosted: Mon Feb 21, 2011 1:06 am 
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Grand Master
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Location: Lethbridge, Alberta, Canada
A187 had a nice cage pattern by Ronnie, based on the Christmas scene, but as far as the Killer Techniques forum is concerned the main interest is in the bottom two rows, specifically the way the cages at R8C4, R8C5 and R8C6 project up from R9 and their cage totals.

In Assassin 187 V2 this is the position after the Prelims

Image

Now we can use
45 rule on R9 3 outies R8C456 = 22 = {589/679}, 9 locked for R8 and N8, clean-up: no 4 in R7C7, no 6 in R8C5
R8C45 and R8C56 cannot total 15 (because they would clash with R89C5, CCC) -> no 7 in R8C46

As Ed has pointed out in this thread, this type of elimination can also be made using IOU.


Last edited by Ed on Mon Feb 21, 2011 8:38 pm, edited 1 time in total.
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 Post subject: Re: CCC
PostPosted: Mon Feb 21, 2011 2:37 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
The CCC in the previous post worked because R8C456 was between 1 and 9 greater than R89C5. In this example from Assassin 187 V4, that situation doesn't exist but instead there's the interesting situation that the hidden cage R8C456 has the same total as the 3-cell cage at R8C6. This can lead to an interesting variant of CCC which doesn't give any immediate eliminations but can still be very useful, as can be seen below.

This is the position after the Prelims, then
45 rule on R9 3 outies R8C456 = 21 = {489/579} (cannot be {678} because no 6,7,8 in R8C5) -> R8C5 = {45}, R8C46 = {789}, 9 locked for R8 and N8, clean-up: no 4,5 in R9C5
and 45 rule on N9 2 innies R7C9 + R9C7 = 14 = {59/68}

Image

Here R8C456 = 21 = {489/579}, 21(3) cage at R8C6 = {489/579/678} and because R8C6 is the crossover and R9C6 "sees" R8C456 it follows that R8C456 and the 21(3) cage at R8C6 must have different combinations. This is clearly useful for cages which only have two or three combinations, less so when there are more possible combinations.

In this particular position we can make more progress, because there's no 7 in R9C7, which does lead to candidate eliminations.

R8C456 and 21(3) cage at R8C6 = {489/579/678} cannot have the same combination because they share cell R8C6 and R9C6 “sees” R8C456 -> at least one of the combinations must contain 7 (and no 7 in R9C7) -> must have 7 in R8C46 + R9C6, locked for N8

and that's not the end of it because there's no 4 in R9C7

Because 7 only in R8C46 + R9C6, either R8C456 or 21(3) cage at R8C6 must be {489} -> must have 4 in R8C5 + R9C6, locked for N9


Last edited by Ed on Mon Feb 21, 2011 8:40 pm, edited 1 time in total.
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