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 Post subject: CCC
PostPosted: Mon Apr 27, 2009 7:31 pm 
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Grand Master
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Combo crossover clashes (CCC) are used fairly often when solving Killers. Two 2-cell cages cannot have the same total if they are in the same nonet, column or row and share a common, or crossover, cell. For this geometry to occur, at least one of the 2-cell cages must be a hidden cage or a split cage.

This example comes from here, in the simplified step at the end of my walkthrough.

Image

14. 45 rule on C12 3(1+2) innies R1C2 + R9C12 = 1 outie R3C3 + 17, IOU R9C12 cannot be 17 = [89]
14a. R9C12 cannot be [69] (because of overlap clash with R89C1 = 15, step 3)
14b. R9C12 not [69/89] -> no 9 in R9C2, R1C2 = 9 (hidden single in C2), R1C3 = 1, clean-up: no 7 in R5C4

Here I used a combo crossover clash between hidden cage R89C1, which totals 15 from an earlier step, and a split cage (2 innies) R9C12 to show that R9C12 cannot total 15 and therefore cannot be [69].

Combo crossover clashes are useful to eliminate combos but on their own they don't often elimination candidates. In this case I also used an IOU, which is discussed here, to make a further combo elimination from R9C12 and therefore the key elimination from R9C2.

In this diagram there is also an example of a combo crossover clash between a normal cage R1C23 and a split cage (2 outies) R13C3. R1C23 = 10 -> R3C13 cannot total 10. This can be used with 45 rule on C12 2 innies R9C12 = 2 outies R13C3 + 7 as a slightly harder way to eliminate R9C12 = 17 = [89].

Both of these examples are crossover clashes within a nonet. When they occur within a column or a row, walkthroughs may refer to them as overlap clashes.

Thanks Ed for providing the diagram! Many thanks also for interesting discussions and suggestions on how to improve this post! I've used a fairly loose definition for split cage, to avoid making the definition of CCC too complicated. Ed takes a narrower view of what a split cage is.


Last edited by Andrew on Wed Nov 18, 2009 2:51 am, edited 1 time in total.

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 Post subject: Re: CCC
PostPosted: Sun May 10, 2009 2:45 am 
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This is an example of a combo crossover clash occurring in a row. It comes from here. I'm still working on JF"F"K6 so I've given the link to the puzzle.

Image

I've only done Prelims, including R34C9 = {12}, locked for C9 with resulting clean-ups, then

1. 45 rule on N2 2 innies R2C4 + R3C6 = 5 = {14/23}, clean-up: no 1,2,3,4 in R2C3

2. 45 rule on N3 2 innies R3C79 = 10 = [82/91]
2a. Min R3C7 = 8 -> max R34C6 = 6, no 6,7,8,9
2b. R3C79 = 10 -> R3C67 cannot be 10 (CCC) -> no 4 in R4C6

There's another CCC in the opposite corner of the diagram. I'll let you find that for yourself as practice! ;)

Thanks again to Ed for providing the diagram!


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 Post subject: Re: CCC
PostPosted: Thu May 28, 2009 10:57 am 
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Thanks Andrew for starting this thread. BTW - I think that for the second example that Andrew gives, IOU is a technically easier way to get that elimination.

ie, "45" on n3, 2 outies r34c6 - 4 = 1 innie r3c9 -> no 4 in r4c6 (IOU). Of course, Andrew's way is not wrong and may be easier to see on those weird nonet i/o configurations.

Now, time to add some examples. First, what I think of as the classic CCC situation.

This is from Jean-Christophe's WT for uA97 which was Afmob's first killer and when Andrew first enjoyed CCC. Even though a few of us used similar moves for this puzzle, I remember really liking the way JC did it compared to the rest of us. I think it was because all permutations for the combinations involved were removed at once.

Image

NOTE: the 17(3)n8 = {179/269/278/467}
1j. 17(3) @ r8c456 <> [962] because it would conflict with h15(2) @ r89c4 = [96] (two 6)
1k. 17(3) @ r8c456 <> {78}[2] because it would conflict with h15(2) @ r89c4 = {78} (two {78})
1l. 17(3) @ r8c456 = [971|764] -> 7 locked @ r8c45 for n8, r8

As Andrew's WT points out, this is the key method of reducing this puzzle.

Second, a harder to see use of CCC that involves two 3-cell cages spread over two nonets but still has the one Crossover cell. This comes from here in A123 "Roulette"

Image

5. "45" n1: 3 innies r2c3 + r3c23 = h17(3)n1 = {179/269/278/359/368/458/467}
5a. the h17(3)n1 and 17(3)n1 have the same cage sum and r3c4 sees all of h17(3) -> each candidate in r2c3 must have a different combination in both cages, otherwise r3c4 would clash with r3c23
5b. -> {179} blocked because no other combination uses 1 and the 1 must be in r2c3!
5c. no 1 in r2c3

Cheers
Ed


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 Post subject: Re: CCC
PostPosted: Wed Jun 10, 2009 11:31 pm 
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Thanks Ed for your excellent examples, both of which broaden the definition of CCC.

My posts gave examples of clashes between two 2-cell cages with a common cell.

In Ed's first example the clash is between a 2-cell and a 3-cell cage which have a common cell.

The second example goes further. Here there are two 3-cell cages with a common cell; the actual combination clash is between the remaining 2 cells of each of these cages.


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 Post subject: Re: CCC
PostPosted: Wed Nov 18, 2009 3:17 am 
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One thing I've noticed is that there are often occasions when the use of a CCC or an IOU can achieve the same result. Here is such an example from A157. Step 2 of manu's walkthrough here started by using three CCCs in sub-steps b, c and d.

Image

2)a) Innies for r6789 : r6c4+r6c5+r6c6 = h17(3)
b) r6c4<>6 since r6c4=6 would force r6c5 to be equal to r5c6 (cage combinations of h17(3)
and 11(2) at r5c6) (in fact, one can use IOU, but I did not see it before I read Afmob's wt, so I did not
use it here)

c) For the same reason as for step b), r6c6<>6
d) r6c5<>6 since r6c5=6 would force r6c4 to be equal to r5c6 (cage combinations of h17(3)
and 11(2) at r5c6)

As manu has commented, these eliminations can also be made using IOUs, see here.

Thanks Ed for providing the diagram!


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 Post subject: Re: CCC
PostPosted: Wed Nov 18, 2009 4:21 am 
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And now for a CCC which I missed, from my walkthrough for Alt A171 here.

Image

In this position after I'd just done

17. 45 rule on N2 4 innies R1C456 + R3C6 = 24 = {1689/2679/3678/4569/4578} (cannot be {2589/3579} because 2,3,5 only in R3C6, cannot be {3489} which clashes with R23C4)
17a. 1,2,3,5 only in R3C6 -> R3C6 = {1235}

udosuk pointed out, in the A171 thread, that I could have continued

R1C456 cannot be 19 (CCC because R1C567 = 19) -> no 5 in R3C6
This would be immediately followed by
Killer quad 1,2,3,4 in R23C4, 16(3) cage and R3C6, locked for N2
R1C7 = 4 (hidden single in R1)

Thanks udosuk for pointing that out. It's a very nice CCC!

Thanks Ed for providing the diagram!


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 Post subject: Re: CCC
PostPosted: Thu Nov 19, 2009 2:39 am 
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Continuing on the theme that some eliminations can be made using either CCC or IOU, here is an example from the A175 series of puzzles here.

Image

This is an example where the 45 rule can be applied to the starting position and then CCC used to make an important candidate elimination; I'll leave you to find it.

This is how I did it in my A175 V0.75 walkthrough:
45 rule on C12 3 outies R127C3 = 19
R2C23 = 11 -> R12C3 cannot be 11, CCC -> no 8 in R7C3

This is one example where using CCC is as simple as using IOU. The alternative way using an IOU is here.


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 Post subject: Re: CCC
PostPosted: Thu Dec 17, 2009 4:27 am 
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This one is a very unusual CCC from udosuk's walkin for A177 here.

Image

3
Innies r789: r7c159=15=[168|276] (6 of r7 locked)
CCC: r7c89+r8c9 can't be [186]
=> 7/2 r7c8 can't be [16], must be {25|34}
=> 7/2 r7c8 & r9c7 form KNP {24} in n9
=> 11/2 r8c7={38|56} & r7c9 form KNP {68} in n9

I hope that udosuk won't mind me explaining the first two lines in a bit more detail

Innies r789: r7c159=15=[168|276] (6 of r7 locked)
If r7c159=[168] => no 1,6 in r7c8 => r7c89+r8c9 can't be [186]
If r7c159=[276] => r7c9 = 6 => r7c89+r8c9 can't be [186]


An alternative way to achieve this breakthrough is to use an unusual killer pair.

Thanks Ed for providing the diagram!


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 Post subject: Re: CCC
PostPosted: Fri Mar 19, 2010 10:45 am 
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Thanks Andrew and Ed for the nice examples of CCC on this thread. :thumbs:

I'd just like to pick up on Ed's first example here:

Ed wrote:
NOTE: the 17(3)n8 = {179/269/278/467}
1j. 17(3) @ r8c456 <> [962] because it would conflict with h15(2) @ r89c4 = [96] (two 6)
1k. 17(3) @ r8c456 <> {78}[2] because it would conflict with h15(2) @ r89c4 = {78} (two {78})
1l. 17(3) @ r8c456 = [971|764] -> 7 locked @ r8c45 for n8, r8

The above logic (taken from J-C's original WT) is OK, but IMHO overloaded. For clarity, here's the grid state again:

Image

As shown in the screenshot, the real CCC elimination is the removal of the 2 in R8C6. The principle is as follows:

"If a 3-cell cell cage of sum N shares a common cell with a 2-cell cage of sum M, then neither of the two non-overlapping cells in the three cell cage can contain the digit (N - M)."

The proof is simple: if one of the two non-overlapping cells in the 3-cell cage were to contain the digit (N - M), then the resulting split cage consisting of its remaining two cells and the other 2-cell cage (with which it would still share a cell) would have the same cage sum and thus correspond to the prohibited simple 2-cell CCC scenario mentioned by Andrew right at the start of this thread.

Thus, in walkthrough terms, the "pure" CCC move could be expressed as follows:

Quote:
1. 17(3) at R8C456 and h15(2) at R89C4 share the common cell R8C4.
1a. -> R8C56 cannot be (17 - 15) = 2 (CCC)
1b. -> no 2 in R8C6

Note that, in order to make this elimination, it is not necessary to list (or even know!) the combinations for this cage, which can be performed as a separate atomic step after the CCC move:

Quote:
2. 17(3) at R8C4 = {179/467} (no 8)
2a. 7 locked in R8C45 for R8 and N8

Note that it is still not necessary to list the permutations (as J-C did), because they do not provide any further eliminations in this case.

Although it is of course valid to simply enumerate the permutations and note the resulting clashes, I feel that this obscures the underlying CCC move mentioned above and is therefore less elegant.

_________________
Cheers,
Mike


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 Post subject: Re: CCC
PostPosted: Sun May 16, 2010 8:41 am 
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As I commented to Mike by PM - his previous post is very, very elegant because of it's simplicity.

This post by contrast, is about Clone CCC. These can get quite complicated. But boy, are they satisfying to find!

Clone CCC seems to be one move I'm finding in lots of puzzles at the moment. Two are on the Large IOD thread here. However, in addition to IOD, this next example of Clone CCC works because an outies cage sees part of the cloned cell-cage. I'll let the WT explain.
From TJK 41 and my alt step 8 here

Image

Quote:
8. "45" on c89: 1 outie r3c7 = 1 innie r8c8
8a. -> r1c8 "sees" all of n4 directly through c8, the 13(3) cage and it sees r3c7 indirectly since r8c8 = r3c7 EXCEPT r3c9, r45c7
8b. since r1c8 cannot equal r3c9 (no common candidates) -> r1c8 must be cloned in one cell of 9(2) at r45c7
8c. -> 9(2) must have 1,3,4 = {18/36/45}(no 2,7)

8d. "45" on n4: 2 outies r1c8 + r4c9 = 9
8e. since this outie cage and the clone celled cage (step 8b) have the same total -> r4c9 equals the other cell of 9(2) at r45c7
8f. since r4c9 sees r4c7 -> r4c9 = r5c7 = (568) (Clone CCC), r4c7 = (134)
Cheers
Ed


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