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 Post subject: Large IOD
PostPosted: Mon Jun 08, 2009 6:22 pm 
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Here is an example (A 153 V2) for which a large difference between innies and outies can be used fruitfully.

Using 45 rule for c89, one obtains : r7c8+r8c8 = 8 + r1c9+r2c9+r3c9 (1)

In fact, the innies-outies difference is so large that digits at r78c8 cannot be equal to digits at r123c9. For instance, we can prove that r7c8 <> r1c9 (the following can be generalized for the other cells of the Innies- outies)

From (1) , we deduce : r7c8 - r1c9 = (8 - r8c8) + (r2c9+r3c9) (2)

On the one hand, 8 - r8c8 >= -1 since r8c8 <= 9
On the other hand, r2c9+r3c9 >= 1+2=3 since r2c9 and r3c9 see each other.

From (2), we deduce : r7c8 - r1c9 >= -1 + 3 = 2 -> r7c8 <> r1c9

Concerning the above puzzle, we get :
1) digit at r7c8 is locked for c9 at cage 14(2) : r7c8=(5689)
2) digit at r8c8 is locked for c9 at r456c9 : no 1,2,3.

We can go further :

3)r67c9 <> 7. Like this : if one of r67c9 was 7, the other one and r7c8 would total 14 with the same combination as 14(2) at
n9 since r7c8 is locked at 14(2), and there would be a clash between 14(2) and 21(3)
4) No 7 at 21(3) at n6 : combination {489}

The puzzle is not cracked yet, but everything becomes smoother from here...


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 Post subject: Re: Large IOD
PostPosted: Sat Jun 13, 2009 10:22 am 
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Thanks Manu for getting this thread started. BTW - I think a technically easier way to do step 2c is to use max./min. then CPE.
ie: 2c. "45" c9: 2 outies r78c8 - 8 = 3 innies r123c9. Max. 2 outies = 17 -> max. 3 innies = 9 (no 7,8,9)
2d. 7 in r6789c9 -> no 7 in r7c8 (CPE)

However, the follow-up step 3 which eliminates 7 from r67c9 is dependant on the cell-cloning made possible by step 2c. It's really nice!

Manu's step 2c got me thinking about generalising some rules about when innies and outies cannot have the same placements because of the IOD (Innie Outie Difference from the 45 Rule).

NOTE: All these assume that if there is more than one innie then they can all see each other. Same with more than one outie.
A. 1 innie and 1 outie
eg, 1 innie -1 = 1 outie; 1 innie + 1 = 1 outie etc
-Cannot have the same placement if the IOD is NOT zero. Only when 1 outie = 1 innie can they be the same.

B. 1 (small) innie and 2 outies
eg. 1 innie + 10 = 2 outies etc
-Cannot have the same placements if the IOD is equal or more than +10.
Proof: If the innie and 1 outie are equal then the remaining outie (max. 9) cannot equal the IOD if it's 10 -> each outie must be different to the one innie

C. 1 (small) innie and 3 outies
eg. 1 innie + 18 = 3 outies etc
-Cannot have the same placements if the IOD is equal to or more than +18
Proof: If the innie and 1 outie are equal then the remaining 2 outies (max. 17) must equal the IOD which it can't -> each outie must be different to the one innie

Small IOD
eg. 1 innie + 2 = 3 outies; 1 innie + 1 = 3 outies
-Cannot have the same placements if the IOD is equal to or less than +2
Proof: If the innie and 1 outie are equal then the remaining two outies (min. 3) must equal the IOD which they can't -> each outie must be different to the one innie.

D. 1 (large) innie and 2 (or more) outies
eg, 1 innie = 2 outies; 1 innie - 1 = 2 outies etc
-Cannot have the same placements if the IOD is equal to or less than 0. Important, this only works if the one innie is equal to or more than the sum of the two outies
Proof: if one innie = one outie then the second outie would have to be zero (or less) which is impossible.

E. 2 innies and 2 outies.
eg, 2 innies - 9 = 2 outies; 2 innies - 10 = 2 outies; 2 innies + 9 = 2 outies etc
-Cannot have the same digit if the IOD is equal to or greater than +-9.
Proof: If one innie and one outie are equal then the remaining outie would have to be 9 greater(or smaller) than one innie, but the biggest difference possible is 9-1 = 8 -> the innies and outies must have different placments if the IOD is +-9

-Cannot have the same placement if any of the cells contains the IOD
Proof: If one cell has the IOD -> 1 cell = 2 cells which means they cannot be equal (see D.)

F. 2 (small) innies and 3 outies
eg, 2 innies + 17 = 3 outies etc
-Cannot have the same placements if the IOD is equal or more than +17.
Proof: If one innie and one outie are equal then the remaining innie (min. 1) + 17 (1+17=18) cannot equal the max. of the 2 remaining outies (9+8 = 17) -> each outie must be different to each innie when the IOD is +17

G. 2 (large) innies and 3 outies (same as Manu's step 2c above)
eg, 2 innies - 7 = 3 outies; 2 innies - 8 = 3 outies etc
-Cannot have the same placement if the IOD is equal to or less than -7. Important, this only works if the two innies sum 7 more than three outies
Proof: if one innie and one outie are equal then the remaining innie (max. 9) has to be 7 greater than the two remaining outies, but the biggest difference possible is 9-(1+2) = 6

-Cannot have the same digit if either of the two innies has the IOD.
Proof: if one innie has the IOD -> the other innie = 3 outies so they cannot contain the same digit.

H. 2 (small) innies and 4 outies
eg, 2 innies + 24 = 4 outies; 2 innies + 25 = 4 outies etc
-Cannot have the same placement if the IOD is equal to or more than +24.
Proof: if one innie and one outie are equal then the remaining innie (min. 1) + 24 has to be greater than the max. of the three remaining outies (9+8+7=24) which is impossible -> each outie must be different to each innie when the IOD is +24

I. 2 (large) innies and 4 outies
eg, 2 innies - 4 = 4 outies; 2 innies - 5 = 4 outies etc
Cannot have the same placement if the IOD is equal to or less than -4. Important, this only works if the two innies sum 4 more than four outies.
Proof: if one innie and one outie are equal then the remaining innie would have to be 4 more than the three remaining outies, but the biggest difference possible is 9-(1+2+3) = 3

J. 3 innies and 3 outies
eg, 3 innies - 15 = 3 outies; 3 innies + 15 = 3 outies; 3 innies + 16 = 3 outies etc
Cannot have the same placement if the IOD is equal to or great than +-15
Proof: if one innie and one outie are equal then the remaining two innies have to be 15 greater than the two remaining outies but the greatest difference possible is (9+8)-(1+2) = 14

Please let me know if anything is wrong or could be clearer. Note that the original post has been expanded.

Cheers
Ed


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 Post subject: Re: Large IOD
PostPosted: Wed May 12, 2010 10:17 am 
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I've enjoyed finding some Clone CCC recently, two of which used Large/Small IOD to work.

They are pretty mind twisting so hope the explanations are clear.

This one is probably the simplest. It's from A188V2 and my walkthrough here.

Image
Quote:
9. "45" on n78: 1 outie r6c4 + 11 = 2 innies r89c6
9a. min. 2 innies = 12 (no 1,2)

Crucial observation. No eliminations yet.
10. r6c4 must be cloned in one cell of 9(2)n8 since Innie Outie Difference (IOD) n78 is greater than 9 (step 9) Large IOD and r6c4 sees all other cells in n8 apart from r89c5.

11. no 3 in r6c4. Like this.
11a. 25(6)n5 = {123469/123478/123568/124567}
11b. all combos with each of 1,2 & 4 must have 1/2/4 in r6c4 so it will not clash with r7c6
11c. the only combo without each of 1,2,4 is {123568} but 3 cannot be in r6c4 since the 6 in this combo would have to be in n8 which would clash with {36} in 9(2)(from step 10. Clone CCC) -> no 3 in r6c4
Note that step 10 relies on the Large IOD in step 9 (see B. in the previous post) to show that the digit at r6c4 cannot be in r89c6.

The second example uses Small IOD!
From my walkthrough for TJK 40 here
Image
Quote:
Now for a Cloned CCC.
7. "45" on r1: 1 outie r2c5 + 2 = 3 innies r1c123
7a. min. 3 innies = 6 -> min. r2c5 = 4
7b. NOTE: no cell at r1c123 can equal r2c5 since if one did, the other 2 cells would have to sum to the IOD of 2 which is impossible
7c. -> the digit at r2c5 must be cloned at one of r1c89 a 13(2) cage (this cloning eliminates 7 from r2c5 but this is not the important bit)
7d. -> the 15(3) at r1c4 cannot contain 2 in r1c45 since it would leave a 13(2) split cage at one of r1c45 and with r2c5, but this would clash with the 13(2) at r1c8 (Clone CCC)
7e. ->no 2 in r1c45
Note that this move relies on the Small IOD from 1 innie and 3 outies (see C. in previous post) to prove that r1c123 cannot have the same placement as r2c5.

Cheers
Ed


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