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 Post subject: IOU
PostPosted: Fri Apr 17, 2009 10:20 am 
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Grand Master
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This is based on the original IOU post found here There are some variations on this technique which will be in follow-up posts.

IOU
If an Innie & Outie are in the same r, c or n, then the 3rd cell must be Unequal to the I/O difference. Hence, IOU.

Image
For example, Cathy's step 11 (Assassin 59) says
Cathy wrote:
11. O-I N7: r6c23 - r9c3 = 8
Since r6c3 (an Outie) is in the same column as r9c3 (an Innie), then the 3rd cell must be Unequal to the difference - in this case 8. From IOU -> r6c2 <> 8.

Another way of seeing this is that if r6c2 was 8 then the difference would be 0 which means that r6c3 = r9c3: which is impossible because they are in the same column.

Similarly, step 13 is
Cathy wrote:
13. O-I N3: r4c78 - r1c7 = 7
Since the difference is 7: IOU -> r4c8 <> 7

Perhaps with a walk-through we could explain the logic of this move the first time and after that just refer to it as IOU. So with Cathy's step 14,
Cathy wrote:
14. O-I N9: r6c78 - r9c7 = 4
then add: IOU -> r6c8 <> 4

Incidently, Assassin 60 has several of this move - including where the 1 innie (of a r or c) and 1 outie (of that r or c) are in the same nonet. This is a bit harder to see compared to the I/O of nonets.

Image
eg, step 11 of Mike's (mhparker) A60 WT says
Mike wrote:
15. I/O difference R12: R3C25 = R2C6 + 9
15a. -> no 9 in R3C2
This is IOU from 2 rows since r2c6 (the innie) and r3c5 (one outie) are in the same nonet -> no 9 in r3c2.

Simple IOU is now a very common move used in many walkthroughs. My impression is that a puzzle that requires IOU would get into the 1.0 rating range. Let me know of any puzzles you can think of that use IOU as the hardest technique and I'll list them here. ............................................

OR, do a "Search" on IOU on these forums and you will find many, many walk-throughs which use it.

Cheers
Ed


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 Post subject: Re: IOU
PostPosted: Mon Apr 20, 2009 9:49 am 
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Here is a much harder IOU to find since it involves a hidden cage. This was originally posted in Jean-Christophe's Walkthrough here in uA97v2.


Image

2d. "45" on n8: four outies r78c37 - 18 = 1 innie r7c5. However, r8c36 is also a hidden 11(2) cage from "45" on r89 -> r7c37 - 7 = r7c5.
1a. Since 1 outie (either of r7c37) can see the one innie r7c5 -> the other outie (the other one of r7c37) cannot equal the in/out difference (7).
1b. in summary, r7c37 cannot have 7 (IOU on n8)

This locks the 7's in r7 into n8 and effectively cracks the puzzle. This is especially neat since IOU on n7 & n9 (again with IOD of 7) eliminated 7 from r7c12 and r7c89 which makes three super productive IOU moves!

Cheers
Ed


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 Post subject: Re: IOU
PostPosted: Mon Apr 27, 2009 7:40 pm 
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IOU's can also be used to eliminate combos rather than single candidates; in this case they eliminate a combo total although sometimes this can lead to a candidate elimination.

This example with a 2-cell combo 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

This IOU eliminates the total 17 from R9C12 because R1C2 and R3C3, which are both in N1, must be different.

This type of IOU combo total elimination may be most useful when the total being eliminated is at or near the upper or lower end of the possible cage totals.

In this case I also used a combo crossover clash, which is discussed here, to make a further combo elimination from R9C12 and therefore the key elimination from R9C2.

Thanks Ed for providing the diagram!


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 Post subject: Re: IOU
PostPosted: Sun May 17, 2009 6:25 am 
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The previous post by Andrew uses 4-cell IOU (3 innies and one outie) in order to eliminate one combination. From memory, I have only seen one occasion where 4-cell IOU on it's own gives an elimination. It comes from the same puzzle and is used by SudokuSolver as discussed here.

Image

4. "45" on c12: 3 innies r1c2 + r9c12 - 17 = 1 outie r3c3
4a. Now Andrew's nice 4-cell IOU move works in a simple way. r9c12 cannot equal the IOD of 17 since that would force r1c2 & r3c3 to be equal -> no 9 in r9c2 since r9c12 could only be [89] = 17.

Cheers
Ed


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 Post subject: Re: IOU
PostPosted: Fri Jun 05, 2009 11:01 pm 
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This example uses 4 outies and 1 innie. It comes from A123 "Roulette" here.

Image

6. 45 rule on C89 4 outies R1469C7 = 1 innie R5C8 + 23
6a. Min R5C8 = 2 -> min R1469C7 = 25
6b. Max R14C7 = 17 -> min R69C7 = 8, no 1 in R6C7, no 1,2 in R9C7
6c. All combinations for R1469C7 totalling at least 25 need at least two of 7,8,9 -> R14C7 = {789}
6d. IOU R149C7 cannot total 23 and no 7 in R9C7 -> max R149C7 = 22 -> R6C7 must be greater than R5C8 -> max R5C8 = 4, min R6C7 = 3

In this case I was able to use the fact that R149C7 cannot total the I-O difference 23 and also cannot total 24 to show that R6C7 must be greater than R5C8.

Thanks Ed for providing the diagram!

Because the cage pattern for A123 is symmetrical, there is another example which you should be able to find easily. The cage pattern is here.


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 Post subject: Re: IOU
PostPosted: Wed Nov 18, 2009 3:31 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. Afmob and I both used IOUs and manu used CCCs. In my walkthrough here I used three IOUs in step 15.

Image

15. 45 rule on R6789 2 outies R5C46 = 1 innie R6C5 + 5, IOU no 5 in R5C46, clean-up: no 6 in R6C46
15a. 45 rule on R6789 2 outies R5C45 = 1 innie R6C6 + 3, IOU no 3 in R5C45, clean-up: no 8 in R6C4, no 6 in R6C5
15b. 45 rule on R6789 2 outies R5C56 = 1 innie R6C4 + 3, IOU no 3 in R5C6, clean-up no 8 in R6C6

I've also posted manu's CCC steps here.

Thanks Ed for providing the diagram!


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 Post subject: Re: IOU
PostPosted: Wed Nov 18, 2009 4:05 am 
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Here is a very different looking example of an IOU, because it used the 45 rule on the diagonal of a Killer-X. It's from my walkthrough for A163 here.

Image

This is the position after step 18. My next step applied the 45 rule for D/ and I analysed the 3 innies R4C6 + R5C5 + R6C4 = 14. However I later found a much more interesting step.

45 rule on D/ 2 innies R5C5 + R6C4 = 1 outie R5C6 + 5, IOU no 5 in R6C4, using a hidden 2-cell cage R45C6 = 9, because 8 has already been placed at R4C7, for the I-O. This IOU is useful because it eliminates one of the combinations for the three innies on D/.

Thanks Ed for providing the diagram!


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 Post subject: Re: IOU
PostPosted: Thu Nov 19, 2009 2:45 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 an IOU used to make an important candidate elimination; I'll leave you to find it.

This is how Afmob did it for A175 V1:
45 rule on C12 2 outies R17C3 = 1 innie R2C2 + 8, IOU no 8 in R7C3

The alternative way using a CCC is here.


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 Post subject: Re: IOU
PostPosted: Mon Mar 15, 2010 2:59 pm 
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Hi folks,

Now that A188 has been and gone, I'd like to share a move with you here that was used by Ed in his walkthrough (here), and which was on my intended solving path when I posted the puzzle. As a basis, I'll take the grid state that would have been reached after Ed's step 8, but deferring his steps 4 and 5:

Image

At this point, we can retrospectively apply Ed's steps 4 and 5:

Ed wrote:
4. "45" on r89: 1 outie r7c1 + 2 = r8c9 = [24/35] = {[3]/[4]..}

5. "45" on r12: 2 innies r2c19 = 7 = [43/52/61] ([34] blocked by step 4.
5a. r2c1 = (456); r2c9 = (123)
5b. no 1 in r3c9

Here, Ed was making clever use of the fact that there were fixed relationships between the digits in R2C1 and R2C9, and between R7C1 and R8C9. But the interesting thing is that it would have been possible to delete the 4 from R2C9 even if this had not been the case! To see why, notice that (using innie/outie difference on R1289) the three innies (R2C19+R8C9, marked in green) sum to the single outie (R7C1, marked in yellow) + 9. We can therefore safely eliminate 4 from R2C9, because if R2C9 were 4, this would force R28C9 to the naked pair {45}, thus summing to nine, requiring R2C1 and R7C1 to be equal, which is impossible because they are peers of each other.

Note that we can make a formal solving technique (IOU4) for this based on the following requirements, which could also be put to good use in more complicated scenarios.

IOU4

  1. There are three innies and one outie (or vice-versa, of course).
  2. The three innies sum to the outie + D.
  3. One of the innies can "see" the outie.
  4. The other two ("satellite") innies can also see each other (but are not required to see the outie).
  5. One of the two satellite innies is a bivalue cell containing the candidates {xy}, where x + y = D.
  6. If the above conditions are met, we can eliminate {xy} from the other satellite innie.


Incidentally, you may also be interested to know my intended follow-up move to this, which I don't think anybody spotted:

Image

Note that the yellow cell, R2C6, can see all of the 4 dark green cells at R1C7+R2C789. Therefore, these 4 cells cannot contain all three of {123}, because that would leave no possibilities for R2C6. Consequently, one of {123} must go in R3C9, the only other possible place in N3. This constrains the 5(2) cage at R23C9 to {23}, which effectively cracks the puzzle.

_________________
Cheers,
Mike


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 Post subject: Re: IOU
PostPosted: Thu Nov 18, 2010 4:26 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
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As I've already commented in this thread, IOUs and CCCs can often be used to get the same result; thanks Ed for reminding me of that. I've just posted Assassin 196 in the CCC thread as good practice for finding CCCs. It's equally good practice for finding IOUs.

If you want some practice in findings IOUs, 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 IOUs you can find. My walkthrough contains 8 of them; look at all the steps where I've used a CCC, the same eliminations can be obtained using IOUs. With hindsight I think are probably more easily found as IOUs although I think the 45s for the CCCs are more obvious.


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