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 Post subject: IOE
PostPosted: Mon Mar 09, 2009 7:00 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
This move was first explained here.

IOE

When the Outies "see" all of a candidate for a house except for the Innies, the 3rd cell must Equal the in/out difference (IOD).
i. For example, "45" n3: 1 outie r4c9 + 5 = 2 innies r3c78
ii. the 1 outie (r4c9) sees all of 4, 8 & 9 for n3 (they can't be in the 8(2)n3), except for the innies r3c78.
iii. so, when 8 is in r4c9, 8 must be in the innies, r3c78. Since 1 innie = 1 outie -> the 3rd cell (in this case, one of r3c78) must equal the IOD for n3 = 5
iv. same thing for 9 since it sees all 9s in n3 except r3c78. So, 9 in r4c9 -> r3c78 = {59}(IOE)
v. of course, this doesn't work for the 5 in r4c9 because it doesn't see all 5s in n3 apart from the innies. One 5 can hide from r4c9 in r1c8.

9. So, "45" n3: 1 outie r4c9 + 5 = 2 innies r3c78
9a. r4c9 = (589). If not 5, it must be 8 or 9 -> 5 locked in r3c78 (IOE).
9b. -> 5 locked in innies of n3 OR r4c9
9c. -> no 5 in common peers in r4c8, r1c7, r2c789, r3c9 & r1c9.

Image

NOTE: IOE is not enough on it's own to make eliminations in this case. Only if r4c9 = (89) would we have been able to lock 5 in r3c78 for n3 & r3 using IOE only.
9e. no 8 in r4c9 (h13(2))

Thanks to Afmob for suggesting putting this in the techniques forum [edit: and to Andrew for some clarifications.].

Cheers
Ed


Last edited by Ed on Tue Apr 14, 2009 5:32 am, edited 2 times in total.

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 Post subject:
PostPosted: Thu Apr 09, 2009 6:25 pm 
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Grand Master
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Joined: Mon Apr 21, 2008 9:44 am
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Location: MV, Germany
I want to present you a second example of IOE. This time there are 2 Outies and only one Innie.

Image

Let's consider the IOD of N3: 5 = R45C7 - R3C9:

R45C7 (Outies) sees all 4,8 of N3 except for R3C9 (Innie), so when one of R45C7 is 4 or 8, then R3C9 = 4 or 8 since it's the only place where they are possible in N3. So in this case R3C9 = R4C7 or R3C9 = R5C7. If we put this in the equation above then either R4C7 or R5C7 must be 5.

Now in this particular example we have R4C7 = (458), so either R4C7 = R3C9 if R4C7 is one of (48) and R5C7 = 5 or R4C7 = 5. So 5 is locked in R45C7 for C7, N6 and 18(4).

If you want to try it out, here is the PS code of my example: 3x3::k:0:1:2:3:4:5:3078:2055:2055:1801:1801:1801:12:13:14:3078:3088:3088:18:19:20:21:22:23:4632:4632:26:27:28:29:30:31:32:4632:34:35:36:37:38:39:40:41:4632:43:44:45:46:47:48:49:50:51:52:53:54:55:56:57:58:59:60:61:62:63:64:65:66:67:68:69:70:71:72:73:74:75:76:77:78:79:80:
Don't forget to set R4C7 = (458)!

By the way, you can also deduce that R3C9 <> 5 using IOU @ C7 and the reasoning above.


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 Post subject: Re: IOE
PostPosted: Tue Apr 14, 2009 10:10 pm 
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Here is an example of IOE alone making eliminations so is simpler than the composite moves in the last two posts. Originally from here

Image

10. "45" on n8: 1 outie r7c3 + 6 = 2 innies r9c45 (adding in the h8(2) in n8) (note that the in/out Difference (IOD) is 6)
10a. r7c3 = (6789). For each of these, when the candidate is in r7c3, the only place for that digit in n8 is in the innies r9c45 -> the IOD (6) is locked in r9c45 (IOE)
10b. 6 locked in r9c45 for r9 & n8
10c. no 1 in r6c5

11. 21(3)n7 = {678}: 7 & 8 locked for r9

I gave this move a 1.50 rating since it uses a hidden cage to get the correct "45" and involves 4 candidates in the IOE. Not easy to spot. Presumably, simpler ones will come along that may feel like a lower rating.

Thanks to Andrew for suggesting this post go here.

Cheers
Ed


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 Post subject: Re: IOE
PostPosted: Tue May 26, 2009 10:22 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
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Location: Sydney, Australia
Here's a classic example of IOE which works from the unique (and very creative) cage pattern for Assassin A153. This example comes from here

Image

1. "45" r89: 4 outies r6c5 + r7c456 = 14
1a. min. r7c456 = 6 -> max. r6c5 = 8

2. "45" n8: r6c5 + 7 = r9c46
2a. since the outie r6c5 sees all cells in n8 except the two innies -> the IOD (7) is locked in r9c46 (IOE)
2b. 7 locked in r9c46 for r9 & n8
2c. no 7 in r6c5 (since r9c46 cannot be [77]
2d. no 9 in r9c46 since the other one of r9c46 = r6c5 (IOE)
2e. no 1 in 8(2)n8

In addition to locking the IOD in the innies (step 2a), two extra dimensions to this technique are shown here. First, the IOD cannot be in the one outie (since the innies cannot be [77] step 2c) and second, there is an implied clone cell since the 1 outie must equal one of the innies (step 2d).

The cage pattern of this puzzle allows this technique to be used again elsewhere. Fun!

NOTE: an alternative way to make these eliminations is using combined cage and combination work - see Andrew's post here.

Cheers
Ed


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