## Could someone walk me through this?I don't get fishies

Post the puzzle or solving technique that's causing you trouble and someone will help
I wrote:I just wonder if anybody has seen an actual situation where the type 3 solves the puzzle but type 4 doesn't?

Theres' nothing to worry about there guys. A type 4 can maybe destroy a type 3 but if there was a type 3 the type 4 reveals a naked subset that lets us do the elimination. In this case the type 4 elimination would reveal a naked triplet in r2c789 to make the same elimination as the type 3 UR would make.

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` *-----------------------------------------------------------*  | 16    3     5     | 2     8     7     | 146   69    149   |  | 18    2     9     | 4     3     6     | 578   78    58    |  | 7     4     68    | 1     9     5     | 68    3     2     |  |-------------------+-------------------+-------------------|  | 249   1     247   | 5     6     48    | 278   789   3     |  | 49    678   3467  | 39    2     48    | 15    6789  15    |  | 5     68    236   | 39    7     1     | 268   4     89    |  |-------------------+-------------------+-------------------|  | 3     9     48    | 6     1     2     | 48    5     7     |  | 24    67    2467  | 8     5     3     | 9     1     46    |  | 68    5     1     | 7     4     9     | 3     2     68    |  *-----------------------------------------------------------* `

So whoever told you to look for type 3 before type 4 didn't quite think it through.

RW
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RW wrote:
I wrote:I just wonder if anybody has seen an actual situation where the type 3 solves the puzzle but type 4 doesn't?

Theres' nothing to worry about there guys. A type 4 can maybe destroy a type 3 but if there was a type 3 the type 4 reveals a naked subset that lets us do the elimination.

My solver solves the following [edit: #461 of the top1465] when Type 3 precedes Type 4. With the techniques reversed, it gets stuck here:
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` .3.|...|... 87.|6.1|9.3 .4.|...|.2. ---+---+--- 2..|8.6|... ..9|...|1.. ...|15.|... ---+---+--- 4..|..2|..5 ...|..8|.1. 628|.1.|734 19    3     256   | 2457  89    47    | 4568  45678 1678 8     7     25    | 6     24    1     | 9     45    3 19    4     56    | 357   89    37    | 568   2     1678-------------------+-------------------+------------------ 2     15    13    | 8     37    6     | 45    4579  79 35    68    9     | 2347  2347  347   | 1     57    68 7     68    4     | 1     5     9     | 3     68    2-------------------+-------------------+------------------ 4     19    137   | 37    367   2     | 68    689   5 35    59    37    | 347   467   8     | 2     1     69 6     2     8     | 9     1     5     | 7     3     4`

I haven't figured out exactly why yet ... but suspect it has something to do with my "Type 4" including the new fan-dangled strong link stuff on the "Type 6 thread".

IOW re-inventing missing UR candidates has gotten more complicated. There was, er ... is ... a UR(37) in r68c13 and another UR(34) in r46c37 with earlier exclusions of r6c3<>7 and r4c3<>4, respectively.

[edit: added starting grid and last sentence.]
Last edited by ronk on Mon May 15, 2006 8:43 am, edited 3 times in total.
ronk
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ronk wrote:My solver solves the following when Type 3 precedes Type 4.

It would be easier to say what's wrong if you told us where you made the reduction. BTW there's a BUG-lite in r13c159 that solves r5c9=8.

RW
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RW wrote:It would be easier to say what's wrong if you told us where you made the reduction.
Sorry, I already needed to be out the door at the time, so I added the two UR reductions to the prior post. What do you think about starting a new thread on ... "How to replace (reinvent) missing UR candidates"?

RW wrote:BTW there's a BUG-lite in r13c159 that solves r5c9=8.
Thanks. My solver picks up the BUG-Lite, but it only handles Type 3 naked pairs right now.
ronk
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What do you think about starting a new thread on ... "How to replace (reinvent) missing UR candidates"?

I think so. On the Solver Forum. I have some questions on Type 3 UR's with hidden sets.

Keith
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ronk wrote:I added the two UR reductions to the prior post. What do you think about starting a new thread on ... "How to replace (reinvent) missing UR candidates"?

A new thread sounds like a good idea. I'd first like to figure out this example to have a good starting point. I can find the UR in r68c13 that actually eliminates both 3 and 7 from r6c3 => r6c3=4 => r4c3<>4. But I must admit that I don't have a clue where your solver would have found the type 3 eliminations in your first test. I guess I can't advance the puzzle to the right stage. Could you post the grid where your solver makes the type 3 elimination(s) that solves the puzzle.

RW
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RW wrote:Could you post the grid where your solver makes the type 3 elimination(s) that solves the puzzle.

As mentioned earlier, the eliminations were due to URs with strong links. The candidate grids at the time were:
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` 19    3     256   | 2457  89    47    | 4568  45678 1678 8     7     25    | 6     24    1     | 9     45    3 19    4     56    | 357   89    37    | 568   2     1678-------------------+-------------------+------------------ 2     15    134   | 8     37    6     | 345   4579  79 35    68    9     | 2347  2347  347   | 1     57    678*37    68   *347   | 1     5     9     | 34    68    2-------------------+-------------------+------------------ 4     19    137   | 37    367   2     | 68    689   5*357   59   *37    | 347   3467  8     | 2     1     69 6     2     8     | 9     1     5     | 7     3     4 For UR(37) in r68c13 with strong links, r6c3<>7 and r8c1<>7       a ab ------- abX | |a | abY        ab Excludes 'a' from both abX and abY cells`

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` 19    3     256   | 2457  89    47    | 4568  45678 1678 8     7     25    | 6     24    1     | 9     45    3 19    4     56    | 357   89    37    | 568   2     1678-------------------+-------------------+------------------ 2     15   *134   | 8     37    6     |*345   4579  79 35    68    9     | 2347  2347  347   | 1     57    68 7     68   *34    | 1     5     9     |*34    68    2-------------------+-------------------+------------------ 4     19    137   | 37    367   2     | 68    689   5 35    59    37    | 347   467   8     | 2     1     69 6     2     8     | 9     1     5     | 7     3     4 For UR(34) in r46c37 with strong links, r4c3<>4 and r4c7<>3 abX        abY |           | |a          |b |           | ab         ab  Excludes 'a' from abX cell and 'b' from abY cell`

The sticking point (repeat from earlier post) is:
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` 19    3     256   | 2457  89    47    | 4568  45678 1678  8     7     25    | 6     24    1     | 9     45    3  19    4     56    | 357   89    37    | 568   2     1678 -------------------+-------------------+------------------  2     15    13    | 8     37    6     | 45    4579  79  35    68    9     | 2347  2347  347   | 1     57    68  7     68    4     | 1     5     9     | 3     68    2 -------------------+-------------------+------------------  4     19    137   | 37    367   2     | 68    689   5  35    59    37    | 347   467   8     | 2     1     69  6     2     8     | 9     1     5     | 7     3     4`

I see where, if the UR(34) missing UR digits were replaced, type 3 exclusion r4c8<>5 could be detected. It wouldn't solve the puzzle (at least according to Simple Sudoku) ... but this is a type of scenario where I'm trying to figure out rules to replace (reinvent) missing UR digits.

Ditto for the UR(37) ... although I see no exclusion in that case.
ronk
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ronk wrote:
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` For UR(37) in r68c13 with strong links, r6c3<>7 and r8c1<>7        a  ab ------- abX  |  |a  |  abY        ab  Excludes 'a' from both abX and abY cells`

I don't know how to express this in a diagram, but there's another quite common feature in this UR also, if abX=b => abY=b (if r6c3=3 => r8c1=3), which eliminates b from abX and solves r6c3=4. But that's not related to the problem.

ronk wrote:My solver solves the following when Type 3 precedes Type 4.

I was interested in how your solver solved the puzzle when Type 3 precedes type 4. As you said, there is a Type 3 in r46c37, but it doesn't solve the puzzle. I wondered if there was another Type 3 elimination your solver found somewhere?

Replacing the candidates here would be pretty hard, because you already solved some corners, which actually makes it a AUS.

vidarino wrote:Type 3 (AUS)?:
There could also be a type 3, but I haven't found any in the wild yet. It should manifest itself as two unsolved corners + a combination of extra candidates, accompanied by a naked subset consisting of the same candidates. All subset candidates can be eliminated from their intersection.

RW
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RW wrote:Replacing the candidates here would be pretty hard, because you already solved some corners ...

With the a's and b's on diagonals -- only a step away from a deadly pattern -- I'm thinking the missing UR digits for this pattern ...
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` a      bX b      aY`

... are pretty obvious. The difficult one to me is ...
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` a      c bX    ab`

(Although it should go without saying, none of the solved cells can be givens.)
ronk
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### A thought

May I interject?

If the ab possibilities are not on a diagonal, they can be on one of four sides. They will be in the same line (7 shared buddies) and possibly in the same block (another 7 shared buddies, 6 of which are different from the buddies in line.)

If the ab possibilities are on one of only two diagonals, they only have 4 shared buddies.

So, you might expect that diagonal versions of Type 2 and Type 3 reductions are relatively infrequent, only 10-15% of the non-diagonal cases.

(Hey, it's only a thought experiment!)

Thanks, guys. This is a great discussion.

Keith
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### Multiple URs

When uniqueness is assumed, replacing candidates which have been eliminated sounds complicated enough to be both interesting and fraught with danger.

It seems to me that the practical approach to multiple URs, for a computer at least, is to make all the eliminations permitted by all the URs present at one fell swoop. Then reinvent nothing. This should make maximum use of the uniqueness assumption, which evidently tends to lose its power as more eliminations are made.

Steve
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ronk wrote:The difficult one to me is ...
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` a      c  bX    ab `

(Although it should go without saying, none of the solved cells can be givens.)

That one is indeed harder, I'm just trying to figure out if it would be possible to construct a situation where reinventing the candidates in such a case could help us. In this puzzle I can't see how it would help.

Steve R wrote:It seems to me that the practical approach to multiple URs, for a computer at least, is to make all the eliminations permitted by all the URs present at one fell swoop. Then reinvent nothing.

I would rather go for a "pool of lost souls" where I would save all candidates removed by uniqueness technique and single digit removal techniques (XY-wing, coloring and other techniques that can remove digits from the intersection of a row and a column). Then have my solver test for uniqueness reductions both with the lost candidates and without them. The AUS should be programmed separately.

RW
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