The New Sudoku Players' Forum

by **daj95376** » Sat Jan 31, 2015 11:59 am

Leren wrote:
Code: Select all
*--------------------------------------------------------------------------------* | 2 168 138 |B1678 B18 4 | 9 3678 5 | | 357 T1-568 138 | 125678 9 125678 | 68 3678 4 | | 57 4 9 | 5678 3 5678 | 2 1 T678 | |--------------------------+--------------------------+--------------------------| | 1 3 5 | 27 27 9 | 4 68 68 | | 8 2 7 | 456 45 56 | 1 9 3 | | 4 9 6 | 3 18 18 | 7 5 2 | |--------------------------+--------------------------+--------------------------| | 59 7 2 | 158 6 158 | 3 4 89 | | 359 158 138 | 24578 45 2578 | 568 2678 6789 | | 6 58 4 | 9 27 3 | 58 27 1 | *--------------------------------------------------------------------------------*

Exocet : r1c4 r1c5 r2c2 r3c9 (1678) r3c8==r2c2: => - 568 r2c2; stte

I am unable to follow your post justifying this exocet. It's my understanding that every digit in the base cells, when considered true using single-digit logic, must force at least one target cell true for that digit. Here is my analysis for "1":

Code: Select all: r1c4=1 -> r6c5=1, r7c6=1 -> X-Wing r28c23 +-----------------------------------+ | . -1 -1 | =1 -1 . | . . . | | . *1 *1 | -1 . -1 | . . . | | . . . | . . . | . 1 . | |-----------+-----------+-----------| | 1 . . | . . . | . . . | | . . . | . . . | 1 . . | | . . . | . =1 -1 | . . . | |-----------+-----------+-----------| | . . . | -1 . =1 | . . . | | . *1 *1 | . . . | . . . | | . . . | . . . | . . 1 | +-----------------------------------+ r1c5=1 -> r6c6=1, r7c4=1 -> X-Wing r28c23 +-----------------------------------+ | . -1 -1 | -1 =1 . | . . . | | . *1 *1 | -1 . -1 | . . . | | . . . | . . . | . 1 . | |-----------+-----------+-----------| | 1 . . | . . . | . . . | | . . . | . . . | 1 . . | | . . . | . -1 =1 | . . . | |-----------+-----------+-----------| | . . . | =1 . -1 | . . . | | . *1 *1 | . . . | . . . | | . . . | . . . | . . 1 | +-----------------------------------+

How do you get past the X-Wing to force r2c2=1?

_

by **Leren** » Sat Jan 31, 2015 8:10 pm

Hi Danny,

This is Champagne's basic definition of an Exocet. The underlining is by me.

The reduced definition uses

A base of 2 unassigned cells in the same region (row, column, box)
A target of 2 unassigned cells in other regions

having the following property (whatever is the process used to prove it.)

if for any digit solution of the base
one at least of the target is occupied by the same digit

then the target can not contain any other digit than the base.

The key words here are "whatever is the process used to prove it". As far I am concerned there is no inference that the process is limited to a single digit expansion.

I remember discussing this with Champagne some time ago and he was OK with it. What I do agree with is that the more complex the proof is, the less desirable it is, which I think Champagne has also said.

When you discussed your template analysis verifying target cells being linked to the base cells you didn't say whether you used single digit or multi digit analysis. From what you are saying here I assume it's a single digit template analysis.

I'm thinking of adding such a process to my solver, not because I''ll detect any more Exocets, but because I'll feel better, because in some cases I will have used a simpler process (than a multi digit expansion) to prove the Exocet.

There is definitely an Exocet in the puzzle, as the solution proves. Have I crossed a line of reasonability in detecting it? That's a matter of taste, but it doesn't contradict the wording of Champagne's definition.

Leren

by **daj95376** » Sat Jan 31, 2015 9:49 pm

Leren wrote:I'm thinking of adding such a process to my solver, not because I''ll detect any more Exocets, but because I'll feel better, because in some cases I will have used a simpler process (than a multi digit expansion) to prove the Exocet.

Yes, I admit that I use a single-digit pattern constraint for finding Exocet, JExocets, and QExocets. Explaining the process using an expanded approach is more than I wish to address.

If you are using a multi-digit expansion to prove your Exocet, then it must be interesting because a full template multi-digit analysis for <1678> still doesn't find eliminations to support your conclusion of r2c2=1.

Code: Select all: c169 Swordfish (232) -7 r238c4,r28c8 -or- r149 Swordfish (222) -7 r238c4,r28c8 <1678> accepted = 28 template combinations (after Swordfish on "7") <1678> -1 r1c23,r2c46 <1678> -6 r1c48,r2c246,r3c9 (forces r1c2=6) <1678> -8 r1c2,r2c46,r8c68 <1678> -5 r7c4 <1678> r1c245,r2c7,r3c9,r4c89,r6c56,r7c4 locked for candidates (i.e., only candidates allowed) after full template multi-digit analysis for <1678> *-----------------------------------------------------------* | 2 6 38 | 178 18 4 | 9 378 5 | | 357 158 138 | 25 9 257 | 68 368 4 | | 57 4 9 | 568 3 5678 | 2 1 78 | |-------------------+-------------------+-------------------| | 1 3 5 | 27 27 9 | 4 68 68 | | 8 2 7 | 456 45 56 | 1 9 3 | | 4 9 6 | 3 18 18 | 7 5 2 | |-------------------+-------------------+-------------------| | 59 7 2 | 18 6 158 | 3 4 89 | | 359 158 138 | 2458 45 257 | 568 26 6789 | | 6 58 4 | 9 27 3 | 58 27 1 | *-----------------------------------------------------------*

r2c2 is not "locked" for <1678>, and the only elimination for r2c2 is "-6". This leaves r2c2=158 after a full template multi-digit analysis.

ADDENDUM

Consider the following Exocet Single-Target Pattern: Base = r1c45, Target = r3c9

Code: Select all: 6r1c4 - r3c46 = 6r3c9 8r1c45 - r3c46 = 8r3c9 7r1c4 - r23c6 = r8c6 - r8c9 = 7r3c9

We can now conclude that r1c45=1 must be true. Thus, justification for the eliminations for "1" shown in my multi-digit results.

_

by **eleven** » Sun Feb 01, 2015 12:39 am

Leren,

i agree with Danny.
If you denote something as an Exocet, where the property, that a candidate of the base cells must occur in a target cell too, is not derived fom a standard JExocet pattern, you have to write it out.
Otherwise we cannot verify it or evaluate, how complicated this solution is.

by **Leren** » Sun Feb 01, 2015 7:55 am

Here is my analysis for the justification of 1 as a valid Exocet digit:

We start here:

Code: Select all: *--------------------------------------------------------------------------------* | 2 168 138 |B1678 B18 4 | 9 3678 5 | | 357 T1568 138 | 125678 9 125678 | 68 3678 4 | | 57 4 9 | 5678 3 5678 | 2 1 T678 | |--------------------------+--------------------------+--------------------------| | 1 3 5 | 27 27 9 | 4 68 68 | | 8 2 7 | 456 45 56 | 1 9 3 | | 4 9 6 | 3 18 18 | 7 5 2 | |--------------------------+--------------------------+--------------------------| | 59 7 2 | 158 6 158 | 3 4 89 | | 359 158 138 | 24578 45 2578 | 568 2678 6789 | | 6 58 4 | 9 27 3 | 58 27 1 | *--------------------------------------------------------------------------------*

1. Set r1c4 = 1 and remove 1 from r2c2. Apply the following eliminations:

Hidden Text: Show

We end up here:

Code: Select all: *--------------------------------------------------------------------------------* | 2 6 3 | 1 8 4 | 9 7 5 | | 357 8 1 | 256 9 2567 | 6 36 4 | | 57 4 9 | 56 3 567 | 2 1 8 | |--------------------------+--------------------------+--------------------------| | 1 3 5 | 7 2 9 | 4 8 6 | | 8 2 7 | 456 45 56 | 1 9 3 | | 4 9 6 | 3 1 8 | 7 5 2 | |--------------------------+--------------------------+--------------------------| | 5 7 2 | 8 6 1 | 3 4 9 | | 9 1 8 | 245 45 25 | 56 26 7 | | 6 5 4 | 9 7 3 | 8 2 1 | *--------------------------------------------------------------------------------*

Contradiction - No 3 in Row 8.

2. Set r1c5 = 1 and remove 1 from r2c2. Apply the following eliminations:

Hidden Text: Show

We end up here:

Code: Select all: *--------------------------------------------------------------------------------* | 2 6 8 | 7 1 4 | 9 3 5 | | 3 5 1 | 268 9 26 | 68 68 4 | | 7 4 9 | 568 3 568 | 2 1 678 | |--------------------------+--------------------------+--------------------------| | 1 3 5 | 2 27 9 | 4 68 68 | | 8 2 7 | 456 45 56 | 1 9 3 | | 4 9 6 | 3 8 1 | 7 5 2 | |--------------------------+--------------------------+--------------------------| | 59 7 2 | 1 6 58 | 3 4 89 | | 359 1 3 | 2458 45 2578 | 568 268 6789 | | 6 8 4 | 9 27 3 | 58 27 1 | *--------------------------------------------------------------------------------*

Contradiction: No 7 in Row 2.

Conclusion : If r1c4 = 1 or r1c5 = 1 then r2c2 = 1.

Leren

by **eleven** » Sun Feb 01, 2015 12:18 pm

Ok, so you needed 2 (singles) contradiction networks to prove the exocet, which you had not mentioned before.

For the first one you don't need the assumption r2c2<>1 to get a contradiction for 1r1c4:
1r1c4, 8r1c5, 3r1c3, 7r1c8, 7r8c9, 9r8c1 => no 3 in c1.
This alone is a much harder step than needed to solve the puzzle.
([Added:] E.g. better as the elimination of only 1r1c4 above you have:
3r1c8=r1c3-r2c1=(3-9)r8c1=(9-7)r8c9=7r3c9 => -7r1c8 -> 7r1c4, stte)

by **daj95376** » Sun Feb 01, 2015 7:17 pm

Leren,

Thanks for the detailed analysis. I was set back a bit by the extent of your use of expansion.

_

by **Leren** » Mon Feb 02, 2015 12:20 am

Hi Danny, as far as I know the term expansion just means following a chain of logic to reach a conclusion based on some assumption (such as a digit being true in the base).

Champagne uses the term in this sense eg see his post here http://forum.enjoysudoku.com/post239361.html?hilit=expansion#p239361 I'm not quite sure who coined the term.

Leren

by **daj95376** » Mon Feb 02, 2015 8:59 pm

Leren wrote:Hi Danny, as far as I know the term expansion just means following a chain of logic to reach a conclusion based on some assumption (such as a digit being true in the base).

Champagne uses the term in this sense eg see his post here http://forum.enjoysudoku.com/post239361.html?hilit=expansion#p239361 I'm not quite sure who coined the term.

Hmmm!!!

I remember being confused by champagne's post (that you reference). He lists most JE patterns being found using "level 4" logic. IIRC, my QExocet solver found all of his JE patterns (and a few additional patterns) using single-digit logic, which he lists as "level 1-3".

Note: champagne seems to stop solving a puzzle after Singles. I don't stop until after Basics. This accounted for most of the additional patterns that my solver found, but there were seven additional patterns that were found in puzzles after only Singles.

_

by **Leren** » Tue Feb 03, 2015 3:15 am

jaj95376 wrote : Hmmm!!!

Double Hmmm!!! OK what about this post by champagnehttp://forum.enjoysudoku.com/post239324.html?hilit=expansion#p239324 where he says, inter alia,

The second one requires the extended mode, so the digit '3' has a full expansion using all rules (in fact not more than pairs and Xwings).

I think what he's saying here is that for the second Exocet referenced he assumes 3 is in the base and uses a "full expansion" ie eliminations using "all rules" etc - to deduce that 3 must be in at least 1 target cell.

Have you got any suggestions other than expansion to describe this process ?

Leren

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