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Jigsaw 43
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nj3h
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PostPosted: Mon Apr 23, 2007 6:34 pm    Post subject: Jigsaw 43 Reply with quote

All,
So many have helped me on the jigsaw topic in the last week or so. I appreciate your efforts. I am having trouble understanding which cells to use to the left or right of the line (if a vertical line). Take the subject puzzle for instance. Near the end of the puzzle there is a hint about LoL on columns 1 and 2. I see many cells from partial blocks to the right and to the left of a vertical line between columns 2 and 3. Great! Now how do I identify which cells compliment each other?

It must really be time to retire so I can spend more time figuring these techniques out. LOL (not law of leftovers that time).

Thanks again for the help.

Regards,
George
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Para
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PostPosted: Mon Apr 23, 2007 9:41 pm    Post subject: Reply with quote

Hi George

You can use this LoL move, but again it is a difficult one. And there is an easier LoL move. Check the LoL on Columns 123 and C789. It makes the easier eliminations.
My general advise is, keep your number of innies and outies as low as possible. With the LoL on C123 you have minimal 3 innies. With LoL on C12 the minimal number of innies and outies is 6.

So for the LoL on column 12. I would suggest this. Try to minimize the number of innies and outies. Nonets 1 and 7 both have 6 cells in C12, nonet 5 has 5 cells and nonet 9 has 1 cell.
Combining Nonets 1 and 7 we have the minimal of 6 innies left in C12.

So we start of with Nonets 1 and 7.
We have 6 outies: from Nonet 1: R123C3 and from Nonet 7: R6C34 + R7C3.
And we have 6 innies: from Nonet 5: R45C12 + R6C1 and from Nonet 9: R9C2.
The elimination:
We know that innie R6C1 = 9. So the outies must contain a 9 too. This is only possible in R2C3 and R7C3. So we can eliminate 9 from all cells that see both R2C3 and R7C3. So R8C3 can't contain a 9.

For SumoCue hints (Ruud can correct me if i am wrong):
I think SumoCue checks all possible LoL eliminations in this order.
1) LoL eliminations using 2 rows/columns
2) LoL eliminations using 3 rows/columns
3) LoL eliminations using 4 rows/columns

It doesn't use the easiest eliminations possible.

greetings

Para

ps. Look at Nonet 3. It has a great naked triple that makes eliminations outside it's nonet.
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Glyn
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PostPosted: Mon Apr 23, 2007 9:52 pm    Post subject: Reply with quote

I think Para has just covered this but here's my comments for what they are worth.

Using the following Nonet numbers

Code:

111222223
111222233
111444333
555543336
555546666
577746666
777444888
779999888
799999888


The state of the grid at the first LOL on C12 is
Code:

.-------.-------.-------.-------.-------.-------.-------.-------.-------.
| 148   | 189   | 2     | 3     | 149   | 7     | 5     | 189   | 6     |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 13478 | 13789 | 13789 | 2     | 149   | 6     | 189   | 158   | 58    |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 6     | 5     | 18    | 9     | 3     | 4     | 18    | 2     | 7     |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 5     | 2     | 6     | 7     | 8     | 9     | 3     | 4     | 1     |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 138   | 138   | 138   | 4     | 2     | 5     | 67    | 67    | 9     |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 9     | 6     | 5     | 1     | 7     | 8     | 4     | 3     | 2     |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 2     | 4     | 789   | 5     | 6     | 1     | 789   | 789   | 3     |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 37    | 379   | 179   | 8     | 159   | 2     | 1679  | 15679 | 4     |
:-------+-------+-------+-------+-------+-------+-------+-------+-------:
| 78    | 179   | 4     | 6     | 159   | 3     | 2     | 15789 | 58    |
'-------'-------'-------'-------'-------'-------'-------'-------'-------'


The first hint Law of Leftovers in Columns 1-2 eliminates the 9 in R8C3. The second eliminates the 9 in R2C5.

I have another way of looking producing these elimination which you may find useful.

Candidate 9 for Nonets 1 and 7 taken together must come from Columns 2 and 3. No other nonet can contain candidate 9 in those columns.
Therefore for Nonet 9 R8C3<>9 and R9C2<>9.

For Nonet 9 candidate 9 is now restricted to Column 5 implying R1C5 and R2C5 <> 9.

Hope this helps.

All the best

Glyn
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nj3h
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PostPosted: Wed Apr 25, 2007 1:14 am    Post subject: Reply with quote

Para and Glyn,

Thanks for the explanations. I may be getting closer. My one last question (for now anyway) is to understand why the innies and outies have to balance out. I have reviewed the graphical examples posted by Para, especially the one in "Jigsaw 32". I recognize that column 1 + column 2 must have two of each number, but I do not see the logic why the innies equal the outies in the C!@ example.

Maybe someday this will click.

Regards,
George
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Jean-Christophe
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PostPosted: Wed Apr 25, 2007 7:05 am    Post subject: Reply with quote

Do you remember the Set Theory you learned at school ? Cool
Here is a diagram which may help



This is the simplistic case of a single row & nonet with one cell in & out.
Both R1 & N1 must include all 9 different elements {A...I}, no more no less.
They share 8 out of these 9 elements at their intersection in green : {B...I} Thus the element {A} is the complement.

Same reasoning for the case of a single row (or column) & nonet with several cells in & out. The complement is a set of elements: a pair, triplet...

For several rows & nonets, extend this reasoning to multisets.
Here are the top 3 rows of Jigsaw # 43


Both R123 & N123 must include three times all 9 elements.
They share all their elements, except for 3 cells : {ABC}

Hope this helps
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nj3h
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PostPosted: Thu Apr 26, 2007 5:07 pm    Post subject: Reply with quote

Jean-Christophe,

Thank you for your excellent reply. I understand the examples you have provided. Your second example has just one portion in n3 (blue ABC) as an outie and one portion of another nonet (yellow ABC) as an innie

My confusion is the fact that using the LoL for c12 is more complex, since it has many innies and outies.

In the C12 LoL, cells r45c12 + r6c1 + r9c2 = r12367c3 + r6c4 according to previous information posted. I am not able to relate this particular situation to your second example.

As an illustration of my confusion, please consider: r123c3. One cell is a 2, the other two cells are unknown 13789 and 18. Why must the value of these cells occur in one of the innies (r45c12 + r6c1 + r9c2) and not in r6c2 + r78c12 + r9c1? That is the concept I am missing.

Regards,
George
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Jean-Christophe
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PostPosted: Thu Apr 26, 2007 6:54 pm    Post subject: Reply with quote

I did use LoL only on R123 (very effective), R789 (effective too), C123 & C789. That's all. I used other techniques to solve the rest. In particular at this stage, R2C89+R3C7 forms a naked triplet on {158} within N3 -> R2C7 <> 8 = 9 which unlocks the puzzle.



I usually do not consider LoL larger than 2 x 4 cells. IMHO using LoL on C12 for this puzzle is pushing the technique too far. Anyhow the principle is the same but more complex since it involves 2 x 6 cells. The green cells are in common. The yellow cells must hold the same digits as the blue cells. In this particular case one has to pay much attention, because a digit could be duplicated in both sets. For example all cells with X could hold the same digit.




Hope this helps
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Para
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PostPosted: Thu Apr 26, 2007 7:13 pm    Post subject: Reply with quote

Hi George


It all depends on which 2 nonets you compare to columns 12.



When you compare Nonets 1 and 5 to columns 12, you get this situation. The light yellow cells are the outies and the green cells are the innies.

So: outies = innies
R12345C3 + R45C4 = R789C1 + R6789C2



When you compare Nonets 1 and 7 to columns 12, you get this situation.

So: outies= innies
R12367C3 + R6C4 = R456C1 + R349C2



When you compare Nonets 5 and 7 to columns 12, you get this situation.

So: outies = innies
R4567C3 + R456C4 = R123C1 + R1239C2

I understand you have trouble with this concept. So i suggest you get acquainted with the easier Law of leftover moves first. Because honestly i never consider these kinds of LoL moves until the easier ones don't give me any eliminations.
For this puzzle, you should look at the Law of Leftover move for Column 123. This is an easier elimination (and also more effective). This one also matches JC's example better(only 3 innies and outies). When you get used to making these LoL moves, it is probably easier to understand the more complicated LoL moves.

greetings

Para
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nj3h
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PostPosted: Thu Apr 26, 2007 8:29 pm    Post subject: Reply with quote

JC and Para,

I really appreciate you trying to help me. I understand that the C12 LoL is a complex implmentation of the rule. I will study in more detail the answers both of you provided this evening.

However, let me quote JC:

=====
"The green cells are in common. The yellow cells must hold the same digits as the blue cells."
=====

This statement is the crux of my confusion. What makes the green cells to be "in common"? Why aren't some other cells "in common"?

If I can understand this, then the blue = yellow follows from all of the other examples provided.

I am still thinking why the 4 "X" cells could be the same.

That is the logical part of this that I just do not see.

Regards,
George
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Jean-Christophe
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PostPosted: Thu Apr 26, 2007 9:36 pm    Post subject: Reply with quote

nj3h wrote:
However, let me quote JC:

=====
"The green cells are in common. The yellow cells must hold the same digits as the blue cells."
=====

This statement is the crux of my confusion. What makes the green cells to be "in common"? Why aren't some other cells "in common"?

If I can understand this, then the blue = yellow follows from all of the other examples provided.


Study the easier examples first.

I'll try to explain it another way...
See Glyn's post for Nonet numbers.

Notice the transparent coloring I used.
In the example with R123, the rows are tinted in yellow
N123 are tinted in blue
yellow & blue = green
Notice : 3 rows and 3 nonets.

For the more complex example with C12
C12 are tinted in yellow
I did choose N17 which are tinted in blue
yellow & blue = green

Para shows other examples choosing different nonets: N15, N17 & N57

Notice: 2 columns and 2 nonets

nj3h wrote:
I am still thinking why the 4 "X" cells could be the same.

That is the logical part of this that I just do not see.

Because of the geometry of the nonets.

I know it's difficult at first and I really suggest you get familiar with easier LoLs.
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Glyn
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PostPosted: Fri Apr 27, 2007 12:51 am    Post subject: Reply with quote

I don't know whether this will help as Para and JC are giving a great set of notes on this.

Law of Leftovers (Cols 1-2)
===========================

If you look at the red line on the Paras' earlier postings you will see that it cuts across 4 nonets. (N1,N5,N7 and N9). Now 4 nonets must contain 4 sets of the digits (1to9).
We know that we must have 2 sets of the digits (1-9) to the left of the line, they are the contents of Columns 1 and 2.
So taken together the parts of the 4 nonets to the right of the line must contain the other 2 sets of digits (1to9).

If we break the set of digits in each Nonet N(I) into two parts lets call them NL(I) and NR(I) (where L and R mean to the left or right of the line).

Now for each Nonet N(I) NL(I)+NR(I) = 1 set of digits (1to9) (For completion of a Nonet).

so we have NR(I) = 1 set of digits (1to 9) - NL(I) (Eqn 1.)


In this case we also have NL(1) + NL(5) + NL(7) + NL(9) = 2 sets of digits (1to9) (For completion of the Columns). (Eqn 2.)

The first of Paras' cell summations used Nonets 1 and 5 to the right of the line and Nonets 7 and 9 to the left of the line.

Now from Eqn 1.

NR(1) + NR(5) = 2sets of digits(1to9) - NL(1) - NL(5) (Here we are splitting the Nonets)

and from Eqn 2.

NL(7) + NL(9) = 2sets of digits(1to9) - NL(1) - NL(5) (Here we are splitting the Columns)

But the right hand sides of the equations are identical so

NR(1) + NR(5) = NL(7) + NL(9) and these are the cells Para enumerated.

Law of Leftovers (Cols 1-3)
===========================

If we move the red line one column to the right then the line no longer cuts Nonet 1.

The equivalent of equation 2 is now NL(5) + NL(7) + NL(9) = 2 sets of digits(1to9) (For completion of the 3 columns - Nonet1)

Para and JC suggested using Nonet 9 to the left and Nonets 5 and 7 to the right as one example.

NR(5) + NR(7) = 2 sets of digits(1to9) - NL(5) - NL(7) (Splitting the Nonets)

We also have from above

NL(9) = 2 sets of digits(1to9) - NL(5) - NL(7) (Splitting the columns)

Again the right hand sides are identical so

NR(5) + NR(7) = NL(9)

The advantage they pointed out is that the sets are much smaller for human solvers to deal with. (Of course a computer has no problem with that at all).

Hope this helps

All the best

Glyn
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Pete
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PostPosted: Fri Apr 27, 2007 2:05 am    Post subject: Reply with quote

Major Major Major

Now I see where the catch 22 came from.

or am I putting the cart before the horse?
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Jean-Christophe
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PostPosted: Fri Apr 27, 2007 1:13 pm    Post subject: Reply with quote

You may also have a look that this page I wrote:
http://jcbonsai.free.fr/sudoku/?p=107
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nj3h
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PostPosted: Sat Apr 28, 2007 10:04 pm    Post subject: Reply with quote

Para, JC, Glyn, and Ruud:

I want to thank each and every one of you for the time you have taken to help me try to understand the LoL technique. Spending some quiet time this afternoon, I was able to grasp the reason behind the C12 LoL. I now understand the idea of the common cells and the two sets that are in play and have to equal each other. The equations really helped me. Once I understood them, then the other comments, drawings, and help in the previous messages made things clear.

I will try not to ask questions for the next fortnight.

Regards,
George
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Ruud
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PostPosted: Sun Apr 29, 2007 10:22 pm    Post subject: Reply with quote

George,

regarding your PM on this particular LoL move, I made a picture that explains the move more clearly than I could do in 1K words.



Legend:

Yellow: cells common to columns 89 and jigsaw pieces 68.
Green: 7 surplus cells in jigsaw pieces 68.
Blue: 7 surplus cells in columns 89.

LoL: In the solution, the blue and the green area must contain the same numbers.

The green area contains digit 8, so the blue area must also contain digit 8. There are 3 candidates. Each of them eliminates 8 from R2C7, as shown by the little red arrows.

This is about as "Jip & Janneke" as I can possibly explain it. (don't ask Rolling Eyes)

cheers,
Ruud
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