The New Sudoku Players' Forum

[daj95376]

If you followed that you should see that this same pattern has another outbreak at this same position.

Code: Select all

*-----------------------*
| .*5 . |*5*5 . | . 5 . |
| . . 5 | 5-5 . | . . . |
| . 5 5 | . . . | . 5 . |
|-------+-------+-------|
| . . . | . . . | 5 . . |
| . . 5 | 5 . . | . . . |
| .*5 . | .*5 . | . . . |
|-------+-------+-------|
| 5 . . | . . . | . . . |
| . . . | . . 5 | . . . |
| . . . | . . . | . . 5 |
*-----------------------*

Sorry, but I didn't follow your explanation. It also doesn't help that the above example is incorrect because [r1c8] would also be a fin cell ... and it doesn't see the elimination [r2c5]<>5.

[wintder]

Yep, I pooched.

Sorry that you didn't follow the first step.

It was a mess also, cleaned up now, I hope.

[Pat]

i beg to differ

in my view,
Either (r5c4=5) or (r5c4<>5 => r5c3=5 => r2c3<>5 => r2c45=5)
is poor notation

the expression in the 2nd parentheses is itself always true

the combined expression "either A or B" is useless

I agree that the 2nd patentheses is always true. In my attempt to justify [r5c3]=5, I introduced too much information.

Is the following acceptable to you?

Either (r5c4=5) or (r5c3=5 => r2c3<>5 => r2c45=5)

-or- the more formal

(5) [r5c4]=[r5c3]-[r2c3]=[r2c45]; => [r1c4]<>5

with its implied either.

hi daj95376


1. no,
Either (r5c4=5) or (r5c3=5 => r2c3<>5 => r2c45=5)
is just as bad as the earlier form.
the proper notation would go like this --

( r5c4=5 | r5c3=5 ) → ( r5c4=5 | r2c3≠5 ) → ( r5c4=5 | r2c45 contains 5 ) → ( r1c4≠5 | r1c4≠5 ) → r1c4≠5

-- very long and cumbersome.


2. as for
(5) [r5c4]=[r5c3]-[r2c3]=[r2c45]; => [r1c4]<>5
-- i wouldn't know -- haven't studied the definitions used in that notation -- but i trust you do know those definitions, so it's probably fine.

[ronk]

the proper notation would go like this --

( r5c4=5 | r5c3=5 ) → ( r5c4=5 | r2c3≠5 ) → ( r5c4=5 | r2c45 contains 5 ) → ( r1c4≠5 | r1c4≠5 ) → r1c4≠5

-- very long and cumbersome.

I'm appreciating NL and AIC notation now more than ever before.

[Pat]

In order to use any technique properly, you must understand WHY the elimination can be made. Just trying to memorize patterns without understanding the logic behind them will lead to confusion.


Here's how the 5s are located in the grid when the elimination occurs:

Code: Select all

*-----------------------*
| . 5 . |-5 5 . | . 5 . |
| . .*5 |*5#5 . | . . . |
| . 5 5 | . . . | . 5 . |
|-------+-------+-------|
| . . . | . . . | 5 . . |
| . .*5 |*5 . . | . . . |
| . 5 . | . 5 . | . . . |
|-------+-------+-------|
| 5 . . | . . . | . . . |
| . . . | . . 5 | . . . |
| . . . | . . . | . . 5 |
*-----------------------*



finned X-wing in r25c34, fin r2c5

In row 5 digit 5 must be either in column 3 or column 4.
In row 2 the 5 must be either in column 3, 4 or 5, r2c4 and r2c5 are both in box 2.
So what are the options here?
Either r5c4=5 or r5c3=5 => r2c3<>5 => r2c45=5.
So either r5c4=5 or r2c45=5.
Any cell that can see all those cells must not be 5
=> r1c4<>5

i probably wouldn't use this explanation --

( r5c4=5 | r5c3=5 ) → ( r5c4=5 | r2c3≠5 ) → ( r5c4=5 | r2c45 contains 5 ) → ( r1c4≠5 | r1c4≠5 ) → r1c4≠5

-- i'd prefer a contradiction-loop, e.g. --

r1c4=5 → r2 5 in c3 → r5 5 in c4 ( conflict in c4 )

[RW]

Wow, this has become a very long discussion on notation... I believe the problem with my explanation was not in the notation, but in the spacing. If I've interpreted Jasper's post correctly, the misunderstanding was that he read "(Either r5c4=5 or r5c3=5) =>r2c3 <>5=>r2c45=5", when it should have been, as daj pointed out, "Either (r5c4=5) or (r5c3=5 =>r2c3 <>5=>r2c45=5)". One final attempt at giving a clear explanation:

In row 5, digit 5 must be in either r5c3 or r5c4.

if r5c4=5 => r1c4<>5
if r5c3=5 => r2c3<>5 => r2c45=5 => r1c4<>5

In either case, r1c4<>5. This better?

btw. Jasper, did you understand the nice loop notation I used for the chain in my first reply to this thread? Just asking this as a survey, would be good to know how helpful it is to use NL notation when replying to a first time poster.

RW

[daj95376]

In row 5, digit 5 must be in either r5c3 or r5c4.

if r5c4=5 => r1c4<>5
if r5c3=5 => r2c3<>5 => r2c45=5 => r1c4<>5

In either case, r1c4<>5. This better?

Much better in my opinion. It's not about notation so much as it's about showing who does what to whom

[Jasper32]

Hi Rw,
Yes, I understood your chain in your first post. I did comment I wouldn't have found it without your pointing it out. The notation was easy enough to follow through. I understood your ending in this post I am responding to in this email. I was amazed at the differences in notations that were posted. It seems while there are these differences, everybody seems to understand their notations and in the long run that is what counts.

BTW, I found the easiest method method was to use the "Pattern Elimination". but everybody's response was much appreciated. I want to thank you, 999_springs, Daj 95376, Pat, Wintder and Ronk for there input. A special thanks to Cec for his private email to me informing me on the procedure to post quotes in my posts.

Regards to all.

[Pat]

In row 5, digit 5 must be in either r5c3 or r5c4.

• r5c4=5 => r1c4<>5
  
• r5c3=5 => r2c3<>5 => r2c45=5 => r1c4<>5

In either case, r1c4<>5.

that's certainly very clear.


and this is a different view of the logic,
in reasonably plain English i hope --

putting a 5 in r1c4
would force the digit 5 for the two rows r25
both into a single column (c3)

[Jasper32]

Thanks Pat....That I can understand!!!! It isn't as concise as Eureka but clarity takes preference over being concise..... at least for me. It seems almost every body who wrote nad a little different twist on the notation. Really doesn't matter as long as you know what you are doing BUT it can be difficult for a newcomer to understand.

Thanks again

[Sudtyro]

BTW, I found the easiest method method was to use the "Pattern Elimination"...

Code: Select all

+-----------------------------------+
|  .  5  .  | -5  5  .  |  .  5  .  |
|  .  .  5  |  5  5  .  |  .  .  .  |
|  .  5  5  |  .  .  .  |  .  5  .  |
|-----------+-----------+-----------|
|  .  .  .  |  .  .  .  |  5  .  .  |
|  .  .  5  |  5  .  .  |  .  .  .  |
|  .  5  .  |  .  5  .  |  .  .  .  |
|-----------+-----------+-----------|
|  5  .  .  |  .  .  .  |  .  .  .  |
|  .  .  .  |  .  .  5  |  .  .  .  |
|  .  .  .  |  .  .  .  |  .  .  5  |
+-----------------------------------+

Eventually you may want to learn about Alternating Inference Chains (AIC) that describe the underlying logic of many “patterns.” For the above grid position, the elimination of interest can be derived from a very simple AIC (also known in this case as a grouped X-chain or a grouped Turbot fish):

(5): r5c4 = r5c3 – r2c3 = r2c45 => r1c4 <> 5.

The “=” and “-” symbols refer, respectively, to strong and weak inference between the 5s in each indicated cell (or cell group). The chain starts and ends with a strong-inference link. Any other candidate that can “see” (weakly link to) both ends of the chain can be eliminated. In this case, (5)r1c4 can “see” (5)r5c4 and both of the 5s in cell group r2c45. (5)r1c4 can therefore be eliminated. [Edit to add: This AIC, which is in Eureka notation, is equivalent to the chain already provided by daj95376 at the end of page 1.]

If interested, you can read up on inference, links and AICs in both Sudopedia and in this forum at:
http://forum.enjoysudoku.com/viewtopic.php?t=3865.

[Jasper32]

Thanks Sudyro.

I was certainly able to follow that chain as you wrote it. I am new at this (my first post) and need all the help I can get. The AIC will sure be a usefull addition to my "toolbox". I will start lookin for it in my puzzles. I have the feeling this could be a "biggie". Thanks again.

[bingham]

This one got a LOT of attention, and I found the learned discussion interesting, but in the end frustrating because I still couldn't finish the puzzle! O.K. I got the logic sequences that led to r1c4<>5 and r6c8<>1, but where do you go from there?

[Pat]

I found the learned discussion interesting, but in the end frustrating because I still couldn't finish the puzzle!

O.K. I got the logic sequences that led to r1c4<>5 and r6c8<>1, but where do you go from there?

[r6c8]=7=[r4c8]=3=[r4c9]-3-[r8c9]=3=[r8c7]=1=[r5c7]-1-[r6c8]
This chain tells us that if r6c8<>7 => r6c8<>1, in any case r6c8<>1.

Code: Select all

 *--------------------------------------------------------------------*
 | 3      1569   7      | 1459   159    48     | 689    569    2      |
 | 168    2      1569   | 1359   159    38     | 689    4      7      |
 | 48     459    459    | 7      2      6      | 389    359    1      |
 |----------------------+----------------------+----------------------|
 | 47     49     2      | 6      48     1      | 5      379    389    |
 | 146    8      14569  | 345    7      34     | 169    2      69     |
 | 167    156    3      | 2      58     9      | 4      167    68     |
 |----------------------+----------------------+----------------------|
 | 5      146    146    | 8      3      7      | 2      169    469    |
 | 2      7      146    | 19     169    5      | 136    8      346    |
 | 9      3      8      | 14     146    2      | 7      16     5      |
 *--------------------------------------------------------------------*



One XY-wing takes care of the rest.

after excluding 1 at r6c8
we have a "hidden single" 1 in that box

and now as RW said,
it will still need an "XY-wing" --

r5c1 is 4 or 6,
forcing a 9 in r4c2 or in r5c9,

therefore excluding 9 in any cell which sees both r4c2 and r5c9,
i.e. exclusions in r4c89 and r5c3.

[bingham]

Oh dear, I try to work only from logic, and XY-wing doesn't explain it!
What is wrong with the assignments r5c1,3,4,6,9 = 4,9,5,3,6 and
r4c8 o9 = 9?