The New Sudoku Players' Forum

by **ArkieTech** » Sun Mar 05, 2017 12:17 am

Code: Select all: *-----------* |...|.3.|...| |..7|2.5|8..| |.8.|649|.5.| |---+---+---| |.56|...|13.| |9.8|...|5.4| |.43|...|98.| |---+---+---| |.6.|478|.1.| |..4|1.3|6..| |...|.6.|...| *-----------*

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by **SteveG48** » Sun Mar 05, 2017 12:51 am

Code: Select all: *-----------------------------------------------------------* | 46 129 5 | 8 3 7 | 24 2469 169 | | 46 39 7 | 2 1 5 | 8 469 369 | | 3-1 8 a12 | 6 4 9 | 237 5 137 | *-------------------+-------------------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 f17 8 | 3 2 16 | 5 e67 4 | |g12 4 3 | 7 5 16 | 9 8 26 | *-------------------+-------------------+-------------------| | 35 6 b29 | 4 7 8 | 23 1 359 | | 58 c27 4 | 1 9 3 | 6 d27 58 | | 1378 137 19 | 5 6 2 | 347 479 3789 | *-----------------------------------------------------------*

(1=2)r3c3 - r7c3 = (2-7)r8c2 = r8c8 - r5c8 = (7-1)r5c2 = 1r6c1 => -1 r3c1 ; stte

Or:

Code: Select all: *-----------------------------------------------------------* | 46 129 5 | 8 3 7 |c24 c2469 169 | | 46 39 7 | 2 1 5 | 8 c469 369 | | 13 8 12 | 6 4 9 | 237 5 137 | *-------------------+-------------------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 1-7 8 | 3 2 16 | 5 d67 4 | | 12 4 3 | 7 5 16 | 9 8 26 | *-------------------+-------------------+-------------------| | 35 6 a29 | 4 7 8 | 23 1 359 | | 58 a27 4 | 1 9 3 | 6 b27 58 | | 1378 a137 a19 | 5 6 2 |b347 b479 3789 | *-----------------------------------------------------------*

(7=1239)b7p3589 - (2|3=479)b9p578 - (9=246)b3p125 - (6=7)r5c8 => -7r5c2 ; stte

by **Alex_Popov_92** » Sun Mar 05, 2017 12:47 pm

Code: Select all: *-----------------------------------------------------------* | 46 129 5 | 8 3 7 | 24 2469 169 | | 46 39 7 | 2 1 5 | 8 469 369 | | 31 8 12 | 6 4 9 | 237 5 137 | *-------------------+-------------------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 17 8 | 3 2 16 | 5 67 4 | | 12 4 3 | 7 5 16 | 9 8 26 | *-------------------+-------------------+-------------------| | 35 6 29 | 4 7 8 | 23 1 359 | | 58 27 4 | 1 9 3 | 6 27 58 | | 138-7 137 19 | 5 6 2 | 347 479 3789 | *-----------------------------------------------------------*

1. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r9c3 => r9c7<>3
2. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r3c1 - r3c7 = (3-2)r7c7 = (2-7)r7c3 = 7r8c2
=> r9c1 <> 7; stte
1. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r9c2 => r9c7<>3
2. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r3c1 - 3r3c7 = (3-2)r7c7 = (2-7)r7c3 = 7r8c2
=> r9c1 <> 7; stte
On 1. 3 in the 3rd member is in red. I think it is necessary.It is not marked red in 2.
On 2. r7c7=3 because r3c7<>3 AND r9c7<>3 and in column 7
there is no other place for 3 except r7c7. I am using the result from 1.
Should I include AND-OR expression?

by **eleven** » Sun Mar 05, 2017 4:48 pm

Alex_Popov_92 wrote:1. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r9c3 => r9c7<>3
2. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r3c1 - r3c7 = (3-2)r7c7 = (2-7)r7c3 = 7r8c2

For Bat:

Code: Select all: *------------------------------------------------* | 46 129 5 | 8 3 7 | b24 2469 169 | | 46 39 7 | 2 1 5 | 8 469 369 | | 13 8 12 | 6 4 9 | 237 5 137 | |------------------+---------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 17 8 | 3 2 16 | 5 67 4 | | 12 4 3 | 7 5 16 | 9 8 26 | |------------------+---------+-------------------| | a35 6 a29 | 4 7 8 | b23 1 b359 | | 58 a27 4 | 1 9 3 | 6 27 58 | | 138-7 a137 a19 | 5 6 2 | b347 479 3789 | *------------------------------------------------*

(7=59)b7p13589-(5|9=7)r179c7,r7c9 => -7r9c1, stte
[corrected typos - thanks Bat]

by **bat999** » Sun Mar 05, 2017 5:13 pm

eleven wrote:...For Bat...

Yes... r179c7,r7c9 is something that doesn't seem to have a name.
My descriptions are a "ALS whose cells are not all in the same house" or a "Subset that is almost locked".
And in this case it is a "AALS whose cells are not all in the same house" or a "Subset that is almost almost locked".

by **eleven** » Sun Mar 05, 2017 9:46 pm

Alex_Popov_92 wrote:
Code: Select all
*-----------------------------------------------------------* | 46 129 5 | 8 3 7 | 24 2469 169 | | 46 39 7 | 2 1 5 | 8 469 369 | | 31 8 12 | 6 4 9 | 237 5 137 | *-------------------+-------------------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 17 8 | 3 2 16 | 5 67 4 | | 12 4 3 | 7 5 16 | 9 8 26 | *-------------------+-------------------+-------------------| | 35 6 29 | 4 7 8 | 23 1 359 | | 58 27 4 | 1 9 3 | 6 27 58 | | 138-7 137 19 | 5 6 2 | 347 479 3789 | *-----------------------------------------------------------*

1. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r9c3 => r9c7<>3
2. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r3c1 - r3c7 = (3-2)r7c7 = (2-7)r7c3 = 7r8c2
=> r9c1 <> 7; stte
1. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r9c2 => r9c7<>3
2. (7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r3c1 - 3r3c7 = (3-2)r7c7 = (2-7)r7c3 = 7r8c2
=> r9c1 <> 7; stte
On 1. 3 in the 3rd member is in red. I think it is necessary.It is not marked red in 2.
On 2. r7c7=3 because r3c7<>3 AND r9c7<>3 and in column 7
there is no other place for 3 except r7c7. I am using the result from 1.
Should I include AND-OR expression?

Alex,

i am already tired of reading and correcting your solutions.

So you did not understand, what an AIC link is.
Don't you read other's posts ?

3r7c1 = 3r9c2 means:
if r7c1 is not 3, then r9c2 is 3
or equivalently
if r9c2 is not 3, then r7c1 is 3.

And (2-7)r7c3 means:
if r7c3 is 2, then it is not 7
or equivalently
if r7c3 is 7, then it is not 2.

This is not true for your link. What you are writing is not an AIC, but a net, i.e. you have to remember things written elsewhere in the chain (here 7r9c1).
You can do it, if you mark this explicitly, e.g.
(*7-8)r9c1 = (8-5)r8c1 = (5-3)r7c1 = *3r9c2 => r9c7<>3 or
(7-8)9c1 = (8-5)r8c1 = (5-3)r7c1 = 3r9c2 => r9c7<>3

Same for the second line. You have to mark, where you use the result from line 1.
Also the end of the second line is wrong:
(2-7)r7c3 = 7r8c2 should be
2r7c3 - (2=7)r8c2

And you did not mark the cells in the grid again. Who do you think, likes to read that ?

by **ArkieTech** » Mon Mar 06, 2017 2:56 am

Code: Select all: *-----------------------------------------------------------* | 46 129 5 | 8 3 7 |c24 2469 169 | | 46 39 7 | 2 1 5 | 8 469 369 | | 13 8 12 | 6 4 9 | 237 5 137 | |-------------------+-------------------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 17 8 | 3 2 16 | 5 67 4 | | 12 4 3 | 7 5 16 | 9 8 26 | |-------------------+-------------------+-------------------| |a35 6 a29 | 4 7 8 |c23 1 b359 | | 58 a27 4 | 1 9 3 | 6 27 58 | |138-7 a137 a19 | 5 6 2 |c347 479 3789 | *-----------------------------------------------------------* [(7=59p13)b7p13589-(59=3)r7c9-(3=7)r179c7]-7r9c1; ste

by **Sudtyro2** » Mon Mar 06, 2017 12:21 pm

ArkieTech wrote:
Code: Select all
*-----------------------------------------------------------* | 46 129 5 | 8 3 7 |c24 2469 169 | | 46 39 7 | 2 1 5 | 8 469 369 | | 13 8 12 | 6 4 9 | 237 5 137 | |-------------------+-------------------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 17 8 | 3 2 16 | 5 67 4 | | 12 4 3 | 7 5 16 | 9 8 26 | |-------------------+-------------------+-------------------| |a35 6 a29 | 4 7 8 |c23 1 b359 | | 58 a27 4 | 1 9 3 | 6 27 58 | |138-7 a137 a19 | 5 6 2 |c347 479 3789 | *-----------------------------------------------------------* [(7=59p13)b7p13589-(59=3)r7c9-(3=7)r179c7]-7r9c1; ste

Thx, Dan, for 'decompressing' Eleven's second node. But, here's a silly question about the first node...
How can the AND'd 59 digits in (7=59p13) both be true if 9r9c3 happens to be the true digit in b7? IOW, it seems
to me that 9r7c9 must be able to see both 9s in b7 for a valid link to the OR'd 59 pair in the second node.

SteveC

by **ArkieTech** » Mon Mar 06, 2017 1:55 pm

Sudtyro2 wrote:
ArkieTech wrote:
Code: Select all
*-----------------------------------------------------------* | 46 129 5 | 8 3 7 |c24 2469 169 | | 46 39 7 | 2 1 5 | 8 469 369 | | 13 8 12 | 6 4 9 | 237 5 137 | |-------------------+-------------------+-------------------| | 27 5 6 | 9 8 4 | 1 3 27 | | 9 17 8 | 3 2 16 | 5 67 4 | | 12 4 3 | 7 5 16 | 9 8 26 | |-------------------+-------------------+-------------------| |a35 6 a29 | 4 7 8 |c23 1 b359 | | 58 a27 4 | 1 9 3 | 6 27 58 | |138-7 a137 a19 | 5 6 2 |c347 479 3789 | *-----------------------------------------------------------* [(7=59p13)b7p13589-(59=3)r7c9-(3=7)r179c7]-7r9c1; ste

it seems
to me that 9r7c9 must be able to see both 9s in b7 for a valid link to the OR'd 59 pair in the second node.

in the als (7=59)b7p13589
[(7=2)r8c2-(2=9)r7c3]-9r9c3]

How should this be notated?

by **SteveG48** » Mon Mar 06, 2017 2:47 pm

ArkieTech wrote:in the als (7=59)b7p13589
[(7=2)r8c2-(2=9)r7c3]-9r9c3]

How should this be notated?

Dan, I don't think any special notation is needed. Once the 7 is eliminated from the ALS, the positions of all remaining candidates are set. I think Steve just overlooked it. However, if we really want to do it, I would write:

7b9p13589 = (5r9c1)&(9r9c3) . For clarity, I would also write - (5|9=3) in the next term.

by **Sudtyro2** » Mon Mar 06, 2017 4:11 pm

Thx, Dan and Steve, for the helpful feedback. Unfortunately, I'm still at sea here...
Using all digits in the ALS, I can write, say, (7123=59)b7p13589. This tells me (I thought) that if both 7s are false then digits 5&9 must be true, but it doesn't tell me which of the two 9s is true. Yes, Dan's second chain does give me that information, but that would seem to result from some sort of extra 'decompression' of the full ALS. That knowledge wouldn't otherwise normally be obvious, at least to me.

SteveC

by **SteveG48** » Mon Mar 06, 2017 6:02 pm

Sudtyro2 wrote:Thx, Dan and Steve, for the helpful feedback. Unfortunately, I'm still at sea here...
Using all digits in the ALS, I can write, say, (7123=59)b7p13589. This tells me (I thought) that if both 7s are false then digits 5&9 must be true, but it doesn't tell me which of the two 9s is true. Yes, Dan's second chain does give me that information, but that would seem to result from some sort of extra 'decompression' of the full ALS. That knowledge wouldn't otherwise normally be obvious, at least to me.

SteveC

Steve, the way you've written it, (7123=59)b7p13589, is not the same as what Dan wrote. Yours means that if (7 AND 1 AND 2 AND 3) is not true, then 5 AND 9 are true in the set. That is correct, but that can happen if any one of the four candidates on the left side of your expression is false. Then, as you've indicated, we don't necessarily know in which cell the 9 is true. What Dan wrote is (7=59)b7p13589 . This means that if 7 specifically is false in the set, then 5 and 9 are both true. What's more, we know the exact position of each of the candidates in the set. p5 must be a 2, so p3 is 9, p9 is 1, p8 is 3, and p1 is 5. That makes the following double link work. Dan emphasizes this by adding p13 to his expression, (7=59p13) to indicate that 5 is in position 1 and 9 is in position 3. (I've never seen this before, and it's not technically necessary, but I think that's the meaning.)

by **Sudtyro2** » Mon Mar 06, 2017 9:30 pm

Good discussion, Steve & Dan ...
I think the issue for me is probably just the notation convention. When I write the designated ALS as, say, (71239=5)b7p13589 or (7=12395)b7p13589, I assume that means simply the 7s and the 5 are the in/out linking digits. But, I often see that sort of single-linking case shortened to (7=5)b7p13589. For double-linking cases, I would write (7123=59)b7p13589 or maybe (7=59)b7p13589, for short. So, you can see the problem evolving...there are still two 9s in the mix. Dan cleverly avoided that disaster with the 'p13' notation...and justified it with his second chain.
That's all new to me, but I will catch up!

SteveC

by **JC Van Hay** » Mon Mar 06, 2017 9:54 pm

SteveG48 wrote:What Dan wrote is (7=59)b7p13589 . This means that if 7 specifically is false in the set (if p5 is not 7 and p8 is not 7), then 5 and 9 are both true. What's more, we know the exact position of each of the candidates in the set. p5 must be a 2, so (p3 is not 2 and) p3 is 9, (p9 is not 9 and) p9 is 1, (p8 is not 1 and) p8 is 3, and (p1 is not 3 and) p1 is 5. That makes the following double link work (so r7c9 is not 9, r7c9 is not 5, r7c9 is 3) and. Dan emphasizes this by adding p13 to his expression, (7=59p13) to indicate that 5 is in position 1 and 9 is in position 3.

... so r7c7 is not 3 and r7c7 is 2, r1c7 is not 2 and r1c7 is 4, r9c7 is not 3; r9c7 is not 4 and r9c7 is 7.

Dan"s wysiwyg comment can be summarized inside the following presentation called a "Triangular Matrix" where the first column contains the derived constraint {7b7p58, 7r9c7} excluding 7r9c1

Code: Select all: 7b7p5 2b7p5 2b7p3 9b7p3 9b7p9 1b7p9 7b7p8 1b7p8 3b7p8 3b7p1 5b7p1 9r7c9 5r7c9 3r7c9 3r7c7 2r7c7 2r1c7 4r1c7 7r9c7 3r9c7 4r9c7

or as a ttt's diagram

Code: Select all: 3r9c7-3r7c9------------------------------------------------- || || | || 5r7c9-(5=3)b7p1-3b7p8 | || || || | || || 1b7p8-(1=9)b7p9-(9=2)b7p3-(2=7)b7p5 | || || || | | || || 7b7p8 | | || || | | || || | | || 9r7c9----------------------------- | || | 4r9c7-(4=2)r1c7-(2=3)r7c7----------------------------------- || 7r9c7

or according to the player preference as ... :roll:

by **SteveG48** » Mon Mar 06, 2017 10:38 pm

Sudtyro2 wrote:Good discussion, Steve & Dan ...

Agreed!

I think the issue for me is probably just the notation convention. When I write the designated ALS as, say, (71239=5)b7p13589 or (7=12395)b7p13589, I assume that means simply the 7s and the 5 are the in/out linking digits.

Steve, I think that's a dangerous assumption. As we've seen, the two expressions do not mean the same thing, even though both are true in this puzzle.

But, I often see that sort of single-linking case shortened to (7=5)b7p13589.

Yes, and it is perfectly correct to write it that way, or to write (7=12359)b7p13589. Either one is correct as a Boolean expression, correct in the puzzle, and establishes the single link. I only avoid the shortened form because some folks find it to be less clear.

For double-linking cases, I would write (7123=59)b7p13589 or maybe (7=59)b7p13589, for short. So, you can see the problem evolving...there are still two 9s in the mix.

Yes, but those two forms, while correct in the puzzle, are not equivalent, and the first doesn't justify the double link.

Dan cleverly avoided that disaster with the 'p13' notation...and justified it with his second chain.

I agree that Dan's notation is clever, but I think it's important to recognize that it does not avoid the disaster. It is clarification only. Dan avoids the disaster by isolating the 7 on the left side of the expression. That assures that 7 is false and establishes the position of all the other candidates. Declaring an ANDed expression to be false only assures that at least one of the arguments is false, and not necessarily the one you want. He could also have written (7=12359) to establish the double link.

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