The New Sudoku Players' Forum

by **ArkieTech** » Thu Jul 16, 2015 12:15 am

Code: Select all: *-----------* |...|.9.|..7| |...|..1|826| |..4|8..|...| |---+---+---| |..3|1..|5..| |.7.|458|.3.| |..5|..6|4..| |---+---+---| |...|..2|9..| |591|7..|...| |4..|.8.|...| *-----------*

Play/Print this puzzle online

by **SteveG48** » Thu Jul 16, 2015 12:30 am

Code: Select all: *-----------------------------------------------------------------------------* | 12368 123568 268 | 26 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 c26 7 | 13 59 59 | *-------------------------+-------------------------+-------------------------| | 268 4 3 | 1 c27 9 | 5 b678 28 | | 1269 7 269 | 4 5 8 |b26 3 129 | | 1289 128 5 | 23 237 6 | 4 1789 1289 | *-------------------------+-------------------------+-------------------------| | 3678 368 678 | 356 1 2 | 9 56 4 | | 5 9 1 | 7 3-6 4 |a26 a68 238 | | 4 236 26 | 9 8 35 | 7 156 135 | *-----------------------------------------------------------------------------*

(6=28)r8c78 - (28=67)b6p24 - (7=26)r34c5 => -6 r8c5 ; stte

by **pjb** » Thu Jul 16, 2015 12:47 am

Code: Select all: 12368 123568 268 | 26 9 35 | 13 4 7 79 35 79 |c35 4 1 | 8 2 6 1236 1236 4 | 8 26 7 | 13 59 59 ------------------------+----------------------+--------------------- 268 4 3 | 1 b27 9 | 5 a678 28 1269 7 269 | 4 5 8 | 26 3 129 1289 128 5 |b23 237 6 | 4 1789 1289 ------------------------+----------------------+--------------------- 3678 368 678 | 36-5 1 2 | 9 a56 4 5 9 1 | 7 36 4 | 26 a68 238 4 236 26 | 9 8 35 | 7 156 135

(5=7)r478c8 - (7=3)r4c5,r6c4 - (3=5)r2c4 => -5 r7c4; stte

Phil

by **Leren** » Thu Jul 16, 2015 3:46 am

Code: Select all: *--------------------------------------------------------------------------------* | 12368 123568 268 | 26 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 b26 7 | 13 59 59 | |--------------------------+--------------------------+--------------------------| | 268 4 3 | 1 b27 9 | 5 S678 28 | | 1269 7 269 | 4 5 8 | 26 3 129 | | 1289 128 5 | 23 237 6 | 4 1789 1289 | |--------------------------+--------------------------+--------------------------| | 3678 368 678 | 356 1 2 | 9 a56 4 | | 5 9 1 | 7 cb36 4 | 26 c68 28-3 | | 4 236 26 | 9 8 35 | 7 a156 a135 | *--------------------------------------------------------------------------------* 3 Petal Death Blossom: Stem Cell r4c8 {678}: (3=6) r7c8, r9c89 - (6) r4c8; (3=7) r348c5 - (7) r4c8; (3=8) r8c58 - (8) r4c8; => - 3 r8c9; stte

Leren

by **daj95376** » Thu Jul 16, 2015 4:16 am

Code: Select all: +--------------------------------------------------------------------------------+ | 12368 123568 268 | 26 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 c26 7 | 13 59 59 | |--------------------------+--------------------------+--------------------------| | 268 4 3 | 1 d27 9 | 5 e678 28 | | 1269 7 269 | 4 5 8 | 26 3 129 | | 1289 128 5 | 23 237 6 | 4 1789 1289 | |--------------------------+--------------------------+--------------------------| | 3678 368 678 | 356 1 2 | 9 5-6 4 | | 5 9 1 | 7 b36 4 | a26 ae68 238 | | 4 236 26 | 9 8 35 | 7 15-6 135 | +--------------------------------------------------------------------------------+ # 79 eliminations remain 6r8c78 = r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=68)r48c8 => -6 r79c8

_

by **bat999** » Thu Jul 16, 2015 2:22 pm

Code: Select all: .--------------------.---------------.------------------. | 12368 123568 268 | 26 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 *26 7 | 13 59 59 | :--------------------+---------------+------------------: | 268 4 3 | 1 *27 9 | 5 *678 28 | | 1269 7 269 | 4 5 8 | 26 3 129 | | 1289 128 5 | 23 237 6 | 4 *1789 1289 | :--------------------+---------------+------------------: | 3678 368 678 | 356 1 2 | 9 56 4 | | 5 9 1 | 7 *36 4 | 26 *68 *238 | | 4 236 26 | 9 8 35 | 7 *156 3-15 | '--------------------'---------------'------------------'

Hi
If r4c8=7 it forces 3 in r9c9.
(7)r4c8 - (7=2)r4c5 - (2=6)r3c5 - (6=3)r8c5 - (3)r8c9 = (3)r9c9

If r4c8<>7 it kills the 1 and 5 at r9c9 to leave only 3 in r9c9.
(7)r4c8 = (7-1)r6c8 = (1)r9c8 - (1)r9c9
~~(7)r4c8 = (68)r48c8 - (6=5)r7c8 - (5)r9c8~~
(7)r4c8 = (68)r48c8 - (6=5)r7c8 - (5)r9c9

=> - 1 r9c9, -5 r9c9; stte

Is there a more compact way to express these three chains to show that r9c9 is 3 no matter whether r4c8 is 7 or not 7?

by **SteveG48** » Thu Jul 16, 2015 3:33 pm

bat999 wrote:
Code: Select all
.--------------------.---------------.------------------. | 12368 123568 268 | 26 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 *26 7 | 13 59 59 | :--------------------+---------------+------------------: | 268 4 3 | 1 *27 9 | 5 *678 28 | | 1269 7 269 | 4 5 8 | 26 3 129 | | 1289 128 5 | 23 237 6 | 4 *1789 1289 | :--------------------+---------------+------------------: | 3678 368 678 | 356 1 2 | 9 56 4 | | 5 9 1 | 7 *36 4 | 26 *68 *238 | | 4 236 26 | 9 8 35 | 7 *156 3-15 | '--------------------'---------------'------------------'

Hi
If r4c8=7 it forces 3 in r9c9.
(7)r4c8 - (7=2)r4c5 - (2=6)r3c5 - (6=3)r8c5 - (3)r8c9 = (3)r9c9

If r4c8<>7 it kills the 1 and 5 at r9c9 to leave only 3 in r9c9.
(7)r4c8 = (7-1)r6c8 = (1)r9c8 - (1)r9c9
(7)r4c8 = (68)r48c8 - (6=5)r7c8 - (5)r9c8

=> - 1 r9c9, -5 r9c9; stte

Is there a more compact way to express these three chains to show that r9c9 is 3 no matter whether r4c8 is 7 or not 7?

You could start by simply eliminating the second chain and writing the third chain:

(7)r4c8 = (68)r48c8 - (6=15)r79c8 - (15)r9c8

by **JC Van Hay** » Thu Jul 16, 2015 4:24 pm

bat999 wrote:Is there a more compact way to express these three chains to show that r9c9 is 3 no matter whether r4c8 is 7 or not 7?

Code: Select all: +--------------------+---------------+-------------------+ | 12368 123568 268 | 26 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 (26) 7 | 13 59 59 | +--------------------+---------------+-------------------+ | 268 4 3 | 1 (27) 9 | 5 (678) 28 | | 1269 7 269 | 4 5 8 | 26 3 129 | | 1289 128 5 | 23 237 6 | 4 1789 1289 | +--------------------+---------------+-------------------+ | 3678 368 678 | 356 1 2 | 9 (56) 4 | | 5 9 1 | 7 (36) 4 | 26 (68) 28(3) | | 4 236 26 | 9 8 35 | 7 (156) -15(3) | +--------------------+---------------+-------------------+

[3r9c9=3r8c9 - (3=627)r834c5 - (7=6851)r4879c8] - (15=3)r9c9

by **daj95376** » Thu Jul 16, 2015 4:33 pm

bat999:

JC posted while I was preparing my reply. Even though they are very similar, I'm going to post it anyway.

Your second and third chains can be replaced with:

Code: Select all: (7=86)r48c8 - (6=153)r79c8,r9c9

Now, what happens to <7> in r4c8 if your first chain is read from r-to-l? Yes, it's now treated as false. Write your first chain from r-to-l -- dropping the (7)r4c8 term -- and append the chain that I wrote above.

Code: Select all: 3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=153)r79c8,r9c9

You now have a discontinuous loop that assumes -3r9c9 and concludes =3r9c9.

RonK taught me this notational technique, of merging two chains into one, many moons ago.

[Edit: changed the ALS expressions.]

by **bat999** » Thu Jul 16, 2015 4:36 pm

SteveG48 wrote:...You could start...

Steve, there was a typo in my original post that I've corrected!
[Third chain should have been ...- (5)r9c9 not ...- (5)r9c8]
[So your suggested chain should be ...- (15)r9c9 not ...- (15)r9c8]

I think I can see this...
...- (6=15)r79c8
Switches off the both 6s in r7c8 and r9c8 and puts 1 and 5 in them, let the reader figure it out for himself which is the 1 and which is the 5.

by **Ngisa** » Thu Jul 16, 2015 5:00 pm

Code: Select all: +------------------+------------+--------------+ | 12368 123568 268 | 2-6 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 26 7 | 13 59 59 | +------------------+------------+--------------+ | 268 4 3 | 1 27 9 | 5 678 28 | | 1269 7 269 | 4 5 8 | 26 3 129 | | 1289 128 5 | 23 237 6 | 4 1789 1289 | +------------------+------------+--------------+ | 3678 368 678 | 356 1 2 | 9 56 4 | | 5 9 1 | 7 36 4 | 26 68 238 | | 4 236 26 | 9 8 35 | 7 156 135 | +------------------+------------+--------------+

Kraken Cell {678}r4c8 => -6r1c4
6r4c8-r5c7=r8c7-r8c5=(6)r7c4 -6r1c4
7r4c8-(7=2)r4c5-(2=6)r3c5 -6r1c4
8r4c8-(8=6)r8c8-r8c5=(6)r7c4 -6r1c4 =>-6r1c4; stte

Clement

by **bat999** » Thu Jul 16, 2015 6:45 pm

daj95376 wrote:
Code: Select all
3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=153)r79c8,r9c9

You now have a discontinuous loop...

"- (6=153)r79c8,r9c9"
From l-to-r this seems to say " We have switched off the 6 in r7c8, now figure out for yourself where the 153 go in r79c8,r9c9".
When you have concluded that the 3 goes in r9c9 you will see from l-to-r the contradiction "If r9c9 is not 3 then r9c9 is 3".
I can just about live with that. 8-)

But from r-to-l it seems weird...
"- (6=153)r79c8,r9c9"
From r-to-l is it supposed to say "If r79c8,r9c9 are not 153 then r7c8 is 6"?
This comma in the notation "r79c8,r9c9" seems to be disguising something.
Shouldn't it be specific that r9c9 is not 3 before reading from r-to-l to see the contradicton "If r9c9 is not 3 then r9c9 is 3"?
Surely we can't say "r9c9 is not 1 and not 3 and not 5". :lol:

Also, if r79c8 are both neither 1 nor 5 they would both be 6! :roll:

by **SteveG48** » Thu Jul 16, 2015 9:59 pm

daj95376 wrote:
Code: Select all
3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=153)r79c8,r9c9

You now have a discontinuous loop that assumes -3r9c9 and concludes =3r9c9.

Or, if you're looking for compact, just drop the first two terms and get:

(3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=153)r79c8,r9c9

This implies -3 r8c9, which is the equivalent of =3 r9c9.

by **daj95376** » Thu Jul 16, 2015 10:03 pm

bat999 wrote:
daj95376 wrote:
Code: Select all
3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=153)r79c8,r9c9

You now have a discontinuous loop...

"- (6=153)r79c8,r9c9"
From l-to-r this seems to say " We have switched off the 6 in r79c8, now figure out for yourself where the 153 go in r79c8,r9c9".
When you have concluded that the 3 goes in r9c9 you will see from l-to-r the contradiction "If r9c9 is not 3 then r9c9 is 3".
I can just about live with that.

Yes, there is a flaw in ALS notation when designating value location. It's left up to the reader to place the values. I tried at one time to get people to list the candidates and cells in as close an alignment as possible. That didn't catch on, so I don't try anymore, either.

But from r-to-l it seems weird...
"- (6=153)r79c8,r9c9"
From r-to-l is it supposed to say "If r79c8,r9c9 are not 153 then one of r79c8 is 6"?
This comma in the notation "r79c8,r9c9" seems to be disguising something.

The comma isn't hiding anything. It's just there so I can list the three cells without resorting to box/position notation.

Shouldn't it be specific that r9c9 is not 3 before reading from r-to-l to see the contradicton "If r9c9 is not 3 then r9c9 is 3"?
Surely we can't say "r9c9 is not 1 and not 3 and not 5".
Also, if r79c8 are both neither 1 nor 5 they would both be 6!

Aaaagh! You caught me in a notational shorthand. Most of us have fallen into the habit of only qualifying ALS terms as they are read l-to-r. To be technically correct, my chain should have read:

Code: Select all: 3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (651=153)r79c8,r9c9

When read from l-to-r, the last term says: if <6> is not in these three cells, then <153> must occupy these cells.

When read from r-to-l, the last term says: if <3> is not in these three cells, then <651> must occupt these cells.

Normally, since this isn't a continuous loop, I'd shorten the notation to:

Code: Select all: 3r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=3)r79c8,r9c9

by **bat999** » Thu Jul 16, 2015 11:24 pm

daj95376 wrote:... Most of us have fallen into the habit of only qualifying ALS terms as they are read l-to-r...

Question, Is it not necessary for all AIC's to be readable both ways?

If we use an AIC to show that one end forces the other and vice versa so that we can use it to eliminate a candidate.
Like this example...

r4c6 can see both ends of the chain, we use that to make an elimination, so we need to be able to read it both ways.

Code: Select all: .------------------.-----------------.---------------. | 4568 5689 1 | 458 57 a78 | 2 3 569 | | 7 56 3 | 2 1 9 | d46 8 456 | | 2458 2589 489 | 458 3 6 | 17 59 17 | :------------------+-----------------+---------------: | 3 16 5 | 9 4 1-7 | e67 2 8 | | 268 4 68 | 3 27 5 | 9 1 67 | | 9 12 7 | 6 8 12 | 5 4 3 | :------------------+-----------------+---------------: | 458 7 489 | 1 59 b38 | c34 6 2 | | 1 3 2 | 7 6 4 | 8 59 59 | | 4568 5689 4689 | 58 259 238 | 134 7 14 | '------------------'-----------------'---------------'

(7)r1c6=(8)r1c6-(8)r7c6=(3)r7c6-(3)r7c7=(4)r7c7-(4)r2c7=(6)r2c7-(6)r4c7=(7)r4c7 => r4c6<>7
From left to right, if r1c6 is not 7 then r4c7 is 7.
From right to left, if r4c7 is not 7 then r1c4 is 7.
r4c6 can see both ends of the chain.

But with the example from this (July 16) thread, we're using the chain differently.

Is it acceptable only to be able to read it from l-to-r in order to see that there is the contradiction "If r9c9 is not 3 then r9c9 is 3" => r9c9=3

Code: Select all: .--------------------.---------------.------------------. | 12368 123568 268 | 26 9 35 | 13 4 7 | | 79 35 79 | 35 4 1 | 8 2 6 | | 1236 1236 4 | 8 *26 7 | 13 59 59 | :--------------------+---------------+------------------: | 268 4 3 | 1 *27 9 | 5 *678 28 | | 1269 7 269 | 4 5 8 | 26 3 129 | | 1289 128 5 | 23 237 6 | 4 1789 1289 | :--------------------+---------------+------------------: | 3678 368 678 | 356 1 2 | 9 *56 4 | | 5 9 1 | 7 *36 4 | 26 *68 *238 | | 4 236 26 | 9 8 35 | 7 *156 3-15 | '--------------------'---------------'------------------'

(3)r9c9 = (3)r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 => r9c9=3
From l-to-r, if r9c9 is not 3 then r9c9 is 3, contradiction.

Is that it?
Make sure that it can be read both ways if we need that property for some reason, otherwise don't worry about it.

PS
My original solution used both states of the 7 in r4c8 to prove r9c9=3, but it has morphed into a chain that merely uses r4c8 as a node.

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