The New Sudoku Players' Forum

[bat999]

... since r3c1 has 3 possibilities..

Think about it this way...
r3c1 has 2 states
Either
r3c1=5
or
r3c1<>5

[Marty R.]

... since r3c1 has 3 possibilities..

Think about it this way...
r3c1 has 2 states
Either
r3c1=5
or
r3c1<>5

Could be. But what I am seeing is that r3c1<>5 may be one of your either/or, but I see two possibilities. If r3c1 is not 5, then it could be 1 or 8. To me the whole idea of chains is that the conclusion is drawn based on the opening premise being true and confirmed if it's false. I just can't see it here because the false has those two possibilities.

[daj95376]

One of those awkward "coloring" scenarios that's so simple in principle, but so difficult in practice.

There are stable six-cell cycles on <5> and <8>, but their overlap generates a conflict.

Code: Select all

 5: Blue/Green coloring   ;   8: Pink/Amber coloring
 +-----------------------------------------------------+
 |  3    4    7    |  1    2    6    |  9    5    8    |
 |  168P 9    16   |  5    3    78A  |  17   2    4    |
 | B158A 2   G15   |  4    78P  9    |  17   3    6    |
 |-----------------+-----------------+-----------------|
 |  2    5    9    |  8    1    3    |  6    4    7    |
 |  16   16   8    |  7    4    2    |  3    9    5    |
 |  7    3    4    |  6    9    5    |  8    1    2    |
 |-----------------+-----------------+-----------------|
 |  9    7    3    |  2    6    4    |  5    8    1    |
 |  4    16  B156  |  9   G58A  18P  |  2    7    3    |
 | G15   8    2    |  3   B57   17   |  4    6    9    |
 +-----------------------------------------------------+
 # 20 eliminations remain

 5: is Blue  in r3c1 and Green in r8c5
 8: if Amber in r3c1 and Amber in r8c5

 If Amber is true for <8>, then it would force a contradiction on <5>.

 =>  -8 r2c6,r3c1,r8c5


_

[bat999]

... If r3c1 is not 5, then it could be 1 or 8....

If r3c1 is not 5 can you see what number will be in r3c5?
(5)r3c1 = (5)r9c1 – (5=7)r9c5 – (7=8)r3c5

[DonM]

Code: Select all

.-----------------.------------.----------.
|  3     4    7   | 1   2   6  | 9   5  8 |
|  168   9    16  | 5   3   78 | 17  2  4 |
| a15-8  2   15   | 4  b78  9  | 17  3  6 |
:-----------------+------------+----------:
|  2     5    9   | 8   1   3  | 6   4  7 |
|  16    16   8   | 7   4   2  | 3   9  5 |
|  7     3    4   | 6   9   5  | 8   1  2 |
:-----------------+------------+----------:
|  9     7    3   | 2   6   4  | 5   8  1 |
|  4     16   156 | 9   58  18 | 2   7  3 |
| b15    8    2   | 3  b57  17 | 4   6  9 |
'-----------------'------------'----------'

(5)r3c1 = (578)r9c15,r3c5 => -8 r3c1; stte

I think these two expressions are equivalent (because the 8 can only be in r3c5)
(5)r3c1 = (5)r9c1 – (5=7)r9c5 – (7=8)r3c5 => -8 r3c1
(5)r3c1 = (578)r9c15,r3c5 => -8 r3c1

The first expression is a true AIC. The second is not, but is rather an attempt at a summary of the logic that arrives at the exclusion. But if read literally, it is implying that if r3c1 is not 5 then r9c1=5 which is correct, but also that r9c5=5 which is not correct.

Of course, I understand what the intention is, but, regardless, any notation should be specifically correct about each cell's status.

[bat999]

... a summary of the logic that arrives at the exclusion...

Exactly.

[DonM]

... a summary of the logic that arrives at the exclusion...

Exactly.

Er, not so exact unfortunately.

[Marty R.]

... If r3c1 is not 5, then it could be 1 or 8....

If r3c1 is not 5 can you see what number will be in r3c5?
(5)r3c1 = (5)r9c1 – (5=7)r9c5 – (7=8)r3c5

I have no question about this one. I don't know what was accomplished. Every time I ask a question it seems to open up a can of worms which makes things worse. I don't know if I received an answer to my original question and I'm not sure what Don's posts are saying. I should do more staying silent and be thought a fool.

[sultan vinegar]

Hi Marty,

If r3c1 is not 5, then lots of things can happen. The thing that is pertinent for this elimination is what then happens to the 5 in column 1, as opposed to your present view which considers what then happens to the 1 and 8 in r3c1. The 5 in column 1 must go in row 9.

[Marty R.]

Thanks, SV. I'm fine with starting out with r3c1<>5. What I'm still not 100% clear on is the solution that I first questioned, whose first term I think means r3c1=5. I didn't think the conclusion was valid based on the opening premise which I think opened the door for three possibilities. This one: (5)r3c1 = (578)r9c15,r3c5 => -8 r3c1; stte

Code: Select all

    .-----------------.------------.----------.
    |  3     4    7   | 1   2   6  | 9   5  8 |
    |  168   9    16  | 5   3   78 | 17  2  4 |
    | a15-8  2   15   | 4  b78  9  | 17  3  6 |
    :-----------------+------------+----------:
    |  2     5    9   | 8   1   3  | 6   4  7 |
    |  16    16   8   | 7   4   2  | 3   9  5 |
    |  7     3    4   | 6   9   5  | 8   1  2 |
    :-----------------+------------+----------:
    |  9     7    3   | 2   6   4  | 5   8  1 |
    |  4     16   156 | 9   58  18 | 2   7  3 |
    | b15    8    2   | 3  b57  17 | 4   6  9 |
    '-----------------'------------'----------'

(5)r3c1 = (578)r9c15,r3c5 => -8 r3c1; stte

I think these two expressions are equivalent (because the 8 can only be in r3c5)
(5)r3c1 = (5)r9c1 – (5=7)r9c5 – (7=8)r3c5 => -8 r3c1
(5)r3c1 = (578)r9c15,r3c5 => -8 r3c1

[sultan vinegar]

Hi Marty, my next guess for what is confusing you is the passenger digits 5 and 7. Would it make more sense if ...

(5)r3c1 = (578)r9c15,r3c5 => -8 r3c1

... was written with the passenger digits 5 and 7 removed ...

(5)r3c1 = (8)r9c15,r3c5 => -8 r3c1

... which translates to English as "if r3c1 is not 5, then at least one of r9c15,r3c5 is 8, but of course the only one of those three cells eligible to hold 8 is r3c5, so all together, if r3c1 is not 5, then r3c5 is 8, hence r3c1 is not 8." Candidates 5 and 7 in the term (578)r9c15, r3c5 play no role, they just go along for the ride.

[Sudtyro2]

Code: Select all

 *--------------------------------------------------*
 | 3    4    7    | 1    2    6    | 9    5    8    |
 | 68-1 9    16   | 5    3    78   | 17   2    4    |
 | 158a 2    15   | 4    78   9    | 17   3    6    |
 |----------------+----------------+----------------|
 | 2    5    9    | 8    1    3    | 6    4    7    |
 | 6-1  16   8    | 7    4    2    | 3    9    5    |
 | 7    3    4    | 6    9    5    | 8    1    2    |
 |----------------+----------------+----------------|
 | 9    7    3    | 2    6    4    | 5    8    1    |
 | 4    16   156  | 9    58   18   | 2    7    3    |
 | 15ab 8    2    | 3    57b  17   | 4    6    9    |
 *--------------------------------------------------*

One last observation (and eliminations) involving bat's four cells...

The CoALS rule can be applied to overlapping ALS a(158)r39c1 and b(157)r9c15. The overlap cell is r9c1, so the overlap digits are the two 1s and the three 5s. The non-overlap digits are the remaining single digits, 7 and 8.

The CoALS rule says that (78 = 15) in the combined ALS, where both 2-digit pairs are AND'd. However, the (78) digits, if true, both see cell (78)r3c5, which then forms a contradiction. So, the remaining (15) digits must be true. No external 5-digit can see all three 5s, but external digits (1)r25c1 can both see the CoALS digits (1)r39c1.

Hence, r25c1<>1; stte

SteveC

[ArkieTech]

Code: Select all

 *--------------------------------------------------*
 | 3    4    7    | 1    2    6    | 9    5    8    |
 | 68-1 9    16   | 5    3    78   | 17   2    4    |
 | 158a 2    15   | 4    78   9    | 17   3    6    |
 |----------------+----------------+----------------|
 | 2    5    9    | 8    1    3    | 6    4    7    |
 | 6-1  16   8    | 7    4    2    | 3    9    5    |
 | 7    3    4    | 6    9    5    | 8    1    2    |
 |----------------+----------------+----------------|
 | 9    7    3    | 2    6    4    | 5    8    1    |
 | 4    16   156  | 9    58   18   | 2    7    3    |
 | 15ab 8    2    | 3    57b  17   | 4    6    9    |
 *--------------------------------------------------*

One last observation (and eliminations) involving bat's four cells...

The CoALS rule can be applied to overlapping ALS a(158)r39c1 and b(157)r9c15. The overlap cell is r9c1, so the overlap digits are the two 1s and the three 5s. The non-overlap digits are the remaining single digits, 7 and 8.

The CoALS rule says that (78 = 15) in the combined ALS, where both 2-digit pairs are AND'd. However, the (78) digits, if true, both see cell (78)r3c5, which then forms a contradiction. So, the remaining (15) digits must be true. No external 5-digit can see all three 5s, but external digits (1)r25c1 can both see the CoALS digits (1)r39c1.

Hence, r25c1<>1; stte

Thanks SteveC -- most helpful

does this say it?

[(78=15)r39c19]-1r25c1; ste

[Sudtyro2]

Thanks SteveC -- most helpful

does this say it?

[(78=15)r39c19]-1r25c1; ste

I didn't think about a chain in this particular CoALS elimination.
However, something should include the contradiction cell (78)r3c5. I'll work on it!

SteveC

[daj95376]

Code: Select all

.-----------------.------------.----------.
|  3     4    7   | 1   2   6  | 9   5  8 |
|  168   9    16  | 5   3   78 | 17  2  4 |
| a15-8  2   15   | 4  b78  9  | 17  3  6 |
:-----------------+------------+----------:
|  2     5    9   | 8   1   3  | 6   4  7 |
|  16    16   8   | 7   4   2  | 3   9  5 |
|  7     3    4   | 6   9   5  | 8   1  2 |
:-----------------+------------+----------:
|  9     7    3   | 2   6   4  | 5   8  1 |
|  4     16   156 | 9   58  18 | 2   7  3 |
| b15    8    2   | 3  b57  17 | 4   6  9 |
'-----------------'------------'----------'

(5)r3c1 = (578)r9c15,r3c5 => -8 r3c1; stte

You win the Blue Ribbon for compressing terms and driving people crazy. _ _


BTW: Everyone will say, "a Locked Set is N values in N cells". But few will take the time to add, "in a house/unit". Don't go to Sudopedia for a definition because it confuses a Locked Set with a Subset ... and they aren't the same when it comes to eliminations.

_