The New Sudoku Players' Forum

[bat999]

- (15=3)r9c9
...but the ending weak link is not required here either...

Yes, I can understand that.
Though with the final term included it makes the contradiction or discontinuous_nice_loop very obvious (if that's the way somebody might prefer to interpret it).
(3)r9c9 = ... (15=3)r9c9

[ronk]

Though with the final term included it makes the contradiction or discontinuous_nice_loop very obvious (if that's the way somebody might prefer to interpret it).
(3)r9c9 = ... - (15=3)r9c9

What if r9c9 contained an additional candidate? r9c9 <> 15 would still hold, but r9c9 = 3 would not.

[bat999]

...What if r9c9 contained an additional candidate?

If r9c9 contained an additional candidate, killing the 15 is OK, but it would not force a 3 into it.
So "r9c9 <> 15 would still hold, but r9c9 = 3 would not" is correct imho.

But that would be a different puzzle!

I'd like to keep all the terms in my solution to retain the contradiction/loop.

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    d26   7  | 13   59     59   |
:--------------------+---------------+------------------:
| 268    4       3   | 1    e27   9  | 5   f678    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   g56     4    |
| 5      9       1   | 7    c36   4  | 26  f68    b238  |
| 4      236     26  | 9     8    35 | 7   g156   a3-15 |
'--------------------'---------------'------------------'

AIC: (3)r9c9 = r8c9 - (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 => - 1 r9c9, - 5 r9c9; stte
To me this says "If r9c9 is not 3 then r9c9 is 3".
Contadiction, need to keep the 3 in r9c9 and eliminate all the others.
=> - 1 r9c9, - 5 r9c9


But I agree that the last term can be dropped in Steve's solution.

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    c27   9  | 5   d678   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   e56    4    |
| 5      9       1   | 7    a36   4  | 26  d68    238  |
| 4      236     26  | 9     8    35 | 7   e156   3-15 |
'--------------------'---------------'-----------------'

AIC: (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 => - 1 r9c9, - 5 r9c9; stte
To me this says...
If r8c5=3 (the trivial stream)
think about it... there can't be a 3 in r8c9 so it must be in r9c9, so neither 1 nor 5 can be in r9c9.
If r8c5<>3 (the second stream)
think about it... through the chain 15 is forced into r79c8, so neither 1 nor 5 can be in r9c9.
=> - 1 r9c9, - 5 r9c9

[daj95376]

But I agree that the last term can be dropped in Steve's solution.

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    c27   9  | 5   d678   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   e56    4    |
| 5      9       1   | 7    a36   4  | 26  d68    238  |
| 4      236     26  | 9     8    35 | 7   e156   3-15 |
'--------------------'---------------'-----------------'

AIC: (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 => - 1 r9c9, - 5 r9c9; stte
To me this says...
If r8c5=3 (the trivial stream)
think about it... there can't be a 3 in r8c9 so it must be in r9c9, so neither 1 nor 5 can be in r9c9.
If r8c5<>3 (the second stream)
think about it... through the chain 15 is forced into r79c8, so neither 1 nor 5 can be in r9c9.
=> - 1 r9c9, - 5 r9c9

I've lost track of the various solutions posted in this thread. HOWEVER ...

When using the forcing chain analogy, the trivial stream restricts the initial, assigned candidates to directly seeing the elimination candidates. You've created a non-trivial stream where the initial assignment indirectly sees the elimination candidates.

Code: Select all

3r8c5 - 3r8c9 = (3-15)r9c9


_

[bat999]

... You've created a non-trivial stream...

I undersand.
The forcing chain alalogy is cool.
Let's put that final term back.
AIC: (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 => - 1 r9c9, - 5 r9c9; stte

[daj95376]

... You've created a non-trivial stream...

I undersand.
The forcing chain alalogy is cool.
Let's put that final term back.
AIC: (3=6)r8c5 - (6=2)r3c5 - (2=7)r4c5 - (7=86)r48c8 - (6=15)r79c8 - (15=3)r9c9 => - 1 r9c9, - 5 r9c9; stte

Wow, this puzzle and its various chains/solutions won't die. The elimination should be: -3 r8c9 .

This is what happens when alternate viewpoints are intermingled.

_