The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |.5.|..6|8..|
 |8..|...|92.|
 |7..|...|..6|
 |---+---+---|
 |...|9..|...|
 |.64|.25|..9|
 |5..|.1.|3..|
 |---+---+---|
 |...|8..|...|
 |.47|..1|...|
 |..3|.9.|..2|
 *-----------*


Play/Print this puzzle online

[Leren]

Code: Select all

*--------------------------------------------------------------*
| 4     5     9      | 2     b37   6      | 8     37    1      |
| 8     3     6      | 1     457   47     | 9     2     57     |
| 7     2     1      | 35    8     9      | 4     35    6      |
|--------------------+--------------------+--------------------|
| 1     7     8      | 9     46    3      | 2     56    45     |
| 3     6     4      | 7     2     5      | 1     8     9      |
| 5     9     2      | 46    1     8      | 3     67    47     |
|--------------------+--------------------+--------------------|
| 9     1     5      | 8    b67    2      | 67    4     3      |
| 2     4     7      | 35-6 b356   1      |a56    9     8      |
| 6     8     3      | 45    9     47     | 57    1     2      |
*--------------------------------------------------------------*

ALS XZ Rule: X = 5, Z = 6: (6=5) r8c7 - (5=6) r178c5 => - 6 r8c4; stte

Leren

[bat999]

Hi
I did this one too.
I arrived at the same pencilmarks as Leren.

Code: Select all

.---------.--------------.------------.
| 4  5  9 | 2    37   6  | 8   37  1  |
| 8  3  6 | 1    457  47 | 9   2   57 |
| 7  2  1 | 35   8    9  | 4   35  6  |
:---------+--------------+------------:
| 1  7  8 | 9    46   3  | 2   56  45 |
| 3  6  4 | 7    2    5  | 1   8   9  |
| 5  9  2 | 46   1    8  | 3   67  47 |
:---------+--------------+------------:
| 9  1  5 | 8    67   2  | 67  4   3  |
| 2  4  7 | 356  356  1  | 56  9   8  |
| 6  8  3 | 45   9    47 | 57  1   2  |
'---------'--------------'------------'

Then see what happens when the two squares in box 2 are not 3.
When r3c4 is not 3 it is 5 and r9c4 is 4 and r9c6 is 7.
When r1c5 is not 3 it is 7 and r7c5 is 6 and r8c4/c5 are 3 and 5... and r9c4 is 4 and r9c6 is 7 again.
After that things fell into place.

[Leren]

bat999 wrote:

Then see what happens when the two squares in box 2 are not 3.
When r3c4 is not 3 it is 5 and r9c4 is 4 and r9c6 is 7.
When r1c5 is not 3 it is 7 and r7c5 is 6 and r8c4/c5 are 3 and 5... and r9c4 is 4 and r9c6 is 7 again.

You can write all this as a discontinuous loop : (4=5) r9c4 - (5=36) r8c45 - (6=7) r7c5 - (7=3) r1c5 - (3=5) r3c4 - (5=4) r9c4 => r9c4 = 4.

Leren

[SteveG48]

Code: Select all

 *--------------------------------------------------*
 | 4    5    9    | 2    37   6    | 8    37   1    |
 | 8    3    6    | 1    457  47   | 9    2    7-5  |
 | 7    2    1    |b35   8    9    | 4   a35   6    |
 *----------------+----------------+----------------|
 | 1    7    8    | 9    46   3    | 2    6-5 f45   |
 | 3    6    4    | 7    2    5    | 1    8    9    |
 | 5    9    2    |d46   1    8    | 3    67  e47   |
 *----------------+----------------+----------------|
 | 9    1    5    | 8    67   2    | 67   4    3    |
 | 2    4    7    | 356  356  1    | 56   9    8    |
 | 6    8    3    |c45   9    47   | 57   1    2    |
 *--------------------------------------------------*


(5)r3c8 = r3c4 - (5=4)r9c4 - r6c4 = r6c9 - (4=5)r4c9 => -5 r2c9,r4c8 ; stte

[bat999]

You can write all this as a discontinuous loop : (4=5) r9c4 - (5=36) r8c45 - (6=7) r7c5 - (7=3) r1c5 - (3=5) r3c4 - (5=4) r9c4 => r9c4 = 4.

Hi
I understand this bit "=> r9c4 = 4", but the rest is difficult for my lizard brain.

And SteveG48's reply...
"(5)r3c8 = r3c4 - (5=4)r9c4 - r6c4 = r6c9 - (4=5)r4c9 => -5 r2c9,r4c8 ; stte"

I think this means...

If r3c8 is 5 then r2c9 and r4c8 are both not 5.
If r3c8(a) is 3 then r3c4(b) is 5 and r9c4(c) is 4 and r6c4(d) is 6 and r6c9(e) is 4 and r4c9(f) is 5. Again r2c9 and r4c8 are both not 5.

[Ngisa]

Code: Select all

+-------+------------+----------+
| 4 5 9 | 2   7-3  6  | 8  h37 1  |
| 8 3 6 | 1   457 47 | 9   2  g57 |
| 7 2 1 | a35  8   9  | 4  35 6  |
+-------+------------+----------+
| 1 7 8 | 9   e46  3  | 2  56 f45 |
| 3 6 4 | 7   2   5  | 1  8  9  |
| 5 9 2 | 46  1   8  | 3  67 47 |
+-------+------------+----------+
| 9 1 5 | 8   d67  2  | 67 4  3  |
| 2 4 7 | 356 356 1  | 56 9  8  |
| 6 8 3 | b45  9   c47 | 57 1  2  |
+-------+------------+----------+


XY-Chain
(3=5)r3c4-(5=4)r9c4-(4=7)r9c6-(7=6)r7c5-(6=4)r4c5-(4=5)r4c9-(5=7)r2c9-(7=3)r1c8 => -3r1c5; stte

[Leren]

bat999 wrote: I understand this bit "=> r9c4 = 4", but the rest is difficult for my lizard brain.

OK, I'll explain the discontinuous loop in words - maybe that will help you:

(4=5) r9c4 - (5=36) r8c45 - (6=7) r7c5 - (7=3) r1c5 - (3=5) r3c4 - (5=4) r9c4 => r9c4 = 4

Suppose r9c4 is not 4. Then it must be 5. (That's the meaning of the = sign in the first term. X = Y means if X is False Y must be True - that's called a Strong link between X and Y)

So r8c45 are both not 5 (That's the meaning of the - sign between the first and second terms. X - Y means if X is True Y must be False - that's called a Weak link between X and Y)

Since r8c45 are both not 5 they must both contain only 36 (They are a Naked Pair so both 3 and 6 must be located in these cells).

So r7c5 is not 6 so it must be 7. So r1c5 is not 7 so it must be 3. So r3c4 is not 3 so it must be 5. So r9c4 is not 5 so it must be 4.

The net result of all this is that if you assume that r9c4 is not 4 you conclude that r9c4 = 4. This is a contradiction, so the original assumption (r9c4 is not 4) must be False. So r9c4 must be 4.

To avoid any confusion the last = sign in => r9c4 = 4 is not a Strong link, it's just a normal equal sign indicating the conclusion.

Hope this helps.

Leren

[daj95376]

Since no one chose to post the BUG solution, ...

Code: Select all

 BUG+3
 +-----------------------------------------------------+
 |  4    5    9    |  2    37   6    |  8    37   1    |
 |  8    3    6    |  1    45+7 47   |  9    2    57   |
 |  7    2    1    |  35   8    9    |  4    35   6    |
 |-----------------+-----------------+-----------------|
 |  1    7    8    |  9    4-6  3    |  2    56   45   |
 |  3    6    4    |  7    2    5    |  1    8    9    |
 |  5    9    2    |  46   1    8    |  3    67   47   |
 |-----------------+-----------------+-----------------|
 |  9    1    5    |  8    67   2    |  67   4    3    |
 |  2    4    7    |  36+5 35+6 1    |  56   9    8    |
 |  6    8    3    |  45   9    47   |  57   1    2    |
 +-----------------------------------------------------+
 # 24 eliminations remain

 6r8c5                         - 6r4c5
 7r2c5 - (7=6)r7c5             - 6r4c5
 5r8c4 - (5=4)r9c4 - (4=6)r6c4 - 6r4c5


I took a long time to accept ALS logic. Here's an example why.

If you perform the first ALS, you get: -6r78c5 -> r7c5=7, r1c5=3, r8c5=5

If you overlay it on the second ALS, you get: =5r8c5 -> r9c4=4, r9c6=7, r7c5=6

Endpoint logic performs the eliminations on 6, but no reference is made about the contradicting assignments to 7 in [box 8].

Code: Select all

 +------------------------------------------------------+
 |  4    5    9    |  2    a37   6    |  8    37   1    |
 |  8    3    6    |  1     457  47   |  9    2    57   |
 |  7    2    1    |  35    8    9    |  4    35   6    |
 |-----------------+------------------+-----------------|
 |  1    7    8    |  9     46   3    |  2    56   45   |
 |  3    6    4    |  7     2    5    |  1    8    9    |
 |  5    9    2    |  46    1    8    |  3    67   47   |
 |-----------------+------------------+-----------------|
 |  9    1    5    |  8   ab67   2    |  67   4    3    |
 |  2    4    7    |  356 ab356  1    |  56   9    8    |
 |  6    8    3    | b45    9   b47   |  57   1    2    |
 +------------------------------------------------------+
 # 24 eliminations remain

 (6=375)r178c5 - (3=4576)r78c5,r9c46  =>  -6 r4c5,r8c4


_

[SteveG48]

And SteveG48's reply...
"(5)r3c8 = r3c4 - (5=4)r9c4 - r6c4 = r6c9 - (4=5)r4c9 => -5 r2c9,r4c8 ; stte"

I think this means...

If r3c8 is 5 then r2c9 and r4c8 are both not 5.
If r3c8(a) is 3 then r3c4(b) is 5 and r9c4(c) is 4 and r6c4(d) is 6 and r6c9(e) is 4 and r4c9(f) is 5. Again r2c9 and r4c8 are both not 5.

That's correct. A summary in words would be "if r3c8 is not a 5 then r4c9 must be a 5, so either way the eliminations take place". Technically, there's a strong link between the 5's in r3c8 and r4c9.

[Leren]

daj95376 wrote : Endpoint logic performs the eliminations on 6, but no reference is made about the contradicting assignments to 7 in [box 8].

Hi Danny, this seemed odd until I realised that:

1. r4c5 = 6 => r7c5 = 7, r1c5 = 3, r8c5 = 5, r9c5 = 4, r9c6 = 7. 2 7's in Box 8 => r4c5 <> 6.

2. r8c4 = 6 => r7c5 = 7, r1c5 = 3, r8c5 = 5, r9c5 = 4, r9c6 = 7. 2 7's in Box 8 => r8c4 <> 6.

So the ALS eliminations reflect an underlying conflict on 7 in Box 8 (if any of the ALS eliminations were True) without being explicit about it. Nothing wrong with that as far as I can see.

Leren

[Leren]

daj95376 wrote:

I took a long time to accept ALS logic. Here's an example why.
If you perform the first ALS, you get: -6r78c5 -> r7c5=7, r1c5=3, r8c5=5
If you overlay it on the second ALS, you get: =5r8c5 -> r9c4=4, r9c6=7, r7c5=6
Endpoint logic performs the eliminations on 6, but no reference is made about the contradicting assignments to 7 in [box 8].

Code: Select all

 +------------------------------------------------------+
 |  4    5    9    |  2    a37   6    |  8    37   1    |
 |  8    3    6    |  1     457  47   |  9    2    57   |
 |  7    2    1    |  35    8    9    |  4    35   6    |
 |-----------------+------------------+-----------------|
 |  1    7    8    |  9     46   3    |  2    56   45   |
 |  3    6    4    |  7     2    5    |  1    8    9    |
 |  5    9    2    |  46    1    8    |  3    67   47   |
 |-----------------+------------------+-----------------|
 |  9    1    5    |  8   ab67   2    |  67   4    3    |
 |  2    4    7    |  356 ab356  1    |  56   9    8    |
 |  6    8    3    | b45    9   b47   |  57   1    2    |
 +------------------------------------------------------+
 # 24 eliminations remain

 (6=375)r178c5 - (3=4576)r78c5,r9c46  =>  -6 r4c5,r8c4

Hi Danny. The more I look at this the less I am inclined to think of it as an ALS move, although the eliminations are valid.

If you assume r78c5 <> 6 in the a cells then r7c5 = 7, r1c5 = 3, r8c5 = 5, and r9c46 are both 4. Since this is a contradiction then one of r78c5 must be 6.

I suppose that you could write this as (6=375)r178c5 - (36=457)r78c5,r9c46 => -6 r4c5,r8c4 but I'd call it a contradiction chain as opposed to an ALS move.

If you assume r78c5 <> 6 in the b cells then r7c5 = 7, r9c6 = 4, r9c4 = 5 and r18c5 are both 3. Again since this is a contradiction then one of r78c5 must be 6.

You could write this as (6=3457)r78c5,r9c46 - (36=75) r178c5 => -6 r4c5,r8c4.

Leren

[daj95376]

Hello Leren,

I accept my chain as valid even though there are annoying internal contradictions. That's sometimes a fact-of-life when an AIC starts with a false assumption. However, it does demonstrate that the full logic isn't always needed if one stops when an internal contradiction appears. A practice that has been frowned upon for a long time. For this grid, I get:

Code: Select all

6r78c5 = ( 7*r7c5, 3r1c5, 5*r8c5 ) - *(75=4)r9c46 (contradiction)  =>  =6 r78c5


Of course, the smartest choice is to use any one of the numerous non-ALS chains that crack the puzzle. Odds are they won't contain an internal contradiction. My choice would be either of these overlapping XY-Chains.

Code: Select all

 (4=7)r2c6 - (7=3)r1c5 - (3=5)r3c4 - (5=4)r9c4              =>  -4 r9c6        (sufficient by itself)
             (7=3)r1c5 - (3=5)r3c4 - (5=4)r9c4 - (4=7)r9c6  =>  -7 r2c6,r7c5   (sufficient by itself)


They add r2c6,r3c4 while discarding r78c5 from the ALS terms. Same number of cells, with simplicity at the expense of SL count.

Maybe simplicity should be valued higher that SL count !!!

_

[JC Van Hay]

I took a long time to accept ALS logic. Here's an example why.

If you perform the first ALS, you get: -6r78c5 -> r7c5=7, r1c5=3, r8c5=5

If you overlay it on the second ALS, you get: =5r8c5 -> r9c4=4, r9c6=7, r7c5=6

Endpoint logic performs the eliminations on 6, but no reference is made about the contradicting assignments to 7 in [box 8].

Code: Select all

 +------------------------------------------------------+
 |  4    5    9    |  2    a37   6    |  8    37   1    |
 |  8    3    6    |  1     457  47   |  9    2    57   |
 |  7    2    1    |  35    8    9    |  4    35   6    |
 |-----------------+------------------+-----------------|
 |  1    7    8    |  9     46   3    |  2    56   45   |
 |  3    6    4    |  7     2    5    |  1    8    9    |
 |  5    9    2    |  46    1    8    |  3    67   47   |
 |-----------------+------------------+-----------------|
 |  9    1    5    |  8   ab67   2    |  67   4    3    |
 |  2    4    7    |  356 ab356  1    |  56   9    8    |
 |  6    8    3    | b45    9   b47   |  57   1    2    |
 +------------------------------------------------------+
 # 24 eliminations remain

 (6=375)r178c5 - (3=4576)r78c5,r9c46  =>  -6 r4c5,r8c4


_

Internal contradiction in a chain should be avoided as a result of a correct analysis !

Code: Select all

+---------+-------------------+------------+
| 4  5  9 | 2     (37)   6    | 8   37  1  |
| 8  3  6 | 1     457    47   | 9   2   57 |
| 7  2  1 | 35    8      9    | 4   35  6  |
+---------+-------------------+------------+
| 1  7  8 | 9     4-6    3    | 2   56  45 |
| 3  6  4 | 7     2      5    | 1   8   9  |
| 5  9  2 | 46    1      8    | 3   67  47 |
+---------+-------------------+------------+
| 9  1  5 | 8     (67)   2    | 67  4   3  |
| 2  4  7 | 35-6  (356)  1    | 56  9   8  |
| 6  8  3 | (45)  9      (47) | 57  1   2  |
+---------+-------------------+------------+


The 5 cells r178c5.r9c46 exclude 6r4c5.r8c4 because each of their 6 solutions contain either 6r7c5 or 6r8c5.

Example of a proof:
r9c4=4 -> r9c6=7, r7c5=6, r18c5=ANP(357) : 3 solutions
or
r9c4=5 -> r178c5=NT(367), r9c6=ANS(47) : 3 solutions
Therefore, either r7c5=6 or r78c5=6 :=> -6r4c5.r8c4

A chain interpretation of the proof :
NT(367)r178c5=5r8c5 - (5=4)r9c4 - (4=7)r9c6 - (7=6)r7c5 :=> -6r4c5.r8c4
or the following ALS-Chain ...
(637=5)r178c5 - (5=476)r9c46.r7c5 :=> -6r4c5.r8c4
or (6==5)r178c5 - (5==6)r9c46.r7c5 :=> -6r4c5.r8c4

The same analysis can be done from any of the 4 other cells.

From r8c5, one would get a different chain :
NQ(4567)r78c5.r9c46=3r8c5 - (3=7)r1c5 - (7=6)r7c5 :=> -6r4c5.r8c4
or the following ALS-Chain ...
(6457=3)r78c5.r9c46 - (3=76)r17c5 :=> -6r4c5.r8c4
or (6==3)r78c5.r9c46 - (3==6)r17c5 :=> -6r4c5.r8c4

[Leren]

JC Van Hay wrote:
or the following ALS-Chain ...
(637=5)r178c5 - (5=476)r9c46.r7c5 :=> -6r4c5.r8c4
or (6==5)r178c5 - (5==6)r9c46.r7c5 :=> -6r4c5.r8c4

or the following ALS-Chain ...
(6457=3)r78c5.r9c46 - (3=76)r17c5 :=> -6r4c5.r8c4
or (6==3)r78c5.r9c46 - (3==6)r17c5 :=> -6r4c5.r8c4

Now that's more like it. In both cases the restricted common digit (5) and (3) cells do not overlap.

Here's another ALS chain for the same eliminations where both the pincer digit (6) cells and the restricted common digit (5) cells do not overlap.

Code: Select all

*--------------------------------------------------------------*
| 4     5     9      | 2    b37    6      | 8     37    1      |
| 8     3     6      | 1     457   47     | 9     2     57     |
| 7     2     1      | 35    8     9      | 4     35    6      |
|--------------------+--------------------+--------------------|
| 1     7     8      | 9     4-6   3      | 2     56    45     |
| 3     6     4      | 7     2     5      | 1     8     9      |
| 5     9     2      |a46    1     8      | 3     67    47     |
|--------------------+--------------------+--------------------|
| 9     1     5      | 8    b67    2      | 67    4     3      |
| 2     4     7      | 35-6 b356   1      | 56    9     8      |
| 6     8     3      |a45    9     47     | 57    1     2      |
*--------------------------------------------------------------*

ALS XZ Rule: X = 5, Z = 6: (6=5) r69c4 - (5=6) r178c5 => - 6 r4c5, r8c4

Leren