The New Sudoku Players' Forum

[ArkieTech]

Code: Select all

 *-----------*
 |1.7|...|.3.|
 |.5.|4..|..6|
 |4..|...|5..|
 |---+---+---|
 |.46|1.3|..5|
 |..5|2.9|7..|
 |9..|5.6|24.|
 |---+---+---|
 |..8|...|..7|
 |7..|..2|.1.|
 |.2.|...|8.4|
 *-----------*


Play/Print this puzzle online

[SteveG48]

Code: Select all

 *--------------------------------------------------------------------*
 | 1     e689    7      | 69     269    5      | 4      3      28     |
 |b3-8    5      239    | 4      239   c78     | 1      79     6      |
 | 4     d689   d239    |d37     12369 c178    | 5      79    d28     |
 *----------------------+----------------------+----------------------|
 | 2      4      6      | 1      7      3      | 9      8      5      |
 |a38     18     5      | 2      4      9      | 7      6      13     |
 | 9      7      13     | 5      8      6      | 2      4      13     |
 *----------------------+----------------------+----------------------|
 | 5      19     8      | 69     169    4      | 3      2      7      |
 | 7      3      4      | 8      5      2      | 6      1      9      |
 | 6      2      19     | 37     139   c17     | 8      5      4      |
 *--------------------------------------------------------------------*


The logic of this one is simple, but I'm bothered by the notation with 8 on both sides of the = sign in term 3. Comments?

8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r2c1 ; stte

[Leren]

Code: Select all

*-----------------------------------------------------------------------*
| 1     a689    7       |a69    a269    5       | 4      3      28      |
| 3-8    5      239     | 4     S239   b78c     | 1      79c    6       |
| 4      689    239     |b37     12369  178     | 5      79     28      |
|-----------------------+-----------------------+-----------------------|
| 2      4      6       | 1      7      3       | 9      8      5       |
| 38     18     5       | 2      4      9       | 7      6      13      |
| 9      7      13      | 5      8      6       | 2      4      13      |
|-----------------------+-----------------------+-----------------------|
| 5      19     8       | 69     169    4       | 3      2      7       |
| 7      3      4       | 8      5      2       | 6      1      9       |
| 6      2      19      | 37     139    17      | 8      5      4       |
*-----------------------------------------------------------------------*

3 Petal Death Blossom: Stem Cell r2c5 {239}; 

8 r2c1 - (8=2) r1c245     - (2) r2c5;

8 r2c1 - (8=3) r2c6, r3c4 - (3) r2c5;

8 r2c1 - (8=9) r2c68      - (9) r2c5; => - 8 r2c1; stte

Leren

[JC Van Hay]

Code: Select all

+--------------+--------------------+---------------+
| 1   689  7   | 69    69(2)  5     | 4  3     (28) |
| 38  5    239 | 4     (239)  -7(8) | 1  (79)  6    |
| 4   689  239 | (37)  12369  17(8) | 5  79    2(8) |
+--------------+--------------------+---------------+
| 2   4    6   | 1     7      3     | 9  8     5    |
| 38  18   5   | 2     4      9     | 7  6     13   |
| 9   7    13  | 5     8      6     | 2  4     13   |
+--------------+--------------------+---------------+
| 5   19   8   | 69    169    4     | 3  2     7    |
| 7   3    4   | 8     5      2     | 6  1     9    |
| 6   2    19  | 37    139    17    | 8  5     4    |
+--------------+--------------------+---------------+
[XYWing(379)r2c58.r3c4=2r2c5-2r1c5=(2-8)r1c9=8r3c9-8r3c6=8r2c6]-(7=8)r2c6; stte

[Marty R.]

Code: Select all

+------------+--------------+---------+
| 1  689 7   | 69 269   5   | 4 3  28 |
| 38 5   239 | 4  239   78  | 1 79 6  |
| 4  689 239 | 37 12369 178 | 5 79 28 |
+------------+--------------+---------+
| 2  4   6   | 1  7     3   | 9 8  5  |
| 38 18  5   | 2  4     9   | 7 6  13 |
| 9  7   13  | 5  8     6   | 2 4  13 |
+------------+--------------+---------+
| 5  19  8   | 69 169   4   | 3 2  7  |
| 7  3   4   | 8  5     2   | 6 1  9  |
| 6  2   19  | 37 139   17  | 8 5  4  |
+------------+--------------+---------+


Play this puzzle online at the Daily Sudoku site

Simple Coloring

8r3c2-r3c9=r1c9-8r1c2=> -8 r5c2

[Marty R.]

The logic of this one is simple, but I'm bothered by the notation with 8 on both sides of the = sign in term 3. Comments?

Steve, when I knew even less about ALS's than I do now (if that's possible), I learned (probably from Don M.) that the lead-in to an ALS (8 in this case) has to see all instances of that number in the ALS.

[pjb]

Code: Select all

 1       689     7      | 69     269    5      | 4      3      28     
 38      5       239    | 4      239  aA78     | 1      79     6     
 4       689    c239    |b37     12369  17-8   | 5      79    d28     
------------------------+----------------------+---------------------
 2       4       6      | 1      7      3      | 9      8      5     
 38      18      5      | 2      4      9      | 7      6      13     
 9       7       13     | 5      8      6      | 2      4      13     
------------------------+----------------------+---------------------
 5       19      8      | 69     169    4      | 3      2      7     
 7       3       4      | 8      5      2      | 6      1      9     
 6       2      C19     | 37     139   B17     | 8      5      4     

(8=7)r2c6 - (7=3)r3c4 - (39=2)r3c3 - (2=8)r3c9 => -8 r3c6; stte
       \                      /
        (7=1)r9c6 - (1=9)r9c3


Phil

[pjb]

Steve
I share your bother. The third term is the naked triple 178, and r3c6 reduces to 8 alone if r2c6 <> 8.


how about

Code: Select all

8r5c1 = r2c1 - (8=7)r2c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r2c1 ; stte
                     \\     /
                     (8)r3c6

or 8r5c1 = r2c1 - (8=7)r2c6* - (7=18)r39c6 - (78=2369)r3c2349* - (69=8)r1c2 => -8 r2c1 ; stte


Phil

[SteveG48]

Steve
I share your bother. The third term is the naked triple 178, and r3c6 reduces to 8 alone if r2c6 <> 8.
Phil

Exactly, and r2c6 and r2c9 reduce to 7 and 1 respectively. The logic is that we have a locked set (not an ALS) before and after eliminating the 8 at r2c6, but we now know the exact location of each candidate. What is a better way to write the logic??

[SteveG48]

The logic of this one is simple, but I'm bothered by the notation with 8 on both sides of the = sign in term 3. Comments?

Steve, when I knew even less about ALS's than I do now (if that's possible), I learned (probably from Don M.) that the lead-in to an ALS (8 in this case) has to see all instances of that number in the ALS.

Usually, yes. But I don't have an ALS here. The set is already locked and I'm wanting to establish that the 7 and 8 are both in box 2. How to properly write it is the problem.

[DonM]

Steve, it's too late at night for me to look into the chain, but for the moment,isn't there a typo?

8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r2c1 ; stte

Should be:
8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r5c2 ; stte

[JC Van Hay]

The logic of this one is simple, but I'm bothered by the notation with 8 on both sides of the = sign in term 3. Comments?

8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r2c1 ; stte

Code: Select all

]+-------------------+--------------------+-------------+
| 1    (689)  7     | 69    269    5     | 4  3   28   |
| 3-8  5      239   | 4     239    (78)  | 1  79  6    |
| 4    (689)  (239) | (37)  12369  (178) | 5  79  (28) |
+-------------------+--------------------+-------------+
| 2    4      6     | 1     7      3     | 9  8   5    |
| 38   18     5     | 2     4      9     | 7  6   13   |
| 9    7      13    | 5     8      6     | 2  4   13   |
+-------------------+--------------------+-------------+
| 5    19     8     | 69    169    4     | 3  2   7    |
| 7    3      4     | 8     5      2     | 6  1   9    |
| 6    2      19    | 37    139    (17)  | 8  5   4    |
+-------------------+--------------------+-------------+


(68=9)r13c2 - 9r3c3=*XY-Chain[(8=7)r2c6-(7=*238)r3c349] - (8=17)r39c6 - (7=8)r2c6 :=> -8r2c1; stte

[DonM]

Code: Select all

 *--------------------------------------------------------------------*
 | 1     e689    7      | 69     269    5      | 4      3      28     |
 |b3-8    5      239    | 4      239   c78     | 1      79     6      |
 | 4     d689   d239    |d37     12369 c178    | 5      79    d28     |
 *----------------------+----------------------+----------------------|
 | 2      4      6      | 1      7      3      | 9      8      5      |
 |a38     18     5      | 2      4      9      | 7      6      13     |
 | 9      7      13     | 5      8      6      | 2      4      13     |
 *----------------------+----------------------+----------------------|
 | 5      19     8      | 69     169    4      | 3      2      7      |
 | 7      3      4      | 8      5      2      | 6      1      9      |
 | 6      2      19     | 37     139   c17     | 8      5      4      |
 *--------------------------------------------------------------------*


The logic of this one is simple, but I'm bothered by the notation with 8 on both sides of the = sign in term 3. Comments?

8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r2c1 ; stte

Exactly, and r2c6 and r2c9 reduce to 7 and 1 respectively. The logic is that we have a locked set (not an ALS) before and after eliminating the 8 at r2c6, but we now know the exact location of each candidate. What is a better way to write the logic??

What's troubling me about the notation is that since r2c6 and (I think you meant) r9c6 would reduce to 7 and 1 respectively and, thus, 8 would be in r3c6, how can one use (78=2369)r3c2349 which implies that (78) is locked in r3c6? Not to mention that there is a simple (8)r2c6=r3c6-r3c9=r1c9-(8)r1c2 which conflicts with the logic from using the ALS above. I guess I question the validity of the ALS to begin with. But I could be missing something.

[SteveG48]

Steve, it's too late at night for me to look into the chain, but for the moment,isn't there a typo?

8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r2c1 ; stte

Should be:
8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r5c2 ; stte

Thanks, Don. Actually, it should be both. I missed one deletion.

[SteveG48]

Code: Select all

 *--------------------------------------------------------------------*
 | 1     e689    7      | 69     269    5      | 4      3      28     |
 |b3-8    5      239    | 4      239   c78     | 1      79     6      |
 | 4     d689   d239    |d37     12369 c178    | 5      79    d28     |
 *----------------------+----------------------+----------------------|
 | 2      4      6      | 1      7      3      | 9      8      5      |
 |a38     18     5      | 2      4      9      | 7      6      13     |
 | 9      7      13     | 5      8      6      | 2      4      13     |
 *----------------------+----------------------+----------------------|
 | 5      19     8      | 69     169    4      | 3      2      7      |
 | 7      3      4      | 8      5      2      | 6      1      9      |
 | 6      2      19     | 37     139   c17     | 8      5      4      |
 *--------------------------------------------------------------------*


The logic of this one is simple, but I'm bothered by the notation with 8 on both sides of the = sign in term 3. Comments?

8r5c1 = r2c1 - (8=178)r239c6 - (78=2369)r3c2349 - (69=8)r1c2 => -8 r2c1 ; stte

Exactly, and r2c6 and r2c9 reduce to 7 and 1 respectively. The logic is that we have a locked set (not an ALS) before and after eliminating the 8 at r2c6, but we now know the exact location of each candidate. What is a better way to write the logic??

What's troubling me about the notation is that since r2c6 and (I think you meant) r9c6 would reduce to 7 and 1 respectively and, thus, 8 would be in r3c6, how can one use (78=2369)r3c2349 which implies that (78) is locked in r3c6? Not to mention that there is a simple (8)r2c6=r3c6-r3c9=r1c9-(8)r1c2 which conflicts with the logic from using the ALS above. I guess I question the validity of the ALS to begin with. But I could be missing something.

You're right, I meant r9c6, not r2c9.

For the rest, if r2c6 is a 7 and r3c6 is an 8, then we eliminate 7 everywhere in box 2 and 8 everywhere on row 3. That locks the set r3c2349 as 2369, with 69 in box 1. That gives the desired result.

Again, what troubles me is the use of the r239c6 set which is already locked, and which requires the 8 on both sides of the = sign in term 3. In retrospect, I would also have re-written the entire chain as:

8r5c1 = r2c1 - (8=178)r239c6 - (78=239)r3c349 - (9=68)r13c2

But that's another issue.