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VH+ 111522

 
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immpy



Joined: 06 May 2017
Posts: 571
PostPosted: Tue Nov 15, 2022 7:39 pm    Post subject: VH+ 111522 Reply with quote
Hello all, enjoy the puzzle.

Code:

+-------+-------+-------+
| . 5 . | . 4 . | . 3 . |
| . . 1 | 3 7 . | . 5 2 |
| . . 3 | . 9 . | 6 . . |
+-------+-------+-------+
| . . . | 6 . . | . . . |
| . . . | 2 3 . | 1 8 5 |
| . . . | . 5 9 | . 2 . |
+-------+-------+-------+
| 6 3 . | . . . | 8 9 . |
| . . 9 | . . . | . . 4 |
| 8 . 4 | . . . | . . . |
+-------+-------+-------+

Play this puzzle online at the Daily Sudoku site

cheers...immp
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TomC



Joined: 30 Oct 2020
Posts: 358
Location: Wales
PostPosted: Wed Nov 16, 2022 4:27 pm    Post subject: Reply with quote
Remove <8> from r6c4 to solve

Code:

+--------------+------------------+-------------+
| 2   5     78 |  1    4     6    | 79  3   789 |
| 9   6     1  |  3    7     8    | 4   5   2   |
| 47  48    3  |  5    9     2    | 6   17  178 |
+--------------+------------------+-------------+
| 35  28    25 |  6   b18   c17   | 379 4   79  |
| 47  9     6  |  2    3     47   | 1   8   5   |
| 13  148  a78 | x478  5     9    | 37  2   6   |
+--------------+------------------+-------------+
| 6   3     25 |  47   12    1457 | 8   9   17  |
| 15  127   9  |  78   1268  3    | 25  167 4   |
| 8   127   4  |  9    126   157  | 25  167 3   |
+--------------+------------------+-------------+

Play this puzzle online at the Daily Sudoku site

If x=8, a=7, b=1, c=7 leads to no <7> in box 6

.
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dongrave



Joined: 06 Mar 2014
Posts: 568
PostPosted: Wed Nov 16, 2022 4:59 pm    Post subject: Reply with quote
My solution was similar to Tom's:
Code:
+--------------+----------------+-------------+
|  2   5    78 | 1    4    6    | 79  3   789 |
|  9   6    1  | 3    7    8    | 4   5   2   |
|  47  48   3  | 5    9    2    | 6   17  178 |
+--------------+----------------+-------------+
|  35  2-8  25 | 6   a18  b17   | 379 4   79  |
| d47  9    6  | 2    3   c47   | 1   8   5   |
|  13  148 e78 | 478  5    9    | 37  2   6   |
+--------------+----------------+-------------+
|  6   3    25 | 47   12   1457 | 8   9   17  |
|  15  127  9  | 78   1268 3    | 25  167 4   |
|  8   127  4  | 9    126  157  | 25  167 3   |
+--------------+----------------+-------------+

(8=1)r4c5-(1=7)r4c6-r5c6=r5c1-(7=8)r6c3 => r4c2 <> 8; stte.
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Clement



Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
PostPosted: Wed Nov 16, 2022 5:22 pm    Post subject: VH+ 111522 Reply with quote
Code:

+-----------------+---------------------+-------------------+
| 2     5      78 | 1      4       6    | 79     3      789 |
| 9     6      1  | 3      7       8    | 4      5      2   |
| 47    478    3  | 5      9       2    | 6      17     178 |
+-----------------+---------------------+-------------------+
| 35   a28    a25 | 6     a18     a17   | 379    4      79  |
| 47    9      6  | 2      3      d4-7  | 1      8      5   |
| 13    148    78 | 478    5       9    | 37     2      6   |
+-----------------+---------------------+-------------------+
| 6     3     b25 | 47     12     c1457 | 8      9      17  |
| 15    127    9  | 78     1268    3    | 25     167    4   |
| 8     127    4  | 9      126     157  | 25     167    3   |
+-----------------+---------------------+-------------------+

(7=1825)r4c6523 - (5)r7c3 = (5-4)r7c6 = (4)r5c6 => - 7r5c6; stte
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dongrave



Joined: 06 Mar 2014
Posts: 568
PostPosted: Sat Nov 19, 2022 9:57 pm    Post subject: Re: VH+ 111522 Reply with quote
Clement wrote:

(7=1825)r4c6523 - (5)r7c3 = (5-4)r7c6 = (4)r5c6 => - 7r5c6; stte


Hi Clement! Could you explain the logic behind your chain to me? I know that I'm missing something. I can see that if r4c6 is not 7, then r5c6 cannot be 7 - but I don't see pincers on the ends of the chain - and I don't see a contradiction. Can't r9c6 still be 7? Or r6c4 still be 7? Thanks in advance!
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Clement



Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
PostPosted: Sun Nov 20, 2022 2:01 pm    Post subject: Re: VH+ 111522 Reply with quote
dongrave wrote:
Clement wrote:

(7=1825)r4c6523 - (5)r7c3 = (5-4)r7c6 = (4)r5c6 => - 7r5c6; stte


Hi Clement! Could you explain the logic behind your chain to me? I know that I'm missing something. I can see that if r4c6 is not 7, then r5c6 cannot be 7 - but I don't see pincers on the ends of the chain - and I don't see a contradiction. Can't r9c6 still be 7? Or r6c4 still be 7? Thanks in advance!
Good questions Dongrave. If at all 7 was the solution in in r4c6 then, 7 cannot be in r5c6. Now we assume that 7 is not the solution in r4c6 i.e (7=1)r4c6 - (1=8)r4c5 - (8=2)r4c2 - (2=5)r4c3, I have combined these steps in one(step a), so, 5 cannot be in r7c3 it goes to eliminate the 4 in r7c6. Now, 4 has no where to go in in column 6 except r5c6 as they are strongly linked in the column i.e when we say 1 is the soluton in r4c6, 4 becomes the solution in r5c6, no chance for 7 to be in r5c6. We don't care whether 7 is the solution in other places.
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dongrave



Joined: 06 Mar 2014
Posts: 568
PostPosted: Sun Nov 20, 2022 5:28 pm    Post subject: Re: VH+ 111522 Reply with quote
Oh! Of course! Regardless of whether 7 is True or False in r4c6, it is False in r5c6! Thanks for the clarification Clement!
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