## Vanhegan extreme May 23, 2013

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### Vanhegan extreme May 23, 2013

Code: Select all
` *-----------* |..2|1.5|7..| |.6.|.3.|.2.| |3..|...|..5| |---+---+---| |9..|876|..2| |.4.|.5.|.9.| |8..|941|..7| |---+---+---| |2..|...|..4| |.1.|.9.|.5.| |..9|4.3|2..| *-----------*`

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dan

ArkieTech

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### Re: Vanhegan extreme May 23, 2013

Code: Select all
`*--------------------------------------------------------------*| 4     9     2      | 1     8     5      | 7     36    36     ||b15    6     158    | 7     3     4      |c18    2     9      || 3     78    178    | 6     2     9      | 148   148   5      ||--------------------+--------------------+--------------------|| 9     35    135    | 8     7     6      | 1345  14    2      ||a167   4     167    | 3     5     2      |d168   9    e68-1   || 8     2     356    | 9     4     1      | 356   36    7      ||--------------------+--------------------+--------------------|| 2     378   3678   | 5     16    78     | 9     178   4      || 67    1     4      | 2     9     78     | 368   5     368    || 567   578   9      | 4     16    3      | 2     178   168    |*--------------------------------------------------------------*`

S Wing: (1) r5c1 = r2c1 - (1=8) r2c7 - r5c7 = r5c9 => -1 r5c9; stte

Leren
Leren

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Joined: 03 June 2012

### Re: Vanhegan extreme May 23, 2013

(18=68)r5c79 = (6)r6c3 - r7c3 = (6-1)r7c5 = r7c8 = (no 1 column 9) => r5c79 = 18
what's the proper way to notate this ?

found an easier way...
(1)r4c3 = (6)r6c3 - r7c3 = (6-1)r7c5 = r7c8 - r9c9 = r5c9 => r5c13, r4c78 <> 1
7b53
2012 Supporter

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Location: New York

### Re: Vanhegan extreme May 23, 2013

7b53 wrote:(18=68)r5c79 = (6)r6c3 - r7c3 = (6-1)r7c5 = r7c8 = (no 1 column 9) => r5c79 = 18
what's the proper way to notate this ?

found an easier way...
(1)r4c3 = (6)r6c3 - r7c3 = (6-1)r7c5 = r7c8 - r9c9 = r5c9 => r5c13, r4c78 <> 1
Hmmm ... In fact : r5c79=16 or 18 or 68 ! Therefore, one could write :
Chain[6] : (18=6)r5c79-6r6c78=6r6c3-6r7c3=(6-1)r7c5=1r7c8-1r9c9=(18)r5c79 :=> r5c79=18
Or ... Chain[5] : 6r6c78=6r6c3-6r7c3=(6-1)r7c5=1r7c8-1r9c9=(1*-8)r5c9=8r5c7 :=> -6r5c79*(=18)
which is 1) easier to understand and 2) shorter than the "easier way" Chain[6].
JC Van Hay

Posts: 719
Joined: 22 May 2010

### Re: Vanhegan extreme May 23, 2013

thanks JC
JC Van Hay wrote:Hmmm ... In fact : r5c79=16 or 18 or 68 ! Therefore, one could write :
Chain[6] : (18=6)r5c79-6r6c78=6r6c3-6r7c3=(6-1)r7c5=1r7c8-1r9c9=(18)r5c79 :=> r5c79=18
Or ... Chain[5] : 6r6c78=6r6c3-6r7c3=(6-1)r7c5=1r7c8-1r9c9=(1*-8)r5c9=8r5c7 :=> -6r5c79*(=18)
which is 1) easier to understand and 2) shorter than the "easier way" Chain[6].

supposed you meant there's actually 3 combinations.

little request...mind skipping a space in between each node...

7b53
7b53
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### Re: Vanhegan extreme May 23, 2013

Code: Select all
`+--------------+---------+--------------+| 4   9   2    | 1 8  5  | 7    36  36  || 15  6   158  | 7 3  4  | 18   2   9   || 3   78  178  | 6 2  9  | 148  148 5   |+--------------+---------+--------------+| 9   35  135  | 8 7  6  | 1345 14  2   || 167 4   167  | 3 5  2  | 168  9   168 || 8   2   356  | 9 4  1  | 356  36  7   |+--------------+---------+--------------+| 2   378 3678 | 5 16 78 | 9    178 4   || 67  1   4    | 2 9  78 | 368  5   368 || 567 578 9    | 4 16 3  | 2    178 168 |+--------------+---------+--------------+`

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DP 14, r34c78. Either r3c78=8 or r4c7 is 35 pseudo cell. Common outcome: r79c8 both=78.
Marty R.

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