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Puzzle 10/02/26 ___ BBDB as VH+ puzzle

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daj95376

Joined: 23 Aug 2008
Posts: 3854

Posted: Fri Feb 26, 2010 10:45 pm Post subject: Puzzle 10/02/26 ___ BBDB as VH+ puzzle

Code:

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

Play this puzzle online at the Daily Sudoku site

storm_norm

Joined: 18 Oct 2007
Posts: 1741

Posted: Sat Feb 27, 2010 3:06 am Post subject:

Code:

+----------------+----------------+------------+
| 4 59 6 | 59 1 7 | 3 2 8 |
| 8 7 29 | 469 3 269 | 1 49 5 |
| 3 1259 1259 | 459 8 259 | 7 49 6 |
+----------------+----------------+------------+
| 19 258 25 | 3 6 58 | 4 7 19 |
| 6 4 3 | 7 29 189 | 28 5 19 |
| 7 1589 159 | 1589 259 4 | 28 6 3 |
+----------------+----------------+------------+
| 19 19 7 | 68 4 68 | 5 3 2 |
| 5 6 8 | 2 7 3 | 9 1 4 |
| 2 3 4 | 159 59 159 | 6 8 7 |
+----------------+----------------+------------+

the following is to show that the 9 in r1c2 can't exist.
going to try and explain this so here goes.
the 49 UR in r23c48 shows that at least one of the other 9's in c4 has to be true.
so just looking at the upper most 9's, we see that they combine to form a finned x-wing in columns 3 and 4. this would eliminate the 9 in r1c2.

Code:

+-------+---------+--------+
| .-9 . |*9 . . | . . . |
| . .*9 | 469 . . | . 49 . |
| . .*9 | 459 . . | . 49 . |
+-------+---------+--------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . .*9 |*9 . . | . . . |
+-------+---------+--------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | . . . | . . . |
+-------+---------+--------+

if those 9's are false, this leaves the 9 in r9c4 lonely and needs somewhere to go. preferably around box 1... hint hint...
aha.. we see a 59 cell in r9c5
then take a look at the 5's in column 6. one of those 5's has to be true.
so via the 59 cell in r9c5 we get this diagram

Code:

+-----------+----------+--------+
| . 59 . | 59 . . | . . . |
| . . 29 | 469 . . | . 49 . |
| . . | 459 . 5 | . 49 . |
+-----------+----------+--------+
| . . 25 | . . 5 | . . . |
| . . . | . . . | . . . |
| . . | . . | . . . |
+-----------+----------+--------+
| . . . | . . . | . . . |
| . . . | . . . | . . . |
| . . . | 9 59 5 | . . . |
+-----------+----------+--------+

Code:

(9)r9c4 - (9=5)r9c5 - (5)r9c6
||
(5)r4c6 - (5=2)r4c3 - (2=9)r2c3
||
(5)r3c6 - (5)r1c4 = (5)r1c2

so all outs to the UR are accounted for and put it all together.
AUR elimination on the UR49 r23c48 ===> r1c2 <> 9

Code:

(9)r1c4
||
(9)r6c4 - (9)r4c3 = (9)r23c3
||
(9)r9c4 - (9=5)r9c5 - (5)r9c6
||
(5)r4c6 - (5=2)r4c3 - (2=9)r2c3
||
(5)r3c6 - (5)r1c4 = (5)r1c2

notice that after I draw it up in xsudo it finds some other eliminations including the fact that the UR 49 is typically used as a type 4 which would already eliminate the 9's in r23c4. again, the main idea was to eliminate the 9 in r1c2, which blasts a hole in it.

leaves w-wing {2,5} r4c6 <> 5 to clean up the mess.

tlanglet

Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

Posted: Sat Feb 27, 2010 3:48 pm Post subject:

Four steps

Quote:

xy-wing 2-59 with vertex in r2c3,
xyz-wing 459 inr3c4, which sets up a
xy-wing -258 with vertex in r4c6,
coloring 9

Ted

arkietech

Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA

Posted: Sat Feb 27, 2010 7:22 pm Post subject:

Three steps

Code:

xyz 459
w-wing 25
*--------------------------------------------------*
| 4 59 6 | 59 1 7 | 3 2 8 |
| 8 7 29 | 6 3 29 | 1 4 5 |
| 3 125 125 | 4 8 25 | 7 9 6 |
|----------------+----------------+----------------|
| 19 25 25 | 3 6 8 | 4 7 19 |
| 6 4 3 | 7 2 19 | 8 5 19 |
| 7 8 19 | 159 59 4 | 2 6 3 |
|----------------+----------------+----------------|
| 19 19 7 | 8 4 6 | 5 3 2 |
| 5 6 8 | 2 7 3 | 9 1 4 |
| 2 3 4 | 159 59 159 | 6 8 7 |
*--------------------------------------------------*
(9)r1c4 = r2c6 - (9=1)r5c6 - r6c4 = (1-9)r6c3 = r2c3; r1c2,r2c6<>9

Danny did I find another m-wing 19?

daj95376

Joined: 23 Aug 2008
Posts: 3854

Posted: Sat Feb 27, 2010 7:40 pm Post subject:

arkietech wrote:

Danny did I find another m-wing 19?

No, a gM-Wing and a W-Wing have three strong links and two values (for starters). Your chain has four strong links. Still, a nice find!

Regards, Danny

Last edited by daj95376 on Sun Feb 28, 2010 6:10 pm; edited 1 time in total

Marty R.

Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

Posted: Sun Feb 28, 2010 5:58 pm Post subject:

I used my typical hodgepodge of moves, including X, XYZ, UR, W, multi- and simple coloring.

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