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immpy
Joined: 06 May 2017
Posts: 571
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 Posted: Sun Apr 03, 2022 5:36 pm Post subject: VH+ 040322 |
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Hello all, enjoy the puzzle.
| Code: |
+-------+-------+-------+
| . . 6 | . 2 . | 7 . 3 |
| . . 8 | 5 . . | . . . |
| 4 2 . | . . . | . . 6 |
+-------+-------+-------+
| . 5 . | . 3 2 | . . . |
| 1 . . | 6 . 4 | . . 9 |
| . . . | 7 9 . | . 6 . |
+-------+-------+-------+
| 8 . . | . . . | . 4 1 |
| . . . | . . 5 | 6 . . |
| 7 . 3 | . 4 . | 9 . . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
cheers...immp |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Mon Apr 04, 2022 3:55 pm Post subject: |
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Thanks for the puzzle immpy! Here's the grid after basics:
| Code: | +-------------+------------------+------------------+
| 5 19 6 | 4 2 189 | 7 189 3 |
| 39 37 8 | 5 167 1679 | 124 129 24 |
| 4 2 17 | 1389 178 13789 | 158 1589 6 |
+-------------+------------------+------------------+
| 6 5 9 | 18 3 2 | 148 178 478 |
| 1 378 27 | 6 5 4 | 238 238 9 |
|g23 f38 4 | 7 9 e18 | 12358 6 258 |
+-------------+------------------+------------------+
| 8 i69 5 | 239 67 3679 | 23 4 1 |
|h29 4 b12 | 1389 c18 5 | 6 378 78 |
| 7 a16* 3 | 128 4 d168 | 9 258 258 |
+-------------+------------------+------------------+
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And then I used the following chain: (6*-1)r9c2=r8c3-(1=8)r8c5-(6*8=1)r9c6-(1=8)r6c6-(8=3)r6c2-(3=2)r6c1-(2=9)r8c1-(9=6)r7c2 contradiction => r9c2 <> 6; stte. |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Thu Apr 07, 2022 1:26 pm Post subject: VH + 040322 |
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| Code: |
+-------------------+------------------------+------------------------+
| 5 19 6 | 4 2 189 | 7 189 3 |
|e39 d37 8 | 5 cb167 cb1679 | 124 129 24 |
| 4 2 17 | 1389 178 13789 | 158 1589 6 |
+-------------------+------------------------+------------------------+
| 6 5 9 | 18 3 2 | 148 178 478 |
| 1 378 27 | 6 5 4 | 238 238 9 |
| 23 38 4 | 7 9 18 | 12358 6 258 |
+-------------------+------------------------+------------------------+
| 8 ga6-9 5 | 239 b67 b3679 | 23 4 1 |
|f29 4 12 | 1389 18 5 | 6 378 78 |
| 7 16 3 | 128 4 168 | 9 258 258 |
+-------------------+------------------------+------------------------+
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(6)r7c2 = Type 4 UR(67)r27c56 – (7)r2c56 = (7-3)r2c2 = (3-9)r2c1 = r8c1 – (9)r7c2 => - 9r7c2; stte |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Fri Apr 08, 2022 2:22 pm Post subject: |
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| Very nice Clement. I like the way that your loop causes then pirouettes into the UR and back round again. |
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Clement
Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
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| Posted: Fri Apr 08, 2022 4:10 pm Post subject: |
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| Mogulmeister wrote: | | Very nice Clement. I like the way that your loop causes then pirouettes into the UR and back round again. |
Thanks. The UR kept appearing but, took me time to realize that it was useful. |
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dongrave
Joined: 06 Mar 2014
Posts: 568
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| Posted: Sat Apr 09, 2022 1:08 am Post subject: |
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| Clement wrote: | | Mogulmeister wrote: | | Very nice Clement. I like the way that your loop causes then pirouettes into the UR and back round again. |
Thanks. The UR kept appearing but, took me time to realize that it was useful. |
Clement! Thanks to you I learned the 6 different types of UR's today! Please keep on posting! |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Tue Apr 12, 2022 10:00 am Post subject: |
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I started looking at this with a contradictory hat on and observed that if you said that r6c1 = 3 then all manner of bad things happen. Using Don's grid (and lettering)
r2c1 = 9 r1c2 = 1 then a = 6 (r9c2) and b = 1 (r8c3) so c = 8 (r8c5) and d < > 6
at the same time
f = 8 (r6c2) and e = 1 (r6c6) and d also < > 1
So  c = 8 and d = 8 and we now have two 8's in box 8 impossibility so r6c1 < > 3 and puzzle solved. |
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Pat
Joined: 23 Feb 2010
Posts: 207
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| Posted: Thu Apr 14, 2022 5:22 am Post subject: |
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| Code: | +-------------+------------------+------------------+
| 5 19 6 | 4 2 189 | 7 189 3 |
| 39 37 8 | 5 167 1679 | 124 129 24 |
| 4 2 17 | 1389 178 13789 | 158 1589 6 |
+-------------+------------------+------------------+
| 6 5 9 | 18 3 2 | 148 178 478 |
| 1 378 27 | 6 5 4 | 238 238 9 |
| 23 38 4 | 7 9 18 | 12358 6 258 |
+-------------+------------------+------------------+
| 8 69 5 | 239 67 3679 | 23 4 1 |
| 29 4 12 | 1389 18 5 | 6 378 78 |
| 7 16 3 | 128 4 168 | 9 258 258 |
+-------------+------------------+------------------+
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yes, seems it needs split attention:
if r8c1 = 2
then
r8c3 = 1
then two things happen:
combined → r9c6 = 1, r6c6 = 8, r6c2 = 3, r6c1 = 2, conflict in c1 |
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Mogulmeister
Joined: 03 May 2007
Posts: 1151
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| Posted: Sun Apr 17, 2022 11:42 am Post subject: |
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Absolutely Pat - when you make r6c1 = 3 which makes r8c1 = 2 as you say, we need to recognise this bifurcation or split.
To put a nice bow on all this contradiction and sorry about the graphics*, I am going to have a go at this in Eureka:
| Code: | 2r8c1-(2=1)r8c3-(1=8)r8c5-(68=1)r9c6-(1=8)r6c6-(8=3)r6c2-(3=2)r6c1-(2)r8c1; r8c1 < > 2; stte
| ^
V |
(1=6)r9c2----------- |
If you plug 2 into r8c1 the loop removes it showing that it can not be 2. A tributary forms at r8c3 and then rejoins the main river at r9c6.
*Diagram only works in full screen mode on a PC or Mac! |
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