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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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 Posted: Sun Apr 06, 2008 7:18 pm Post subject: Another ... |
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... puzzle with a bouncing ER. M3768664 (80)
| Code: |
+------------+--------------+-----------------+
| 6 1 5 | 4 2 9 | 378 378 38 |
| 3 9 4 | 1678 78 17 | 1256 25 12 |
| 78 28 27 | 16 5 3 | 16 4 9 |
+------------+--------------+-----------------+
| 5 37 9 | 78 478 6 | 2348 1 2348 |
| 1 4 8 | 5 3 2 | 9 6 7 |
| 2 37 6 | 178 9 147 | 348 38 5 |
+------------+--------------+-----------------+
| 478 28 1 | 3 47 5 | 247 9 6 |
| 47 5 237 | 9 6 8 | 12347 237 1234 |
| 9 6 37 | 2 1 47 | 34578 3578 348 |
+------------+--------------+-----------------+
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Play this puzzle online
(a) There are at least two ERs here. One of them 'bounces' through 3 hinges. It's not necessary though.
(b) I noticed in a recent puzzle posted by Storm_Norm that Johan & Ravel both ignored possible UR eliminations. They weren't necessary - just made things easier for most of us, more interesting for J & R. However, I think that this puzzle will be seriously difficult unless you use a UR of type > 4. (I've been wrong before, & may be wrong here.) The whole puzzle isn't that easy anyway, tho' there's no need for really serious AICs / extended Medusa. |
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Johan
Joined: 25 Jun 2007
Posts: 206
Location: Bornem Belgium
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| Posted: Sun Apr 06, 2008 11:27 pm Post subject: |
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Taking a closer look at the {78}*UR in R24C45, either R2C4=16 or R4C5=4 to avoid the 78DP.
R2C4=16 => R46C4=78 => R4C5=4 => R7C5=7
R4C5=4 => R7C5=7
which means that R7C5 must be <7> solving the puzzle.
| Code: | +------------+--------------+-----------------+
| 6 1 5 | 4 2 9 | 378 378 38 |
| 3 9 4 |*1678*78 17 | 1256 25 12 |
| 78 28 27 | 16 5 3 | 16 4 9 |
+------------+--------------+-----------------+
| 5 37 9 |*78 *478 6 | 2348 1 2348 |
| 1 4 8 | 5 3 2 | 9 6 7 |
| 2 37 6 | 178 9 147 | 348 38 5 |
+------------+--------------+-----------------+
| 478 28 1 | 3 47 5 | 247 9 6 |
| 47 5 237 | 9 6 8 | 12347 237 1234 |
| 9 6 37 | 2 1 47 | 34578 3578 348 |
+------------+--------------+-----------------+ |
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ravel
Joined: 21 Apr 2006
Posts: 536
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| Posted: Mon Apr 07, 2008 2:13 pm Post subject: |
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Nice!
I could not see that, because i immediately placed an 8 in r2c5 because of the strong links 
A kite kills 4 in r7c7 and opens a w-wing 47 in r8c1 and r9c6 with the strong link for 4 in r7.
This gives a 3 in r8c3, but not more.
| Code: | *--------------------------------------------------*
| 6 1 5 | 4 2 9 | 378 378 38 |
| 3 9 4 | 67 8 17 | 256 25 12 |
| 78 28 27 | 16 5 3 | 16 4 9 |
|--------------+---------------+-------------------|
| 5 37 9 | 78 47 6 | 2348 1 2348 |
| 1 4 8 | 5 3 2 | 9 6 7 |
| 2 37 6 | 178 9 147 | 348 38 5 |
|--------------+---------------+-------------------|
| 478 28 1 | 3 47 5 | 27 9 6 |
| 47 5 27 | 9 6 8 | 12347 237 1234 |
| 9 6 3 | 2 1 47 | 4578 578 48 |
*--------------------------------------------------* |
Here i didn't find anything nice, but a new game/method, which i would call memory Medusa. This special kind of Medusa coloring does not need any markers, its easy to do it in the head, similar to xy-chains.
From the bivalue cells, look, if a number is forced to go to another one via a (grouped) strong link. Now try to find a way from the other number, such that it forces numbers, that see those in the target cell. They can be eliminated then.
The 1 in r2c9 jumps to r8c7 via r8c9.
The 2 in r2c9 forces r8c8=2 (via r2c8), r7c7=7 and r9c789=458.
Eliminate 247 from r8c7.
The 2 in r7c7 jumps to r4c9.
The 7 in r7c7 goes to r4c5, r4c2=3 and r4c4=8
Eliminate 38 from r4c9.
The 5 in r2c8 jumps to r9c7 via r9c8.
The 2 leads to r2c9=1, r3c7=6, r3c4=1, r2c6=7, r9c6=4, r9c9=8
Eliminate 48 in r9c7.
The 6 in r3c4 jumps to r2c7.
The 1 leads to r2c6=7, r9c6=4, r7c5=7, r7c7=2.
Eliminate 2 in r2c7.
To write this down probably took more time than doing it, including the tries without success.
Other puzzles might not be that memory Medusa friendly. I only tried Victors last puzzle and found a single, but useful elimination.
Of course its very similar to the "strong link chains" like:
r8c7=1 or (r3c7=1 => r2c9=2 => r8c8=2) ==> r8c7<>2
or
r8c9=1 or (r2c9=1 => r3c4=1 => r46c4=78 => r4c5=4 => r7c5=7 => r7c8=2) ==> r8c9<>2
Not very elegant that all, but for me a way to finish some puzzles without resorting to Medusa. |
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Victor
Joined: 29 Sep 2005
Posts: 207
Location: NI
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| Posted: Mon Apr 07, 2008 11:26 pm Post subject: |
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I too put in the 8 in r2c5 & didn't see Johan's nice (and interesting) elimination. I won't bother saying how I did it - pretty boring really. You could have killed the 4 in r8c7 as well (which was necessary for my solution) - simple colouring or an ER based on the 4s in c1: I like the way you can bounce the top one through 3 mirrors/hinges.
ANd I like the idea of memory Medusa. (I've realised that one of the reasons why I'm personally a little resistant to any kind of colouring is because of a sort of innate laziness.) |
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Asellus
Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA
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| Posted: Tue Apr 08, 2008 7:28 pm Post subject: |
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Johan's UR solution is indeed nice! It is interesting to note that it can be seen as an AIC placement of <4> in r4c5. First, however, note that the 78 UR creates a strong inference between the <4> in r4c5 and the {16} locked pair in r23c4. In Eureka: (4)r4c5=({16})r23c4. This is our starting point:
(4)r4c5=({16})r23c4-(1={78})r46c4-({78}=4)r4c5; r4c5=4
This is very compact! |
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