How long does it make sense to bang one's head!?

Post the puzzle or solving technique that's causing you trouble and someone will help

How long does it make sense to bang one's head!?

Postby georgers » Sat Jun 11, 2005 9:22 pm

It's been 5 days.... I've sold other very hards but this one.... *!$**!£$

Anyone prepared to waste their time and save me from concussion?

6*7 2** 3**
4** 8** **6
59* 76* ***

9** 6** 8**
17* *** 642
**6 1** **9

*** *81 *63
861 **2 **5
**9 **6 1**

Any hint gratefully accepted: the x-wings I've found didn't seem to help -- perhaps I still haven't grasped quite how they work.

Thanks in advance...
georgers
 
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Postby Animator » Sat Jun 11, 2005 9:43 pm

There are two intresting groups of numbers in box 3 (one of 3 numbers (5, 7, 9) and one of 4 numbers (1, 2, 4, 8)).

I suggest you take a really good look at box 3. (I'm not sure if you are familiar with the technique that is required or not... if you are not then I (or someone else) will explain it a bit more)


When you are done with that you need to look at column 8. You will find a similar thing in it (same technique, different numbers).

When you found it then you can fill in two cells in column 8
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Thank you very much!

Postby georgers » Sun Jun 12, 2005 9:23 am

I am amazed that what was so opaque for so long for me (at least it felt like forever) was clearly so easily apparent to you. Thank you -- your hint resolved matters speedily.

Makes me want to ask whether YOU ever get stuck, and whether there are any techniques you use for spotting X-wings, multiple possibles etc.

Or is it just practice and confidence?

Anyway, many thanks again: the ringing in my ears is already lessening!

George
(georgers1@gmail.com)
georgers
 
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Re: Thank you very much!

Postby joolslee » Sun Jun 12, 2005 12:54 pm

georgers wrote:Makes me want to ask whether YOU ever get stuck, and whether there are any techniques you use for spotting X-wings, multiple possibles etc.


Hi georgers - I've been wondering the same thing about Animator :o) Also Animator, I've been wondering what the initials underneath your username mean? Can't see them now because I'm writing this lol ... but I think they're MVP.
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Postby simes » Sun Jun 12, 2005 2:24 pm

MVP = Most Valuable Professional.
See https://mvp.support.microsoft.com/gp/mvpintro
Last edited by simes on Sun Dec 11, 2011 9:59 am, edited 1 time in total.
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Postby joolslee » Sun Jun 12, 2005 3:43 pm

Hey thank you for that explanation and link Simes - a well deserved accolade to Animator.
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Postby axis » Sun Jun 12, 2005 7:26 pm

Well, I'm just noodling around with this puzzle as well since the problem was posted. Animator, you said look at box 3 to solve it, I'm not sure what technique you are referring to to use to derive a number. Could you explain this please? Thanks.
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Postby joolslee » Sun Jun 12, 2005 9:54 pm

Hi Axis, as no one else has come forward yet, I think the technique referred to is explained at http://www.simes.clara.co.uk/programs/sudokutechnique9.htm
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Postby simes » Sun Jun 12, 2005 10:10 pm

Axis,

Look at where the numbers 5, 7 and 9 can go in both box 3 and column 8. Then think about the effects on the candidates for other cells in the box and column.
Last edited by simes on Sun Dec 11, 2011 10:00 am, edited 1 time in total.
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Postby axis » Sun Jun 12, 2005 11:20 pm

Okay, I'm still lost. I understand disjoint subsets but I'm not too sure on reducing candidates from the cells that hold the unique subset. I believe in this puzzles case, the 5,7,9 in block 3 was the triple candidates for the cells r1c8, r2c7, r2c8. However, what confuses me is why this works when r1c8 doesn't hold all three just two of them?

Also, assuming I did understand the above, to work it out gives me a 7 in r8c8 and a 2 in r9c8? Is this the two cells in column 8 I should find? Bear in mind that if it is I didn't figure it out really because I still dont see why the reducing in block 3 works without the 7 in r1c8? Thanks.
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Postby georgers » Mon Jun 13, 2005 11:08 am

axis wrote:Okay, I'm still lost. I understand disjoint subsets but I'm not too sure on reducing candidates from the cells that hold the unique subset. I believe in this puzzles case, the 5,7,9 in block 3 was the triple candidates for the cells r1c8, r2c7, r2c8. However, what confuses me is why this works when r1c8 doesn't hold all three just two of them?


HI!
Failing a response from my elders and betters, let's have a shot at saying how I tackled this:
I actually started with the 1,2,4,8 set because I couldn't, at first, see how the 5,7,9 set worked. There are 4 numerals so 4 locations need to be part of this set --- and further locations with one or more of the constituent numbers can have those numbers deleted. The 4 locations are r1c9, r3c7, r3c8,r3c9. So the other 2's in that square and, crucially, the 8 in r1c8 can all be removed.
That leaves as possibles 5 & 9 in r1c8, and 5,7,9 in r2c7 and r2c8: the grouping Animator was talking about.
After that repeated application of the same concept elsewhere led to the solution.
Hope that helps...
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