Fiendish No. 160

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Fiendish No. 160

Postby MNJim » Mon Jun 06, 2005 3:34 am

I actually put this in another forum, but I think it should go here, so I'll repeat:

Hi all,

I've been lurking in the forum for a while, and have enjoyed all of your comments. I do the Times puzzle from home when it's posted online, and about 95% of the time I can get through it. However, I'm stuck on the June 2nd puzzle, and was hoping someone could explain the next step to me. Here's where I'm at:

563 4*8 *1*
*79 *** 5**
**8 5*7 *3*

*3* **1 **5
7** 264 ***
8** **5 *2*

*1* 7*9 ***
3*4 *** 971
987 1*3 *5*

Now, I set this up in the step-by-step program, and it says the next step is to place a 9 at 4,4. I don't see this - it looks to me like that particular spot can be either an 8 or a 9. Help?? Thank you...
MNJim
 
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Postby obtuse » Mon Jun 06, 2005 4:26 am

When I play this game from your position I found that the next cell to fall was r4c5. Look for pairs in c5 and look at the 9 in square 2.
obtuse
 
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Postby MNJim » Mon Jun 06, 2005 6:06 am

obtuse wrote:When I play this game from your position I found that the next cell to fall was r4c5. Look for pairs in c5 and look at the 9 in square 2.


I'm probably being "obtuse" here, but I'm still not getting it. Square 2 has to have the 9 in either 5,1 or 5,3. I'm not sure what you mean by "look for pairs." The cell you're talking about (r4c5) I've narrowed down to either 7 or 8, but I can't make any headway after that.
MNJim
 
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Postby obtuse » Mon Jun 06, 2005 8:38 am

I think I was just living up to my name.
I just tried to follow my own advice and got confused.

So I tried again and got this;

You can rule out 3 as an option in r6c7 amd r6c9
You can rule out 8 as an option in r4c7 and r4c8

This give a "pair" for 3 and 8 in c7
i.e. 3 and 8 only appear as an option in r5c7 and r7c7. Thus you can rule out all other possibilities for these cells.

This leaves r6c7 as the only possible location for a 1 in square 6.

i think...
obtuse
 
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Postby SteveF » Mon Jun 06, 2005 11:42 am

As an alternative, consider where 6 can go in boxes 2, 7 and 8.

Then have a look at boxes 3 and 9, to see if this limits where a 6 can go in these two.

Then have a look at column 8 and you should see something that will help.
SteveF
 
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Postby MNJim » Mon Jun 06, 2005 3:09 pm

SteveF wrote:As an alternative, consider where 6 can go in boxes 2, 7 and 8.

Then have a look at boxes 3 and 9, to see if this limits where a 6 can go in these two.

Then have a look at column 8 and you should see something that will help.


Of course, I should have seen that. I'd limited where a 6 can go in squares 3 and 9, and didn't put them together to place the 6 in square 6. I'll run with that, thanks.

Obtuse: "You can rule out 3 as an option in r6c7 amd r6c9
You can rule out 8 as an option in r4c7 and r4c8

This give a "pair" for 3 and 8 in c7
i.e. 3 and 8 only appear as an option in r5c7 and r7c7. Thus you can rule out all other possibilities for these cells."

But the 8 can also appear in r5c8 and r7c8, so I think that rules out the pair you're referring to. Am I missing it?

Thanks, I'll see where I can go with this now...
MNJim
 
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Postby MNJim » Mon Jun 06, 2005 3:16 pm

OK, the placement of the 6 in column 8 did it for me - I should have seen that! Took about 3 minutes to finish the puzzle after that. Thanks for the help!
MNJim
 
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Postby chardir » Thu Jun 09, 2005 6:35 pm

Hi,

I've been stuck on this one for a couple of days but I've got to a slightly different position from MNJim:

*6* 4*8 *1*
**9 *** 5**
**8 5*7 *3*

*3* **1 **5
**1 264 ***
8** **5 12*

*1* 7*9 ***
**4 *** 9*1
98* 1*3 *5*

Can anyone give me any hints?
chardir
 
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Postby Animator » Thu Jun 09, 2005 7:10 pm

Ok,

Take a good look at the number 3.

Where can it go in box 9?
Where can it go in column 3?


When you are done with that you can take a close look at the number 5.
Animator
 
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Postby chardir » Fri Jun 10, 2005 7:59 am

Cheers, I'd clearly been staring at that one for too long!
chardir
 
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Postby tabber » Thu Jun 16, 2005 10:47 am

To go back to the original position at the top, I was stuck here myself for over an hour (spread over 4 or so train journeys). I actually solved it a different way in the end, which I thought I'd share. I will however go back and try the other suggestions.

For me, I had failed to spot that box 5 must have an 8 in the top row, and a 3 in the bottom row (because the middle row is already filled in, and box 4 has 8 and 3 in the opposite rows). Don't know how I missed this, that is fairly basic. Anyway, this means you can exclude 3 or 8 as candidates in column 7, rows 4 and 6, and this leaves you a 3/8 pair in column 7, rows 5 and 7. Once you have removed all the other candidates except 3/8 from these two cells, row 6, column 7 must be a 1. Fell very quickly after that.

Cheers,
Tabber.
tabber
 
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