Very Hard - from Times Website

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Postby Pappocom » Wed Mar 30, 2005 3:17 pm

shakers wrote:I'd love Wayne to comfirm whether this would count as a logical deduction, or T&E...

I don't think I understand Nigel's suggestion ... which makes me think it must be T&E! Sorry, Nigel.

The technique you're looking for is one which I call "X-Wings" - named for the Star Wars fighters. That may not help you find it, but after you've found it you will know immediately why I call it that.:!:

You're looking at the technique and it's looking back at you. You just have to see it.

- Wayne
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Postby Guest » Wed Mar 30, 2005 3:19 pm

I know what you mean - I've been staring at this puzzle for weeks now - I'm convinced I have a blind spot somewhere, but staring more doesn't seem to help!

Wayne - Thanks. I think! It'll get me thinking at least, but I'm not sure the force is strong enough in this one!
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Postby Guest » Wed Mar 30, 2005 3:26 pm

Pappocom wrote:
shakers wrote:I'd love Wayne to comfirm whether this would count as a logical deduction, or T&E...

I don't think I understand Nigel's suggestion ... which makes me think it must be T&E! Sorry, Nigel.

The technique you're looking for is one which I call "X-Wings" - named for the Star Wars fighters. That may not help you find it, but after you've found it you will know immediately why I call it that.:!:

You're looking at the technique and it's looking back at you. You just have to see it.

- Wayne


I think I have to agree as I cannot replicate what I did, at the moment.

However as I did complete (Byluck or judgment) I can see what you mean about X wings
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Postby shakers » Wed Mar 30, 2005 3:27 pm

NigelG wrote:Well I have started again, to try and help you and now I can't do it.

Must have hit brain fade time.

Is it normal to blank like this:(


Yup! I solved it last night, and cannot follow my own clues to solve it again today!

I think it's time to quote the words of the immortal Homer Simpson: "D'oh!"
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Postby Guest » Wed Mar 30, 2005 3:31 pm

I've just got it - that's beautiful. The diagonals must also have a complete set of 1-9. Now why did I never think of that before? I've solved it now. Thanks.:D
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Postby shakers » Wed Mar 30, 2005 3:31 pm

Pappocom wrote:I don't think I understand Nigel's suggestion ... which makes me think it must be T&E! Sorry, Nigel.


If r1c1 is 7 then r8c1 is 1 and r8c4 is 1 and r4c4 is 3 and r3c4 is 7, r1c6 must be 6. Unsurprisingly if you take r1c1 to be 1 it all fails.

I'm glad you consider that to be T&E though!
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Postby lil_miss » Wed Mar 30, 2005 3:32 pm

no..... diagonals dont have a 1-9 at all!!!! how did u manage that?
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Postby shakers » Wed Mar 30, 2005 3:34 pm

IJ wrote:I've just got it - that's beautiful. The diagonals must also have a complete set of 1-9. Now why did I never think of that before? I've solved it now. Thanks.:D


No they don't! I've tried that before! LOL!
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Postby Guest » Wed Mar 30, 2005 4:39 pm

No, you're right. I've just been out and thinking about why this might be true and concluded that it wasn't! Certainly, looking at other puzzles, this is not always true. In fact, not even in this case, now I've checked. Fit's nicely with Wayne's clue though! Back to the drawing board then.
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Postby Guest » Wed Mar 30, 2005 4:58 pm

meetings get in the way of everything.....

To explain how I worked out the position of my 7s, and to ask opinion of whether this is T&E ?

you can logically work out that there are only two positions for the 7s in columns 1 and 6,

and there are also only two positions for 7s in rows 7 and 8.

the 7s can't go r6c3 and r7c4 cos the 7s for columns 1 & 6 would both have to go in row 1.......

it's hard to explain in words but easier to see on paper


7 * * | * * 7 | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * * | * * * | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * 7 | * * 7 | * * *
7 * * | 7 * * | * * *
* * * | * * * | * * *

Jim
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Postby Guest » Wed Mar 30, 2005 5:07 pm

Apologies for repeat postings - spotted an error in my references and tried to edit, but it just kept postin !!!

......meetings get in the way of everything.....

To explain how I worked out the position of my 7s, and to ask opinion of whether this is T&E ?

you can logically work out that there are only two positions for the 7s in columns 1 and 6,

and there are also only two positions for 7s in rows 7 and 8.

the 7s can't go r7c3 and r8c4 cos the 7s for columns 1 & 6 would both have to go in row 1.......

it's hard to explain in words but easier to see on paper


7 * * | * * 7 | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * * | * * * | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * 7 | * * 7 | * * *
7 * * | 7 * * | * * *
* * * | * * * | * * *

Jim
Guest
 

Postby Guest » Wed Mar 30, 2005 5:56 pm

Hi Jim, I don't quite follow - I think these 6 cells have at least two solutions:

7 * * | * * 6 | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * * | * * * | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * 7 | * * 3 | * * *
1 * * | 7 * * | * * *
* * * | * * * | * * *

and
1 * * | * * 6 | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * * | * * * | * * *
* * * | * * * | * * *
* * * | * * * | * * *
-------|--------|------
* * 1 | * * 3 | * * *
7 * * | 1 * * | * * *
* * * | * * * | * * *

In fact there are many more, so I don't think it can count as a logical deducton from these facts that r1c1 must be a 1.

I think the best you can conclude is that r1c1, r7c3 and r8c4 are the same, but not whether they are 1 or 7.
Guest
 

Postby Guest » Wed Mar 30, 2005 6:07 pm

yeah, but you can't put the 6 and the 3 in column 6 as you've shown, cos you've no where left for your 7 to go in column 6 (7 can only go in c6r1 or c6r7)
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Postby Guest » Wed Mar 30, 2005 6:16 pm

I knew it was difficult to explain....

the 7s in r7 and r8 work in two pairs....

i.e. they have to go in r7c3 and r8c4

or

r7c6 and r8c1.

this is because there are only two positions for the 7s on each of these rows.

all I've shown is that you can't have them in r7c3 and r8c4 because this would force the 7s for columns 1 and 6 to both be in row 1 !!!!

Therefore, the 7s must be r7c6 and r8c1, which forces a 3 into r8c6 and so on, and therefore solves the puzzle....

sorry if I can't make it clear - I can see it in my mind but it's hard to put forward. I'm still not sure if it's not trial and error though, just using two pairs of numbers.

Cheers, Jim
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Postby shakers » Wed Mar 30, 2005 6:21 pm

I can see it being argued as T&E, but I think I'm coming down on the logic side of the explanation.

What I'm trying to do is to think of a way to describe the method in wider terms...

If you join up the two pairs of 7s in rows 7 and 8 you certainly get something which if you squint hard enough could look like an X-Wing fighter!:D
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