Times Fiendish no. 59 (very old puzzle)

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Times Fiendish no. 59 (very old puzzle)

Postby george-no1 » Mon May 30, 2005 2:02 pm

I was looking through a few old papers, and I found a Su Doku that I hadn't done before, and got stuck after 20 minutes. This has never happened before! I'm probably being impatient, but could someone look through the archives and give me a bit of help please!
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Postby Animator » Mon May 30, 2005 2:07 pm

You really want us to first search where the archives are, then try to establish a date-range in which to look (you only said the number, without any reference to the date), and then look the puzzle?

Perhaps it would be easier/better/more helpful if you post the solution you have so far...
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Postby george-no1 » Mon May 30, 2005 2:12 pm

In my frustration, I rubbed it all out because I had decided that I must have done something wrong. I'll have another look at it but I doubt I'll get anywhere - if not I'll post what I have.

Thanks anyway,

George
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Postby george-no1 » Mon May 30, 2005 3:52 pm

OK I might be wrong because I did this in a hurry but so far I have

**1 7*9 4**
2*4 *6* **8
*73 *4* *16

*2* *54 *3*
3** *1* *49
*4* *7* *6*

412 **7 69*
7** *2* *84
538 496 1**

Its not very much but most cells only have 2 or 3 possible numbers. I'm sure there's lots of X-Wings but I'm not good with techniques like that. Also, I heard that X-Wings are never necessary in Times Puzzles.

George
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Postby Animator » Mon May 30, 2005 4:05 pm

First, look at r3c7, decide what can go in there.

Then take a close look at box 2. Now you can remove one candidate from r3c7.

Then take a close look at boxes 6 and 9. Now you can remove another candidate.

And that should result in only one candidate for r3c7.
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Postby Guest » Mon May 30, 2005 6:31 pm

Thank you!!

I solved it very quickly after your hint. However, I didn't solve it by logic. I'm afraid I only had two candidates for r3c7 anyway, 5 and 9, and I couldn't work out how to get that down to just one. When you told me to look at that cell, though, something in my mind was telling me that it absolutely had to be a 9. So, I tried it, and it worked perfectly.

My question is, though, how did you work out (using logic) that r3c7 was a 9 and not a 5?

You are definitely a different class to me - I can do some 'Fiendish' puzzles easily, but others of the same standard get me stuck.

Thanks again,

George
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Postby Animator » Mon May 30, 2005 7:27 pm

Well, look at box 6 and box 9, there was a reason why I added them in my hint :)

Box 6 can have the number 5 in the cells: r5c7, r6c7 and r6c9

Box 9 can have the number 5 in the cells: r7c9 and r8c7.

Box 6 nor Box 9 have the number 5 as candidate for column 8, this means that box 3 has to have it's number 5 in column 8. This means you can remove it as a candidate from r3c7 (it's in box 3 and in column 7)
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Postby Guest » Tue May 31, 2005 8:55 am

OK Thank you I see what you mean now.

George
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Postby george-no1 » Tue May 31, 2005 9:03 am

Sorry I keep forgetting to log in - when it says 'Guest' that was me.

George:)
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