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31 Dec, Monster

 
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Sue
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PostPosted: Sun Jan 01, 2006 5:44 am    Post subject: 31 Dec, Monster Reply with quote

Hi All,
I've just started doing the monsters.
I've tried the 31st Dec 3 times, and each time, towards the end, I've doubled up a number!
Do you think I'm just losing concentration and making simple mistakes?
I tend to do them all in one sitting, and do get pretty tired after a while.
Or, is anyone else having trouble with this one?
Thanks
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blasr
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PostPosted: Sun Jan 01, 2006 12:35 pm    Post subject: 31st Dec Reply with quote

You are not alone! I am asking a friend for help - even when I key in to Draw my solution so far
1 2 ? 6 7 4 ? ? ?
? 7 ? 5 8 3 ? 1 2
? ? 3 1 9 2 6 ? ?

? 5 ? 9 6 1 2 ? ?
9 6 7 2 3 5 1 8 4
3 1 2 7 4 8 5 9 6

? 9 1 4 ? ? 3 ? ?
? 3 ? 8 ? 9 ? 6 1
? ? ? 3 1 ? ? ? 9

I am offered a number I do not understand, row 7, col 9 = 8
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dotdot



Joined: 07 Dec 2005
Posts: 29
Location: oberseen

PostPosted: Sun Jan 01, 2006 5:05 pm    Post subject: Reply with quote

If you want that to be your next cell, you need the following preparation:


The values 2 and 5 in col4 and col6 force col5 to have these values in box 8, i.e. r7..8c5 are a {2,5} pair.

The 7s in row1 and row2 force col7 to have this value in box9, excluding 7 from the other free cells in box9.

But with 7 excluded, r7c8 is {2,5}. This pairs up with the {2,5} in r7c5 to exclude 2 and 5 from the rest of row7.

So not only 7 but also 5 is excluded from r7c9.
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blasr
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PostPosted: Mon Jan 02, 2006 9:09 am    Post subject: Reply with quote

Oh deary me, must be the flu! I was being very stupid; many thanks dotdot! I should have spotted that 7s were possible in box 9 only in col7 as you say and hence the rest falls into place!
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boweasel
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PostPosted: Tue Jan 03, 2006 9:18 pm    Post subject: Reply with quote

I must either have a real bad case of the flu, or am simply taking the word "stupid" to new levels...

For columns 8 and 9, I have 7s as a possibility in r3, r4, and r7, as well as (r9,c8). I understand the bit about the pair of (2,5)s in column 5, box 8, but am completely in the dark about how that has anything to do with the box 9 values.

Could someone please explain how "The 7s in row1 and row2 force col7 to have this value" [(2,5), I suppose] "in box9, excluding 7 from the other free cells in box9."

Still a newbie, so be patient, please.

Bo
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Steve R



Joined: 24 Oct 2005
Posts: 289
Location: Birmingham, England

PostPosted: Tue Jan 03, 2006 10:03 pm    Post subject: December 31 – very hard Reply with quote

Bo:

I'm not sure why this is in the monster thread, but I hope I can help with your question.

Check the top right-hand box. The 7s already present further left in rows 1 and 2 force 7 to occupy column 8 or column 9 in this box.

In the middle right-hand box, the only places left are in column 8 and column 9. So the top two right-hand boxes between them account for the 7’s in both column 8 and column 9.

The only column open to a 7 in the bottom right-hand box is column 7.

Steve
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dotdot



Joined: 07 Dec 2005
Posts: 29
Location: oberseen

PostPosted: Tue Jan 03, 2006 10:47 pm    Post subject: slow replay Reply with quote

Quote:
I understand the bit about the pair of (2,5)s in column 5, box 8

Ok, that was the first intermediate step.
Its result is (partially) used later on when talking about r7c5.

Quote:
but am completely in the dark about how that has anything to do with the box 9 values

Not directly, it doesn't.
The further intermediate steps deal with this. And near the end they refer to r7c5, knowing it to be {2,5}, which the first step established.

Quote:
please explain how "The 7s in row1 and row2 force col7 to have this value" [(2,5), I suppose] "in box9

This this is a local this and means to refer to the value 7.
Looking at col7 in isolation, we could suppose it could have its 7 in any one of the remaining free cells, two at the top and two at the bottom.
But looking at col7 together with row1 and row2, shows us that the top two free cells in col7 can't hold a 7; row1 and row2 each already have a 7.

Thus it is the one of the bottom two cells of col7 which contain the 7.
Because both these cells are in box9, they will, between them, contribute a 7 to box9. This implies that none of the other cells in box9 are allowed have a 7.

That was the second step.
It was only talking about 7s.

The third step excludes a 5 from r7c9, but not directly.
It uses the exclusion of 7 from r7c8 (from the second step) to see that r7c8 is left with only {2,5} as possibilities.
Then, knowing r7c5 (from the first step) to be {2,5} as well, we can assert that these two cells, between them, contribute a 2 and a 5 to row7. This implies that none of the other cells in row7 are allowed to have a 2 or a 5.


So the first step was a way of finding out about r7c5. It also found out about r8c5 but we didn't use that while working towards r7c9.
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David Bryant



Joined: 29 Jul 2005
Posts: 559
Location: Denver, Colorado

PostPosted: Wed Jan 04, 2006 6:32 pm    Post subject: Re: 31 Dec, Monster Reply with quote

Sue wrote:
Hi All,
... I've tried the 31st Dec 3 times, and each time, towards the end, I've doubled up a number!
... Or, is anyone else having trouble with this one?
Thanks

I don't ordinarily work the "monster" puzzles -- they're very time-consuming.

Anyway, I took a stab at this one, and the first time through I ended up with no place to put a "9" in row 12. So then I went back and retraced my steps, and figured out that I had made an error with about 40 cells still to be resolved. It may not be the same mistake you made, but I looked at column 4 a little cross-eyed and thought I saw a triplet {5, 8, 9} in r2c4, r5c4, & r10c4. So then I put a "4" in r12c4 ... that cell is actually a "9" in this puzzle.

I had confused an "8" with a "B" -- they look a lot like each other. Happily the puzzle was easily solved once I got that "9" in the right spot.

I'm not sure if this will help, Sue ... if you like, please post the position where you first realized you were off track, and maybe someone can help you figure it out. dcb
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Sue
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PostPosted: Sun Jan 08, 2006 11:32 am    Post subject: Reply with quote

Thanks David,
I actually did it 4 times before I finally cracked it.
I think I was just going cross eyed by not taking a break.
Every time I redid I , I found the previous mistatke, and it was just careless!
Fun, though I haven't done any since, think I OD'd.
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