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04 March 2009 @ 08:57 am
KenKen for 3/4/09 - The Great Divide  
Aside from one of the initial KenKen releases in the London Times (#18 I believe, which had a long outer border 44+ in ~14 cells), the next largest region I've ever seen in a KenKen is a 5 (one time) and then some 4's in some tough ones but mostly just 3 and 2 and 1. This may come from the fact the first 2 books were hand-written and then the rest were computer-generated with a not yet interesting puzzle generator that doesn't like taking things to a different level (or at least is scared of pentominoes).

I find a lot of the potential logical interest in KenKen would come from exploring larger shapes that interact with others in different ways. A commenter suggested no operations made a harder puzzle. Well, in this grid, there is one large operation but it is certainly not what's going on there in a now record 16 cell region.

This is a Wednesday, 6x6 Mediumer puzzle (if Easy and Hard aren't acceptable descriptions for KenKen elsewhere, then Medium isn't an acceptable description for me here).

Rules: Same as normal KenKen, fill one to six in each row/column so each digit is used exactly once. The numbers in the upper-left of each bolded region indicate the value for some operation (+,-,*,/) applied within each cell of that region but the identity of the operation is left as an exercise for the reader.
thedan on March 4th, 2009 05:22 pm (UTC)
devjoedevjoe on March 4th, 2009 06:02 pm (UTC)
This is unsolvable. The large region can only reach 759375 as a product of five 3s, five 5s, and six 1s. Thus, in each row and each column where it has three cells, those cells must contain 1, 3, and 5, forcing the other cells in the row or column to contain even digits. These constraints force the remaining 3 and 5 to go into the other corners in some order. However, the three-cell region with clue 8 cannot be fulfilled with a single 3 or 5 and two even digits; if it was a sum, it would be odd, and if it was a product, it would contain a factor of 3 or 5.

Highlight below for a solution assuming that 8 is supposed to be a 9:

motrismotris on March 4th, 2009 06:08 pm (UTC)
Maybe this is why I need a really good editor looking at my KenKen. My path didn't require working out the product and I totally got the math wrong. The image has changed.
Mike Selinkerselinker on March 4th, 2009 06:27 pm (UTC)
Huh. My confidence in your KenKen PuzzleCraft is waning.

motrismotris on March 4th, 2009 06:42 pm (UTC)
If you're volunteering to check tomorrow's crazy puzzle, just tell me that's what you want to do. Because I can totally use a test-solver.
Mike Selinkerselinker on March 4th, 2009 06:45 pm (UTC)
Thomas, if I could handle puzzles like this, why would I need you to maintain the illusion that I can give advice on how to make any kind of puzzle?
Adam R. Woodzotmeister on March 5th, 2009 09:42 pm (UTC)
<shameless whoring of arguable talent>
I would be perfectly happy to assist either of you with virtually anything puzzle-related.
</shameless whoring of arguable talent>

- ZM
thedan on March 4th, 2009 06:27 pm (UTC)
I was just about to say this; somehow when I solved it I wrote down the same wrong factorization you used, but when my friend Tom did it, he proved there was a contradiction the same way Joe did.
motrismotris on March 4th, 2009 07:13 pm (UTC)
It turns out my theme was busted too. Reposted another time to get the right grid shape.

What lesson have we learned? I really shouldn't write puzzles right after waking up as I forgot / is division, not \ and couldn't even use my graphing calculator to get 3^6*5^5 right. So, the position of tester is currently open and I'm now a day ahead of my constructions (themes are already out there) so there will be no delay in the coming puzzles.
(Anonymous) on November 24th, 2010 07:20 pm (UTC)
devjoe is incorrect. It has a very lovely solution, with the 6 3s all in a diagonal from bottom left to top right,, the 5 5s above each 3 and the 5 1s below each 3. Treat the top left group to be a X with a 1 2 4, the two groups with a 2 solution as -, the rest + and bob's your uncle
devjoedevjoe on November 24th, 2010 07:29 pm (UTC)
Re: devjoe
Mr, Anonymous,

If you read all the comments on this post, you will see that the original broken problem has been replaced with a different, solvable one, and in another comment I already solved the corrected one.
carljohanr on March 4th, 2009 10:16 pm (UTC)
Search and you will find..
That's definitely the biggest region I have seen too, but at the cost of the size of other regions... as is only natural. Other creators of puzzles (like djape) have created several puzzles with larger regions (region sizes of 6 to 8 are common). So I guess it really depends on the target reader as well.
(Anonymous) on March 4th, 2009 11:12 pm (UTC)
Vaguely entertaining, much more so than anything I've seen in the Times here - although having to look up prime factorisations to kick start everything seems a teeny bit artificial.

On the other hand, if you do that by hand, your title 'The Great Divide' becomes excellent - reflecting accurately how you solve the puzzle whilst underlining a long and cumbersome calculation!

motrismotris on March 4th, 2009 11:17 pm (UTC)
You don't. This is why I had the wrong value there initially. Take away that big number and let me mark it like this: "its odd and its a multiplication region". The value is immaterial.

My solution involved identifying lots of obviously even cells to figure out the forced identities of the two non-divide corners (the only odds not in the great divide). I don't like factorization problems at all. Thanks for pointing out the alternate meaning to the title though. That explains why no one else commented I had the / as a \ when I first posted it. I'd not concerned that possibility, but then I just wanted to write a themed puzzle where odds and evens were identifiable quickly and then you got down to business.
(Anonymous) on March 4th, 2009 11:17 pm (UTC)
p.s. the one flaw with leaving the operations blank is that division and subtraction are not associative. On t'other hand, I've seen some truly evil puzzles which have been made of only 2-cages. Maybe something for Friday - though I think it'd be tougher to add a notably human touch (as opposed to something a computer could generically generate) to something like that...
devjoedevjoe on March 4th, 2009 11:50 pm (UTC)
OK, solution for the corrected version.

The large region can only be the product of six 3s, five 5s, and five 1s. This means that the middle four rows and middle four columns have 1,3,5 in the large region, and only even numbers in the other regions. This forces the remaining 1 and 5 into the other corners. There's no way for the 3-cell corner to reach a result of 8 with 5 and two even numbers, so the 1 goes there. Repeatedly ensuring no repeated digits in the rows and columns forces the 3s along the diagonal and 5s and 1s into parallel stripes.

The 8 with a 1 can only be a product of 1,2,4, and the 2-cell regions with 2 can only be 2,4 or 4,6. So the other top-row cell in the 8 is a 2. The 3-cell regions with 12 made of all even numbers can only be the sum of 2,4,6, so the numbers in the 8 force the 12 nearby.

Now you need to switch to the other corner where the 15 must be the sum of 4,5,6. Again, the 4 in the last column is needed in the 2-clue, so it goes on the bottom row. This again forces the 12 and the remaining numbers easily fill in.

motrismotris on March 5th, 2009 01:16 am (UTC)
Thanks for posting this solution path for the corrected puzzle, and apologies again for having the error in the one you worked on first.
lunchboylunchboy on March 5th, 2009 04:42 am (UTC)
Very nice. If you specify that each operation is used at least once, then there's no ambiguity about which of the two potential operators is used to produce the 2 in the rightmost column.