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09 March 2009 @ 08:29 am
This is not really a puzzle  
Difficulty Easier 4x4 KenKen:

So this is probably not really a puzzle. As I've been exploring "extremes" of KenKen space, it is fairly obvious to me that 2 region KenKen are plentiful but almost always uninteresting. In an n x n Latin Square, you need to only specify the right T(n-1) squares, T(n) being the triangular number 1+2+...+n, to specify the whole Latin Square. Well, this isn't that hard with smallest or largest sums/products. What was interesting to me was to split the two regions to have the same value, which this and only this puzzle in arithmetic seems to be able to do (with region redrawing obviously possible).
(Anonymous) on March 9th, 2009 04:59 pm (UTC)
Interesting puzzle
Thank you for posting this puzzle. I, too, was wondering about the fun-ness of interesting shaped puzzles like this one.
devjoedevjoe on March 9th, 2009 07:08 pm (UTC)
You're right. It's not a puzzle, it's an answer to the puzzle "How can I make a Kenken with only two regions and a unique solution?"
motrismotris on March 9th, 2009 07:27 pm (UTC)
Well, to us its not a puzzle. For some KenKen solvers, this would still take some puzzle-y steps. I've changed my mind on posting the largest possible "multiplication of equal valued regions" thing because it actually is a good puzzle. You'll see it by 6x6 puzzle space in the week.
(Anonymous) on March 9th, 2009 09:08 pm (UTC)
Try replacing the 20 by 576, if you want to kill another 60 seconds
(or less).

Also, have you found a 7-9 split for the 4x4?

Gerhard Paseman, 2009.03.09
(Anonymous) on March 10th, 2009 12:59 am (UTC)
That doesn't have a unique solution. If you put 3 1s in the big region it does, but there are many solutions with all 4 1s in the big region.
(Anonymous) on March 10th, 2009 04:41 pm (UTC)
Oops. Missed that. Now I no longer expect a 7-9 even split to exist.
Gerhard Paseman.
motrismotris on March 10th, 2009 09:14 pm (UTC)
No, I don't suspect 7-9 or 8-8 splits of the regions can exist either. The extra cell(s) force an either/or choice back into a set which the T(n-1)-T(n) split will not.
thedan on March 9th, 2009 07:18 pm (UTC)
Perhaps the first time I've solved one of these in my head. :)
Adam R. Wood: butasanzotmeister on March 11th, 2009 04:51 pm (UTC)
Damn it, now I feel I have to. - ZM
Adam R. Wood: butasanzotmeister on March 11th, 2009 04:53 pm (UTC)
Got it. That wasn't as bad as I'd thought. Of course, I didn't prove uniqueness... - ZM
Adam R. Wood: butasanzotmeister on March 11th, 2009 04:56 pm (UTC)
Got that. That was even easier.

...So does doing all that quickly in my head, but failing to realize how easy it was, mean that I'm smart or that I'm stupid? Actually, in keeping with the theme, the answer is, "If you have to ask, Adam..."

I'll stop commenting now and go hide in a corner, or something. - ZM
grandpascorpion on March 10th, 2009 12:14 am (UTC)
I enjoyed it just the same :)