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motris
motris
14 March 2009 @ 11:09 am
3/14/09 - A Whole Lot Of Nothing  
Not a lot of people have been commenting during this second week of "regular" KenKen, so I thought I'd shake things up today for the Saturday puzzle. I like exploring variations, so here I'm going to break the standard formula a little bit, adding nothing to the puzzle, to make a very hard but original challenge:

7x7 "Zero KenKen"



Rules: Fill in the cells with digits from 0 to 6 so that each digit appears a single time in each row and column. The numbers in the upper-left corner of each bolded region indicate the value of an operation (+, -, x, /) applied to the cells in that region. For multi-cell subtraction and division, the largest number is used first and all other numbers are subtracted/divided from it. Division by zero is not allowed.
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11 comments | Leave a comment
 
( 11 comments — Post a new comment )
[info]ericberlin on March 14th, 2009 07:41 pm (UTC)
Pre-attempt reaction: Oy.
(Anonymous) on March 14th, 2009 08:12 pm (UTC)
You can sometimes divide by zero (if we want to explore satisfactory limitations on operations once more haha), provided the zero arises from a finite product of linear factors. It's either that, or introducing homogeneous coordinates onto kenken space. You'd have to have infinity into your puzzles though. That would certainly be an eye-opening variant!

Anyhow enough silliness, I look forward to giving this a go. This certainly looks like kenken with a pronounced flavour!

Tom.C
(Anonymous) on March 14th, 2009 08:14 pm (UTC)
p.s. with reference to a 0 score in tennis being called as "love" and the Led Zep song, this might have an amusing alternative title :)
(Anonymous) on March 15th, 2009 01:00 pm (UTC)
Having finished this, I have to say this is by far my favourite of your series. I think this 0 idea is definitely the one to explore further - for example 0s in addition cages would certainly shake things up.

Tom.C
motris[info]motris on March 15th, 2009 03:55 pm (UTC)
I wanted to start "simpler" and have the 0's identifiable first. I definitely agree that 0's in addition (or subtraction) cages would complicate things and I may revisit this style throughout the next week (although I have some other small changes I'd like to sample).
(Anonymous) on April 22nd, 2009 06:47 pm (UTC)
Dividing by zero is undefined and is not equivalent to infinity. The only way you get to divide by zero is if you either redefine division or if zero isn't actually the null element of the field...and I don't think anyone suggested doing either of those.
devjoe[info]devjoe on March 14th, 2009 10:33 pm (UTC)
Here goes nothing!

Because of the way division is defined for these puzzles, it can't give a zero result, so 0s are either subtraction or multiplication, except for the single cell which is simply a 0, and two-cell 0s must be multiplication. So we know that the two-cell 0 next to the given 0 has a 0 on top, the two two-cell 0s in rows 2-3 take the 0s in those rows and columns, and the two-cell 0s at the upper right and lower left have 0s in the corners.

That leaves row 4 and column 3 as the only ones without a 0, so we put one there, making the large 0 region a product. There are no 0s left for the 3-cell and 4-cell 0 regions, so they must be subtraction, with only 6-1-2-3 in some order possible for the 4-cell. The 12 is too large to be a sum, so it must be a product, and with only 4 and 5 possible in the corner, it must be 3x4. This puts a 5 in the 0 region in the upper right.

With 1, 2, 5, and 6 available, the 3-cell 0 in the bottom row can only be 6-5-1. This puts a 2 into the region to its left.

The 24 must be 4x6 in some order. The 20 must be a product 1x4x5 or 2x5x2, and with 5 used in row 2, the 5 must go into row 3, and the product must be 1x4x5 with the 4 in column 4.

Now there are 4 cells in column 5 which contain 1, 4, 5, or 6: the two cells of the 24, the cell in the 20, and the bottom cell. The 15 thus cannot be the sum 4+5+6, so it must be the product 1x3x5.

Now consider the 3-cell 0 in column 6. It must be a subtraction, but which one. It cannot contain a 3, nor can it contain both 5 and 1, so it can only be 6-4-2. This puts three 6s in the last three columns in rows 3-6, so there can only be one more 6 in these rows. This also forces the 5 in this column into the top row; the 0 region is 0x1, and the 15 can be filled in. The bottom row can also be filled. This also makes the 6 2x3 in some order, and the 0 region in column 5 is 0x2. This accounts for all the 2s in rows 3-6, and forces the one in the last column into row 3, and this puts two 3s into rows 4-6, leaving only one more.

Now consider the 2-cell regions 8, 4, and 9, which cannot contain a 2. The 8 can only be 3+5, using the last 3 available. The 9 can only be 4+5, which puts 5s in both rows. The 4 can then only be 1x4. Now the top row can be filled in by elimination, and the only place for a 4 in row 3 is in the 24. Then the rest can be filled in by elimination.

4621350
2364105
6035412
1502643
5143026
3450261
0216534
.the .greg[info]lardarsegreg on March 14th, 2009 11:14 pm (UTC)
I'm actually rather surprised that you didn't do anything for Pi Day...
lunchboy[info]lunchboy on March 16th, 2009 05:44 pm (UTC)
Nice. I assumed the 0-shape in the upper left was where the solving path was going to end, but then I liked that it got filled in earlier than that to resolve placements elsewhere.
Robert Hutchinson[info]ertchin on March 18th, 2009 01:15 am (UTC)
I haven't been commenting because I'm horribly behind on reading LJ. But I'm catching up, and I especially liked this one.
motris[info]motris on March 18th, 2009 01:28 am (UTC)
My guess is a lot of people are a bit behind, so I could honestly slip to once every other day and that would still be ok.

I really liked how this zero KenKen played out and bet a couple more, that actually use the 0 in + and - regions as well, would have good depth.