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21 March 2009 @ 09:25 am
3/21/09 - Last Hurrah  
9x9 KenKen*

*(use digits 1 to 9, multi-cell division and subtraction are possible)

Beware! While this has sudoku-like boxes, there is no 3x3 region constraint. I hope that this, like my "Here Comes The Sun" puzzle from two weeks ago, is a very fun but challenging puzzle for you. Out of all my puzzles, while the theming on other days of the week has been stronger visually, the Saturday puzzles are the three I'll come back to and appreciate the most.

This will be my last "daily" KenKen for awhile. I have a couple other puzzle projects that in the next two months need a lot more attention than I've been giving them. I may post an occasional idea here - I still have some concepts I want to explore - but don't count on too many in any given week or month. I can report, for those who have enjoyed these puzzles, that there is a high probability I will be writing a book of these puzzles sometime in the future. Perhaps when I start working on that project, and have some "slush" puzzles from my writing, I'll begin posting puzzles regularly here again, but that won't be until this summer.
( 6 comments — Leave a comment )
(Anonymous) on March 22nd, 2009 02:08 am (UTC)
Can I claim a 1get here :p

I loved how this was so counter-intuitive, pretending to look like sudoku and then totally bitch-slapping you if you started thinking along those lines.

The intitial work-in to the puzzle is very, very nice. In fact (with the 0-kenken coming in a fairly distant second) it has to be *the* highlight of your series. Well according to this ken-ken sceptic anyway, heh.

Thanks for sharing this series, it's generally been good fun solving (and arguing!)

Robert Hutchinsonertchin on March 22nd, 2009 03:07 am (UTC)
This one definitely gets a "hurrah" from me. Let's see if I can write a solving description that is at all followable.

The 23 on the bottom row must be a sum of 6,8,9. The 50 to the left can only either be a sum of 7,8,8,9,9,9 or a product of 1,1,1,2,5,5. The 9 within the 23+ forces the latter, with diagonals of 1,1,1, 5,5, and 2.

The 45 in the sixth row can only be a product of 5 and 9, ordered 9,5 due to the 5 below.

The bottom row now has 2,5,1, then some order of 6,8,9, leaving 3,4,7 to go into the 22 on the right. This makes the region a sum. The two possible sets for the remaining squares are 2,2,4 and 2,3,3. Both of these force a 2 into the rightmost column, which means that the 50 at the top cannot be a product due to the conflict with the placement of the 2. It is instead a sum, with the diagonals of 9,9,9, 8,8, and 7.

The 15 in the sixth row cannot be a product (the 5 is already placed on that row), so it is a sum, forced to be 8,7 by previous placement. This lets us place the 7 in the lower left of the 22+. This now eliminates putting 2,3,3 in the upper three squares (conflict with the 3 below it), so that's (left to right top to bottom) 2,2,4,4,3.

The 17 in the lower right must be a sum (no factors), and since it now cannot have any 9s, 8s, or 7s, it must be 5,6,6. The 16 just to its left cannot be a product of 2 and 8 (conflict to the right), so it's a sum of 9 and 7 in some order. This lets us place the 9 in the leftmost square of the 23+ below.

We now have enough 9s in the grid to know that there is one in the 2 region on the lower left, forcing that to be a difference. The two digits being subtracted must be 3,4 (2,5 and 1,6 both have conflicts). The two 8 digits in rows 7 and 8 must now go, one each, into the 5 and 1 regions, making both of those differences. We can now finish the 23+ below as 9,6,8. Rows 7 and 8 will now all fill in once we get a conflict established higher in the grid.

The 32 in the upper left is a sum, as a product would require using at least one 1, which would conflict below. The 15 next to it is also a sum, since it cannot use 5 as a factor for the same reason. The 18 in the first column is also a sum, as its factors are 2,3,3 and you can't combine any of those without leaving a conflict with the 1 below.

Due to other 9 placements, we can put a 9 in the top of that 18+, leaving 6 and 3 as the only other possible entries. This leaves 4,7,8 to go directly above it, and in that order going down. Due to other 8 placements, we know that 8s go into both of the 1 and 6 regions in the middle left, making both of those differences of 8,7 and 8,2 in some order.

The top of the 32+ must now have 6 and 3 next to the 4, as any other (smaller) digits cause a conflict in the remaining box when summing. This lets us put a 4 in the "center" of the upper left region, which will make rows 7 and 8 below now fall into place. This establishes a 3 below, making the top of the 32+ above 4,6,3.

The 15+ must now contain 7,6,2 in some order. Due to the placement of the remaining 7s in the grid, we can put a 7 in the lower box of the 3-region in the middle, making it a difference with a 4 in the upper box.

(pause for a tremendous amount of cogitation)

The first 3-region going down in the 4th column now must contain a 2, no matter what the other number is. This, plus the 2 that will go into the 15+ to its left, means no 2 can go into the 12-region in the upper right. With only the digits 1345 available for the 12-region, the only possible combinations left are either a product of 1,3,4 or a sum of 3,4,5. Either way, we can place a 4 in the lower-left square of this region.

(Broke the character limit! To be continued.)
Robert Hutchinsonertchin on March 22nd, 2009 03:09 am (UTC)

This (whew) lets us place the remaining two 4 digits in the center--at the top of the 8-region in column 6 and the right of the 2-region in row 6. The 8-region must now be the product of 4 and 2, which forces the 2-region to its left to be the difference of 6 and 4. Now we can put 9.6.3 in the middle of the first column, and 6,5,1 in the middle of the last column.

The 2 digit for the 7th column must now go into the 3-region, and the 2 digit for the 5th column goes in the top square. Now the 6 digit in the 6th column must go in the 6-region (evil puzzle!), making that a division of 6 and 1, which lets us fill in 1,2,5 to finish out the top row above. The 5-digit in the 5th column can only go into the 2-region, making a difference of 5,3. Filling in the rest of the grid is (at least compared to the previous description) pretty basic elimination, so please forgive me for not showing my work there.


Hopefully I didn't leave anything out of my logic, or (a bigger danger for me) make an unwarranted leap. Great, great puzzle.
motrismotris on March 22nd, 2009 03:12 am (UTC)
On quick reading (with the ever highlighting mouse), your walkthrough looks perfectly thorough and correct. Glad you enjoyed this challenge.
Robert Hutchinsonertchin on March 22nd, 2009 03:30 am (UTC)
Heh--I just realized that I left out placing the given 9 in the center. Which is appropriate, as I didn't even see it for a couple of minutes when solving.
(Anonymous) on June 11th, 2010 12:28 pm (UTC)

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