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motris
motris
20 March 2009 @ 09:00 am
3/20/09 - What?  
7x7 Mystery KenKen -



Rules: Fill in each row and column with a set of seven different non-negative integers such that each integer appears a single time in each row/column. The values in the upper-left of the bold-regions must be satisfied by after applying the indicated operation to the digits inside the cells.


Last week I introduced "Zero KenKen", a simple change to the formulation that seems to have some interesting depth to it. While there are a ton of other variations out there to do (from crypto forms to ones that use different mathematical operations), one thing that seems potentially interesting to me, and a logical extension of the addition of zero, is to use a variable set of digits (given to solvers) since certain digit sets will be better for all the operations than 1 to n. For example, in a 4x4 puzzle, the set {1,2,3,6} seems to be an improvement worth exploring as I can do many different multi-cell subtractions and divisions alongside multiplications and additions. If I gave you a 4x4 puzzle, and gave you the digits {1,2,3,6}, the resulting puzzle seems fair even if you have to relearn some forced sums.

Well, I'm not happy introducing this concept of different digit sets with a simple puzzle where I give you the digits. I want to introduce the concept with a blog-level (and less-commercial) puzzle where you have to figure out what digits get used yourself. If you can find the integers from 0 to ??? that are needed for the puzzle, you can maybe solve this Mystery KenKen.
 
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( 13 comments — Post a new comment )
[info]ericberlin on March 20th, 2009 05:04 pm (UTC)
That was great. Your puzzles make me feel smart.
zundevil[info]zundevil on March 20th, 2009 05:13 pm (UTC)
This is *almost* what I was going to propose or, heaven forbid, actually write. Not the cool ? design, but maybe something that only had a rule that you could not repeat any digits/integers in a row or column. Mine included the ability to have different sets of digits for each row/column, although I don't know how possible that is to construct. Especially for me.
motris[info]motris on March 20th, 2009 05:20 pm (UTC)
If you had a ground rule like numbers from 1-9 are used, then something 6x6 to 7x7 with "whatever digits you want, just don't repeat them" would be constructable, but it might be hard to make it very interesting. The pro is that a mostly filled column/row has several possibilities left. The con is that a mostly filled column/row has several possibilities left. Either the cage breakdowns are obvious, or the construction is super meticulous, or you might veer into non-uniqueness space.

I guess an overarching constraint like "use four occurrences of each digit" in a 6x6 puzzle could work, to give it a little more form at the end, but a lot of potential interest before getting there.
Robert Hutchinson[info]ertchin on March 20th, 2009 10:48 pm (UTC)
That bent my brain a little. I did one of two what-ifs, thought I proved it impossible, went with the other one, found *it* impossible, and then had my way in by going back and checking my work.

I may have possibly made this one a bit harder, at least at the outset, by not assuming that all the entries would be one digit.
motris[info]motris on March 20th, 2009 10:54 pm (UTC)
To be honest, I wanted people to think it could go to, say, 11 or something like that, without actually going there yet. I could have even said just "integers" in the description, and likely still had a unique puzzle, but allowing negative numbers also felt less elegant than I wanted for the first example so I gave that helpful constraint to help in the id-ing of digits
acroarcs: Arc lamp[info]acroarcs on March 20th, 2009 11:36 pm (UTC)
I narrowed the possible set of integers down to three possibilities before having to brute force those three. (It turned out to be the last set, of course.) Although since it was the same problem that eliminated the bad two, there was probably a higher order conceptual issue I didn't quite grab hold of, and I'm left wondering what that was.
motris[info]motris on March 21st, 2009 12:07 am (UTC)
The way to identify all the numbers is basically this:The right column gives you an overall sum of 31. The left column has an outie which must be 4. There is another 4+ in that row which, with a nonnegative constraint, must be 1 and 3. Summing up the bottom 2 rows you have 56 without the dot of the question mark and a single cell that is in the 14+. This means the sum of the dot and that single cell is 6. The dot cannot be 1, 3, 4 from numbers already in that row, or 2 (as this would put a 4 in the same column as the 4) so it could be 0, 5, or 6 with the other digit being 6, 1, or 0, giving the other 14+ cage number as 8, 13, 14.

In the case with 0/14 in the 14+, you now have 6 {0,1,3,4,6,14} of the seven digits specified with a total sum of 28 leaving 3 for the missing number which is not possible. In the 1/13 case, you have 5 {1,3,4,5,13} of the seven digits specified with a total sum of 26 with 5 to go. However, there are no two digits that can join the set to leave 31 total so that is also no go.

So the 14+ must be 6/8 and the dot of the question mark is the dot-like number 0. The remaining digit is 9.
cyrebjr: Numbers[info]cyrebjr on March 21st, 2009 12:49 am (UTC)
I'd like to ask how you feel about using the word "spectrum" for the set of allowable numbers. Because that's how I'm about to use it. What follows is a slightly simpler alternative to your second paragraph.

When we conclude that R6C2 is one of 0, 5, or 6, we know that R5C2 equals 14, 13, or 8. But if 13 or 14 is part of the spectrum, the only places the can go in R67 are both C7, which aren't both possible. That leaves 8 for R5C2.
motris[info]motris on March 21st, 2009 12:55 am (UTC)
Sure, the terminology is fine and the explanation suffices and is indeed simpler. I guess I'm a bit more of a "big picture" guy so, whenever I think of an assignment, I go back to the overall constraint before I look for potential contradictions in the puzzle itself.
Robert Hutchinson[info]ertchin on March 22nd, 2009 01:23 am (UTC)
Ahhh. I looked at the bottom two rows, but failed to learn the proper lesson, only noting that the top number in the +14 would be 8 greater than the dot of the question mark.

Having established 1, 3, and 4 on the bottom row, I looked at the +3 and realized it had to be either 0,3 or 1,2. Using a 0 left the other digits as 6,8,9, while using a 2 left several possibilities. I proceeded to do some math incorrectly from there, as previously described.
devjoe[info]devjoe on March 21st, 2009 12:52 am (UTC)
Oh, this looks like a killer. :-)

For starters, we know from the last column that the seven numbers add to 31. The sum on the three regions in the first column then tells us that the bottom of the second column is a 4. So the top two numbers in that column add to 10.

Now the five sums that lie entirely in the bottom two rows add to 56. The whole two rows add to 62, so the bottom of the 14 and the dot of the ? add to 6. Those two cells cannot be 2 and 4 since both are in line with the 4, so they are both 3, or 1 and 5, or 0 and 6, since we're told they are non-negative.

There is also a two-cell 4 in the bottom row. With 4 already appearing in the row, this can only be 1+3. That means that there must be a 0; if the two digits from the last paragraph are 1 and 5, the 1 must go in the 14, making the other cell 13, and 13 + 1 through 6 is more than 31. These two cells cannot both be 3s now, so the only other option is that they are 0 and 6.

Suppose that that 14 is 13+1. Then the 13 in the last column must go into the 14, and its other cell is a 1, but we've already placed 1s in both rows in this case. So this case is impossible, and the bottom of the 14 in column 2 and the dot of the ? are 0 and 6 in some order. And our set of numbers is now 0, 1, 3, 4, 6, and two other numbers. The last two add to 17. If the bottom of the 14 is a 0, then there is a 14, but then the last number would have to be another 3, so this is not the case. The bottom of the 14 is a 6, so the top is an 8, and the set of numbers is 0, 1, 3, 4, 6, 8, 9.

Now it's much easier. The only way to fill the remaining 11 of the 15 in the lower left corner is 8+3, which must go in with the 3 on top. The 14 in the lower right is 8,6, and the 10 in column 3 is 1,9. Then the 14 in column 5 can also be filled as 8+6. With 1 and 9 gone, the three-cell 14 must be 8+6+0 which can only be entered in that order.

A number of the other regions are now limited to specific sums. 3=0+3, 5=1+4, 7=3+4, 13=4+9, and the three-cell 19=9+6+4. This forces the 7 to be entered with 4 on top; the rest of column 3 and the rest of the 19 can be entered. The second row can be filled by elimination, and the top of column 2 is a 9, so the 15 in the upper right can be filled.

The three-cell 13 must be 9+4+0 in that order. The 5+ and 13+ in rows 4-5 both have 4s, so 6 rows and columns have 4s, and the remaining one is in the middle of row 3. Now the 8 in the middle column must go on top, and the rest can be filled by elimination.

6938014
0146983
9084361
4361890
1803649
3619408
8490136
zebraboy3[info]zebraboy3 on March 21st, 2009 06:47 pm (UTC)
Someone different route to the integer set:




I should've, but didn't, notice the constraint on the bottom two rows, so my method of arriving at the set of numbers was a bit more roundabout.

1) R7C2=4, R7C5,6=1,3 as described above.

2) Given that C2R1,2 must =10, there are 10 possible sets of integers that can fill this column.

3) 3 sets can be eliminated because they do not contain both a 1 and 3

4) In all sets, the greater of the digits used in the C2 14 cage can only fit in C7 into the 14 cage also; thus, those two cages have the same digits. Therefore, R7C2,7=14 and R7C1,3=4.

5) Only one set of integers from above contains the appropriate digits to fill C1R6,7 and C3R6,7

From there, a pretty straightforward solve.

Very, very nice.
zebraboy3[info]zebraboy3 on March 21st, 2009 06:48 pm (UTC)
rrr, "somewhat", not "someone". Obviously this hurt my brain.