if 7+6 = 113 and
9+1= 810 and
5+3 = 28 and
9+6 = 315 then what does 7+2 = ?

Paul K’s answer gives ALL the possible combinations, regardless of cages, since the rules of how the numbers can be arranged in the grid will still apply. The cages and the operations they contain simply guide the player with number placement. These are determined by the KenKen Generator Programs after the numbers have been filled, and depend on a difficulty level. The possible combinations of numbers that may be chosen from is quite limited to the grid’s size, as explained by several contributors to this thread. Perhaps you would enjoy rewriting the code for one if actually playing has ceased to be diverting.

RESULTS OF CODE:

Grid size (n): 2
Number of grids: 2

Grid size (n): 3
Number of grids: 12

Grid size (n): 4
Number of grids: 576

Grid size (n): 5
Number of grids: 161280

Grid size (n): 6
Number of grids: 812851200

Grid size (n): 7
Number of grids: 61479419904000

Say our field is 3×3

123
312
231

After that you shuffle the vertical lines and you get something like that:

231
123
312

After that you shuffle the horizontal lines:

312
231
123

You now have a random kenken. For each operation you have 3! possibilities. The total permutations of a 3×3 kenken is thereby (3!)^2

And to generalise it for a n*n kenken:

(n!)^2

cheers,

wintix

For Ken Ken squares of size 2×2, 3×3, and 4×4, I find 2, 12, and 864 possibilities, respectively. So, I agree with a number of your responders. My goal was to work this out for 6×6, but at present I have not even finished the 5×5 case. The bigger the NxN value the more branch points there are, and there does not seem to be any obvious pattern to it that you can write down. I would like to see the formula for NxN if anybody knows it. It would be very hard to get it by empirical extrapolation from the three cases I solved.

P. Kunasz

According to wikipedia there is not a general solution to this problem for a nxn square – and 4×4 is about the last case where it is feasible to work it out in a few minutes.

Now based on column conflicts alone, there are 2 possible values for each value in the third row, meaning that there are 16 possible permutations that would not violate the column restrictions. 24*9*16 = 3456, Mr. L’s initial answer. However, you might see that intuitively this doesn’t make much sense, since row 3 would have more permuations than row two, despite having more constraints – this answer does not consider the row restrictions in row 3!

So let’s consider what is known about row 3. We know the fourth element is defined by the first 3, so only has one possibility. We know the first element in the row only has column restraints so has 2 possible values for each permutation of rows 1 and 2.

Element 2 of row 3 can have 1 or 2 possible values depending on the columnn restraints from row 1 and 2 and the row restraint provided by the first element in row 3. Similarly, element 3 can have 0 (over constrainted), 1 or 2 possibilities depending on the column restraints of row 1 and 2 and the row restraints of the first two elements in row 3. By inspection I saw that 3 of the permissable permuations for row 2 would result in 4 permissable permutations in row 3, while the other 6 permuations in row 2 would only result in 2 permissable permuations in row 3 (i.e. the row is full specified by the first element). So the total number of permutations for a 4×4 Ken Ken is (24*3*4)+(24*6*2)=576