Ooh, I miss these!
I think I might post a similar thing (or two), but maybe in Euler's Avenue; it's not really Computer Math. I had a Descrete Mathematics course this semester, and the prof showed us some very cool proofs involving partitions, mostly combinatorial!
Yeah, basically what I did is partition the length into 1's and 2's, then find the number of ways to permute each one and to insert the one step backwards. (I.e., the same thing as you, phro, except I did not actually permute anything, just counted the permutations.)
Hello MathisFun forummers,
I stumbles accross this excercise:
and I was thinking why should I rationalize it?
You are NOT rationalizing it; you are doing the reverse!
To rationalize is to eliminate the square-root signs in the denominator. Here you want to put the square-root signs back in the denominator!
And why do you that? Because it's the only way to solve limit problems of this sort!
Not exactly true...
Maybe a bit more complicated than the anti-rationalization method, but much more generalizable.
You are usign the highest sum you can find in the grid. In the 2x2 case, that is 6. If you arrange the numbers any other way, you'll get a higher maximum sum. For example:
For the second, 4x4 grid, the highest sum inside it is 19. To prove that this is the minimal highest sum, we would need to prove that for all other arrangements of numbers from 1 to 16 into the grid, there will be two adjacent numbers which sum to 19 or higher.
You look at all possible fillings of the grid and for each filling you see what the maximum sum of adjacent ellements is. Then you find the minimum of those numbers.
For example, for n=1, you are filling a 2x2 grid with numbers 1,2,3,4. The lowest adjacent sum, which is 6, can be obtained with the filling:
For n=2, a filling with S=19 *can* be found, but I am not sure if there is a lower one (probably not):