Random array generator

Question

Given 'n', 'm', 'k', 'x' and 'y' integer values...
I have a numeric ArrayList with 'n' positions and I need to create 'k' other arrays using the values in the first array and with 'm' positions. How can I did it ensuring that the sum of the numbers is 'x' with a maximum margin of error of 'y' and the arrays to be as different as possible between them?

I will use this in a test generator to randomize the questions. The numbers represent the difficult of the questions. When I tried to do it I randomize situations and checked if they were correct, but that is very slow. Someone knows a better way to do this?

Very confusing question. It is not clear at all what you want... consider serious revision. — Menelaos, Apr 28 '13 at 11:29

score 1 · Answer 1 · answered Apr 28 '13 at 11:34

From Your description it sounds like a variation of Discrete Knapsack Problem. Basically You search for several solutions of a modifies DKP - if there will be more that k of them You can remove additional ones, if less - You can permute that ones You obtained to generate some more.

The naive implementation would be searching solutions of DKP from n = x-y to x+y, and then processing them as described above, it could be really slow though. You might obtain some better solution asking on Mathematics Stack Exchange.

score 0 · Accepted Answer · answered Apr 28 '13 at 11:51

You have some less than n! / (m! . (n-m)!) acceptable solutions, out of which to pick as most differing solutions.

The possible candidate solutions adhere to an optimal cost of square of deviation to y.
For a fixed number of possible solutions pick as definitive solution where the difference to prior accepted definitive solutions is minimal: sum of difficulties of same entries. (This is just locally optimal, but should do.)

Sort the n# list on decreasing difficulty. Iterate for m# sublists in principle for n! / (m! . (n-m)!).

Change take candidates with respect to the allowed range: skipping/failing on those out of range.

score 0 · Answer 3 · edited May 23 '17 at 11:52

First sort your array.then find value of the element that is closest to X/m that i call mid.

find m number closest to mid.or use this idea

So set source point equal mid:

for (int i=0; i < n ; i++)

{ a[i]=a[i]-mid }

now you need m numbers that the sum of the numbers is zero,see link1&link2&link3 maybe useful. for better guidance please explain that what it is your mean that the arrays to be as different as possible between them.

Random array generator

3 Answers3