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I've searched all over the place and I can't seem to find a solution to my problem. I have an array that contains how many times my random walk lands on a specified location. Now I need to find the sum of this array but for some reason the sum always gives 1000:( 

int[] hx;
hx = new int[1000];
int hx_sum = 0;
for (int i = -500; i < 500; i++) {
   hx[i+500] = totals.get(i);
}
for(int w = 0; w < 1000; w++){
   hx_sum += hx[w];
}
System.out.print(hx_sum);
}

The result of my hx_sum is ALWAYS = 1000.

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17
  • 1
    For better help sooner, post an SSCCE. Commented Nov 28, 2013 at 15:29
  • what is returned by totals.get(i), that's what's filling your array. Commented Nov 28, 2013 at 15:30
  • 1
    No, it's not because the numbers that come out of it fill your array and those are the numbers that are summed to get your total Commented Nov 28, 2013 at 15:31
  • 1
    @user2966573 No it's relevant. Please add this informations. Commented Nov 28, 2013 at 15:31
  • 1
    Maybe we can get at this another way. You plan to use this total for something. What? Almost certainly, you don't want the total because it is just NBR_WALKS, but maybe there is some other function of the hash map that would mean what you want. Commented Nov 28, 2013 at 15:46

1 Answer 1

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4

You do 1000 steps (NBR_WALK), and in each step you increment one point in your hash map. The map was initialized with all 0 before.

Quick question: What is the sum of 1000 increments?

Do you see the error? It is not really an error, it is just mathematically certain that no different result is possible.

If you don't see it, I suggest you take a dice, throw it 10 times, and increment each time the array element corresponding to the number (which you can draw on paper). Initialize the array with all 0 before you start. Then compute the sum of the array. It can only ever be 10.

This could look like:

Dice      1s         2s      3s     4s     5s    6s
  1      x
  5      x                                 x
  3      x                   x             x
  1      xx                  x             x

Now count the xs in the last line. Its 4, because you were rolling the dice 4 times, and every time you added 1 x. Also, in each line, there is exactly one more x then in the preceding line. Thus, after 1000 rounds you will have 1000 x.

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8 Comments

It's all functioning exactly like it should, but I need to be able to have my program calculate the sum of the array, all the slots added together.
Yes, @user2966573. Think about it: if - by chance - you landed always on the same spot, what would be the total for that spot? What would be the total for all other slots?
I'm sorry mate, but when I print out my totals.get(i) i get exactly what I'm looking for, I just want all these numbers added together.
Here's another way to see the issue. You initialize totals with all values 0. The only thing that changes a value is the line totals.put(total_value, totals.get(total_value) + 1);. You do that exactly once on each iteration of a loop. What will the total value be after one iteration? After 2 iterations? After 3 iterations? After 1000 iterations?
Still another way: Suppose the sum of all totals is n. Then you do 1 step and increment exactly 1 slot. What is the sum of the array now?
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