What is the proof of of (N–1) + (N–2) + (N–3) + ... + 1= N*(N–1)/2 [closed]_问答_开发者

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I got this formula from a data structure book in the bubble sort algorithm.

I know that we are (n-1) * (n times), but why the division by 2?

Can anyone please explain this to me or give the detailed proof for it.

Thank you

(N-1) + (N-2) +...+ 2 + 1 is a sum of N-1 items. Now reorder the items so, that after the first comes the last, then the second, then the second to last, i.e. (N-1) + 1 + (N-2) + 2 +... The way the items are ordered now you can see that each of those pairs is equal to N (N-1+1 is N, N-2+2 is N). Since there are N-1 items, there are (N-1)/2 such pairs. So you're adding N (N-1)/2 times, so the total value is N*(N-1)/2.

Start with the triangle...

    *
   **
  ***
 ****

representing 1+2+3+4 so far. Cut the triangle in half along one dimension...

     *
    **
  * **
 ** **

Rotate the smaller part 180 degrees, and stick it on top of the bigger part...

    **
    * 

     *
    **
    **
    **

Close the gap to get a rectangle.

At first sight this only works if the base of the rectangle has an even length - but if it has an odd length, you just cut the middle column in half - it still works with a half-unit-wide twice-as-tall (still integer area) strip on one side of your rectangle.

Whatever the base of the triangle, the width of your rectangle is (base / 2) and the height is (base + 1), giving ((base + 1) * base) / 2.

However, my base is your n-1, since the bubble sort compares a pair of items at a time, and therefore iterates over only (n-1) positions for the first loop.

Try to make pairs of numbers from the set. The first + the last; the second + the one before last. It means n-1 + 1; n-2 + 2. The result is always n. And since you are adding two numbers together, there are only (n-1)/2 pairs that can be made from (n-1) numbers.

So it is like (N-1)/2 * N.

See triangle numbers.

I know that we are (n-1) * (n times), but why the division by 2?

It's only (n - 1) * n if you use a naive bubblesort. You can get a significant savings if you notice the following:

After each compare-and-swap, the largest element you've encountered will be in the last spot you were at.
After the first pass, the largest element will be in the last position; after the k^th pass, the k^th largest element will be in the k^th last position.

Thus you don't have to sort the whole thing every time: you only need to sort n - 2 elements the second time through, n - 3 elements the third time, and so on. That means that the total number of compare/swaps you have to do is (n - 1) + (n - 2) + .... This is an arithmetic series, and the equation for the total number of times is (n - 1)*n / 2.

Example: if the size of the list is N = 5, then you do 4 + 3 + 2 + 1 = 10 swaps -- and notice that 10 is the same as 4 * 5 / 2.

Sum of arithmetical progression

(A1+AN)/2*N = (1 + (N-1))/2*(N-1) = N*(N-1)/2

This is a pretty common proof. One way to prove this is to use mathematical induction. Here is a link: http://zimmer.csufresno.edu/~larryc/proofs/proofs.mathinduction.html

Assume n=2. Then we have 2-1 = 1 on the left side and 2*1/2 = 1 on the right side.

Denote f(n) = (n-1)+(n-2)+(n-3)+...+1

Now assume we have tested up to n=k. Then we have to test for n=k+1.

on the left side we have k+(k-1)+(k-2)+...+1, so it's f(k)+k

On the right side we then have (k+1)*k/2 = (k^2+k)/2 = (k^2 +2k - k)/2 = k+(k-1)k/2 = kf(k)

So this have to hold for every k, and this concludes the proof.

Here's a proof by induction, considering N terms, but it's the same for N - 1:

For N = 0 the formula is obviously true.

Suppose 1 + 2 + 3 + ... + N = N(N + 1) / 2 is true for some natural N.

We'll prove 1 + 2 + 3 + ... + N + (N + 1) = (N + 1)(N + 2) / 2 is also true by using our previous assumption:

1 + 2 + 3 + ... + N + (N + 1) = (N(N + 1) / 2) + (N + 1) = (N + 1)((N / 2) + 1) = (N + 1)(N + 2) / 2.

So the formula holds for all N.