Exercise 2.2.2

Consider sorting $n$ numbers in an array $A$ by first finding the smallest element of $A$ and exchanging it with the element in $A[1]$. Then find the second smallest element of $A$, and exchange it with $A[2]$. Continue in this manner for the first $n - 1$ elements of $A$. Write pseudocode for this algorithm, which is known as selection sort. What loop invariants does this algorithm maintain? Why does it need to run for only the first $n - 1$ elements, rather than for all $n$ elements? Give the best-case and the worst-case running times of selection sort in $\Theta$-notation.


  for i = 1 to A.length - 1
      min = i
      for j = i + 1 to A.length
          if A[j] < A[min]
              min = j
      temp = A[i]
      A[i] = A[min]
      A[min] = temp

Loop invariants

At the start of each iteration of the outer for loop, the subarray $A[1..i - 1]$ contains the smallest $i - 1$ elements of the array, sorted in nondecreasing order.


At the start of each iteration of the inner for loop, $A[min]$ is the smallest number in the subarray $A[i..j - 1]$.

Why $n - 1$ elements?

In the final step, the algorithm will be left with two elements to compare. It will store the smaller one in $A[n-1]$ and leave the larger in $A[n]$. The final one will be the largest element of the array, since all the previous iteration would have sorted all but the last two elements (the outer loop invariant). If we do it $n$ times, we will end up with a redundant step that sorts a single-element subarray.

Running times

In the best-case time (the array is sorted), the body of the if is never invoked. The number of operations (counting the comparison as one operation) is:

$$ (n - 1)(\frac{n + 2}{2} + 4) $$

In the worst-case time (the array is reversed), the body of the if is invoked on every occasion, which doubles the number of steps in the inner loop, that is:

$$ (n - 1)(n + 6) $$

Both are clearly $\Theta(n^2)$.