Question

Silly-Sort(A,i,j)
if A[i] > A[j]
then exchange A[i] and A[j];
if i+1 >= j
then return;
k = floor(j-i+1)/3);
Silly-Sort(A,i,j-k);
Silly-Sort(A,i+k,j);
Silly-Sort(A,i,j-k);
(a) Argue (by induction) that if n is the length of A, then Silly-
Sort(A,1,n) correctly sorts the input array A[1...n]
(b) Give a recurrence relation for the worst-case run time of Silly-Sort
and a tight bound on the worst-case run time
(c) Compare this worst-case runtime with that of insertion sort, merge
sort, heapsort and quicksort.

Answer 1

A.

By induction:
For the base case let n = 2. The first two lines of the algorithm will check if
the two elements are sorted; if not, it exchanges them (and now they are sorted).
The algorithm returns after the following if statement. Thus Silly-Sort sorts
correctly for n = 2.
Assume Silly-Sort correctly sorts an input array of size 2n/3 (you may also
assume that Silly-Sort correctly sorts an array of size k, for any 1 ≤ k < n).
Let A[1..n] be an input array of size n. By the induction hypothesis the first call to
Silly-Sort(A, i, j−k) correctly sorts the first 2n/3 elements, so that the elements
1 . . .n/3 are less than elements (n+1)/3 . . . 2n/3. The call to Silly-Sort(A, i, j−
k) correctly sorts the last 2n/3 elements, so that the elements (n + 1)/3 . . . 2n/3

are less than elements 2(n + 1)/3 . . .n. Also, after this call to Silly-Sort the
elements 2(n + 1)/3 . . .n (the last n/3 elements) are the largest n/3 elements in A.
The last call to Silly-Sort(A, i, j − k) sorts correctly (by induction hypothesis)
the elements 1 . . . 2n/3. These sorted elements are less than elements 2(n+1)/3 . . .n.
Thus the array A of size n is sorted.

b. Give a recurrence for the worst-case running time of Silly-Sort and a tight
asymptotic (Θ-notation) bound on the worst-case running time.
Solution:

T(n) = 3T(2n3) + Θ(1)

= . . .

= Θ(n^log3/2 3)

= Θ(n^2.7...).

c.
Insertion-Sort: Θ(n^2)
Merge-Sort: Θ(n lg n)
Heapsort: Θ(n lg n)
Quicksort: Θ(n^2)
Silly-Sort: Θ(n^2.7...)