Question

Provide a most efficient divide-and-conquer algorithm for determining the smallest and second smallest values in a given unordered set of numbers. Provide a recurrence equation expressing the time complexity of the algorithm, and derive its exact solution in the number of comparisons. For simplicity, you may assume the size of the problem to be an exact power of a the number 2

Answer 1

The efficient divide-and-conquer algorithm follows a tournament-like approach, leading to n + log2(n) - 2 comparisons for finding both the smallest and second smallest numbers in a power-of-2 sized unordered set.

Step-by-step explanation:

The question deals with creating an efficient divide-and-conquer algorithm to find the smallest and second smallest values in an unordered set of numbers. Assuming the set size is a power of 2, the algorithm implemented will successively pair up elements, compare, and then carry forward the smaller ones along with second smallest candidates in a tournament-like fashion until the smallest two are determined. A potential recurrence relation for this algorithm could be T(n) = 2T(n/2) + 2, where T(n) is the number of comparisons to find the smallest and second smallest of n elements.

In this case, the exact solution to the recurrence relation is shown by unfolding it:

On the first level of recursion (n elements), we do n/2 comparisons to find the smallest of each pair.

On the second level (n/2 elements), we do n/4 comparisons to continue the tournament.

Continuing this process, we find that we make 2 comparisons at each level of the tournament for the second smallest.

Since there are log2(n) levels in the tournament (as the problem size halves at each step), we do a total of n - 2 comparisons for the smallest element and an additional log2(n) comparisons for the second smallest, leading us to the final count of n + log2(n) - 2 comparisons.

Answer 2

Check the explanation

Step-by-step explanation:

Let algorithm be Called minSecondMin(A, p, q) where A is input array with starting index p and ending index q. q-p+1 = n

pseudoCode in C style:-

minSecondMin (A,p,q)// suppose it takes T(n) time

{

if(p +1 == q)

{

if(a[p]>a[q])// here one comparison is done

{

min = a[q]

secondMin = a[p]

}

else

{

min = a[p]

secondMin = a[q]

}

}

else

{

m = floor (p+q/2)

(min1, secondMin1) = minSecondMin (A, p, m)// it takes T(n/2)

(min2, secondMin2) = minSecondMin (A, m+1,q)//it takes T(n/2)

if(min1 > min2)// here 1 comparison

{

min = min2

if(min1>secondMin2)//here 1 comparison

secondMin = secondMin2

else

secondMin = min1

}

else

{

min = min1

if(min2>secondMin1)//here 1 comparison

secondMin = secondMin1

else

secondMin = min2

}

}

}

Recurrence relation for time;-

T(n) = O(1) , if n = 2

= 2T(n/2) + c , if n> 2

Recurrence relation for number of comparisons:-

Let number of comparison = C(n)

Then C(n) = 1 , if n = 2

= 2C(n/2) + 2 , if n > 2

C(n) = 2C(n/2) + 2

Solving using substitution method we get

= 2(2C(n/4) + 2) + 2

=
`2^2`C(n/4) +
`2^2` + 2

=
`2^3` C(n/8) +
`2^3` +
`2^2` + 2

Continuing this k times

=
`2^k`C(n/
`2^k`) +
`2^k` +
`2^{k-1` +
`2^{k-2` + ...+
`2^2` + 2

It will stop when n/
`2^k`= 2 so k = log​​​​2​​​n - 1

=
`2^{log_(2)n - 1} C(1) + 2(2^(k-1+1) - 1)`

=
`2^{log_(2)(n/2)} + 2^(k+1) - 2`

=
`n/2 + 2^{log_(2)n} -2`

= n/2 + n - 2

= 3n/2 - 2

So, required number of comparisons = 3n/2 - 2