Question

1. Order following function by growth rate: N, √N, N^1.5, N log (N), log (log (N)), log (N) log (N), N², 2^N, 200, N^N

2. Give a useful Θ (big Theta) estimation for each of following function t(n).

a. t(n) = 122 * 212

b. t(n) = 2log2(n2) + log4(n ) + (log2 n) 2 + (log2 (202)) 2

c. t(n) = 3t(n/2) + n

d. t(n) = 3t(n/2) + (n+1)(n-1)

e. t(n) = 4t(n/2) + (n2 + n-1)

f. t(n) is the runtime of following function,

public static int f1(int n){

int mid = n/2;

for (int i = mid; i >= 0; i--) System.out.println(i);

for (int i = mid + 1; i <= n; i++) System.out.println(i); return mid;

}

g. t(n) is the runtime of following function, public static int f2(int n){

if (n < 1) return 1; //update from original int mid = n/2;

mid = f2(mid);

for (int i = 30; i > 0; i /= 3){

System.out.println(i );

}

return mid; }

h. t(n) is the runtime of following function, public static int f3(int n){

for (int i = n; i >= 0; i--){

for (int j = 0, j <= i + i; j++)

for (int k = n; k > 0; k /= 3) System.out.println(i * j + k);

}

return n;

}

i. t(n) is the runtime of following function,

public static int f4(int [] a, int start, int end){

int ans = 0;

if (start >= end) ans = a[start]; else {

int mid = (start + end) / 2;

int x = f4(a, start, mid);

int y = f4(a, mid + 1, end);

print(a, start, end); //print each element in a from start to end if (x < y) ans = x;

else ans = y;

}

return ans;

}

public static void print(int [] a, int s, int e){

for (int i = s; i <= e; i++) System.out.println(i); }

j. t(n) the run time of the following method: The method removeLast for a doubly linked list that has size n

Answer 1

1. 200 < O(log(log n )) < O(logn) < O(log n log n) <O(√n )< O(n) < O(1.5N) < O(2N) < O(nlog n)< O(N2) < O(NN)

While for some cases 200 would not be the least but constant complexity algorithm is always preferred over those which depends on N(number of elements)

2. a)An algorithm is said to be constant time (also written as O(1) time) if the value of T(n) is bounded by a value that does not depend on the size of the input(N)

Hence t(n) = Θ(1)

2.b) t(n) = 2log2(n2) + log4(n ) + (log2 n) 2 + (log2 (202)) 2

T(n)=Θ (log n²) ,since 2log ₂ (n2) would be the greatest degree in above expression.

2.c)t(n) = 3t(n/2) + n

T(n)= Θ(n) ,since highest power of n Above is 1.

2.d)t(n) = 3t(n/2) + (n+1)(n-1)

Highest power here is n ² as (n+1)(n-1) gives us n^2 -1

T(n)= Θ(n ² )

2.e)

t(n) = 4t(n/2) + (n2 + n-1)

Although highest power is n^2 but time dependent highest power is n in term 4t(n/2) ,hence time complexity

T(n)=Θ(n^2)

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