I'm trying to find out the Complexity of the given program. Suppose we have;
int a = θ;
for (i=θ; i<n; i++){
for(j = n; j>i; j--)
{
a = a + i + j;
}
}
Complexity: O(N*N)
Explanation:
The code runs total no of times
`= N + (N – 1) + (N – 2) + … 1 + 0
= N * (N + 1) / 2
= 1/2 * N^2 + 1/2 * N
O(N^2) times`
Related
I need to find time and space complexity for this java code for recursive calculation of the determinant of a matrix:
public int determinant(int[][] m) {
int n = m.length;
if(n == 1) {
return m[0][0];
} else {
int det = 0;
for(int j = 0; j < n; j++) {
det += Math.pow(-1, j) * m[0][j] * determinant(minor(m, 0, j));
}
return det;
}
}
public int[][] minor(final int[][] m, final int i, final int j) {
int n = m.length;
int[][] minor = new int[n - 1][n - 1];
int r = 0, s = 0;
for(int k = 0; k < n; k++) {
int[] row = m[k];
if(k != i) {
for(int l = 0; l < row.length; l++) {
if(l != j) {
minor[r][s++] = row[l];
}
}
r++;
s = 0;
}
}
return minor;
}
Help: Determine the number of operations and memory consumption of the algorithm with respect to n, and after dividing by n^2 you will get the desired result.
I'm confused. I calculate the time complexity as the sum of the input size (n^2) and the number of steps, and the space complexity as the input size. (?) So its O(n^2) but I don't think I'm doing it right. And why should I divide it (help is from the teacher).
Can someone explain to me how to calculate this in this case?
Let us see. For an input matrix of size n, denote the time complexity as T(n).
So, for a matrix of size n:
if n = 1, we have the answer right away: T(1) = O(1);
otherwise, we loop over j for a total of n times, and for each j, we:
construct a minor in O(n^2) (the minor function), and then
run the function recursively for that minor: this one takes T(n-1) time.
Putting it all together, we have T(1) = O(1) and T(n) = n * (n^2 + T(n-1)).
To understand what is going on, write down what is T(n-1) there:
T(n) = n * (n^2 + T(n-1)) =
= n * (n^2 + (n-1) * ((n-1)^2 + T(n-2)))
And then, do the same for T(n-2):
T(n) = n * (n^2 + T(n-1)) =
= n * (n^2 + (n-1) * ((n-1)^2 + T(n-2))) =
= n * (n^2 + (n-1) * ((n-1)^2 + (n-2) * ((n-2)^2 + T(n-3)))) = ...
Then we write what is T(n-3), and so on.
Now, open the brackets:
T(n) = n * (n^2 + (n-1) * ((n-1)^2 + (n-2) * ((n-2)^2 + T(n-3)))) =
= n^3 + n * (n-1) * ((n-1)^2 + (n-2) * ((n-2)^2 + T(n-3)))) =
= n^3 + n * (n-1)^3 + n * (n-1) * (n-2) * ((n-2)^2 + T(n-3)))) =
= n^3 + n * (n-1)^3 + n * (n-1) * (n-2)^3 + n * (n-1) * (n-2) * T(n-3)))) =...
As we can see going further, the highest term among these, in terms of n, will be
T(n) = n * (n-1) * (n-2) * ... * 3 * 2 * 1^3, which is just n! (n-factorial).
The above is about time complexity.
To address space complexity, consider how recursion uses memory.
Note that it is much different from what happens to the time.
The reason is that, at each point of time, we are in some single particular branch of the recursion tree, and at this time other branches don't use the memory they need.
Therefore, unlike with time, we can't just add up the total memory used by all recursive calls.
Instead, we have to consider all branches of the recursion tree, and find the one that uses maximum memory.
In this particular case, it is just any deepest branch of the recursion.
Denote the memory consumption by M(n) where n is the matrix size.
if n = 1, we have the answer right away: M(1) = O(1);
otherwise, we loop over j for a total of n times, and for each j, we:
construct a minor that takes O(n^2) memory (the minor function), and then
run the function recursively for that minor: this one takes M(n-1) memory.
Note that the loop does not accumulate memory used by minors.
Instead, when we construct the minor for next j, the minor for previous j is not needed anymore.
Thus we have M(n) = n^2 + M(n-1).
Again, writing down what is M(n-1), then M(n-2), and so on gets us to
M(n) = n^2 + (n-1)^2 + (n-2)^2 + ... + 3^2 + 2^2 + 1^2, which is O(n^3) with some constant factor we need not care about.
So, by the above reasoning, the answer is: T(n) = O(n!) and M(n) = O(n^3).
What's the hint about, with dividing by n^2, I don't have a clue, sorry!
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How do I calculate the big-O time complexity of the following nested for loop with dependent indices:
void function1 (int n)
{
int x = 0;
for (int i = 0; i <= n/2; i+=3)
for (int j = i; j <= n/4; j+=2)
x++;
}
Complexity of the code is defined as how many times your code will be executed for a given n.
There are two ways to do it.
Simulation: Run the code for different value of n and find out the values. In this case this is equivalent to the final value of x.
Theoretical:
Let's first check for each i how many times your code runs:
Using the arithmetic progression formula (a_n = a_1 + (k-1)*d):
i=0 => n/4 = 0 + (k-1)*2 => n/8 + 1 times
i=3 => n/4 = 3 + (k-1)*2 => (n-12)/8 + 1 times
i=6 => n/4 = 6 + (k-1)*2 => (n-24)/8 + 1 times
i=9 => n/4 = 9 + (k-1)*2 => (n-36)/8 + 1 times
Let's check the last i's now:
i=n/4 => n/4 = n/4 + (k-1)*2 => 1 times
i=n/4 - 3 => n/4 = (n/4-3) + (k-1)*2 => 3/2 + 1 times
i=n/4 - 6 => n/4 = (n/4-6) + (k-1)*2 => 6/2 + 1 times
So total number of times inner loop will be running is:
= (1) + (3/2 + 1) + (6/2 + 1) + (9/2 + 1) ... + ((n-12)/8 + 1)+ (n/8 + 1)
=> (0/2 + 1) + (3/2 + 1) + (6/2 + 1) + (9/2 + 1) ... + ((n-12)/8 + 1)+ (n/8 + 1)
Can be written as:
=> (0/2 + 3/2 + 6/2 + ... (n-12)/8 + n/8) + (1 + 1 + 1 ... 1 + 1)
Let's assume there are total P terms in the series:
Let's find out P:
n/8 = (0/2) + (P-1)*(3/2) => P = (n+12)/12
Now summing up the above series:
= [(P/2) (0/2 + (P-1) * 3/2)] + [P]
= P(3P+1)/4
= (n+12)(3(n+12)+12)/(4*12*12)
= (n^2 + 28n + 96)/192
So the final complexity of the code is
= (number of operation in each iteration) * (n^2 + 28n + 96)/192
Now look at the term (n^2 + 28n + 96)/192 For a very large n this will be close to ~n^2
Following is the complexity comparison:
Linear scale was difficult to analyse to I plotted log scale. Though for small n you don't see the complexity converging to n^2.
Using a very relax approach one can say that:
for (int i = 0; i <= n/2; i+=3){
for (int j = i; j <= n/4; j+=2) {
x++;
}
}
in the same as :
for (int i = 0; i <= n/4; i+=3){
for (int j = i; j <= n/4; j+=2) {
x++;
}
}
since with i > n/4 the inner loop will not execute. Moreover, to simplify the math you can say that the code is approximately the same as:
for (int i = 0; i < n/4; i+=3){
for (int j = i; j < n/4; j+=2) {
x++;
}
}
since the context is big-O it does not make a difference for the calculation of the upper-bound of the double loop. The number of iterations of a loop of the form:
for (int j = a; j < b; j += c)
can be calculated using the formula (b - a) /c. Hence, the inner loop will run approximately ((n/4) - i) / 2) times, or n/8 - i/2 times.
The outer-loop can be thought as running from k=0 until n/12. So with both loops we have
the summation of [k=0 to n/12] of (n/8 - 3k/2),
which is equivalent to
the summation [k=0 to n/12] of n/8 - the summation [k=0 to n/12] of 3k/2.
Hence,
(N^2) / 96 - the summation [[k=0 to n/12] of 3k/2
which is approximately (n^2) / 192. Therefore, the upper bound is O (n^2).
What is the time complexity and tilde for the loop below?
for (int i = N/2; i < N; i++) {
for (int j = i; j < N; j++) {
doSomething(i, j);
}
}
I think that it runs N/2 + (N/2 + 1) + (N/2 + 2) + ... + (N-1) times, but how do I get it's time complexity and tilde?
For example - if N = 100, the loop will run 50 + 51 + 52 + 53 + ... + 99 times.
I am assuming doSomething(i, j); is not iterating all the elements between i and j; if this is the case, the complexity of this algorithm is O(N^2).
The outer loop for (int i = N/2; i < N; i++) { will execute O(N) times, cause N/2 is actually constant value.
The inner loop in worst case will execute N times (or N - i times) too, this will also merge with previous O(N).
Therefore, overall time complexity will be O(N^2) in worst case scenario.
The inner loop is executed:
N/2-1 times for i = N/2,
N/2-2 times for i = N/2+1
....
1 time for i = N-2
therefore the total time for the inner loop is :
(N/2-1) + (N/2-2) + .... (N/2-k) where k = N/2 - 1
= N/2*k - (1 + 2 + ... + k)
= N/2*(N/2-1) - (N/2-1)(N/2)/2
= N/2(N/2 - 1 - N/4 + 1/2)
= N/2(N/4 - 1/2)
= N^2/8 - N/4
Hence the order of growth of the code is of N^2
If you consider tilde notation which is defined as :
"∼g(n) to represent any quantity that, when divided by f(n), approaches 1 as n grows" from here, you can see that ~g(n) = ~N^2/8 because as N grows (N^2/8)/(N^2/8-N/4) approaches 1.
I struggle to define the running time for the following algorithm in O notation. My first guess was O(n), but the gap between the iterations and the number I apply isn't steady. How have I incorrectly defined this?
public int function (int n )
{
if ( n == 0) {
return 0;
}
int i = 1;
int j = n ;
while ( i < j )
{
i = i + 1;
j = j - 1;
}
return function ( i - 1) + 1;
}
The while is executed in about n/2 time.
The recursion is executed passing as n a value that is about half of the original n, so:
n/2 (first iteration)
n/4 (second iteration, equal to (n/2)/2)
n/8
n/16
n/32
...
This is similar to a geometric serie.
Infact it can be represented as
n * (1/2 + 1/4 + 1/8 + 1/16 + ...)
So it converges to n * 1 = n
So the O notation is O(n)
Another approach is to write it down as T(n) = T(n/2) + n/2 + 1.
The while loop does n/2 work. Argument passed to next call is n/2.
Solving this using the master theorem where:
a = 1
b = 2
f = n/2 + 1
Let c=0.9
1*(f(n/2) + 1) <? c*f(n)
1*(n/4)+1 <? 0.9*(n/2 + 1)
0.25n + 1 <? 0.45n + 0.9
0 < 0.2n - 0.1
Which is:
T(n) = Θ(n)
I deduced the time complexity of bubble sort in its best case according to the mothod used in book ALGORITHMS 2.2. But the answer turned out to be O(n^2).
Here's my derivation, hope anyone can help me find out where is wrong:
public void bubbleSort(int arr[]) {
for(int i = 0, len = arr.length; i < len - 1; i++) {
for(int j = 0; j < len - i - 1; j++) {
if(arr[j + 1] < arr[j])
swap(arr, j, j + 1);
}
}
}
Statements cost times
i = 0,len = arr.length c1 1
i < len - 1 c2 n
i++ c3 n - 1
j = 0 c4 n - 1
j < len - i - 1 c5 t1(i=0) + t1(i=1) + ... + t1(i = n-2)
j++ c6 t2(i=0) + t2(i=1) + ... + t2(i = n-2)
arr[j + 1] < arr[j] c7 t3(i=0) + t3(i=1) + ... + t3(i = n-2)
swap(arr, j, j + 1) c8 t4(i=0) + t4(i=1) + ... + t4(i = n-2)
T(n) = c1 + c2n + c3(n - 1) + c4(n - 1) + c5t5 + c6t6 + c7t7 + c8t8
= c1 + c2n + c3(n - 1) + c4(n - 1) + c5[t1(i=0) + t1(i=1) + ... + t1(i = n-2)] + c6[t2(i=0) + t2(i=1) + ... + t2(i = n-2)] + c7[t3(i=0) + t3(i=1) + ... + t3(i = n-2)] + c8[t4(i=0) + t4(i=1) + ... + t4(i = n-2)];
in its best cast, the sequence is already positive before sorting. Then t8 sould be 0.
T(n) = c1 + c2n + c3(n - 1) + c4(n - 1) + c5[t1(i=0) + t1(i=1) + ... + t1(i = n-2)] + c6[t2(i=0) + t2(i=1) + ... + t2(i = n-2)] + c7[t3(i=0) + t3(i=1) + ... + t3(i = n-2)]
The time complexity is O(n^2)
Your implementation
public void bubbleSort(int arr[]) {
for(int i = 0, len = arr.length; i < len - 1; i++) {
for(int j = 0; j < len - i - 1; j++) {
if(arr[j + 1] < arr[j])
swap(arr, j, j + 1);
}
}
}
lacks the control whether there was any swap in the inner loop, and the breaking out of the outer loop if there wasn't.
That control makes it possible that the best case (an already sorted array) is O(n), since then there are no swaps in the inner loop when it runs the first time.
public void bubbleSort(int arr[]) {
boolean swapped = true;
for(int i = 0, len = arr.length; swapped && i < len - 1; i++) {
swapped = false;
for(int j = 0; j < len - i - 1; j++) {
if(arr[j + 1] < arr[j]) {
swap(arr, j, j + 1);
swapped = true;
}
}
}
}
The best case for bubble sort is when the elements are already sorted.
The usual implementation gives O(n^2) time complexity for best, average, worst case.
We can modify the bubble sort by checking if array is sorted or not(a swap would indicate an unsorted array) at every iteration.
As soon as the array is found to be sorted(if no swap occurs) control exits from loops or loop continues to execute till length-1.
And same is true for insertion sort as well!
I am not sure what are you counting. In general, when you are talking about comparison sort algorithms you should count the number of comparisons made. Bubble sort is regarded as such. In this case the algorithm you presented is O(n^2).
If you count the number of swaps its O(1) or maybe even one could say O(0). It is however rare to analyze Bubble sort like that.
You can, however very easily improve Bubble to get O(N) on best case. E.g by introducing a flag swap_was_made. If its false at the end of inner for you can finish. On best case it will cut complexity to O(N) (one inner for loop). In case of fair even distribution it cuts the expected or average complexity to O(N^2/2) ... But please double check me on it I might be wrong. Didn't do the math here.