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)
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 stumbled upon a loop for which I am not sure what the time complexity is. It is the following loop:
for(i = 1; i <= n^2; i++){
for(j = 1; j <= i; j++) {
//some elementary operation
}
}
I would have argued that the outer for-loop runs in n^2 and the inner for loop would also run in n^2 as for every iteration of the outer-loop we do n^2 - (n^2 - 1), n^2 - (n^2 - 2),..., n^2. Am I totally going in the wrong direction here?
So the time complexity would be in n^4
The number of operation will be :
1 + 2 + 3 + 4 + 5 + ... + n²
Which is equal to (n² * (n² - 1)) / 2.
The Big O notation is O(n^4). You are correct.
It's a simple arithmetic progression happening here.
Every new iteration is bigger by 1.
Outer loop will do n^2 operations which will result for following sequence:
1 + 2 + 3 + ... + n + ... + n^2 = n^2 (n^2+1) / 2 = O(n^4)
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`