So many loops, I stuck at counting how many times the last loop runs.
I also don't know how to simplify summations to get big Theta. Please somebody help me out!
int fun(int n) {
int sum = 0
for (int i = n; i > 0; i--) {
for (int j = i; j < n; j *= 2) {
for (int k = 0; k < j; k++) {
sum += 1
}
}
}
return sum
}
Any problem has 2 stages:
You guess the answer
You prove it
In easy problems, step 1 is easy and then you skip step 2 or explain it away as "obvious". This problem is a bit more tricky, so both steps require some more formal thinking. If you guess incorrectly, you will get stuck at your proof.
The outer loop goes from n to 0, so the number of iterations is O(n). The middle loop is uncomfortable to analyze because its bounds depend on current value of i. Like we usually do in guessing O-rates, let's just replace its bounds to be from 1 to n.
for (int i = n; i > 0; i--) {
for (int j = 1; j < n; j *= 2) {
perform j steps
}
}
The run-time of this new middle loop, including the inner loop, is 1+2+4+...+n, or approximately 2*n, which is O(n). Together with outer loop, you get O(n²). This is my guess.
I edited the code, so I may have changed the O-rate when I did. So I must now prove that my guess is right.
To prove this, use the "sandwich" technique - edit the program in 2 different ways, one which makes its run-time smaller and one which makes its run-time greater. If you manage to make both new programs have the same O-rate, you will prove that the original code has the same O-rate.
Here is a "smaller" or "faster" code:
do n/2 iterations; set i=n/2 for each of them {
do just one iteration, where you set j = i {
perform j steps
}
}
This code is faster because each loop does less work. It does something like n²/4 iterations.
Here is a "greater" or "slower" code:
do n iterations; set i=n for each of them {
for (int j = 1; j <= 2 * n; j *= 2) {
perform j steps
}
}
I made the upper bound for the middle loop 2n to make sure its last iteration is for j=n or greater.
This code is slower because each loop does more work. The number of iterations of the middle loop (and everything under it) is 1+2+4+...+n+2n, which is something like 4n. So the number of iterations for the whole program is something like 4n².
We got, in a somewhat formal manner:
n²/4 ≤ runtime ≤ 4n²
So runtime = O(n²).
Here I use O where it should be Θ. O is usually defined as "upper bound", while sometimes it means "upper or lower bound, depending on context". In my answer O means "both upper and lower bound".
Related
Although, I found pretty good replies to the same question! However, I want time complexity equation of the following piece of code
sum = 0; c1=11
for (i=0; i<N; i++) c2=6
for (j=0; j<N; j++) c2=6
for (j=0; j<N; j++) c2=7
sum += arr[i][j] c3=2
While each statement has a cost associated with it, I require complete time complexity equation and its answer.
Regards
The comments section got quite long so I am going to write up an answer summarizing everything.
Measuring Time Complexity
In Computer Science, we measure time complexity by the number of steps/iterations your algorithm takes to evaluate.
So if you have a simple array of length n and you go through this array only once, say to print all the elements, we say that this algorithm is O(n) because the time is takes to run will grow proportionally to the size of the array you have, thus n
You can think of Big-O O(..) as a higher order function that compares other functions. if we say f(x) = O(n) it means that you function grows at most as fast as y=n thus linearly. This means that if you were to plot these functions on a graph, there would be a point c x = c after which the graph of n will always be on top of f(x) for any given x > c. Big-O signifies upper bound of a function in terms of another function.
So let's look at your original question and what it means to be constant time. Say we have this function
def printFirst5(arr: Array[Int]) = {
for(i =0 ;i < 5; i++){
println(arr[i])
}
}
This is what we call a constant time algorithm. Can you see why? Because no matter what array you pass into this (as long as it has at least 5 elements), it will only go through the first 5 elements. You can pass it an array of length 100000000000 you can pass it an array of length 10 it doesn't matter. In each case it will only look at the first 5 elements. Meaning this function printFirst5 will never go above the line y=5 in terms of time complexity. These kind of functions are denoted O(1) for constant time.
Now, finally, let's look at you edited version. (I am not sure what you are trying to do in your example because it is syntactically wrong, so I will write my own example)
def groupAllBy3(array: Array[Int]) = {
for(i=0; i < array.length; i++){
for(j=0; j < array.length; j++){
for(k=0; k< array.length; k++{
println(s"$array[i], $array[j], $array[k]")
}
}
}
}
This functions time complexity is O(N3). Why? Let's take a look.
The innermost loop will go through N elements for every j
How many js are there? Well there will be N js for every i.
How many is are there? N many.
So in total we get numberof-i * numberof-j * numberof-k = N * N * N = O(N^3)
Just to make sure you understand this correctly, let's take a look at another example. What would happen if these loops weren't nested? If we had:
def printAllx3(array: Array[Int]) = {
for(i=0; i < array.length; i++){
println(s"array[i]")
}
for(j=0; j < array.length; j++){
println(s"array[j]")
}
for(k=0; k< array.length; k++{
println(s"array[k]")
}
}
What is the case here?
The first loop goes through N elements, the second loop goes through N elements, the third loop goes through N elements. But they don't depend on each other in terms of iterations so we get N + N + N = 3N = O(N)
Do you see the difference?
With all due respect, I believe you are missing some of the fundamentals of what time complexity is & how we measure it. There is only so much I can explain here, I highly recommend you do some reading on the subject and ask any further questions you don't understand.
Hope this helps
These programs do the calculation ∑𝑖=0 𝑎𝑖 𝑥
I am trying to figure out big O calculations. I have done alot of study but I am having a problem getting this down. I understand that big O is worst case scenario or upper bounds. From what I can figure program one has two for loops one that runs for the length of the array and the other runs to the value of the first loop up to the length of the array. I think that if both ran the full length of the array then it would be quadratic O(N^2). Since the second loop only runs the length of the length of the array once I am thinking O(NlogN).
Second program has only one for loop so it would be O(N).
Am I close? If not please explain to me how I would calculate this. Since this is in the homework I am going to have to be able to figure something like this on the test.
Program 1
// assume input array a is not null
public static double q6_1(double[] a, double x)
{
double result = 0;
for (int i=0; i<a.length; i++)
{
double b = 1;
for (int j=0; j<i; j++)
{
b *= x;
}
result += a[i] * b;
}
return result;
}
Program 2
// assume input array a is not null
public static double q6_2(double[] a, double x)
{
double result = 0;
for (int i=a.length-1; i>=0; i--)
{
result = result * x + a[i];
}
return result;
}
I'm using N to refer to the length of the array a.
The first one is O(N^2). The inner loop runs 1, 2, 3, 4, ..., N - 1 times. This sum is approx N(N-1)/2 which is O(N^2).
The second one is O(N). It is simply iterating through the length of the array.
Complexity of a program is basically number of instructions executed.
When we talk about the upper bound, it means we are considering the things in worst case which should be taken in consideration by every programmer.
Let n = a.length;
Now coming back to your question, you are saying that the time complexity of the first program should be O(nlogn), which is wrong. As when i = a.length-1 the inner loop will also iterate from j = 0 to j = i. Hence the complexity would be O(n^2).
You are correct in judging the time complexity of the second program which is O(n).
After looking at this question, this article, and several other questions, I have still not been able to find a general way to determine the computational complexity of algorithms with looping variables dependent on the parent loop's variable. For example,
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
for (int k = i; k < j; k++) {
//one statement
}
}
}
I know that the first loop has a complexity of n, but the inner loops are confusing me. The second loop seems to be executed n-i times and the third loop seems to be executed j-i times. However, I'm not sure how to turn this into a regular Big-O statement. I don't think I can say O(n(n-i)(j-i)), so how can I get rid of the i and j variables here?
I know this is something on the order of n^3, but how can I show this? Do I need to use series?
Thanks for your help!
(If you were wondering, this is from a brute force implementation of the maximum sum contiguous subsequence problem.)
First loop hits N items on average.
Second loop hits N / 2 items on avarage
Third loop hits N / 4 items on average
O(N * N / 2 * N / 4) is about O((N^3)/8) is about O(N^3)
I have been practicing analyzing algorithms lately. I feel like I have a pretty good understanding of analyzing non-recursive algorithms but I am unsure, and have just begun to embark on a full understanding of recursive algorithm as well. Although, I have not had a formal check on my methods and if what I have been doing is indeed correct
Would it be too much to ask if someone could check a few algorithms that I have implemented and analyzed and see if my understanding is along the right lines or if I am completely off.
here:
1)
sum = 0;
for (i = 0; i < n; i++){
for (j = 0; j < i*i; j++){
if (j % i == 0) {
for (k = 0; k < j; k++){
sum++;
}
}
}
}
My analysis of this one was O(n^5) due to:
Sum(i = 0 to n)[Sum(j = 0 to i^2)[Sum(k = 0 to j) of 1]]
which evaluated to:
(1/2)(n^5/5 + n^4/2 + n^3/3 - n/30) + (1/2)(n^3/3 + n^2/2 + n/6) + (1/2)(n^3/3 + n^2/2 + n/6) + n + 1.
Hence it is O(n^5)
Is this correct as an evaluation of the summations of the loops?
a triple summation. I have assumed that the if statement will always pass for worse case complexity. Is this a correct assumption for worst case?
2)
int tonyblair (int n, int a) {
if (a < 12) {
for (int i = 0; i < n; i++){
System.out.println("*");
}
tonyblair(n-1, a);
} else {
for (int k = 0; k < 3000; k++){
for (int j = 0; j < nk; j++){
System.out.println("#");
}
}
}
}
My analysis of this algorithm is O(infinity) due to the infinite recursion in the if statement if it is assumed to be true, which would be the worst case. Although, for pure analysis, I analyzed if this were not true and the if statement would not run. I then got a complexity of O(nk) due to:
Sum(k = 0 to 3000)[Sum(j = 0 to nk) of 1]
which then evaluated to nk(3001) + 3001. Hence is O(nk), where k is not discarded due to it controlling the number of iterations of the loop.
Number 1
I can't tell how you've derived your formula. Usually adding terms happens when there are multiple steps in an algorithm, such as precomputing data and then looking up values from the data. Instead, nested for loops implies multiplication. Also, the worst case is the best case for this snippet of code, because given a value of n, sum will be the same at the end.
To find the complexity, we want to find the number of times that the inner loop is evaluated. Summations are often easy to solve if they go from 1 to n, so I'm going to drop the 0s from them later on. If i is 0, the middle loop won't run, and if j is 0, the inner loop won't run. We can rewrite the code equivalently as:
sum = 0;
for (i = 1; i < n; i++)
{
for (j = 1; j < i*i; j++)
{
if (j % i == 0)
{
for (k = 0; k < j; k++)
{
sum++;
}
}
}
}
I could make my life harder by forcing the outer loop to start at 2, but I'm not going to. The outer loop now runs from 1 to n-1. The middle loop runs based on the current value of i, so we need to do a summation:
The middle for loop always goes to (i^2 - 1), and j will only be divisible by i for a total of (i - 1) times (i, i*2, i*3, ..., i*(i-2), i*(i-1)). With this, we get:
The middle loop then executes j times. The j in our summation is not the same as the j in the code though. The j in the summation represents each time the middle loop executes. Each time the middle loop executes, the j in the code will be (i * (number of executions so far)) = i * (the j in the summation). Therefore, we have:
We can move the i to in-between the two summations, as it is a constant for the inner summation. Then, the formula for the sum of 1 to n is well known: n*(n+1)/2. Because we are going to n - 1, we must subtract n out. This gives:
The summations for the sum of squares and the sum of cubes are also well known. Keeping in mind that we are only summing to n-1 in both cases, we must remember to subtract n^3 and n^2, respectively, and we get:
This is obviously n^4. If we solve it all the way, we get:
Number 2
For the last one, it is in fact O(infinity) if a < 12 because of the if statement. Well, technically everything is O(infinity), because Big-O only provides an upper bound on runtime. If a < 12, it is also omega(infinity) and theta(infinity). If only the else runs, then we have the summation from 1 to 2999 of i*n:
It's very important to notice that the summation from 1 to 2999 is a constant (it's 4498500). No matter how large a constant is, it's still a constant, and not dependent on n. We will end up throwing it out of the runtime calculations. Sometimes, when a theoretically fast algorithm has a large constant, it is practically slower than other algorithms that are theoretically slow. One example I can think of is Chazelle's linear time triangulation algorithm. No one has ever implemented it. In any case, we have 4498500 * n. This is theta(n):
I am reading some information on time complexity and I'm quite confused as to how the following time complexities are achieved and if there is a particular set of rules or methods for working this out?
1)
Input: int n
for(int i = 0; i < n; i++){
print("Hello World, ");
}
for(int j = n; j > 0; j--){
print("Hello World");
}
Tight: 6n + 5
Big O: O(n)
2)
Input: l = array of comparable items
Output: l = array of sorted items
Sort:
for(int i = 0; i < l.length; i++){
for(int j = 0; j < l.length; j++){
if(l{i} > l{j}){
} }
Swap(l{i},l{j});
}
return ls;
Worst Case Time Complexity: 4n2 +3n+2 = O(n2)
For a given algorithm, time complexity or Big O is a way to provide some fair enough estimation of "total elementary operations performed by the algorithm" in relationship with the given input size n.
Type-1
Lets say you have an algo like this:
a=n+1;
b=a*n;
there are 2 elementary operations in the above code, no matter how big your n is, for the above code a computer will always perform 2 operations, as the algo does not depend on the size of the input, so the Big-O of the above code is O(1).
Type-2
For this code:
for(int i = 0; i < n; i++){
a=a+i;
}
I hope you understand the Big-O in O(n), as elementary operation count directly depend on the size of n
Type-3
Now what about this code:
//Loop-1
for(int i = 0; i < n; i++){
print("Hello World, ");
}
//Loop-2
for(int i = 0; i < n; i++){
for(int j = 0; j < n; j++) {
x=x+j;
}
}
As you can see loop-1 is O(n) and loop-2 is O(n^2). So it feel like total complexity should be O(n)+O(n^2). But no, the time complexity of the above code is O(n^2). Why? Because we are trying to know the fair enough count of elementary operations performed by the algorithm for a given input size n, which will be comparatively easy to understand by another person. With this logic, O(n)+O(n^2) become O(n^2), or O(n^2)+O(n^3)+O(n^4) become O(n^4)!
Again, you may ask: But how? How all the lower power of Big-O become so insignificant as we add it with a higher power of Big-O, that we can completely omit them (lower powers) when we are describing the complexity of our algorithm to another human?
I will try show the reason for this case: O(n)+O(n^2)=O(n^2).
Lets say n=1000 then the exact count for O(n) is 1000 operations and the exact count for O(n^2) is 1000*1000=1000000, so O(n^2) is 1000 time bigger than O(n), which means your program will spend most of the execution time in O(n^2) and thus it is not worth to mention that your algorithm also has some O(n).
PS. Pardon my English :)
In the first example, the array has n elements, and you go through these elements Twice. The first time you start from index 0 until i, and the second time you start from index n to 0. So, to simplify this, we can say that it took you 2n. When dealing with Big O notation, you should keep in mind that we care about the bounds:
As a result, O(2n)=O(n)
and O(an+b)=O(n)
Input: int n // operation 1
for(int i = 0; i < n; i++){ // operation 2
print("Hello World, "); // Operation 3
}
for(int j = n; j > 0; j--) // Operation 4
{
print("Hello World"); //Operation 5
}
As you can see, we have a total of 5 operations outside the loops.
Inside the first loop, we do three internal operations: checking if i is less than n, printing "Hello World", and incrementing i .
Inside the second loop, we also have three internal operations.
So, the total number of of opetations that we need is: 3n ( for first loop) + 3n ( second loop) + 5 ( operations outside the loop). As a result, the total number of steps required is 6n+5 ( that is your tight bound).
As I mentioned before, O( an +b )= n because once an algorithm is linear, a and b do not have a great impact when n is very large.
So, your time complexity will become : O(6n+5) =O(n).
You can use the same logic for the second example keeping in mind that two nested loops take n² instead of n.
I will slightly modify Johns answer. Defining n is one constant operation, defining integer i and assigning it to 0 is 2 constant operations. defining integer j and assigning with n is another 2 constant operations. checking the conditions for i,j inside for loop,increment,print statement depends on n so the total will be 3n+3n+5 which is equal to 6n+5. Here we cannot skip any of the statements during execution so its average case running time will also be its worst case running time which is O(n)