I have an algorithm exam.. and am a bit not great in the loops time complexity :s I just started to get the basics of it..
I have this while loop
i=2
while (i<n)
{
i=i*i
x=x+1
}
I believe that the solution must be like:
(i) will run from 2 to 2k where k = 2i
every time it execute the statement 1 time..
so 1+1+1+.. , this means 1*2k
and from here I can't continue..
the second question guys.. please recommend a site or sth that I can practice some more of these.. searched but didn't find :s
The loop runs as long as the i is less than n. In other words, you need to find k such that 22k < n. ==> k = log2log2n. So the loop iterates for log2log2n times and at each iteration it performs 2 operations (multiplication and addition) - these operations take O(1) time.
==> The time to execute the loop is equal to log2log2n * O(1) ==> total complexity is O(loglogn).
Related
I am trying to do this problem out of a book and am struggling to understand the answer.
for (i = 0; i < N; ++i) {
if ((i % 2) == 0) {
outVal[i] = inVals[i] * i
}
}
here's how I was breaking it down:
I=0 -> executes 1 time
I < n and ++I each execute once every iteration. so 1n+1n = 2n.
the if statement contains 2 operands, so now we are at 4n+1.
the contents of the if statement only executes n/2 times, so we are at 4n+1+n/2
however, big O drops those terms off, leaving us with N as the answer
Here's what I don't get: the explanation for the answer of my problem says this:
outVal[i] = inVals[i] * i; executes every other loop iteration, so the total number of operations include: 1 operation at the start of the loop, 3 operations every loop iteration, 1 operation every other loop iteration, and 1 operation for the final loop condition check.
how are there only 3 operations in the loop? I counted 4 as stated above. Please let me know the rationale behind this.
The complexity is measured by the time/space you take to accomplish a task. i<N and ++i do not take time dependant of your space variable N (the length of the loop).
You must not add how many times an operation is done and sum all of them - you must, instead, choose the one who takes more time or space, as that's the algorithm bottleneck. In a loop, msot of the operations run equal times, so we use the length of the loop as its complexity space or time.
The loop will run N times, so that's its complexity -> O(n)
Inside the loop, the if scope will run N/2 times, as you correctly said -> O(n/2)
But those runs are already added to the first loop iterations. You will not add it since there are no external iterations.
So, the complexity of the algorithm is O(n).
Regarding the operations, the 3 are:
Checking I
Adding 1 to I
The if condition
All of them are done in every iteration.
I'm trying to understand the time complexity of insertion sort. I got stuck at while loop. I'm unable to understand how many times while loop executes
InsertionSort(A)
for j = 2 to A.length
key = A[j]
i = j - 1
while i>0 and A[i]>key
A[i+1] = A[i]
i = i - 1
A[i+1] = key
I know that for loop executes n+1 times and every statement in the loop execute n times
while loop also executes n times
But, what I don't understand is "How many times statements under while loop executes for both worst and best cases?"
In the worst case, A is sorted in descending order, which means that for the j'th entry, the inner loop will run j times (give or take a "+1" or "-1"...). Happily, there is a formula for that: as Gauss famously found out spontaneously and under duress, summing up all numbers from 1 to n yields a result of n*(n+1)/2.
As we only care about complexity and not actual values, we can leave the constant and multiplicative factors off and end up with O(n^2).
Tongue-in-cheek aside, the fact that there is a loop within a loop is a strong indication for O(n^2) when the inner loop count is bounded linearly - which it is here.
Best case, with A already sorted in ascending order, the inner loop will be entered not at all, and overall complexity will be O(n).
The average case depends heavily on what your expected "unorderedness" looks like. For example, the sort will behave greatly if your list is basically always sorted already, and there are only very few, very local switchups.
When analyzing algorithms, I don't usually have issues with for loop while I do have problems analyzing the running time of algorithms that involves while loop. I'll show an example to express my question.
The following one is a part of the coin-change algorithm (well-known algorithm for most of us) that I am studying right now:
counter = 0
s = 0
while(s <= n/100)
s = s+1
t1 = n- 100*s
h = 0
while(h <= t1/50)
h = h +1
t2 = t1 - 50*h
...
Can someone explain the best way of knowing the running time of such algorithms that have nested while loops?
The while loop is just another way of writing a for loop so assessing complexity should not be any different.
Here, the outer while loop runs n times in complexity world (proportional to n in real world) since s increases by 1 for each iteration and it runs until s reaches a value proportional to n.
The inner loop which I assume you are a little confused about, runs t1 times (again in complexity world) where t1 = n - 100s. Now you are thinking the algorithm is O(n^2) but t1 decreases in each iteration so the inner loop runs fewer number of times for each subsequent iteration and may be its not O(n^2).
t1 is different for each iteration so the entire set of iterations will run for: n - 100 + n - 200 + n - 300 + .... 0 times. Because the numbers of terms in this series is proportional to n, the summation will have a n squared term and for reporting complexity all lower order terms are ignored so you need not worry about what the rest of the numbers sum to. This algorithm is O(n^2).
The trick is to ignore constant and lower order terms at each step and it becomes easy !
Please excuse the intentional verbosity
Here is a small program excerpt:
for i=1 to n
j= 0;
while(j<=n);
j=j+1;
If I have to find the complexity(Big O) of this code:
I'll first count how many times the inner loop will execute which in this case is n+1 times because from 1 to n, it is n times and since j is 0, it adds to the while looping. So in total n+1 times for the while loop.
The number of times the outer for loop will execute is n times because from 1 to n, the total count is n.
Hence the grand total is n+1+n is 2n+1.
Discarding all constants it's big O(n).
Is it correct? The web page from where I found this example says the outer loop will run n(n+1)/2 times. I didn't understand this. Please help!
No.
For each value i is getting (and there are n of those), you run the while (inner) loop n+1 times (j=0,j=1,...j=n).
Thus, the total number of times the line j=j+1 is being executed is n*(n+1), which is in O(n^2)
sum = 0;
for(int i = 0; i < N; i++)
for(int j = i; j >= 0; j--)
sum++;
From what I understand, the first line is 1 operation, 2nd line is (i+1) operations, 3rd line is (i-1) operations, and 4th line is n operations. Does this mean that the running time would be 1 + (i+1)(i-1) + n? It's just these last steps that confuse me.
To analyze the algorithm you don't want to go line by line asking "how much time does this particular line contribute?" The reason is that each line doesn't execute the same number of times. For example, the innermost line is executed a whole bunch of times, compared to the first line which is run just once.
To analyze an algorithm like this, try identifying some quantity whose value is within a constant factor of the total runtime of the algorithm. In this case, that quantity would probably be "how many times does the line sum++ execute?", since if we know this value, we know the total amount of time that's spent by the two loops in the algorithm. To figure this out, let's trace out what happens with these loops. On the first iteration of the outer loop, i == 0 and so the inner loop will execute exactly once (counting down from 0 to 0). On the second iteration of the outer loop, i == 1 and the inner loop executes exactly twice (first with j == 1, once with j == 0. More generally, on the kth iteration of the outer loop, the inner loop executes k + 1 times. This means that the total number of iterations of the innermost loop is given by
1 + 2 + 3 + ... + N
This quantity can be shown to be equal to
N (N + 1) N^2 + N N^2 N
--------- = ------- = --- + ---
2 2 2 2
Of these two terms, the N^2 / 2 term is the dominant growth term, and so if we ignore its constant factors we get a runtime of O(N2).
Don't look at this answer as something you should memorize - think of all of the steps required to get to the answer. We started by finding some quantity to count, and then saw how that quantity was influenced by the execution of the loops. From this, we were able to derive a mathematical expression for that quantity, which we then simplified. Finally, we took the resulting expression and determined the dominant term, which serves as the big-O for the overall function.
Work from inside-out.
sum++
This is a single operation on it's own, as it doesn't repeat.
for(int j = i; j >= 0; j--)
This loops i+1 times. There are several operations in there, but you probably don't mean to count the number of asm instructions. So I'll assume for this question this is a multiplier of i+1. Since the loop contents is a single operation, the loop and its block perform i+1 operations.
for(int i = 0; i < N; i++)
This loops N times. So as before, this is a multiplier of N. Since the block performs i+1 operations, this loop performs N(N+1)/2 operations in total. And that's your answer! If you want to consider big-O complexity, then this simplifies to O(N2).
It's not additive: the inner loop happens once for EACH iteration of the outer loop. So it's O(n2).
By the way, this is a good example of why we use asymptotic notation for this kind of thing -- depending on the definition of "operation" the exact details of the count could vary pretty widely. (Like, is sum++ a single operation, or is it add sum to 1 giving temp; load temp to sum?) But since we know that all that can be hidden in a constant factor, it's still going to be O(n2).
No; you don't count a specific number of operations for each line and then add them up. The entire point of constructions like 'for' is to make it possible for a given line of code to run more than once. You're supposed to use thinking and logic skills to figure out how many times the line 'sum++' will run, as a function of N. Hint: it runs once for every time that the third line is encountered.
How many times is the second line encountered?
Each time the second line is encountered, the value of 'i' is set. How many times does the third line run with that value of i? Therefore, how many times will it run overall? (Hint: if I give you a different amount of money on several different occasions, how do you find out the total amount I gave you?)
Each time the third line is encountered, the fourth line happens once.
Which line happens most often? How often does it happen, in terms of N?
So guess what interest you is the sum++ and how many time you execute it.
The final stat of sum would give you that answer.
Actually your loop is just:
Sigma(n) n goes from 1 to N.
Which equal to: N*(N+1) / 2 This give you in big-o-notation O(N^2)
Also beside the name of you question there is no worst case in you algorithm.
Or you could say that the worst case is when N goes to infinity.
Using Sigma notation to represent your loops: