I am looking for some clarification in working out the time efficiency of an Algorithm, specifically T(n). The algorithm below is not as efficient as it could be, though it's a good example to learn from I believe. I would appreciate a line-by-line confirmation of the sum of operations in the code:
Pseudo-code
1. Input: array X of size n
2. Let A = an empty array of size n
3. For i = 0 to n-1
4. Let s = x[0]
5. For j = 0 to i
6. Let sum = sum + x[j]
7. End For
8. Let A[i] = sum / (i+1)
9. End For
10. Output: Array A
My attempt at calculating T(n)
1. 1
2. n
3. n
4. n(2)
5. n(n-1)
6. n(5n)
7. -
8. n(6)
9. -
10. 1
T(n) = 1 + n + n + 2n + n^2 - n + 5n^2 + 6n + 1
= 6n^2 + 9n + 2
So, T(n) = 6n^2 + 9n + 2 is what I arrive at, from this I derive Big-O of O(n^2).
What errors, if any have I made in my calculation...
Edit: ...in counting the primitive operations to derive T(n)?
Your result O(n^2) is correct and is given by the two nested loops. I would prefer the derivation like
0 + 1 + 2 + + (n-1) = (n-1)n/2 = O(n^2)
that follows from observing the nested loops.
I'm not really sure on your methodology but O(n^2) does seem to be correct. At each iteration through the first loop you do a sub loop of the previous elements. Therefore you're looking at 1 the first time 2 the second then 3 then... then n the final time. This is equivalent to the sum from 1 to n which gives you complexity of n^2.
Related
T(n) = 2n^2 + n + 1
I understand the 2n^2 and 1 parts, but I am confused about the n.
test = 0
for i in range(n):
for j in range(n):
test = test + i*j
This really depends on how your professor/book really break down the operation cost but I think we can kind of figure it out from here. Let's break 2n^2 + n + 1 down. The n^2 comes from the two loops.
for i in range(n):
for j in range(n):
The 2 coefficient comes from the two operations presumably. Note: this alone is just constant time complexity AKA O(1)
test = test + i * j
The initial calculation of range(n) could cost n (being the + n in your calculation). Then second call to range(n) could be optimized to use a cached value.
Finally it could be the test = 0 statement in the beginning could be the + 1. This could total in 2n^2 + n + 1. However, the worse case time complexity to this is still O(n^2).
so I got this algorithm I need to calculate its time complexity
which goes like
for i=1 to n do
k=i
while (k<=n) do
FLIP(A[k])
k = k + i
where A is an array of booleans, and FLIP is as it is, flipping the current value. therefore it's O(1).
Now I understand that the inner while loop should be called
n/1+n/2+n/3+...+n/n
If I'm correct, but is there a formula out there for such calculation?
pretty confused here
The more exact computation is T(n) \sum((n-i)/i) for i = 1 to n (because k is started from i). Hence, the final sum is n + n/2 + ... + n/n - n = n(1 + 1/2 + ... + 1/n) - n, approximately. We knew 1 + 1/2 + ... + 1/n = H(n) and H(n) = \Theta(\log(n)). Hence, T(n) = \Theta(n\log(n)). The -n has not any effect on the asymptotic computaional cost, as n = o(n\log(n)).
Lets say we want to calculate sum of this equation
n + n / 2 + n / 3 + ... + n / n
=> n ( 1 + 1 / 2 + 1 / 3 + ..... + 1 / n )
Then in bracket ( 1 + 1 / 2 + 1 / 3 + ... + 1 / n ) this is a well known Harmonic series and i am afraid there is no proven formula to calculate Harmonic series.
The given problem boils down to calculate below sum -Sum of harmonic series
Although this sum can't be calculated accurately, however you can still find asymptotic upper bound for this sum, which is approximately O(log(n)).
Hence answer to above problem will be - O(nlog(n))
What is the time complexity for the following function?
for(int i = 0; i < a.size; i++) {
for(int j = i; j < a.size; i++) {
//
}
}
I think it is less than big O n^2 because we arent iterating over all of the elements in the second for loop. I believe the time complexity comes out to be something like this:
n[ (n) + (n-1) + (n-2) + ... + (n-n) ]
But when I solve this formula it comes out to be
n^2 - n + n^2 - 2n + n^2 - 3n + ... + n^2 - n^2
Which doesn't seem correct at all. Can somebody tell me exactly how to solve this problem, and where I am wrong.
That is O(n^2). If you consider the iteration where i = a.size() - 1, and you work your way backwards (i = a.size() - 2, i = a.size - 3, etc), you are looking at the following sum of number of iterations, where n = a.size.
1 + 2 + 3 + 4 + ... + n
The sum of this series is n(n+1)/2, which is O(n^2). Note that big-O notation ignores constants and takes the highest polynomial power when it is applied to a polynomial function.
It will run for:
1 + 2 + 3 + .. + n
Which is 1/2 n(n+1) which give us O(n^2)
The Big-O notation will only keep the dominant term, neglecting constants too
The Big-O is only used to compare algorithms on the same variation of a problem using the same complexity analysis standard, if and only if the dominant terms are different.
If the dominant terms are the same, you need to compare Big-Theta or Time complexity, which will show minor differences.
Example
A
for i = 1 .. n
for j = i .. n
..
B
for i = 1 .. n
for j = 1 .. n
..
We have
Time(A) = 1/2 n(n+1) ~ O(n^2)
Time(B) = n^2 ~ O(n^2)
O(A) = O(B)
T(A) < T(B)
Analysis
To visualize how we got 1 + 2 + 3 + .. n:
for i = 1 .. n:
print "(1 + "
sum = 0
for j = i .. n:
sum++
print sum") + "
will print the following:
(1+n) + (1+(n-1)) + .. + (1+3) + (1+2) + (1+1) + (1+0)
n+1 + n + n-1 + .. + 3 + 2 + 1
1 + 2 + 3 + .. + n + n+1
1/2 n(n+1) + (n+1)
1/2 n^2 + 1/2 n + n + 1
1/2 n^2 + 3/2 n + 1
Yes, the number of iterations is strictly less than n^2, but it's still Θ(n^2). It will eventually be greater than n^k for any k<2, and it will eventually be less than n^k for any k>2.
(As a side note, computer scientists often say big-O when they really mean big-theta (Θ). It's technically correct to say that almost every algorithm you've seen has O(n!) running time; all reasonably algorithms have running times that grow no more quickly than n!. But it's not really useful to say that the complexity is O(n!) if it's also O(n log n), so by some kind of Gricean maxim we assume that when someone says an algorithm's complexiy is O(f(x)) that f(x) is as small as possible.)
I have an algorithm with the following pseudocode:
R(n)
if(n = 1)
return 1
else
return(R(n-1) + 2 * n + 1)
I need to setup a recurrence relation for the number of multiplications carried out by this algorithm and solve it.
Is the following right?
R(1) = 0
R(n) = R(n-1) + n^2
You are performing only one multiplication per step. Therefore, the relation will be:
R(n) = R(n-1) + 1
In the algorithm as shown, R(n) is calculated by adding R(n-1) to 2*n+1. If 2*n is calculated using a multiplication, there will be one multiplication per level of recursion, thus n-1 multiplications in the calculation of R(n).
To compute that via a recurrence, let M(n) be the number of multiplications used to compute R(n). The recurrence boundary condition is M(1) = 0 and the recurrence relation is M(i) = M(i-1) + 1 for i>1.
Errors in writing “R(1) = 0; R(n) = R(n-1) + n^2” as the recurrence for the number of multiplications include:
• R() already is in use as a function being computed, hence re-using R() as the number of multiplications is incorrect
• Each level of recursion in the algorithm adds one multiplication, not n² multiplications.
Note, R(n) = 1 + 5 + 7 + ... + 2n+1 = 1 + 3 + 5 + 7 + ... + 2n+1 - 3 = n²-3; that is, function R(n) returns the value n²-3.
I am trying to find complexity of Fibonacci series using a recursion tree and concluded height of tree = O(n) worst case, cost of each level = cn, hence complexity = n*n=n^2
How come it is O(2^n)?
The complexity of a naive recursive fibonacci is indeed 2ⁿ.
T(n) = T(n-1) + T(n-2) = T(n-2) + T(n-3) + T(n-3) + T(n-4) =
= T(n-3) + T(n-4) + T(n-4) + T(n-5) + T(n-4) + T(n-5) + T(n-5) + T(n-6) = ...
In each step you call T twice, thus will provide eventual asymptotic barrier of:
T(n) = 2⋅2⋅...⋅2 = 2ⁿ
bonus: The best theoretical implementation to fibonacci is actually a close formula, using the golden ratio:
Fib(n) = (φⁿ – (–φ)⁻ⁿ)/sqrt(5) [where φ is the golden ratio]
(However, it suffers from precision errors in real life due to floating point arithmetics, which are not exact)
The recursion tree for fib(n) would be something like :
n
/ \
n-1 n-2 --------- maximum 2^1 additions
/ \ / \
n-2 n-3 n-3 n-4 -------- maximum 2^2 additions
/ \
n-3 n-4 -------- maximum 2^3 additions
........
-------- maximum 2^(n-1) additions
Using n-1 in 2^(n-1) since for fib(5) we will eventually go down to fib(1)
Number of internal nodes = Number of leaves - 1 = 2^(n-1) - 1
Number of additions = Number of internal nodes + Number of leaves = (2^1 + 2^2 + 2^3 + ...) + 2^(n-1)
We can replace the number of internal nodes to 2^(n-1) - 1 because it will always be less than this value :
= 2^(n-1) - 1 + 2^(n-1)
~ 2^n
Look at it like this. Assume the complexity of calculating F(k), the kth Fibonacci number, by recursion is at most 2^k for k <= n. This is our induction hypothesis. Then the complexity of calculating F(n + 1) by recursion is
F(n + 1) = F(n) + F(n - 1)
which has complexity 2^n + 2^(n - 1). Note that
2^n + 2^(n - 1) = 2 * 2^n / 2 + 2^n / 2 = 3 * 2^n / 2 <= 2 * 2^n = 2^(n + 1).
We have shown by induction that the claim that calculating F(k) by recursion is at most 2^k is correct.
You are correct that the depth of the tree is O(n), but you are not doing O(n) work at each level. At each level, you do O(1) work per recursive call, but each recursive call then contributes two new recursive calls, one at the level below it and one at the level two below it. This means that as you get further and further down the recursion tree, the number of calls per level grows exponentially.
Interestingly, you can actually establish the exact number of calls necessary to compute F(n) as 2F(n + 1) - 1, where F(n) is the nth Fibonacci number. We can prove this inductively. As a base case, to compute F(0) or F(1), we need to make exactly one call to the function, which terminates without making any new calls. Let's say that L(n) is the number of calls necessary to compute F(n). Then we have that
L(0) = 1 = 2*1 - 1 = 2F(1) - 1 = 2F(0 + 1) - 1
L(1) = 1 = 2*1 - 1 = 2F(2) - 1 = 2F(1 + 1) - 1
Now, for the inductive step, assume that for all n' < n, with n ≥ 2, that L(n') = 2F(n + 1) - 1. Then to compute F(n), we need to make 1 call to the initial function that computes F(n), which in turn fires off calls to F(n-2) and F(n-1). By the inductive hypothesis we know that F(n-1) and F(n-2) can be computed in L(n-1) and L(n-2) calls. Thus the total runtime is
1 + L(n - 1) + L(n - 2)
= 1 + 2F((n - 1) + 1) - 1 + 2F((n - 2) + 1) - 1
= 2F(n) + 2F(n - 1) - 1
= 2(F(n) + F(n - 1)) - 1
= 2(F(n + 1)) - 1
= 2F(n + 1) - 1
Which completes the induction.
At this point, you can use Binet's formula to show that
L(n) = 2(1/√5)(((1 + √5) / 2)n - ((1 - √5) / 2)n) - 1
And thus L(n) = O(((1 + √5) / 2)n). If we use the convention that
φ = (1 + √5) / 2 ≈ 1.6
We have that
L(n) = Θ(φn)
And since φ < 2, this is o(2n) (using little-o notation).
Interestingly, I've chosen the name L(n) for this series because this series is called the Leonardo numbers. In addition to its use here, it arises in the analysis of the smoothsort algorithm.
Hope this helps!
t(n)=t(n-1)+t(n-2)
which can be solved through tree method:
t(n-1) + t(n-2) 2^1=2
| |
t(n-2)+t(n-3) t(n-3)+t(n-4) 2^2=4
. . 2^3=8
. . .
. . .
similarly for the last level . . 2^n
it will make total time complexity=>2+4+8+.....2^n
after solving the above gp we will get time complexity as O(2^n)
The complexity of Fibonacci series is O(F(k)), where F(k) is the kth Fibonacci number. This can be proved by induction. It is trivial for based case. And assume for all k<=n, the complexity of computing F(k) is c*F(k) + o(F(k)), then for k = n+1, the complexity of computing F(n+1) is c*F(n) + o(F(n)) + c*F(n-1) + o(F(n-1)) = c*(F(n) + F(n-1)) + o(F(n)) + o(F(n-1)) = O(F(n+1)).
The complexity of recursive Fibonacci series is 2^n:
This will be the Recurrence Relations for recursive Fibonacci
T(n)=T(n-1)+T(n-2) No of elements 2
Now on solving this relation using substitution method (substituting value of T(n-1) and T(n-2))
T(n)=T(n-2)+2*T(n-3)+T(n-4) No of elements 4=2^2
Again substituting values of above term we will get
T(n)=T(n-3)+3*T(n-4)+3*T(n-5)+T(n-6) No of elements 8=2^3
After solving it completely, we get
T(n)={T(n-k)+---------+---------}----------------------------->2^k eq(3)
This implies that maximum no of recursive calls at any level will be at most 2^n.
And for all the recursive calls in equation 3 is ϴ(1) so time complexity will be 2^n* ϴ(1)=2^n
The O(2^n) complexity of Fibonacci number calculation only applies to the recursion approach. With a few extra space, you can achieve a much better performance with O(n).
public static int fibonacci(int n) throws Exception {
if (n < 0)
throws new Exception("Can't be a negative integer")
if (n <= 1)
return n;
int s = 0, s1 = 0, s2 = 1;
for(int i= 2; i<=n; i++) {
s = s1 + s2;
s1 = s2;
s2 = s;
}
return s;
}
I cannot resist the temptation of connecting a linear time iterative algorithm for Fib to the exponential time recursive one: if one reads Jon Bentley's wonderful little book on "Writing Efficient Algorithms" I believe it is a simple case of "caching": whenever Fib(k) is calculated, store it in array FibCached[k]. Whenever Fib(j) is called, first check if it is cached in FibCached[j]; if yes, return the value; if not use recursion. (Look at the tree of calls now ...)