I am looking for an efficient algorithm of the problem, for any N find all i and j such that N=i^j.
I can solve it of O(N^2) as follows,
for i=1 to N
{
for j=1 to N
{
if((Power(i,j)==N)
print(i,j)
}
}
I am looking for better algorithm(or program in any language)if possible.
Given that i^j=N, you can solve the equation for j by taking the log of both sides:
j log(i) = log(N) or j = log(N) / log(i). So the algorithm becomes
for i=2 to N
{
j = log(N) / log(i)
if((Power(i,j)==N)
print(i,j)
}
Note that due to rounding errors with floating point calculations, you might want to check j-1 and j+1 as well, but even so, this is an O(n) solution.
Also, you need to skip i=1 since log(1) = 0 and that would result in a divide-by-zero error. In other words, N=1 needs to be treated as a special case. Or not allowed, since the solution for N=1 is i=1 and j=any value.
As M Oehm pointed out in the comments, an even better solution is to iterate over j, and compute i with pow(n,1.0/j). That reduces the time complexity to O(logN), since the maximum value for j is log2(N).
Here is a method you can use.
Lets say you have to solve an equation:
a^b = n //b and n are known
You can find this using binary search. If, you get a condition such that,
x^b < n and (x+1)^b > n
Then, no pair (a,b) exists such that a^b = n.
If you apply this method in range for b from 1..log(n), you should get all possible pairs.
So, complexity of this method will be: O(log n * log n)
Follow these steps:
function ifPower(n,b)
min=1, max=n;
while(min<max)
mid=min + (max-min)/2
k = mid^b, l = (mid + 1)^b;
if(k == n)
return mid;
if(l == n)
return mid + 1;
if(k < n && l > n)
return -1;
if(k < n)
max = mid - 1;
else
min = mid + 2; //2 as we are already checking for mid+1
return -1;
function findAll(n)
s = log2(n)
for i in range 2 to s //starting from 2 to ignore base cases, powers 0,1...you can handle them if required
p = ifPower(n,i)
if(b != -1)
print(b,i)
Here, in the algorithm, a^b means a raised to power of b and not a xor b (its obvs, but just saying)
Related
In pseudo-code:
j = 5;
while (j <= n) {
j = j* j* j * j;
}
What is the time complexity of this code?
It is way shorter than O(logn), is there even any reason to go lower than that?
Let's trace through the execution of the code. Suppose we start with initial value j0:
0. j ← j0
1. j ← j0^4
2. j ← [j0^4]^4 = j0^(4^2)
3. j ← [j0^(4^2)]^4 = j0^(4^3)
4. j ← [j0^(4^3)]^4 = j0^(4^4)
...
m. j ← [j0^(4^(m-1))]^4 = j0^(4^m)
... after m loops.
The loop terminates when the value exceeds n:
j0^(4^m) > n
→m > log(4, log(j0, n))
Thus the time complexity is O(m) = O(log log n).
I used help from MathSE to find out how to solve this. The answer is same as another one by #meowgoesthedog, but I understand it the following way:
On every iteration, the value of j is going to increase by its own 4th power. Or, we can look at it from the side of n, that on every iteration n is going to reduce by its 4th root. Hence, the recurrence will look like:
T(n) = 1 + T(n1/4)
For any integer k, with 24k + 1 <= n <= 24k + 1, the recurrence will become:
T(n) = 1 + k
if we go on to assume that the 4th root will always be an integer. It won't matter if it is not as the constant of +/- 1 will be ignored in the Big-O calculation.
Now, since the assumption of 4th root being an integer simplifies things for us, we can try to solve the following equation:
n = 24k,
with the equation yielding k = (Log(Log(n)) - Log(2))/Log(4).
This implies that O(T(n)) = O(Log(Log(n))).
If you had code that looks like this, what would the big O be? I'm uncertain as to how if statements affect big O.
n = some arbitrary number
for(i = 0; i < n; i++)
for(j = 0; j < n; j++)
if(i <= j)
for(k = i; k <= j; k++)
//do some simple operation
y = x+1
else
//do some simple operation
y = y+1
I'm not considering compiler optimizations. I know this is somewhere between O(n^2) and O(n^3) but am not sure as the if statement does not always execute the inner most loop.
O(N * N * N) which we can just say is O(N^3)
First Loop happens N times.
Second Loop happens N times.
Those multiply together to get O(N^2)
Out of all the possible N^2 loops, the third loop would run about half the time, which is O(N/2) which is equivalent to O(N).
And that's how you get O(N * N * N) or O(N^3)
In fact you can count (almost exactly) how many operations you do:
i: 0 to n-1 = N operations
x
j: 0 to n-1 = N operations
x
only when i<=j, from i to j, or another task O(1)
The other task gives you NxN operations, then O(NxN)
that is, if you invert
for every j (0 to n-1) : N operations
then for every i from 0 to j, you do an operation from i to j, that is j-i+1
exactly the same as every 0 to j operations. Then you have (j+1)x(j+2)/2 operations.
then finally, you get Sum (j+1)x(j+2)/2 from 0 to N which is
1/2 ((N+1)x(N+2)/2 + (N+1)^3/3+(N+1)^2/2+(N+1)/6) operations, so O(N^3)
Perhaps, I forgot some +/-1
You can analyse your algorithm using Sigma notation:
From this, it's obvious the time complexity will depend on cubic n terms, and hence your algorithm is in O(n^3).
This is O(N^3).
Proof: http://www.wolframalpha.com/input/?i=sum+sum+%28j+-+i+%2B+1%29%2C+j+%3D+i+to+n+-+1%2C+i+%3D+0+to+n+-+1
Last cycle runs (j - i + 1) times.
How to find this sum manually?
This is not a rocket math.
Try to read about https://en.wikipedia.org/wiki/Telescoping_series
Just to save time it's easier to use wolframalpha for that purpose.
Consider the following function:
int testFunc(int n){
if(n < 3) return 0;
int num = 7;
for(int j = 1; j <= n; j *= 2) num++;
for(int k = n; k > 1; k--) num++;
return testFunc(n/3) + num;
}
I get that the first loop is O(logn) while the second loop gives O(n) which gives a time complexity of O(n) in total. But due to the recursive calls I thought the time complexity would be O(nlogn), but apperantly it is only O(n). Can anyone explain why?
The recursive call pretty much gives the following for the complexity(denoting the complexity for input n by T(n)):
T(n) = log(n) + n + T(n/3)
First observation as you correctly noted is that you can ignore the logarithm as it is dominated by n. Now we are only left with T(n) = n + T(n/3). Try writing this up to 0 for instance. We have:
T(n) = n + n/3 + n/9+....
You can easily prove that the above sum is always less than 2*n. In fact better limits can be proven but this one is enough to state that overall complexity is O(n).
For procedures using a recursive algorithm such as the following:
procedure T( n : size of problem ) defined as:
if n < base_case then exit
Do work of amount f(n) // In this case, the O(n) for loop
T(n/b)
T(n/b)
... a times... // In this case, b = 3, and a = 1
T(n/b)
end procedure
Applying the Master theorem to find the time complexity, the f(n) in this case is O(n) (due to the second for loop, like you said). This makes c = 1.
Now, logba = log31 = 0, making this the 3rd case of the theorem, according to which the time complexity T(n) = Θ(f(n)) = Θ(n).
How can I calculate time complexity of following piece of code? Suppose m is close to n. What I got is f(n) = 2*f(n-1). So time complexity is f(n) = O(2^n). Am I right?
int uniquePaths(int m, int n) {
if (m < 1 || n < 1) return 0;
if (m == 1 && n == 1) return 1;
return uniquePaths(m - 1, n) + uniquePaths(m, n - 1);
}
There is some hand-waving involved in what follows, but I think it's essentially correct.
Every leaf in the call tree will contribute 1 to the total result, so the number of leaves is uniquePaths(m,n). Since uniquePaths(m,n) == "m+n-2 choose n-1", when m and n are similar the execution time of your algorithm will be roughly the central binomial coefficient "2n choose n", which is in O(4^n).
I just resolve a problem but i don't have the solution of that so i kindly ask you if you can confirm if my solution is correct or not
int h=1; int cont = 0;
for (j = 2^N; j>1; j = j/2) {
h = h * 2;
for (i =1; i < j; i = i*2)
for (k=2; k<h; k++)
cont ++;
}
I must find the complexity of that portion of code in BIGTHETA.
So, i analyze that the third cycle grow up in that way
k -> linear until = h (h grow up like 2^w) - So the complexity is log n.
About the second, the first cycles' limit is 0 so i think that the complexity is log n.
About the first 2^N = 2^N-1 so the complexity is n
The total complexity is n * log n
You can proceed formally, step by step, using Sigma notation (I skipped some steps, but feel free to ask for more details if necessary):