i need algorithm for generate random complex number please help i know how generate random number but random complex number confuse me
I would simply generate two random numbers and use one for the real part and one for the imaginary part.
Generate 2 random numbers (x, y) (use the built-in rand/rnd/random class from your environment's libraries), where x is the real part and y is the imaginary part.
Create a complex number class (with a constructor that takes a real and imaginary parameter)
Use the 2 random numbers from step 1 to create a complex number, x + i y
1.Generate 2 vector of numbers say one is real_vector and another is imaginary_vector of size say MAX_SIZE to be generated randomly with differrent seeds.
2.Random shuffle the numbers in vectors(real_vector+imaginary_vector) using any distribution let us say use of std::random_shuffle(uniform distribution).
3.randomly generate a index and apply modulo operator for MAX_SIZE and select index from first array that will provide an real part of ur random number.
4.use step 3 to get imaginary part of your random number.
5.Create a complex number using number got from step 3 and step 4 and store in a container.
6.go to step 3 and check if you want any more complex number;if no then break;
Related
i am generating random numbers currently like this :
rand.sample(range(1, network_ip_node_size))
this generates a random number from 1 to x and is working fine
however what i am trying to do is say i have a list (list_of_nodes) for example, i want to generate a random number, but say have a probability of 20% of being a number already contained within the list_of_nodes
what is the most effecient way of achieving this ?
This question already has answers here:
Getting N random numbers whose sum is M
(9 answers)
Closed 9 years ago.
Are there any STL functions that allow one to create a vector with random numbers that add up to 1? Ideally, this would be dependent on the size of the vector, so that I can make the vector size, say, 23 and this function will populate those 23 elements with random numbers between 0 and 1 that all add up to 1.
One option would be to use generate to fill the vector with random numbers, then using accumulate to sum up the values, and finally dividing all the values in the vector by the sum to normalize the sum to one. This is shown here:
std::vector<double> vec(23);
std::generate(vec.begin(), vec.end(), /* some random source */);
const double total = std::accumulate(vec.begin(), vec.end(), 0.0);
for (double& value: vec) value /= total;
Hope this helps!
No, but you can do this easily with the following steps:
Fill the vector with random float values, say 0 to 100.
Calculate the sum.
Divide each value by the sum.
There are certainly lots of standard functions to generate random numbers. To get the normalization to happen, you'll want to do that after you've generated all the numbers. (For instance, you might generate the numbers, then divide them all by their sum.) Note that you probably won't have uniformly-distributed numbers at that point, if it matters.
This depends on the kind of distribution of random numbers that you want. One approach (which has been suggested in another answer) is to just generate some random numbers, then divide them each by their total sum.
Another approach is to make a list of random numbers from the interval [0, 1), then sort them. You can then take the differences between consecutive numbers (adding 0 and 1 to the beginning and end of your list respectively). These differences will naturally sum up to 1. So, for example, let's say you picked 3 random numbers and they were: {0.38, 0.05, 0.96}. Let's add 0 and 1 to this list and then sort it:
{0, 0.05, 0.38, 0.96, 1}
Now let's take the differences:
{0.05, 0.33, 0.58, 0.04}
If you add these up, they sum to 1. If you don't understand why this works, imagine you have a piece of rope of length 1 and you use a knife to cut it some random distance from the end (without moving the pieces apart as you cut it). Naturally all the pieces will add up to the original length. That's exactly what's happening here.
Now, like I said, this approach will give you a different distribution of random numbers than the divide by sum method, so don't consider them to be the same!
I am trying to calculate number of ways of composition of a number using numbers 1 and 2.
This can be found using fibonacci series where F(1)=1 and F(2)=2 and
F(n)=F(n-1)+F(n-2)
Since F(n) can be very large I just need F(n)%1000000007.To speed up the process I am using fibonacci exponentiation .I have written two codes for the same problem(both are almost similar).But one of them fails for large numbers.I am not able to figure out which one is correct ?
CODE 1
http://ideone.com/iCPEyz
CODE 2
http://ideone.com/Un5p2S
Though I have a feeling first one should be correct.I am not able to figure what would happen when there is a case like when we are multiplying say a and b and value of a has already exceeded the upper limit of a and when we multiply this by b ,then how sure can I be that a*b is correct. As per my knowledge if a value is above its data type limits then the value starts again from the lowest value like in below example.
#include<iostream>
#include<limits.h>
using namespace std;
int main()
{
cout<<UINT_MAX<<endl;
cout<<UINT_MAX+2;
}
Output
4294967295
1
"Overflow" (you don't really call it that for unsigneds, they wrap around) of unsigned n-bit types will preserve values modulo 2^n only, not modulo an arbitrary modulus (how could they? Try to reproduce the steps with pen and paper). You therefore have to make sure that no operation ever goes over the limits of your type in order to maintain correct results mod 100000007.
I have a homework problem which i can solve only in O(max(F)*N) ( N is about 10^5 and F is 10^9) complexity, and i hope you could help me. I am given N sets of 4 integer numbers (named S, F, a and b); Each set of 4 numbers describe a set of numbers in this way: The first a successive numbers, starting from S included are in the set. The next b successive numbers are not, and then the next a numbers are, repeating this until you reach the superior limit, F. For example for S=5;F=50;a=1;b=19 the set contains (5,25,45); S=1;F=10;a=2;b=1 the set contains (1,2,4,5,7,8,10);
I need to find the integer which is contained in an odd number of sets. It is guaranteed that for the given test there is ONLY 1 number which respects this condition.
I tried to go trough every number between min(S) and max(F) and check in how many number of sets this number is included, and if it is included in an odd number of sets, then this is the answer. As i said, in this way I get an O (F*N) which is too much, and I have no other idea how could I see if a number is in a odd number of sets.
If you could help me I would be really grateful. Thank you in advance and sorry for my bad English and explanation!
Hint
I would be tempted to use bisection.
Choose a value x, then count how many numbers<=x are present in all the sets.
If this is odd then the answer is <=x, otherwise >x.
This should take time O(Nlog(F))
Alternative explanation
Suppose we have sets
[S=1,F=8,a=2,b=1]->(1,2,4,5,7,8)
[S=1,F=7,a=1,b=0]->(1,2,3,4,5,6,7)
[S=6,F=8,a=1,b=1]->(6,8)
Then we can table:
N(y) = number of times y is included in a set,
C(z) = sum(N(y) for y in range(1,z)) % 2
y N(y) C(z)
1 2 0
2 2 0
3 1 1
4 2 1
5 2 1
6 2 1
7 2 1
8 2 1
And then we use bisection to find the first place where C(z) becomes 1.
Seems like it'd be useful to find a way to perform set operations, particularly intersection, on these sets without having to generate the actual sets. If you could do that, the intersection of all these sets in the test should leave you with just one number. Leaving the a and b part aside, it's easy to see how you'd take the intersection of two sets that include all integers between S and F: the intersection is just the set with S=max(S1, S2) and F=min(F1, F2).
That gives you a starting point; now you have to figure out how to create the intersection of two sets consider a and b.
XOR to the rescue.
Take the numbers from each successive set and XOR them with the contents of the result set. I.e., if the number is currently marked as "present", change that to "not present", and vice versa.
At the end, you'll have one number marked as present in the result set, which will be the one that occurred an odd number of times. All of the others will have been XORed an even number of times, so they'll be back to the original state.
As for complexity, you're dealing with each input item exactly once, so it's basically linear on the total number of input items -- at least assuming your operations on the result set are constant complexity. At least if I understand how they're phrasing things, that seems to meet the requirement.
It sounds like S is assumed to be non-negative. Given your desire for an O(max(F)*N) time boundary you can use a sieving-like approach.
Have an array of integers with an entry for each candidate number (that is, every number between min(S) and max(F)). Go through all the quadruples and add 1 to all array locations associated with included numbers represented by each quadruple. At the end, look through the array to see which count is odd. The number it represents is the number that satisfies your conditions.
This works because you're going under N quadruples, and each one takes O(max(F)) or less time (assuming S is always non-negative) to count the included numbers. That gives you O(max(F)*N).
I have to pick an element from an ascending array. Smaller elements are considered better. So if I pick an element from the beginning of the array it's considered a better choice. But at the same time I don't want the choice to be deterministic and always the same element. So I'm looking for
a random numbers generator that produces numbers in range [0, n], but
the smaller the number is, the more chance of it being produced.
This came to my mind:
num = n;
while(/*the more iteration the more chance for smaller numbers*/)
num = rand()%num;
I was wondering if anyone had a better solution.
I did look at some similar questions but they have details about random number generation generally. I'm looking for a solution to this specific type of random number generation, either an algorithm or a library that provides it.
Generate a Random number, say x, between [0,n) and then generate another Random floating point number, say y, between [0,1]. Then raise x to the power of y and use floor function, you'll get your number.
int cust(int n)
{
int x;
double y, temp;
x = rand() % n;
y = (double)rand()/(double)RAND_MAX;
temp = pow((double) x, y);
temp = floor(temp);
return (int)temp;
}
Update: Here are some sample results of calling the above function 10 times, with n = 10, 20 and 30.
2 5 1 0 1 0 1 4 1 0
1 2 4 1 1 2 3 5 17 6
1 19 2 1 2 20 5 1 6 6
Simple ad-hoc approach that came to my mind is to use standard random generators, but duplicate indices. So in the array:
0, 0, 0, 1, 1, 2, 3
odds are good that smaller element will be taken.
I dont' know exactly what do you need. You can also define your own distribution or maybe use some random number generation libraries. But suggested approach is simple and easy to configure.
UPDATE2: You don't have to generate array explicitly. For array of size 1000, you can generate random number from interval: [0,1000000] and then configure your own distribution of selected values: say, intervals of length 1200 for smaller values (0-500) and intervals of length 800 for larger (500-1000). The main point that this way you can easily configure the probability and you don't have to re-implement random number generator.
Use an appropriate random distribution, e.g. the rounded results of an exponential distribution. Pick a distribution that fits your needs, document the distribution you used, and find a nice implementation. If code under the GNU Public License is an option, use the excellent GNU Scientific Library (GSL), or try Boost.Random.
Two tools will solve many random distribution needs
1) A uniform random number generator which you have
2) and a function which maps uniform values onto your target distribution
I've gotta head to the city now, but I'll make note to write up a couple of examples with a drawing when I get back.
There are some worthwhile methods and ideas discussed in this related question (more about generating a normal pseudo random number)