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Number of independent trials is N, probability of success is p. I want to calculate
Probability of m consecutive successes.
Probability of m or more consecutive successes.
The numbers are very large, so the algorithm should be highly optimized.
N = 877646440
m = 79279,
p = 6204/6205 (or 0.999838839645447....)
I seem to have the answer on mathematical SE where I originally started this question. https://math.stackexchange.com/questions/1888887/easily-calculable-minimum-probability-for-m-or-more-consecutive-outcomes/1889372#1889372
I will implement that solution and update the questions.
Edit: I have gotten the answer on the mathematical SE question and implemented the solution.
Thanks
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In A* algorithm, if g=0 and h=0 then what will be the result of f?
I know f(x)=g(x)+h(x). So it is true that f(x) will be zero?
f(x) would be 0.
But this should hardly ever occur.
g(x)=0 means you had no costs to reach x (should only be the case for the starting point)
h(x)=0 means the heuristics says that the costs to reach the goal from x costs not more than 0 (means that you are at the goal)
so f(x)=0 should only be possible if you start at the goal.
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Suppose we are given n intervals [xi;yi] for 1 <= i <= n, which together cover an interval [x0;y0]. Each interval [xi;yi] has a positive cost ci. Give an efficient algorithm to find a subset of intervals with minimum total cost which together still cover the interval [x0;y0].
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In the interval covering problem, we are given n intervals
[s1,t1), [s2,t2), ···, [sn,tn)
such that
S i∈[n][si,ti) = [0,T).
The goal of the problem is to return a smallest-size set
S ⊆ [n]
such that
S i∈S[si,ti) = [0,T).
Design a greedy algorithm for this problem.
A greedy algorithm could be devised as follows. As long as there is a point p in [0,T) which is not contained in one of the already selected intervals, select an interval [s_i,t_i) , which must exist, since the union of all [s_i,t_i) is [0,T) as stated in the requirements. As the set of intervals [s_i,t_i) is finite, this procedure must terminate.
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I need help with the following problem:
for i <- 5 to m do P(i)
where P(i) is executed (m-i) times provided m >= 3
I realize that this is the summation
But I'm not sure exactly how to calculate the run time from this. Any suggestions?
If you count complexity of P(i) as constant, then you have just loop m - 5 times - it will give you complexity O(m)
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This question is a practice problem on interviewstreet.com.
Find number of integral solution of 1/x+1/y=1/N! for a given N
For N=1 answer is 1.
I tried to solve this questions , but cant predict from where to start. I am not from math background.
I am looking for the approach , how should I proceed towards the solution.
Is there any direct formula for this?
Try to solve it as a iterative problem. All the solutions of N-1 are also valid for N. The only uncovered solutions are where x and y are both not divisible by N, which should be easier to count.