The Karger's algorithm relates to graph theory where G=(V,E) is an undirected graph with |E| edges and |V| vertices. The objective is to find the minimum number of cuts in edges in order to separate G into two disjoint graphs. The algorithm is randomized and will, in some cases, give the minimum number of cuts. The more number of trials, the higher probability that the minimum number of cuts will be obtained.
The Karger's algorithm will succeed in finding the minimum cut if every edge contraction does not involve any of the edge set C of the minimum cut.
The probability of success, i.e. obtaining the minimum cut, can be shown to be ≥ 2/(n(n-1))=1/C(n,2), which roughly equals 2/n^2 given in the question.Given: EACH randomized trial using the Karger's algorithm has a success rate of P(success,1) ≥ 2/n^2.
This means that the probability of failure is P(F,1) ≤ (1-2/n^2) for each single trial.
We need to estimate the number of trials, t, such that the probability that all t trials fail is less than 1/n.
Using the multiplication rule in probability theory, this can be expressed as
P(F,t)= (1-2/n^2)^t < 1/n
We will use a tool derived from calculus that
Lim (1-1/x)^x as x->infinity = 1/e, and
(1-1/x)^x < 1/e for x finite.
Setting t=(1/2)n^2 trials, we have
P(F,n^2) = (1-2/n^2)^((1/2)n^2) < 1/e
Finally, if we set t=(1/2)n^2*log(n), [log(n) is log_e(n)]
P(F,(1/2)n^2*log(n))
= (P(F,(1/2)n^2))^log(n)
< (1/e)^log(n)
= 1/(e^log(n))
= 1/n
Therefore, the minimum number of trials, t, such that P(F,t)< 1/n is t=(1/2)(n^2)*log(n) [note: log(n) is natural log]
Use the Pyth. Theorem to calculate the total distance flown by the plane in 3 1/2 hours:
d = sqrt(750^2 + 750^2) = 750sqrt(2) miles
Now divide this distance by the total time (3 1/2 hours):
750sqrt(2) miles
---------------------- = 303 miles per hour average
3.5 hours
If their equal(Same amount)
Answer:
Step-by-step explanation:
All other things being equal, spinner A is fine. It is a fair spinner.
Spinner B is not fair. Players 1 and 3 have only 1 number each. Player 2 on the other hand, has 2 numbers that work for him. If player 2 puts in two dollars and players 1 and 3 one dollar each, that should even up the odds. Now you want it just to be fair. So I think player 2 has to put up 2 dollars and players 1 and 3 each put up one. The pot is 4 dollars each time it is played.
Spinner 3 is not fair either. Player one has 2 chances. Player 2 has 3 chances and player 3 has but one chance. There are 6 chances in all. Player 1 should put up 2 dollars to play player 1 should put up 3 dollars and player 1 should put up 1 dollar.
