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]
When raising a fraction to a power, we can raise the numerator and denominator to the power. To find x, we need to find the cube root of both 125 and 27. That is, the numbers that when multiplied by themseelves 3 times result in 125 and 27.
![\sqrt[3]{125} = 5](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B125%7D%20%3D%205%20)
![\sqrt[3]{27} =3](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B27%7D%20%3D3)
So x =
Answer:
Can't answer because we don't know the distance between start position and Middelburg, as well as the number of stops along the way
Step-by-step explanation:
Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)
first factor into (x-r1)(x-r2)... form
p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2
so
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real
baseically
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the root 1 and has multiplity 1
343in3 is the answer to this math question
