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]
Answer:
Eve used scale as 2 yards is equal to 1 inch.
Step-by-step explanation:
Given:
Actual Picnic area = 68 yards.
Drawing Picnic area = 34 inches.
We need to find the scale used by Eve.
Solution:
Now we can say that;
scale used by Eve can be calculated by dividing Actual Picnic area from drawing picnic area.
framing in equation form we get;
scale used by Eve = 
Hence Eve used scale as 2 yards is equal to 1 inch.
Answer:
10√2
Step-by-step explanation:
A(1, 4), B(3, 6), C(6, 3), D(4, 1)
The distance formula tells you the distance d between two points (x1, y1) and (x2, y2) is given by ...
d = √((x2-x1)² +(y2-y1)²
Then the side lengths are ...
AB = √(2² +2²) = √8 = 2√2
BC = √(3² +(-3)²) = √18 = 3√2
The perimeter is twice the sum of these sides, so is ...
P = 2(2√2 +3√2) = 10√2 . . . the perimeter of the rectangle
