Show that 8π^2 β 5π + 7 = π(π^3)

To show that 8π^2 β 5π + 7 = π(π^3), we need to find constants C and k such that for all values of n greater than k, the inequality 8π^2 β 5π + 7 β€ Cπ^3 holds true.

Letβs simplify the equation:

8π^2 β 5π + 7 β€ Cπ^3

Rearranging the terms:

8π^2 β 5π + 7 β Cπ^3 β€ 0

Now, letβs take the limit as n approaches infinity:

lim(n->β) (8π^2 β 5π + 7 β Cπ^3) β€ 0

Since the limit is less than or equal to zero, we can conclude that 8π^2 β 5π + 7 = π(π^3).

Determine whether each of these functions is bounded by O(n). Say yes or no.

a) π(π) = 3Yes, it is bounded by O(n) because it is a constant function.

b) π(π) = π^2 + π + 12Yes, it is bounded by O(n) because the highest degree term is n^2, which grows slower than n^3.

c) π(π) = (π)Yes, it is bounded by O(n) because it is a linear function.

d) π(π) = 17π + 7Yes, it is bounded by O(n) because it is a linear function.

e) π(π) = 15π log πYes, it is bounded by O(n log n) because it grows slower than n^3.

f) π(π) = (π/2)Yes, it is bounded by O(n) because it is a linear function.

Arrange the following functions in ascending order of growth rate.

π3(π) = (log π)^3π4(π) = βπ log ππ2(π) = π^100 + π^3 log ππ6(π) = π^99 + π^98π1(π) = 1.5^π + 50π7(π) = (π!)^2π5(π) = 10^π

Solve the following recurrences using Master Theorem.

a) π(π) = 16π (π/4)+ πβπThe recurrence relation can be written as π=16, π=4, and π=1/2.Since logβ16 = 2, and π = 1/2 < logβ16, we apply case 1 of the Master Theorem.The time complexity of this recurrence relation is Ξ(nΒ²βn).

b) π(π) = π (π/5)+ 10The recurrence relation can be written as π=1, π=5, and π=0.Since logβ
1 = 0, and π = 0 = logβ
1, we apply case 2 of the Master Theorem.The time complexity of this recurrence relation is Ξ(log n).

In the divide and conquer closest pair of points, we sorted the coordinates (x or y). Once sorted, are we ready to start computing distances?

No, once the coordinates are sorted, we need to divide the points into smaller subsets and recursively compute the closest pair of points within each subset. Sorting the coordinates is a preprocessing step that allows us to efficiently divide the points and apply the divide and conquer strategy. After dividing the points and finding the closest pairs within each subset, we need to merge the results and compare distances to find the overall closest pair of points. So, sorting the coordinates alone does not complete the computation of distances.

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