Bounded functions

Show that 8𝑛^2 − 5𝑛 + 7 = 𝑂(𝑛^3)
To show that 8𝑛^2 − 5𝑛 + 7 = 𝑂(𝑛^3), we need to find constants C and k such that for all values of n greater than k, the inequality 8𝑛^2 − 5𝑛 + 7 ≤ C𝑛^3 holds true.
Let’s simplify the equation:
8𝑛^2 − 5𝑛 + 7 ≤ C𝑛^3
Rearranging the terms:
8𝑛^2 − 5𝑛 + 7 – C𝑛^3 ≤ 0
Now, let’s take the limit as n approaches infinity:
lim(n->∞) (8𝑛^2 − 5𝑛 + 7 – C𝑛^3) ≤ 0
Since the limit is less than or equal to zero, we can conclude that 8𝑛^2 − 5𝑛 + 7 = 𝑂(𝑛^3).
Determine whether each of these functions is bounded by O(n). Say yes or no.
a) 𝑓(𝑛) = 3Yes, it is bounded by O(n) because it is a constant function.
b) 𝑓(𝑛) = 𝑛^2 + 𝑛 + 12Yes, it is bounded by O(n) because the highest degree term is n^2, which grows slower than n^3.
c) 𝑓(𝑛) = (𝑛)Yes, it is bounded by O(n) because it is a linear function.
d) 𝑓(𝑛) = 17𝑛 + 7Yes, it is bounded by O(n) because it is a linear function.
e) 𝑓(𝑛) = 15𝑛 log 𝑛Yes, it is bounded by O(n log n) because it grows slower than n^3.
f) 𝑓(𝑛) = (𝑛/2)Yes, it is bounded by O(n) because it is a linear function.
Arrange the following functions in ascending order of growth rate.
𝑓3(𝑛) = (log 𝑛)^3𝑓4(𝑛) = √𝑛 log 𝑛𝑓2(𝑛) = 𝑛^100 + 𝑛^3 log 𝑛𝑓6(𝑛) = 𝑛^99 + 𝑛^98𝑓1(𝑛) = 1.5^𝑛 + 50𝑓7(𝑛) = (𝑛!)^2𝑓5(𝑛) = 10^𝑛
Solve the following recurrences using Master Theorem.
a) 𝑇(𝑛) = 16𝑇 (𝑛/4)+ 𝑛√𝑛The recurrence relation can be written as 𝑎=16, 𝑏=4, and 𝑐=1/2.Since log₄16 = 2, and 𝑐 = 1/2 < log₄16, we apply case 1 of the Master Theorem.The time complexity of this recurrence relation is Θ(n²√n).
b) 𝑇(𝑛) = 𝑇 (𝑛/5)+ 10The recurrence relation can be written as 𝑎=1, 𝑏=5, and 𝑐=0.Since log₅1 = 0, and 𝑐 = 0 = log₅1, we apply case 2 of the Master Theorem.The time complexity of this recurrence relation is Θ(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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