Answer:
Answer "C.flat tip"
Explanation:
I dont know why this is on here but here you go:
A square shaft is commonly found on flathead screwdrivers, also known as flatted screwdrivers or slotted screwdrivers.
Answer:
C
Explanation:
because flat tip have a square surface
Perform an “average case” time complexity analysis for Insertion-Sort, using the given proposition
and definition. I have broken this task into parts, to make it easier.
Definition 1. Given an array A of length n, we define an inversion of A to be an ordered pair (i, j) such
that 1 ≤ i < j ≤ n but A[i] > A[j].
Example: The array [3, 1, 2, 5, 4] has three inversions, (1, 2), (1, 3), and (4, 5). Note that we refer to an
inversion by its indices, not by its values!
Proposition 2. Insertion-Sort runs in O(n + X) time, where X is the number of inversions.
(a) Explain why Proposition 2 is true by referring to the pseudocode given in the lecture/textbook.
(b) Show that E[X] = 1
4n(n − 1). Hint: for each pair (i, j) with 1 ≤ i < j ≤ n, define a random indicator
variable that is equal to 1 if (i, j) is an inversion, and 0 otherwise.
(c) Use Proposition 2 and (b) to determine how long Insertion-Sort takes in the average case.
a. Proposition 2 states that Insertion-Sort runs in O(n + X) time, where X is the number of inversions.
b. The expected number of inversions, E[X], E[X] = 1/4n(n-1).
c. In the average case, Insertion-Sort has a time complexity of approximately O(1/4n²).
How to calculate the information(a) Proposition 2 states that Insertion-Sort runs in O(n + X) time, where X is the number of inversions. To understand why this is true, let's refer to the pseudocode for Insertion-Sort:
InsertionSort(A):
for i from 1 to length[A] do
key = A[i]
j = i - 1
while j >= 0 and A[j] > key do
A[j + 1] = A[j]
j = j - 1
A[j + 1] = key
b. The expected number of inversions, E[X], can be calculated as follows:
E[X] = Σ(i,j) E[I(i, j)]
= Σ(i,j) Pr((i, j) is an inversion)
= Σ(i,j) 1/2
= (n(n-1)/2) * 1/2
= n(n-1)/4
Hence, E[X] = 1/4n(n-1).
(c) Using Proposition 2 and the result from part (b), we can determine the average case time complexity of Insertion-Sort. The average case time complexity is given by O(n + E[X]).
Substituting the value of E[X] from part (b):
Average case time complexity = O(n + 1/4n(n-1))
Simplifying further:
Average case time complexity = O(n + 1/4n^2 - 1/4n)
Since 1/4n² dominates the other term, we can approximate the average case time complexity as:
Average case time complexity ≈ O(1/4n²)
Therefore, in the average case, Insertion-Sort has a time complexity of approximately O(1/4n²).
Learn more about proposition on
https://brainly.com/question/30389551
What are the two instructions needed in the basic computer in order to set the E flip-flop to 1?
Answer:
Load and save instructions. The specific instructions may vary depending on the computer`s architecture, but the general process is to load the desired value into a register and store it in a flip-flop. Below is an example of a hypothetical assembly procedure.
Load Instruction: Load the value 1 into a register.
"""
LOAD R1, 1
"""
Store Instruction: Store the value from the register into the flip-flop.
"""
STORE R1, FlipFlop
"""
weird question but this might help
how to make (All capital) enhancement in Ms word
while creating a style
Answer:
keyword involved
Explanation:
simple select the desired text and click SHIFT + F3