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Natali5045456 [20]
3 years ago
13

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routin

e that we covered in class on the quicksort slides. Consider another almost-best case for quicksort, in which the pivot always splits the arrays 1/3: 2/3, i.e., one third is on the left, and two thirds are on the right, for all recursive calls of Deterministic Quicksort. (a) Give the runtime recurrence for this almost-best case. (b) Use the recursion tree to argue why the runtime recurrence solves to Theta (n log n). You do not need to do big-Oh induction. (c) Give a sequence of 4 distinct numbers and a sequence of 13 distinct numbers that cause this almost-best case behavior. (Assume that for 4 numbers the array is split into 1 element on the left side, the pivot, and two elements on the right side. Similarly, for 13 numbers it is split with 4 elements on the left, the pivot, and 8 elements on the right side.)
Engineering
1 answer:
juin [17]3 years ago
3 0

Answer:

Answer for the question:

Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routine that we covered in class on the quicksort slides. Consider another almost-best case for quicksort, in which the pivot always splits the arrays 1/3: 2/3, i.e., one third is on the left, and two thirds are on the right, for all recursive calls of Deterministic Quicksort. (a) Give the runtime recurrence for this almost-best case. (b) Use the recursion tree to argue why the runtime recurrence solves to Theta (n log n). You do not need to do big-Oh induction. (c) Give a sequence of 4 distinct numbers and a sequence of 13 distinct numbers that cause this almost-best case behavior. (Assume that for 4 numbers the array is split into 1 element on the left side, the pivot, and two elements on the right side. Similarly, for 13 numbers it is split with 4 elements on the left, the pivot, and 8 elements on the right side.)

is given in the attachment.

Explanation:

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Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.
likoan [24]

Answer:

P<u>rogram:</u>

# Enter Numbers #

number1 = int(input("Enter number: " ))

print("Largest: " + string(number1))

#for num 2 #

number2 = int(input("Enter a number: "))

if number2 > number1:

 print("Largest: " + string(number2))

else:

 print("Largest: " + string(num1))

#for num 3 #

number3 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3)))  

#for num 4 #

number4 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4)))

#for num 5 #

number5 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5)))

#for num 6 #

number6 = int(input("Enter a number: "))

print("Largest: " + string(max(number1, number2, number3, number4, number5, number6)))        

# END #

4 0
3 years ago
Fill in the blank to correctly complete the statement below.
frutty [35]

Answer:

The invention of the pendulum-driven ___<u>clocks</u>___ in the 1600s paved the way for a new industrial era.

4 0
3 years ago
The electric motor exerts a torque of 800 N·m on the steel shaft ABCD when it is rotating at a constant speed. Design specificat
kodGreya [7K]

Answer:

d= 4.079m ≈ 4.1m

Explanation:

calculate the shaft diameter from the torque,    \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}

Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).

r = Radius of the shaft.

T = Twisting Moment or Torque.

J = Polar moment of inertia.

C = Modulus of rigidity for the shaft material.

l = Length of the shaft.

θ = Angle of twist in radians on a length.  

Maximum Torque, ζ= τ ×  \frac{ π}{16} × d³

τ= 60 MPa

ζ= 800 N·m

800 = 60 ×  \frac{ π}{16} × d³

800= 11.78 ×  d³

d³= 800 ÷ 11.78

d³= 67.9

d= \sqrt[3]{} 67.9

d= 4.079m ≈ 4.1m

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