代写算法作业，计算代码的时间复杂度。
Purpose
The purpose of this assignment is to practice your analysis of algorithm
complexity. It is recommended that this assignment be printed, and that the
work be done right on it. This assignment is to be submitted in paper form to
the assignment boxes, just to the left of the elevators, on the second floor
of the Goldberg CS Building. For each of the algorithms below:
- Derive the cost function for the algorithm. Be sure to show your work like we do in lectures.
- State the complexity of the algorithm in Big-Oh.
- Prove that the derived cost function is in the stated order (big-Oh).
Problem 0: Reverse
input: vals[n]
output: vals reversed
for i = 0 ... n/2
t = vals[i]
vals[i] = vals[n-i-1]
vals[n-i-1] = t
return vals
—|—
Problem 1: Multiplication
input: A[n][n] and B[n][n] // 2D arrays of size n x n
output: C[n][n]
C = array[n][n] // Assume O(1) cost to create 2D array
for i = 0 ... n
for j = 0 ... n
C[i][j] = 0
for k = 0 ... n
C[i][j] = C[i][j] + A[i][k] x B[k][j]
return C
—|—
Problem 2: Edge Detection
input: A[n][n], F[3][3] // 2D arrays
output: E[n][n]
E = array[n][n] // Assume O(1) basic operation to create 2D array
for x = 1 ... n - 1
for y = 1 ... n - 1
E[x][y] = 0
for k = -1 ... 2
for l = -1 ... 2
E[x][y] = E[x][y] + A[x+k][y+l] x F[k+1][l+1]
return E
—|—
Problem 3: Sorting
input: vals[n]
output: sorted vals[n]
for i = n - 1 ... 0
for j = 0 ... i
if vals[j] ] vals[j+1]
v = vals[j]
vals[j] = vals[j+1]
vals[j+1] = v
return vals
—|—
Problem 4: Longest Common Subsequence
Based on code from
https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
input: X[n], Y[n]
output: length of longest common sequence between X and Y
C = Array[n][n]
for i = 0 … n
C[i,0] = 0
C[0,i] = 0
for i = 1 … n
for j = 1 … n
if X[i] == Y[j]
C[i,j] = C[i-1,j-1] + 1
else if C[i,j-1] < C[i-1,j]
C[i,j] = C[i-1,j]
else
C[i,j] = C[i,j-1]
return C[n,n]
—|—
Problem 5: Sorting, again
input: vals[n]
output: sorted vals[n]
for i = 1 ... n
for j = i ... 1
if vals[i] >= vals[i-1]
break
v = vals[i]
vals[i] = vals[i-1]
vals[i-1] = v
return vals
—|—
Problem 6: Factorization
input: n-bit integer X // see not below the algorithm
output: list of prime factors and their multiplicity
acc = X
for i = 2 ... sqrt( X )
count = 0
while acc % i == 0
acc = acc / i
count = count + 1
if count > 0
print i (count)
—|—
Note: Recall that the range of values of an n-bit integer is 0 … 2^n.
Problem 7: Pivot
input: vals[n]
output: vals divided by a pivot at val[0]
low = 1
high = n - 1
while low <= high
if vals[0] <= vals[high]
t = vals[high]
vals[high] = vals[low]
vals[low] = t
low = low + 1
else
high = high - 1
t = vals[high]
vals[high] = vals[0]
vals[0] = t
—|—
Bonus Problem: Sorting, Once more with feeling!
input: vals[n]
output: sorted vals[n]
sort( vals[n] )
function sort( V[n] ):
if n > 2
m=n/2
sort( V[0:m] )
sort( V[m,n] )
A = array[n] // assume 1 operation to create array
i=0
j=m
k=0
while i < m and j < n
if V[i] [<= V[j]
A[k] = V[i]
i=i+1
else
A[k] = V[j]
j=j+1
k=k+1
if j < n
i=j
for l = k ... n
A[l] = V[i]
i=i+1
for l = 0 ... n
V[l] = A[l]
return V
—|—
Hints and Suggestions
- Print this assignment and show your work right on it.
- Feel free to use O(1) instead of specific constants when analyzing the algorithms.
- Be sure to show your work!
Grading
Each problem will be graded out of 10 points:
- 4 marks for deriving the correct cost function. The cost function should be of the correct order, and it should be clear how it was derived from the algorithm. One mark for the correct answer, three marks for showing your work.
- 2 marks for identifying the correct order of the function (big Oh). One mark for the correct order, and one mark for correct notation. I.e., no multiplicative constants, or multiple terms.
- 4 marks for justifying your answer. Follow the steps discussed in lecture. Approximately, 1 mark per step.
What to Hand In
Submit you assignment in hardcopy (paper form) to the submission box, which is
located to the left of the elevators, on the second floor of the CS Goldberg
Building.
