5115 CPSC Homework Set 2
Algorithm Analysis and Design Due Wednesday, September 28, 2005
DO NOT PLACE ANY WORK OR ANSWERS ON THIS QUESTION SHEET. WRITE ALL OF YOUR ANSWERS ON A SEPARATE SHEET OF PAPER. 1. Problem 4 on page 91 of the textbook. "Hypothesize a likely efficiency class of an algorithm based on the following empirical observations of its basic operation's count."
Size Count 1000 11966 2000 24303 3000 39992 4000 53010 5000 67272 6000 78692 7000 91274 8000 113063 9000 129799 10000 140538
Answer: The easy way to get this is to take the ratio of Count (10000) / Count (1000). This value is 140538 / 11966 = 11.74 10. One could do a plot of the data.
Page 91, Problem 4
160000
140000
120000
100000 Count
80000
Series1
60000
40000
20000
0 0 2000 4000 6000 Size 8000 10000 12000
The plot above was done in MS-Excel. It strongly suggests a linear function.
2. Problem 8 on page 91 of the textbook. "The averagecase efficiency of Euclid's algorithm on inputs of size n can be measured by the average number of divisions Davg(n) made by the algorithm for computing gcd(n, 1), gcd(n, 2), ..., gcd(n, n). For example, Davg(5) = (1 + 2 + 3 + 2 + 1) / 5 = 1.8 Produce a scatterplot of Davg(n) and indicate the algorithm's likely average efficiency class." Limit your work to computing Davg(3) through Davg(10). Answer: Here is my copy of Euclid's algorithm, adapted from the algorithm on page 5 of the textbook. Algorithm Euclid (M, N) While (N 0) Do Q=M/N // Integer division. We count this operation. R=MQN M=N N=R End While Return M End Euclid Here are the numbers I got 3 1.33 4 1.25 5 1.80 6 1.33 7 2.00 8 1.88 9 1.89 10 1.80 11 2.55 12 1.75
Here is my plot. It might be either linear or logarithmic. Probably it has no good class. My best guess for T(N) would be the Euler Number (N) the number of integers less than N that are relative prime to N.
Division Count for Euclid
3.00
2.50
2.00 Average Count
1.50
Series1
1.00
0.50
0.00 0 2 4 6 Input Value 8 10 12 14
3. Problem 4 (a, b, & c) on page 102 of the textbook. " a. Design a bruteforce algorithm for computing the value of a polynomial p(x) = anxn + an-1xn-1 + ... + a1x + a0 at a given point x0 and determine its worst-case efficiency class. b. If the algorithm you design is (n2), design a linear algorithm for this problem. c. Is it possible to design an algorithm with better than linear efficiency?" You may want to do some online research for this one. Any reasonable answer for 4 c) will get a good grade. ANSWER: a) Here is a (N2) solution. Algorithm Silly (A[0 .. N], X) P = A[0] For J = 1 to N Do XJ = 1 For K = 1 to J Do XJ = XJ X P = P + A[J] XJ Next J Return P
// Compute XJ.
Here is a ( N log(N) ) solution. It uses the optimal algorithm for generating an arbitrary positive integer power of a number. Call this F(X, J) for XJ. Algorithm LessSilly (A[0 .. N], X) P = A[0] For J = 1 to N Do P = P + A[J] F(X, J) Next J Return P Here is another linear algorithm. It depends on computing a sequence of powers of X, thus X1, X2, X3, X4, ....., XN. Algorithm NotBad (A[0 .. N], X) P = A[0] XJ = 1 For J = 1 to N Do XJ = XJ X P = P + A[J] XJ Next J b) Horner's rule (textbook, page 227) is the answer to this one. Algorithm Horner (A[0 .. N], X) P = A[N] For J = (N 1) Down To 0 Do P = X P + A[J] Return P c) Every algorithm must use each of the (N + 1) elements in A[0 .. N], so NO, it is not possible to design an algorithm with better than linear efficiency.
4. Problem 8 on page 102 of the textbook. The entire problem reads "Sort list the E, X, A, M, P, L, E in alphabetical order by bubble sort." ANSWER: Here is the Bubble Sort algorithm, adapted from page 100 of the textbook. Algorithm BubbleSort ( A [ 0 .. (N 1)] ) For I = 0 to (N 2) Do For J = 0 to (N 2 I) Do If A[J + 1] < A[J] Then Swap A[J] and A[J + 1] Here is the data set K A[K] I=0 J = 0 to 5 J=0 J=1 K A[K] J=2 K A[K] J=3 K A[K] J=4 K A[K] J=5 K A[K] A[1] < A[0] A[2] < A[1] 0 E 1 A "X" < "E" "A" < "X" 2 X "M" < "X" 1 A 2 M "P" < "X" 1 A 2 M "L" < "X" 1 A 2 M "E" < "X" 1 A 2 M 3 P 3 P YES 4 L 3 P 3 X YES 4 X YES 4 L 3 M YES 4 P NO YES. Swap A[1] and A[2] 4 P 5 L 6 E N = 7, so N 2 = 5 0 E 1 X 2 A 3 M 4 P 5 L 6 E
A[3] < A[2] 0 E
Swap A[2] and A[3] 5 L 6 E
A[4] < A[3] 0 E
Swap A[3] and A[4] 5 L 6 E
A[5] < A[4] 0 E
Swap A[4] and A[5] 5 X 6 E
A[6] < A[5] 0 E
Swap A[5] and A[6] 5 E 6 X
Note that the largest value, "X", has "bubbled up" to the top.
Here is the array after the first pass K A[K] I=1 J = 0 to 4 J=0 K A[K] J=1 J=2 J=3 K A[K] J=4 K A[K] A[1] < A[0] 0 A 1 E "A" < "E" 2 M "M" < "E" "P" < "M" "L" < "P" 1 E 2 M "E" < "P" 1 E 2 M 3 L 3 L YES 4 E 3 P NO NO YES 4 P Swap A[3] and A[4] 5 E 6 X YES 4 L Swap A[0] and A[1] 5 E 6 X 0 E 1 A 2 M 3 P 4 L 5 E 6 X
A[2] < A[1] A[3] < A[2] A[4] < A[3] 0 A
A[5] < A[4] 0 A
Swap A[4] and A[5] 5 P 6 X
The second largest value, "P", has "bubbled up" to second place. I=2 J = 0 to 3 J=0 J=1 J=2 K A[K] J=3 K A[K] A[1] < A[0] A[2] < A[1] A[3] < A[2] 0 A 1 E "E" < "A" "M" < "E" "L" < "M" 2 L "E" < "M" 1 E 2 L 3 E 3 M YES 4 M NO NO YES 4 E
Swap A[2] and A[3] 5 P 6 X
A[4] < A[3] 0 A
Swap A[3] and A[4] 5 P 6 X
The value "M" has "bubbled up" to third place.
Here is the array now, after the third pass. K A[K] I=3 J = 0 to 2 J=0 J=1 J=2 K A[K] I=4 J = 0 to 1 J=0 J=1 I=5 J=0 J = 0 A[1] < A[0] We have completed the sort. "E" < "A" NO A[1] < A[0] A[2] < A[1] "E" < "A" "E" < "E" NO NO A[1] < A[0] A[2] < A[1] A[3] < A[2] 0 A 1 E "E" < "A" "L" < "E" "E" < "L" 2 E 3 L NO NO YES 4 M 0 A 1 E 2 L 3 E 4 M 5 P 6 X
Swap A[2] and A[3] 5 P 6 X
5. a) Give a definition of "stable sort"; i.e., what does it mean for a sort to be stable? b) Is bubble sort stable? You may need to do some online research for this one. For part b), you may quote any source or give an example. ANSWER: According to the text (see page 20), a sort is stable if "it preserves the relative order of two elements in its input." The idea of stable sorts arose from the idea of major key and minor key in data processing. The most common example is a data set in which the person's last name is the major key and the person's first name is a minor key. Here we would sort by minor key only when the values of the major key are the same. Bubble sort can be shown to be stable. Nobody proved the claim, but it was quite satisfactory to quote a web source. An example would have been useful only if bubble sort were not stable. Since it is stable, examples are of no use.