Fw: [試題] 103上 蕭旭君 演算法設計與分析 期末考+解答
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作者: rod24574575 (天然呆) 看板: NTU-Exam
標題: [試題] 103上 蕭旭君 演算法設計與分析 期末考+解答
時間: Sun May 17 20:22:57 2015
課程名稱︰演算法設計與分析
課程性質︰必修
課程教師：蕭旭君
開課學院：電資學院
開課系所︰資工系
考試日期（年月日）︰2015.01.16
考試時限（分鐘）：180
試題 :
Instructions
˙ This is a 3-hour close-book exam.
˙ There are 7 questions worth of 100 points plus 35 bonus points. The
maximum of your score is 100, so you can allocate your time and select
problems based on certain optimal strategies of your own.
˙ For any of the questions or sub-questions except bonus ones, you get 1/5
of the credit when leaving it blank and get 0 when the answer is
completely wrong. You get partial credit when the answer is partially
correct.
˙ Please write clearly and concisely; avoid giving irrelevant information.
˙ You have access to the appendix on the last page.
˙ Tips: read every question first before start writing. Exam questions are
not all of equal difficulty, move on if you get stuck.
˙ Please print your name, student ID, and page number on every answer page.
Problem 1: Short Answer Questions (30 points)
Answer the following questions and briefly justify your answer to receive full
credit. Clear and concise (e.g., in one or two sentences) explanations are
appreciated.
(a) (3 points) True or False: If P=NP, every NP-complete problem can be solved
in polynomial time.
Ans: True.
Since NPC = NP ∩ NPhard ⊂ NP = P. NPC can be solved in polynomial
time.
(b) (3 points) True or False: If A ≦p B and there is an O(n^3) algorithm
for B, then there is an O(n^3) algorithm for A.
Ans: False.
For example, it may take Ω(n^4) to transform an instance in A to an
instance in B.
(c) (3 points) True or False: If A can be reduced to B in O(n^2) time and there
is an O(n^3) algorithm for B, then there is an O(n^3) algorithm for A.
Ans: False.
The input size of two problem may be different. Let input size of A
be n_1, and input size of B be n_2. Since A can be reduce to B in
O(n^2), n_2 may equals c*(n_1)^2 where c is constant. Although there's
an O(n^3) algorithm for B, in this case it means that B could be done
in O((n_2)^3) = O((c*(n_1)^2)^3) = O((n_1)^6).
(d) (3 points) True or False: It is more efficient to represent a sparse graph
using an adjacency matrix than adjacency lists.
Ans: False.
The space complexity of adjacency matrix is O(V^2) and adjacency list
is O(V+E). Since E << V^2 in sparse graph, adjacency list is generally
more efficient than adjacency matrix.
(e) (3 points) True or False: Kruskal's algorithm and Prim's algorithm always
output the same minimum spanning tree when all edge weights are distinct.
Ans: True.
Since all edge weights are distinct, the minimum spanning tree must
be unique.
(f) (3 points) True or False: A weakly connected directed graph with n vertices
and n edges must have a cycle. (A directed graph is weakly connected if the
corresponding undirected graph is connected.)
Ans: False.
The graph below is a weakly connected directed graph with 3 vertices
and 3 edges but is acyclic.
╭─────────╮
↓ │
┌─┐ ┌─┐ ┌─┐
│１│←─│２│←─│３│
└─┘ └─┘ └─┘
(g) (4 points) Describe a computational problem that is not in NP, that is,
cannot be verified in polynomial time.
Ans: Halting problem. Since halting problem is undecidable, it's not in NP.
(h) (4 points) Bloom filters support worst-case O(1) time membership testing
at the cost of false positives. Explain the meaning of false positives
here.
Ans: Membership test returns YES even though it is not a member.
(i) (4 points) Construct an example demonstrating that plurality is not
strategyproof. Recall that plurality is a voting rule in which each voter
awards one point to top candidate, and the candidate with the most points
wins, and a voting rule is strategyproof if a voter can never benefit from
lying about his or her preferences.
Ans: ┌─────┬──┬──┬──┐
│Preference│ 1st│ 2nd│ 3rd│
├─────┼──┼──┼──┤
│Voter 1 │ Ａ │ Ｃ │ Ｂ │
├─────┼──┼──┼──┤
│Voter 2 │ Ａ │ Ｂ │ Ｃ │
├─────┼──┼──┼──┤
│Voter 3 │ Ｂ │ Ｃ │ Ａ │
├─────┼──┼──┼──┤
│Voter 4 │ Ｃ │ Ａ │ Ｂ │
├─────┼──┼──┼──┤
│Voter 5 │ Ｃ │ Ａ │ Ｂ │
└─────┴──┴──┴──┘
By lying about his preference, Voter 3 can ensures his 2nd choice is
elected rather than his 3rd choice, so plurality is not strategyproof.
Problem 2: Maximum Independent Sets (15 points)
An independent set of a graph G = (V, E) is a subset V' ⊆ V of vertices
such that each edge in E is incident on at most one vertex in V'. The
independent-set problem is to find a maximum-size independent set in G. We
denote by A_opt and A_dec the optimization problem and the decision problem
of the independent-set problem, respectively.
(a) (3 points) Describe the corresponding decision problem, A_dec.
Ans: Given a graph G = (V, E) and a thershold k, if there exists an
independent set V' where |v'| ≧ k. (Or |V'| = k is acceptable)
(b) (3 points) Show that A_dec ≦p A_opt.
Ans: Given G = (V, E) and k as the input of A_dec, we output the result of
A_dec by A_dec = (|A_opt(G)| ≧ k)? true : false. Because if the size
of the maximum-size independent set is bigger than k, we can find an
independent set V' s.t. |V'| ≧ k. This reduction takes constant time
because the reduction only runs a if statement on the result of A_opt.
(c) (3 points) Show that A_opt ≦p A_dec.
Ans: Given G = (V, E) as the input of A_opt, we output the result of A_opt
by run the following algorithm
┌────────────────────────────────┐
1│for i = |V| .. 0 │
2│ if Decision(V, i) is true │
3│ return i │
└────────────────────────────────┘
where function Decision is the algorithm solving A_dec. Because the
maximum-size independent set will meet the highest threshold
descending from |V| to 0, the first k that the decision problem
answer yes is the size we want. This reduction takes O(|V|) for
reduction, which is a polynomial time reduction. (Binary search on k
is acceptable, complexity = O(log|V|)).
(d) (6 points) Although the related decision problem is NP-complete, finding
an independent set of maximum size on certain special graphs can be done
in polynomial time. Suppose G is an n-by-m grid, express the maixmum size
of independent sets as a function of n and m.
Ans: ceil(n*m/2), or any equivalent function.
Problem 3: Approximation Algorithms (20 points)
in Homework 5, we learned how to reduce 3-CNF-SAT to k-CNF-SAT for any integer
k > 3 in polynomial time. Answer the following related questions:
(a) (10 points) The MAX-k-CNF-SAT problem is similar to the k-CNF-SAT problem,
except that it aims to return an assignment of the variables that maximize
the number of clauses evaluating to 1. Design a randomized
(2^k / (2^k - 1))-approximation algorithm for k-CNF-SAT (k > 3). Your
algorithm should run in polynomial time. To simplify this question, you
can assume that no clause contains both a variable and its negation.
Ans: Just randomly assign every variable to be true / false. The processing
time is O(n), where n is the number of the variable, of course bound
by three times the number of the clauses. Consider every true clauses
in the optimal solution. The clause will be false only if all the
variables in this clause is assigned to be false. As a result, the
expected bias ratio is (2^k)/(2^k - 1).
(b) (10 points) Your friend Ada claimed that, for any k, she can design a
(2^k / (2^k - 1))-approximation algorithm for the MAX-3-CNF-SAT problem.
Given a k and a formula in the 3-CNF form, her algorithm first reduces the
formula into the k-CNF form, and then solves the MAX-k-CNF-SAT problem
(as described in part (a)).
Do you agree with her? Justify your answer.
Ans: No. The solution of the MAX-k-CNF-SAT problem does not imply a
MAX-3-CNF-SAT problem. Since the clauses expanded from the same clause
in 3-CNF need to be all true if there exist a solution make this the
original clause true. Any solution talking about the dependency of the
clauses of the expanded MAX-k-CNF-SAT problem is acceptable.
Problem 4: Negative-Weight Edges (20 points)
(a) (10 points) Explain why Dijkstra's algorithm (Figure 2) may return
incorrect results when the graph contains negative-weight edges. Provide
a small (no more than 4 vertices) counterexample as well.
Ans: Course slide Graph 3 page 45.
(b) (10 points) Johnson has learned a powerful technique called reweighting
in Homework 4. Given a graph G = (V, E) and a weight function w, this
reweighting technique constructs a new weight function
(w^)(u, v) = w(u, v) + h(u) - h(v) such that (w^)(u, v) ≧ 0 for all
(u, v) ∈ E.
Prove that the following statement is true for any function h that maps
vertices to real numbers: For any pair of vertices u,v ∈ V, a path p is a
shortest path from u to v using the weight function w if and only if p is
a shortest path from u to v using weight function (w^).
PS. (w^)為w上方有^的符號
Ans: Let p = (v_0, v_1, ..., v_k) be any path from vertex v_0 to
vertex v_k. Since p has (w^)(p) = w(p) + h(v_0) - h(v_k), and both
h(v_0) and h(v_k) do not depend on the path, if one path from v_0 to
v_k is shorter than another using weight function w, then it is also
shorter using (w^). Thus, w(p) = σ(v_0, v_k) if and only if
(w^)(p) = (σ^)(v_0, v_k). (ref. HW4 prob2.)
Problem 5: Degree-Constrained Minimum Spanning Tree (15 points)
In class we have learned how to find a minimum spanning tree on a graph in
polynomial time.
Now let's consider a variant of minimum spanning trees called
degree-constrained minimum spanning trees. The decision version of the
degree-constrained minimum spanning tree problem is defined like this: Given
a graph G, a weight function w over edges, a weight limit W, and a degree
constraint b, is there a spanning tree whose total weight is at most W and the
degree of each vertex in the spanning tree is at most b?
(a) (5 points) Recall that in Prim's algorithm (Figure 3), we greedily grow a
spanning tree from an arbitrary root vertex by adding an edge with the
minimal weight such that one end of the edge is connected to the tree and
the other end is not. Your friend proposed a slightly modified greedy
algorithm to solve the degree-constrained minimum spanning tree problem.
In this modified algorithm, an additional check degree(u) < b is conducted
at Line 9 in Figure 3.
Provide a counterexample to show that such a greedy algorithm is incorrect.
Ans: Let b = 2, W = 300, A is source.
This greedy algorithm: (1+2+100 = 103) Correct answer: (1+3+4 = 8)
┌─┐ 1 ┌─┐ ┌─┐ 1 ┌─┐ ┌─┐ 1 ┌─┐
│Ａ├──┤Ｂ│ │Ａ├──┤Ｂ│ │Ａ├──┤Ｂ│
└┬┘ ├┬┘ └┬┘ ├┬┘ └┬┘ ├┬┘
│ 2 ╱ │ │ 2 ╱ │ │ 2 ╱ │
3│ ╱ │4 → 3│ ╱ │4 3│ ╱ │4
│ ╱ │ │ ╱ │ │ ╱ │
┌┴┤ ┌┴┐ ┌┴┤ ┌┴┐ ┌┴┤ ┌┴┐
│Ｃ├──┤Ｄ│ │Ｃ├──┤Ａ│ │Ｃ├──┤Ｄ│
└─┘ 100└─┘ └─┘ 100└─┘ └─┘ 100└─┘
↓
┌─┐ 1 ┌─┐ ┌─┐ 1 ┌─┐
│Ａ├──┤Ｂ│ │Ａ├──┤Ｂ│
└┬┘ ├┬┘ └┬┘ ├┬┘
│ 2 ╱ │ │ 2 ╱ │
3│ ╱ │4 ← 3│ ╱ │4
│ ╱ │ │ ╱ │
┌┴┤ ┌┴┐ ┌┴┤ ┌┴┐
│Ｃ├──┤Ｄ│ │Ｃ├──┤Ｄ│
└─┘ 100└─┘ └─┘ 100└─┘
(b) (10 points) Prove that this problem is NP-complete.
Hint: The Hamiltonian Path problem is a known NP-complete problem.
Ans: (1) If we have a solution, you can add all edges and check it is not
bigger than W in O(E) time. Then counting all degree of vertex
and check it is not bigger than b in O(E+V) time. So it is in NP.
(2) If we have a Hamiltonian Path problem with graph G. Let b = 2,
W = |V-1| and all edge in G have weight 1. If we can solve it in
Degree-Constrained Minimum Spanning Tree, we can also solve
origin problem in Hamiltonian Path. So it is in NP-hard.
(3) Combine (1) and (2), this problem is in NP-complete!
Problem 6: Bitonic TSP (15 bonus points)
In the euclidean traveling-salesman problem, we are given a set of n points in
the plane, and we wish to find the shortest closed tour that connects all n
points. Figure 1(a) shows the solution to a 7-point problem. The general
problem is NP-hard, and its solution is therefore believed to require more
than polynomial time.
J. L. Bentley has suggested that we simplify the problem by restricting our
attention to bitonic tours, that is, tours that start at the leftmost point,
go strictly rightward to the rightmost point, and then go strictly leftward
back to the starting point. Figure 1(b) shows the shortest bitonic tour of the
same 7 points. In this case, a polynomial-time algorithm is possible.
Describe an O(n^2)-time algorithm for determining an optimal bitonic tour.
You may assume that no two points have the same x-coordinate and that all
operations on real numbers take unit time.
Hint: Scan left to right, maintaining optimal possibilities for the two parts
of the tour using the concept of dynamic programming.
Figure 1: http://i.imgur.com/pnZZEAG.png
Ans:
First sort the points from left to right. O(n*log(n))
Denote: (1) the sorted points as P_1, P_2, ..., P_n
(2) path P_(i, j), where i ≦ j, includes all points before P_j, start
at some point P_i, goes strictly left to P_1 and go strictly right
to P_j. b[i, j] is the cost of P_(i, j).
We have the following formula:
(1) b[1, 2] = dis(P_1, P_2)
(2) b[i, j] = b[i, j-1] + dis(P_(j-1), P_i) for i < j-1
(3) b[j-1, j] = min_(i≦k＜j-1) {b[k, j-1] + dis(P_k, P_j)}
Consider the (j-1)-th point in the path P_(i, j). It can be on the right going
path, i.e. the immediate predecessor of node j. If that so, the optimal
substructure can be found easily, since the path from i to j-1 must be shortest
and is one of P(i, j-1).
Or it can be on the left going path, i.e. i (It can only be the right-most
node). Then we can enumerate the immediate predecessor of node j and get
b[j-1, j] = min_(i≦k＜j-1) {b[k, j-1] + dis(P_k, P_j)} and we store the
immediate predecessor of P_j in r to reconstruct the tour.
┌────────────────────────────────────┐
1│BitTSP(P[N]) │
2│ sort(P) │
3│ for j = 3 .. N │
4│ for i = 1 .. j-2 │
5│ b[i][j] = b[i, j-1] + dis(P_(j-1), Pj) │
6│ r[i][j] = j-1 │
7│ b[j-1, j] = ∞ │
8│ for k = 1 .. j-2 │
9│ q = b[k][j-1] + dis(P_k, P_j) │
10│ if q < b[j-1][j] │
11│ b[j-1][j] = q │
12│ r[j-1][j] = k │
13│ b[n][n] = b[n-1][n] + dis(P_(n-1), P_n) │
14│ return b, r │
15│PRINT-TOUR(r, n) │
16│ print P_n, P_(n-1) │
17│ k = r[n-1][n] │
18│ PRINT-PATH(r, k, n-1) │
19│ print P_k │
20│PRINT-PATH(r, i, j) │
21│ if i < j │
22│ k = r[i][j] │
23│ if k != i │
24│ print P_k │
25│ if k > 1 │
26│ PRINT-PATH(r, i, k) │
27│ else k = r[j][i] │
28│ if k > 1 │
29│ PRINT-PATH(r, k, j) │
30│ print P_k │
└────────────────────────────────────┘
Problem 7: Dynamic Binary Search (bonus 20 points)
Binary search of a sorted array takes logarithmic search time, but the time to
insert a new element is linear in the size of the array. We can improve the
time for insertion by keeping several sorted arrays. Specifically, suppose
that we wish to support SEARCH and INSERT on a set of n elements. Let
k = ceil(log2(n+1)), and let the binary representation of n be
＜n_(k-1), n_(k-2), ..., n_0＞ We have k sorted arrays A_0, A_1, ..., A_(k-1),
where for i = 0, 1, ..., k-1, the length of array A_i is 2^i. Each array is
either full or empty, depending on whether n_i = 1 or n_i = 0, respectively.
The total number of elements held in all k arrays is therefore
Σ(i = 0 to k-1) (n_i * 2^i) = n. Although each individual array is sorted,
elements in different arrays bear no particular relationship to each other.
Hint: To simplify the setting, you can compute the running time based on the
number of comparisons only, ignoring memory allocation, copy, and free.
(a) (5 points) Describe how to perform the SEARCH operation for this data
structure. Analyze its worst-case running time.
Ans: SEARCH operation can be performed by searching all arrays A_i for
which n_i = 1. In each array, binary search is used. The total
running time in worst case is Σ(i = 0 to k) O(n_i * log(2^i)) =
O(k*log(n)) = O((log(n))^2).
(b) (5 points) Describe how to perform the INSERT operation. Analyze its
worst-case running time.
Ans: First put the new element in A_0. If this bucket overflows, then
merge(sort) the new element and old element in A_0, and try to put
them in A_1. In addition, set n_0 as the appropriate value. If A_0
overflows and n_1 = 1 before insertion, then merge these two arrays
and try to put the merged array into A_2. Continue this until an
empty bucket (with n_i = 0) is found. Set the value of n after each
try of putting elements. In the worst case, all n_i = 1 and each
merge takes 2^i time. Thus the worst-case running time is
Σ(i = 0 to k) 2^i = O(n).
(c) (10 points) Analyze the amortized running time of INSERT.
Ans: Inserting a total of n elements actually cost O(n*log(n)) time as
follows. When a bit flips from 0 to 1 (A_i from empty to full), we
merely sets the pointer, hence costs θ(1); When i-th bit flips from
1 to 0, we perform merge takes θ(2^i) time. By aggregate method,
the i-th bit will be flipped at most O(n/(2^i)) times, hence the
amortized cost is O(Σ(i = 0 to k) (n/(2^i) * O(2^i)) = O(n*log(n)).
Appendix
DIJKSTRA(G, w, s)
1 INITIALIZE-SINGLE-SOURCE(G, s)
2 S = ψ
3 Q = G.V
4 while Q != ψ
5 u = EXTRACT-MIN(Q)
6 S = S∪{u}
7 for each vertex v ∈ G.Adj[u]
8 RELAX(u, v, w)
Figure 2: Dijkstra's algorithm.
MST-PRIM(G, w, r)
1 for each u ∈ G.V
2 u.key = ∞
3 u.π = NIL
4 r.key = 0
5 Q = G.V
6 while Q != ψ
7 u = EXTRACT-MIN(Q)
8 for each v ∈ G.Adj[u]
9 if v ∈ Q and w(u, v) < v.key
10 v.π = u
11 r.key = w(u, v)
Figure 3: Prim's algorithm.
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