Assignment 4. (Due on Dec 2. 2:30 p.m.)

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• Assignment 4. (Due on Dec 2. 230 p.m.)
• This time, Prof. Yao and I can explain the
  questions, but we will NOT tell you how to solve
  the problems.
• Question 1. (20 points) Give an O(n2) dynamic
  algorithm to find the longest monotonically
  increasing subsequence of a sequence of n
  numbers. (Assume that each integer appears once
  in the input sequence of n numbers)
• Example Consider sequence 1,8, 2,9, 3,10, 4, 5.
  Both subsequences 1, 2, 3, 4, 5 and 1, 8, 9, 10
  are monotonically increasing subsequences.
  However, 1,2,3, 4, 5 is the longest.

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• Assignment 4. (Due on Dec 2. 230 p.m.)
• Question 2. (40 points) Give an O(n2) dynamic
  algorithm to find the longest monotonically
  increasing subsequence of a sequence of n
  numbers, where an odd number in the
  monotonically increasing subsequence must be
  followed by an even number.
• Example Consider sequence 1,7,8, 2,9, 3,10, 4,
  5. Both subsequences 1, 2, 3, 4, 5 and 1, 8, 9,
  10 are monotonically increasing subsequences.
  However, 1,2,3, 4, 5 is not the one that we are
  looking for since 5 is not followed by an even
  number.

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• Assignment 4. (Due on Dec 2. 230 p.m.)
• Question 3 (40 points) Let T be a rooted binary
  tree, where each internal node in the tree has
  two children and every node (except the root) in
  T has a parent. Each leaf in the tree is
  assigned a letter in ?A, C, G, T. Figure 1
  gives an example. Consider an edge e in T. If
  every end of e is assigned a letter, then the
  cost of e is 0 if the two letters are identical
  and the cost is 1 if the two letters are not
  identical. The problem here is to assign a
  letter in ? to each internal node of T such that
  the cost of the tree is minimized, where the
  cost of the tree is the total cost of all edges
  in the tree. Design a polynomial-time dynamic
  programming algorithm to solve the problem.

4

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• Lemma 26.8 If the Edmonds-Karp algorithm is run
  on a flow network G(V,E) with source s and sink
  t, then for all vertices v?V-s, t, the
  shortest-path distance ?f(s,v) in the residual
  network Gf increases monotonically with each flow
  augmentation.
• Proof We prove it by contradiction.
• Let f and f be the flows for the k-th and
  (k1)-th iterations, the augmentation between
  them is the first that decrease some
  shortest-path distance.
• Let v be the vertex with minimum ?f(s,v) such
  that ?f(s,v)lt ?f(s,v).

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• Let P s -- u-gtv be the shortest path in Gf.
• Thus, (u,v) ?Ef and ?f(s,u) ?f(s,v)-1.
• By the choice of v, ?f(s,u)? ?f(s,u).
• Thus, (u,v) ?Ef.
• (Otherwise, we have
• ?f(s,v)? ?f(s,u)1 ? ?f(s,u)1 ?f(s,v). )
• How can we have (u,v) ?Ef and (u,v) ?Ef. ?
• The augmentation must have increased the flow
  form v to u.

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• Since E-K algorithm always uses shortest path,
  edge (v, u) is the last edge in the path from s
  to u. So,
• ?f(s,v) ?f(s,u)-1 ? ?f(s,u)-1 ?f(s,v)-2.
• Contradiction to ?f(s,v)lt ?f(s,v).

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• Theorem 26.9. If Edmonds-Karp algorithm is run on
  a flow network G(V,E) with source and sink t,
  then the total number of flow augmentations
  performed by the algorithm is O(VE).
• Proof Each augmentation has a critical edge (u,
  v),
• where c(u,v) is minimum among all edges in the
  shortest path.
• We show that each edge (u,v) can be a critical
  edge for at most O(V) times.
• Consider a critical edge (u,v) for flow f.
• ?f(s,v) ?f(s,u)1.
• (u,v) will disappear after that path is used for
  augmentation.

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• (u,v) will appear again after an augmentation
  with (v,u) in the shortest path. Let f be the
  flow in G when this event occurs, we have
• ?f(s,u) ?f(s,v)1.
• Thus, ?f(s,u) ?f(s,v)1? ?f(s,v)1 ?f(s,u)2.
• Since the length of the shortest path is at most
  V-1. Thus, each edge (u,v) can be a critical
  edge for at most O(V) times.
• Total number of augmentations is O(VE).
• Each iteration takes at most O(E) time, the total
  time is O(VE2).

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• Maximum k-Clustering
• Problem Given a set of points V in the plane (or
  some other metric space), find k points c1, c2,
  .., ck such that for each v in V,
• min i1, 2, , k d(v, si) ? d
• and d is minimized.

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• Fasthest-point clustering algorithm
• Step 1 arbitrarily select a point in V as c1.
• Step 2 let i2.
• Step 3 pick a point ci from V c1, c2, ,
  ci-1 to maximize min c1ci, c2ci,,ci-1
  ci.
• Step 4 ii1
• Step 5 repeat Steps 3 and 4 until ik.

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• Theorem Farthest-point clustering algorithm has
  ratio-2.
• Proof Let c k1 be an point in V that maximize
• ?min c1ck1, c2ck1,,ck ck1.
• Since two of the k1 points must be in the same
  group, ? lt2opt.
• Since for any v in V,
• min c1v, c2v,,ck vlt ? (based on the
  alg.), so the algorithm has ratio-2.