Dynamic Programming

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• Dynamic Programming

?? ???? ????????????
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Dynamic Programming

• ? divide-and-conquer ???, ????????????.
• ? divide-and-conquer????????????????, ?????.

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?????? (Rod cutting problem)
??????N(??)?????, ?i??????,??pi?????i??????????.
???????????????????
?
?? i 1 2 3 4 5 6 7 8 9 10
??pi 1 5 8 9 10 17 17 20 24 30
N7, 734???17 716223 ???18.
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?????? (Rod cutting problem)

• ??????????????k?,?
• Ni1i2ik
• rNpi1pik ----???
• rN max i1..n pirN-i,
• r00.
• CUT-ROD(p, n)
• if n0 return 0
• q-99999999
• for i1 to n
• q max(q, piCUT-ROD(p, n - i))
• return q

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?????? (Rod cutting problem)

• Memoized-CUT-ROD(p, n)
• let r0,..n be a new array
• for i0 to n
• ri -9999999
• return Memoized-CUT-ROD-AUX(p, n, r)

Time O(n2), why?

• Memoized-CUT-ROD-AUX(p, n, r)
• if rngt0 return rn
• if n0 q0
• else q-9999999
• for i0 to n
• q max(q, piMemoized-CUT-ROD-AUX(p,
  n-i, r))
• rn q
• return q

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• BOTTOM-UP-CUT-ROD(p, n)
• let r0,..n be a new array
• r00
• for j1 to n
• q-9999999
• for i 1 to j
• qmax(q, pi rj-i)
• rjq
• return rn

• EXTENDED-BOTTOM-UP-CUT-ROD(p, n)
• let r0,..n and s0..n be a new arrays
• r00
• for j1 to n
• q-9999999
• for i 1 to j
• if q lt pi rj-i)
• q pi rj-i)
• sji
• rjq
• return r and s

• PRINT-CUT-ROD-SOLUTION(p, n)
• (r, s) EXTENDED-BOTTOM-UP-CUT-ROD(p, n)
• while n gt 0
• print sn nn - sn

?? i 1 2 3 4 5 6 7 8 9 10
??pi 1 5 8 9 10 17 17 20 24 30
ri 1 5 8 10 13 17 18 22 25 30
si 1 2 3 2 2 6 1 2 3 10
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Crazy eights puzzle

• Given a sequence of cards c0, c1,,cn-1,
  e.g. 7H, 6H, 7D, 3D, 8C, JS,..
• Find the longest subsequence ci1, ci2,,
  cik, (i1lt i2ltlt ik), where cij and cij1
  have the same suit or rank or one has rank 8.--
  match

• Let Ti be the length of the longest subsequence
  
• starting at ci.
• Ti 1 max Tj ci and cj have a match
  and j gtn
• Optimal solution max Ti.

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?????? (??)
??????? ?A1, A2, , An?, ???? Ai ???? pi?1? pi,
???? A1A2An ?????, ???? scalar ?????????.
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??????(?)
(A1(A2(A3A4)))? A1?(A2A3A4) ? A2?(A3A4) ? A3?A4
cost 13534 58934
89334 2210 15130
9078 26418
(A1(A2(A3A4))), costs 26418 (A1((A2A3)A4)),
costs 4055 ((A1A2 )( A3A4)), costs 54201
((A1(A2 A3))A4), costs 2856 ((( A1A2)A3)A4),
costs 10582
A1 ? A2 ? A3 ? A4 13 5 89 3 34
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Catalan Number
For any n, ways to fully parenthesize the
product of a chain of n1 matrices binary
trees with n nodes. permutations generated
from 1 2 n through a stack. n pairs of
fully matched parentheses. n-th Catalan Number
C(2n, n)/(n 1) ?(4n/n3/2)
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???
A1 ? A2 ? A3 ? A4
(A1(A2(A3A4))) (A1((A2A3)A4)) ((A1A2 )(
A3A4)) ((A1(A2 A3))A4) ((( A1A2)A3)A4)
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??????(??1)

• If T is an optimal solution for A1, A2, , An

T
k
T2
T1
1, , k
k1, , n

• then, T1 (resp. T2) is an optimal solution for
  A1, A2, , Ak (resp. Ak1, Ak2, , An).

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?????? (??2)

• Let mi, j be the minmum number of scalar
  multiplications needed to compute the product
  AiAj , for 1 ? i ? j ? n.
• If the optimal solution splits the product AiAj
  (AiAk)?(Ak1Aj), for some k, i ? k lt j, then
• mi, j mi, k mk1, j pi?1 pk pj .
  Hence, we have

mi, j mini ? k lt jmi, k mk1, j pi?1
pk pj 0 if i j
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ltP0, P1,, Pngt

• MATRIX-CHAIN-ORDER(P)
• n p.length -1
• let m1..n, 1..n and s1..n-1, 2..n be new
  tables
• for i 1 to n mi, i0
• for l 2 to n
• for i 1 to n l 1
• ji l- 1
• mi, j ?
• for k i to j-1
• q mi, k mk1, j Pi-1PkPj
• if qltmi, j
• mi, j q si, j k
• return m and s
• Time O(n3)

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(No Transcript)
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?????? (??)

• Consider an example with sequence of dimensions
  lt5,2,3,4,6,7,8gt

mi, j mini ? k lt jmi, k mk1, j pi?1
pk pj
1 2 3 4 5 6 1 0 30 64 132 226 348
2 0 24 72 156 268 3 0
72 198 366 4 0 168 392 5
0 336 6 0
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• Constructing an optimal solution
• Each entry si, jk records that the optimal
  parenthesization of AiAi1Aj splits the product
  between Ak and Ak1
• Ai..j?(A i..si..j )(A si..j1..j)

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?????? (??)
mi, j mini ? k lt jmi, k mk1, j pi?1
pk pj si, j a value of k that gives the
minimum
s 1 2 3 4 5 6 1 1 1 1 1 1 2 2 3 4 5
3 3 4 5 4 4 5 5
5
1,6
A1(((( A2A3)A4)A5) A6)
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?????? (??)
mi, j mini ? k lt jmi, k mk1, j pi?1
pk pj

• To fill the entry mi, j, it needs ?(j?i)
  operations. Hence the execution time of the
  algorithm is

Time ?(n3) Space ?(n2)
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• MEMOIZED-MATRIX-CHAIN(P)
• n p.length -1
• let m1..n, 1..n be a new table
• for i 1 to n
• for j i to n
• mi, j ?
• return LOOKUP-CHAIN(m, p, 1, n)

ltP0, P1,, Pngt

• Lookup-Chain(m, P, i, j)
• if mi, j lt ? return mi, j
• if ij mi, j 0
• else for ki to j-1
• q Lookup-Chain(m, P, i, k)
• Lookup-Chain(m, P, k1,
  j) Pi-1PkPj
• if q lt mi, j mi, j q
• return mi, j time O(n3) space
  ?(n2)

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??? DP ??????

1. Characterize the structure of an optimal
   solution.
2. Derive a recursive formula for computing the
   values of optimal solutions.
3. Compute the value of an optimal solution in a
   bottom-up fashion (top-down is also applicable).
4. Construct an optimal solution in a top-down
   fashion.

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Elements of Dynamic Programming

• Optimal substructure (a problem exhibits optimal
  substructure if an optimal solution to the
  problem contains within it optimal solutions to
  subproblems)
• Overlapping subproblems
• Reconstructing an optimal solution
• Memoization

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• Given a directed graph G(V, E) and vertices u, v
  ?V
• Unweighted shortest path Find a path from u to v
  consisting the fewest edges. Such a path must be
  simple (no cycle).
• Optimal substructure? YES.
• Unweighted longest simple path Find a simple
  path from u to v consisting the most edges.
• Optimal substructure? NO.
• q ? r ? t is longest but q ? r
• is not the longest between q and r.

w
v
u
q
r
s
t
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Printing neatly ??

• Given a sequence of n words of lengths l1,
  l2,,ln, measured in characters, want to
  print it neatly on a number of lines, of which
  each has at most M characters. .
• If a line has words i through j, iltj, and there
  is exactly one space between words, the cube of
  number of extra space characters at the end of
  the line is
• Bi, j (M j i - (lili1lj))3.
  
• Want to minimize the sum over all lines (except
  the last line) of the cubes of the number of
  extra space at the ends of lines.
• Let ci denote the minimum cost for printing
  words i through n.
• ci min iltjltip (cj1 Bi,j), where p is
  the maximum number of words starting from i-th
  word that can be fitted into a line.

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Longest Common Subsequence (??)
Given two sequences X ltx1, x2, , xmgt and Y
lty1, y2, , yngt find a maximum-length common
subsequence of X and Y.
?1Input ABCBDAB BDCABA C.S.s
AB, ABA, BCB, BCAB, BCBA Longest BCAB, BCBA,
Length 4
? 2 vintner writers
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Step 1 Characterize longest common subsequence

• Let Z lt z1, z2, , zkgt be a LCS of X ltx1,
  x2, , xmgt and Y lty1, y2, , yngt.
• If xm yn, then zk xm yn and ltz1, z2, ,
  zk?1gt is a LCS of ltx1, x2, , xm?1gt and lty1, y2,
  , yn?1gt.
• If zk ? xm, then Z is a LCS of ltx1, x2, , xm?1gt
  and Y.
• If zk ? yn, then Z is a LCS of X and lty1, y2, ,
  yn?1gt.

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Step 2 A recursive solution

• Let Ci, j be the length of an LCS of the
  prefixes Xi ltx1, x2, , xigt and Yj lty1, y2,
  , yjgt, for 1 ? i ? m and 1 ? j ? n. We have

Ci, j 0 if i 0, or j 0 Ci?1, j?1
1 if i , j gt 0 and xi yj max(Ci, j?1,
Ci?1, j) if i , j gt 0 and xi ? yj
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Step 3 Computing the length of an LCS

• LCS-Length(X,Y)
• m X.length
• n Y.length
• let b1..m, 1..n and c0..m, 0..n be new
  tables
• for i 1 to m ci, 0 0
• for j 1 to n do c0, j 0
• for i 1 to m do
• for j 1 to n do
• if xi yj
• ci,j ci-1,j-11 bi,j ?
• else if ci-1, j ? ci, j-1
• ci, j ci-1, j bi, j ?
• else ci, j ci, j-1 bi, j
  ?
• return b, c

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Step 4 Constructing an LCS

• Print-LCS(b, X, i, j)
• if i0 or j0 return
• if bi,j ?
• Print-LCS(b, X, i-1, j-1)
• print xi
• else if bi,j ?
• Print-LCS(b, X, i-1, j)
• else Print-LCS(b, X, i, j-1)

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Longest Common Subsequence (???)
Ci, j 0 if i 0, or j 0 Ci?1, j?1
1 if i , j gt 0 and xi yj max(Ci, j?1,
Ci?1, j) if i , j gt 0 and xi ? yj
???LCS BCBA
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Optimal Polygon Triangulation
v0
v6
v1
v5
v2
v3
v4
Find a triangulation s.t. the sum of the weights
of the triangles in the triangulation is
minimized.
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Optimal Polygon Triangulation (??1)

• If T is an optimal solution for ?v0, v1, , vn?

• then, T1 (resp. T2) is an optimal solution for
  ?v0, v1, , vk? (resp. ?vk, vk1, , vn?), 1 ?
  k lt n.

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Optimal Polygon Triangulation (??2)

• Let ti, j be the weight of an optimal
  triangulation of the polygon ?vi?1, vi,, vj?,
  for 1 ? i lt j ? n.
• If the triangulation splits the polygon into
  ?vi?1, vi,, vk? and ?vk, vk1, ,vn? for some
  k, then
• ti, j ti, k tk1, j w(?vi?1 vk vj)
  . Hence, we have

ti, j mini ? k lt jti, k tk1, j
w(?vi?1 vk vj) 0 if i j
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Optimal binary search trees

• k(k1,k2,,kn) of n distinct keys in sorted order
  (k1ltk2ltlt kn)
• Each key ki has a probability pi that a search
  will be for ki
• Some searches may fail, so we also have n1 dummy
  keys d0,d1,,dn, where d0ltk1, knltdn and
  kiltdiltki1 for i1,,n-1.
• For each dummy key di, we have a probability qi
  that a search will correspond to di .

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Optimal binary search trees??
i 0 1 2 3 4 5
pi 0.15 0.1 0.05 0.1 0.2
qi 0.05 0.1 0.05 0.05 0.05 0.1
Expected search cost 2.8
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• Goal Given a set of probabilities, want to
  construct a binary search tree whose expected
  search cost is smallest .
• Step 1 The structure of an optimal BST.
• Consider any subtree of a BST, which must
  contain contiguous keys ki,,kj for some 1?i?j?n
  and must also have as its leaves the dummy keys
  di-1,,dj .
• Optimal-BST has the property of optimal
  substructure.

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• Step 2 A recursive solution
• Find an optimal BST for the keys ki,,kj, where
  j?i-1, i?1, j?n. (When ji-1, there is no actual
  key, but di-1)
• Define ei,j as the expected cost of searching
  an optimal BST containing ki,,kj. Want to find
  e1,n.
• For dummy key di-1, ei,i-1qi-1 .
• For j?i, need to select a root kr among ki,,kj .
• For a subtree with keys ki,,kj, denote

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• ei,jpr(ei,r-1w(i,r-1))(er1,jw(r1,j))
• ei,r-1er1,jw(i,j)
• Thus
• ei,j
• Step 3 Computing the expected search cost of an
  optimal BST
• Use a table w1..n1,0..n for w(i,j)s .
• wi,i-1qi-1
• wi,jwi,j-1pjqj

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• OPTIMAL-BST(p,q,n)
• let e1..n1, 0..n, w1..n1, 0..n and
  root1..n, 1..n be new tables
• for i1 to n1
• ei,i-1qi-1
• wi,i-1qi-1
• for l1 to n
• for i1 to n-l1
• jil-1
• ei,j?
• wi,jwi,j-1pjqj
• for r i to j
• tei, r-1er1, j wi,
  j
• if t lt ei,j
• ei,jt
• rooti,j r
• return e and root
• 
  ?(n3)