Dynamic Programming. Preview. Dynamic Programming. Dynamic Programming. Dynamic Programming (Example: Fibonacci Sequence)

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1 /24/27 Prevew Fbonacc Sequence Longest Common Subsequence Dynamc programmng s a method for solvng complex problems by breakng them down nto smpler sub-problems. It s applcable to problems exhbtng the propertes of overlappng sub-problems whch are only slghtly smaller. Dynamc programmng Ths method s applcable when the subproblems are not ndependent.(.e., when subproblems share sub-subproblems.) dynamc-programmng method solves every sub-subproblem ust once and then saves ts answer n a table. The dynamc programmng method s often appled to optmzaton problems. When developng a dynamc programmng algorthm, we typcally follow a sequence of four steps:. Characterze the structure of an optmal soluton. 2. Recursvely defne the values of an optmal soluton. 3. Compute the value of an optmal soluton n a bottom-up fashon. 4. Construct an optmal soluton from computed nformaton. (Fbonacc Sequence) nt fb(nt n) { f n == or n == return ; else return fb(n-) + fb(n-2); Notce that f we call, say, fb(5), we produce a call tree that calls the functon on the same value many dfferent tmes. In partcular, fb(2) s calculated three tmes from scratch. In larger examples, even more values of fb,.e., subproblems, are recalculated, resultng n an exponental tme algorthm.. fb(5) 2. fb(4) + fb(3) 3. (fb(3) + fb(2)) + (fb(2) + fb()) 4. ((fb(2) + fb()) + (fb() + fb())) + ((fb() + fb()) + fb()) 5. (((fb() + fb()) + fb()) + (fb() + fb())) + ((fb() + fb()) + fb()) (Example: Fbonacc Sequence) We can mprove recurrence verson of Fbonacc sequence by usng dynamc programmng dea. Now, suppose we have a smple map obect, m, whch maps each value of fb that has already been calculated to ts result. The resultng functon requres only O(n) tme nstead of exponental tme (but requres O(n) space): map M (key, value) M() = ; M() = ; nt fb(nt n) { f map M does not contan key n else return M(n)= fb(n-)+ fb(n-2); return M(n)

2 /24/27 Gven a sequence X=(x, x 2,, x m ), another sequence Z=(z, z 2,, z k ) s a subsequence of X f there exsts a strctly ncreasng sequence (, 2,, k ) of ndces of X such that for all =, 2,, k, we have x = z. Ex) Z = (B, C, D, B) s a subsequence of X = (, B, C, B, D,, B) wth correspondng ndex sequence (2, 3, 5, 7). Gven two sequences X and Y, we say that a sequence Z s a common subsequence of X and Y f Z s a subsequence of both X and Y. Ex) Wth X = (, B, C, B, D,, B) Y = (B, D, C,, B, ), Common subsequence of both X and Y are (, B, ), (B, C, B), (B, C, ), (B, D,, B), longest-common-subsequence problem (LCS) Input: Gven two sequences X = (x, x 2,, x m ), and Y = (y, y 2,, y n ) Output: Fnd a maxmum-length common subsequence of X and Y. Defnton) Gven a sequence X = (x, x 2,, x m ), we defne s prefx of X, for =,,, m, as X = (x, x 2,, x ) Ex) X = (, B, C, B, D,, B), then X 4 = (, B, C, B) X s empty sequence Theorem: Optmal substructure of an LCS Let X = (x, x 2,, x m ) and Y = (y, y 2,, y n ) be sequences, and Z = (z, z 2,, z k ) be any LCS of X and Y.. f x m = y n, then z k = x m = y n and Z k- s an LCS of X m- and Y n-. 2. If x m y n, then z k x m mples that Z s an LCS of X m- and Y. 3. If x m y n, then z k y n mples that Z s an LCS of X and Y n- From the theorem, we can dvde LCD calculatons as follows:. When x m = y n, we must fnd an LCS of X m- and Y n-. 2. When x m y n, we need to solve two subproblems n order to fnd out whch one s maxmum.. LCS of X m- and Y n. 2. LCS of X m and Y n- 2

3 /24/27 recursve soluton to the LCS problem: Let us defne c[, ] to be the length of an LCS of the sequences X and Y. If ether = or =, the LCS has length. c[, ] c[, ] max( c[, ], c[, ]) f or, f, and x y f, and x y LCS_Length(X, Y) { m = length of X n = length of Y (longest for = common to m { subsequence) c[, ] = for = to n c[, ] = for = to m { for = to n { f x = y { c[, ] = c[-, -] + b[, ] = /* up and left */ else f c[-, ] c[, -] { c[, ] = c[-, ] b[, ] = /* up */ else { c[, ] = c[, -] b[, ] = /* left */ /* end of nner for loop */ /* end of outer for loop */ dynamc programmng soluton that computes the soluton bottom up return c and b 3

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8 /24/ Least Common Subsequence (LCS) s BCB Gven a sequence (chan) of n matrces < 2 n > to be multpled. We wsh to compute the product 2 3 n. We can multply two matrces and B only f the number of columns of s equal to the number of rows of B. For example, f s an n m matrx and B s an m p matrx than the resultng matrx C s n p matrx. The tme to compute C s domnated by the number of scalar multplcatons, whch s n m p. We want to mnmze the number of scalar multplcatons!! Consder a matrx chan < 2 3 > where : 2 : 5 3 : 5 5 We can multply these matrces two dfferent ways (( 2 ) 3 ): ( 5) + ( 5 5) = 75 ( ( 2 3 )): ( 5 5) +( 5) = 75 ykes! 8

9 /24/27 The matrx-chan multplcaton problem Gven a matrx chan < 2 n > of n matrces, where for =, 2,, n, matrx has dmensons p - p, fully parenthesze the product 2 3 n n a way that mnmzes the number of scalar multplcatons. Characterze problem Defne m(, ) = cost of computng +... Snce = p - p and + = p p + m(, +) = p - p p k k k k... m(, k) m(k+, ) k s a dvde pont for optmzaton where k k = p - p k We can defne a recursve equaton... k k k k... k+.. =p k p f m(, ) mn m (, k) m( k, ) p pk p f k We need nvestgate for all k between and -to fnd the mnmum sze m (, ) for all m(,2) sze 2 m(2,3) m(3,4) ( m(,) m(2,3) 385) m(,3) mn 78 k,2 ( m(,2) m(3,3) 3 25 sze3 ( m(2,2) m(3,4) 8 2 4) m(2,4) mn 4 k 2,3 ( m(2,3) m(4,4) m(,2) sze 2 m(2,3) m(3,4) m(,2) sze 2 m(2,3) m(3,4) sze 4 : m(,4) ( m(,) m(2,4) 38 4) mn ( m(,2) m(3,4) 3 2 4) 2 k,2,3 ( (,3) (4,4) 3 5 4) m m 2 ( m(,) m(2,3) 385) m(,3) mn 78 k,2 ( m(,2) m(3,3) 3 25 sze3 ( m(2,2) m(3,4) 8 2 4) m(2,4) mn 4 k 2,3 ( m(2,3) m(4,4)

10 /24/27 procedure MnMult(p) begn for tolength of chan do m(, ) for d to (length of for to (length of d; end of MnMult chan -) do chan - d) do m(, ) mn { m(, k) c( k, ) p k - p p k sze = + sze 2 = + n sze 3 = +2 n sze r = + (n ) n (n ) = n - # of m(, ) s s O(n 2 ) For computng each m(, ): Total complexty s O(n 3 ) O(n) tme slghtly modfed verson of the algorthm MTRIX-CHIN-ORDER(p) n = p.length - 2 let m[..n,..n] and s[..n, 2..n] be new tables 3 for ( = to n) 4 m[, ] = 5 for (l = 2 to n) // l s the chan length 6 for ( = to n - l + ) 7 = + l 8 m[, ] = // nfnty or a really bg number 9 for (k = to ) q = m[, k] + m[k +, ] + p - p k p f (q < m[, ]) 2 m[, ] = q 3 s[, ] = k 4 return m and s lthough the soluton has been determned, t s not really n an easy to nterpret form. Table s can be used to accomplsh ths. PRINT-OPTIML-PRENS(s,, ) f == 2 prnt 3 else 4 prnt ( 5 PRINT-OPTIML-PRENS(s,, s[, ]) 6 PRINT-OPTIML-PRENS(s, s[, ] +, ) 7 prnt )