Dynamic Programming

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Dynamic Programming

• Irena Pevac
• CS463

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Dynamic Programming

• Introduced in 1950 by Bellman as a general method
  for optimizing control systems.
• It is technique for solving problems with
  overlapping subproblems. Most often subproblems
  are recursive calls to several instances of
  simpler problem of the same type.

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Fibonacci

• F(0)0
• F(1)1
• F(n)F(n-1)F(n-2), for ngt1
• Recursive method recomputes the same F(k) many
  times. It works in ?(an) where a is golden ratio
  (1sqrt(5))/2.
• Dynamic programming version eliminates
  recomputing by storing previously computed
  results.

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Fibonacci

• F00
• F11
• for i 2 to n
• FiFi-1Fi-2
• We use one-dimmensional array, fill in the values
  of subproblems, and use those values for future
  calculations.

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Longest Common Subsequence (LCS)

• Let x1,x2,xm and y1,y2,yn be two lists.
• Subsequence is sublist containing subset of
  items from the original list that appear in the
  same order, but do not have to be consecutive.
  
• For example, x1,x4,x5,x8 is a subsequence of
  x1,x2,x3.x9, but x1,x5,x2 is not.
• LCS is longest subsequence of both lists.
• X abcabba Y cbabac One LCS is baba and
  the other is cbba.

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Recursive solution for LCS

• Let x1,x2,xm and y1,y2,yn be two lists.
• If xm!yn than LCS cannot include both xm and yn.
• So, it must be either
• An LCS of x1,x2,xm-1 and y1,y2,yn
• or
• An LCS of x1,x2,xm and y1,y2,yn-1
• If xmyn than LCS includes xm that is
  concatenated to the tail of LCS of x1,x2,xm-1
  and y1,y2,yn-1

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Length L of LCS

• If (mn0) // both lists are empty, LCS is ?
• L(0,0) 0
• Else if (m0 or n0)
• L(m,n)0
• Else if mgt0 and ngt0 and xm!yn
• L(m,n)max (L(m,n-1), L(m-1,n))
• Else if mgt0 and ngt0 and xmyn
• L(m,n)1 L(m-,n-1)
• This has exponential running time

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Dynamic Programming Solution for LCS

• If we build two-dimensional array to store L(i,j)
  for the various values of i and j then we avoid
  recomputing them. This brings time down to O(mn).
• Let x1,x2,xm and y1,y2,yn be stored in arrays
  x and y.

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LCS Table

• for (j 0 jltn j)
• L0j0
• for (i1 iltmi)
• Li00
• for ( j1 jltn j)
• if (xi ! yi)
• if (Li-1jgtLij-1)
• LijLi-1j
• else
• LijLij-1
• else
• Lij 1 Li-1j-1

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Example cbabac abcabba
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Recovery of an LCS

• Table provides matching pairs of elements that
  form one of the LCSs.
• We should find the path through the table,
  beginning in the upper right corner.
• If we are in row i and column j, check if
  xiyj. If so, take xi to be part of the LCS
  ahead of all elements of LCS found so far. After
  that move to row i-1 and column j-1.
• If xi and yj are different Lij must equal
  at least one of Li-1j or Lij-1.
• If Lij Li-1j we shall move our path one
  row down.
• If Lij Lij-1 we shall move our path one
  column left.
• We stop when we arrive to the left lower corner.

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Binomial Coefficients

• (ab)n C(n, 0) anb0 C(n,1) an-1b1
• C(n, 2) an-1b2 ...
• C(n, k)an-kbk
• C(n, n)a0bn
• Where C(n, k) is n!/ (k! (n-k)! )

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Properties of Binomial Coefficients

• C(n,0)1
• C(n,n)1
• C(n,k)C(n-1,k-1)C(n-1,k) for ngtkgt0
• Recurrence for C(n,k) uses overlapping
  subproblems. So, let us use dynamic programming
  technique. We record the values of binomial
  coefficients in a table of n1 rows and k1
  columns.

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Computing C(n,k)
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Computing C(n,k)
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Algorithm for C(n,k)

• for( row 0 rowltn row) // first column
  items are 1s
• Crow01
• for row 1 to k
• for col1 to row-1
• Crowcol Crow-1col-1
  Crow-1col
• Crowrow 1 // last item in a
  row is 1
• for row k1 to n
• for col1 to k
• Crowcol Crow-1col-1
  Crow-1col

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Cost of the Dynamic Algorithm

• Basic operation addition
• First k1 rows 0..k form a triangle.
• Remaining n-k rows k1..n form a rectangle
• Number of additions to determine C(n,k) is