Algorithms Chapter 15 Dynamic Programming - Rod

AlgorithmsChapter 15 Dynamic Programming - Rod B98570104廖翎妤 B98570107方敏 B98570110李紫綾 B98570135邱郁庭

Outline • Rod Cutting • Recursive top – down implementation • Using dynamic programming for optimal rod cutting • Subproblem graphs • Reconstructing a solution • Exercise

Rod - cutting • Given a rod of length n inches and a table of prices pi for i = 1, 2, …, n. • Determine the maximum revenue rn obtainable by cutting up the rod and selling the pieces.

Example • Consider the case when n=4. • The optimal strategy is part (c). With length=2, value=10. 9 1 8 5 5 (a) (b) (c) 1 1 5 1 5 1 8 1 (e) (f) (d) 1 1 1 1 5 1 1 (h) (g)

Conclusion • If an optimal solution cuts the rod into k pieces, for some 1 ≦k ≦n, then an optimal decomposition n = i1+ i2+…+ ik<7=2+2+3>of the rod into pieces of lengths i1 , i2 ,…, ikprovides maximum corresponding revenuern = pi1 + pi2 + …+ pik<r7 = p2+p2+p3 = 5+5+8 = 18> • More generally,rn = max (pn , r1+rn-1 , r2+rn-2 ,…, rn-1+r1) • Simpler solution, rn = max (pi + rn-i) <r7 = p2+r5 = 5+13 = 18> 1 ≦i ≦n

Recursive top – down implementation • 假設將鐵條切割成ｋ段 N = i1 + i2 + … + ik r [ N ] = p [ i1 ] + … + p [ ik] ----總價格 r [ N ] = max i=1..n { p [ i ] + r [ N – i ] } r[0]=0 • CUT – ROD ( p , n ) if n = = 0 return 0 q = - ∞ for i = 1 to n q = max( q , p [ i ] + CUR – ROD ( p , n – i ) ) return q

4 2 1 0 3 2 1 0 1 0 0 T ( n ) = 1 + Σn-1j=0 T ( j ) T ( n ) = 2n CUT-RODexplicitlyconsidersallthe2n-1possiblewaysofcuttinguparodoflengthn. 1 0 0 0 0

Using dynamic programming for optimal rod cutting • 算出子問題的答案，並將結果記下來，若再遇到重複的子問題，就不必重複計算，也因此能提高效率，但要多花一些記憶體來檢查此時要算的子問題是不是已經算過了。 • 使用dynamic programming 的演算法有兩種做法: • 1. top-down with memoization • 2. bottom-up method • top-down with memoization和bottom-up method都是Θ(n^2)

Top-down with memoization • 此作法為遞迴，先檢查子問題是否有算過，若沒算過，就先算再將答案記下來（給之後可能會重複出現的子問題使用），若有算過，就將之前算過的答案拿出來使用。 • Memoized-Cut-Rod(p, n) • let r[0..n] be a new array • for i = 0 to n • r[i] =-∞ • return Memoized-Cut-Rod-Aux(p,n,r)

Top-down with memoization • Memoized-Cut-Rod-Aux(p,n,r) • if r[n] ≥ 0 • return r[n] • if n == 0 • q = 0 • else q = -∞ • for i = 1 to n • q = max(q, p[i] + Memoized-Cut-Rod-Aux(p,n-i,r)) • r[n] = q • return q

Bottom-up method • 按照子問題的大小，從最小的問題做到最大的問題。因較大的問題的最佳解須包含子問題的最佳解。 • Bottom-Up-Cut-Rod(p,n) • let r[0..n] be a new array • r[0] = 0 • for j = 1 to n • q = -∞ • for i = 1 to j • q = max(q, p[i] + r[j-i]) • r[j] = q • return r[n]

Subproblem graphs 4 • Dynamic-programming problem • Ex: rod-cutting problem 假設棍子長度 n=4，有圖上這些切割方式。 • Bottom-up method • top-down method可視為「depth-first search」 3 2 1 0

Reconstructing a solution • BOTTOM-UP-CUT-ROD V.S EXTENDED-BOTTOM-UP-CUT-ROD • EXTENDED-BOTTOM-UP-CUT-ROD(p,n) 1 Let r[0..n] and s[0..n] be new arrays 2 r[0]=0 3 for j=1 to n 4 q= -∞ 5 for i=1 to j 6 if q<p[i]+r[j-i] 7 q= p[i]+r[j-i] 8 s[j]=I 9 r[j]=q 10 return r and s

Reconstructing a solution • PRINT-CUT-ROD-SOLUTION(p,n) 1 (r,s) = EXTENDED-BOTTOM-UP-CUT-ROD(p,n) 2 while n>0 3 print s[n] 4 n=n-s[n]

The End THANK YOU VERY MUCH!!

Algorithms Chapter 15 Dynamic Programming - Rod