1. Algorithms Chapter 15 Dynamic Programming - Rod
B 廖翎妤
B 方 敏
B 李紫綾
B 邱郁庭

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

3. 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.
Length i 1 2 3 4 5 6 7 8 9 10
Price pi 17 20 24 30

4. 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)

5. 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 ,…, ik provides maximum corresponding revenue rn = pi1 + pi2 + …+ pik <r7 = p2+p2+p3 = = 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

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

7. 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

8. 4 2 1 3
T ( n ) = 1 + Σn-1j=0 T ( j )
T ( n ) = 2n
CUT-ROD explicitly considers all the 2n-1 possible ways of cutting up a rod of length n.

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

10. 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)

11. 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)

12. 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

13. 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]

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

15. Subproblem graphs Dynamic-programming problem Ex: rod-cutting problem4
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

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

17. 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= -∞
for i=1 to j
if q<p[i]+r[j-i]
q= p[i]+r[j-i]
s[j]=I
9 r[j]=q
10 return r and s

18. 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] i 1 2 3 4 5 6 7 8 9 10
r[i] 13 17 18 22 25 30
s[i]

19. The End
THANK YOU VERY MUCH!!