# Algorithms Chapter 15 Dynamic Programming

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```					       Algorithms
Chapter 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.

Length i   1   2    3   4   5    6    7    8    9    10

Price pi   1   5    8   9   10   17   17   20   24   30
Example

 Consider the case when n=4.
9                         1             8                             5         5

(a)                               (b)                                         (c)

8                 1       1        1            5                 1           5         1

(d)                             (e)                                   (f)

1           1         1       1
5                 1       1

(h)
(g)

 The optimal strategy is part (c). With length=2, value=10.
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 = 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
Outline

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

 假設將鐵條切割成ｋ段
N = i 1 + i 2 + … + 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

3                    2                  1            0

2           1   0   1             0             0

T ( n ) = 1 + Σn-1j=0 T ( j )
1       0       0       0
T ( n ) = 2n

CUT-ROD explicitly considers all
the 2n-1 possible ways of cutting up a
0                               rod of length n.
Outline

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

 Rod Cutting
 Recursive top – down implementation
 Using dynamic programming for optimal rod cutting
 Subproblem graphs
 Reconstructing a solution
 Exercise
Subproblem graphs
4

 Dynamic-programming problem
 Ex:
3
rod-cutting problem
假設棍子長度 n=4，
有圖上這些切割方式。
2

 Bottom-up method
 top-down method可視為              1

「depth-first search」
0
Outline

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

i     0   1   2   3   4    5    6    7    8    9    10

r[i]   0   1   5   8   10   13   17   18   22   25   30

s[i]   0   1   2   3   2    2    6    1    2    3    0
The End

THANK YOU VERY MUCH!!

```
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