# Dynamic Programming for the confused : Rod cutting problem

I have always struggled with optimization problems. They can be hard to wrap your mind around from just the code. Hence, from my learning , I decided to write series of posts dealing with classic dynamic programming problems. Rod cutting problem is a classic optimization problem which serves as a good example of dynamic programming.

### What is the problem ?

Rod cutting problem is very much related to any real-world problem we face. You have a rod of some size and you want to cut it into parts and sell in such a way that you get the maximum revenue out of it. Now, here is the catch, prices of different size of pieces are different and it is a possibility that a cutting into smaller pieces can fetch more revenue than selling a bigger piece, so a different strategy is needed.
Before that, lets state the problem more formally,
You are given a rod of size n >0, it can be cut into any number of pieces k (k ≤ n). Price for each piece of size i is represented as p(i) and maximum revenue from a rod of size i is r(i) (could be split into multiple pieces). Find r(n) for the rod of size n.

The size vs price table

### Thought process

One can see that the problem distills down to the fact : where the cuts will be ? Which one provides the maximum revenue r(4)?
4 rods of size 1 (achieved by 3 cuts) = 4 x p(1) = 4 x 1 = 4
2 rods of size 1 + 1 rod of size 2 (achieved by 2 cuts) = 2 x p(1) + 1 x p(2) = 2 x 1 + 5 = 7
2 rods of size 2 (achieved by 1 cut)= 2 x p(2) = 2 x 5 = 10
1 rod of size 1 + 1 rod of size 3 (achieved by 2 cuts)= 1 x p(1) + 1 x p(3) = 1 + 8= 9
original rod of size 4 (achieved by no cuts)= 1 x p(4) = 9
Thus, maximum revenue possible is 10, can be achieved by making a cut at size=2, splitting the original rod into two rods of size 2 with no further cuts in any of them.
Generalizing it,
Max revenue r(n) on rod of size i=n, can be achieved by :

1. Starting with size i=1 and going up to size i=n repeat the following 2 & 3 :
2. Deciding whether a cut at i has to be made or not (results in 2 different cases)
3. Repeating from 1 for rest of the rod (n-i)
4. Calculating r(n) in every case at the end
5. Max value among all calculated r(n) is the answer. Additional storage is required to track the cuts made. Recursion is an ideal candidate for this. Find recursive code here What is the time complexity ? This can be answered by asking how many ways a cut can be made ? In a rod of size n, 1 cut can be made in (n-1)C(1) ways 2 cuts = (n-2) C(2) ways k cuts = (n-k)C(k) ways Totaling it all : ways to do 1 cut + ways to do 2 cuts + ….. ways to do k cuts ..+ ways to do n-1 cuts Source : http://www.cs.uml.edu/~kdaniels/courses/ALG_503_F12/DynamicRodCutting.pdf Thus, time complexity is exponential here. There should be a better way of doing things. One trick is to use recursion with memoization i.e memo taking(explained here). However, still it remains an exponential operation which is infeasible for large values of n. Enter Dynamic Programming. ### Solving with Dynamic Programming ### The problem already shows optimal substructure and overlapping sub-problems. r(i) = maximum revenue achieved by applying 0, 1, …..(i-1) cuts respectively to a rod. r(i) = max { p(i), p(1)+r(i-1), p(2)+r(i-2)….p(i-1)+r(1) } In other words, r(i) can be reached by either having no cuts, only p(i) or adding a cut at 1 and adding it to r(i-1) i.e. previously solved problem of max revenue of rod size i-1(which may have multiple more cuts). Similarly, cutting at 2 and solving for r(i-2). So, r(i) is dependent on previously computed values for smaller sizes than i. But, unlike greedy, it analyzes them all to take a decision. In fact, you go back and compare on every value of i, this is an overlapping sub-problem with optimal value of each sub-problem contributing to next one, thus, optimal substructure. ### Calculations ### > r(1) = p(1) = 1 (cut at size 1 i.e. no cut) > r(2) = max{ p(2), p(1)+r(1)} = max(5, 2) = 5 (cut at size 2 i.e. no cut) > r(3) = max{p(3), p(1)+r(2), p(2)+r(1)} = max(8, 1+5, 5+1) = 8 (cut at size 3 i.e. no cut) > r(4) = max{p(4), p(1)+r(3), p(2)+r(2), p(3)+r(1)} = max(9, 1+8, 5+5, 8+1) = 10 (cut at size 2) > r(5) = max{p(5), p(1)+r(4), p(2) +r(3), p(3)+r(2), p(4)+r(1)} = max(10, 1+10, 5+8, 8+5, 9+1) = 13 (cut at size 2) and so on …

Completed table :

Completed table
Final answer for r(5) = 13.
Now, how to find no. of cuts it took to reach 13 ?
r(5) has cut value at 2, that means one cut was made at 2.
For the remaining 5–2 = 3 size part , lets check r(3). r(3) cut value is 3 ie it was taken as whole with no cuts.
Thus, we only have a cut at size 2.
Maximum revenue for rod of size 5 can be achieved by making a cut at size 2 to split it into two rods of size 2 and 3.
This non-recursive approach is bottom up one. Its time complexity is :

T(n) = T(n-1) + T(n-2) +…..+T(1) + 1
T(1)=1
T(2) = T(1) + 1= 2
T(3) = T(2) + T(1)+ 1 = 2 + 1 + 1
T(n) = n + n-1 + n-2 + … in an Arithmetic Progression
T(n) = O(n²)

Compared to O(2n), O(n²) is much better.
In this way, one can develop intuitions to solve optimization problems. Not every problem of optimization can be solved this way, but it provides a good starting point.   