[Algorithms] Dynamic programming for solving the rod cutting problem

Summary

This video by Badri Adhikari explores how dynamic programming can be employed to solve the rod cutting problem, emphasizing the importance of solving each subproblem only once to optimize performance. Dynamic programming's approach, whether top-down or bottom-up, involves memorizing solutions to prevent repeated computations, thus achieving efficiency. The video elaborates on memoization and the bottom-up strategy, discussing their implementation differences and computational complexities. It further delves into obtaining optimal revenue from rod cutting while addressing computational nuances.

Highlights

• Introducing dynamic programming for rod cutting, focusing on optimizing subproblem resolution. 🪓
• Understanding top-down versus bottom-up approaches: memoization vs iterative solutions. 🔍
• Top-down memoization improves recursive designs by adding a results cache, enhancing efficiency. 💾
• Bottom-up strategy begins with the simplest problems, solving upwards to complex ones—enabling efficiency by avoiding recursion. 🔨
• The importance of selecting optimal cutting points to achieve maximum revenue alongside solving for maximum revenue itself. 💡
• Constructing subproblem graphs allows for visual understanding of interdependencies and problem size estimation. 📊

Key Takeaways

• Dynamic programming optimizes the rod cutting problem by solving each subproblem only once, either through memoization (top-down) or a bottom-up approach. 🧩
• The memoization approach improves recursive solutions by storing solved instances, reducing computation time significantly. 🔄
• Bottom-up dynamic programming starts solving from the smallest subproblems, often leading to better performance due to elimination of recursion. 📈
• Both strategies aim to provide not just the maximum revenue but also the specific points at which to cut the rod optimally. ✂️
• Understanding the dependency of subproblems via subproblem graphs aids in visualizing the problem-solving pathway. 🗺️

Overview

Ah, the classic rod cutting problem! This time, it's all about dynamic programming and how it transforms this seemingly knotty issue into a smooth ride. Badri Adhikari walks us through how solving a rod of length 4 requires us to also solve for lengths 3, 2, and 1. Dynamic programming ensures we tackle each subproblem only once! What a win for efficiency, right?

Now, onto the strategies: We have the top-down approach, using memoization. Think of it as a supercharged recursive method — storing solutions in a table, so we never need to retrace our steps. Then, we have the bottom-up approach, a clever technique that starts from the simplest problems and builds its way up. This method often wins due to avoiding recursive overheads. Both give the same powerful results in terms of time complexity but choose your weapon!

Finally, it's not just about making the most bucks; it’s also about knowing where to slice and dice. Badri emphasizes that, with dynamic programming, we must track not only the revenue but the optimal cut points. Also, using subproblem graphs provides a nifty visualization of the entire operation, ensuring you see the big picture without missing the detail. Cut smart, and thrive!

[Algorithms] Dynamic programming for solving the rod cutting problem Transcription

• 00:00 - 00:30 let us see how dynamic programming can be used to solve the rod cutting problem that is to find an optimal rod cutting here is a sub sub problem tree for the rod cutting problem for a rod of length equals four if you remember we know that solving the optimal revenue for a rod of length four involves solving for length three two one and further three needs to be solved
• 00:30 - 01:00 for rod of length two one zero and so on the key idea behind applying dynamic programming to the rod cutting problem is we arrange the sub problems in such a way that we only solve one sub problem just once so in this case here's one sub problem to solve a rod of length 2 here's the same sub problem
• 01:00 - 01:30 that is being solved twice so the idea is we want to update our code or change our algorithm such that these two sort problems are only solved once solved once only that is exactly what we would like to do with dynamic programming now during computation say for example we are trying to calculate the optimal revenue for a lot of length 4 and then we happen to solve this so if
• 01:30 - 02:00 this for r2 is solved then what we would like to do is during computation when we are calculating other things we simply want to look this solution up look it up and use a already solved solution instead of recomputing it so that's the core idea behind how we can apply dynamic programming to solve the
• 02:00 - 02:30 road cutting problem now there are two approaches to implement or to design our dynamic programming solution one is the top down approach so this is the same approach as the recursive approach where we start by trying to solve the problem for a large value of l in this case we want to solve for r4 then we length equals to 4 and then we break it down to l3
• 02:30 - 03:00 l2 and so on and then we further break it down so this design is similar to the recursive design so recursive um design but then we simply want to attach the idea of dynamic programming that is a memoization in other words just memorizing um what uh what the already solved solutions are the next approach is to adjust the opposite that is the bottom up approach where we
• 03:00 - 03:30 instead of solving directly for l equals to four we start with uh uh a solution that starts by solving for equals to zero first and then one two three and so on so in other words just opposite to the way uh a recursive solution would solve it so here's the top-down dynamic programming approach to solve the rod cutting
• 03:30 - 04:00 problem so this is simply an improved version of the recursive solution we also call this the memoized rod cutting algorithm now you might notice that this is not memorized but it is memoized but we could say that memoized basically refers to memorizing the solution in a table so that we could use the solution as needed so to simply add this memoization to the recursive algorithm
• 04:00 - 04:30 all that we do is add a initialization code to initialize this table for to maintain the solutions for a various length so if this is my length and this is my optimal solution optimal solution or in this case revenue i initially we can initialize this to be a very large negative number saying that this is unsolved so what i do is create an array
• 04:30 - 05:00 called r which i initialize to minus infinite for every element and then this is where the main program starts happening by taking p which we earlier had in the recursive solution but now we also pass um r an array that maintains the best known solutions for sub problems so far so rest is all same just like the regular the standard recursive broadcast algorithm the algorithm takes pn now it
• 05:00 - 05:30 additionally takes this pre-solved table of solutions and then we simply check if the table already has the solution if the solution is there we don't do anything else and we simply return the solution otherwise everything else is same so we once again um check it initialize the [Music] the revenue to some very um large negative number and then we try to cut every possible place
• 05:30 - 06:00 at one two three four and so on and then recursively caught the rod caught algorithm um again now one additional difference at the end is once we have found a solution what we want to do is maintain the solution in a table right so that the stable um contains solutions um that further will not be resolved again now this this um table r that contains solution must be passed by reference so that
• 06:00 - 06:30 subsequent function calls keep changing the same the same table now if we compare the recursive top-down approach versus the dynamic programming top-down approach that is the memoized recursive solution you will see that the only difference is that before trying to solve the rod card problem for a length of n we check if the solution already
• 06:30 - 07:00 exists in a table is it it already exists in a table right this is checking and then if it doesn't we update the table so this is where we update um the table otherwise the these two solutions are basically the same now to solve the road cutting problem instead of going top down that is
• 07:00 - 07:30 trying to solve the largest length rod first and then going down to solve the smaller length we can actually do the opposite that is go bottom up in other words we first try to solve the problem for a rod of a very small length and then gradually grow up this is exactly what the bottom-up broadcasting algorithm does so once again we initialize or we create an array called r where we want to
• 07:30 - 08:00 maintain our solution for a rod of all the lengths up to n right so for r equals 1 to n equals 1 2 3 all the way to n we want to maintain a table with all the corresponding solutions now we initialize that for a rod of length zero the revenue will be zero and then we run the loop to calculate the revenue for a rod of all the lengths from one all the way to n so first we calculate
• 08:00 - 08:30 the revenue for a lot of length one then in the second time um for when j equals two we calculate the revenue for the rod of length two and here um this assignment updates the table each time so once we have solved for j equals 1 we update the table and we say that we have calculated the revenue for the route of length 1 and so on and the inner for loop here since we have double followed what the inner for loop does is it looks for all the possible ways to
• 08:30 - 09:00 cut the rod that is for a rod of length let's say 4 we want to go from i one to j that is we want to cut at the first place second place third place and so on so every time we make a cut we notice that we don't need to call the rod cutting algorithm again what we simply do is look up this table and pull the value down from the table now since we are solving a sub problem i that is always
• 09:00 - 09:30 here this the sub problem i that we have inside is always smaller than the sub problem j that we are trying to solve we will always find that our this this table r j minus i has a solution that we are looking for so in this way we calculate the optimal revenue for a lot of length one then we calculate for um sorry zero then we calculate for one and then for two and then for three and then for four in a bottom-up fashion
• 09:30 - 10:00 so this is how the bottom-up rod cutting algorithm works now to briefly compare the two different approaches of dynamic programming the first one is the memoization approach that is improvisation of the recursive approach and the next one is the bottom-up approach what we see is that both of them have a similar asymptotic running time of around big theta n square now it's any square because
• 10:00 - 10:30 if you if you look at the recursion tree for the memoized rod card that will also be evident um but for the bottom up approach we see that we have a double for loop right so the first for loop goes from one to n second goes from um one to j so this becomes like an the number of times the inner for loop runs is like an arithmetic series 1 plus 2 plus 3 essentially if we sum the total number of calculations that the algorithm has to do it will eventually turn out to be in the
• 10:30 - 11:00 order of big theta of n squared now even though for some people the recursive algorithm might sound intuitive recursion is something that is not encouraged in in many applications um for many reasons and also it turns out that the bottom-up approach actually um slightly outperforms the top-down approach by a constant factor uh probably because it doesn't need to initialize your stack and all those things so typically
• 11:00 - 11:30 the bottom up version with the double for loop type implementations are usually preferred now the dynamic programming solutions that we looked so far only give us the optimal revenue maximum revenue r j for a rod of length j but we know that the solution that we need to generate is not just the revenue ri but also the answer to
• 11:30 - 12:00 where to cut the first piece off so for an input table like this which is our problem right problem our solution should look something like this where um this i is the length of the rod and then what we have ri is the optimal revenue that we are going to make with that given rod length and si is the first place where we should cut the rod here's an example if i have a rod of
• 12:00 - 12:30 length 4 the maximum revenue i can make is 10 and the first place i should cut the rod is at after two so if i have a lot of length four if i cut it at two then um that is the optimal cut i can make to make the most amount of um the highest revenue and the maximum revenue i can make is ten dollars so the solution that we saw so far only give us these numbers that is the maximum revenue
• 12:30 - 13:00 they don't give us the um the optimal place to make the first um cut returning such a solution is not extremely difficult um say for example if we are talking about the bottom-up solution then all we need to do is change the max to this if condition so that max goes to if condition so that any any time we observe an improvement
• 13:00 - 13:30 we notice that we make a note of where we actually have the case of better revenue and then save it into an array sj now this array s j or s n maintains the list of indices to make the optimal cut now when applying dynamic programming to a problem we should understand primarily the set of sub-problems involved that is
• 13:30 - 14:00 how many unique sub-problems do i have in my problem that need to be solved essentially what we want to do is sort of get an intuitional understanding of how big this table may look like also what we want to understand is how do the sub problems depend on each other and the best way to represent these two information that is how big the sub problems are how many sub problems we have
• 14:00 - 14:30 and how they depend on each other one really good way to visualize this is using something known as sub problem graphs and here's an example of a sub problem graph for the rod cutting problem with a rod of length l equals four so you can see that to solve a rod of length 4 we need to have a solution for a rod of length 3 represented by this arrow rod of length 2 row of length 1 and dot of length 0. similarly
• 14:30 - 15:00 to solve a rod of length three it depends on the solution to the load of length two one and zero and so on so this is more or less like a reduced or collapsed version of the recursion tree where it is much more clear how many unique sub problems we have and how the sub problems depend on each other