Dynamic Programming Series - Introduction
Ankur Sheel Updated on ・7 min read
This article has been cross-posted from here
Dynamic Programming is one of those techniques that every programmer should have in their toolbox. But, it is also confusing for a lot of people. For a long time, I struggled to get a grip on how to apply Dynamic Programming to problems. Most articles that I could find on the internet, gave the final dynamic programming solution without actuallyÂ showing theÂ approach taken to arrive at the final solution.
In this article, I will show the benefits of using a Dynamic Programming approach to solving problems with an example. In the end, IÂ will show some steps you can use to find a Dynamic Programming solution.Â Hopefully, after reading this article, you will find Dynamic Programming simple and intuitive.
What is Dynamic Programming?
Dynamic programming is an efficient method for solving specific types of complicated computational problems. TheseÂ problems are generally those that can be broken down into smaller overlapping sub-problems. It can be characterised basically as recursion along with memoization. Memoization is the ability to save the results of specific states to reuse later.
Profiling
To prove that we are improving our solution, we need statistics that we can compare. I will be using google benchmarkÂ to help profile our solutions. The benchmark will look like this
Benchmark Name | Running Time | Iterations/sec | Items/sec |
---|
- Benchmark Name: The name of the benchmark. It will be the format FunctionName/value passed in.
- Running Time: The time it took for our function to return a result.
- Iterations/sec: The number of times the function could be invoked in 1 second.
- Items/sec:Â The number of items that were processed in 1 second.
While 2 and 3 will give usÂ an indication of the time complexity of the function, 4 will give us the space complexity.
Example: Fibonacci Series
The classic example to explain dynamic programming is the Fibonacci computationÂ which can be formalised as follows
Fibonacci(n) = 0; if n = 0
Fibonacci(n) = 1; if n = 1
Fibonacci(n) = Fibonacci(n-1) + Fibonacci(n-2) ; if n >=2
Naive Recursive Approach
The Fibonacci sequence can easily be solved by the following recursive method:
long Fibonacci(long n)
{
if (n == 0)
{
return 0;
}
if (n == 1)
{
return 1;
}
return Fibonacci(n - 2) + Fibonacci(n - 1);
}
On running the above code and profiling it on my machine, I get:
Benchmark Name | Running Time | Iterations/sec | Items/sec |
---|---|---|---|
FibonacciRecursive/10 | 0 ms | 2488889 | 17.0685M |
FibonacciRecursive/20 | 0 ms | 21333 | 303.026k |
FibonacciRecursive/30 | 8 ms | 179 | 3.60887k |
FibonacciRecursive/40 | 997 ms | 1 | 40 |
FibonacciRecursive/50 | 124510 ms | 1 | 0.40282 |
Although this method returns almost instantly for n <= 30, it takes a little less than aÂ second for n = 40 and approximately 2 minutes for n = 50. Why is the amount of running time increasing so rapidly? This can be explained easily by following the executionÂ stack. Let's do this for n = 6 to keep it simple. The following image shows the sequence of calls that are made. Fibonacci Series : From Wikimedia Commons
Looking at the image, we can see that to calculate fibonacci(6), we calculateÂ fibonacci(5) 1Â time,Â fibonacci(4) 2Â times_,Â Â _fibonacci(3) 3Â times,Â fibonacci(2) 5 times_,Â fibonacci(1)Â 8 times_ andÂ Â fibonacci(0) 5 times. Throughout the call-stack, we are repeatedly computing values that we have already computed. This amount of duplicated work being done keeps on increasing as n becomes larger.Â You must have realised that this solution is not at all scalable. If you are thinking that there has to be a better way, you are correct.
Top-Down Recursive approach with Memoization
The 1st step to improving the above solution is to add memoization i.e to store the previously computed values in some kind of a data structure. Although you can use any data structure that you like, for the purposes of this example, I will use a map.
#define MOD 1000000007
long FibonacciMemonized(long n)
{
std::map<long, long> computedValues;
computedValues.insert(make_pair(0, 1));
computedValues.insert(make_pair(1, 1));
return FibonacciMemonized(n, computedValues);
}
long FibonacciMemonized(long n, std::map<long, long>& computedValues)
{
if (computedValues.find(n) != computedValues.end())
{
return computedValues[n];
}
long newValue = (FibonacciMemonized(n - 1, computedValues) + FibonacciMemonized(n - 2, computedValues)) % MOD;
computedValues.insert(make_pair(n, newValue));
return newValue;
}
The topÂ method is our main method. All it does is add the 2 base cases to a map and then calls the bottom method with the map as one of the arguments. This bottom method is our recursive method. In this method, we check if the map contains the computed value. If it does.Â then we just return that value, otherwise, we compute the value for n-1 andÂ n-2. We mod the result usingÂ 1000000007 to avoid overflows. Before, returning their sum, we store the value in our map. How better is this version?Â Let's look at the benchmark results
Benchmark Name | Running Time | Iterations/sec | Items/sec |
---|---|---|---|
FibonacciMemonized/1000 | 0 ms | 4073 | 3.60284M |
FibonacciMemonized/5000 | 2 ms | 896 | 2.90891M |
FibonacciMemonized/10000 | 3 ms | 407 | 2.82288M |
FibonacciMemonized/15000 | 5 ms | 242 | 2.66937M |
FibonacciMemonized/20000 | 7 ms | 187 | 2.65432M |
We can see that we have reduced the amount of time drastically. Even for n = 20000, the result is instantaneous.Â However, there is a problem with this approach. Can you spot it? If you said memory usage, you are absolutely correct. Â Although the new version is much faster, it is still a recursive algorithm. And, the problem with recursive algorithms is that each recursive call takes some space on the stack. A high enough n, and we run the risk of running out of memory. Let's see why this happens with an example whereÂ n = 100. Because, we don't have the result when we start, we call the method recursively for 999, 998, 997 ... 1. At that point, we have all the computed results in our map. Now, as we return from ourÂ recursive functions, we just lookup the value in the table and return it. So, even though we have reduced the number of recursive calls,Â we still make n recursive calls before getting our initial result. This can easily be seen by comparing the iteration/seconds between this and the previous algorithm. Let's try something better.
Bottom Up Approach withÂ Dynamic Programming
In the previous approach, our main problem was the recursive nature of our algorithm. Let's see if we can get rid of it by using an iterative approach. How do we do this? Instead of starting from the final value, we will start with the smallest value ofÂ n and build up the results.
#define MOD 1000000007
long FibonacciDP(long n)
{
if (n == 0)
{
return 0;
}
if (n == 1)
{
return 1;
}
long* results = new long[n + 1];
results[0] = 0;
results[1] = 1;
for (int i = 2; i <= n; i++)
{
results[i] = (results[i - 1] + results[i - 2]) % MOD;
}
long value = results[n];
delete[] results;
return value;
}
In the above function, we have an array ofÂ n+1 to store the results. We initialise the array for our base cases of n=0Â and n=1Â **and then start iterating from **2 to n. At each step, we can use the 2 previously computed values and finally return the result. Lets again look at the benchmark results to see how does this approach do?
Benchmark Name | Running Time | Iterations/sec | Items/sec |
---|---|---|---|
FibonacciDP/100000 | 1 ms | 1906 | 130.711M |
FibonacciDP/600000 | 5 ms | 280 | 111.456M |
FibonacciDP/1100000 | 10 ms | 145 | 110.626M |
FibonacciDP/1600000 | 14 ms | 100 | 112.249M |
FibonacciDP/2100000 | 18 ms | 81 | 110.448M |
Even whenÂ n=210,000, this approach returns almost instantly. At the same time, since, thisÂ algorithm is not recursive in nature, we have drastically reduced the amount of space that is required. We can see this by comparing the items/secÂ which is decreasing much slower even thoughÂ n is increasing more rapidly than before. Now you may be thinking, that since we have got a linear time complexity and linear space complexity, this is the end. In most cases, that would be true. But, in this case, we can optimise our solution further.
Bottom Up Approach withÂ Dynamic Programming (Optimised)
In the last algorithm, the amount of space required is proportional toÂ n. This is because we are storing all the results. But, weÂ don't need to store all of them. Let's get rid of the array by using just 3 variables - 2 to store the previous results and 1 to store the current result.
#define MOD 1000000007
long FibonacciDPOptimized(long n)
{
if (n == 0)
{
return 0;
}
if (n == 1)
{
return 1;
}
long n1 = 0;
long n2 = 1;
long current = 0;
for (int i = 2; i <= n; i++)
{
current = (n1 + n2) % MOD;
n1 = n2;
n2 = current;
}
return current;
}
Although this algorithm is doing exactly the same thing as the previous one, we have reduced our space complexity to be constant since the amount of space needed is no longer dependent onÂ n. Again the benchmark results for comparison
Benchmark Name | Running Time | Iterations/sec | Items/sec |
---|---|---|---|
FibonacciDPOptimized/100000 | 0 ms | 2987 | 202.569M |
FibonacciDPOptimized/600000 | 3 ms | 498 | 207.242M |
FibonacciDPOptimized/1100000 | 5 ms | 280 | 202.138M |
FibonacciDPOptimized/1600000 | 7 ms | 187 | 205.188M |
FibonacciDPOptimized/2100000 | 10 ms | 128 | 205.0708M |
Here, we can see that althoughÂ **nÂ **is increasing, the items/sec is more or less the same. This is the best we can do and no further optimisations are possible.
Conclusion
From the above example, we can see that we only need to identifyÂ overlapping subproblems and then avoid duplicated work by caching the computed results. To recap, we can use these steps to find a dynamic programming approach to our problem
- Find the overlapping subproblem.
- Start with a recursive solution
- Modify the recursive solution to use a top-down memoized version.
- Remove the recursion by making it an iterative solution.
- If you don't need to keep all the previous results, keep only the ones that are required.
If you would like to run the code yourself, you can find the code hereÂ
Hopefully, this article has removed the mystery around Dynamic Programming. In the next few articles in the series, we will look at some of the more common problems that can be solved by Dynamic Programming.and use the above steps to come up with a solution. Have you tried Dynamic Programming before? How was your experience? Let me know in the comments.
You forgot that C++ has tail call optimization. This means you can run it in constant space:
this is similar to the for loop
You are right about the tail call optimisation. However, that would have beaten the purpose of gearing this article towards beginners and also keeping it language agnostic. The main issue I have had with other articles (and that I wanted to avoid here) is that either they skip intermediate steps or use tricks which are not conducive to understanding the technique itself.
You left in the
#define MOD 1000000007
and% MOD
which isn't part of the explanation. Though I recall see computing tests that have this requirement on them. :)I do mention the reason in the Top-Down Recursive approach with Memoization section. Re-iterating it here for your convenience We mod the result using 1000000007 to avoid overflows.
Wonderful post :-) Cleared up a lot of concepts for me.
Thanks Suraj.
Good article! I think C++ isn't the right language for it, though (syntactic noise, difficulty reproducing, the numbers are too large for it to handle).
Thanks Josh.
I feel that since it is a technique, it is not limited to a language. I guess it might be more accessible if it was in one of the newer languages. But, it's better to stick with a language you are comfortable with and I don't need to figure out a whole new benchmarking system :)
I could have written it in pseudocode but then the benchmarking would have been extremely difficult.
What would you suggest be done differently?
Oh, further optimization is possible. For example, with matrix exponents ;)
Haha. You got me there. I should have specified no further optimizations from a DP perspective :)