What is DP
Dynamic programming is a programming paradigm for a massive class of problems. It solves the problem by partitioning it into smaller subproblems and solving it to avoid double computing the same results. Thus, the optimal solution of the main problem depends on the optimal solution of its subproblems.
The simplest (even classic) example is solving the Fibonacci Numbers problem - find the n-th Fibonacci number. As you know, each next Fibonacci number is a sum of the previous two Fibonacci numbers. So, if we have some function fib(n) that returns n-th Fibonacci numbers, we can implement it like this:
12def fib(n): return fib(n-1) + fib(n-2)
Thus, to solve the problem of the n-th Fibonacci number, you should solve the fib(n-1) and fib(n-2) subproblems first. But we can solve both of these subproblems in the same way.
You can note that this recursion has no bottom yet, so we have to add the stop condition:
1234def fib(n): if n <= 1: # Bottom return n return fib(n-1) + fib(n-2)
You'll find some info about the main DP properties in the next two chapters.
Swipe to start coding
- Following the example, implement the function
fib(n). - Make function calls to check how it works.
The function calls are already in the task code; do not change them. Edit the fib(n) function only.
Solution
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Dynamic programming is a programming paradigm for a massive class of problems. It solves the problem by partitioning it into smaller subproblems and solving it to avoid double computing the same results. Thus, the optimal solution of the main problem depends on the optimal solution of its subproblems.
The simplest (even classic) example is solving the Fibonacci Numbers problem - find the n-th Fibonacci number. As you know, each next Fibonacci number is a sum of the previous two Fibonacci numbers. So, if we have some function fib(n) that returns n-th Fibonacci numbers, we can implement it like this:
12def fib(n): return fib(n-1) + fib(n-2)
Thus, to solve the problem of the n-th Fibonacci number, you should solve the fib(n-1) and fib(n-2) subproblems first. But we can solve both of these subproblems in the same way.
You can note that this recursion has no bottom yet, so we have to add the stop condition:
1234def fib(n): if n <= 1: # Bottom return n return fib(n-1) + fib(n-2)
You'll find some info about the main DP properties in the next two chapters.
Swipe to start coding
- Following the example, implement the function
fib(n). - Make function calls to check how it works.
The function calls are already in the task code; do not change them. Edit the fib(n) function only.
Solution
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