Introduction To Dynamic Programming: Beginner

Edward Chu

January 2, 2023

Origin post from Dumitru — a topcoder member

What is a dynamic programming, how can it be described?

A DP is an algorithmic technique which is usually based on a recurrent formula and one (or some) starting states. A sub-solution of the problem is constructed from previously found ones. DP solutions have a polynomial complexity which assures a much faster running time than other techniques like backtracking, brute-force etc.

Now let’s see the base of DP with the help of an example:

Given a list of N coins, their values (V1, V2, … , VN), and the total sum S. Find the minimum number of coins the sum of which is S (we can use as many coins of one type as we want), or report that it’s not possible to select coins in such a way that they sum up to S.

Now let’s start constructing a DP solution:

First of all we need to find a state for which an optimal solution is found and with the help of which we can find the optimal solution for the next state.

What does a “state” stand for?

It’s a way to describe a situation, a sub-solution for the problem. For example a state would be the solution for sum i, where i≤S. A smaller state than state i would be the solution for any sum j, where j<i. For finding a state i, we need to first find all smaller states j (j<i) . Having found the minimum number of coins which sum up to i, we can easily find the next state – the solution for i+1.

How can we find it?

It is simple – for each coin j, V_j ≤ i, look at the minimum number of coins found for the i-V_j_sum (we have already found it previously). Let this number be m. If m+1 is less than the minimum number of coins already found for current sum i, then we write the new result for it.

For a better understanding let’s take this example:

Given coins with values 1, 3, and 5. And the sum S is set to be 11.

First of all we mark that for state 0 (sum 0). We have found a solution with a minimum number of 0 coins. We then go to sum 1. First, we mark that we haven’t yet found a solution for this one (a value of Infinity would be fine). Then we see that only coin 1 is less than or equal to the current sum. Analyzing it, we see that for sum 1-V1= 0 we have a solution with 0 coins.

Because we add one coin to this solution, we’ll have a solution with 1 coin for sum 1. It’s the only solution yet found for this sum. We write (save) it.

Then we proceed to the next state – sum 2. We again see that the only coin which is less or equal to this sum is the first coin, having a value of 1. The optimal solution found for sum (2-1) = 1 is coin 1. This coin 1 plus the first coin will sum up to 2, and thus make a sum of 2 with the help of only 2 coins. This is the best and only solution for sum 2. Now we proceed to sum 3. We now have 2 coins which are to be analyzed – first and second one, having values of 1 and 3. Let’s see the first one.

There exists a solution for sum 2 (3 – 1) and therefore we can construct from it a solution for sum 3 by adding the first coin to it. Because the best solution for sum 2 that we found has 2 coins, the new solution for sum 3 will have 3 coins. Now let’s take the second coin with value equal to 3.

The sum for which this coin needs to be added to make 3, is 0. We know that sum 0 is made up of 0 coins. Thus we can make a sum of 3 with only one coin: 3. We see that it’s better than the previous found solution for sum 3 , which was composed of 3 coins. We update it and mark it as having only 1 coin. The same we do for sum 4, and get a solution of 2 coins: –1+3. And so on.

Pseudocode:

Set Min[i] equal to Infinity for all of i
Min[0]=0

For i = 1 to S
For j = 0 to N – 1
If (Vj<=i AND Min[i-Vj]+1<Min[i])
Then Min[i]=Min[i-Vj]+1

Output Min[S]

Having understood the basic way a DP is used, we may now see a slightly different approach to it. It involves the change (update) of best solution yet found for a sum i, whenever a better solution for this sum was found. In this case the states aren’t calculated consecutively.

Let’s consider the problem above. Start with having a solution of 0 coins for sum 0. Now let’s try to add first coin (with value 1) to all sums already found. If the resulting sum t will be composed of fewer coins than the one previously found – we’ll update the solution for it. Then we do the same thing for the second coin, third coin, and so on for the rest of them.

For example, we first add coin 1 to sum 0 and get sum 1. Because we haven’t yet found a possible way to make a sum of 1 – this is the best solution yet found, and we mark S[1]=1. By adding the same coin to sum 1, we’ll get sum 2, thus making S[2]=2. And so on for the first coin. After the first coin is processed, take coin 2 (having a value of 3) and consecutively try to add it to each of the sums already found. Adding it to 0, a sum 3 made up of 1 coin will result. Till now, S[3] has been equal to 3, thus the new solution is better than the previously found one.

We update it and mark S[3]=1. After adding the same coin to sum 1, we’ll get a sum 4 composed of 2 coins. Previously we found a sum of 4 composed of 4 coins; having now found a better solution we update S[4] to 2. The same thing is done for next sums – each time a better solution is found, the results are updated.