# The Coin Change Problem

The problem

How many ways can you make change for a dollar?

You have four denominations available. You can use quarters, dimes, nickels, and pennies, and you have all the coins that you might need. So, you could make change for a dollar with 100 pennies. You could use four quarters. You could use three quarters, two dimes, and one nickel. You could use one quarter, two dimes, four nickels, and 35 pennies.

There are a lot of ways to make change for a dollar. How do you find out the total number of ways?

In general, given a list of denominations D and an amount of money M, how do you find the total number of ways to make change for M?

To answer that question, we’re going to apply dynamic programming. First we’ll take a top-down approach. Then we’ll take a bottom-up approach.

The top-down approach

This is the top-down approach in a nutshell: If the given problem instance is trivial, then solve it; if it’s not trivial, then cut it down to size.

Given an instance of the problem, we continually reduce the size of the problem instance until we reach subinstances that are trivial to solve. To produce a solution to the original problem instance, we combine all of the solutions to the subinstances.

Two parameters define the coin change problem: an amount of money (M) and a list of coin denominations (D). So, you can reduce a problem instance in two ways: decrease M or shorten D. We’ll do both.

We divide a nontrivial problem instance into these two subinstances:

(1) Find the number of ways to make change for M using the first denomination in D
(2) Find the number of ways to make change for M without using the first denomination in D

To illustrate the procedure, let’s say that M = 100 and D = [quarters, dimes, nickels, pennies].

If you use the first denomination in D, then you put in 25 cents. Now you only have to make change for \$0.75. So, you reduce the size of the problem instance because you reduce the size of its first parameter, M.

Let’s say that you don’t use the first denomination in D. So, you’ve effectively reduced D to [dimes, nickels, pennies]. So, you reduce the size of the given problem instance by reducing the second parameter of the problem.

You continue dividing problem instances until you reach trivial subinstances. The coin change problem is trivial in these instances:

-You must make change for no amount of money. There is only one way to solve that problem: You use no coins.
-You must make change for a negative amount of money. There are zero ways to do that.
-You must make change for some amount of money, but there are no denominations to choose coins from. There are no ways to do that, either.

The subproblems are the same type of problem that the original problem is, because the subproblems require you to find the number of ways to make change for a given amount of money, too. So, we apply the same problem-solving procedure to the subproblems that we do to the original problem. Consequently, it is natural to use a recursive algorithm, like it was when I implemented a solution to the Towers of Hanoi problem. So, I implemented the algorithm in Scheme.

The bottom-up approach

We just examined a top-down approach to the coin change problem. We applied a divide-and-conquer strategy. We started with a problem instance and divided it into smaller and smaller subinstances, working our way down until we arrived at problem instances that are trivial to solve.

We can also employ a bottom-up approach, where we start with the solutions to the smallest subinstances of the problem and continually build on them until we get up to a solution to the entire problem instance.

The top-down, recursive approach typically solves a lot of the same subinstances many times. The bottom-up approach does not. So, the bottom-up approach is typically a lot faster. In the specific case of the coin change problem, the bottom-up approach is O(mn), where m is the number of coin denominations and n is the amount of money (in cents). In contrast, if you don’t memoize the recursive algorithm, it runs in exponential time.

As I have in the past, I used dynamic programming to design just such a bottom-up strategy. Here is a Python program that implements it. The program also features a naive recursive solution and a memoized recursive solution.

In the past, I also implemented a top-down, recursive algorithm and then designed and implemented a bottom-up algorithm. The algorithms find the maximum independent set of a tree. Here is the Python program that implements the bottom-up algorithm. Here is the Python program that implements the top-down algorithm.

The algorithms that solve the maximum independent set problem and the coin change problem have a feature in common: The use of inclusion and non-inclusion to neatly divide a problem instance into just two subinstances.

The maximum independent set of a tree either includes the root of the tree or it doesn’t include the root of the tree. Exploiting that fact, you construct the independent set that includes the root and the independent set that doesn’t include the root, and then you take the larger of the two.

Likewise, when you count the total number of ways that you can make change for some amount of money, you count the number of ways that you can make change for it if you include the first denomination on the list and the number of ways if you don’t include that denomination.

I have found that many dynamic programming approaches feature this technique. For example, an approach to the discrete knapsack problem features it, too: To compute the maximum value that the knapsack can contain, you compute the value when you include the current item and when you don’t include the current item, and then you take the larger of the two. I have implemented that approach, too.

MIT professor Erik Demaine has said that an essential part of dynamic programming is guessing. In his opinion, dynamic programming is “careful brute force”. If you don’t know the solution to some problem, then try every possible solution and take a best one. And that is what is going on in the above procedures. The technique that all of these procedures deploy is just a clever way to partition the set of all possible solutions. Because of the law of excluded middle, a solution either includes an element or doesn’t include it. So, we only have to process two possibilities.

# Miscellaneous Implementations

Over the past year, I have implemented several algorithms and data structures. I decided to do a brief round-up:

I implemented insertion sort for lists of numbers in Python.

I implemented insertion sort for linked lists in Python.

I implemented insertion sort for linked lists in Java.

I implemented prime factorization in Java.

I implemented binary search in Python.

I implemented tries in Python.

# Q-Learning

In reinforcement learning, an agent takes actions within an environment. When an agent acts, a reward function assigns feedback to the agent that either positively or negatively reinforces the agent’s action. The agent’s goal is to use this feedback to learn an optimal policy.

For any given state of the environment, a policy tells the agent an action to take when the environment is in that state. An optimal policy tells the agent the best action to take. Of course, there are several different ways to define what a best action is. So, let’s say that an optimal policy maximizes the sum of discounted rewards over time.

Say that Q(s, a) is the reward received from taking action a from state s, plus the discounted value of following the optimal policy thereafter. So, if the agent knows Q, then it can execute the optimal policy. Whenever the agent is in state s, the optimal policy is to take the action a that maximizes Q(s, a). So, the agent’s task is to learn Q.

To learn Q, create a table Q[s][a] and initialize the entries to zero. Then, over and over again, have the agent take actions from states and receive the ensuing rewards. Update Q[s][a] each time. Q[s][a] will converge to the true value of Q as the number of times that each state-action pair is visited converges to infinity.

I implemented Q-learning, and I applied it to a simple environment. The code is up at my GitHub.

References:

Mitchell, Tom. Machine Learning. McGraw-Hill, 1997.

# Turing Machines

In a previous post, I gave an informal explanation of the P=NP? problem. In that post, I described both deterministic and nondeterministic Turing machines. A couple of months ago, I decided to implement the deterministic Turing machine in Python.

I also implemented a special deterministic Turing machine that decides whether a given positive integer is divisible by 4, just like the hypothetical Turing machine that I describe in my previous post.

I include a Bash script that tests that implementation. The script inputs every multiple of 4, ranging from 4 to 40, into the Turing machine program. The program passes every test.

# Ranked Pairs and “The Wire”

A few days ago, a redditor asked the people of /r/TheWire to rank the five seasons of the HBO television show “The Wire”, and several people posted their rankings. I wanted to take their individual rankings and uncover one aggregate ranking that captured the overall preferences of the entire group of posters. In other words, I wanted to implement a preferential voting scheme.

I was aware of Arrow’s theorem, which says that no voting scheme can satisfy a certain small set of reasonable criteria when voters are ranking three or more alternatives. However, I was also aware that a number of countries employ preferential voting systems. So, I didn’t give up, and I searched for a preferential voting system that would fit the bill.

I found one that worked: the ranked pairs voting scheme. In the ranked pairs voting system, you first count all the times that voters ranked Alternative X ahead of Alternative Y, for all pairs of alternatives (X, Y). If voters ranked Alternative X ahead of Alternative Y more often, then Alternative X is a majority over Alternative Y.

Next you sort the majorities. For example, say that voters ranked Alternative X ahead of Alternative Y three times and that they ranked Alternative A ahead of Alternative B five times. Then the latter majority comes before the former majority in the sort.

But what if the list of majorities creates a cycle? For example, say that we have [(A, B), (B, C), (C, A)]. Then A is a majority over B and B is a majority over C. But C is a majority over A. So, by transitivity, A is a majority over A. But an alternative X can be a majority over an alternative Y only if X was ranked ahead of Y, and A can’t be ranked ahead of itself.

So, here’s how you resolve that issue. You treat the list of majorities like a directed acyclic graph where the dag includes a directed edge from X to Y if X is a majority over Y. And you just go down the sorted list of majorities. For each majority, you try to add a corresponding edge to the dag. If adding that edge would produce a cycle, then you don’t add it. If it wouldn’t produce a cycle, then you do add it, and that majority gets locked in to the final list of majorities.

The winner of the election is the source of the dag, i.e., the vertex that doesn’t have any incoming edges. But I didn’t want to know just the winner. I wanted to know the aggregate ranking. So, I sorted the alternatives by the number of outgoing edges that their corresponding vertices have in the dag. The number of outgoing edges of a vertex corresponds to the number of alternatives that an alternative is a majority over.

I collected the data on the night of September 1. I simply scrolled down the list of comments and recorded the rankings that commenters reported. The only complication was that a couple of commenters expressed indifference between some seasons. So, for example, one person wrote, “4 > 1 > 3/2 > 5”. In such a case, I ranked seasons in the order that they were written. So, in this particular case, I placed Season 3 before Season 2.

Then I ran the program, and the result was 4 > 1 > 2 > 3. Season 5 didn’t even make the list.

You can find my program up at my GitHub.

Further references:

# Swapping Cards

Earlier today I searched through the ACM-ICPC archive of problems that have been used in ACM-ICPC Regionals and World Finals, looking for some interesting problems to solve. I found one that reminded me of another ACM-ICPC problem that I solved. You can read the full description of it, but I’ll put it in a nutshell:

You’re given a stack of playing cards. You can swap adjacent cards. You want to find the minimum number of swaps it takes to sort the cards so that all the diamonds and hearts in the stack appear before all the spades and clubs.

I realized that you are searching for a certain state of the stack of cards. Because you want to find that state in the minimum number of swaps, you can perform a version of breadth-first search to solve the problem.

I remembered the Marbles in Three Baskets problem, another ACM-ICPC problem that I solved. I applied the same basic technique and philosophy of problem-solving to this problem that I outline in the post about the Marbles in Three Baskets problem.

You can find my solution to this problem at my GitHub.

# The Sieve of Eratosthenes

Given an integer N > 1, you want to find all of the prime numbers from 1 to N. For example, if N = 11, then you get {2, 3, 5, 7, 11}.

The Sieve of Eratosthenes is a simple way to solve this problem. The basic idea behind it is this: Construct the list of all integers from 1 to N, and remove the composite numbers.

It is easy to construct the list of all integers from 1 to N. So, the question is: How do you remove the composite numbers?

It is a fact of number theory that every composite number k has a prime factor that is less than or equal to sqrt(k). To see this, consider any composite number k = mn. If both m > sqrt(k) and n > sqrt(k), then k < mn, which contradicts the fact that k = mn. So, k must have a prime factor that is less than or equal to sqrt(k).

So, every composite number from 1 to N must have a prime factor that is less than or equal to sqrt(N). So, just iterate over each prime number from 1 to sqrt(N) and remove all of its multiples in the list. In that way you remove all of the composite numbers from the list.

I implemented the Sieve of Eratosthenes in C, and I implemented the Sieve of Eratosthenes in Python. Both of my implementations are up at my GitHub.

References: