Often I'll hear about how you can optimise a for loop to be faster or how switch statements are faster than if statements. Most computers have over 1 core, with the ability to support multiple threads. Before worrying about optimising for loops or if statements try to attack your problem from a different angle.

Divide and Conquer is one way to attack a problem from a different angle. Don't worry if you have **zero **experience or knowledge on the topic. This article is designed to be read by someone with very little programming knowledge.

I will explain this using 3 examples. The first will be a simple explanation. The second will be some code. The final will get into the mathematical core of divide and conquer techniques. (Don't worry, I hate maths too).

## What is divide and conquer? 🌎

Divide and conquer is where you divide a large problem up into many smaller, much easier to solve problems. The rather small example below illustrates this.

We take the equation "3 + 6 + 2 + 4" and cut it down into the smallest set of equations, which is [3 + 6, 2 + 4]. It could also be [2 + 3, 4 + 6]. The order doesn't matter, as long as we turn this one long equation into many smaller equations.

Let’s say we have 8 numbers:

$$4 + 6 + 3 + 2 + 8 + 7 + 5 + 1$$

We want to add them all together. We first divide the problem into 8 equal sub-problems. We do this by breaking the addition up into individual numbers.

$$4 \; 6 \; 3 \; 2 \; 8 \; 7 \; 5 \; 1$$

We then add 2 numbers at a time.

Then 4 numbers into 8 numbers which is our resultant.

Why do we break it down to individual numbers at stage 1? Why don't we just start from stage 2? Because while this list of numbers is even if the list was odd you would need to break it down to individual numbers to better handle it.

A divide and conquer algorithm tries to break a problem down into as many little chunks as possible since it is easier to solve with little chunks. It does this with recursion.

The technique is, as defined in the famous Introduction to Algorithms by Cormen, Leiserson, Rivest, and Stein, is:

- Divide

If the problem is small, then solve it directly. Otherwise, divide the problem into smaller subsets of the same problem.

2. Conquer

Conquer the smaller problems by solving them recursively. If the sub-problems are small enough, recursion is not needed and you can solve them directly.

**Recursion is when a function calls itself.** It's a hard concept to understand if you've never heard of it before. This page provides a good explanation. In short, a recursive function is one like this:

```
n = 6
def recur_factorial(n):
if n == 1:
return n
else:
return n * recur_factorial(n-1)
print(recur_factorial(n))
```

I'll explain the code in a second.

3. Combine

Take the solutions to the sub-problems and merge them into a solution to the original problem.

With the code from above, some important things to note. The Divide part is also the recursion part. We divide the problem up at `return n * recur_factorial(n-1)`

.

The `recur_factorial(n-1)`

part is where we divide the problem up.

The conquer part is the recursion part too, but also the if statement. If the problem is small enough, we solve it directly (by returning n). Else, we perform `return n * recur_factorial(n-1)`

.

Combine. We do this with the multiplication symbol. Eventually, we return the factorial of the number. If we didn't have the symbol there, and it was `return recur_factorial(n-1)`

it wouldn't combine and it wouldn't output anything similar to the factorial. (It'll output 1, for those interested).

We'll explore how divide and conquer works in some famous algorithms, Merge Sort and the solution to the Towers of Hanoi.

### Merge Sort 🤖

Merge Sort is a sorting algorithm. The algorithm works as follows:

- Divide the sequence of n numbers into 2 halves
- Recursively sort the two halves
- Merge the two sorted halves into a single sorted sequence

In this image, we break down the 8 numbers into separate digits. Just like we did earlier. Once we've done this, we can begin the sorting process.

It compares 51 and 13. Since 13 is smaller, it puts it in the left-hand side. It does this for (10, 64), (34, 5), (32, 21).

It then merges (13, 51) with (10, 64). It knows that 13 is the smallest in the first list, and 10 is the smallest in the right list. 10 is smaller than 13, therefore we don't need to compare 13 to 64. We're comparing & merging two **sorted **lists.

In recursion we use the term *base case* to refer to the absolute smallest value we can deal with. With Merge Sort, the base case is 1. That means we split the list up until we get sub-lists of length 1. That's also why we go down all the way to 1 and not 2. If the base case was 2, we would stop at the 2 numbers.

If the length of the list (n) is larger than 1, then we divide the list and each sub-list by 2 until we get sub-lists of size 1. If n = 1, the list is already sorted so we do nothing.

Merge Sort is an example of a divide and conquer algorithm. Let's look at one more algorithm to understand how divide and conquer works.

### Towers of Hanoi 🗼

The Towers of Hanoi is a mathematical problem which compromises 3 pegs and 3 discs. This problem is mostly used to teach recursion, but it has some real-world uses. The number of pegs & discs can change.

Each disc is a different size. We want to move all discs to peg C so that the largest is on the bottom, second largest on top of the largest, third largest (smallest) on top of all of them. There are some rules to this game:

- We can only move 1 disc at a time.
- A disc cannot be placed on top of other discs that are smaller than it.

We want to use the smallest number of moves possible. If we have 1 disc, we only need to move it once. 2 discs, we need to move it 3 times.

The number of moves is a power of 2 minus 1. Say we have 4 discs, we calculate the minimum number of moves as $2^4 = 16 - 1 = 15$.

To solve the above example we want to store the smallest disc in a buffer peg (1 move). See below for a gif on solving Tower of Hanoi with 3 pegs and 3 discs.

Notice how we need to have a buffer to store the discs.

We can generalise this problem. If we have n discs: move n-1 from A to B recursively, move largest from A to C, move n-1 from B to C recursively.

If there is an even number of pieces the first move is always into the middle. If it is odd the first move is always to the other end.

Let's code the algorithm for ToH, in pseudocode.

```
function MoveTower(disk, source, dest, spare):
if disk == 0, then:
move disk from source to dest
```

We start with a base case, `disk == 0`

. `source`

is the peg you're starting at. `dest`

is the final destination peg. `spare`

is the spare peg.

```
FUNCTION MoveTower(disk, source, dest, spare):
IF disk == 0, THEN:
move disk from source to dest
ELSE:
MoveTower(disk - 1, source, spare, dest) // Step 1
move disk from source to dest // Step 2
MoveTower(disk - 1, spare, dest, source) // Step 3
END IF
```

Notice that with step 1 we switch `dest`

and `source`

. We do not do this for step 3.

With recursion, we know 2 things:

- It always has a base case (if it doesn't, how does the algorithm know to end?)
- The function calls itself.

The algorithm gets a little confusing with steps 1 and 3. They both call the same function. This is where multi-threading comes in. You can run steps 1 and 3 on different threads - at the same time.

Since 2 is more than 1, we move it down one more level again. So far you've seen what the divide and conquer technique is. You should understand how it works and what code looks like. Next, let's learn how to define an algorithm to a problem using divide and conquer. This part is the most important. Once you know this, it'll be easier to create divide and conquer algorithms.

### Fibonacci Numbers 🐰

We can find Fibonacci numbers in nature. The way rabbits produce is in the style of the Fibonacci numbers. You have 2 rabbits that make 3, 3 rabbits make 5, 5 rabbits make 9 and so on.

The numbers start at 1 and the next number is the current number + the previous number. Here it’s 1 + 0 = 1. Then 1 + 1 = 2. 2 + 1 = 3 and so on.

We can describe this relation using a recursion. A recurrence is an equation which defines a function in terms of its smaller inputs. Recurrence and recursion sound similar and are similar.

With Fibonacci numbers if n = 0 or 1, it results in 1. Else, recursively add f(n-1) + f(n -2) until you reach the base case. Let's start off by creating a non-recursive Fibonacci number calculator.

We know that if n = 0 or 1, return 1.

```
def f(n):
if n == 0 or n == 1:
return 1
```

The Fibonacci numbers are the last two numbers added together.

```
def f(n):
if n == 0 or n == 1:
return 1
else:
fibo = 1
fibroPrev = 1
for i in range (2, n):
temp = fibo
fibo = fibo + fiboPrev
fiboPrev = temp
return fibo
```

Now we've seen this, let's turn it into recursion using a recurrence.

$$ F(n) = \begin{cases} n, \text{If n = 0 or 1} \end{cases}$$

When creating a recurrence, we always start with the base case. The base case here is if n == 0 or 1, return n.

If we don't return n, but instead return 1 this leads to a bug. For example, F(0) would result in 1. When it should result in 0.

Next, we have the formula. If n isn't 0 or 1, what do we do? We calculate F(n - 1) + F(n - 2). In the end, we want to merge all the numbers together to get our final result. We do this using addition.

$$ F(n) = \begin{cases} n, \text{If n = 0 or 1} \\ F(n - 1) + F(n - 2), \; \text{if n > 1} \end{cases}$$

This is the formal definition of the Fibonacci numbers. Normally, recurrences are used to talk about the running time of a divide and conquer algorithm. It's actually a good tool to create a divide and conquer algorithm.

```
def F(n):
if n == 0 or n == 1:
return n
else:
return F(n-1)+F(n-2)
```

With knowledge of divide and conquer, the above code is cleaner and easier to read.

We often calculate the result of a recurrence using an execution tree. Computer overlords 🤖 don't need to do this, but it's useful for humans to see how your divide and conquer algorithm works. For F(4) this looks like:

n is 4, and n is larger than 0 or 1. So we do f(n-1) + f(n-2). We ignore the addition for now. This results in 2 new nodes, 3 and 2. 3 is larger than 0 or 1 so we do the same. Same for 2. We do this until we get a bunch of nodes which are either 0 or 1. We then add all the nodes together. 1 + 1 + 0 + 0 + 1 = 3, which is the right answer.

## Greedy vs Divide & Conquer vs Dynamic Programming

Greedy vs Divide & Conquer vs Dynamic Programming | ||
---|---|---|

Greedy |
Divide & Conquer |
Dynamic Programming |

Optimises by making the best choice at the moment | Optimises by breaking down a subproblem into simpler versions of itself and using multi-threading & recursion to solve | Same as Divide and Conquer, but optimises by caching the answers to each subproblem as not to repeat the calculation twice. |

Doesn't always find the optimal solution, but is very fast | Always finds the optimal solution, but is slower than Greedy | Always finds the optimal solution, but could be pointless on small datasets. |

Requires almost no memory | Requires some memory to remember recursive calls | Requires a lot of memory for memoisation / tabulation |

## Conclusion 📕

Once you've identified how to break a problem down into many smaller pieces, you can use concurrent programming to execute these pieces at the same time (on different threads) speeding up the whole algorithm.

Divide and conquer algorithms are one of the fastest and perhaps easiest ways to increase the speed of an algorithm and are useful in everyday programming. Here are the most important topics we covered in this article:

- What is divide and conquer?
- Recursion
- MergeSort
- Towers of Hanoi
- Coding a divide and conquer algorithm
- Recurrences
- Fibonaccinumbers

The next step is to explore multi-threading. Choose your programming language of choice and Google, as an example, "Python multi-threading". Figure out how it works and see if you can attack any problems in your own code from this new angle.

You can also learn about how to solve recurrences (finding out the asymptotic running time of a recurrence), which is the next article I'm going to write. If you don't want to miss it, or you liked this article do consider subscribing to my email list 😁✨