Converting hex to binary in the brain

This is a tutorial on hex which is very useful if you are ever going to read low-level code or program low-level things like network protocols or microcontrollers. I use a real project that I worked on to showcase all this, namely a matrix of 9 LEDs.

You should be able and understand why people put hex in the code instead of raw binary (if it exists for that programming language). There are very specific reasons for doing this and since converting from hex to binary is so damn easy, there is no excuse for you to not be able and do it in your brain.

Binary and LED patterns

I was building a trivial LED matrix the other day for an MBED microcontroller (think Arduino-like). The usual problem is that my brain is faulty so I do all sorts of things in the wrong way. I take this blog as the opportunity to make up for what I learn just to make sure that I won’t forget (and ofcourse to teach others if they are interested).

So my task was to achieve some patterns with 9 LEDs. Notice that it doesn’t matter how the microcontroller was connected etc. since here I am only dealing with how bits and bytes fit into low level programming. My first pattern was a rotating star that you can see below:

This star is made out of two LED patterns: a star and a cross.

The 1s and 0s simply mean if the LED at a position should be turned on or off. Now, when we’re dealing with low level things the minimum unit of information that can be sent is 1 byte (= 8 bits). In our case we have 9 LEDs so we need however a minimum of 2 bytes (= 16 bits). The above examples become the below binaries:

Star:  0000000101010101
Cross: 0000000010111010

Now, the problem is that when we deal with low level programming, most low level languages (C, C++ etc.) don’t let you write numbers as binary in your code. You can’t write printf(0b101) for example. You need separate libraries if you want to do that and that would be fine for our case. But imagine if there was a matrix of 100 LEDs. Someone reading printf(001001010101101010101010101010101110101001011100101) would just get lost in the 0s and 1s. That’s one of the big reasons hex is used – it’s super minimal.

Binaries as integers

At first when I wanted to create a star, I simply converted each binary to an integer and just put it in my code. Below you can see a simplified version of my code.

..
#define STAR  341
#define CROSS 186

int main() {
    while (1) {
        leds.write(CROSS)
        sleep(1)
        leds.write(STAR)
        sleep(1)
    }
}

The way I found those integers was by simply using Python. It is a rather trivial procedure:

>>> 0b101010101
341
>>> 0b10111010
186

Notice that I omit the extra 0s since they don’t add any value just like 000150 is always going to be 150 no matter how many zeros you add at front.

Binaries as hex

The code I used, worked fine. The problem with this solution is that it’s impossible to have a clue what an integer is in binary – and when we deal with low-level programming that matters most of the times. In our case for example each single 1 and 0 controls one LED. Being able to figure out fast the binary of a number in this case is very important.

For example say you find the code snippet below somewhere:

#define STAR1 341
#define STAR2 186

How can you tell if it’s the STAR1 or STAR2 that looks like an ‘X’? It’s just impossible. And what if there were many more stars or if the LED matrix was huge? Then it would be a nightmare to understand the code. That’s where hex comes in handy.

The good thing with hex is that someone can see the hex and decode it to binary in his brain. So say we had the above example but instead of integers had hex values:

#define STAR1 0x155
#define STAR2 0xba

A skilled programmer would directly see 0001 0101 0101 and 0000 1011 1010 with no effort. And he wouldn’t either need to decode the whole number to find out. Watching just the last hex digit of each STAR would give him (or us) a hint about which STAR is which.

It’s time we become that skilled programmer, don’t you think?

Hex to binary in da brain

Fortunately it is very simple to convert hex to binary in the brain. You simply have to understand that each hex number is made out of 4 bits since we need a max of 4 bits to represent the largest number in base 16 (which is the character ‘F’). So 0xF is 0b1111. (Notice that putting 0x in front denotes that the number is in hexadecimal represation and putting 0b denotes the binary representation accordingly.)

The procedure of binarizing a hex is simple:

1. Find the binary of each hex character
2. Place 0s in front of each binary (from above) so we always have 4 digits
3. Squeeze them all together as if they were strings

So for example:

F   is       1111
5   is       0110
FF  is  1111 1111
55  is  0110 0110
5F  is  0110 1111
F5  is  1111 0110

Hopefully you get the hang of it. The question is.. what happens if we have 0x102? This might seem tricky since we get three very simple binaries: 1, 0 and 10. But as I said, if you add the 0s in front before you squeeze them together, you should get the correct value – in this case 1 0000 0010!

Also you need to memorise a few things to be able and do all this. I have written the bare minimum below:

Binary     Decimal
   1    =    1
  10    =    2
 100    =    4
1000    =    8
1010    =    A
1111    =    F   

Then it’s quite easy to find in brain all the rest. For example to find the binary of B we can simply think that A is 1010, and that since B is just one step ahead, we add 1 to it and thus get 1011. Or to find 5 we simply have to add 1 to 4 which becomes 100+1=101. And so on.

This should also make it clear what the command chmod 777 in Linux does.

Big fat hex stars

The below is more like an exercise to test what we’ve learned. It should be rather simple to find the hex of the star below.

It might seem overwhelming, but the only thing you need to do is go in groups of 4s and write down each hex value.

Grouping in 4bit groups:

Decoding the above becomes 8388A08388A0 which is WRONG.

Row 1: 8 3..
..
Row 7: ..A 0 (and a remaining 1?!)

This was actually a trap to teach you the hard way that we should always start from the last digit. In this case in the end we are in a situation where we have an orphan digit 1. We can’t work with that since we need 4 digits to make a hex number.

The right way is to always start from the end. This is for all numbers no matter if they are represented in binary, octal, hex, decimal or whatever – as long as they are numbers, always start from the last digit and you’ll do fine. The reason is that when you finally get to the last number you can add as many zeros as you like (or need) without altering the value of the whole thing.

So the correct grouping is this (starting from bottom-right):

And then we just start from the bottom and get 1051141051141! Notice that in the end we again have a single 1 (at the top left this time), but this time we can add as many zeroes as we want since adding zeros in front of a number doesn’t change its vallue.

We can also validate what we got with Python:

>>> bin(0x1051141051141)
'0b1000001010001000101000001000001010001000101000001'

Traversing binary trees

It can be hard to remember how to traverse trees by name. It can also be hard to understand the difference between traversing in different ways(postorder, preorder, inorder, level order, etc). I try here to make it easier for the reader to understand the different ways of traversing a binary tree. Intuition is the key to remembering(and visualization ofcourse).

I use the words parents and children for the elements in the tree instead of nodes, root, branches and leafs. I think those words are easier to describe what is going on without getting anyone too confused.

Recursive traversal

I want to firstly point out that this method is commonly called “depth/height traversal”. However I use the word recursion as it seems more appropriate to me.
There are three ways to traverse a tree recursively. The only real difference between them is the order we visit the parent node.

• Pre order ——- here we visit the parent in the beginning
• In order ——— here we visit the parent second
• Post order —— here we visit the parent lastly

Because recursion is hard to grasp and even harder to have a visual insight of the goings, we will use a simple example to begin with. We start by examining, with all three methods, this binary tree:

Pre order

Here the path we follow to traverse is:

1. Parent
2. Left child
3. Right child

Applied to our example tree:

So if we use pre-order to traverse the example binary tree, then we would get the values(by order we visited them):

4, 2, 7

In order

Here the path we follow to traverse is:

1. Left child
2. Parent
3. Right child

Applied to our example tree:

So if we use in-order to traverse the example binary tree, then we would get the values(by order we visited them):

2, 4, 7

Post order

Here the path we follow to traverse is:

1. Left child
2. Right child
3. Parent

Applied to our example tree:

So if we use in-order to traverse the example binary tree, then we would get the values(by order we visited them):

2, 7, 4

The bigger picture

All this seems simple for a tree with only three nodes. But what happens when we have a huge tree with multiple nodes? Which node would we visit first?

The answers are rather simple but there are not good resources actually to give a good visualization of how a recursive traversing can look like.

So let’s start by taking this big tree:

The trick is to start by seeing see the whole structure as a tree with three nodes. In all three traversing methods we start with the node on the top of the tree: the root node. We thus see that as the parent in the beginning of the traversing. All the rest of the nodes on the left is considered the left child and everything on the right of it is considered the right child.

The bigger complex tree has been compressed to a three-node tree. We can now start traversing it with whatever -order traversal method we want.

I will demonstrate how I would go with in-order traversal. Read this line by line.

1. First I visit the left child: node b
   1. I now get into this subtree and visit the left child: 5
   2. Then I go to the parent node: 2
   3. Then I go to the right child: 9
2. Then I visit the parent: node a
   1. Here there is no parent or children. The only thing I can do is to return the value of the single node: 4
3. Then I visit the right child: node c
   1. I now get into this subtree and visit the left child: 6
   2. Then I go to the parent node: 7
   3. Then I go to the right child: 1

Note that 1, 2, 3 are traversal of the a, b, c nodes. All nested traversals are traversals for the nodes inside a, b and c.

An even more complicated tree could look like bellow. However exactly the same principles are valid.

This looks a lot like a fractal, doesn’t it? That’s because fractals are actually based on recursion. The only difference with this example is that the recursion is not endless but instead is made of three different levels: the outer level where we see three cyan circles, the level where we see three green circles and the level where we see three nodes(inside each green circle).

Traversal by Breadth

All methods discussed earlier are actually what is called traversal by height or depth. I am not so sure as to why we call it that instead of traversal by recursion as the real difference between depth traversal and breadth traversal is:

• Height traversal uses recursion.
• Breadth traversal is linear, going from one node to the next as you see the whole tree.

Traverse by level

Traversing by breadth or by level(which is more intuitive) is the simplest method to traverse a tree. You can see the nodes of an arbitrary tree grouped into different levels depending on how close to the root they are. In the bellow tree I have numbered the different level of nodes:

Node with value 4(the root) belongs to level 1. Nodes 2 and 7 belong to level 2. Lastly, nodes 5, 6, 9 and 1 belong to level 3.

To traverse the tree with traversal by level we just go from left to right on each level jumping from node to node like this:

Thus we traverse the nodes in this order:

4, 2, 7, 5, 9, 6, 1

If the tree would be more complicated or not complete(missing nodes at some level) then we just jump over the missing nodes and get to the next one. An example follows bellow.

The nodes visited in order:

4, 2, 7, 5, 9, 6, 1, 26, 13, 10, 8, 3