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### Browsed byTag: algebra

Mailbox Answers: Calculating New Parity After an Overwrite

## Mailbox Answers: Calculating New Parity After an Overwrite

I recently did some work for Mr. Howard Marks, an independent analyst and founder of Deep Storage on the subject of data protection and data loss. He e-mailed me with a question regarding calculating the new parity for a stripe of data on a storage system.

Let us consider the case of a $10+1$ RAID 5 set with a strip size of 64KB. When an application performs a 4KB write the system must:

1. Read the 64KB strip that contains the 4KB to be overwritten into a memory buffer
2. Modify the memory buffer with the data to be written
3. Read however much other data as would be needed to recalculate the parity strip for this stripe
4. Write the new data and new parity strip (Min 1 4KB write, 1 64KB write)

When we casually talk about this condition we say the RAID controller would need to read all 10 data strips in the stripe so it can XOR all ten together as part of step 4. I, however have been thinking about XOR and think that rather than requiring N+1 total I/Os I can get it down to three.

If $P$, the parity strip, already contains

$D_1 \oplus D_2 \oplus D_3 \oplus D_4 \oplus D_5 \oplus D_6 \oplus D_7 \oplus D_8 \oplus D_9 \oplus D_{10}$

and we’re overwriting part of $D_4$ couldn’t I [do the following]:

1. Read the existing $D_4$ into a variable $D'_4$.
2. Modify the data into $D_4$.
3. Calculate the changes as $D_4\oplus D_{4}'$ into variable $C$
4. Read the parity strip $P$
5. Calculate new parity strip as $P=P \oplus C$

In short, the answer is yes. We’re going to prove that here, because I think this is a great exercise to really show off the power of XOR. We’ve explored the operation here and began our discussion of coding theory by looking at maximum likelihood decoding. Let’s take a brief review of the XOR (or mod 2) operation:

## XOR is just modulo 2 addition

Let’s call generic binary words $D_{j}$. That is, each $D_{j}$ is simply a string of 1s and 0s of whatever length $l$ we feel like setting. So a binary word $D_{j} = d_{1}^{(j)}d_{2}^{(j)}\ldots d_{l}^{(j)}$ consists of binary bits given by the lowercase1 $d_{i}$.  XOR operation works bit-by-bit, and will be denoted by $\oplus$:

\begin{aligned}D_{j} \oplus D_{k} &= (d_{1}^{(j)}d_{2}^{(j)}\ldots d_{l}^{(j)})\oplus d_{1}^{(k)}d_{2}^{(k)}\ldots d_{l}^{(k)}\\&= (d_{1}^{(j)} \oplus d_{1}^{(k)})(d_{2}^{(j)}\oplus d_{2}^{(k)})\ldots (d_{l}^{(j)}\oplus d_{l}^{(k)})\end{aligned}

For a quick numerical example, suppose $D_{j} = 1011$ and $D_{k} = 0010$. Then

$$D_{j} \oplus D_{k} = 1011 \oplus 0010 = (1\oplus 0)(0\oplus 0)(1\oplus 1)(1\oplus 0)$$

Remember, too, that XOR is addition modulo 2, so we add the bits together, then divide by 2 and take the remainder. So, in particular, $1\oplus 1 = 0$ because 1+1 leaves a remainder of 0 when divided by 2. So,

$$D_{j} \oplus D_{k} = 1001$$

## Back to the question

Mr. Marks’ question can be stated mathematically in the following way (and I’m going to generalize it to any finite amount of words of any length XORed together, because that’s what mathematicians do):

Suppose $P = D_{1} \oplus D_{2} \oplus \ldots \oplus D_{j} \oplus \ldots \oplus D_{K}$ for some $K$, and one word (say $D_{j}$) is modified to give $D_{j}'$. Let $C$ be the binary word that represents the changes between $D_{j}$ and $D_{j}'$. That is,

$C = D_{j} \oplus D_{j}'$

(Note: XOR as an operation to identify differences in binary words is one of the more elegant features. If all the bits in two words are the same, then bitwise XORing would always give the binary word of all 0s. Only when two bits are different is their XOR result a 1.) Call $P'$ the new XOR sum with $D_{j}'$ substituted for $D_{j}$. So

$P' := D_{1}\oplus D_{2}\oplus \ldots \oplus D_{j}'\oplus \ldots \oplus D_{K}.$

Then does $P'= P \oplus C$?

### Numerical example

Whenever I’m seeking to prove a statement, I always “play” with an example. Now, simply finding an example that fits the statement doesn’t constitute proof. But playing with explicit numbers can often yield a way to prove the statement in general. Plus, we can deepen our understanding by really “seeing the theorem in action,” as opposed to just manipulating symbols via logic.

Let’s just test this with a sum of 3 words to make our lives easier. Let $D_{1} = 110$, $D_{2} = 010$, and $D_{3} = 101$. Then

$$P = D_{1} \oplus D_{2} \oplus D_{3} = 110 \oplus 010 \oplus 101 = 001$$

Now suppose we change $D_{2}$ to $D_{2}' = 101$. First, the new sum $P'$ is given by

$P' = 110 \oplus 101 \oplus 101 = 110$

Now, the change in $D_{2}$ and $D_{2}'$ is given by

$$C = 010 \oplus 101 = 111$$

Notice that all three positions changed. Each position that is different has a 1.Let’s see if $P \oplus C = P'$ $P \oplus C = 001 \oplus 111 = 110$

Success! Now, this doesn’t mean we’re done. One example doesn’t constitute proof. We have to show this is true for any finite number of binary words of any  length.

### Time to prove this statement is true

So, let $D_{1},...,D_{K}$ be binary words of generic length $l$. Choose one word $D_{j}$ and modify it to form the new word $D_{j}'$. Let $C = D_{j} \oplus D_{j}'$ denote the change vector. Then

\begin{aligned}P^{\prime}&=D_{1}\oplus D_{2}\oplus\ldots D^{\prime}_{j}\oplus\ldots D_{K}\end{aligned}

Now, let’s note that $C = D_{j}\oplus D^{\prime}_{j}$ tells us which positions changed between the two. Another way to look at it is that $C$ is the word you need to XOR with $D_{j}$ to get to the new $D^{\prime}_{j}$. That is, $D^{\prime}_{j} = D_{j} \oplus C$.2. Now, let’s plug in the new expression for $D_{j}'$ into $P'$:

\begin{aligned}P^{\prime}&=D_{1}\oplus D_{2}\oplus\ldots D^{\prime}_{j}\oplus\ldots D_{K}\\&=D_{1}\oplus D_{2}\oplus\ldots (D_{j} \oplus C)\oplus\ldots D_{K}\end{aligned}

Now, we know from this post that XOR is a commutative operation. Coupled with the associative property3, we can actually rearrange the order of the XORing to put $C$ last.

(You’ve done this with regular arithmetic all the time. $5 + 1 + 6 + 3$ can be rearranged and grouped into $(6+3+1) + 5 = 10 +5 = 15$. Commutativity and associativity combined allow this to happen with any operation.)

So,

\begin{aligned}P^{\prime}&=D_{1}\oplus D_{2}\oplus\ldots D^{\prime}_{j}\oplus\ldots D_{K}\\&=D_{1}\oplus D_{2}\oplus\ldots (D_{j} \oplus C)\oplus\ldots D_{K}\\&=(D_{1}\oplus D_{2}\oplus\ldots D_{j}\oplus\ldots D_{K})\oplus C\end{aligned}

But wait, that last thing in parenthesis is exactly $P$. Therefore,

\begin{aligned}P^{\prime}&=D_{1}\oplus D_{2}\oplus\ldots D^{\prime}_{j}\oplus\ldots D_{K}\\&=D_{1}\oplus D_{2}\oplus\ldots (D_{j} \oplus C)\oplus\ldots D_{K}\\&=(D_{1}\oplus D_{2}\oplus\ldots D_{j}\oplus\ldots D_{K})\oplus C\\&= P \oplus C\end{aligned}

Since we showed this for any generic number of binary words added together, and allowed to binary words to be any length, we’ve proven the statement.

## Bonus: Multiple modifications

What if we modified more than one word in our original sum $P$? It’s a pretty simple extension to run through the proof again with multiple modified words and show that if we have multiple $C^{\prime}$s, one for each modified word, we can perform the same substitution and show that the new $P^{\prime}$ is simply the old $P$ XORed with all of the change vectors. Alternatively, you could XOR all the change vectors first into one big change vector, then XOR that with your original $P$ to compute the new $P^{\prime}$. If you want to verify it for yourself formally, simply follow the same steps we did above for one modified word. You’ll just be performing the same type of substitution multiple times to account for each modification.

## Conclusion

Mr. Marks brought this up because he was seeking a way to compute the new parity strip in a more efficient way (with fewer arithmetic steps) than simply following the definition. You can absolutely “brute force” your way to calculating the new parity strip. Companies and startups are always concerned about “scalability”. Sure, you won’t notice the time different between 10 extra things added together. But what about 10 million? 1 billion? More than that? None of those numbers are infeasible for the amount of calculations we perform on data now. In those cases, the brute force method of simply using the definition starts to cause performance problems. It was worth taking the time to “be clever” and search for a nice, elegant way that cuts down the number of operations necessary to calculate a new parity strip. It took a little upfront work, but the result speaks loudly for itself.

Group Theory, XOR, and Binary Codes: Introducing Coding Theory

## Group Theory, XOR, and Binary Codes: Introducing Coding Theory

Binary codes and bit-wise operations are fundamental in computer science. Whatever device you are running today works because of binary codes, and the bit-wise operations AND, OR, NOT, and XOR. (A fun exercise, prove to yourself that each of these rules meets the definition of an operation. If you need a refresher, check out this post.) We can also look at binary codes and the bit-wise operations in a more algebraic way.

Let’s continue exploring abstract algebra via Pinter’s delightful algebra book. The previous post discussed operations on sets, and explored concatenation as an example. Here, we will build on this and look at a set $G$ combined with an operation $\ast$ that satisfies three axioms. This operation/set combination is a special algebraic object called a group, and has been studied extensively. We’ll look at a specific group to illustrate the concept: binary codes with the XOR operation.1 (Note: we also looked at modulo arithmetic on finite groups of integers here as another example of a group.)

## Some quick terminology and setup

We currently (until the advent of quantum or DNA computers, or something else entirely) transmit information by coding it into binary strings – strings of 1s and 0s. We call these binary words. 100111 is a binary word, as are 001, 1, and 10110. A general binary word can have any length we want.

When we send these binary words via some communication channel, we always run the risk of errors in transmission. This means we need to devise some way to detect and correct transmission errors in binary codes. One way to do this is the XOR operation.

Let $\mathbf{a} = a_{1}a_{2}\ldots a_{n}$ and $\mathbf{b} = b_{1}b_{2}\ldots b_{n}$  be binary words of length $n$, where $a_{i},b_{i} \in \{0,1\}$ for $i=1,...,n$. We define the XOR operation, which we will denote by the symbol $\oplus$ on each bit. For two bits $a_{i}, b_{i}$ in a word,

$$a_{i}\oplus b_{i} = (a_{i} + b_{i}) \bmod 2$$

(Now would be a good time to review your modulo arithmetic if you’ve forgotten.)

Then for two words $\mathbf{a}$ and $\mathbf{b}$, the XOR operation is done bit-wise (component by component). That is,

$$\mathbf{a} \oplus \mathbf{b} = (a_{1}\oplus b_{1})(a_{2}\oplus b_{2})\ldots (a_{n}\oplus b_{n})$$

As a quick example,

$$110010 \oplus 110001 = 000011$$

Notice that the result of the XOR operation shows us the positions in which $\mathbf{a}$ and $\mathbf{b}$ differ. Another way to look at it is if $\mathbf{a}$ was transmitted, and $\mathbf{b}$ was received, there was an error in the last two positions. We can call the result of XORing two binary words the error pattern.

## Showing Binary Words with the XOR operation is a group

Now we’ll look at the set of all binary words of length $n$, called $\mathbb{B}^{n}$ coupled with the operation XOR.2 We want to show that this set along with the XOR operation forms an algebraic structure called a group. A group is one of the most basic algebraic structures we can study. We will be exploring all sorts of things we can do and build with groups in future posts, so first we need to define a group.

### What’s a group?

From Pinter (1982)

A group is a set $G$ coupled with an operation $\ast$, denoted $\langle G, \ast \rangle$ that satisfies the following three axioms:

(G1: Associativity of the operation) The operation $\ast$ is associative. (See the previous post for a review of associativity.)

(G2: Existence of an identity element) There is an identity element inside the set $G$ that we will call $e$ such that for every element $g \in G$, $e\ast g = g\ast e =g$

(G3: Existence of an inverse for every element) For every element $g \in G$, there is a corresponding element $g^{-1} \in G$ such that $g\ast g^{-1} = g^{-1}\ast g = e$

All three properties were discussed in the previous post. It’s important to note that a group is a set and an operation. If we change one, then we either have a different group, or we lose the group classification. Real numbers under addition are a group, as are nonzero real numbers under multiplication, but those are two different groups. (This will be important when we get to rings and fields.) Integers under addition are also a group, but a different one than real numbers under addition.

### Let’s prove $\langle\mathbb{B}^{n}, \oplus \rangle$ is a group.

Showing a set and operation is a group is pretty algorithmic, in a sense. We just have to show that all three axioms are satisfied.3

(G1): Associativity

This one will be a bit tedious. We have to show associativity for words of any length $n$. But fear not. Since XOR of words is done bit wise, we can exploit that and first show associativity for binary words of length 1, then “scale it up”, if you will.

In this case, for words of length 1, we just have to brute-force it. We have to show that for any $a,b,c \in \mathbb{B}^{1}$, that

$$(a\oplus b) \oplus c = a \oplus (b \oplus c)$$

So,

\begin{aligned} 1\oplus (1\oplus 1) = 1 \oplus 0 = 1 &\text{ and } (1 \oplus 1) \oplus 1 = 0 \oplus 1 = 1\\ 1\oplus (1 \oplus 0) = 1\oplus 1 = 0 &\text{ and } (1\oplus 1) \oplus 0 = 0 \oplus 0 = 0\\ &\vdots\end{aligned}

Continue in this fashion until you have tried all combinations. (It does work out.)

Now that we have that this is true for words of length 1, we just use the definition of XOR operation on words of length 1 to “scale up” to words of length n. Since the operation is done component-wise, and it is associative on each component, it is associative on the whole word. We’ll show this formally now:

Let $\mathbb{a,b,c} \in \mathbb{B}^{n}$. So $\mathbf{a} = a_{1}a_{2}\ldots a_{n}$, $\mathbf{b} = b_{1}b_{2}\ldots b_{n}$, and $\mathbf{c} = c_{1}c_{2}\ldots c_{n}$. Then

\begin{aligned}\mathbf{a}\oplus (\mathbf{b} \oplus \mathbf{c}) &= a_{1}a_{2}\ldots a_{n}\oplus [(b_{1}\oplus c_{1})(b_{2}\oplus c_{2})\ldots (b_{n}\oplus c_{n})]\\ &= (a_{1} \oplus (b_{1} \oplus c_{1}))(a_{2} \oplus (b_{2} \oplus c_{2}))\ldots (a_{n} \oplus (b_{n} \oplus c_{n}))\\&= ((a_{1} \oplus b_{1})\oplus c_{1})((a_{2}\oplus b_{2})\oplus c_{2})\ldots ((a_{n}\oplus b_{n})\oplus c_{n})\\&= (\mathbf{a} \oplus \mathbf{b})\oplus \mathbf{c}\end{aligned}

That third equality holds because we already showed that XOR was bit-wise associative. The last equality just recalls what it means to XOR two binary words. With that, we have shown associativity of the XOR operation.

(G2): Existence of an identity element

When we want to show that a group has an identity element, we must actually find a candidate and show it meets the criterion. Here is (frustratingly, sometimes), where intuition and experience tend to play the biggest role. But, as my mother always told me when I was frustrated at my middle school math homework: “Make it look like something you’ve seen before”. XOR is bitwise addition, just with a twist (add then mod out by 2). So let’s start by considering the identity element for addition: 0. We’re looking at binary words of length $n$, so a good candidate for our identity element $e$ would be a string of $n$ 0s. But does it fit?

Any word

\begin{aligned}a_{1}a_{2}\ldots a_{n} \oplus 000\ldots 0 &= (a_{1}\oplus 0)(a_{2} \oplus 0)\ldots (a_{n} \oplus 0)\\&= a_{1}a_{2}\ldots a_{n}.\end{aligned}

You can check also that $0 \oplus \mathbf{a} = \mathbf{a}$, and thus our candidate is a match!4

(G3): Existence of an inverse element for every element in the set

This one is a little bit trickier. We need to be able to find any generic binary word, and show that there is another binary word such that when we XOR them together, we get the sequence of all 0s. (Computer science friends are ahead on this one.)

Think back to how we looked at the XOR operation as a form of error checking. If we XORed two words together, there was a 1 in every position in which they differ, and a 0 in every position in which they were identical.

Therefore, we come to the interesting conclusion that every element is its own inverse! If you XOR an element with itself, you will get a sequence of 0s. With our error checking interpretation, this makes perfect sense. We know the communication line transmits perfectly if we can XOR the sent and received word and get all 0s every time.

Conclusion: every element is its own inverse, and every element in the group is, well, in the group, so we have satisfied the third axiom.

Therefore, we have a group, fellow math travelers!

## Bonus round: showing that $\langle \mathbb{B}^{n}, \oplus \rangle$ is an abelian group

Notice that we are missing one property that all of you likely take for granted in regular arithmetic: being able to add and multiply in any order you like. That is, you take for granted that 2+3 = 3+2. This property is called commutativity, and groups with this bonus property are called abelian groups. Not all groups are abelian. Square matrices under matrix multiplication come to mind. However, we will show that $\langle \mathbb{B}^{n}, \oplus \rangle$ is an abelian group.

This actually can be done quite simply by exploiting two things we already know: modulo arithmetic and bit-wise operations. We know that XOR is basically bit wise addition modulo 2. Addition is absolutely commutative, and modding out by 2 doesn’t change that, since we add in the regular ol’ way, then divide by 2 and take the remainder.

That means that XORing binary words of length 1 is commutative. Now, since XORing is done bit-wise, which we noted while proving associativity, we can use that same reasoning again (in fact, it looks almost exactly the same), to conclude that XOR is indeed a commutative operation, and thus we have an abelian group.5

## Conclusion

We explored the definition of a group using a common operation and set from computer science. We needed this foundation to be able to study more. The next post in the coding theory and algebra series will take an in depth look at maximum likelihood decoding. This decoding process is a way to attempt to decode and correct errors in transmission when the communication channel is noisy (and most are, somewhat).

Abstract algebra is a powerful branch of mathematics that touches many more things than most of us realize. Now that we know binary words under the XOR operation is a group, we can start extending this and studying it as a structure, or skeleton, rather than focusing on the individual elements. The elements and operation are really just the drywall or stucco on a house. The algebraic structure is the skeleton of beams inside the house. We learn a lot more by studying the structure than the covering, which is why it is important to enter that abstract realm. We will uncover similarities in strange places, and applications we didn’t know were possible.

Concatenation as an Operation

## Concatenation as an Operation

Mathematics is like any activity, sport, or skill: it must be honed and practiced. With that in mind, I have been bolstering up my abilities in algebra with a fantastic book A Book of Abstract Algebra, by Charles C. Pinter.1.

As I go through the chapters, I will be posting and discussing selected relevant exercises that have applications. This first post is a short one on operations and concatenation. I discussed operators here, but will revisit the topic.

## What’s an operation?

First, we define an operation. From Pinter:

Definition (operation): An operation on a given set $A$, often denoted $\ast$ is a rule which assigns a pair of elements $(a,b)$ in $A$ to exactly one element denoted $a\ast b$ in $A$

Pinter stresses several things about this definition that should not just be read and glossed over. First, $a\ast b$ has to be defined for every single pair of elements in the set $A$. Some obvious examples include multiplication on integers, or addition on real numbers.

Some rules that try to sneak by, but are not operations including division on the real numbers. Why does this fail? Remember that 0 is a real number. I can divide 0 by anything I like, but can I divide anything by 0? No. A real number divided by 0 is not defined.2.

Secondly, the rule sends a pair of elements to exactly one element. Mathematically, this means a rule must be well defined to be an operation. It can’t take the same pair of elements and get two possible answers.

Lastly, a rule must be closed to be an operation. Whatever $(a,b)$ gets mapped to must also be in the set $A$. Here again, addition of real numbers is closed. Adding two real numbers yields a real number. Division also becomes our counterexample here, but let’s look at division on the space of just integers. Dividing 2 by 3 is certainly defined, but the result is not an integer; it’s a rational number.3.

## On to Concatenation

Operations are not just defined on numbers, and they don’t necessarily have to just be your standard arithmetic examples (addition, subtraction, multiplication). We can define operations on any space we want, provided we satisfy the definition. Here we will talk about concatenation as an operation on the space of sequences of symbols. If you’re a programmer, you probably assume we’re looking at binary sequences, but we concatenate English words too. (We just call these compound words, like “lifetime” or “backbone”.)

Let’s call $A$ the alphabet. If we are living in binary, then $A = \{0,1\}$. If we are speaking of English letters, the alphabet is $A = \{a,b,c,\ldots,z\}$. Now we can call $A^{\ast}$ the set of all sequences of symbols in the alphabet $A$.4.

Now we will define an operation on $A^{*}$: concatenation. Many of you already know what this is, so we’ll formally define it here:

If a and b are two sequences in $A^{*}$, where $\mathbf{a} = a_{1}a_{2}\ldots a_{n}$, and $\mathbf{b} = b_{1}b_{2},\ldots b_{m}$ (and each $a_{i}, b_{i}$ are drawn from the alphabet), then the concatenation of  and is given by

$$\mathbf{ab} = a_{1}a_{2}\ldots a_{n}b_{1}b_{2}\ldots b_{m}$$

That is, just stick b to the end of a. Quick example, if we live in binary, then for $\mathbf{a} = 1010$ and $\mathbf{b} = 001$, then

$$\mathbf{ab} = 1010001$$

Let’s also note that there is an empty sequence or NULL sequence $\lambda$ that consists of nothing at all. It’s pretty easy to check that concatenation meets the definition of an operation. Pinter asks us to do three simple things here:

### 1. Prove that the operation defined above is associative.

Associativity is the property of an operation that allows us to group however we like. Addition is associative:

$$(1+2) + 3 = 1+ (2+3)$$

In general, to show associativity of an operation, we need to show that for $\mathbf{a,b,c} \in A^{*}$,

$$(\mathbf{ab})\mathbf{c} = \mathbf{a}(\mathbf{bc})$$

First, we will define three generic sequences in our $A^{*}$. Let

\begin{aligned}\mathbf{a} &= a_{1}a_{2}\ldots a_{m}\\\mathbf{b} &=b_{1}b_{2}\ldots b_{n}\\\mathbf{c} &= c_{1}c_{2}\ldots c_{p}\end{aligned}

Notice here that I did not specify an alphabet, and that the length of $\mathbf{a,b}$ and $\mathbf{c}$ are different. We need to keep this as general as possible to prove the statement. Every restriction we place (same length, specific alphabet, etc) weakens the argument.

Now we will show associativity formally:

\begin{aligned}(\mathbf{ab})\mathbf{c}&=(a_{1}a_{2}\ldots a_{m}b_{1}b_{2}\ldots b_{n})c_{1}c_{2}\ldots c_{p}\\&= a_{1}a_{2}\ldots a_{m}b_{1}b_{2}\ldots b_{n}c_{1}c_{2}\ldots c_{p}\end{aligned}

Here we can see that we can put parentheses any way we want to, and it doesn’t affect the end word. That is,

\begin{aligned}(\mathbf{ab})\mathbf{c}&=(a_{1}a_{2}\ldots a_{m}b_{1}b_{2}\ldots b_{n})c_{1}c_{2}\ldots c_{p}\\&= a_{1}a_{2}\ldots a_{m}b_{1}b_{2}\ldots b_{n}c_{1}c_{2}\ldots c_{p}\\&=a_{1}a_{2}\ldots a_{m}(b_{1}b_{2}\ldots b_{n}c_{1}c_{2}\ldots c_{p})\\&=\mathbf{a}(\mathbf{bc})\end{aligned}

So we’ve concluded that the grouping in which you perform the operation on several elements doesn’t matter, and thus we have concluded that concatenation is associative.

### 2. Explain why the operation is not commutative.

commutative operation is an operation where the order of the elements in the operation doesn’t matter. That is, for an operation $\ast$, $a\ast b = b\ast a$. Examples of commutative operations are addition and multiplication on real numbers ( 2+5 = 5+2, for example). An example of a noncommutative operation is matrix multiplication. In general, the order in which you multiply matrices matters. As an explicit illustration,

$$\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}1& 1\\ 1 & 2\end{bmatrix} =\begin{bmatrix} 3 & 5 \\ 2 & 3\end{bmatrix}\neq \begin{bmatrix}2 & 3\\3 & 4\end{bmatrix}=\begin{bmatrix}1& 1\\ 1 & 2\end{bmatrix}\begin{bmatrix}1&2\\1&1\end{bmatrix}$$

We can see now why concatenation is not commutative. If you append $\mathbf{b}$ to $\mathbf{a}$, you will definitely not get the same result as appending $\mathbf{a}$ to $\mathbf{b}$.  “Townhome” and “hometown” are certainly two different words, but the two pieces: “home” and “town” are the same. Switching the order of concatenation gave a different result. Since I have produced a single counter example, concatenation cannot be commutative in general.5

### 3. Prove there is an identity element for this operation.

An identity element is an element that lives in $A^{*}$ that, when applied via the operation in question to any other element in the set, in any order, returns that other element. Formally, $\mathbf{e} \in A^{*}$ is an identity element for operation $\ast$ if and only if $\mathbf{ea} = \mathbf{ae} = \mathbf{a}$ for every single element in $\mathbf{A^{*}}$.

Some examples:

• the identity element for regular addition on the real numbers is 0. 0 + x = x + 0 = x for any real number x
• the identity element for multiplication on the real numbers is 1.

For concatenation, we seek an element that, when concatenated with any other element, in any order, returns that element. Proving existence has several strategies. The simplest one (in theory) is to find a candidate element, and show it meets the criteria.6. Here, since concatenation involves essentially “compounding” two things together, the only way we could keep an element “unchanged” is to concatenate nothing to it. The NULL element $\lambda$ (which is certainly a word, just a NULL word. Computer scientists are very familiar with this concept) fits our bill here. Try it out. If you take nothing and concatenate a word to it, you just get that word. Conversely, concatenating nothing to a given word doesn’t change the word. So the NULL element is our identity element.

## Conclusion

Operations and rules aren’t equivalent. A rule is anything we make up for some purpose, but an operation has to meet specific criteria to be called one. This post was meant to show that operations are not restricted to just the space of numbers that we deal with every day, but that we can look at spaces of objects, matrices, functions, words, sequences…anything we like. In addition, we can define operations, as long as we satisfy the definition. This represents the power of abstract algebra: we can take structure we thought only belongs to a very restrictive space (like numbers, or even matrices), and look at other things, like concatenation, in a different light.

Like Clockwork: Modulo Addition and Finite Groups of Integers

## Like Clockwork: Modulo Addition and Finite Groups of Integers

Modulo arithmetic always scared me in college. (Honestly, it’s still not something I can do as easily as integration or spotting pdfs hidden inside icky integrals.) Then when you add abstract algebra (or modern algebra as some call it) and group theory on top of it, the whole topic can become intimidating. Many books on algebra don’t slow down enough to really examine the examples they give, so I’m going dive in and explore some of these. Algebra is a powerful branch of mathematics, with ripple effects all throughout mathematics, engineering, experimental design, and computer science. It’s worth study and understanding, and the topics that scare us tend to lose their fangs after we get to know them a bit.

Most of us do modulo arithmetic more often than we think, especially if we look at an analog clock to tell time. For example, if I asked you to convert 14:00 from military time into “regular” time, you’d respond “2 pm” pretty quickly. But how did you do it?

You got that answer because you went all the way around the clock and then had 2 remaining. This is modulo addition — in this case $\mod 12$.

Modulo addition is regular addition with an extra couple steps: dividing by your “base” $n$ (on a clock $n = 12$) and then grabbing the remainder1. We write mathematically that for integers $a$,$b$, and $n$, that

$$a \equiv b \bmod n \Leftrightarrow b-a= k\cdot n \text{ for an integer } k$$

You read this as  $a$ is equivalent to $b$ modulo $n$ if $b-a$ is a multiple (integer multiple) of $n$. Another way to write it is to say that

$$a \equiv b \bmod n \Leftrightarrow b = k\cdot n + a \text{ for an integer } k$$

In other words, if $b$ can be expressed as a multiple of $n$ plus some extra amount $a$, then $a$ is equivalent to $b$ modulo $n$.

Back to our clock example, obviously $2 \neq 14$, but $2\equiv 14 \bmod 12$, because

$$14 = 1\cdot 12 + 2$$

Here, our $b, 14$ can be expressed as an integer multiple of 12 ($k=1$), plus some extra $a=2$, so we say that $2\equiv 14 \bmod 12$.

That $n$ matters. If I changed the $n$, that statement doesn’t hold true anymore. That is, 2  is not equivalent to $14 \mod 5$ because the remainder when you divide 14 by 5 is 4. Next, let’s study the set of possible remainders when we divide an integer by a given $n$

## The Finite Groups $\mathbb{Z}_{n}$

Since you’re super clever, I bet you’re already one step ahead. If we look at dividing any number by 12, the remainder can only be the numbers 0 – 11. Why? Well if the remainder is 12, then it’s a multiple of 12. The remainder of any integer, no matter how big it is, when dividing by $n$ has to be somewhere in the set $\{0,1,2,\ldots, n-1\}$

Let’s look at a few with $n=12$. As an exercise, verify each of these on your own:

$$1\equiv 13 \bmod 12 \text{ because } 13 = 1\cdot 12 + 1$$ $$0\equiv 24 \bmod 12 \text{ because } 24 = 2\cdot 12 + 0$$ $$11\equiv 35 \bmod 12 \text{ because } 35 = 2\cdot 12 + 11$$ $$0\equiv 48 \bmod 12 \text{ because } 48 = 4\cdot 12 + 0$$ $$3\equiv 123 \bmod 12 \text{ because } 123 = 10\cdot 12 + 3$$

That $k$ is telling us how many times we’ve gone around the clock. So we’re always stuck on the clock, no matter how many times we go around, which means that the remainder can’t be any larger than 11, or we’ve started another cycle around the clock. The set $\{0,1,2,\ldots, n-1\}$ forms an algebraic structure called a group when we pair the set with the operation modulo addition.

### So what’s a group?

group is a set paired with an operation that has to have 3 certain properties. The operation doesn’t always have to be addition, or modulo addition, and the set doesn’t have to be integers. Operations can be addition, multiplication, function composition, or some weird one you define. But to pair an operation with a set, we have to check to make sure some rules are followed:

• First, your operation has to fit the formal definition of an operation. An operation is defied mathematically as a rule that takes any two elements in a set $A$ and returns a unique element that is still in $A$. This means that you can’t create a rule that lets you escape the set. If my set is integers, and my rule is division, then it’s not an operation on that set because the result of $3/4$ isn’t an integer. I escaped the set. The uniqueness requirement is what it means to be well-defined. If my rule is that I take two integers $a$ and $b$ and return the number whose square is the product $ab$, then $2\cdot 8 = 16$ could return either 4 or -4. That’s not well-defined. The rule doesn’t guarantee that a unique element is returned.

Assuming our operation really is an operation (and modulo addition on the integers is. If you don’t believe me, try to prove it for yourself.), then we can move on. We have the set (integers modulo $n$) and the operation (modulo addition). Now, to meet the definition of a group, the set and operation must follow these three axioms

1. The operation is associative. That is, the grouping or order shouldn’t matter when I go to execute the operation on 3 elements. That is, $$[(a+b) + c] \bmod n \equiv [a + (b+c)]\bmod n.$$
2. There is an identity element in my set. In other words, there has to be some element in my set such that performing the operation (in this case, modulo n) on any other element in the set return that other element, and the order doesn’t matter. If you take some element and operate on it using the identity, or if you take the identity element and operate on it using that other element, you would always get that other element. 2
3. Every element in the set has to have an inverse. This means that for every element in my group, there has to be another element such that performing the operation on the element and its proposed inverse in any order returns the identity element we talked about in (2).3

We have a special way of writing the set of the integers modulo n: $\mathbb{Z}_{n} = \{0,1,\ldots,n-1\}$. We’ll do a specific example and verify that $\mathbb{Z}_{3}$ under modulo addition is a group.

### $\mathbb{Z}_{3}$ and operation tables

We can examine groups via operation tables, which are a nice visual. Think of your multiplication tables when you were young. The elements are placed on the rows and columns, and the entries are the result of applying the operation to the two elements. The row entry goes first, then the column entry. For $\mathbb{Z}_{3}$, we know the elements are $\{0,1,2\}$, so the operation table looks like this:

$$\begin{array}{l|rrr}\color{purple}{+}&\color{purple}{0} &\color{purple}{1} & \color{purple}{2}\\\color{purple}{0} & 0 & 1 & 2\\\color{purple}{1} & 1 & 2 & 0\\\color{purple}{2} & 2 & 0 & 1\end{array}$$

So, reading the table, the first entry is row 1 + column 1 ($\bmod 3\!$), which is

$$(0 + 0) \bmod 3 \equiv 0 \bmod 3$$

Another one is the row 2 + column 3:

$$(1 + 2) \bmod 3 = 3 \bmod 3 \equiv 0\bmod 3$$

Why? Well, after we add, we have to divide by 3 and take the remainder. As an exercise, make sure you can generate this table yourself. It gives good practice, and if you can comfortably generate this, then you’re doing great!

### Now, let’s verify that $\mathbb{Z}_{3}$ under modulo addition is a group

We know that modulo addition is an operation, so let’s see if combining it with $\mathbb{Z}_{3}$ fits the definition of a group. This means we have to verify the three axioms, starting with the first:

Axiom 1: Associativity:

We need to show the following to satisfy the first axiom:

$$[(a+b) + c] \bmod 3 \equiv [a + (b+c)]\bmod 3.$$

First, let’s take any three unspecified elements of $\mathbb{Z}_{3}$. Call the $a, b,$ and $c$. We don’t specify these because we have to keep things general in order to show that it doesn’t matter which three elements we pick.4 We are going to start with $[(a+b) + c]\bmod 3$ and manipulate it to show that it’s equivalent to $[a+(b+c)]\bmod 3$. To differentiate between regular old addition and modulo addition (modulo 3 in this case), I am going to use the notation $+_{3}$ when I mean addition modulo 3.

First, by the definition of modulo arithmetic,

$$[(a+b)+c]\bmod 3 = (a+b)\bmod 3 +_{3} c\bmod 3$$

The right hand side of the above is the same thing as using regular addition to add $a, b,$ and $c$ together first, then take that result modulo 3. 5 So we write that

\begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\end{aligned}

Look inside the parentheses. Now we’re back in the nice, safe world of regular addition, and we know we can group and add any ol’ way we want to inside those parentheses. So let’s do that. We know that

$$(a+b+c) = (a+b) + c = a + (b+c).$$

Grab that last equality.6 Then we can write

\begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\\&=[a+(b+c)]\bmod 3\end{aligned}

Hmm, now it looks like we can “descend” the same ladder we walked up to start with. If $[(a+b)+c]\bmod 3 = (a+b)\bmod 3 +_{3} c\bmod 3,$ then it also holds that $[a+(b+c)]\bmod 3 = a\bmod 3 +_{3} (b+c)\bmod 3.$7 \begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\\&=[a+(b+c)]\bmod 3\\&= a \bmod 3 +_{3} (b+c) \bmod 3\end{aligned}

Last step. We are now basically done. Use the definition of modulo arithmetic one more time to finish.

\begin{aligned}[(a+b)+c]\bmod 3 &= (a+b)\bmod 3 +_{3} c\bmod 3\\&= (a+b+c)\bmod 3\\&=[a+(b+c)]\bmod 3\\&= a \bmod 3 +_{3} (b+c) \bmod 3\\&\equiv [a+(b+c)]\bmod 3\end{aligned}

Axiom 2: Existence of an identity

This one is nice and easy here. We already did the work by calculating out that operation table. Which element when added to any other element returns the other element? Does the order matter? Use the multiplication table to verify that 0 is the identity element of $\mathbb{Z}_{3}$

Axiom 3: Every element has to have an inverse

We also did the setup for this one too. Look at each element (there are only 3). Each element needs to have a corresponding pair that you can modulo add to it to retrieve the identity element $0 \mod 3$. Clearly, 08 is its own inverse.9 $(0+0) \bmod 3 = 0\mod 3$. Also,

$$(1+2)\bmod 3 \equiv (2+1) \bmod 3 \equiv 0 \bmod 3.$$

So 1 and 2 are inverses of each other. Therefore, every element has an inverse, and we’ve proven that $\mathbb{Z}_{3}$ is a group under the operation modulo addition.

## I read all the way down to here, so what should I take from this?

All the things we just did were to learn a couple things:

1. There is more than one kind of addition. Modulo addition is very commonly used, with the typical example being clock reading. We also use it when we program computers (because machines operate in binary, which is actually $\mathbb{Z}_{2}$. It also shows up in many programming applications, like calendars (also use modulo addition) and the storage mapping function. (For a fantastic explanation of the “why I should care”, check out I Programmer’s article on SMF and the article on the mod function. )
2. Group theory can simplify our lives by giving us fewer things to study. If I want to study the effect dividing by 3 has on numbers, I can just study $\mathbb{Z}_{3}$, because dividing by 3 can only have, well, 3 possible options: a remainder of 0, 1, or 2. Now I don’t have to study all the integers in this situation to get the information I need.

This article will be just the start of various explorations of group theory. Here I gave an introduction to a specific type of group, the integers modulo n, because it will turn out that the structure and behavior of these groups are actually the same as other groups that look much harder to study, like symmetries of regular shapes. For example, $\mathbb{Z}_{3}$ has the same mathematical structure as all the lines of symmetry through an equilateral triangle.10. This has vast uses in chemistry and physics, studying the shapes of molecules.

Abstract algebra is a vastly useful subject, despite its reputation for living “in the clouds.” Since I myself need to deepen my understanding of this remarkable branch of mathematics, join me on the journey and stay tuned for more posts exploring notions in group theory. Perhaps as I become better friends with algebra, it will appear more useful and beautiful to you as well.