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Month: August 2017

Taking Things for Granted: Elementary Properties of Groups

Taking Things for Granted: Elementary Properties of Groups

We take a lot of things for granted: electricity, gas at the pump, and mediocre coffee at the office. Many concepts in basic algebra are also taken for granted, such as cancellation of terms, and commutativity. This post will revisit some basic algebra (think solving for x), but with some of those things we took for granted removed.

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Theory of Coding, Episode 2: Maximum-Likelihood Decoding

Theory of Coding, Episode 2: Maximum-Likelihood Decoding

The introduction to coding theory in this post will now allow us to explore some more interesting topics in coding theory, courtesy of Pinter’s A Book of Abstract Algebra. We’ll introduce the notion of a code, informations, and parity check equations. Most communication channels are noisy to some extent, which means that a transmitted codeword may have one or more errors. If the recipient of a codeword attempts to decode a message, he needs to be able to determine first if there was a possible error in the code, and then correct it. This will lead us to one method called maximum likelihood decoding.

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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.
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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.
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