Problem 5 - Applied Discrete Structures

Question

Consider the following functions on the set of bit strings of length \(4 .\) In these definitions, addition is done modulo \(2,\) so that \(1+1=0 .\) Which of these functions has an inverse? For those that have an inverse, what is it? For those that don't explain why. (a) \(f_{1}\left(b_{1} b_{2} b_{3} b_{4}\right)=b_{2} b_{3} b_{4} b_{1}\) (b) \(f_{2}\left(b_{1} b_{2} b_{3} b_{4}\right)=b_{4} b_{3} b_{2} b_{1}\) (c) \(f_{3}\left(b_{1} b_{2} b_{3} b_{4}\right)=\left(b_{1}+b_{2}\right)\left(b_{2}+b_{3}\right)\left(b_{3}+b_{4}\right)\left(b_{4}+b_{1}\right)\) (d) \(f_{4}\left(b_{1} b_{2} b_{3} b_{4}\right)=b_{1}\left(b_{1}+b_{2}\right)\left(b_{1}+b_{2}+b_{3}\right)\left(b_{1}+b_{2}+b_{3}+b_{4}\right)\)

Step-by-Step Solution

Answer

Functions \(f_1\) and \(f_2\) have inverses. Functions \(f_3\) and \(f_4\) do not have inverses.

Step 1: Analyzing Function f_1

Function \(f_1\) rearranges the input bits by rotating the bit positions such that the first bit becomes the last, and the rest are shifted left by one position: \(f_1(b_1 b_2 b_3 b_4) = b_2 b_3 b_4 b_1\). To find the inverse, reverse the operation. This inverse function must translate \(b_2 b_3 b_4 b_1\) back to \(b_1 b_2 b_3 b_4\). Hence, the inverse function \(f_1^{-1}(b_2 b_3 b_4 b_1) = b_4 b_1 b_2 b_3\). Since every different input produces a unique output, \(f_1\) is bijective, and thus invertible.

Step 2: Analyzing Function f_2

Function \(f_2\) reverses the order of the bits, producing \(f_2(b_1 b_2 b_3 b_4) = b_4 b_3 b_2 b_1\). To find the inverse, reverse the order again. Therefore, the inverse function \(f_2^{-1}(b_4 b_3 b_2 b_1) = b_1 b_2 b_3 b_4\). Each input results in a unique output, so \(f_2\) is bijective and invertible.

Step 3: Analyzing Function f_3

Function \(f_3\) transforms the input bits using modulo 2 addition: \(f_3(b_1 b_2 b_3 b_4) = (b_1+b_2)(b_2+b_3)(b_3+b_4)(b_4+b_1)\). Consider two identical input sequences of all zeroes \((0000)\) and all ones \((1111)\); both result in the output \(0000\). Since different inputs can yield the same output, \(f_3\) is not injective (not one-to-one) and hence does not have an inverse function.

Step 4: Analyzing Function f_4

For function \(f_4\), compute each component: \(f_4(b_1 b_2 b_3 b_4) = b_1(b_1+b_2)(b_1+b_2+b_3)(b_1+b_2+b_3+b_4)\). If \(b_1=0\), the output is trivially \(0000\) irrespective of the other bits, as the expression is multiplied by zero. Hence, many different inputs will map to the same output, making \(f_4\) non-injective (many-to-one), without an inverse.

Step 5: Conclusion

Functions \(f_1\) and \(f_2\) have inverses, specifically \(f_1^{-1}(b_2 b_3 b_4 b_1) = b_4 b_1 b_2 b_3\) and \(f_2^{-1}(b_4 b_3 b_2 b_1) = b_1 b_2 b_3 b_4\). Functions \(f_3\) and \(f_4\) do not have inverses, as they are not injective.

Key Concepts

Inverse Functions

In mathematics, an inverse function essentially reverses the effect of the original function. Think of it as a way to "undo" the function's output back to the original input. For a function to have an inverse, it must be both injective (one-to-one) and surjective (onto). This ensures that each element in the output has a unique and distinct element in the input.
In simpler terms, if you perform a function and get an output, applying its inverse should lead you back to the initial input. For instance, if a function swaps the position of bits or rearranges them in some fashion, then its inverse will have to complete the reversing operation.
In the context of the exercise above, function \(f_1\) changes the order of the bit string, and by undoing this operation, we get back the original arrangement of bits, giving \(f_1\) its inverse. Similarly, \(f_2\) reverses the bit sequence, and its own reversal acts as its inverse. Functions without unique mappings in their input-output pairs do not possess inverses as demonstrated with \(f_3\) and \(f_4\).

Bit Strings

A bit string is a sequence composed of bits, which are binary digits, typically represented by the numbers 0 and 1. In digital contexts, they form the basis for data encoding and are extensively used in computer systems and networks.

• They can represent a variety of information, from numerical values to characters and data packets.
• In the exercises above, bit strings of length 4 are utilized, where each bit can be toggled between 0 and 1.

The length of bit strings determines their complexity and potential variation. For example, a 4-bit string can have 2⁴ or 16 possible unique combinations, ranging from 0000 to 1111.
The transformations applied in the functions in the exercise involve rearranging these bit strings, which illustrates how operations can alter the sequence but still stay within the boundaries of binary logic. These operations highlight the fundamental manipulation capabilities of bit strings.

Modulo Operations

Modulo operation is a mathematical operation that returns the remainder after division of one number by another. It's commonly expressed as \(a \, \text{mod} \, b\), where \(a\) is divided by \(b\), and the result is the remainder.
In binary contexts, especially with bit strings, modulo 2 operations are frequently used:

• With modulo 2, the operation wraps numbers back to a starting point once they reach 2. Hence, \(1+1\) in modulo 2 becomes 0, as illustrating a circle where after 1 comes 0.
• This properties can simplify various operations on binary numbers, particularly for parity bits, and it is utilized heavily in digital computing for error detection and correction.

In the exercise, functions like \(f_3\) utilize modulo addition to transform bits. This use of modulo highlights how binary arithmetic can create patterns or gain specific properties in bit manipulation, without the traditional carry flow of regular arithmetic.

Injective Functions

An injective function, also known as a one-to-one function, is one where each element of the function's domain maps to a unique element in its codomain. In other words, no two different inputs can map to the same output. This unique pairing ensures that every output corresponds to exactly one input.
For a function to be invertible, it must necessarily be injective. This is because if two different inputs produce the same output, it's impossible to discern which input led to that output when trying to apply an inverse function.
In the problem, the lack of injectivity in functions \(f_3\) and \(f_4\) becomes evident. They are unable to pair each input uniquely with an output — this is highlighted by different bit strings producing identical outputs. Without this one-to-one relationship, those functions cannot possess an inverse. Therefore, understanding injective properties is crucial for determining function reversibility.