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Chapter 6: Problem 262

(Gray codes) There are \(2^{n}\) sequences of length \(n\) consisting of 0 s and 1 s. Prove that, for each \(n \geqslant 2,\) these sequences can be arranged in a cyclic list such that any two neighbouring sequences (including the last and the first) differ in exactly one coordinate position.

Short Answer

Expert verified
Binary sequences for any length \( n \geq 2 \) can be arranged cyclically in Gray code.

Step by step solution

01

Understanding Gray Codes

Gray code is a binary numeral system where two successive values differ in only one bit. We need to prove that for any length \( n \), we can arrange all possible binary sequences in such a way that each sequence differs from the next by exactly one bit.
02

Base Case for Induction

Start by verifying the base case, \( n = 2 \), where sequences are 00, 01, 11, and 10. These can be arranged cyclically so that each pair of successive sequences differs by one bit: 00 -> 01 -> 11 -> 10 -> 00.
03

Inductive Hypothesis

Assume for a given \( n = k \), sequences of length \( k \) can be arranged in a cyclic list in Gray code format. This will be our inductive hypothesis.
04

Constructing for n = k+1

To construct a Gray code sequence of length \( k+1 \), begin with the Gray code sequence of length \( k \) as the first half, and then mirror it while flipping the first bit to create the second half. This ensures that the boundary conditions as well as the cyclic condition are maintained.
05

Ensuring Cyclic Continuity

In the transition from the end of the constructed list to the beginning, ensure that the first sequence and last sequence differ by one bit. This holds because the mirroring process inherently flips only one bit through the change from 0 to 1 or vice versa.
06

Conclusion of Induction

Since the construction works for \( n = k+1 \) based on assumption for \( n = k \), by mathematical induction, the Gray code arrangement holds for any \( n \geq 2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binary Sequences
Binary sequences are fundamental to the idea of Gray codes. In binary sequences, each number is represented purely by the digits 0 and 1. These are the building blocks of binary representation in the computational world. For instance, sequences of length 2 are 00, 01, 10, and 11. The challenge in dealing with binary sequences for Gray codes is arranging them such that each subsequent sequence differs in exactly one digit position.

This concept is significant when you want to ensure minimal disruption in signal processing, coding, and digital electronics where only small variations are practical. Gray codes effectively handle changes, ensuring only a single bit changes at a time, unlike the typical binary sequence where multiple bits may change during successive numbers.
Inductive Hypothesis
The inductive hypothesis is a critical component in mathematical induction. It involves assuming a statement is true for some arbitrary case, let's say for a sequence of length \( n = k \). This assumption forms a strong basis to prove that the statement holds true for the next case, \( n = k+1 \).

In the context of Gray codes, the inductive hypothesis assumes we can cyclically order sequences of length \( n = k \). With this hypothesis, the task is then to show that this property will still hold true for a sequence one bit longer. By assuming the ability to correctly arrange shorter sequences, it lays the foundation for building and verifying the correctness of sequences of greater lengths.
Cyclic List
A cyclic list in Gray codes means that when you arrange all sequences linearly, the list wraps around to form a loop. Therefore, the last sequence in this list connects back to the first one.

Ensuring a cyclic arrangement implies proving that the transition from the end of the sequence back to the beginning still respects the Gray code condition: differing by only one bit. This is accomplished through a clever design using mirror inversion of the sequence accumulation process, ensuring continuity and adherence to the Gray code principal even at the cycle's point of closure.
Mathematical Induction
Mathematical induction is a methodical way to prove statements for all natural numbers. It involves two critical steps: first proving a statement for an initial case (usually \( n = 2 \) for Gray codes), and then using the inductive hypothesis to prove that if it holds for an arbitrary case \( n = k \), it must hold for the next case, \( n = k+1 \).

In proving Gray codes, mathematical induction ensures that, given a correct sequence arrangement for a smaller set, the arrangement holds as sequences grow larger. Through establishing the truth of this for small cases, and using induction to extend it outward, one can prove the correctness of Gray code sequence crafting comprehensively.

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Most popular questions from this chapter

Let \(x\) be any real number \(\geqslant-1\). Prove by induction that $$ (1+x)^{n} \geqslant 1+n x, \text { for all } n \geqslant 1 $$

(a) Prove by induction that \(n\) points on a straight line divide the line into \(n+1\) parts. (b)(i) By experimenting with small values of \(n,\) guess a formula \(R_{n}\) for the maximum number of regions which can be created in the plane by an array of \(n\) straight lines. (ii) Prove by induction that \(n\) straight lines in the plane divide the plane into at most \(R_{n}\) regions. (c)(i) By experimenting with small values of \(n,\) guess a formula \(S_{n}\) for the maximum number of regions which can be created in 3-dimensions by an array of \(n\) planes. (ii) Prove by induction that \(n\) planes in 3 -dimensions divide space into at most \(S_{n}\) regions.

When John von Neumann \((1903-1957)\) was seriously ill in hospital, a visitor tried (rather insensitively) to distract him with the following elementary mathematics problem. Have you heard the one about the two trains and the fly? Two trains are on a collision course on the same track, each travelling at \(30 \mathrm{~km} / \mathrm{h}\). A super-fly starts on Train \(A\) when the trains are 120 \(\mathrm{km}\) apart, and flies at a constant speed of \(50 \mathrm{~km} / \mathrm{h}-\) from Train \(A\) to Train \(B\), then back to Train \(A\), and so on. Eventually the two trains collide and the fly is squashed. How far did the fly travel before this sad outcome?

(a) Are 3,7,31,211 all prime? (b) Is \(2 \times 3 \times 5 \times 7 \times 11+1\) prime? (c) Is \(2 \times 3 \times 5 \times 7 \times 11 \times 13+1\) prime? We have already met two excellent historical examples of the dangers of plausible pattern-spotting in connection with Problem 118. There you proved that: "if \(2^{n}-1\) is prime, then \(n\) must be prime." You then showed that \(2^{2}-1,2^{3}-1,2^{5}-1,2^{7}-1\) are all prime, but that \(2^{11}-1=2047=23 \times 89\) is not. This underlines the need to avoid jumping to (possibly false) conclusions, and never to confuse a statement with its converse. In the same problem you showed that: "if \(a^{b}+1\) is to be prime and \(a \neq 1,\) then \(a\) must be even, and \(b\) must be a power of \(2 . "\) You then looked at the simplest family of such candidate primes namely the sequence of Fermat numbers \(f_{n}\) : $$ f_{0}=2^{1}+1=3, f_{1}=2^{2}+1=5, f_{2}=2^{4}+1=17, f_{3}=2^{8}+1=257, f_{4}=2^{16}+1 $$ It turned out that, although \(f_{0}, f_{1}, f_{2}, f_{3}, f_{4}\) are all prime, and although Fermat (1601-1665) claimed (in a letter to Marin Mersenne \((1588-1648))\) that all Fermat numbers \(f_{n}\) are prime, we have yet to discover a sixth Fermat prime! There are times when a mathematician may appear to guess a general result on the basis of what looks like very modest evidence (such as noticing that it appears to be true in a few small cases). Such "informed guesses" are almost always rooted in other experience, or in some unnoticed feature of the particular situation, or in some striking analogy: that is, an apparent pattern strikes a chord for reasons that go way beyond the mere numbers. However those with less experience need to realise that apparent patterns or trends are often no more than numerical accidents. Pell's equation (John Pell (1611-1685)) provides some dramatic examples. \- If we evaluate the expression " \(n^{2}+1\) " for \(n=1,2,3, \ldots,\) we may notice that the outputs \(2,5,10,17,26, \ldots\) never give a perfect square. And this is to be expected, since the next square after \(n^{2}\) is $$ (n+1)^{2}=n^{2}+2 n+1 $$ and this is always greater than \(n^{2}+1\). However, if we evaluate \(" 991 n^{2}+1 "\) for \(n=1,2,3, \ldots,\) we may observe that the outputs never seem to include a perfect square. But this time there is no obvious reason why this should be so - so we may anticipate that this is simply an accident of "small" numbers. And we should hesitate to change our view, even though this accident goes on happening for a very, very, very long time: the smallest value of \(n\) for which \(991 n^{2}+1\) gives rise to a perfect square is apparently

The sequence of Fibonacci numbers $$ 0,1,1,2,3,5,8,13, \ldots $$ is defined by its first two terms \(F_{0}=0, F_{1}=1,\) and by the recurrence relation: $$ F_{n+2}=F_{n+1}+F_{n} \text { when } n \geqslant 0 $$ Prove by induction that, for all \(n \geqslant 0\) $$ F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}}, \text { where } \alpha=\frac{1+\sqrt{5}}{2} \text { and } \beta=\frac{1-\sqrt{5}}{2} $$
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