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Chapter 8: Problem 17

a) Find a recurrence relation for the number of ternary strings of length \(n\) that do not contain consecutive symbols that are the same. b) What are the initial conditions? c) How many ternary strings of length six do not contain consecutive symbols that are the same?

Short Answer

Expert verified
a(n) = 2 \times a(n-1) for n >= 3 with initial conditions a(1) = 3, a(2) = 6. Hence, a(6) = 96.

Step by step solution

01

- Define the Problem and Variables

Consider a ternary string where each character can be '0', '1', or '2'. Define a function, say, \(a(n)\), to represent the number of ternary strings of length \(n\) that do not have consecutive identical symbols.
02

- Analyze Base Case

When \(n = 1\), any single character from '0', '1', or '2' is valid, so \(a(1) = 3\) because there are three valid strings of length 1.
03

- Further Examine Base Cases

When \(n = 2\), the first and second characters must be different from each other. Since the first character has 3 choices and the second one has 2 choices (different from the first), there are \(3 \times 2 = 6\) valid strings of length 2. Thus, \(a(2) = 6\).
04

- Establish the Recurrence Relation

To form a valid ternary string of length \(n\), the last character must be different from the second-to-last character. Therefore, for any string of length \(n-1\), there are 2 possible choices for the \(n\)-th character (different from the \(n-1\)-th character). This implies that \(a(n) = 2 \times a(n-1)\) for \(n \geq 3\).
05

- Summarize the Initial Conditions

From our earlier analysis, we know the initial conditions are \(a(1) = 3\) and \(a(2) = 6\).
06

- Calculate the Number of Valid Strings for Length 6

Using the recurrence relation \(a(n) = 2 \times a(n-1)\), calculate as follows:\(a(3) = 2 \times a(2) = 2 \times 6 = 12\),\(a(4) = 2 \times a(3) = 2 \times 12 = 24\),\(a(5) = 2 \times a(4) = 2 \times 24 = 48\),\(a(6) = 2 \times a(5) = 2 \times 48 = 96\).Thus, \(a(6) = 96\).

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Key Concepts

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

Ternary Strings
When studying discrete mathematics, 'ternary strings' are sequences composed of three different symbols: typically '0', '1', and '2'. Unlike binary strings (which use two symbols), ternary strings can form more complex patterns.
Understanding ternary strings helps in topics like coding theory, where symbols can represent multiple states or conditions.
It's important to define how many valid patterns exist under given constraints. For example, calculating how many ternary strings of length n do not contain consecutive same symbols requires careful analysis and formulation of a recurrence relation.
Initial Conditions
In any recurrence relation, initial conditions are crucial as they serve as the base or starting point for generating further values. For the problem of finding ternary strings without consecutive identical symbols, we define initial conditions as follows:
- For a string of length 1, any single character '0', '1', or '2' is valid. Thus, there are 3 possible strings: {0, 1, 2}.
This gives us the first initial condition: \(a(1) = 3\)
- For a string of length 2, the two characters must differ. The first character has 3 choices, and the second has 2 (different from the first). Thus, there are 3 x 2 = 6 possible strings: {01, 02, 10, 12, 20, 21}. So, the second initial condition is: \(a(2) = 6\)
With these conditions, one can use a recurrence relation to find the number of valid strings for larger n.
Consecutive Symbols
Avoiding consecutive identical symbols in ternary strings leads us to interesting constraints. Given any valid ternary string of length \(n-1\), appending a valid character to form length \(n\) must respect the non-consecutive constraint.
Let's delve deeper:
- Consider x(n-1) = valid string of length (n-1).
- To form x(n), the nth character must differ from the (n-1)th character. Since there are 2 valid choices for this, we multiply the number of valid (n-1) strings by 2.
This forms the recurrence relation: \(a(n) = 2 \times a(n-1)\)
Therefore, using this relation and initial conditions, one can compute valid ternary strings avoiding consecutive symbols for any length n.

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Most popular questions from this chapter

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