To derive the recurrence relation for the number of ternary strings of length $n$ that do not contain two consecutive $0$s or two consecutive $1$s, let's break it down:

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1. Understanding the Problem

A ternary string consists of the digits $\{0, 1, 2\}$. To meet the condition:

• Two consecutive $0$s (e.g., $00$) are forbidden.
• Two consecutive $1$s (e.g., $11$) are forbidden.
• There are no restrictions on the digit $2$.

We aim to determine the total number of such strings of length $n$ and develop a recurrence relation based on their formation.

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2. Base Cases

For small values of $n$:

• $n = 1$: Any single-digit ternary string satisfies the condition. Thus: $a_1 = 3 \quad \text{(strings: 0, 1, 2)}$

• $n = 2$: The strings of length 2 that satisfy the condition can be enumerated:

  • Start with $0$: Followed by $1$ or $2$ (not $0$): $01, 02$
  • Start with $1$: Followed by $0$ or $2$ (not $1$): $10, 12$
  • Start with $2$: Followed by $0, 1,$ or $2$: $20, 21, 22$

  Total: $6$ strings, so: $a_2 = 6$

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3. Forming the Recurrence Relation

For $n \geq 3$, consider how a valid ternary string of length $n$ can be constructed:

Case 1: The string ends in $0$

• The preceding digit cannot be $0$, so the first $n-1$ digits must end in $1$ or $2$.
• The number of such strings is $a_{n-1}$, since all valid strings of length $n-1$ can add $0$ without violating the rule.

Case 2: The string ends in $1$

• The preceding digit cannot be $1$, so the first $n-1$ digits must end in $0$ or $2$.
• The number of such strings is also $a_{n-1}$, for the same reasoning as Case 1.

Case 3: The string ends in $2$

• There is no restriction on the preceding digit for $2$, so any valid string of length $n-1$ can have $2$ appended to it.
• The number of such strings is $a_{n-1} + a_{n-2}$, because:
  • If the second-to-last digit is $2$, the remaining $n-2$ digits must form a valid string.
  • If the second-to-last digit is $0$ or $1$, the remaining $n-1$ digits must form a valid string.

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4. Combining the Cases

The total number of valid strings of length $n$ is the sum of all three cases: $a_n = 2a_{n-1} + a_{n-2}, \quad \text{for } n \geq 2$

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5. Final Answer

The recurrence relation for the number of ternary strings of length $n$ that do not contain two consecutive $0$s or two consecutive $1$s is: $a_n = 2a_{n-1} + a_{n-2}, \quad \text{with } a_1 = 3 \text{ and } a_2 = 6.$

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Would you like a step-by-step example to confirm the recurrence relation?

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Relative Questions:

1. What are the implications of removing restrictions on $2$ in the ternary string problem?
2. Can you derive the recurrence relation if all digits (0, 1, 2) have restrictions on consecutive occurrences?
3. How does this recurrence compare to similar binary string problems?
4. Can this recurrence be solved explicitly using characteristic equations?
5. What happens to the relation if $n = 0$ is considered?

Tip:

When solving recurrence relations, always verify the base cases against the recurrence formula to ensure consistency.