Math Problem Statement

"Find a recurrence relation for the number of ternary

strings of length n that contain two consecutive sym-

bols that are the same." answer: "a) Let a_n be the number of ternary strings that contain two consecutive symbols that are the same. We will develop a recurrence relation for a_n by exploiting the symmetry among the three symbols. In particular, it must be the case that a_n/3 such strings start with each of the three symbols. Now let us see how we might specify a string of length n satisfying the condition. We can choose the first symbol in any of three ways. We can follow this by a string that starts with a different symbol but has in it a pair of consecutive symbols; by what we have just said, there are 2a_{n-1}/3 such strings. Alternatively, we can follow the initial symbol by another copy of itself and then any string of length n-2; there are clearly 3^{n-2} such strings. Thus the recurrence relation is a_n = 3 \cdot ((2a_{n-1}/3) + 3^{n-2}) = 2a_{n-1} + 3^{n-1}. It is valid for all n \geq 2." explain answer

Solution

The problem involves finding a recurrence relation for the number of ternary strings of length nn that contain at least two consecutive symbols that are the same. Let’s break the explanation into clear steps:


1. Define the recurrence relation:

  • Let ana_n represent the number of ternary strings of length nn that meet the condition of containing at least two consecutive identical symbols.
  • The goal is to develop a recurrence formula that expresses ana_n in terms of an1a_{n-1} and possibly other terms.

2. Exploiting symmetry:

Since the problem involves ternary strings (using the symbols {0,1,2}\{0, 1, 2\}), we can use symmetry:

  • For any string, it is equally likely to start with 00, 11, or 22. Thus, an3\frac{a_n}{3} of the strings start with each symbol.

3. Constructing valid strings of length nn:

Case 1: The first two symbols are different.

  • If the first symbol is xx, the second symbol is yy (where xyx \neq y), the remaining n1n-1 characters must include at least one pair of consecutive symbols.
  • This reduces the problem to finding the number of such strings of length n1n-1, which is represented by an1a_{n-1}.
  • Since there are two choices for yy (distinct from xx), there are 2an13\frac{2a_{n-1}}{3} strings of this type.

Case 2: The first two symbols are the same.

  • If the first two symbols are identical (e.g., x,xx, x), then the rest of the string (of length n2n-2) can be any valid ternary string, whether or not it contains two consecutive symbols.
  • The total number of ternary strings of length n2n-2 is 3n23^{n-2}, since each position can be filled in 33 ways independently.

4. Combining cases to form the recurrence relation:

The total number of valid strings of length nn is the sum of the contributions from both cases: an=3(2an13+3n2)a_n = 3 \cdot \left( \frac{2a_{n-1}}{3} + 3^{n-2} \right) Simplify: an=2an1+3n1a_n = 2a_{n-1} + 3^{n-1}


5. Validity of the recurrence relation:

  • The relation holds for all n2n \geq 2, as the cases above cover all possibilities of how a string of length nn can start and satisfy the condition.

Summary of reasoning:

  • The recurrence is built by first dividing the strings based on the first two symbols (different or same).
  • For different symbols, the problem reduces to strings of length n1n-1 (scaling symmetry).
  • For identical symbols, the rest of the string can be freely chosen, contributing 3n23^{n-2} strings.

This step-by-step division leads to the recurrence: an=2an1+3n1.a_n = 2a_{n-1} + 3^{n-1}.


Would you like to see specific examples of strings or further details about ternary string construction?


Expanding Questions

  1. How do we compute ana_n explicitly for small values of nn?
  2. Can we generalize this recurrence to kk-ary strings (e.g., strings with kk symbols)?
  3. How does the solution change if we forbid strings with three consecutive identical symbols instead?
  4. What is the closed-form solution (if any) for ana_n?
  5. How do combinatorics and recurrence relations apply in real-world problems?

Tip

Recurrence relations are a powerful tool for describing problems where the solution for larger cases depends on smaller cases. Visualizing the first few cases often helps in understanding the underlying pattern.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Recurrence Relations
Ternary Strings

Formulas

Recurrence relation: a_n = 2a_{n-1} + 3^{n-1}
Total number of ternary strings: 3^n

Theorems

Principle of Inclusion-Exclusion
Basic Recurrence Relation Derivation

Suitable Grade Level

Grades 11-12 and early college