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 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 represent the number of ternary strings of length that meet the condition of containing at least two consecutive identical symbols.
- The goal is to develop a recurrence formula that expresses in terms of and possibly other terms.
2. Exploiting symmetry:
Since the problem involves ternary strings (using the symbols ), we can use symmetry:
- For any string, it is equally likely to start with , , or . Thus, of the strings start with each symbol.
3. Constructing valid strings of length :
Case 1: The first two symbols are different.
- If the first symbol is , the second symbol is (where ), the remaining characters must include at least one pair of consecutive symbols.
- This reduces the problem to finding the number of such strings of length , which is represented by .
- Since there are two choices for (distinct from ), there are strings of this type.
Case 2: The first two symbols are the same.
- If the first two symbols are identical (e.g., ), then the rest of the string (of length ) can be any valid ternary string, whether or not it contains two consecutive symbols.
- The total number of ternary strings of length is , since each position can be filled in ways independently.
4. Combining cases to form the recurrence relation:
The total number of valid strings of length is the sum of the contributions from both cases: Simplify:
5. Validity of the recurrence relation:
- The relation holds for all , as the cases above cover all possibilities of how a string of length 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 (scaling symmetry).
- For identical symbols, the rest of the string can be freely chosen, contributing strings.
This step-by-step division leads to the recurrence:
Would you like to see specific examples of strings or further details about ternary string construction?
Expanding Questions
- How do we compute explicitly for small values of ?
- Can we generalize this recurrence to -ary strings (e.g., strings with symbols)?
- How does the solution change if we forbid strings with three consecutive identical symbols instead?
- What is the closed-form solution (if any) for ?
- 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
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