Math Problem Statement
"Find a recurrence relation for the number of bit strings of length n that contain three consecutive 0s." answer: " Let an be the number of bit strings of length n containing three consecutive 0’s. In order to construct a bit string of length n containing three consecutive 0’s we could start with 1 and follow with a string of length n−1 containing three consecutive 0’s, or we could start with a 01 and follow with a string of length n−2 containing three consecutive 0’s, or we could start with a 001 and follow with a string of length n − 3 containing three consecutive 0’s, or we could start with a 000 and follow with any string of length n − 3. These four cases are mutually exclusive and exhaust the possibilities for how the string might start. From this analysis we can immediately write down the recurrence relation, valid for all n ≥ 3: an = an−1 + an−2 + an−3 + 2n−3 ." explain answer
Solution
To derive and understand the recurrence relation for the number of bit strings of length that contain three consecutive 's, let's break it down step by step:
Variables and Context
Let represent the number of bit strings of length that contain three consecutive 's. The goal is to find how relates to smaller cases of , leading to a recurrence relation.
Case Analysis
To construct a bit string of length that contains three consecutive 's, consider the different ways the string can start:
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Start with "1":
- The remaining string (of length ) must still contain three consecutive 's.
- There are such bit strings.
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Start with "01":
- The remaining string (of length ) must still contain three consecutive 's.
- There are such bit strings.
-
Start with "001":
- The remaining string (of length ) must still contain three consecutive 's.
- There are such bit strings.
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Start with "000":
- The first three consecutive 's are already present, so the remaining bits can be any bit string.
- There are such bit strings because each bit has 2 possible values (0 or 1).
Recurrence Relation
Since these four cases are mutually exclusive and cover all possibilities for the start of the bit string, we can add them up:
Validity of the Relation
This recurrence is valid for , since for , no bit string can contain three consecutive 's.
Intuition Behind the Formula
- The first three terms () account for strings that start with "1", "01", or "001" and rely on smaller subproblems to satisfy the condition of having three consecutive 's.
- The last term () directly handles the case where the string already begins with "000", ensuring any continuation of the bit string is valid.
This reasoning ensures the recurrence relation is both correct and comprehensive.
Let me know if you'd like further clarification or additional examples! Here are 5 follow-up questions to consider:
- How would you compute the initial conditions () for this recurrence relation?
- How would the recurrence relation change if the goal was to avoid three consecutive 's?
- Can you express explicitly (closed-form) using this recurrence?
- How does the recurrence relation differ if we wanted four consecutive 's instead?
- What is the computational complexity of solving using this recurrence iteratively?
Tip: When solving recurrence relations, always ensure initial conditions are correctly determined to match the recurrence logic.
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Math Problem Analysis
Mathematical Concepts
Recurrence Relations
Combinatorics
Bit Strings
Formulas
a_n = a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3}
Theorems
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Suitable Grade Level
Grades 11-12