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 nn that contain three consecutive 00's, let's break it down step by step:

Variables and Context

Let ana_n represent the number of bit strings of length nn that contain three consecutive 00's. The goal is to find how ana_n relates to smaller cases of nn, leading to a recurrence relation.


Case Analysis

To construct a bit string of length nn that contains three consecutive 00's, consider the different ways the string can start:

  1. Start with "1":

    • The remaining string (of length n1n-1) must still contain three consecutive 00's.
    • There are an1a_{n-1} such bit strings.
  2. Start with "01":

    • The remaining string (of length n2n-2) must still contain three consecutive 00's.
    • There are an2a_{n-2} such bit strings.
  3. Start with "001":

    • The remaining string (of length n3n-3) must still contain three consecutive 00's.
    • There are an3a_{n-3} such bit strings.
  4. Start with "000":

    • The first three consecutive 00's are already present, so the remaining n3n-3 bits can be any bit string.
    • There are 2n32^{n-3} 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:

an=an1+an2+an3+2n3a_n = a_{n-1} + a_{n-2} + a_{n-3} + 2^{n-3}


Validity of the Relation

This recurrence is valid for n3n \geq 3, since for n<3n < 3, no bit string can contain three consecutive 00's.


Intuition Behind the Formula

  • The first three terms (an1,an2,an3a_{n-1}, a_{n-2}, a_{n-3}) account for strings that start with "1", "01", or "001" and rely on smaller subproblems to satisfy the condition of having three consecutive 00's.
  • The last term (2n32^{n-3}) 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:

  1. How would you compute the initial conditions (a0,a1,a2a_0, a_1, a_2) for this recurrence relation?
  2. How would the recurrence relation change if the goal was to avoid three consecutive 00's?
  3. Can you express ana_n explicitly (closed-form) using this recurrence?
  4. How does the recurrence relation differ if we wanted four consecutive 00's instead?
  5. What is the computational complexity of solving ana_n 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