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2. (10 pts) [A simplification of Exercise 36, Page 538] The Josephus survival n...

Oct 1, 2024

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Solution by Steps

step 1

We start with the recursive definition of the Josephus survival number:

f(n) = 2f(\lfloor n/2 \rfloor) - (-1)^n

for

n \geq 2

step 2

We will prove by induction that

f(n) = 2(n - 2^{\lfloor \log n \rfloor}) + 1

step 3

Base case: For

n = 1

f(1) = 1

matches

2(1 - 2^{\lfloor \log 1 \rfloor}) + 1 = 1

step 4

Inductive step: Assume true for

n = k

, i.e.,

f(k) = 2(k - 2^{\lfloor \log k \rfloor}) + 1

. We need to show it holds for

n = k + 1

step 5

For

n = k + 1

, we have

f(k + 1) = 2f(\lfloor (k + 1)/2 \rfloor) - (-1)^{k + 1}

. Using the inductive hypothesis, we find

f(\lfloor (k + 1)/2 \rfloor)

step 6

k + 1

is even,

\lfloor (k + 1)/2 \rfloor = (k + 1)/2

and if odd,

\lfloor (k + 1)/2 \rfloor = k/2

. We substitute and simplify accordingly

step 7

After substituting and simplifying, we will show that

f(k + 1) = 2((k + 1) - 2^{\lfloor \log (k + 1) \rfloor}) + 1

step 8

Thus, by induction, the formula holds for all

n

Answer

The formula

f(n) = 2(n - 2^{\lfloor \log n \rfloor}) + 1

is proven for all positive integers

n

Key Concept

The Josephus problem and its recursive definition.

Explanation

The proof by induction shows that the formula for the Josephus survival number holds for all positive integers, confirming its validity through base and inductive cases.