CBMS-2024-07

CBMS-2024A-07

题目来源：Question 7 日期：2024-07-31 题目主题：CS-算法-迭代函数分析

解题思路

这个问题涉及迭代函数的分析，主要考察了对数函数的性质和迭代过程的理解。我们需要仔细分析每个子问题，利用函数迭代和对数的性质来解决。问题 (1) 和 (2) 为问题 (3) 的解答铺垫，因为它们帮助我们理解 $f^{\ast }$ 函数的行为。

对于问题 (3)，我们需要更仔细地分析 $f(f^{\ast }(n))$ 和 $f^{\ast }(f(n))$ 的关系。这需要我们对不同范围的 n 进行分类讨论，因为 $f(n)={\log }_{2}n$ 的行为在不同的 n 值范围内会有所不同。

Solution

1. Proving $f^{\ast }(n)=\lceil {\log }_{2}n\rceil$ when $f(n)=\frac{n}{2}$

To prove this, we need to show that $\lceil {\log }_{2}n\rceil$ is the minimum number of iterations needed to reduce $n$ to a value less than or equal to 1.

Let $k=\lceil {\log }_{2}n\rceil$. By definition, $2^{k-1}<n\le 2^k$.

Applying $f$ iteratively $k$ times:

$f^{(k)}(n)=\frac{n}{2^k}\le \frac{2^k}{2^k}=1$

This shows that $k$ iterations are sufficient to reduce $n$ to a value $\le 1$.

Now, let’s consider $k-1$ iterations:

$f^{(k-1)}(n)=\frac{n}{2^{k-1}}>\frac{2^{k-1}}{2^{k-1}}=1$

This shows that $k-1$ iterations are not enough to reduce $n$ to a value $\le 1$.

Therefore, $k=\lceil {\log }_{2}n\rceil$ is the minimum number of iterations needed, proving that $f^{\ast }(n)=\lceil {\log }_{2}n\rceil$.

2. Finding minimum $n$ such that $f^{\ast }(n)=5$ when $f(n)={\log }_{2}n$

We need to find the smallest $n$ such that it takes exactly 5 iterations of ${\log }_2$ to reduce $n$ to a value $\le 1$.

Let’s work backwards:

$f^{(5)}(n)\le 1$

$f^{(4)}(n)>1$

${\log }_2({\log }_2({\log }_2({\log }_2({\log }_{2}n))))\le 1$

$2^1=2\le {\log }_2({\log }_2({\log }_2({\log }_{2}n)))$

Continuing to apply $2^x$:

$2^2=4\le {\log }_2({\log }_2({\log }_{2}n))$

$2^4=16\le {\log }_2({\log }_{2}n)$

$2^{16}=65536\le {\log }_{2}n$

$2^{65536}\le n$

Therefore, the minimum value of $n$ that satisfies $f^{\ast }(n)=5$ is $2^{65536}$.

3. Comparing $f(f^{\ast }(n))$ and $f^{\ast }(f(n))$ when $f(n)={\log }_{2}n$

Let’s analyze these two expressions by considering different ranges of n:

1. $f(f^{\ast }(n))$: This applies ${\log }_2$ to the number of iterations needed to reduce $n$ to $\le 1$.
2. $f^{\ast }(f(n))$: This finds the number of iterations needed to reduce ${\log }_{2}n$ to $\le 1$.

Let’s consider different cases:

Case 1: $1<n\le 2$

• In this case, $f^{\ast }(n)=1$, because one application of ${\log }_2$ is sufficient to reduce $n$ to $\le 1$.
• $f(f^{\ast }(n))=f(1)=0$
• $f^{\ast }(f(n))=f^{\ast }({\log }_{2}n)=0$, because $0<{\log }_{2}n\le 1$
• Therefore, when $1<n\le 2$, $f(f^{\ast }(n))=f^{\ast }(f(n))=0$

Case 2: $2<n\le 4$

• In this case, $f^{\ast }(n)=2$, because two applications of ${\log }_2$ are needed to reduce $n$ to $\le 1$.
• $f(f^{\ast }(n))=f(2)=1$
• $f^{\ast }(f(n))=f^{\ast }({\log }_{2}n)=1$, because $1<{\log }_{2}n\le 2$
• Therefore, when $2<n\le 4$, $f(f^{\ast }(n))=f^{\ast }(f(n))=1$

Case 3: $n>4$

• In this case, $f^{\ast }(n)\ge 3$
• $f(f^{\ast }(n))={\log }_2(f^{\ast }(n))$
• $f^{\ast }(f(n))=f^{\ast }({\log }_{2}n)=f^{\ast }(n)-1$

For $n>4$, we can prove that $f(f^{\ast }(n))<f^{\ast }(f(n))$:

Let $k=f^{\ast }(n)$.

$f(f^{\ast }(n))={\log }_{2}k<k-1=f^{\ast }(n)-1=f^{\ast }(f(n))$

This inequality holds because for any integer $k\ge 3$, ${\log }_{2}k<k-1$.

Therefore, we can conclude:

• When $1<n\le 2$: $f(f^{\ast }(n))=f^{\ast }(f(n))$
• When $2<n\le 4$: $f(f^{\ast }(n))=f^{\ast }(f(n))$
• When $n>4$: $f(f^{\ast }(n))<f^{\ast }(f(n))$

In summary, $f(f^{\ast }(n))\le f^{\ast }(f(n))$ for all $n>1$, with strict inequality when $n>4$.

知识点

函数迭代对数性质天花板函数不等式

难点思路

问题 (2) 中，从 $f^{\ast }(n)=5$ 反推 $n$ 的最小值需要仔细思考迭代过程。我们需要从最后一步开始，逐步应用指数函数来得到最终结果。

解题技巧和信息

1. 在处理迭代函数时，正向和反向思考都很重要。有时从最终状态反推初始值会更容易。
2. 对数函数和指数函数的性质在这类问题中经常用到，熟悉它们的特性和关系很重要。
3. 在比较复杂函数的大小关系时，考虑极限情况和通用情况都很重要。
4. 在分析复杂函数关系时，分类讨论是一个强有力的工具。它可以帮助我们处理不同范围内的特殊情况。
5. 对于涉及对数和整数的问题，特别注意整数边界情况（如 2 的整数次幂）往往很重要。
6. 在证明不等式时，有时需要考虑具体的数值范围，而不仅仅是一般情况。

重点词汇

• iterative function 迭代函数
• ceiling function 天花板函数
• logarithm 对数
• exponentiation 指数运算
• inequality 不等式

参考资料

1. Concrete Mathematics: A Foundation for Computer Science (2nd Edition) by Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Chapter 3: Integer Functions
2. Introduction to Algorithms (3rd Edition) by Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein, Chapter 3: Growth of Functions