Linear Sieve Method for Prime Numbers

本文介绍了一种优化的素数筛选算法,通过避免重复删除合数来提高效率。利用每个合数都能分解为两个素数乘积的特性,仅需删除特定形式的合数即可找出所有素数。

Problem description:When we calculate for prime numbers with a sieve method,we delete so many numbers which is not necessary repeatly.For instance,there is a number which consists of 3x7x17x23,and we delete it when we delete the multiples of 3 as we delete the same number when we delete the multiples of 7,17,and 23.Please write a program that will not do these jobs more than once.
   
Thinking: There is a factorization theorem:every composite number could be decomposed into the multiplication of some primer numbers.Hence,the number can be decomposed in the form of (both of p andq are prime numbers and p < q).Therefore,what we need to remove is:,,...and,i=1,2,3.....The value of p and q is the numbers which are not removed currently and in a sequence from small to large.It is easy to write the program.

 

#include <stdio.h>
  #define MAX 1000
  #define null1 0
  #define NEXT(x)  x=next[x]
  #define REMOVE(x) {   previous[next[x]]=previous[x];   \
                        next[previous[x]]=next[x];       \
                    }
  
  #define INITIAL(n)  { unsigned long i;                    \
                        for(i=2;i<=n;i++)                   \
                            previous[i]=i-1,next[i]=i+1;    \
                        previous[2]=next[n]=null1;           \
                      }
  
  int main()
  {
      unsigned long previous[MAX+1]={0};
      unsigned long next[MAX+1]={0};
      unsigned long prime,fact,i,mult;
      unsigned long n;
      unsigned long count=0;
      
      scanf("%lu",&n);
  
      INITIAL(n); //initial the array
  
      for(prime=2;prime*prime<=n;NEXT(prime))
      {
          for(fact=prime;prime*fact<=n;NEXT(fact)) 
          {
              for(mult=prime*fact;mult<=n;mult*=prime) 
                  REMOVE(mult);
          }
      }
      for(i=2;i!=null1;NEXT(i))
          printf("%lu ",i),count++;
      printf("\nThe sum of the prime numbers is %lu\n",count);
  }

Reference material: C语言名题精选百则技巧篇 in Chinese.

\documentclass[12pt]{article} \usepackage{amsmath, amssymb} \usepackage{graphicx} \usepackage{geometry} \usepackage{setspace} \usepackage{caption} \usepackage{fancyhdr} \usepackage{titlesec} \geometry{a4paper, margin=1in} \onehalfspacing \titleformat{\section}{\large\bfseries}{\thesection}{1em}{} \titleformat{\subsection}{\normalsize\bfseries}{\thesubsection}{1em}{} \title{Sieve of Eratosthenes} \author{Zhang Hongwei} \date{December 2, 2025} \begin{document} \maketitle \begin{abstract} The Sieve of Eratosthenes is one of the oldest and most efficient algorithms for finding all prime numbers up to a given limit \( n \). This paper presents a comprehensive overview of its historical background, algorithmic principles, step-by-step execution, time and space complexity analysis, and comparisons with optimized variants such as the segmented sieve and Euler's linear sieve. With clear illustrations and mathematical derivations, this work aims to provide both beginners and practitioners with a solid understanding of this classical method in number theory and computer science. \end{abstract} \section{Introduction} Prime numbers have fascinated mathematicians for centuries due to their fundamental role in number theory and modern cryptography. An efficient way to generate all primes not exceeding a given integer \( n \) is essential in various computational tasks. Around 200 BCE, the Greek mathematician Eratosthenes devised an elegant algorithm—now known as the \textit{Sieve of Eratosthenes}—that systematically eliminates composite numbers from a list of integers, leaving only the primes. This algorithm remains widely used today due to its simplicity and effectiveness for small to medium ranges (typically \( n \leq 10^7 \)). It operates by iteratively marking the multiples of each discovered prime starting from 2. The unmarked numbers that remain are precisely the primes. Formally, given a positive integer \( n \), the goal is to output all prime numbers \( \leq n \). The time complexity is \( O(n \log \log n) \), and the space complexity is \( O(n) \), making it highly practical for many applications including education, primality testing pre-processing, and cryptographic key generation. \section{Algorithm Principle} A \textbf{prime number} is a natural number greater than 1 that has no positive divisors other than 1 and itself. A \textbf{composite number} has at least one additional divisor. The core idea of the Sieve of Eratosthenes is simple: \begin{quote} Start from the smallest prime, mark all its multiples as composite; move to the next unmarked number (which must be prime); repeat until \( \sqrt{n} \). \end{quote} Two critical optimizations make this method efficient: \subsection*{Why start marking from \( p^2 \)?} For a prime \( p \), any multiple \( k \cdot p \) with \( k < p \) must have already been marked by a smaller prime factor of \( k \). For example, \( 6 = 2 \times 3 \) is eliminated when processing \( p = 2 \). Thus, we begin marking from \( p^2 \), avoiding redundant operations. \subsection*{Why stop at \( \sqrt{n} \)?} Every composite number \( \mathrm{num} \) has at least one prime factor \( \leq \sqrt{\mathrm{num}} \). Suppose otherwise: let \( \mathrm{num} = a \times b \), where both \( a > \sqrt{\mathrm{num}} \) and \( b > \sqrt{\mathrm{num}} \). Then: \[ a \times b > \sqrt{\mathrm{num}} \cdot \sqrt{\mathrm{num}} = \mathrm{num}, \] which contradicts \( a \times b = \mathrm{num} \). Hence, at least one factor must be \( \leq \sqrt{\mathrm{num}} \). Therefore, checking primes up to \( \sqrt{n} \) suffices to eliminate all composites \( \leq n \). \section{Algorithm Steps} Let us illustrate the process for \( n = 100 \): \begin{enumerate} \item Create a list of integers from 2 to 100. \item Initialize a boolean array \texttt{prime[0..100]} with all values set to \texttt{True}. \item Set \texttt{prime[0]} and \texttt{prime[1]} to \texttt{False} (not primes). \item Let \( p = 2 \). If \texttt{prime[p]} is \texttt{True}, mark all multiples \( \geq p^2 = 4 \) as \texttt{False}. \item Find the next \( p \) such that \texttt{prime[p] == True}, and repeat step 4. \item Stop when \( p > \sqrt{100} = 10 \). \item All indices \( i \) with \texttt{prime[i] == True} are prime. \end{enumerate} % 第一幅图 \begin{figure}[h!] \centering \includegraphics[width=0.8\linewidth]{Flowchart.jpg} \caption{Visualization of the sieving process: multiples of 2, 3, 5, and 7 are progressively removed. Remaining numbers are primes.} \label{fig:sieve} \end{figure} As shown in Figure~\ref{fig:sieve}, after eliminating multiples of 2, 3, 5, and 7, the remaining unmarked numbers are the primes below 100. \section{Complexity Analysis} \subsection{Time Complexity} Each prime \( p \) requires approximately \( \frac{n}{p} \) operations to mark its multiples. The total number of operations is roughly: \[ T(n) \approx \sum_{\substack{p \leq \sqrt{n} \\ p\ \text{prime}}} \frac{n}{p} = n \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \cdots + \frac{1}{p_k} \right), \] where \( p_k \) is the largest prime \( \leq \sqrt{n} \). From analytic number theory, the sum of reciprocals of primes up to \( m \) grows asymptotically as \( \log \log m \). Setting \( m = \sqrt{n} \), we get: \[ \log \log \sqrt{n} = \log \left( \frac{1}{2} \log n \right) = \log \log n - \log 2. \] Thus, the overall time complexity is: \[ O(n \log \log n). \] \subsection{Space Complexity} We need a boolean array of size \( n+1 \), so the space complexity is \( O(n) \). \section{Variants and Comparison} \begin{table}[h!] \centering \caption{Comparison of prime-finding algorithms} \label{tab:comparison} \begin{tabular}{|l|c|c|l|} \hline \textbf{Algorithm} & \textbf{Time} & \textbf{Space} & \textbf{Use Case} \\ \hline Trial Division (per number) & $O(\sqrt{n})$ & $O(1)$ & Single prime check \\ Standard Sieve of Eratosthenes & $O(n \log \log n)$ & $O(n)$ & Small-medium scale \\ Segmented Sieve & $O(n \log \log n)$ & $O(\sqrt{n})$ & Large-scale ($n > 10^8$) \\ Euler’s Linear Sieve & $O(n)$ & $O(n)$ & High-performance batch \\ \hline \end{tabular} \end{table} \subsection{Improved Versions} \begin{itemize} \item \textbf{Segmented Sieve}: Divides the range into blocks, reducing memory usage to \( O(\sqrt{n}) \), suitable for very large \( n \). \item \textbf{Euler's Sieve (Linear Sieve)}: Ensures every composite is marked exactly once using the smallest prime factor, achieving \( O(n) \) time but with higher constant factors and poorer cache behavior. \end{itemize} \section{Conclusion and Outlook} The Sieve of Eratosthenes stands as a timeless example of algorithmic elegance and efficiency. Despite being over two millennia old, it remains relevant in modern computing. Its intuitive logic makes it ideal for teaching, while its performance suits real-world applications within moderate limits. \textbf{Recommendations:} \begin{itemize} \item Use standard sieve for $ n \leq 10^7 $ \item Apply trial division for individual checks \item Employ segmented sieve for $ n > 10^8 $ \item Consider Euler’s sieve for optimal speed in competitive programming \end{itemize} Future enhancements may involve parallelization across CPU cores or GPU acceleration. Nevertheless, the original sieve continues to serve as a foundational tool in algorithm design and number theory exploration. \end{document} 将代码进行修改 尽量减少AI痕迹
12-03
\documentclass[12pt]{article} \usepackage{amsmath, amssymb} \usepackage{graphicx} \usepackage{geometry} \usepackage{setspace} \usepackage{caption} \usepackage{titlesec} % 页面设置 \geometry{a4paper, margin=1in} \onehalfspacing % 调整章节标题格式 \titleformat{\section}{\large\bfseries}{\thesection}{1em}{} \titleformat{\subsection}{\normalsize\bfseries}{\thesubsection}{1em}{} % 论文信息 \title{Sieve of Eratosthenes} \author{Zhang Hongwei} \date{December 2, 2025} \begin{document} \maketitle \begin{abstract} This paper describes the Sieve of Eratosthenes, an ancient algorithm for identifying all prime numbers up to a given limit $ n $. The method works by iteratively marking the multiples of each prime starting from 2. We outline its procedure, justify key optimizations, analyze time and space complexity, and compare it with modern variants. A flowchart is included to illustrate the execution process. \end{abstract} \section{Introduction} Finding all primes less than or equal to $ n $ is a basic problem in number theory. While checking individual numbers for primality can be done by trial division, generating many primes efficiently requires a different approach. The Sieve of Eratosthenes, attributed to the Greek mathematician Eratosthenes in the 3rd century BCE, provides a simple and effective solution. It avoids expensive divisibility tests by eliminating composite numbers through multiplication: once a number is identified as prime, all of its multiples are marked as non-prime. Given a positive integer $ n $, the algorithm produces all primes $ \leq n $. Its time complexity is $ O(n \log \log n) $, and it uses $ O(n) $ memory. This makes it practical for $ n $ up to several million on modern computers. \section{Basic Idea} A prime number has no divisors other than 1 and itself. The sieve exploits the fact that every composite number must have at least one prime factor not exceeding its square root. Starting with a list of integers from 2 to $ n $, we proceed as follows: \begin{itemize} \item Mark 2 as prime, then mark all multiples of 2 greater than $ 2^2 = 4 $ as composite. \item Move to the next unmarked number (3), mark it as prime, and eliminate multiples starting from $ 3^2 = 9 $. \item Repeat this process for each new prime $ p $ until $ p > \sqrt{n} $. \end{itemize} After completion, all unmarked numbers are prime. \subsection*{Why start from $ p^2 $?} Any multiple of $ p $ less than $ p^2 $, say $ k \cdot p $ where $ k < p $, would have already been marked when processing smaller primes. For example, $ 6 = 2 \times 3 $ is removed during the pass for 2. Thus, there's no need to revisit these values. \subsection*{Why stop at $ \sqrt{n} $?} If a number $ m \leq n $ is composite, it can be written as $ m = a \cdot b $, with $ 1 < a \leq b $. Then: \[ a^2 \leq a \cdot b = m \leq n \quad \Rightarrow \quad a \leq \sqrt{n}. \] So $ m $ must have a prime factor $ \leq \sqrt{n} $. Therefore, scanning beyond $ \sqrt{n} $ is unnecessary. \section{Implementation Steps} Consider $ n = 100 $. We use a boolean array \texttt{prime[0..100]}, initialized to \texttt{true}. Set \texttt{prime[0]} and \texttt{prime[1]} to \texttt{false}. \begin{enumerate} \item Start with $ p = 2 $. Since \texttt{prime[2]} is true, mark $ 4, 6, 8, \dots, 100 $ as false. \item Next, $ p = 3 $ is unmarked. Mark $ 9, 15, 21, \dots $ (odd multiples $ \geq 9 $). \item $ p = 4 $ is already marked; skip. \item $ p = 5 $ is prime. Mark $ 25, 35, 45, \dots $ \item $ p = 7 $: mark $ 49, 77, 91 $ \item $ p = 11 > \sqrt{100} $, so stop. \end{enumerate} All indices $ i \geq 2 $ where \texttt{prime[i] == true} are prime. \begin{figure}[h!] \centering \includegraphics[width=0.7\linewidth]{Flowchart.jpg} \caption{Flowchart of the Sieve of Eratosthenes algorithm} \label{fig:flowchart} \end{figure} Figure~\ref{fig:flowchart} shows the control flow: initialization, loop over $ p $ from 2 to $ \sqrt{n} $, and marking multiples starting at $ p^2 $. \section{Complexity Analysis} \subsection{Time Usage} For each prime $ p \leq \sqrt{n} $, we mark about $ n/p $ elements. Summing over such $ p $: \[ T(n) \approx n \sum_{\substack{p \leq \sqrt{n} \\ p\ \text{prime}}} \frac{1}{p}. \] It is known from number theory that the sum of reciprocals of primes up to $ x $ grows like $ \log \log x $. So: \[ \sum_{p \leq \sqrt{n}} \frac{1}{p} \sim \log \log \sqrt{n} = \log(\tfrac{1}{2}\log n) = \log \log n + \log \tfrac{1}{2} \approx \log \log n. \] Hence, total time is $ O(n \log \log n) $. \subsection{Memory Requirement} The algorithm requires one boolean value per integer from 0 to $ n $, leading to $ O(n) $ space usage. \section{Variants and Practical Considerations} \begin{table}[h!] \centering \caption{Common methods for generating primes} \label{tab:methods} \begin{tabular}{|l|c|c|l|} \hline Method & Time & Space & Remarks \\ \hline Trial division (single number) & $O(\sqrt{n})$ & $O(1)$ & Simple, slow for batches \\ Standard sieve & $O(n \log \log n)$ & $O(n)$ & Good for $ n \leq 10^7 $ \\ Segmented sieve & $O(n \log \log n)$ & $O(\sqrt{n})$ & Reduces memory usage \\ Linear sieve (Euler) & $O(n)$ & $O(n)$ & Faster in theory, more complex \\ \hline \end{tabular} \end{table} In practice, the standard sieve performs well due to good cache behavior and low constant factors. For very large $ n $, segmented versions divide the range into blocks processed separately. The linear sieve improves asymptotic time by ensuring each composite is crossed off exactly once using its smallest prime factor, but the overhead often negates benefits for moderate inputs. \section{Conclusion} The Sieve of Eratosthenes remains a fundamental tool in algorithm design. Its simplicity allows easy implementation and teaching, while its efficiency supports real-world applications in cryptography, number theory, and data processing. Although newer algorithms exist, the original sieve continues to be relevant—especially when clarity and reliability matter more than marginal speed gains. With minor improvements, it scales well within typical computational limits. \section{References} \begin{thebibliography}{9} \bibitem{knuth} Donald E. Knuth. \textit{The Art of Computer Programming, Volume 2: Seminumerical Algorithms}. 3rd Edition, Addison-Wesley, 1997. ISBN: 0-201-89684-2. (See Section 4.5.4 for discussion of prime number sieves.) \bibitem{hardy} G. H. Hardy and E. M. Wright. \textit{An Introduction to the Theory of Numbers}. 6th Edition, Oxford University Press, 2008. ISBN: 978-0-19-921986-5. (Chapter 1 discusses prime numbers and includes historical notes on Eratosthenes.) \bibitem{pomerance} Carl Pomerance. \newblock “A Tale of Two Sieves.” \newblock \textit{Notices of the American Mathematical Society}, vol.~43, no.~12, pp.~1473–1485, December 1996. Available online: \url{https://www.ams.org/journals/notices/199612/199612FullIssue.pdf#page=1473} \bibitem{crandall} Richard Crandall and Carl Pomerance. \textit{Prime Numbers: A Computational Perspective}. 2nd Edition, Springer, 2005. ISBN: 978-0-387-25282-7. (A detailed treatment of sieve methods including Eratosthenes and segmented variants.) \bibitem{eratosthenes-original} Thomas L. Heath (Ed.). \textit{Greek Mathematical Works, Volume II: From Aristarchus to Pappus}. Harvard University Press (Loeb Classical Library), 1941. ISBN: 978-0-674-99396-7. (Contains surviving fragments and references to Eratosthenes’ work in ancient sources.) \end{thebibliography} \end{document} 修改错误 ,并且增加字数在2000字左右
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12-03
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