알고리즘 읽기 노트

Sieve of Eratosthenes

The Sieve of Eratosthenes finds all primes up to n by marking multiples of each discovered prime as composite.

유형알고리즘카테고리Math깊이초급일반적인 복잡도O(n log log n)

짧은 면접 답변

이 멘탈 모델부터 시작하세요

The Sieve of Eratosthenes finds all primes up to n by marking multiples of each discovered prime as composite.

간단한 설명

불필요한 전문용어 없이 동작 방식 이해하기

Start with all candidates marked prime.

For each prime p up to √n, mark multiples starting at p²; smaller multiples were already handled.

실행

알고리즘을 연속된 의사결정으로 읽기

State the mathematical identity/property.

Choose numeric representation and constraints.

Translate the property into an iterative/recursive computation and analyze precision.

기억해 둘 작은 예제

Read a tiny input by hand and write down the state after every important operation. For Sieve of Eratosthenes, the goal is to make the invariant visible rather than to memorize code.

읽기 쉬운 JavaScript 구현JavaScript

function primesUpTo(n) {
  const prime = Array(n + 1).fill(true);
  prime[0] = prime[1] = false;
  for (let p = 2; p * p <= n; p++) {
    if (!prime[p]) continue;
    for (let x = p * p; x <= n; x += p) prime[x] = false;
  }
  return prime.flatMap((isPrime, i) => isPrime ? [i] : []);
}

이 코드를 읽는 방법

문법을 외우지 마세요. 각 줄에서 바뀌는 상태를 따라가고 위에서 설명한 불변식과 연결하세요.

복잡도

O(n log log n)

Do not quote this complexity mechanically. Ask which operation or input dimension it describes, then derive the cost from the loops, recursion depth, data-structure operations and any preprocessing assumptions.

언제 사용하는가

적합한 경우

Use the direct mathematical property when it removes unnecessary brute force.

흔한 실수

보통 어디에서 문제가 생기는가

Ignoring floating-point precision or integer range.
Using a formula outside its valid assumptions.

면접에서 설명하는 방법

전문적인 답변 구조

Mention O(n log log n) time and O(n) space, and explain why marking can start from p².

문제→핵심 아이디어→불변식 / 상태→Big-O→트레이드오프

빠른 면접 체크

What problem does Sieve of Eratosthenes solve?

The Sieve of Eratosthenes finds all primes up to n by marking multiples of each discovered prime as composite.

복잡도에 대해 무엇을 언급해야 하나요?

Its typical cost here is O(n log log n), but a complete answer should connect that cost to the representation, input constraints and the dominant operations.

가장 흔한 실수는 무엇인가요?

Ignoring floating-point precision or integer range.

계속 읽기Number theoryNumerical precisionRecurrences

function primesUpTo(n) { const prime = Array(n + 1).fill(true); prime[0] = prime[1] = false; for (let p = 2; p * p <= n; p++) { if (!prime[p]) continue; for (let x = p * p; x <= n; x += p) prime[x] = false; } return prime.flatMap((isPrime, i) => isPrime ? [i] : []); }