47 lines
No EOL
1.8 KiB
Text
47 lines
No EOL
1.8 KiB
Text
Algorithm to calculate the first 1000 prime numbers using the Sieve of Eratosthenes:
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1. Initialize a list `is_prime` of size `n` (where `n` is an upper bound estimate for the 1000th prime) with all entries set to `True`. The value of `n` can be estimated using the approximation `n ≈ 1000 * log(1000 * log(1000))`.
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2. Set `is_prime[0]` and `is_prime[1]` to `False` since 0 and 1 are not prime numbers.
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3. Start with the first prime number, `p = 2`.
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4. Mark all multiples of `p` (starting from `p^2`) as `False` in the `is_prime` list.
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5. Find the next number in the list that is still `True` and set `p` to this number.
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6. Repeat steps 4 and 5 until `p^2` is greater than `n`.
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7. Collect all indices `i` where `is_prime[i]` is `True`. These indices are the prime numbers.
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8. Return the first 1000 prime numbers from this list.
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Python implementation:
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python
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import math
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def sieve_of_eratosthenes(limit):
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is_prime = [True] * limit
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is_prime[0] = is_prime[1] = False
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for start in range(2, int(math.sqrt(limit)) + 1):
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if is_prime[start]:
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for multiple in range(start*start, limit, start):
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is_prime[multiple] = False
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return [num for num, prime in enumerate(is_prime) if prime]
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# Estimate the upper bound for the 1000th prime
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n = int(1000 * math.log(1000 * math.log(1000)))
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# Get the first 1000 primes
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primes = sieve_of_eratosthenes(n)[:1000]
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# Validate the list of primes
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assert len(primes) == 1000
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assert all(primes[i] < primes[i+1] for i in range(len(primes) - 1))
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assert all(all(primes[i] % primes[j] != 0 for j in range(i)) for i in range(1, len(primes)))
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print(primes)
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This algorithm efficiently calculates the first 1000 prime numbers using the Sieve of Eratosthenes. The validation checks ensure that the list contains exactly 1000 primes, that they are in ascending order, and that each number is indeed a prime. |