Anthropic let the academics in, and everything fell apart within a week
Travis Wheeler
@wheelerlab.org
434 followers 230 following 324 posts
Professor, University of Arizona Algorithms, Machine Learning, Data Resources Computational Genomics, Drug Discovery, etc. wheelerlab.org
![To compute the number of pairs of primes with a difference that is also a prime, in subquadratic runtime, we need an efficient approach that avoids brute-force comparisons of all pairs. Here’s a step-by-step outline of an optimized method:
Optimized Method: Two-Pointer Approach with a Prime Set
1. Generate Primes Efficiently: Use a sieve (like the Sieve of Eratosthenes) to generate all primes up to the given range (e.g., 100,000). Store these primes in a list (primes) and a set (prime_set) for O(1) lookups.
2. Sort Primes (if needed): The list of primes is inherently sorted by generation, so no additional sorting is needed.
3. Two-Pointer Iteration:
• Use two pointers (i and j) to iterate through the sorted prime list.
• Initially, set i = 0 and j = 1.
• For every prime[i], increment j until prime[j] - prime[i] exceeds the largest possible difference.
• For every valid difference prime[j] - prime[i], check if the difference exists in prime_set (O(1) lookup).
• Count the pairs when the difference is a prime.
4. Complexity Analysis:
• Generating primes takes O(N log(log(N))) using the sieve.
• The two-pointer approach iterates through the list in O(n) time since each pointer moves only forward.
• Checking for primality of the difference is O(1) using the set.
Overall complexity: O(N log(log(N)) + n), where N is the maximum number in the range and n is the number of primes.](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:jz5tbtw7wukgafhuojdefslr/bafkreihlpzzykbfmlrygikelstmplm3y7rnsk7sfjof36kcaguwo2qf7ly)
![# Optimized approach: Use a set for fast lookup of primes and limit iteration
large_primes_set = set(large_primes)
# Iterate only once through pairs of primes
prime_pairs_with_prime_difference_large = [
(p, q) for i, p in enumerate(large_primes) for q in large_primes[i+1:]
if (q - p) in large_primes_set
]
# Number of such pairs
len(prime_pairs_with_prime_difference_large)](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:jz5tbtw7wukgafhuojdefslr/bafkreidywixayrap5pkjkcdbywwi43syi6mxgzry52xmt5i2qbgqggu2em)
![First this lists the code
# List of all primes between 1 and 100
primes = list(primerange(1, 101))
# Find pairs where the difference is also a prime
prime_pairs_with_prime_difference = [
(p, q) for p in primes for q in primes
if p < q and (q - p) in primes
]
prime_pairs_with_prime_difference
Then it gives the results:
1. (2, 5)
2. (2, 7)
3. (2, 13)
4. (2, 19)
5. (2, 31)
6. (2, 43)
7. (2, 61)
8. (2, 73)
9. (3, 5)
10. (5, 7)
11. (11, 13)
12. (17, 19)
13. (29, 31)
14. (41, 43)
15. (59, 61)
16. (71, 73)](https://cdn.bsky.app/img/feed_thumbnail/plain/did:plc:jz5tbtw7wukgafhuojdefslr/bafkreiec7xpn2pdcjc76jngfmsc6smlz34crd4dyxsrklhgw6dyvsu7ipi)
