A 14-year-old wins award for program that calculates antiprime numbers

Originally published at: A 14-year-old wins award for program that calculates antiprime numbers | Boing Boing

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I’ll be 40 this month, and I’ve juuuust about nailed “pants first, then shoes” so, each at our own pace.

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Oh shit, really? Are you sure?

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One useful lesson I’ve learned from wearing minimalist footwear is that it’s way easier to violate this rule.

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You’re onto something…

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That’s why I like minimalist pantwear

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I didn’t know this definition. It means that 1 and 2 are highly composite numbers, despite neither being composite…2 is of course prime and 1 is a unit (since it needs to be ignored for unique factoring). I guess fair enough, it feels weird to specifically exclude them, but it’s weird to have them too.

Kudos of course to Sankaran.

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You need some of those pants with the expandable leg holes. It removes the frustration of having to start over when you accidentally start with the shoes.

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Oh yeah? Let’s just see what this 14yo has to say about Anti Optimus Prime!

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Isn’t this how Crisis on Infinite Earths started?

[quote=“DonatellaNobody, post:7, topic:208188, full:true”]

Also: culottes.

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Any of yall wanna earn 25 grand for your kid, or maybe just tired of jokes about shoes and pants? Here’s my solution:

``````import math
# find all antiprimes less than <limit>
def find_antiprimes(limit):
# list of primes
primes = []
# dictionary of composite numbers and their factors:
#   {n: {2: <power of two>, 3: <power of three>, etc}, etc}
composites = {}
factor_count = 0
antiprimes = {}
for n in range (2, limit):
# factorize n
if len(primes) == 0:
primes.append(n)
composites[n] = {n: 1}
else:
for p in primes:
# if no divisors found by p > sqrt(n), n is a prime
if p > math.sqrt(n):
primes.append(n)
composites[n] = {n: 1}
break
if n % p == 0:
# n == p * d
composites[n] = {}
d = int(n/p)
# copy the factor dictionary
for c in composites[d]:
composites[n][c] = composites[d][c]
# update the factor count for p
if composites[n].get(p) == None:
composites[n][p] = 1
else:
composites[n][p] += 1
break
# count the factors
factors = 1
for c in composites[n]:
factors *= (composites[n][c] + 1)
if factors > factor_count:
# antiprime is found
factor_count = factors
antiprimes[n] = factors
print ("antiprime " + str(n) + " has " + str(factors) + " factors.")
``````
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I work in laboratory automation and I can appreciate the anti-prime nature of 24 / 48 / 96 / 384 / 1536.

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