Hacker News Reader

Comment by mindhash | Hacker News Reader

on a tangent, one way could be assign a number code to each alphabet. Add the numbers that occur in the strings. IF the sum matches, they are anagrams.

View on HN

I don't see how the sum would be unique to a particular combination of letters.

“ad” = “bc”

You’re right.

I briefly mused about just summing the ASCII codes for each letter in the strings. But quickly discarded the idea for this reason :)

Works if you assign prime numbers to each letter and multiply instead. So a=2, b=3, etc.

To get decent anagram lengths and complexity, implement the numbers as a dict of repetitions of primes, and implement the multiplication by summing the repetitions. ;-)

In which case you can just compare the dicts without performing the multiplication (which happens to be the costliest part for arbitrary-precision integers).

Exactly.

What if a=1, b=10, c=100 - etc? Assuming the strings were English words...

You would have to make sure the sum of any combination of all characters was unique. For example, if the code was the character number, a=1, b=2, etc, both "abc" and "bbb" would have the same sum.

So I think something silly like: character_code = len(string)*len(alphabet)^character_index should work.

You need to multiply the numbers, (and they must be prime) not add them.

a=1, b=2, c=3, ...

ac = 1 + 3 = 4 bb = 2 + 2 = 4

I guess the numbers should be primes... 4+1=3+2

10 = 5 + 5 = 7 + 3

that's right... even prime isn't enough

Prime is ok if it’s multiplied (there is one and only one prime factorization of a number).... but that can get to absurdly large numbers.

Still, consider the word ‘abe’ with a = 2, b = 3, c = 5, d = 7 and e = 11.

abe would then be 2^1 * 3^1 * 5^0 * 7^0 * 11^1. ‘abba’ would be 2^2 * 3^2. Each anagram would have a distinct value.