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Comment by bawolff | Hacker News Reader

So 10 8's is mythic. That's only 2^40 (probably less since the 8's dont have to be in a specific position in the hash). Based on https://gist.github.com/epixoip/a83d38f412b4737e99bbef804a27... you should be able to bruteforce a mythic amulet in 48 seconds with a high end gpu from 2016.

Doesn't sound that mythic to me.

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I wrote a CUDA kernel to look for some. My 3080 managed 3 billion/second including population counting the 8s. After 48 seconds it had spat out three 10-eights and one 11-eights amulet:

  Here are some excellent hex digits: 251d5b059cefc6f3

which hashes to

  b282a850c34ebfbfe4d41797aebc567988888888888312220e834356a26b65fd

Of course, it gets a lot harder if the goal is poetry, and not just jamming hex chars into a string.

Can you please share how you got the 2^40 number? I've been trying to think of how to figure out the odds of these.

Odds of 4 hex 8's in a row given a 4 digit string is (1/16)^4.

Odds of 4 hex 8's in a row given a 5 digit string is number of ways to arrange 8s in the first 4 digits (1) times 16 possible 5th digits plus number of ways to arrange 8s in the last 4 digits (1) times 16 possible 1st digits, all divided by the number of possible arrangements (16^5)

So 8888X or X8888 is (2 * 16) / (16^5)?

And then 8888XX or X8888X or XX8888 is...

(16^2 + 16^2 + 16^2) / (16^6) ???

I did 10 8's (mythic!) at the beggining of the hash (2^4)^10 = 2^40.

I didnt account they could start at any position, so the actual number is probably

(64-10)/2^40 ≈ 1 / 2^35

We should also subtract strings longer than 10 that were double counted. However i think the probability of such things is negligible relative to 2^35.

If instead we were doing 4 8's i think it would be: (64-4)/((2^4)^4) = 60/2^16 ≈ 2^10

I've always been bad at calc probabilities so i may have messed this up.

> Odds of 4 hex 8's in a row given a 5 digit string is number of ways to arrange 8s in the first 4 digits (1) times 16 possible 5th digits plus number of ways to arrange 8s in the last 4 digits (1) times 16 possible 1st digits, all divided by the number of possible arrangements (16^5)

You're double counting "88888". But then again so am i.

Mythic is nine 8's. They didn't put a name for ten.

What's poetic about brute forcing it?

Nobody here has been making them by hand.

Sure, but the worthwhile ones are actual prose.

If you can brute force 9 8's from only English words (or whichever language), I'd still call that impressive.

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