(In short, allowing all unicode characters makes this trivially easy... HN gets rid of some of the unicode weirdness though so I had to put it on Pastebin. I assume this matches the 'load bearing' criteria as I only use variable width spaces rather than additional characters.)
Not putting it on the blockchain though because I don't hate the planet.
> I assume this matches the 'load bearing' criteria as I only use variable width spaces
I think you might be stretching the rules slightly with that stuff.
Using "medium mathematical space" (U205F) instead of an ascii space seems pretty clearly to be using a special character for the purpose of manipulating the hash.
I'd much rather amulets be created not by messing with invisible codepoint differences, but rather by making more typographically distinct entries and picking out ones that are correct.
When I saw "generated", I thought "have a thesaurus suggest word swaps", not "manipulate invisible unicode characters"
Adding data to a blockchain doesn't use extra energy... As in, probably in the order of posting it to HN. The next block will be mined whether or not your amulet is in it.
More likely however is that the code you used to generate this amulet was worse for the environment.
#!/bin/bash
COUNT=1
while true ; do
COUNT=$((COUNT+1))
SHA=`echo "I adopted $COUNT puppies." | sha256sum | egrep 8888`
if [ "$SHA" != "" ] ; then
echo "$SHA I adopted $COUNT puppies."
fi
done
Sample output:
4f3e14ded07eda16a7fb57c42aebf1f97ef67acba4980bf472a8e188887c7726 - I adopted 6484 puppies.
57c3e9fd05f06a705206d38888b317b5eb596d35c6f308b909a440e2c8e391c8 - I adopted 6627 puppies.
4624f6bb4c7d3d8328888495c1422a1891b609b81abda3ff4d70bedd1c4f6cdd - I adopted 12631 puppies.
9c0be05548139888804d3799ad5729ea424b0487487a6b18803f1aa5746c1904 - I adopted 15663 puppies.
c60fc3f97f62ee07665325e235faa05fefc445aee7d356a9d81f58888c475147 - I adopted 15861 puppies.
Bonus points if the number of puppies adopted is a palindromic prime. [1]
This concept is philosophically interesting, because an infinitely-powerful deity could compose text with a hash value of all 8's, and anyone could verify that it was authored by someone with more computing power than the universe.
The fact that this is possible, but hasn't happened, may imply that we have never received a message from such an entity.
There was a religious apologetics claim, widely presented and debated in the 1990s, that there were statistical anomalies in the Torah to show that it was really written by God: https://en.wikipedia.org/wiki/Bible_code
However, the measurement technique was somewhat underdetermined, which is a problem akin to p-hacking (as you could look for many possible anomalies and only publish the ones that you successfully find).
> anyone could verify that it was authored by someone with more computing power than the universe
I never thought of this that way, but that's an excellent way to put it! (Although you also need to assume that the hash is extremely strongly preimage-resistant, which we don't have good mathematical techniques to prove today.)
One challenge for this: what is the canonical text encoding and canonicalization, and what is the canonical hash function?
This script [0] takes any amulet and adds various combinations of unicode white space to the end to create a rarer version without visually changing the original. It definitely goes against the spirit of white space being "load bearing", but was fun to make anyway.
If you accept the questionable premise that the people that write short summaries used in IMDB are writing poetry, then the summary of "Road Trip" (tt7328966):
Two friends set out on a road trip but one of them has a few questions he needs answered.
It didn't play for me, but this is fantastic and brings me back to the 90's / early 00's era of web design with midis and sound effects. I miss that web.
Music and sound effects add to the mood and aesthetic of presentation. It's not so strange, either. TikTok leans on it heavily and people are loving it.
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.
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
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)
[+] [-] Closi|4 years ago|reply
> Hello Hackernews, This is one. Whats so tough?
47d751f8964d717320b888888b81db0a8a35e79f528549f0e9dba13e0e4d6c4c
(In short, allowing all unicode characters makes this trivially easy... HN gets rid of some of the unicode weirdness though so I had to put it on Pastebin. I assume this matches the 'load bearing' criteria as I only use variable width spaces rather than additional characters.)
Not putting it on the blockchain though because I don't hate the planet.
[+] [-] TheDong|4 years ago|reply
I think you might be stretching the rules slightly with that stuff.
Using "medium mathematical space" (U205F) instead of an ascii space seems pretty clearly to be using a special character for the purpose of manipulating the hash.
I'd much rather amulets be created not by messing with invisible codepoint differences, but rather by making more typographically distinct entries and picking out ones that are correct.
When I saw "generated", I thought "have a thesaurus suggest word swaps", not "manipulate invisible unicode characters"
[+] [-] jamesrom|4 years ago|reply
More likely however is that the code you used to generate this amulet was worse for the environment.
[+] [-] SquibblesRedux|4 years ago|reply
[1] https://en.wikipedia.org/wiki/Palindromic_prime
[+] [-] PaulDavisThe1st|4 years ago|reply
7c31d274888888f28e37743d4c554e5fffdb7df35a1291bc83119e31743a233b - I adopted 43653 puppies. 5888888c44ce19109d9021679ca8023314cb06a0b48fe9f57f34d2f83ee5a971 - I adopted 321255 puppies. b4b4f1bdf07da6049c7aa172504e8f7464d13333508888889743ebd9136f0345 - I adopted 521564 puppies.
It's hard to buy the classification of rarity outlined in TFA.
[+] [-] PaulDavisThe1st|4 years ago|reply
88888eb74ed90df21487f33795e8af9f578b6b5e359db71e4ac0b2bd8b1115c8 - I adopted 34534 puppies. 7c31d274888888f28e37743d4c554e5fffdb7df35a1291bc83119e31743a233b - I adopted 43653 puppies. e88888508aeaea73f45925e9af3e95890ec15362ce0b0244f56d271f0ba8ce8e - I adopted 53527 puppies.
probably more coming soon ...
[+] [-] unknown|4 years ago|reply
[deleted]
[+] [-] jffry|4 years ago|reply
[+] [-] unknown|4 years ago|reply
[deleted]
[+] [-] davidcollantes|4 years ago|reply
[+] [-] wgetch|4 years ago|reply
8c29152333c388888888961e813457759ca87a8b54078779f8cbdefedf6401d2
You picture an infinite and essentially captivated dimension.
a8888ff4ddb3e7c0c38d30a8b4184f78d85888806f67ab73ca64c5ad51cae6b8
[+] [-] Drblessing|4 years ago|reply
[+] [-] SquibblesRedux|4 years ago|reply
[+] [-] puzzlingcaptcha|4 years ago|reply
so do we start with markov chain generators or jump straight to gpt-3?
[+] [-] LeoPanthera|4 years ago|reply
shuf -rn 5 /usr/share/dict/words | tr '\n' ' '
[+] [-] 0xbadcafebee|4 years ago|reply
[+] [-] p1mrx|4 years ago|reply
The fact that this is possible, but hasn't happened, may imply that we have never received a message from such an entity.
[+] [-] schoen|4 years ago|reply
However, the measurement technique was somewhat underdetermined, which is a problem akin to p-hacking (as you could look for many possible anomalies and only publish the ones that you successfully find).
> anyone could verify that it was authored by someone with more computing power than the universe
I never thought of this that way, but that's an excellent way to put it! (Although you also need to assume that the hash is extremely strongly preimage-resistant, which we don't have good mathematical techniques to prove today.)
One challenge for this: what is the canonical text encoding and canonicalization, and what is the canonical hash function?
[+] [-] amicin|4 years ago|reply
https://suricrasia.online/unfiction/basilisk/
[+] [-] Closi|4 years ago|reply
[+] [-] fwip|4 years ago|reply
[+] [-] unicodepepper|4 years ago|reply
[deleted]
[+] [-] blopker|4 years ago|reply
[0] https://github.com/blopker/amulet_finder
[+] [-] mfabbri77|4 years ago|reply
«...i'm dancing in a nutshell»
https://opensea.io/assets/0x2a2127753653f6210d26f5b470738bf1... (EPIC)
We do not do it...
https://opensea.io/assets/0x2a2127753653f6210d26f5b470738bf1... (LEGENDARY)
[+] [-] banana_giraffe|4 years ago|reply
[+] [-] banana_giraffe|4 years ago|reply
Two friends set out on a road trip but one of them has a few questions he needs answered.
Is an "epic" Amulet (7 8s in a row)
I think I need to stop now.
[+] [-] hectim|4 years ago|reply
https://twitter.com/Killeen__/status/1387194611068588034
didn't cheat with trailing spaces haha
> first and foremost
> I avocado toast
---------
> what the fuck is an NFT
> can someone please explain
> waste io
[+] [-] airstrike|4 years ago|reply
[+] [-] snypher|4 years ago|reply
[+] [-] fxtentacle|4 years ago|reply
2c27e38aaba003380ea55c4c674aa2f3f17743481c107dc388888884a689353b
This is an epic amulet!
[+] [-] metalelf0|4 years ago|reply
de la quartana, c’ha già l’unghie smorte, 07103f886e0b6cd70c39da95aa9048888232bd40cab66554775314ba54be086e
Maggiore aperta molte volte impruna 48888078da7fcab4688ed1cdb6672b9ad95a9c477a7863458837f376fd7b3797
‘Per che non reggi tu, o sacra fame 58b2731a241fa27f1e64c88887c44e3698185a427d7cf533cea6771122262745
la fede, sanza qual ben far non basta. 2ed62851d5e0f10d452c9830100872f9b0298888a63d5732bb95dd3f92fd6fd2
Tu dunque, che levato hai il coperchio 6389e4ba0c7c4b9fbb0087d9cc26ce8e35ba695e0dda56daa4b42d76838888ef
e per magrezza e per voler leggera. 99dccbc91227e117588884213598c4ae4c761537ad22285cc7b21ffa6b3a7bee
tosto che ’l vostro viso si nascose». 6d4924f638b4daf8888bb1bae05ecf0a5db66fc9d10ebaf9f93dbb809f19e52e
disfrenata saetta, quanto eramo bc4bf75c28572855818888a4d11c05c3aab85b4121b73f19599521705d5d3b6e
La pena dunque che la croce porse ffdf0fc7d4f37198888924624c85095676a0132398c8131322e8bd4b9414b3e6
ciascun di quei candori in sù si stese 4dd8888bfa1ea4ddea85e87565e77c3965334bfd86eaf704f84fd42c847aa9da
La forma general di paradiso bac7ba4fc96a04ba8409ba288889741a4463d14f45c9d288d95e380f8fad5ada
[+] [-] redcodenl|4 years ago|reply
[+] [-] cjlm|4 years ago|reply
8⃣ = = = = = = : 37f52a074751239fd88888116130302ddb5074212bee34ecff0f1db531f43199
edit: welp, the unicode was borked. It's the number 8, scorpion, amulet, eight-ball, octopus, stop sign, spider
[+] [-] nicksdjohnson|4 years ago|reply
[+] [-] xXx_uwu_xXx|4 years ago|reply
[+] [-] berniemadoff69|4 years ago|reply
[0] https://text.bargains/media/coin.mp3
[+] [-] echelon|4 years ago|reply
Music and sound effects add to the mood and aesthetic of presentation. It's not so strange, either. TikTok leans on it heavily and people are loving it.
I want a return to sound on the web.
[+] [-] dane-pgp|4 years ago|reply
Never enough time, only enough weighting
[+] [-] cjlm|4 years ago|reply
My rarest I've found is probably 🇸🇲 [1]
[0] https://sourcetarget.email/editions/29/ [1] https://opensea.io/assets/0x2a2127753653f6210d26f5b470738bf1...
[+] [-] wppick|4 years ago|reply
[+] [-] bawolff|4 years ago|reply
Doesn't sound that mythic to me.
[+] [-] jffry|4 years ago|reply
[+] [-] eihli|4 years ago|reply
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) ???
[+] [-] BelenusMordred|4 years ago|reply
[+] [-] mpalmer|4 years ago|reply
[+] [-] unknown|4 years ago|reply
[deleted]