4 ms·

Twitter Bot Finds Anagrams of Twitter Statuses

113 points by nateguchi 12y ago

Igglyboo 12y ago
"Mustache got thicker" vs "git checkout hamster" This bot is pretty funny. Anyone know how it works? I'm assuming it just sorts the string and puts it in a hashmap/table and looks for collisions.
- breyten 12y ago
  The source is on github : https://github.com/cmyr/anagramatron https://github.com/cmyr/anagramatron
- subir 12y ago
  This particular one had me in splits (:
- duiker101 12y ago
  I found "another math genius" vs "He ain't smart enough." pretty funny
- solutionyogi 12y ago
  I thought this was deep: Go All out or Die trying. v/s R u really going to do it?
- MiguelVieira 12y ago
  Probably just takes tweets, canonicalizes them, and then hashes them based on a 26-length vector of character counts. For every new tweet, it looks for old tweets with the same character count.
bussiere 12y ago
Nice one :) I love when people use programming to play with words. Sad it's english only. I may work on a french version. But really nice idea.
- cmyr 12y ago
  hey, author here: It's english only mostly because of volume, and because I review results. Making a french language version would mostly just be a matter of hosting. If you're interested let me know, I'd love to help you out.
  - bussiere 12y ago
    maybe as a side project. But i have some friends who makes rap and there is some diamonds that i've found with your bot : = I want to see this world change. Let's see what I can do right now = And : = you have destroyed me do you deserve my hate? = i keep it in mind but it will not be before six months. I will make a pull or notify with github, i code also in python. In french we have some software to find rimes. Putting the finding in a database could be a nice addition, i could help to compose text. You have my admiration for the idea and the execution ...
madcaptenor 12y ago
I'm a bit surprised that there are anagrams to be found. It's easy to find them if they exist, but there's no guarantee at all that there actually should be collisions.
- DanBC 12y ago
  I'm not a statistician. Is it really that surprising? English has plenty of redundancy; Twitter statuses have limited length. What's surprising to me is the niceness of the found anagrams. "another math genius" / "he ain't smart enough".
  - madcaptenor 12y ago
    I am a statistician. Maybe I should sit down and actually do some calculations.
    - sillysaurus3 12y ago
      Please do. I'd be interested.
  - denzil_correa 12y ago
    > What's surprising to me is the niceness of the found anagrams. That's because they are manually curated [0] Q: Is this manually curated? A: Mostly for issues of volume ( there are a lot of variations of 'goooood mooornnniinng!', there are a lot of spam bots posting subtely different versions of the same message, etc) the bot doesn't automatically post every anagram it finds. Essentially there's an iphone client that reviews matches, which are manually approved or rejected. [0] https://github.com/cmyr/anagramatron https://github.com/cmyr/anagramatron
- sirclueless 12y ago
  It's actually extremely likely. The chance that any two statuses are anagrams is miniscule, and even the chance that a particular status has an anagram among all other statuses is probably small, but the chances that there are no collisions at all is tiny. See a description of the Birthday Paradox[1] for the mathematics behind this. For example, if you put 70 people in a room, there is a 99.9% chance that two people share a Birthday. [1]: http://en.wikipedia.org/wiki/Birthday_problem http://en.wikipedia.org/wiki/Birthday_problem
fjcaetano 12y ago
This is delightfully ironic
- bengali3 12y ago
  Oily Shirted Filch Linguist
joopxiv 12y ago
I find it interesting that they're manually approving the hits, because, as they indicate, most hits are (nearly) identical. It shouldn't be too difficult to solve this automatically though. Identical hits can be discarded very easily. The ones that only have a few words or letters reversed can be detected with some kind of similarity algorithm.
- jcampbell1 12y ago
  I had a look at the source code, and it does quite a bit of filtering, particularly around making sure the words are unique, and there is a primitive character comparison algorithm. The code could be simplified by using Python's set() and improved by doing a copy'n'paste on a Levenshtein function.
  - cmyr 12y ago
    oh hey yea that would've been useful. ^_^

113 points by nateguchi 12y ago

"Mustache got thicker" vs "git checkout hamster" This bot is pretty funny. Anyone know how it works? I'm assuming it just sorts the string and puts it in a hashmap/table and looks for collisions.

breyten 12y ago
The source is on github : https://github.com/cmyr/anagramatron https://github.com/cmyr/anagramatron
subir 12y ago
This particular one had me in splits (:
duiker101 12y ago
I found "another math genius" vs "He ain't smart enough." pretty funny
solutionyogi 12y ago
I thought this was deep: Go All out or Die trying. v/s R u really going to do it?
MiguelVieira 12y ago
Probably just takes tweets, canonicalizes them, and then hashes them based on a 26-length vector of character counts. For every new tweet, it looks for old tweets with the same character count.

bussiere 12y ago

Nice one :) I love when people use programming to play with words. Sad it's english only. I may work on a french version. But really nice idea.

cmyr 12y ago
hey, author here: It's english only mostly because of volume, and because I review results. Making a french language version would mostly just be a matter of hosting. If you're interested let me know, I'd love to help you out.
- bussiere 12y ago
  maybe as a side project. But i have some friends who makes rap and there is some diamonds that i've found with your bot : = I want to see this world change. Let's see what I can do right now = And : = you have destroyed me do you deserve my hate? = i keep it in mind but it will not be before six months. I will make a pull or notify with github, i code also in python. In french we have some software to find rimes. Putting the finding in a database could be a nice addition, i could help to compose text. You have my admiration for the idea and the execution ...

madcaptenor 12y ago

I'm a bit surprised that there are anagrams to be found. It's easy to find them if they exist, but there's no guarantee at all that there actually should be collisions.

DanBC 12y ago
I'm not a statistician. Is it really that surprising? English has plenty of redundancy; Twitter statuses have limited length. What's surprising to me is the niceness of the found anagrams. "another math genius" / "he ain't smart enough".
- madcaptenor 12y ago
  I am a statistician. Maybe I should sit down and actually do some calculations.
  - sillysaurus3 12y ago
    Please do. I'd be interested.
- denzil_correa 12y ago
  > What's surprising to me is the niceness of the found anagrams. That's because they are manually curated [0] Q: Is this manually curated? A: Mostly for issues of volume ( there are a lot of variations of 'goooood mooornnniinng!', there are a lot of spam bots posting subtely different versions of the same message, etc) the bot doesn't automatically post every anagram it finds. Essentially there's an iphone client that reviews matches, which are manually approved or rejected. [0] https://github.com/cmyr/anagramatron https://github.com/cmyr/anagramatron
sirclueless 12y ago
It's actually extremely likely. The chance that any two statuses are anagrams is miniscule, and even the chance that a particular status has an anagram among all other statuses is probably small, but the chances that there are no collisions at all is tiny. See a description of the Birthday Paradox[1] for the mathematics behind this. For example, if you put 70 people in a room, there is a 99.9% chance that two people share a Birthday. [1]: http://en.wikipedia.org/wiki/Birthday_problem http://en.wikipedia.org/wiki/Birthday_problem

fjcaetano 12y ago

This is delightfully ironic

bengali3 12y ago
Oily Shirted Filch Linguist

joopxiv 12y ago

I find it interesting that they're manually approving the hits, because, as they indicate, most hits are (nearly) identical. It shouldn't be too difficult to solve this automatically though. Identical hits can be discarded very easily. The ones that only have a few words or letters reversed can be detected with some kind of similarity algorithm.

jcampbell1 12y ago
I had a look at the source code, and it does quite a bit of filtering, particularly around making sure the words are unique, and there is a primitive character comparison algorithm. The code could be simplified by using Python's set() and improved by doing a copy'n'paste on a Levenshtein function.
- cmyr 12y ago
  oh hey yea that would've been useful. ^_^