Tracking down Twitter retweets... a fuzzy approach -

I know that tracking changes in this fashion:

source - & gt; User1 - & gt; User2

is not possible because Twitter only provides you the original tweet and no one is retweeted. Based on this .. do you think it is possible to track retweets based on the followers of the twitter? For example, what I want to do, get a retett, find all the users who have tweeted it again, and then find out how the user is connected to the person who originally tweeted it. If B is not a follower of the provider, but has tweeted a tweet, can it be that a person B, who he follows, is a follower of the tweet operator and B dismissed it so that eventually See it?

Thanks in advance for any feedback.

I'm currently seeing how the user interacts with them timeline, and this is such a problem That's what I have also considered. The problem is that a tweet is not always the result of appearing on a user's timeline. Users can often re-tweet the public streams, trending topics, or the Tweets found in the "Explore" tab.

I think it is possible that your approach can work, but you will need to apply some filters, which type of users you see, you have to limit the range of followers who see it, because many Popular people come from popular users. Crawling 10,000 users to track a rivet is not really an option to consider API ranges.

A more viable option might be to track a reverse from some users who had retweeted it. For example, you have user A, which is user A, which has copied a tweet generated from user b. You can consider all users of User A, which is a tweet for User B (which includes User B) and has recovered any candidate retired. Find the users who follow the user until you reach B It looks deeper than the first search (or fourth-before, depending on the implementation details). However, if you delete any user with search, it is private to be involved in a problem. In this case, you will not be able to see your information and the vacancy has broken.

I know that this is not a black and white answer, but hopefully it helps a bit. This is an interesting question.