Addresses can be grouped as belonging to the same user (or, better, same "account") by grouping together the addresses that participate as inputs to a single transaction, and doing it transitively. This has been done in the recent paper by Shamir and Ron, by some previous researchers and can be done by znort's parser. I know this is not 100 % accurate but I'm fine with it.

What I'm looking for is a site that displays the results of such grouping graphically, possibly with transactions between accounts as well.

I've read about such a site on bitcointalk.org, but I can't find the topic any more. I'm looking for the url of the (now possibly defunct) site or the bitcointalk topic describing it.

  • Wouldn't that be dependent on being able to associate a transaction with a geographic location? Commented Oct 27, 2012 at 22:50

3 Answers 3


The tool is called Blockviewer. The website, blockviewer.com, no longer exists. However, the source code is available on Github: https://github.com/thallium205/BitcoinVisualizer. Thus, you might be able to download the source code, compile it, and run it on your own machine.

You can also read the original blog post introducing some of the ideas behind it.


I found the bitcointalk topic:


  • If anyone knows what happened to blockviewer.com (the site's offline now), or if there are any similar applications/sites, I'm still interested.
    – mazi
    Commented Oct 24, 2012 at 18:40
  • 1
    It is still being developed. The latest is found here: toolongdidntread.com Commented Oct 27, 2012 at 22:45

The above-mentioned site, blockviewer.com, is now down.

There should be some type of search/querying engine that allows you to find the addresses associated with a given address based on the classic user-generating algorithm.

Anecdotally, my university might be hosting such an engine by the end of the summer. We're also looking at Javascript-based visualization engines to host the data through.

Sadly, however, it seems that no such visualizer (or even public data source) currently exists, as far as I can tell.

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