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Can Darkcoin's Darksend (I think this is called coinjoin, but I am unsure) be implemented on any cryptocurrency (without a fork), or must the feature be present when the coin is first released?
Can Darkcoin's Darksend be implemented on top of Bitcoin?
Answers
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Yes, DarkSend is one specific implementation of CoinJoin, and CoinJoin can implemented directly on top of Bitcoin without a modification of the protocol. So no fork is necessary.
The Darkcoin Whitepaper does not mention anything that would deviate from the CoinJoin idea and thus require a modification of Bitcoin. (However, the paper is not very detailed and I did not look at the source code, so I cannot provide an answer with 100% certainty.)
CoinJoin has been originally proposed by Gregory Maxwell. The main advantage of CoinJoin to avoid that coins can be stolen while mixing them with other users. It does not change Bitcoin, it runs on top of it, i.e., it only provides a special way to create perfectly valid Bitcoin transactions that can be submitted to the network. Of course, users need a special client to participate in the protocol. For users with normal clients, the created transaction is just a normal Bitcoin transaction that appears in the blockchain.
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Can you perhaps explain what "Darksend" is? I looked at http://www.darkcoin.io/ and there is a PDF that explains how to enable Darksend, but nothing about what it is or what it does.
– Greg Hewgill – 2014-03-29T23:38:33.653This doesn't directly answer your question, but note that there is Zerocoin, a similar protocol designed for Bitcoin.
– Tony – 2014-03-30T00:37:08.417Read the darksend documentation http://darkcoin.io/downloads/DarkSendDocumentation.pdf && whitepaper http://darkcoin.io/downloads/DarkcoinWhitepaper.pdf
Note, darksend is in early beta stage, actually beta V5
@Tony Would implementing ZeroCoin require a fork of the Bitcoin Blockchain, or could wallets be released that support this protocol while still being compatible with everyone who doesn't upgrade their wallet? I would guess that the answer to my original question is the same. – DJG – 2014-03-30T18:59:11.320