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Topic: [ANN][DASH] Dash (dash.org) | First Self-Funding Self-Governing Crypto Currency - page 5491. (Read 9723733 times)

sr. member
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legendary
Activity: 1946
Merit: 1100
Leading Crypto Sports Betting & Casino Platform
LOL what has Darkcoin done to these guys. I just don't understand the hatred. It is unbelievable.

Not one person has asked in the XC thread about the closed source code. It's like all the "right" crowd has been paid proper. Right now the focus is to bring down DRK I guess. Then the guys will go after XC because let's face it, Dan has openly accepted his fascination for everything related to DRK. Capitalism at it's best.

hero member
Activity: 1302
Merit: 502
snip

Interesting, I must have missed the 40% gain post at some point but I'm not a miner so I don't keep up with the latest news on that front.
Thanks for your posts!


full member
Activity: 196
Merit: 100


Hmm X11 Coin hey, let me check out their website http://www.x11coin.com  Grin

Redirect to Darkcoin website... nice.
I was like


did you just

and you were like


haha
hey look theres internetape Tongue
hero member
Activity: 700
Merit: 500
You're right, that is my point.

I just did... using crc32 as an example.  You can create your examples using another collision attackable hash like say, md5.
Thank you. I have learned with minimal pain.
Absolutely.  Sharing our ideas, knowledge, and code, is what makes these communities what they are.  
Thank-you for being receptive to new information - far too many people aren't these days.

And for everyone: I'm not saying X11 is broken.  It's just no more secure then using a single hashing algorithm, IMO.
hero member
Activity: 560
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www.OroCoin.co
If the 5th algo in the chain is found to be broken, you'd still have to come up with an input of the 4th algo that generates the output you need to feed into the 5th algo; so you'd have to come up with an input for the 3rd algo that generates the correct output to feed into the 4th algo to generate the output you need for the 5th algo; and then you'd have to come up with an input for the 2nd algo that generates the correct output to feed into the 3rd algo.... and so on.  So basically you'd need rainbow tables for all of the preceding algorithms in the chain, in order to do anything meaningful.
This is what I was originally thinking. Rainbow tables do exist... But this does make it far-fetched... To put a number on it is more than 1, less than 11.

The argument now becomes circular.

Yes, more than one potential attack vector is worse than just one. But, if that 1 vector is exposed, you are bork just as bad...

Bork or not bork is the resulting boolean interest and I'm still not convinced that 11 hashes = 11 vectors via baddw's example.

Some hashes have had their bork streamlined to break in seconds on a Core i5...
hero member
Activity: 560
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www.OroCoin.co
You're right, that is my point.

I just did... using crc32 as an example.  You can create your examples using another collision attackable hash like say, md5.
Thank you. I have learned with minimal pain.
hero member
Activity: 700
Merit: 500
Ill have to give it to them, those x11 guys know how to do their bullshitting and marketing. Peoole with small brains will fall for it. Luckily we have big brains, thats why we own darkcoin.
You have brains large enough to trust a closed-source centralized anonymizing system?  Erm... I think your small brain might be getting the words "small" and "big" mixed up here.

It seems this would only be a risk if the very last hashing function were compromised. Even then you would need to know how to generate the correct input to the last hashing function using the other hashing functions. It seems inherently more secure to me.
if hash1(x) has collisions, then so does hash2(hash1(x)) and hash4(hash3(... and so on, until we reach the full hashing stack, which also has collisions. Similarly, if hash2, hash3, etc, or hash11 have collisions, then so does X11(x). Simply put, if there's a collision attack for any hash#(x), then the same attack applies to X11(x).

This is the same baseless declaration with more words. You still failed to make the correlation or demonstrate a causation.

Why would this result in what is essentially a cascade failure? Why would an attack on Math A result in a failure of all other Math B to Math K to be Math anymore?

And on top of it, it's still an if...

I think you're fundamentally wrong. Collision on hash A does not break Hash B.
Really?  This is very concerning that you seem to think this...  None of what I am saying is baseless - for some reason you aren't even running simple tests?

Simple example:
sha256(crc32("plumless")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49
sha256(crc32("buckeroo")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49

crc32 has known collisions and is essentially broken in that sense, and so sha256(crc32()) is as well, and so blake256(kekkack(sha256(crc32(etc)) has known collisions.

Yes, this is an "if" situation... if a collision attack is discovered against any of the 11 sha3 candidate algorithms, that collision attack works for the whole chain.  So, 11 chained algorithms are inherently less secure in regards to collision resistance vs 1 algorithm.

It's really only a problem for Darkcoin if the very first algorithm in the chain is attacked.  (Notice that you're using crc32 as the first algo in every example.)  There is still an input value that is put into the first algorithm, and the first algorithm produces a good result, then that result must be fed through the other algorithms in the chain.

If the 5th algo in the chain is found to be broken, you'd still have to come up with an input of the 4th algo that generates the output you need to feed into the 5th algo; so you'd have to come up with an input for the 3rd algo that generates the correct output to feed into the 4th algo to generate the output you need for the 5th algo; and then you'd have to come up with an input for the 2nd algo that generates the correct output to feed into the 3rd algo.... and so on.  So basically you'd need rainbow tables for all of the preceding algorithms in the chain, in order to do anything meaningful.

Your example falls apart if the algorithm ordering is crc32(sha256(input)).
hero member
Activity: 700
Merit: 500
Your point is that there are 11 possible attack vectors instead of just one;

IF a collision attack can be applied to even one of the algos

IF collision attack compromises the full stack, as you say it does.

Demonstrate the part in bold. Everything else is proven and makes sense. Show us.
You're right, that is my point.

I just did... using crc32 as an example.  You can create your examples using another collision attackable hash like say, md5.

phzi is right you plebs. It is statistically speaking less secure but it does have a benefit - lower power usage.
This is such a useless arguement. Has any "collision" attacks been discovered for sha256? If not, then it's 99.9999999% likely that there won't be any found out for Any other algorithms. Useless convo here...
lol.
phzi, don't even bother.
Heh, I resisted the urge to respond to this guy...

The thing about X11's lower power consumption, is that it's kinda a crock of shit.  It only uses less power because the OpenCL code is horribly optimized.  Properly optimized CL implementation will likely have similar power consumption to scrypt (or at least, the margins will be much decreased).  We've already seen 40%+ improvements just by calling functions differently and not even modifying the cl code.
Rux
legendary
Activity: 1291
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https://crypto.ba
any of you non-belivers, just try Darksend feature and report us the feeling ...

Wink UNKNOWN my favorite word
hero member
Activity: 1302
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phzi is right you plebs. It is statistically speaking less secure because it has a higher probability of one day being exploited, however it does have a benefit - lower power usage.



This is such a useless arguement. Has any "collision" attacks been discovered for sha256? If not, then it's 99.9999999% likely that there won't be any found out for Any other algorithms. Useless convo here...

lol.
phzi, don't even bother.
legendary
Activity: 1414
Merit: 1001
To weird to live To rare to die
Buy at Cryptsy now this guy believes

 014-05-28 18:16:20   Buy   0.01327600   1200.00000000   15.93120000
hero member
Activity: 518
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I just sold the last of my XC and placed an order for a few more DRK, TBH though guys I said earlier, XC is an attractive coin, as is DRK, lots of coins dropped recently to fund the DRK rise... A lot of this is just BC moving about.. Would not surprise me if in 12 hours DRK is back in the 160 + range... As XC carries on rising it is starting to look expensive and DRK is looking cheap....
hero member
Activity: 504
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eidoo wallet
hero member
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www.OroCoin.co
It seems this would only be a risk if the very last hashing function were compromised. Even then you would need to know how to generate the correct input to the last hashing function using the other hashing functions. It seems inherently more secure to me.
if hash1(x) has collisions, then so does hash2(hash1(x)) and hash4(hash3(... and so on, until we reach the full hashing stack, which also has collisions. Similarly, if hash2, hash3, etc, or hash11 have collisions, then so does X11(x). Simply put, if there's a collision attack for any hash#(x), then the same attack applies to X11(x).

This is the same baseless declaration with more words. You still failed to make the correlation or demonstrate a causation.

Why would this result in what is essentially a cascade failure? Why would an attack on Math A result in a failure of all other Math B to Math K to be Math anymore?

And on top of it, it's still an if...

I think you're fundamentally wrong. Collision on hash A does not break Hash B.
Really?  This is very concerning that you seem to think this...  None of what I am saying is baseless - for some reason you aren't even running simple tests?

Simple example:
sha256(crc32("plumless")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49
sha256(crc32("buckeroo")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49

crc32 has known collisions and is essentially broken in that sense, and so sha256(crc32()) is as well, and so blake256(kekkack(sha256(crc32(etc)) has known collisions.

Yes, this is an "if" situation... if a collision attack is discovered against any of the 11 sha3 candidate algorithms, that collision attack works for the whole chain.  So, 11 chained algorithms are inherently less secure in regards to collision resistance vs 1 algorithm.
Your point is that there are 11 possible attack vectors instead of just one;

IF a collision attack can be applied to even one of the algos

IF collision attack compromises the full stack, as you say it does.

Demonstrate the part in bold. Everything else is proven and makes sense. Show us.
hero member
Activity: 504
Merit: 500
eidoo wallet
Ill have to give it to them, those x11 guys know how to do their bullshitting and marketing. Peoole with small brains will fall for it. Luckily we have big brains, thats why we own darkcoin.
You have brains large enough to trust a closed-source centralized anonymizing system?  Erm... I think your small brain might be getting the words "small" and "big" mixed up here.

It seems this would only be a risk if the very last hashing function were compromised. Even then you would need to know how to generate the correct input to the last hashing function using the other hashing functions. It seems inherently more secure to me.
if hash1(x) has collisions, then so does hash2(hash1(x)) and hash4(hash3(... and so on, until we reach the full hashing stack, which also has collisions. Similarly, if hash2, hash3, etc, or hash11 have collisions, then so does X11(x). Simply put, if there's a collision attack for any hash#(x), then the same attack applies to X11(x).

This is the same baseless declaration with more words. You still failed to make the correlation or demonstrate a causation.

Why would this result in what is essentially a cascade failure? Why would an attack on Math A result in a failure of all other Math B to Math K to be Math anymore?

And on top of it, it's still an if...

I think you're fundamentally wrong. Collision on hash A does not break Hash B.
Really?  This is very concerning that you seem to think this...  None of what I am saying is baseless - for some reason you aren't even running simple tests?

Simple example:
sha256(crc32("plumless")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49
sha256(crc32("buckeroo")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49

crc32 has known collisions and is essentially broken in that sense, and so sha256(crc32()) is as well, and so blake256(kekkack(sha256(crc32(etc)) has known collisions.

Yes, this is an "if" situation... if a collision attack is discovered against any of the 11 sha3 candidate algorithms, that collision attack works for the whole chain.  So, 11 chained algorithms are inherently less secure in regards to collision resistance vs 1 algorithm.

This is such a useless arguement. Has any "collision" attacks been discovered for sha256? If not, then it's 99.9999999% likely that there won't be any found out for Any other algorithms. Useless convo here...
hero member
Activity: 658
Merit: 500
The Buck Stops Here.
This is exactly what happened to Blackcoin and WHitecoin. XC is the new Whitecoin - a complete and utter rip off, a shameful clone to snag the greedy - Look where Whitecoin is now....





So, essentially you're saying XC is a copy/clone of Darkcoin. It's ok bro, I can already see that from it's name, X11Coin....

Not at all. XC uses the multi path paradigm as an extension of the dual path paradigm, similar to the TOR onion routing protocol. As well as offering encryption and far greater decentralisation of it's autonomous xnodes. And that's just the start. XC will blow DRK out of the water.

These bolded words sound amazing please show me the code.
legendary
Activity: 1344
Merit: 1001
This is exactly what happened to Blackcoin and WHitecoin. XC is the new Whitecoin - a complete and utter rip off, a shameful clone to snag the greedy - Look where Whitecoin is now....



XC offers encrypted transactions. And the coming xnodes will be clustered and completely decentralised... no master node type structure.

I collated these points from the XC developer of things to look forward to in the future, to build upon our already solid foundation:

The mixer technology will be multi-path across a cluster of xnodes which runs over the encrypted protocol, so it won't be a central server, it would be similar to a distributed cluster, and the data is encapsulated inside that encrypted tunnel to prevent off-the wire sniffing (or off-the-air* for the mobile platform)...  DRK didn't "invent" supernodes or the concept, and while some of concepts are similar and/or overlap, we are on a different path than DRK and as we refine the platform, it will be clear why our design, implementation and infrastructure is superior as we will have autonomous supernodes using PKI that can scale up and out to become a distributed anonymous application platform

This is why the multi-path paradigm network I built back in 2005 still functions to do this day and scaled up to meet the demands of 2014... think back to 2005, granted we had the internet, but not the mobile technology - and that is a key part of my background is mobility that will be leveraged in the android anonymous POS wallet

(http://eprint.iacr.org/2009/385.pdf
http://arxiv.org/ftp/arxiv/papers/1208/1208.3022.pdf)
(The principles for this concept come from a design that I originally leveraged back in 2005,[multi-path paradigm] the system still runs to do this day handling 50k to 100k users per day )

If your waiting for NSA proof anonymity - well that is the goal of our REV 2 release..

No, mixing is only part of it, XC now contains a completely new transaction protocol layer that is encrypted and will be leveraged as a platform for all sorts of concepts.... and it will leverage multi-path paradigm topology (which again is a custom developed concept based on dual-path paradigm)So mixing is just the first application to run on top of this new protocol........ Don't want this key point to get lost --- mixing is just the first application, the list is long and the potential is huge, and since the foundation has been built and deployed, it is now time to build on that foundation.... with some revolutionary concepts and idea's.....

Also the infrastructure for this will support numerous other applications, such as encrypted messaging, P2P exchange..etc granted we can only develop new features so fast, but we are putting the foundation for a multi-use, multi-purpose platform

You seem to be promoting XC, but your signature is promoting Monero.

What one of the two do you prefer?

Should i invest in both?

I think that they are both good coins and want to expand into these.

Monero is the cryptonote coin with the biggest community and their ring signature approach offers greater anonymity baked into their protocol rather DRK's way of bolting it onto the existing Bitcoin protocol. Indeed Evan himself initially wanted to adopt Monero's rings and said this would offer substantially improved privacy. But last I read he had realised the scale of this task (someone who supports DRK will scream about Monero blockchain bloat but really that's not much different that DRK sending transactions through 10 nodes to gain similar levels of privacy).

Monero's main problem is it lacks a user friendly wallet GUI and the community infrastructure is obviously behind that of a bitcoin clone which is why Mintpal and Cryptsy haven't taken it on board.

XC, I've spoken about already. Economically it offers a great investment because very few new coins are produced via mining now the PoW phase is over. Unlike most coins which face tons of newly mined coins being dumped on the market each day for profit, XC won't suffer from this downward price pressure. Additionally, I see XC blast past the $60 million market cap barrier if things go well in the next 3 weeks.

DRK has good qualities and I have a lot of respect for the DRK community. I believe all privacy centric coins will fare well in the coming months.
hero member
Activity: 560
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www.OroCoin.co
Ill have to give it to them, those x11 guys know how to do their bullshitting and marketing. Peoole with small brains will fall for it. Luckily we have big brains, thats why we own darkcoin.
He does actually have a valid point burried in all that buzzword crap. But does it matter? I'm not so sure...
hero member
Activity: 700
Merit: 500
Ill have to give it to them, those x11 guys know how to do their bullshitting and marketing. Peoole with small brains will fall for it. Luckily we have big brains, thats why we own darkcoin.
You have brains large enough to trust a closed-source centralized anonymizing system?  Erm... I think your small brain might be getting the words "small" and "big" mixed up here.

It seems this would only be a risk if the very last hashing function were compromised. Even then you would need to know how to generate the correct input to the last hashing function using the other hashing functions. It seems inherently more secure to me.
if hash1(x) has collisions, then so does hash2(hash1(x)) and hash4(hash3(... and so on, until we reach the full hashing stack, which also has collisions. Similarly, if hash2, hash3, etc, or hash11 have collisions, then so does X11(x). Simply put, if there's a collision attack for any hash#(x), then the same attack applies to X11(x).

This is the same baseless declaration with more words. You still failed to make the correlation or demonstrate a causation.

Why would this result in what is essentially a cascade failure? Why would an attack on Math A result in a failure of all other Math B to Math K to be Math anymore?

And on top of it, it's still an if...

I think you're fundamentally wrong. Collision on hash A does not break Hash B.
Really?  This is very concerning that you seem to think this...  None of what I am saying is baseless - for some reason you aren't even running simple tests?

Simple example:
sha256(crc32("plumless")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49
sha256(crc32("buckeroo")) = sha256("4DDB0C25") = b7186aaec033aa3fb05e6b41444618f30e299f3879dde29d31f055f64fd8be49

crc32 has known collisions and is essentially broken in that sense, and so sha256(crc32()) is as well, and so blake256(kekkack(sha256(crc32(etc)) has known collisions.

Yes, this is an "if" situation... if a collision attack is discovered against any of the 11 sha3 candidate algorithms, that collision attack works for the whole chain.  So, 11 chained algorithms are inherently less secure in regards to collision resistance vs 1 algorithm.
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