What interesting people we have here in btt community  |
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In the passage I addressed in the previous post, if we replace the side or service in debt with the tx receiver in our context and the lender with tx sender, my first post scheme can be implemented using a similar merkele-tree of accounts using modular numbers instead of tx bytes codes explained in dacoinminster's spec paper, that handles all transactions (except buying mastercoin with bitcoin) off the bitcoin blockchain.
This will apparently resolve the blockchain bloat problem effectively.
PS: also as an additional security-protection layer, it can be settled for this off-chain merkele-tree os to send a message including the list of all accounts (> 0) balances as an on-blockchain message to the exudus address every 24 hour(or any other period) once. But this one can be neglected if looks bloating the blockchain size as well (which seems not)
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Forgive my ignorance but could you explain what 'off blockchain script-messaging' is in this context?
The idea is better explained when we use the off-chain transaction methodology (with embedded modular numbers instead of Jr.Willet's bytes code as transactions messages, which I can explain and formulate later on) using merkele-sum trees of accounts. Here I include a passage from https://en.bitcoin.it/wiki/Off-Chain_TransactionsAuditing
In addition to hacks, currently no trusted third party payment systems in Bitcoin provide any way for users to determine if the services actually hold the Bitcoins they claim to hold. Conventionally banks and payment processors are audited regularly by third-parties - because Bitcoin is based on cryptography auditing can be done in a cryptographically provable way.
Gregory Maxwell has proposed[3] to use merkle-sum trees of accounts to audit funds held by third parties. Each account with the service is assigned a number, such as a SHA256 digest, and those digests are formed into a merkle tree. Additionally for every node in the tree the sum of the account balances on both leaves is computed, and that sum becomes part of the data hashed by the parent node. The tip of the tree is then the sum of all balances for all accounts.
The service proves they control the Bitcoins they claim to by signing statements with the private keys capable of spending transaction outputs present on the blockchain, and in addition regularly sign statements attesting what is the current tip of the account merkle-sum tree.
Clients check that their account is included in that tree by regularly demanding proof, in the form of a merkle path, that their account leads to the claimed tip. Any discrepancy is evidence of fraud on by the service, or at least poor record-keeping. |
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I know it might be a bit late and also off-topic, but why don't you guys try to use the off blockchain script-messaging based possibility to include and handle all transactions of the higher levels (expect the one when somebody sends bitcoins to buy some mastercoins)
Since the number of mastercoins are constants (well there will be 10% gradually being added, but at any known interval the total is a constant number), we can use a list of all mastercoin wallets with > 0 ballance, plus a modular arithmetic way of encoding every and each tx in a standard binary number (chiness remainder theorem can produce these numbers easily) to be dealt and then to be added and modified in the wallets list.
As the decentralized handling of this scheme may be questioned, there could be two ways:
1- Using several gradually increasing number of info wallets, as seeders, to just keep the wallets balance list+ recent script-based tx.
2-Making use of random double-spending-check functioning nodes, that is: for each tx, two or three random online wallet>0 will be chosen to check the tx correctness due to their list and if they all permit it, tx can be finalized and included in their balance lists (and new one be informed to others as well)
*Or rather an optimal combination of 1,2.
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Nice design and taste, please just don't change that block background to some typical white neutral in future  Lovely, thank you! |
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You have to admit though that it would have been cool to have such an M. Pretty slick proof.
Yes it would have been a Mr. pretty proof had it worked that way! By the way are you planning to launch at this coming October? And if not, then what's your schedule plan? |
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It seems the proposed corpus fails due to a fairly trivial structure property I hadn't considered. I was busy building a bounding argument for generating p1,p2 and r off of a given n and hadn't even considered that they are either both congruent to 1 or 3 creating the obvious problem (4k1 +1)^r + (4k2 +1)^r cannot sum to a number that is divisible by 4. Feels pretty idiotic on my part honestly.
Don't worry these kinds of mistakes are too common for busy minds, and frequently happens to all of us once in a while again and again Wish a good fortune for your current project |
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I can tell you 100% surely that there is no way to deduce from 2n=p1+p2^r the existence of such m (actually when n is an odd number then p1 and p2 should both be either in the 4k+1 or both in the 4k-1 form and that will result in m^r not divisible by 4 while m is an even number, obviously impossible) But yes this off-topic discussion may be a noise for the general audience and let's quit it here |
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@bytemaster: sorry to guess wrongly, actually I have seen you discussing technical stuffs with bitfreak in his thread, that made me first guess that you are the math guy, so I directed my first post to you. Then Charles came in and by the real name from coindesk news I got to know he is the math guy, and then by the well known math selfishness, I supposed the creation should be his work  @Charles: unfortunately your trick is not correct and won't work for solving Goldbach's conjecture, sorry for extending this off-topic here, but I try to explain why: you want to prove for a particular n>1 there are two primes p1 and p2 so that 2n=p1+p2. By Chen's result we know there is an r (r<c where c is an upperbound) so that there are prime numbers p1 and p2 that 2n=p1+p2^r. So here you suppose there could be an integer m so that p1^r+p2^r=m^r BUT STOP, WHY?! Since p1 and p2 and r are fixed from the previous equation for 2n and not arbitrary, you will not certainly be able to deduce the existence of an m such that p1^r+p2^1=m^r for general (but fixed for each n) p1, p2 and r, other than when r=1, but that is exactly what you want to prove then after by the help of this trick |
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Goldbach is actually one of the hardest problems to resolve in number theory in my opinion. It has some of the deepest connections possible in prime distribution and explaining how factorization changes with addition, which is why Hardy and Littlewood spent so much time analyzing it.
I actually came up with a corpus using Wiles work alongside some insights from Tao to resolve the Goldbach conjecture in a fairly creative way, but it's beyond my current mathematical ability to explore. I would need about ten years and access to some great minds.
I do not really see how Wiles methods or results could help solving Goldbach's conjecture (but that's the adventurous beauty of math discovery to bring in unexpected relations) but Tao's approach and in general ergodic methods are powerful tools in analytic number theory! Did you participate in the Quixote project coding too? I know bytemaster is a skilled c++ developer, but I was wondering to know if you just designed the idea and algorithms, or also participated in coding process |
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Yeah changing the world money system seems to be much easier than solving Goldbach's conjecture, and Goldbach's has to be much easier than some other number theory conjectures Hungerford's doesn't include much about homological algebra and nothing about group representations. Lang's is again dry but self-sufficient. Shafarevich also has a general book about Algebra that I like very much. |
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I thought you are speaking about his Algebraic Number Theory book.. His Diophantine Geometry one is a very nice treatment! Actually Cassels and Frohlich is much older I suspect, 1967, but much more modern in presentation. Serge Lang was very Borabakist and dry in style. I had a short glance in Nathanson's lectures years ago, though not a careful reading, but will have a more careful look into that paper for sure! Actually I've got some nice crypto-economic project-ideas too recently, but unfortunately I am not enough good and patient in coding  I wish you great success with your current project |
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Serge Lang book is a bit out of date, there are much better books like Cassels and Frohlich one (as editors)
Analytic methods are used and play a crucial rule in Algebraic Number Theory as well, for me algebraic means a framework to search into things more systematized.
I used to work a bit in Langlands and also a bit in class field theory in general, but had a small project in elliptic curves as well.
Cryptography I didn't like much before (preferred more pure proof-based theoretical stuffs) but since I got more time and also got familiar with bitcoin, I get more interests in the field everyday.
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I heard you are a math guy, curious to know which field of math? I, myself, specialized in Algebraic Number Theory and nice to see math people around! |
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bytemaster, I have got a question:
If value of the system belongs to bitshares and these bitshares will be mining from 0 (no-premine) then,
how will you get back some value (or bitshares) in compensation for you idea-design and coding-effort-time that you've invested and contributed to this project?
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Hey Guys! I would like to sell my previously purchased (for an idea) and unused domain cryptoltd.com Please pm me with your offers or post them here Thank you! |
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Hey Guys! I would like to sell my previously registered (for an idea) domain cryptoltd.com Please pm me your offers Thank you! |
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I've been under the weather these past days and bound to bed. I have done some initial work and I will be releasing some source code as soon I've fought of this flu Sorry to ask this, a long thread: are you coding in c++ or java or perhaps python? I am thinking about joining the force but am not sure yet |
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"nothing changed Sad"
obviously, it's you who are not reading all the posts, pal...
admin already stated lately, that things have been changed!
but you are too ignorant, living your little life, i understand...
Oh ignorant, little life, how do you know all that kid? You should think your words before they come out from your big mouth. Aren't you one of the typical guys usually and only rambling after the free money everywhere, the little worm-like freerollers.. I think like movies there should be some kind of filtering and ban for age groups as teenagers don't need to be present here! I take a break, got tired of yesterday registered (only 10 or 15 posted) troll mouths! |
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i wasn't asking for anything - i was responding to Dabs post and explaining what turbo means;
guy! - read the subject of the matter first before posting, okay!? peace, bro!
You didn't perceive the main point of my post, okay 13 posted jr. is too young, I understand. |
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