Posts

Object Code Checking; what constitutes a back door in encryption?

 If I want to certify object code but don't want to share the source, ONE idea would be to provide an interpretive script that duplicates it. Encryption would be an example. My idea was: if I can write C++ code that decrypts text enciphered by my python3 implementation, that would certify it while capitalizing on faster execution, yet defeating efforts to brute force cryptograms.  Specifically in encryption, the rule is: nothing secret except the message and the password. Ethically, derived key material represents a grey area. If I use a C++ object from an unknown source, and this object decrypts what Python has enciphered AND Python decrypts what the object has enciphered, then the object is valid - as long as I check EVERY message. A back door treats some passwords (or messages) differently. As a learned rule, this helps me think more clearly about it.

Cryptocurrency: A non-mathematical take.

 Cryptocurrency is NOT a suitable medium of exchange for the 21st century. Cryptocurrency is conceptually, non-counterfeitable coins attached to a resilient blockchain. By associating the blockchain entry with a jpg, we get the idea of an NFT. These are worth what someone will pay for them in the same way as a painting. However, the coins are subject to failings. 1. Cryptomining of the currency (computing new coins) is inflationary. 2. Cryptowallets can be summarily deleted, by either a hacker or a despotic government. A Backup wallet is a workaround, but by the mechanism you reconnect the wallet to the blockchain, you can introduce a mathematical attack. 3. Marked bills do not name criminals. They identify the holder by possession or transaction participation. Each cryptocoin contains literally EVERY transaction it has ever been associated with as part of the coin. So arguably EVER coin acts as a marked bill. 4. No one regulates the capitalization of the blockchain. If by some art...

A Simple Open DRM

We love to hate DRM, because we associate it with degraded quality or loss of data control. Sometimes we associate it with loss of privacy when we register.  Here is a formative "Open DRM." The kernel of the system is going to be the idea that a salted hash is a member of a class of some kind. If I start with a MEMBER md5 hash and run 200,000 iterations of the correct salt, I hit a control value predictably.  The potential problem arises not from providing 199,999 different possible keys, but from providing a registered customer with his own valid replacement, on that rare occasion when he proves he has purchased it, and we must re-present his original key. Standing behind the warranty is what makes us reputable. To preserve his privacy AND the security of our system, we do not want to store the enumerated value on the server.  However, we can reasonably suppose that we can manage to keep a secret value secure. I propose that the DBM or LLC choose a passphrase of some kin...

Newton's Method for nth root, as an integer arithmetic for Computers

 The actual values of my previous post are truly astronomical, and when I reviewed my python3 implementation, I found I re-invented the wheel differently on paper. Here is my python3 code, with comments: It is slightly more conservative with actual values. By experiment, it gives very good square roots for 3 - 8, at 4 iterations ############## main program ############################### # # This script attempts to calculate the nth root of k # ######################################################### k = int(input("Enter the number we are taking the root of: ")) n = int(input("Enter the power of the root: ")) i = int(input("How many iterations would you like to try? ")) a = k b = n ######################################################### # #   The formula for the Convergence Method is: x_k+1 = x_k - {[x_k^n]- K #                                             ...

Newton's Method of converging on the nth root of C, represented as an Integer computation:

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 Newton's method is known to converge slowly, and I do not have an associated margin of error at this time. Newton proved: x(k+1) = x(k) - [x(k)^n + C]/[n * x(k)^n-1] converges on the nth root of C. My first step is to rewrite x(k) as (a/b)(k) or a(k)/b(k) From this we obtain: (a/b)(k+1) = (a/b)(k) - [(a/b)(k)^n - C / n * (a/b)(k) ^ (n-1)] Focusing on x(k), we begin: I do not know how to use the editor that does notation.

A Number Theory Observation

Six sided dice can result in random numbers, when the face value is converted to Base 10, DESPITE displaying no "0." I suspect that Number Theory, as a class, would have taught me more than some finite number of Arithmetic tricks. It is my understanding that Number Theory investigates how these apparent coincidences are true using Algebra. It has been brought to my attention that, IF Erasthmus's Sieve is correct, THEN the modified Sieve in the previous post is also correct, as follows: Erasthmus used a positive integer number line, and stepped by the relevant n. In the sieve I uncovered (seeking a way to improve compactness in computer RAM) I start with odd integers, as a number line, and as a practical matter strike by 2*n. Explicitly, if we start with 3 and strike every third value, for example, we step over an even number in between each odd number skipped, resulting (algebraically) in striking by 2*3, or 6. Under this constraint, 3 being odd, we strike all ODD multip...

Modified Erasthmus's Sieve Demo

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 I wrote before, asking if the modification on Erasthmus's Sieve I discovered was still Erasthmus's Sieve. I am convinced it is not the Sieve of Erastosthenes, since the odd numberline is not a geometric progression. I allude to it as a Modified Erasthmus's Sieve, while open to correction. Earlier I demonstrated n < 100. Although it took me a long time to present the current entry, I did understand that n < 100 was not sufficient. Below I present a clearer picture of how it develops, using n<=361, or 19^2. I don't know enough Number Theory to offer Proof. In the first slide, I enumerate 2 literal, and develop the odd number line up to 361. I then start with 3 as the next unstruck number, and blindly strike by 3s. The result leaves NO MULTIPLES OF 3 UNSTRUCK. In the second slide, I start with the next unstruck number (5) and blindly strike by that number. The result leaves NO MULTIPLES OF 5 UNSTRUCK. By observation of experiment, multiples of 3 x 5 (15) are doub...