Proof isn't always possible

ECM on 100 digits (prescribed threshold) I had to retract my claim of “a practical description,” because I failed to account for the necessary computation of probability for automatic SIQS. I hope others will be interested enough to compensate this potential lack. ten sided dice—x5 or x20 each roll (to remove order as a concern.) It takes time to record or enter x25 digits. x20 digits takes a little time, but there is “action,” in between. After data entry 300 curves on one core—ruling out factors < 10^25.  Note: for complete factoring, any count of factors, each < 20^25. Although summary SIQS is plausible any time after 300 curves, it is desirable to run fully 2100 curves for the 100 digit range, before coercing SIQS. In this game we describe 2400 curves. SIQS as a mathematical choice only accounts for two (2) unique factors. Full number should be reduced to two factors (at least as a statistical probability) before attempting SIQS. 2100 - 300 =1,800  1,800 / 30...

Yesterday I piloted the idea of using vector matrices to defeat salting. I noted that the salt need not be in a small range; this might still fall to investigations of multiple instances. Whatever variety of possibilities, the step from one to the next in a series will be constant. If we defeat salting, we might defeat the effective benefit of hashing altogether. I advance a possible solution from the memory of an embarrassing mistake. If my salt has become compromised, I can seek a remedy by introducing variations between steps. This is not what we expect from a deterministic hash, so what am I suggesting?  In the algorithm, the stepping is by one pass of the hash algorithm. Although it is inefficient, I can vary the stepping by taking two hexadecimal numerals from a fixed offset within my hash, converting them to an integer, and choosing to step by a number of hash passes equal to that integer. Because the hash is deterministic, the resulting list would always be the same, bu...

To envision a potential attack on OpenDRM, we modify a dictionary attack against the salt. We see dictionary attacks on passwords, and rainbow table projects against hashes. In OpenDRM, we salt the progress from md5 hash to md5 hash by iteration.  In the algorithm, we specify “object code looking stuff.” This category does not lend itself to any dictionary attack, and while you could possibly rule out large swaths of pronounceable instances, you would still be left with a lot of entries that iterate randomly.  Part two, securing the extant article against my new attack, is that the LENGTH of the salt is NOT specified, leaving no structure to construct a dictionary against it. Attack: Hitherto, Red Team has attacked salts by calculating creatively chosen instances, and seeking matches in rainbow tables. My idea is to contrive some idea of a generalized vector pointing from hash0 to hash1. The improvement of this strategy is that creatively modifying the vector might res...

Compromise The Missionary To play Compromise The Missionary, hold out all 7’s. Shuffle three 7’s and place them in a small draw pile face down. Hold 7 of spades face up. Deal five cards each.  Assessing Red suits at positive integers, and Black suits as negative integers, use your cards to contrive the sum of seven as many ways as possible. Score each successful summation as 5 points on a graph of potential X’s. Represent the first two sums as the crossbars of the “X.” Score successive summations as a dot in a side of the “X.” So each completed X adds up to 30 points (like many domino games). Cards used in the summation(s) go into a discard pile. Draw to fill a hand of five cards. If you choose, you can neglect to devise a summation. Discard at least one card in the discard pile and draw to fill a hand of five cards. Whenever a player plays an Ace, he draws from the 7’s pile. If he plays the Ace of Spades, he automatically collects the 7 of Spades. The hand ends with scoring the su...

This is an open question. It represents a shortcut to the cryptanalyst between multiple rounds of a cipher algorithm. Represent the cipher as y = f(c) in two dimensions.  For superior ciphers, f(c) is NOT A GROUP. Question: can we devise a y = Ef(c) in three (3) dimensions such that Ef(c) IS A GROUP? Then we apply some creative calculation f(c) and Ef(c) to derive z1 and z2. Then we attempt to use z2 to compute (for example) Ef(z2) == f(f(f(f(f(z1))))), effectively skipping 4 rounds. Is there a way to CALCULATE, or DERIVE these formulae?

The construction of the Schroedinger’s Cat paradox finds application in the discussion of Intellectual Property . If an idea was the Cat, then measurement would be analogous to evaluating the idea for merit, and the quality of life would be analogous to fitness or marketability  of the idea under consideration.