What is that all about?

• Generate All Keys Using Armstrong Axiom Algorithm 2
• Generate All Keys Using Armstrong Axiom Algorithm For Mac
• Generate All Keys Using Armstrong Axion Algorithm List
• Generate All Keys Using Armstrong Axiom Algorithm For Beginners

Is there a correct way to generate a symmetric key? Ask Question Asked 4 years, 5 months ago. Every paper I've seen describing the algorithm never show a way to generate a key, but show the available key sizes. This leads me to a question: How should I generate a key as someone who is implementing an algorithm? Armstrong axioms are a complete set of inference rules or axioms, introduced and developed by William W. Armstrong in 1974. The inference rules are sound which is used to test logical inferences of functional dependencies. The axiom which also refers to as sound is used to infer all the functional dependencies on a relational database. (C#) Generate Encryption Key. This could be a 'single-use' key that is // derived from a secure key exchange algorithm using RSA. // They do so using asymmetric encryption algorithms (public/private keys). It is not // required to use a key exchange algorithm to achieve the goal of having both sides // in possession of the same secret key.

The task is to generate Armstrong Numbers from 1 up to the length of N decimal digits.

Generate All Keys Using Armstrong Axiom Algorithm 2

Armstrong number (aka Narcissistic number) of length N digits is a number which is equal to the sum of its digits each in power of N. For example: 153 = 1^3 + 5^3 + 3^3 = 3 + 125 + 27 = 153

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Brute force algorithm

There is an obvious bruteforce algorithms that:

0. Pre-generation of all powers i^j, where i is a digits, and j is possible length from 1 to N - this is necessary for all solutions
1. For each integer i from 1 to K
2. Divides i by digits
3. Calculate power of each digit
4. Sum up those powers
5. If this sum is equal to i - add it to the result list

Implementation: ArmstrongNumbersBruteforce.java

It can be improved by parallel calculation of sum of digit powers to the number generation.

Implementation: ArmstrongNumbersBruteforceOpt.java

Hash Approach - Divide At Impera

There is another interesting idea of bruteforce approach improvement.

1. Divide a number for two equal parts. In case of an odd N first part will be a bit longer. For example, if N=7, the number will be divide like XXXXYYY, where XXXX the first part (4 decimal digits), and YYY the second part with 3 digits.
2. Generate all integers i of the second part (in our example there will be integers from 001 to 999).
3. Calculate p equal to sum of digits in power of N.
4. Add to some hash the following pair {p-i, i}. For example, for i=725, p=7^7+2^7+5^7=901796. We add pair {901071, 725}.
5. Generate all integers i of the first part without leading zeros (in our example there will be integers from 1000 to 9999).
6. Calculate p equal to sum of digits in power of N.
7. Check if hash has a key of (i*10^(N/2)-p). For example, i=1741, thus p=1^7 + 7^7 + 4^7 + 1^7=839929. We look for key (1741000 - 839929) = (901071). OMG! It exists!!!
8. In case that key exists we unite the Armstrong number from two parts and add it to the result list. 1741000 + 725 = 1741725

One addition, is that we cannot store simply (key, value), we need to store multiple values, for example to be able to generate 370 and 371.

Implementation: ArmstrongNumbersHash.java

Multi Sets Approach

We may note that for each multi-set of digits, like [1, 1, 2, 4, 5, 7, 7] there is only one sum of powers, which in its turn may either be or be not represented by the digits from set. In the example 1^7 + 1^7 + 2^7 + 4^7 + 5^7 + 7^7 + 7^7 = 1741725, which can be represented by the digits and thus is an Armstrong number.

We may build an algorighm basing on this consideration.

1. For each number length from 1 to N
2. Generate all possible multi-sets of N digits
3. For each multi-set calculate sum of digits^N
4. Check if it's possible to represent the number we got on step 4 with the digits from the multi-set
5. If so - add the number to the result list

Complexity assestment The number of cases calculated for each length N is equal to the number of combinations (N + 9, 9) = (N+9)!/(9!N!). Thus for all Ns less than 10 we will generate 92,377 cases. For N<20: 20,030,009 cases.

Implementation: ArmstrongNumbersMultiSetLong.java

With optimizations:

• For long: ArmstrongNumbersMultiSetLongOpt.java
• For BigInteger: ArmstrongNumbersMultiSetBigIntegerOpt.java

Benchmarking

Generate All Keys Using Armstrong Axiom Algorithm For Mac

Generate All Keys Using Armstrong Axion Algorithm List

Let's compare the algorithms performance for different numbers of length N. I did the tests with my MacBook Pro.

Generate All Keys Using Armstrong Axiom Algorithm For Beginners

Clear win of the multi-set algorithm!