PasswordCrack is a language which should be used in generating alphanumeric combinations. It’s compiler(the only one at this moment, but I plan to write one in C in the future) is written in Go.
./passwordCrack <pc-doc-to-compile>Statements consist of either a literal or a variable, or combinations of both. A literal is for example “zoidberg” or “password”, and it’s evaluated to “zoidberg”, respectively “password”.
{password}
{%someVarExample%}
{password%years%}
Variables are declared by the following expression:
%variable_name%
This is followed by all the rules that one wants to define in that variable. In the end one must specify the end of a set of rules that need to be attributed to a variable, by repeating the opening expression. IMPORTANT: There cannot exist nested variables! Example of a variable declaration:
%dogBreeds% p^ug (golden retriever/labrador) corgi %dogBreeds%
Example of a variable use:
{Rex the %dogBreeds%}
Explanation The code from the above is going to output Rex the + all the possible outcomes of all the rules defined in the variables dogBreeds. This means it’s going to generate:
Rex the pug Rex the Pug
Because of the first rule
Rex the golden retriever Rex the labrador
Because of the second rule
Rex the corgi
Because of the third rule
Rules are the heart and soul of passwordCrack. A rule takes a literal and modifies it by a set of deriving rules creating all the possible outcomes when derivated. Rules cannot contain variables inside them.
Literal
corgi
This one is just a literal, so it’s gonna evaluate to “corgi”
^
p^ug
The or opposite means that the character(or group of characters) behind it can either be in the state they currenly are or in the opposite. In this case the rule it’s gonna evaluate to [“pug”, “Pug”]. And in this case:
L^abrador
it’s gonna evaluate to [“Labrador”, “labrador”]
() The begin group and end group operators have the role of grouping two or more characters togheter in order to apply the operation to them as a group. Any operation that can be applied to a single character can also be applied to a group. For example:
(REX)^ the (dog)?
It’s gonna evaluate to: [“REX the dog”, “REX the “, “rex the dog”, “rex the “]. There is a special operation that can be applied in a grouped set of charachters, and that one is the or ( / ) operation. This one means that the rule should evaluate to either the left side of the operator or the the right side. For example
Rex the (funny/cute) dog
it’s gonna evaluate to [“Rex the funny dog”, “Rex the cute dog”]
? The one or zero operator means that the character/group of characters behind the operator can be ommited.(Therefor we can have one or zero instances of it). For example:
pass(word)?
it’s gonna evaluate to [“password”, “pass”].
/+ The till now or plus operator means that the rule should either evaluate to everything that can be evaluated behind this sign or to everything behind this sign plus the rest of the rule. For example:
password1/+2/+/3+4
it’s gonna evaluate to [“password1”, “password12”, “password123”, password1234”].
?^ or ^? If there wouldn’t be a special case built in the compiler for this scenario whenever the user would try to apply this both operations the compiler would either apply the rule to a character we don’t want to derive or it would create a double of one of the outcomes.
IMPORTANT: all the characters listed above as operators are reserved characters, so to use them one should proceed by placing a backslach behind them. Keep in mind, that the blackslash itself is a reserved character as well.
pass(w\ord/phrase)?
The rule from above it’s going to give us an error, as backslash needs a reserved character afterwards.
pass(w\\ord/phrase)?
The rule from above it’s going to evaluate to [“passw\ord”, “passphrase”, “pass”].
pass\(w\\ord\/phrase\)\?
But the rule from above it’s going to evaluate to “pass(w\ord/phrase)?”.