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expression evaluation with right-associativity in java

I am trying to solve a problem in which I have to solve a given expression consisting of one or more initialization in a same string with no operator precedence (although with bracketed sub-expressions). All the operators have right precedence so I have to evaluate it from right to left. I am confused how to proceed for the given problem. Detailed problem is given here : http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=108

I'll give you some ideas to try:
First off, you need to recursively evaluate inside brackets. You want to do brackets from most nested to least nested, so use a regex that matches brackets with no ) inside of them. Substring the result of the computation into the part of the string the bracketed expression took up.
If there are no brackets, then now you need to evaluate operators. The reason why the question requires right precedence is to force you to think about how to answer it - you can't just read the string and do calculations. You have to consider the whole string THEN start doing calculations, which means storing some structure describing it. There's a number of strategies you could use to do this, for example:
-You could tokenize the string, either using a scanner or regexes - continually try to see if the next item in the string is a number or which of the operators it is, and push what kind of token it is and its value onto a list. Then, you can evaluate the list from right to left using some kind of case/switch structure to determine what to do for each operator (either that, or each operator is associated with what it does to numbers). = itself would address a map of variable name keys to values, and insert the value under that variable's key, and then return (to be placed into the list) the value it produced, so it can be used for another assignment. It also seems like - can be determined as to whether it's subtraction or a negative number by whether there's a space on its right or not.
-Instead of tokenization, you could use regexes on the string as a whole. But tokenization is more robust. I tried to build a calculator based on applying regexes to the whole string over and over but it's so difficult to get all the rules right and I don't recommend it.
I've written an expression evaluating calculator like this before, so you can ask me questions if you run into specific problems.

Elegant way to do variable substitution in a java string

Pretty simple question and my brain is frozen today so I can't think of an elegant solution where I know one exists.
I have a formula which is passed to me in the form "A+B"
I also have a mapping of the formula variables to their "readable names".
Finally, I have a formula parser which will calculate the value of the formula, but only if its passed with the readable names for the variables.
For example, as an input I get
String formula = "A+B"
String readableA = "foovar1"
String readableB = "foovar2"
and I want my output to be "foovar1+foovar2"
The problem with a simple find and replace is that it can be easily be broken because we have no guarantees on what the 'readable' names are. Lets say I take my example again with different parameters
String formula = "A+B"
String readableA = "foovarBad1"
String readableB = "foovarAngry2"
If I do a simple find and replace in a loop, I'll end up replacing the capital A's and B's in the readable names I have already replaced.
This looks like an approximate solution but I don't have brackets around my variables
How to replace a set of tokens in a Java String?

That link you provided is an excellent source since matching using patterns is the way to go. The basic idea here is first get the tokens using a matcher. After this you will have Operators and Operands
Then, do the replacement individually on each Operand.
Finally, put them back together using the Operators.

A somewhat tedious solution would be to scan for all occurences of A and B and note their indexes in the string, and then use StringBuilder.replace(int start, int end, String str) method. (in naive form this would not be very efficient though, approaching smth like square complexity, or more precisely "number of variables" * "number of possible replacements")
If you know all of your operators, you could do split on them (like on "+") and then replace individual "A" and "B" (you'd have to do trimming whitespace chars first of course) in an array or ArrayList.

A simple way to do it is
String foumula = "A+B".replaceAll("\\bA\\b", readableA)
.replaceAll("\\bB\\b", readableB);

Your approach does not work fine that way
Formulas (mathematic Expressions) should be parsed into an expression structure (eg. expression tree).
Such that you have later Operand Nodes and Operator nodes.
Later this expression will be evaluated traversing the tree and considering the mathematical priority rules.
I recommend reading more on Expression parsing.

Matching Only
If you don't have to evaluate the expression after doing the substitution, you might be able to use a regex. Something like (\b\p{Alpha}\p{Alnum}*\b)
or the java string "(\\b\\p{Alpha}\\p{Alnum}*\\b)"
Then use find() over and over to find all the variables and store their locations.
Finally, go through the locations and build up a new string from the old one with the variable bits replaced.
Not that It will not do much checking that the supplied expression is reasonable. For example, it wouldn't mind at all if you gave it )A 2 B( and would just replace the A and B (like )XXX 2 XXX(). I don't know if that matters.
This is similar to the link you supplied in your question except you need a different regular expression than they used. You can go to http://www.regexplanet.com/advanced/java/index.html to play with regular expressions and figure out one that will work. I used it with the one I suggested and it finds what it needs in A+B and A + (C* D ) just fine.
Parsing
You parse the expression using one of the available parser generators (Antlr or Sable or ...) or find an algebraic expression parser available as open source and use it. (You would have to search the web to find those, I haven't used one but suspect they exist.)
Then you use the parser to generate a parsed form of the expression, replace the variables and reconstitute the string form with the new variables.
This one might work better but the amount of effort depends on whether you can find existing code to use.
It also depends on whether you need to validate the expression is valid according to the normal rules. This method will not accept invalid expressions, most likely.

Combining (OR) arbitrary regular expressions

tl;dr Is there a way to OR/combine arbitrary regexes into a single regex (for matching, not capturing) in Java?
In my application I receive two lists from the user:
list of regular expressions
list of strings
and I need to output a list of the strings in (2) that were not matched by any of the regular expressions in (1).
I have the obvious naive implementation in place (iterate over all strings in (2); for each string iterate over all patterns in (1); if no pattern match the string add it to the list that will be returned) but I was wondering if it was possible to combine all patterns into a single one and let the regex compiler exploit optimization opportunities.
The obvious way to OR-combine regexes is obviously (regex1)|(regex2)|(regex3)|...|(regexN) but I'm pretty sure this is not the correct thing to do considering that I have no control over the individual regexes (e.g. they could contain all manners of back/forward references). I was therefore wondering if you can suggest a better way to combine arbitrary regexes in java.
note: it's only implied by the above, but I'll make it explicit: I'm only matching against the string - I don't need to use the output of the capturing groups.

Some regex engines (e.g. PCRE) have the construct (?|...). It's like a non-capturing group, but has the nice feature that in every alternation groups are counted from the same initial value. This would probably immediately solve your problem. So if switching the language for this task is an option for you, that should do the trick.
[edit: In fact, it will still cause problems with clashing named capturing groups. In fact, the pattern won't even compile, since group names cannot be reused.]
Otherwise you will have to manipulate the input patterns. hyde suggested renumbering the backreferences, but I think there is a simpler option: making all groups named groups. You can assure yourself that the names are unique.
So basically, for every input pattern you create a unique identifier (e.g. increment an ID). Then the trickiest part is finding capturing groups in the pattern. You won't be able to do this with a regex. You will have to parse the pattern yourself. Here are some thoughts on what to look out for if you are simply iterating through the pattern string:
Take note when you enter and leave a character class, because inside character classes parentheses are literal characters.
Maybe the trickiest part: ignore all opening parentheses that are followed by ?:, ?=, ?!, ?<=, ?<!, ?>. In addition there are the option setting parentheses: (?idmsuxU-idmsuxU) or (?idmsux-idmsux:somePatternHere) which also capture nothing (of course there could be any subset of those options and they could be in any order - the - is also optional).
Now you should be left only with opening parentheses that are either a normal capturing group or a named on: (?<name>. The easiest thing might be to treat them all the same - that is, having both a number and a name (where the name equals the number if it was not set). Then you rewrite all of those with something like (?<uniqueIdentifier-md5hashOfName> (the hyphen cannot be actually part of the name, you will just have your incremented number followed by the hash - since the hash is of fixed length there won't be any duplicates; pretty much at least). Make sure to remember which number and name the group originally had.
Whenever you encounter a backslash there are three options:
The next character is a number. You have a numbered backreference. Replace all those numbers with k<name> where name is the new group name you generated for the group.
The next characters are k<...>. Again replace this with the corresponding new name.
The next character is anything else. Skip it. That handles escaping of parentheses and escaping of backslashes at the same time.
I think Java might allow forward references. In that case you need two passes. Take care of renaming all groups first. Then change all the references.
Once you have done this on every input pattern, you can safely combine all of them with |. Any other feature than backreferences should not cause problems with this approach. At least not as long as your patterns are valid. Of course, if you have inputs a(b and c)d then you have a problem. But you will have that always if you don't check that the patterns can be compiled on their own.
I hope this gave you a pointer in the right direction.

Recursively verifying if a string is a valid prefix expression?

I'm rather new to the community but I've seen some helpful posts on here so I thought I'd ask.
I've got a homework question that asks us to recursively check whether a given string is a valid prefix expression given by the two following rules (standard):
Variables (a-z) are prefix expressions
If O is a binary operator and F and E are prefix expressions, OFE
Now, I kind of get the evaluation and have looked at the prefix-to-infix algorithms, but I can't for the life of me figure out how to implement just the evaluation methods (as I only need to check if it's valid, so not +a-b for example).
I know most of the implementation for these problems is done using stacks but I don't see how I would do it recursively here... some help would be tremendously appreciated.

Think of it this way. (I'm not going to write the code, since that's what you need to learn).
You want to check if a certain string is a prefix expression, so you have a function:
boolean isPrefix(string)
Now, there's two way that string could be a prefix:
It's a character from a-z
It's in the format O(prefix)(prefix)
So first, you check if the string has a length of one and is between a-z, and if so, the answer is yes.
Next you can check if the string starts with an O. If it does, you need to test the rest of the string to see if it is composed of two prefix expressions (FE).
So you start iterating from 1 to length, and passing each substring (0->i, i->length) into isPrefix(). If both substrings are also valid prefix expressions, the answer is yes.
Otherwise, the answer is no.
That's pretty much it, but the implementation, however, is up to you.

I'm not sure I entirely understand the point of this, but I imagine you should have some method like checkPrefixIn(String s) that looks at only part of the given String, returns true if it is only a prefix, false if it is only an operator (or invalid character), or the return value of checkPrefixIn(partOfS), where partOfS is a substring of the input s

Regex for checking > 1 upper, lower, digit, and special char

^.(?=.{15,})(?=.\d)(?=.[a-z])(?=.[A-Z])(?=.[!##$%^&+=]).*$
This is the regex I am currently using which will evaluate on 1 of each: upper,lower,digit, and specials of my choosing. The question I have is how do I make it check for 2 of each of these? Also I ask because it is seemingly difficult to write a test case for this as I do not know if it is only evaluating the first set of criteria that it needs. This is for a password, however the requirement is that it needs to be in regex form based on the package we are utilizing.
EDIT
Well as it stands in my haste to validate the expression I forgot to validate my string length. Thanks to Ken and Gumbo on helping me with this.
This is the code I am executing:
I do apologize as regex is not my area.
The password I am using is the following string "$$QiouWER1245", the behavior I am experiencing at the moment is that it randomly chooses to pass or fail. Any thoughts on this?
Pattern pattern = Pattern.compile(regEx);
Matcher match = pattern.matcher(password);
while(match.find()){
System.out.println(match.group());
}
From what I see if it evaluates to true it will throw the value in password back to me else it is an empty string.

Personally, I think a password policy that forces use of all three character classes is not very helpful. You can get the same degree of randomness by letting people make longer passwords. Users will tend to get frustrated and write passwords down if they have to abide by too many password rules (which make the passwords too difficult to remember). I recommend counting bits of entropy and making sure they're greater than 60 (usually requires a 10-14 character password). Entropy per character would depend roughly on the number of characters, the range of character sets they use, and maybe how often they switch between character sets (I would guess that passwords like HEYthere are more predictable than heYThEre).
Another note: do you plan not to count the symbols to the right of the keyboard (period, comma, angle brackets, etc.)?
If you still have to find groups of two characters, why not just repeat each pattern? For example, make (?=.\d) into (?=.\d.*\d).
For your test cases, if you are worried that it would only check the first criteria, then write a test case that makes sure each of the following passwords fails (because one and only one of the criteria is not met in each case): Just for fun I reversed the order of expectation of each character set, though it probably won't make a difference unless someone removes/forgets the ?= at some future date.
!##TESTwithoutnumbers
TESTwithoutsymbols123
&*(testwithoutuppercase456
+_^TESTWITHOUTLOWERCASE3498
I should point out that technically none of these passwords should be acceptable because they use dictionary words, which have about 2 bits of entropy per character instead of something more like 6. However, I realize that it's difficult to write a (maintainable and efficient) regular expression to check for dictionary words.

Try this:
"^(?=(?:\\D*\\d){2})(?=(?:[^a-z]*[a-z]){2})(?=(?:[^A-Z]*[A-Z]){2})(?=(?:[^!##$%^&*+=]*[!##$%^&*+=]){2}).{15,}$"
Here non-capturing groups (?:…) are used to group the conditions and repeat them. I’ve also used the complements of each character class for optimization instead of the universal ..

If I understand your question correctly, you want at least 15 characters, and to require at least 2 uppercase characters, at least 2 lowercase characters, at least 2 digits, and at least 2 special characters. In that case you could it like this:
^.*(?=.{15,})(?=.*\d.*\d)(?=.*[a-z].*[a-z])(?=.*[A-Z].*[A-Z])(?=.*[!##$%^&*+=].*[!##$%^&*+=]).*$
BTW, your original regex had an extra backslash before the \d

I'm not sure that one big regex is the right way to go here. It already looks far too complicated and will be very difficult to change in the future.
My suggestion is to structure the code in the following way:
check that the string has 2 lower case characters
return failure if not found or continue
check that the string has 2 upper case characters
return failure if not found or continue
etc.
This will also allow you to pass out a return code or errors string specifying why the password was not accepted and the code will be much simpler.