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How to transfer an outside recursion program into a non-recursive form (using stack not CPS)？

there are many questions about how to convert recursive to non-recursive, and I also can convert some recursive programs to non-recursive form
note: I use an generalized way (user defined Stack), because I think it is easy to understand, and I use Java, so can not use GOTO keyword.
Things don't always go so well, when I meet the Backtracking, I am stuck. for example, The subset problem. and my code is here: recursive call with loop
when i use user defined Stack to turn it to non-recursive form. I do not know how to deal with the loop (in the loop existing recursive call).
I googled found that there is many methods such as CPS. and I know there is an iterative template of subset problem. but i only want to use user defined Stack to solve.
Can someone provide some clues to turn this kind of recursive(recursive with loop) to non-recursive form(by using user defined Stack, not CPS etc..) ?
here is my code recursive to non-recusive(Inorder-Traversal), because of there is no loop with recursive call, so i can easily do it. also when recursive program with a return value, I can use a reference and pass it to the function as a param. from the code, I use the Stack to simulated the recursive call, and use "state" variable to the next call point(because java does not allow using GOTO).
The following is the information I have collected. It seems that all of them does not satisfy the question I mentioned(some use goto that java not allowed, some is very simple recursive means that no nested recursive call or recursive call with loop ).
1 Old Dominion University
2 codeproject
----------------------------------Split Line--------------------------------------
Thks u all. after when I post the question... It took me all night to figure it out. here is my solution: non-recursive subset problem solution, and the comment of the code is my idea.
To sum up. what i stuck before is how to deal with the foo-loop, actually, we can just simply ignore it. because we are using loop+stack, we can do a simple judgment on whether to meet the conditions.

On your stack, have you thought about pushing i (the iteration variable)?
By doing this, when you pop this value, you know at which iteration of the loop you were before you pushed on the stack and therefore, you can iterate to the next i and continue your algorithm.

Non-negative numbers only for simplicity. (Also no IntFunction.)
The power function, as defined here, is a very simple case.
int power(int x, int exponent) {
if (exponent == 0) {
return 1;
} else if (exponent % 2 == 0) {
int y = power(x, exponent /2);
return y * y;
} else {
return x * power(x, exponent - 1);
}
}
Now the stack is there to do in the reverse order to a partial result, what you did in recursion with the result.
int power(final int x, int exponent) {
Stack<Function<Integer, Integer>> opStack = new Stack<>();
final Function<Integer, Integer> square = n -> n * n;
final Function<Integer, Integer> multiply = n -> x * n;
while (exponent > 0) {
if (exponent % 2 == 0) {
exponent /= 2;
opStack.push(square);
} else {
--exponent;
opStack.push(multiply);
}
}
int result = 1;
while (!opStack.isEmpty()) {
result = opStack.pop().apply(result);
}
return result;
}
An alternative would be to "encode" the two branches of if-else (odd/even exponent) by a boolean:
int power(final int x, int exponent) {
BooleanStack stack = new BooleanStack<>();
while (exponent > 0) {
boolean even = exponent % 2 == 0;
stack.push(even);
if (even) {
exponent /= 2;
} else {
--exponent;
}
}
int result = 1;
while (!stack.isEmpty()) {
result *= stack.pop() ? result : x;
}
return result;
}
So one has to distinghuish:
what one does to prepare the recursive arguments
what one does with the partial results of the recursive calls
how one can merge/handle several recursive calls in the function
exploit nice things, like x being a final constant
Not difficult, puzzling maybe, so have fun.

Find all valid words when given a string of characters (Recursion / Binary Search)

I'd like some feedback on a method I tried to implement that isn't working 100%. I'm making an Android app for practice where the user is given 20 random letters. The user then uses these letters to make a word of whatever size. It then checks a dictionary to see if it is a valid English word.
The part that's giving me trouble is with showing a "hint". If the user is stuck, I want to display the possible words that can be made. I initially thought recursion. However, with 20 letters this can take quite a long time to execute. So, I also implemented a binary search to check if the current recursion path is a a prefix to anything in the dictionary. I do get valid hints to be output however it's not returning all possible words. Do I have a mistake here in my recursion thinking? Also, is there a recommended, faster algorithm? I've seen a method in which you check each word in a dictionary and see if the characters can make each word. However, I'd like to know how effective my method is vs. that one.
private static void getAllWords(String letterPool, String currWord) {
//Add to possibleWords when valid word
if (letterPool.equals("")) {
//System.out.println("");
} else if(currWord.equals("")){
for (int i = 0; i < letterPool.length(); i++) {
String curr = letterPool.substring(i, i+1);
String newLetterPool = (letterPool.substring(0, i) + letterPool.substring(i+1));
if(dict.contains(curr)){
possibleWords.add(curr);
}
boolean prefixInDic = binarySearch(curr);
if( !prefixInDic ){
break;
} else {
getAllWords(newLetterPool, curr);
}
}
} else {
//Every time we add a letter to currWord, delete from letterPool
//Attach new letter to curr and then check if in dict
for(int i=0; i<letterPool.length(); i++){
String curr = currWord + letterPool.substring(i, i+1);
String newLetterPool = (letterPool.substring(0, i) + letterPool.substring(i+1));
if(dict.contains(curr)) {
possibleWords.add(curr);
}
boolean prefixInDic = binarySearch(curr);
if( !prefixInDic ){
break;
} else {
getAllWords(newLetterPool, curr);
}
}
}
private static boolean binarySearch(String word){
int max = dict.size() - 1;
int min = 0;
int currIndex = 0;
boolean result = false;
while(min <= max) {
currIndex = (min + max) / 2;
if (dict.get(currIndex).startsWith(word)) {
result = true;
break;
} else if (dict.get(currIndex).compareTo(word) < 0) {
min = currIndex + 1;
} else if(dict.get(currIndex).compareTo(word) > 0){
max = currIndex - 1;
} else {
result = true;
break;
}
}
return result;
}

The simplest way to speed up your algorithm is probably to use a Trie (a prefix tree)
Trie data structures offer two relevant methods. isWord(String) and isPrefix(String), both of which take O(n) comparisons to determine whether a word or prefix exist in a dictionary (where n is the number of letters in the argument). This is really fast because it doesn't matter how large your dictionary is.
For comparison, your method for checking if a prefix exists in your dictionary using binary search is O(n*log(m)) where n is the number of letters in the string and m is the number of words in the dictionary.
I coded up a similar algorithm to yours using a Trie and compared it to the code you posted (with minor modifications) in a very informal benchmark.
With 20-char input, the Trie took 9ms. The original code didn't complete in reasonable time so I had to kill it.
Edit:
As to why your code doesn't return all hints, you don't want to break if the prefix is not in your dict. You should continue to check the next prefix instead.

Is there a recommended, faster algorithm?
See Wikipedia article on "String searching algorithm", in particular the section named "Algorithms using a finite set of patterns", where "finite set of patterns" is your dictionary.
The Aho–Corasick algorithm listed first might be a good choice.

Parse an Expression to its components and sub components

I need to parse an expression such as: neg(and(X,Y))
I need it to come out with the Abstract Stack Machine Code Such as for the example above:
LOAD X;
LOAD Y;
EXEC and;
EXEC neg;
But for now the machine code is not an issue, how can i parse / break up my input string of an expression into all its sub expressions?
I have tried to find the first bracket and then concat from that to the last bracket but that then gives isuess if you have a inner expression?
code that i have tried: (please not it is still very much in the development phase)
private boolean evaluateExpression(String expression) {
int brackets = 0;
int beginIndex = -1;
int endIndex = -1;
for (int i = 0; i < expression.length(); i++) {
if (expression.charAt(i) == '(') {
brackets++;
if (brackets == 0) {
endIndex = i;
System.out.println("the first expression ends at " + i);
}
}
if (expression.charAt(i) == ')') {
brackets--;
if (brackets == 0) {
endIndex = i;
System.out.println("the first expression ends at " + i);
}
}
}
// Check for 1st bracket
for (int i = 0; i < expression.length(); i++) {
if (expression.charAt(i) == '(') {
beginIndex = i;
break;
}
}
String subExpression = expression.substring(beginIndex, endIndex);
System.out.println("Sub expression: " + subExpression);
evaluateExpression(subExpression);
return false;
}
I am just looking for a basic solution, It only has to do: and, or, neg

The expressions you are trying to parse are actually making a Context Free Language, which can be represented as a Context Free Grammer.
You can create a context free grammer that represents this language of expressions, and use a CFG parser to parse it.
One existing java tool that does it (and more) is JavaCC, though it could be an overkill here.
Another algorithm to parse sentences using a CFG is CYK, which is fairly easy to program and use.
In here, the CFG representing the available expressions are:
S -> or(S,S)
S -> and(S,S)
S -> not(S)
S -> x | for each variable x
Note that though this is relatively simple CFG - the language it describes is irregular, so if you were hoping for regex - it's probably not the way to go.

Actually if you want your parser to be strong enough to deal with most cases, you would like to use a tokenizer(java has a implemented tokenizer class) to token the string first, then try to recognize each expression, storing operands and operators in a tree structure, then evaluate them recursively.
If you only want to deal with some simple situations, remember to use recursion, that is the core part~

Parsing things like this is typically done using syntax trees, using some type of preference for order of operations. An example for what you have posted would be as follows:
Processing items left to right the tree would be populated like this
1arg_fcall(neg)
2arg_fcall(and)
Load Y
Load X
Now we can recursively visit this tree bottom to top to get
Load X
Load Y
EXEC and //on X and Y
EXEC neg //on result of and

Changing recursive method to iterative

i have recrusive function which works fine. The problem is it gives stackoverflow error when the number of lines are huge. I want to put it in iterative, probably using a for loop. Need some help in doing it.
private TreeSet validate(int curLine, TreeSet errorSet) {
int increment = 0;
int nextLine = 0;
if (curLine == lines.length || errorSet.size() != 0) {
return errorSet;
} else {
String line = lines[curLine];
//validation starts. After validation, line is incremented as per the requirements
increment = 1 //As per requirement. Depends on validation results of the line
if (increment > 0) {
try{
Thread.currentThread().sleep(100);
}catch(Exception ex){
System.out.println(ex);
}
nextLine = (curLine + increment);
validate(nextLine, errorSet);
}
}
return errorSet;
}
Poster's description of the method:
The method does validates textlines, these lines has instructions of how much line has to be skipped, if the line is valid. So, if the line is valid that many of lines will be skipped using the increment. if the line is not valid increment will be 0.

I'm not sure why this was ever recursive in the first place. This is perfectly suited for the use of a FOR loop. use something like so:
private TreeSet validate(int curLine, TreeSet errorSet) {
int increment = 0;
if (errorSet.size() != 0)
return errorSet;
for (int curLine = 0; curLine < lines.Length; curLine += increment)
{
// put your processing logic in here
// set the proper increment here.
}
}
If the increment is always going to be 1, then you can just use curr++ instead of curLine += increment

for(String line : lines) {
// validate line here
if(!errorSet.isEmpty()) {
break;
}
}

The solution for your problem could be simple for loop or while, with logical expression for stop condition. Typically we use for loop when we have to pass through all elements of Iterable or array. In case when we are not aware how many loops we are going to do we use a while loop. Advantage of for loop over while, is that we for free have localized variables so we ca not use them out side of the loop, therefore we reduce possibility to have some bug.
You problem is that you have to break the program on two conditions:
When errorSet is not empty.
When the array of lines have no longer items.
As contradiction, we can say that your program should continue:
Until errorSet is empty,
and until line number is smaller than array size where they are stored.
This provide us to simply expression
errorSet.isEmpty()
lineNumber < lines.length()
We can combine them using logical operator && and use as a stop rule in for loop.
for(int lineNumber= 0; errorSet.isEmpty() && lineNumber< lines.length(); lineNumber++) {
//code to operate
}
Note:
Typically for logical expression is used operator &&, that assure that every part of the logical expression is evaluated. An alternative for this is &, that in case of false do not operate longer and return false. We could be tempted to use this operator for this expression but i would be bad idea. Because when we would traversed all lines without error code will generate IndexOutOfBoundException, if we switch the places then we would not have any optimization as first expression would be evaluated same number of times.

Check whether a string is parsable into Long without try-catch?

Long.parseLong("string") throws an error if string is not parsable into long.
Is there a way to validate the string faster than using try-catch?
Thanks

You can create rather complex regular expression but it isn't worth that. Using exceptions here is absolutely normal.
It's natural exceptional situation: you assume that there is an integer in the string but indeed there is something else. Exception should be thrown and handled properly.
If you look inside parseLong code, you'll see that there are many different verifications and operations. If you want to do all that stuff before parsing it'll decrease the performance (if we are talking about parsing millions of numbers because otherwise it doesn't matter). So, the only thing you can do if you really need to improve performance by avoiding exceptions is: copy parseLong implementation to your own function and return NaN instead of throwing exceptions in all correspondent cases.

From commons-lang StringUtils:
public static boolean isNumeric(String str) {
if (str == null) {
return false;
}
int sz = str.length();
for (int i = 0; i < sz; i++) {
if (Character.isDigit(str.charAt(i)) == false) {
return false;
}
}
return true;
}

You could do something like
if(s.matches("\\d*")){
}
Using regular expression - to check if String s is full of digits.
But what do you stand to gain? another if condition?

org.apache.commons.lang3.math.NumberUtils.isParsable(yourString) will determine if the string can be parsed by one of: Integer.parseInt(String), Long.parseLong(String), Float.parseFloat(String) or Double.parseDouble(String)
Since you are interested in Longs you could have a condition that checks for isParsable and doesn't contain a decimal
if (NumberUtils.isParsable(yourString) && !StringUtils.contains(yourString,".")){ ...

This is a valid question because there are times when you need to infer what type of data is being represented in a string. For example, you may need to import a large CSV into a database and represent the data types accurately. In such cases, calling Long.parseLong and catching an exception can be too slow.
The following code only handles ASCII decimal:
public class LongParser {
// Since tryParseLong represents the value as negative during processing, we
// counter-intuitively want to keep the sign if the result is negative and
// negate it if it is positive.
private static final int MULTIPLIER_FOR_NEGATIVE_RESULT = 1;
private static final int MULTIPLIER_FOR_POSITIVE_RESULT = -1;
private static final int FIRST_CHARACTER_POSITION = 0;
private static final int SECOND_CHARACTER_POSITION = 1;
private static final char NEGATIVE_SIGN_CHARACTER = '-';
private static final char POSITIVE_SIGN_CHARACTER = '+';
private static final int DIGIT_MAX_VALUE = 9;
private static final int DIGIT_MIN_VALUE = 0;
private static final char ZERO_CHARACTER = '0';
private static final int RADIX = 10;
/**
* Parses a string representation of a long significantly faster than
* <code>Long.ParseLong</code>, and avoids the noteworthy overhead of
* throwing an exception on failure. Based on the parseInt code from
* http://nadeausoftware.com/articles/2009/08/java_tip_how_parse_integers_quickly
*
* #param stringToParse
* The string to try to parse as a <code>long</code>.
*
* #return the boxed <code>long</code> value if the string was a valid
* representation of a long; otherwise <code>null</code>.
*/
public static Long tryParseLong(final String stringToParse) {
if (stringToParse == null || stringToParse.isEmpty()) {
return null;
}
final int inputStringLength = stringToParse.length();
long value = 0;
/*
* The absolute value of Long.MIN_VALUE is greater than the absolute
* value of Long.MAX_VALUE, so during processing we'll use a negative
* value, then we'll multiply it by signMultiplier before returning it.
* This allows us to avoid a conditional add/subtract inside the loop.
*/
int signMultiplier = MULTIPLIER_FOR_POSITIVE_RESULT;
// Get the first character.
char firstCharacter = stringToParse.charAt(FIRST_CHARACTER_POSITION);
if (firstCharacter == NEGATIVE_SIGN_CHARACTER) {
// The first character is a negative sign.
if (inputStringLength == 1) {
// There are no digits.
// The string is not a valid representation of a long value.
return null;
}
signMultiplier = MULTIPLIER_FOR_NEGATIVE_RESULT;
} else if (firstCharacter == POSITIVE_SIGN_CHARACTER) {
// The first character is a positive sign.
if (inputStringLength == 1) {
// There are no digits.
// The string is not a valid representation of a long value.
return null;
}
} else {
// Store the (negative) digit (although we aren't sure yet if it's
// actually a digit).
value = -(firstCharacter - ZERO_CHARACTER);
if (value > DIGIT_MIN_VALUE || value < -DIGIT_MAX_VALUE) {
// The first character is not a digit (or a negative sign).
// The string is not a valid representation of a long value.
return null;
}
}
// Establish the "maximum" value (actually minimum since we're working
// with negatives).
final long rangeLimit = (signMultiplier == MULTIPLIER_FOR_POSITIVE_RESULT)
? -Long.MAX_VALUE
: Long.MIN_VALUE;
// Capture the maximum value that we can multiply by the radix without
// overflowing.
final long maxLongNegatedPriorToMultiplyingByRadix = rangeLimit / RADIX;
for (int currentCharacterPosition = SECOND_CHARACTER_POSITION;
currentCharacterPosition < inputStringLength;
currentCharacterPosition++) {
// Get the current digit (although we aren't sure yet if it's
// actually a digit).
long digit = stringToParse.charAt(currentCharacterPosition)
- ZERO_CHARACTER;
if (digit < DIGIT_MIN_VALUE || digit > DIGIT_MAX_VALUE) {
// The current character is not a digit.
// The string is not a valid representation of a long value.
return null;
}
if (value < maxLongNegatedPriorToMultiplyingByRadix) {
// The value will be out of range if we multiply by the radix.
// The string is not a valid representation of a long value.
return null;
}
// Multiply by the radix to slide all the previously parsed digits.
value *= RADIX;
if (value < (rangeLimit + digit)) {
// The value would be out of range if we "added" the current
// digit.
return null;
}
// "Add" the digit to the value.
value -= digit;
}
// Return the value (adjusting the sign if needed).
return value * signMultiplier;
}
}

You can use java.util.Scanner
Scanner sc = new Scanner(s);
if (sc.hasNextLong()) {
long num = sc.nextLong();
}
This does range checking etc, too. Of course it will say that "99 bottles of beer" hasNextLong(), so if you want to make sure that it only has a long you'd have to do extra checks.

This case is common for forms and programs where you have the input field and are not sure if the string is a valid number. So using try/catch with your java function is the best thing to do if you understand how try/catch works compared to trying to write the function yourself. In order to setup the try catch block in .NET virtual machine, there is zero instructions of overhead, and it is probably the same in Java. If there are instructions used at the try keyword then these will be minimal, and the bulk of the instructions will be used at the catch part and that only happens in the rare case when the number is not valid.
So while it "seems" like you can write a faster function yourself, you would have to optimize it better than the Java compiler in order to beat the try/catch mechanism you already use, and the benefit of a more optimized function is going to be very minimal since number parsing is quite generic.
If you run timing tests with your compiler and the java catch mechanism you already described, you will probably not notice any above marginal slowdown, and by marginal I mean it should be almost nothing.
Get the java language specification to understand the exceptions more and you will see that using such a technique in this case is perfectly acceptable since it wraps a fairly large and complex function. Adding on those few extra instructions in the CPU for the try part is not going to be such a big deal.

I think that's the only way of checking if a String is a valid long value. but you can implement yourself a method to do that, having in mind the biggest long value.

There are much faster ways to parse a long than Long.parseLong. If you want to see an example of a method that is not optimized then you should look at parseLong :)
Do you really need to take into account "digits" that are non-ASCII?
Do you really need to make several methods calls passing around a radix even tough you're probably parsing base 10?
:)
Using a regexp is not the way to go: it's harder to determine if you're number is too big for a long: how do you use a regexp to determine that 9223372036854775807 can be parsed to a long but that 9223372036854775907 cannot?
That said, the answer to a really fast long parsing method is a state machine and that no matter if you want to test if it's parseable or to parse it. Simply, it's not a generic state machine accepting complex regexp but a hardcoded one.
I can both write you a method that parses a long and another one that determines if a long can be parsed that totally outperforms Long.parseLong().
Now what do you want? A state testing method? In that case a state testing method may not be desirable if you want to avoid computing twice the long.
Simply wrap your call in a try/catch.
And if you really want something faster than the default Long.parseLong, write one that is tailored to your problem: base 10 if you're base 10, not checking digits outside ASCII (because you're probably not interested in Japanese's itchi-ni-yon-go etc.).

Hope this helps with the positive values. I used this method once for validating database primary keys.
private static final int MAX_LONG_STR_LEN = Long.toString(Long.MAX_VALUE).length();
public static boolean validId(final CharSequence id)
{
//avoid null
if (id == null)
{
return false;
}
int len = id.length();
//avoid empty or oversize
if (len < 1 || len > MAX_LONG_STR_LEN)
{
return false;
}
long result = 0;
// ASCII '0' at position 48
int digit = id.charAt(0) - 48;
//first char cannot be '0' in my "id" case
if (digit < 1 || digit > 9)
{
return false;
}
else
{
result += digit;
}
//start from 1, we already did the 0.
for (int i = 1; i < len; i++)
{
// ASCII '0' at position 48
digit = id.charAt(i) - 48;
//only numbers
if (digit < 0 || digit > 9)
{
return false;
}
result *= 10;
result += digit;
//if we hit 0x7fffffffffffffff
// we are at 0x8000000000000000 + digit - 1
// so negative
if (result < 0)
{
//overflow
return false;
}
}
return true;
}

Try to use this regular expression:
^(-9223372036854775808|0)$|^((-?)((?!0)\d{1,18}|[1-8]\d{18}|9[0-1]\d{17}|92[0-1]\d{16}|922[0-2]\d{15}|9223[0-2]\d{14}|92233[0-6]\d{13}|922337[0-1]\d{12}|92233720[0-2]\d{10}|922337203[0-5]\d{9}|9223372036[0-7]\d{8}|92233720368[0-4]\d{7}|922337203685[0-3]\d{6}|9223372036854[0-6]\d{5}|92233720368547[0-6]\d{4}|922337203685477[0-4]\d{3}|9223372036854775[0-7]\d{2}|922337203685477580[0-7]))$
It checks all possible numbers for Long.
But as you know in Java Long can contain additional symbols like +, L, _ and etc. And this regexp doesn't validate these values. But if this regexp is not enough for you, you can add additional restrictions for it.

Guava Longs.tryParse("string") returns null instead of throwing an exception if parsing fails. But this method is marked as Beta right now.

You could try using a regular expression to check the form of the string before trying to parse it?

A simple implementation to validate an integer that fits in a long would be:
public static boolean isValidLong(String str) {
if( str==null ) return false;
int len = str.length();
if (str.charAt(0) == '+') {
return str.matches("\\+\\d+") && (len < 20 || len == 20 && str.compareTo("+9223372036854775807") <= 0);
} else if (str.charAt(0) == '-') {
return str.matches("-\\d+") && (len < 20 || len == 20 && str.compareTo("-9223372036854775808") <= 0);
} else {
return str.matches("\\d+") && (len < 19 || len == 19 && str.compareTo("9223372036854775807") <= 0);
}
}
It doesn't handle octal, 0x prefix or so but that is seldom a requirement.
For speed, the ".match" expressions are easy to code in a loop.