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I need to decrypt a file efficiently [closed]

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I am trying to decrypt an encrypted file with unknown key - the only thing I know about it is that the key is an integer x, 0 <= x < 1010 (i.e. a maximum of 10 decimal digits).
public static String enc(String msg, long key) {
String ans = "";
Random rand = new Random(key);
for (int i = 0; i < msg.length(); i = i + 1) {
char c = msg.charAt(i);
int s = c;
int rd = rand.nextInt() % (256 * 256);
int s2 = s ^ rd;
char c2 = (char) (s2);
ans += c2;
}
return ans;
}
private static String tryToDecode(String string) {
String returnedString = "";
long key;
String msg = reader(string);
for (long i = 0; i <= 999999999; i++) {
System.out.println("decoding message with key + " + i);
key = i;
System.out.println("decoding with key: " + i + "\n" + enc(msg, key));
}
return returnedString;
}
I expect to find the plain text
The program works very slowly, is there any way to make it more efficient?

You can use Parallel Array Operations added in JAVA 8 if you are using Java 8 to achive this.
The best fit for you would be to use Spliterator
public void spliterate() {
System.out.println("\nSpliterate:");
int[] src = getData();
Spliterator<Integer> spliterator = Arrays.spliterator(src);
spliterator.forEachRemaining( n -> action(n) );
}
public void action(int value) {
System.out.println("value:"+value);
// Perform some real work on this data here...
}
I am still not clear about your situation. Here some great tutorials and articles to figure out which parallel array operations of java 8 is going to help you ?
http://www.drdobbs.com/jvm/parallel-array-operations-in-java-8/240166287
https://blog.rapid7.com/2015/10/28/java-8-introduction-to-parallelism-and-spliterator/

First things first: You can't println billions of lines. This will take forever, and it's pointless - you won't be able to see the text as it scrolls by, and your buffer won't save billion of lines so you couldn't scroll back up later even if you wanted to. If you prefer (and don't mind it being 2-3% slower than it otherwise would be), you can output once every hundred million keys, just so you can verify your program is making progress.
You can optimize things by not concatenating Strings inside the loop. Strings are immutable, so the old code was creating a rather large number of Strings, especially in the enc method. Normally I'd use a StringBuilder, but in this case a simple character array will meet our needs.
And there's one more thing we need to do that your current code doesn't do: Detect when we have the answer. If we assume that the message will only contain characters from 0-127 with no Unicode or extended ASCII, then we know we have a possible answer when the entire message contains only characters in this range. And we can also use this to further optimize, as we can then immediately discard any message that has a character outside of this range. We don't even have to finish decoding it and can move on to the next key. (If the message is of any length, the odds are that only one key will produce a decoded message with characters in that range - but it's not guaranteed, which is why I do not stop when I get to a valid message. You could probably do that, though.)
Due to the way random numbers are generated in Java, anything in the seed above 32 bits is not used by the encoding/decoding algorithm. So you only need to go up to 4294967295 instead of 9999999999. (This also means the key that was originally used to encode the message might not be the key this program uses to decode it, since 2-3 keys in the 10 digit range will produce the same encoding/decoding.)
private static String tryToDecode4(String msg) {
String returnedString = "";
for (long i=0; i<=4294967295l; i++)
{
if (i % 100000000 == 0) // This part is just to see that it's making progress. Remove if desired for a small speed gain.
System.out.println("Trying " + i);
char[] decoded = enc4(msg, i);
if (decoded == null)
continue;
returnedString = String.valueOf(decoded);
System.out.println("decoding with key: " + i + " " + returnedString);
}
return returnedString;
}
private static char[] enc4(String msg, long key) {
char[] ansC = new char[msg.length()];
Random rand = new Random(key);
for(int i=0;i<msg.length();i=i+1)
{
char c = msg.charAt(i);
int s = c;
int rd = rand.nextInt()%(256*256);
int s2 = s^rd;
char c2 = (char)(s2);
if (c2 > 127)
return null;
ansC[i] = c2;
}
return ansC;
}
This code finished running in a little over 3 minutes on my machine, with a message of "Hello World".
This code will not work well for very short messages (3-4 characters or less.) It will not work if the message contains Unicode or extended ASCII, although it could easily be modified to do so if you know the range of characters that might be in the message.

Print out Yijing Hexagram Symbols

I encountered a problem while coding and I can't seem to find where I messed up or even why I get a wrong result.
First, let me explain the task.
It's about "Yijing Hexagram Symbols".
The left one is the original and the right one is the result that my code should give me.
Basically every "hexagram" contains 6 lines that can be either diveded or not.
So there are a total of
2^6 = 64 possible "hexagrams"
The task is to calculate and code a methode to print all possible combinations.
Thats what I have so far :
public class test {
public String toBin (int zahl) {
if(zahl ==0) return "0";
if (zahl ==1 ) return "1";
return ""+(toBin( zahl/2)+(zahl%2));
}
public void show (String s) {
for (char c : s.toCharArray()){
if (c == '1'){
System.out.println("--- ---");
}
if(c=='0'){
System.out.println("-------");
}
}
}
public void ausgeben (){
for(int i = 0 ; i < 64; i++) {
show (toBin(i));
}
}
}
The problem is, when I test the 'show'-methode with "10" I get 3 lines and not 2 as intended.
public class runner {
public static void main(String[] args){
test a = new test();
a.ausgeben();
a.show("10");
}
}
Another problem I've encoutered is, that since I'm converting to binary i sometimes have not enough lines because for example 10 in binary is 0001010 but the first "0" are missing. How can I implement them in an easy way without changing much ?
I am fairly new to all this so if I didn't explain anything enough or made any mistakes feel free to tell me.

You may find it easier if you use the Integer.toBinaryString method combined with the String.format and String.replace methods.
String binary = String.format("%6s", Integer.toBinaryString(zahl)).replace(' ', '0');
This converts the number to binary, formats it in a field six spaces wide (with leading spaces as necessary), and then replaces the spaces with '0'.

Well, there are many ways to pad a string with zeros, or create a binary string that is already padded with zeros.
For example, you could do something like:
public String padToSix( String binStr ) {
return "000000".substring( 0, 5 - binStr.length() ) + binStr;
}
This would check how long your string is, and take as many zeros are needed to fill it up to six from the "000000" string.
Or you could simply replace your conversion method (which is recursive, and that's not really necessary) with one that specializes in six-digit numbers:
public static String toBin (int zahl) {
char[] digits = { '0','0','0','0','0','0' };
int currDigitIndex = 5;
while ( currDigitIndex >= 0 && zahl > 0 ) {
digits[currDigitIndex] += (zahl % 2);
currDigitIndex--;
zahl /= 2;
}
return new String(digits);
}
This one modifies the character array ( which initially has only zeros ) from the right to the left. It adds the value of the current bit to the character at the given place. '0' + 0 is '0', and '0' + 1 is '1'. Because you know in advance that you have six digits, you can start from the right and go to the left. If your number has only four digits, well, the two digits we haven't touched will be '0' because that's how the character array was initialized.
There are really a lot of methods to achieve the same thing.

Your problem reduces to printing all binary strings of length 6. I would go with this code snippet:
String format = "%06d";
for(int i = 0; i < 64; i++)
{
show(String.format(format, Integer.valueOf(Integer.toBinaryString(i))));
System.out.println();
}
If you don't wish to print leading zeros, replace String.format(..) with Integer.toBinaryString(i).

Scala: Function test a string for unique char

Solved! Solution at the bottom.
I'm porting some Java code to Scala for fun and I trapped into a pretty nifty way of bit-shifting in Java. The Java code below takes a String as input and tests if it consists of unique characters.
public static boolean isUniqueChars(String str) {
if (str.length() > 256)return false; }
int checker = 0;
for (int i = 0; i < str.length(); i++) {
int val = str.charAt(i) - 'a';
if ((checker & (1 << val)) > 0) return false;
checker |= (1 << val);
}
return true;
Full listing is here: https://github.com/marvin-hansen/ctci/blob/master/java/Chapter%201/Question1_1/Question.java
How the code exactly works is explained here:
How does this Java code which determines whether a String contains all unique characters work?
Porting this directly to Scala doesn't really work so I'm looking for a more functional way to re-write the stuff above.
I have tried BigInt & BitSet
def isUniqueChars2(str : String) : Boolean =
// Java, char's are Unicode so there are 32768 values
if (str.length() > 32768) false
val checker = BigInt(1)
for(i <- 0 to str.length){
val value = str.charAt(i)
if(checker.testBit(value)) false
checker.setBit(value)
}
true
}
This works, however, but without bit-shifting and without lowercase assumption.
Performance is rather unknown ....
However, I would like to do a more functional style solution.
Thanks to user3189923 for the solution.
def isUniqueChars(str : String) = str.distinct == str
That's it. Thank you.

str.distinct == str
In general, method distinct preserves order of occurrence after removing duplicates. Consider
implicit class RichUnique(val str: String) extends AnyVal {
def isUniqueChars() = str.distinct == str
}
and so
"abc".isUniqueChars
res: Boolean = true
"abcc".isUniqueChars
res: Boolean = false

How about:
str.toSet.size == str.size
?

Efficiently parse single digit arithmetic expression

How would you efficiently (optimizing for runtime but also keeping space at a minimum) parse and evaluate a single digit arithmetic expression in Java.
The following arithmetic expressions are all valid:
eval("-5")=-5
eval("+4")=4
eval("4")=4
eval("-7+2-3")=-8
eval("5+7")=12
My approach is to iterate over all elements, keeping track of the current arithmetic operation using a flag, and evaluate digit by digit.
public int eval(String s){
int result = 0;
boolean add = true;
for(int i = 0; i < s.length(); i++){
char current = s.charAt(i);
if(current == '+'){
add = true;
} else if(current == '-'){
add = false;
} else {
if(add){
result += Character.getNumericValue(current);
} else {
result -= Character.getNumericValue(current);
}
}
}
return result;
}
Is this the only optimal solution? I have tried to use stacks to keep track of the arithmetic operator, but I am not sure this is any more efficient. I also have not tried regular expressions. I only ask because I gave the above solution in an interview and was told it is sub-optimal.

This seems a bit more compact. It certainly requires fewer lines and conditionals. The key is addition is the "default" behavior and each minus sign you encounter changes the sign of what you want to add; provided you remember to reset the sign after each addition.
public static int eval(String s){
int result = 0;
int sign = 1;
for(int i = 0; i < s.length(); i++){
char current = s.charAt(i);
switch (current)
{
case '+': break;
case '-': sign *= -1; break;
default:
result += sign * Character.getNumericValue(current);
sign = 1;
break;
}
}
return result;
}
As a note, I don't think yours produces correct results for adding a negative, e.g., "4- -3". Your code produces 1, rather than the correct value of 7. On the other hand, mine allows expressions such as "5+-+-3", which would produce the result 8 (I suppose that's correct? :). However, you didn't list validation as a requirement and neither of us are checking for sequential digits, alpha characters, white space, etc. If we assume the data is properly formatted, the above implementation should work. I don't see how adding data structures (such as queues) could possibly be helpful here. I'm also assuming just addition and subtraction.
These test cases produce the following results:
System.out.println(eval("1+2+3+4"));
System.out.println(eval("1--3"));
System.out.println(eval("1+-3-2+4+-3"));
10
4
-3

You need to lookup up 'recursive descent expression parser' or the Dijkstra shunting-yard algorithm. Your present approach is doomed to failure the moment you have to cope with operator precedence or parentheses. You also need to forget about regular expressions and resign yourself to writing a proper scanner.

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.