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Compare letter in java

I want to compare every letter on file 2 with file 1.
example :
file 1 : my name
file 2 : mi n#mes
i want to get the number of difference is 3, on file 2 : (i, #,and s).
Can you help me
Here is my code
public float getCER(String originalteks,String extractteks){
int end=0;
int start=0;
int different_char=0;
if(originalteks.length()!=extractteks.length()){
different_char=Math.abs(originalteks.length()-extractteks.length());
}
while(start<end){
if(originalteks.charAt(start)!=originalteks.charAt(start++))
different_char++;//jumlah diferent chart
}
return (float) different_char/originalteks.length();
}
And it's only counting the number of characters, not the different characters.

The following implementation tests for the total difference you need and is able to handle strings with different length, by comparing the shorter string to each substring of the longer up to the maximum offset of their difference. From those differences the smallest is chosen. Of course, if handleOffset is false, then we limit ourselves to only the start of the string and adding the difference to the result;
public int getCER(String originalteks,String extractteks, boolean handleOffset){
String shorter = originalteks;
String longer = extractteks;
if (shorter.length() > longer.length()) {
shorter = extractteks;
longer = originalteks;
}
int[] differences = new int[handleOffset ? (longer.length() - shorter.length + 1) : 1];
for (int i = 0; i < differences.length; i++) differences[i] = 0;
for (int i = 0; i < minLength; i++) {
for (j = 0; j < differences.length; j++) {
if (shorter.charAt(i) !== longer.charAt(i + j)) {
differences[j]++;
}
}
}
int min = shorter.length() + 1;
for (int i = 0; i < differences.length; i++) {
if (differences[i] < min) min = differences[i];
}
if (!handleOffset) min += longer.length() - shorter.length();
return min;
}

This should work for you. I just comment my changes within the example.
public int getCER(String originalteks,String extractteks){
int end;
int different_char=0;
//define the shorter end
if(originalteks.length < extractteks.length)
end = originalteks.length();
else
end = extractteks.length();
//no if needed -> same length, diff will be 0
different_char=Math.abs(originalteks.length()-extractteks.length());
for(int start = 0; start < end; start++){
if(originalteks.charAt(start)!=extractteks.charAt(start))
different_char++;//jumlah diferent chart
}
return different_char;
}

Numbers too big for variables

I am to find the last ten digits of 1^1 + 2^2 + 3^3.. + 1000^1000.
Is there any way to find this out with pure logic? I think you can't store a number that big.
This question is from a math competition, but I thought of trying to do this in Java.

You don't need to store number that big, you just need the last ten digits. You can store this in a long.
An efficient way to calculate large powers is to multiply and the squares e.g. 19^19 = 19 * 19^2 * 19 ^ 16 = 19 * 19 ^ 2 * 19^2^2^2^2. When you have value which is greater than 10^10 you can truncate the last 10 digits.
BTW the last ten digits of 1000^1000 is 0000000000 and when your add this to your sum, it's the same as adding zero ;)
Edit: While you don't have to use BigInteger, it is simpler to write.
BigInteger tenDigits = BigInteger.valueOf(10).pow(10);
BigInteger sum = BigInteger.ZERO;
for (int i= 1; i <= 1000; i++) {
BigInteger bi = BigInteger.valueOf(i);
sum = sum.add(bi.modPow(bi, tenDigits));
}
sum = sum.mod(tenDigits);
modPow is more efficient than pow with mod seperately as it doesn't have to calculate very large numbers, only the result of the mod.

You could use BigIntegers...
public static void main(String[] args) {
BigInteger acc = BigInteger.ZERO;
for (int k = 1; k <= 1000; k++) {
BigInteger pow = BigInteger.valueOf(k).pow(k);
acc = acc.add(pow);
}
System.out.println(acc);
}

I believe the problem comes from Project Euler, so it's not just a math problem; it should require some computation as well. I don't know how it could be solved with pencil and paper other than by duplicating the calculations a computer might make. I can't see much in the way of a purely mathematical solution. Mathematics can help us optimize the code, however.
To raise a^n, find the binary expansion of n:
n = n_k x 2^k + n_(k-1) x 2^(k-1) + ... + n_0 x 2^0
where n_i = 0 or 1 are the binary digits of n with the zeroth digit on the right. Then
a^n = a^(n_k x 2^k) x a^(n_(k-1) x 2^(k-1)) x ... x a^(n_0 x 2^0).
We can ignore any factors where n_i = 0, since the factor is then a^0 = 1. The process can be written as an algorithm which is O(log n) time and O(1) space (see below).
Next, as a challenge, in order to avoid the use of BigInteger, we can break the calculation into two parts: finding the answer mod 2^10 and finding the answer mod 5^10. In both cases the numbers in the relevant ranges and products of numbers in the relevant ranges fit into longs. The downside is that we have to use the Chinese Remainder Theorem to recombine the results, but it's not that hard, and it's instructive. The hardest part of using the Chinese Remainder Theorem is finding inverses mod m, but that can be accomplished in a straightforward manner using a modification of the Euclidean algorithm.
Asymptotic running time is O(n log n), space is O(1), and everything fits into a few long variables, no BigInteger or other sophisticated library required.
public class SeriesMod1010 {
public static long pow(long a,long n,long m) { // a^n mod m
long result = 1;
long a2i = a%m; // a^2^i for i = 0, ...
while (n>0) {
if (n%2 == 1) {
result *= a2i;
result %= m;
}
a2i *= a2i;
a2i %= m;
n /= 2;
}
return result;
}
public static long inverse(long a, long m) { // mult. inverse of a mod m
long r = m;
long nr = a;
long t = 0;
long nt = 1;
long tmp;
while (nr != 0) {
long q = r/nr;
tmp = nt; nt = t - q*nt; t = tmp;
tmp = nr; nr = r - q*nr; r = tmp;
}
if (r > 1) return -1; // no inverse
if (t < 0) t += m;
return t;
}
public static void main(String[] args) {
long twoTo10 = 1024;
long sum210 = 0;
for (long i=1; i<=1000; i++) {
sum210 += pow(i,i,twoTo10);
sum210 %= twoTo10;
}
long fiveTo10 = 9_765_625;
long sum510 = 0;
for (long i=1; i<=1000; i++) {
sum510 += pow(i,i,fiveTo10);
sum510 %= fiveTo10;
}
// recombine the numbers with the Chinese remainder theorem
long tenTo10 = 10_000_000_000L;
long answer = sum210 * inverse(fiveTo10,twoTo10) * fiveTo10
+ sum510 * inverse(twoTo10,fiveTo10) * twoTo10;
answer %= tenTo10;
System.out.println(answer);
}
}

use BigIntegers :
import java.math.BigInteger;
public class Program {
public static void main(String[] args) {
BigInteger result = new BigInteger("1");
BigInteger temp = new BigInteger("1");
BigInteger I;
for(int i = 1 ; i < 1001 ; i++){
I = new BigInteger(""+i);
for(int j = 1 ; j < i ; j++){
temp = temp.multiply(I);
}
result = result.multiply(temp);
temp = new BigInteger("1");
}
System.out.println(result);
}
}

It can be solved without BigInteger, because you need to store only 10 last digits on every addition or multiplication operation, using % to avoid overflow:
int n = 1000;
long result = 0;
long tenDigits = 10_000_000_000L;
for (int i = 1; i <= n; i++) {
long r = i;
for (int j = 2; j <= i; j++) {
r = (r * i) % tenDigits;
}
result += r;
}
return result % tenDigits;
Complexity is O(N^2), supposed that multiplication runs in constant time.
Answer: 9110846700.

The decimal base uses 0...9 (10 digits) to represent digits, a number that is in the second position right to left represents Digits * base.length^l2rPosition. Using this logics you can create a class that "pretty much does what your primary school teacher told you to, back when we used paper to calculate stuff, but with a baseN number and base-to-base conversions" I have done this class fully functional in C#, but I don't have time to translate it completely to java, this is about the same logics behind java.math.BigInteger. (with less performance I bet for I used a lot of lists >_>" No time to optimize it now
class IntEx {
ArrayList<Integer> digits = new ArrayList<>();
long baseSize = Integer.MAX_VALUE+1;
boolean negative = false;
public IntEx(int init)
{
set(init);
}
public void set(int number)
{
digits = new ArrayList<>();
int backup = number;
do
{
int index = (int)(backup % baseSize);
digits.add(index);
backup = (int) (backup / baseSize);
} while ((backup) > 0);
}
// ... other operations
private void add(IntEx number)
{
IntEx greater = number.digits.size() > digits.size() ? number : this;
IntEx lesser = number.digits.size() < digits.size() ? number : this;
int leftOvers = 0;
ArrayList<Integer> result = new ArrayList<>();
for (int i = 0; i < greater.digits.size() || leftOvers > 0; i++)
{
int sum;
if (i >= greater.digits.size())
sum = leftOvers;
else if(i >= lesser.digits.size())
sum = leftOvers + greater.digits.get(i);
else
sum = digits.get(i) + number.digits.get(i) + leftOvers;
leftOvers = 0;
if (sum > baseSize-1)
{
while (sum > baseSize-1)
{
sum -= baseSize;
leftOvers += 1;
}
result.add(sum);
}
else
{
result.add(sum);
leftOvers = 0;
}
}
digits = result;
}
private void multiply(IntEx target)
{
ArrayList<IntEx> MultiParts = new ArrayList<>();
for (int i = 0; i < digits.size(); i++)
{
IntEx thisPart = new IntEx(0);
thisPart.digits = new ArrayList<>();
for (int k = 0; k < i; k++)
thisPart.digits.add(0);
int Leftovers = 0;
for (int j = 0; j < target.digits.size(); j++)
{
int multiFragment = digits.get(i) * (int) target.digits.get(j) + Leftovers;
Leftovers = (int) (multiFragment / baseSize);
thisPart.digits.add((int)(multiFragment % baseSize));
}
while (Leftovers > 0)
{
thisPart.digits.add((int)(Leftovers % baseSize));
Leftovers = (int) (Leftovers / baseSize);
}
MultiParts.add(thisPart);
}
IntEx newNumber = new IntEx(0);
for (int i = 0; i < MultiParts.size(); i++)
{
newNumber.add(MultiParts.get(i));
}
digits = newNumber.digits;
}
public long longValue() throws Exception
{
int position = 0;
long multi = 1;
long retValue = 0;
if (digits.isEmpty()) return 0;
if (digits.size() > 16) throw new Exception("The number within IntEx class is too big to fit into a long");
do
{
retValue += digits.get(position) * multi;
multi *= baseSize;
position++;
} while (position < digits.size());
return retValue;
}
public static long BaseConvert(String number, String base)
{
boolean negative = number.startsWith("-");
number = number.replace("-", "");
ArrayList<Character> localDigits = new ArrayList<>();
for(int i = number.toCharArray().length - 1; i >=0; i--) {
localDigits.add(number.charAt(i));
}
// List<>().reverse is missing in this damn java. -_-
long retValue = 0;
long Multi = 1;
char[] CharsBase = base.toCharArray();
for (int i = 0; i < number.length(); i++)
{
int t = base.indexOf(localDigits.get(i));
retValue += base.indexOf(localDigits.get(i)) * Multi;
Multi *= base.length();
}
if (negative)
retValue = -retValue;
return retValue;
}
public static String BaseMult(String a, String b, String Base)
{
ArrayList<String> MultiParts = new ArrayList<>();
// this huge block is a tribute to java not having "Reverse()" method.
char[] x = new char[a.length()];
char[] y = new char[b.length()];
for(int i = 0; i < a.length(); i++) {
x[i] = a.charAt(a.length()-i);
}
for(int i = 0; i < b.length(); i++) {
y[i] = a.charAt(a.length()-i);
}
a = new String(x);
b = new String(y);
// ---------------------------------------------------------------------
for (int i = 0; i < a.length(); i++)
{
ArrayList<Character> thisPart = new ArrayList<>();
for (int k = 0; k < i; k++)
thisPart.add(Base.charAt(0));
int leftOvers = 0;
for (int j = 0; j < b.length(); j++)
{
// Need I say repeated characters in base may cause mayhem?
int MultiFragment = Base.indexOf(a.charAt(i)) * Base.indexOf(b.charAt(j)) + leftOvers;
leftOvers = MultiFragment / Base.length();
thisPart.add(Base.charAt(MultiFragment % Base.length()));
}
while (leftOvers > 0)
{
thisPart.add(Base.charAt(leftOvers % Base.length()));
leftOvers = leftOvers / Base.length();
}
char[] thisPartReverse = new char[thisPart.size()];
for(int z = 0; z < thisPart.size();z++)
thisPartReverse[z] = thisPart.get(thisPart.size()-z);
MultiParts.add(new String(thisPartReverse));
}
String retValue = ""+Base.charAt(0);
for (int i = 0; i < MultiParts.size(); i++)
{
retValue = BaseSum(retValue, MultiParts.get(i), Base);
}
return retValue;
}
public static String BaseSum(String a, String b, String Base)
{
// this huge block is a tribute to java not having "Reverse()" method.
char[] x = new char[a.length()];
char[] y = new char[b.length()];
for(int i = 0; i < a.length(); i++) {
x[i] = a.charAt(a.length()-i);
}
for(int i = 0; i < b.length(); i++) {
y[i] = a.charAt(a.length()-i);
}
a = new String(x);
b = new String(y);
// ---------------------------------------------------------------------
String greater = a.length() > b.length() ? a : b;
String lesser = a.length() < b.length() ? a : b;
int leftOvers = 0;
ArrayList<Character> result = new ArrayList();
for (int i = 0; i < greater.length() || leftOvers > 0; i++)
{
int sum;
if (i >= greater.length())
sum = leftOvers;
else if (i >= lesser.length())
sum = leftOvers + Base.indexOf(greater.charAt(i));
else
sum = Base.indexOf(a.charAt(i)) + Base.indexOf(b.charAt(i)) + leftOvers;
leftOvers = 0;
if (sum > Base.length()-1)
{
while (sum > Base.length()-1)
{
sum -= Base.length();
leftOvers += 1;
}
result.add(Base.charAt(sum));
}
else
{
result.add(Base.charAt(sum));
leftOvers = 0;
}
}
char[] reverseResult = new char[result.size()];
for(int i = 0; i < result.size(); i++)
reverseResult[i] = result.get(result.size() -i);
return new String(reverseResult);
}
public static String BaseConvertItoA(long number, String base)
{
ArrayList<Character> retValue = new ArrayList<>();
boolean negative = false;
long backup = number;
if (negative = (backup < 0))
backup = -backup;
do
{
int index = (int)(backup % base.length());
retValue.add(base.charAt(index));
backup = backup / base.length();
} while ((backup) > 0);
if (negative)
retValue.add('-');
char[] reverseRetVal = new char[retValue.size()];
for(int i = 0; i < retValue.size(); i++)
reverseRetVal[i] = retValue.get(retValue.size()-i);
return new String(reverseRetVal);
}
public String ToString(String base)
{
if(base == null || base.length() < 2)
base = "0123456789";
ArrayList<Character> retVal = new ArrayList<>();
char[] CharsBase = base.toCharArray();
int TamanhoBase = base.length();
String result = ""+base.charAt(0);
String multi = ""+base.charAt(1);
String lbase = IntEx.BaseConvertItoA(baseSize, base);
for (int i = 0; i < digits.size(); i++)
{
String ThisByte = IntEx.BaseConvertItoA(digits.get(i), base);
String Next = IntEx.BaseMult(ThisByte, multi, base);
result = IntEx.BaseSum(result, Next, base);
multi = IntEx.BaseMult(multi, lbase, base);
}
return result;
}
public static void main(String... args) {
int ref = 0;
IntEx result = new IntEx(0);
while(++ref <= 1000)
{
IntEx mul = new IntEx(1000);
for (int i = 0; i < 1000; ++i) {
mul.multiply(new IntEx(i));
}
result.add(mul);
}
System.out.println(result.toString());
}
}
Disclaimer: This is a rough translation/localization from a C# study, there are lots of code omitted. This is "almost" the same logics behind java.math.BigInteger (you can open BigInteger code on your favorite designer and check for yourself. If may I be forgetting a overloaded operator behind not translated to java, have a bit of patience and forgiveness, this example is just for a "maybe" clarification of the theory.
Also, just a sidenote, I know it is "Trying to reinvent the wheel", but considering this question has academic purpose I think its fairly rasonable to share.
One can see the result of this study on gitHub (not localized though), I'm not expanding that C# code here for its very extensive and not the language of this question.

This gives the correct answer without excess calculations. A Long is sufficient.
public String lastTen() {
long answer = 0;
String txtAnswer = "";
int length = 0;
int i = 1;
for(i = 1; i <= 1000; i++) {
answer += Math.pow(i, i);
txtAnswer = Long.toString(answer);
length = txtAnswer.length();
if(length > 9) break;
}
return txtAnswer.substring(length-10);
}

Selection sorting java array

I'm doing a project in which I must generate an array list the size of a user input(in this case i chose 4), with random numbers between -1000 and 1000. Then I have to have it do a selection sort and display both the unsorted numbers in output1 and the sorted numbers in output2 Heres what I have thus far
ArrayList <Integer> unSortedNumbers = new ArrayList <Integer>();
Integer [] numberSort;
...
private void SortActionPerformed(java.awt.event.ActionEvent evt) {
String input, sortedNumberOutput = "";
int int1, int2 = 0, min = -1000, max = 1000, j, minimum, temp = 0;
input = Input.getText();
int1 = Integer.parseInt(input);
Random number = new Random();
while (int2 < int1) {
for (int i = 0; i < int1; i++) {
int randomInt = number.nextInt(max - min + 1) + min;
unSortedNumbers.add(randomInt);
int1--;
}
}
numberSort = new Integer[unSortedNumbers.size()];
unSortedNumbers.toArray(numberSort);
for (int i = 0; i < numberSort.length; i++) {
sortedNumberOutput += numberSort[i] + (i != numberSort.length ? "," : "");
}
if (Selection.isSelected() && Ascending.isSelected()) {
for (int i = 0; i < numberSort.length - 1; i++) {
minimum = numberSort[i];
for (j = i + 1; j <= numberSort.length - 1; j++) {
if (minimum > numberSort[j]) {
numberSort[temp] = numberSort[i];
numberSort[i] = numberSort[j];
numberSort[j] = numberSort[temp];
}
}
}
}
Output1.setText("Unsorted Numbers " + unSortedNumbers);
Output2.setText("Sorted Numbers " + numberSort);
unSortedNumbers.clear();
numberSort = null;
}
So when I run that, the unSortedNumbers are displayed properly in output1, but instead of displaying the sorted numbers in output2, it displays this :
Sorted Numbers [Ljava.lang.Integer;#7a279c
I'm not sure why this is happening, my could is probably wrong somewhere, If you can help, thank you!

What you are seeing is the result of the default toString() method being called on an array object. [Ljava.lang.Integer tells you it is an array of Integers and the #7a279c gives you the hex string of the hashcode. Do as ZouZou suggests and use Arrays.toString(unSortedNumbers);

How to convert big floats to int

I wrote a program in java and php. In which a loop runs 64 times. And keep adding n to n:
Java Code:
public static void main(String[] args) {
double n = 1;
double p = 1;
for(int i = 1;i <= 64;i++){
n = n + n;
p = p + n;
}
System.out.println(p);
}
PHP code:
<?php
$n = 1;
$p = 0;
for($i = 1;$i <= 64;$i++){
$n = $n + $n;
$p = $p + $n;
}
echo($p);
?>
And the output of both of these is:
3.6893488147419E+19
Now I want to know is it possible to convert this big float to int? if yes, Then how. In both languages.

I would use the BigInteger type,
public static void main(String[] args) {
BigInteger n = BigInteger.ONE;
BigInteger p = BigInteger.ONE;
for(int i = 1;i <= 64;i++){
n = n.add(n);
p = p.add(n);
}
System.out.println(p);
}
Output is
36893488147419103231
Edit Based on your comment, you really wanted something more like The Legend of the Chessboard -
BigInteger n = BigInteger.ONE;
BigInteger p = BigInteger.ZERO;
BigInteger TWO = new BigInteger("2");
for (int i = 1; i <= 64; i++) {
StringBuilder sb = new StringBuilder();
sb.append("For square #: " + i);
sb.append(", Grains on square: " + n);
p = p.add(n);
n = n.multiply(TWO);
sb.append(", Running Total: " + p);
System.out.println(sb.toString());
}

The number is too large to fit in a long. To get the closest integral approximation, convert the double to a BigInteger by way of a BigDecimal:
BigDecimal bd = BigDecimal.valueOf(p);
BigInteger bi = bd.toBigInteger();
However, to get an exact result, perform all the calculations using BigIntegers:
import static java.math.BigInteger.ONE;
BigInteger n = ONE;
BigInteger p = ONE;
for (int i = 1; i <= 64; i++) {
n = n.add(n);
p = p.add(n);
}
System.out.println(p);
The difference between the approximate and exact values is:
36893488147419103000
36893488147419103231

Java: Math.round(float), PHP: round(value,precision) However you will precision

Converting binary to decimals and characters

I want to convert binary to decimals and characters like this:
11010 --> 1101 + 0(parity bit) -->decimals= 11 --> char ";"
10101 --> 1010 + 1 -->decimals= 5 --> char "5"
.
.
public class stringek {
String bitek = "1101010101001000001000001";
String[] bits;
String four;
char par;
int parity;
String digits;
int n = 0;
int b;
int kurens;
int decimalis;
int digit;
public stringek() {
this.kurens = 0;
bits = new String[200];
for (int i = 0; i < 25; i += 5) {
bits[n] = bitek.substring(i, i + 5);
n++;
}
for (int i = 0; i < n; ++i) {
int j = 0;
four = bits[i].substring(j, j + 4);
for (int p = 0; p < 4; ++p) {
b = Integer.parseInt(four.substring(p));
kurens += b;
}
par = bits[i].charAt(j+4);
//System.out.print(par);
parity = par-'0';
decimalis = Integer.parseInt(four, 2);
digit = decimalis + 48;
if ((kurens + parity) % 2 == 0) {
System.out.println("Binarys: "+four+"-"+par+" = "+"'"+(char)digit+"'"+" Decimalis:"+decimalis+" Parity <INVALID> ");
}
else{
System.out.println("Binarys: "+four+"-"+par+" = "+"'"+(char)digit+"'"+" Decimalis:"+decimalis+" Parity <VALID> ");
}
}
}
}
but my program results this:
Binarys: 1101-0 = '=' Decimalis:13 Parity <INVALID>
Binarys: 1010-1 = ':' Decimalis:10 Parity <VALID>
Binarys: 0010-0 = '2' Decimalis:2 Parity <INVALID>
Binarys: 0001-0 = '1' Decimalis:1 Parity <INVALID>
Binarys: 0000-1 = '0' Decimalis:0 Parity <VALID>
Can anyone help me to resolve? I have to say cause in my case all Parity is VALID, but I don't know why here some Parity is Invalid (I know cause the results from if give me this results, but I want to know how to resolve to be VALID when is valid and INVALID when is really invalid). thanks

public String[] splitStringEvery(String s, int interval) {
int arrayLength = (int) Math.ceil(((s.length() / (double)interval)));
String[] result = new String[arrayLength];
int j = 0;
int lastIndex = result.length - 1;
for (int i = 0; i < lastIndex; i++) {
result[i] = s.substring(j, j + interval);
j += interval;
} //Add the last bit
result[lastIndex] = s.substring(j);
return result;
}

You wouldn't use String.split() or a StringTokenizer
Use a for loop that increments by 5, checking against length of your string
Use String.substring() to extract the 5 character strings.
To compute the length of the target array you need, you'll need to divide your string length by 5. A Better idea is to use a List<String>.

Use the Guava Libraries Splitter object, specifically the fixedLength(...) method which does exactly what you're trying to do.
Splitter splitter = Splitter.fixedLength(5);
Iterable<String> tokens= splitter.split(myVeryLongString);