We Keep Coding

How to print an integer with commas every 'd' digits, from right to left

I had to write a program that will receive an int 'n' and another one 'd' - and will print the number n with commas every d digits from right to left.
If 'n' or 'd' are negative - the program will print 'n' as is.
I although had to make sure that there is no commas before or after the number and I'm not allowed to use String or Arrays.
for example: n = 12345678
d=1: 1,2,3,4,5,6,7,8
d=3: 12,345,678
I've written the following code:
public static void printWithComma(int n, int d) {
if (n < 0 || d <= 0) {
System.out.println(n);
} else {
int reversedN = reverseNum(n), copyOfrereversedN = reversedN, counter = numberLength(n);
while (reversedN > 0) {
System.out.print(reversedN % 10);
reversedN /= 10;
counter--;
if (counter % d == 0 && reversedN != 0) {
System.out.print(",");
}
}
/*
* In a case which the received number will end with zeros, the reverse method
* will return the number without them. In that case the length of the reversed
* number and the length of the original number will be different - so this
* while loop will end the zero'z at the right place with the commas at the
* right place
*/
while (numberLength(copyOfrereversedN) != numberLength(n)) {
if (counter % d == 0) {
System.out.print(",");
}
System.out.print(0);
counter--;
copyOfrereversedN *= 10;
}
}
}
that uses a reversNum function:
// The method receives a number n and return his reversed number(if the number
// ends with zero's - the method will return the number without them)
public static int reverseNum(int n) {
if (n < 9) {
return n;
}
int reversedNum = 0;
while (n > 0) {
reversedNum += (n % 10);
reversedNum *= 10;
n /= 10;
}
return (reversedNum / 10);
}
and numberLength method:
// The method receives a number and return his length ( 0 is considered as "0"
// length)
public static int numberLength(int n) {
int counter = 0;
while (n > 0) {
n /= 10;
counter++;
}
return counter;
}
I've been told that the code doesn't work for every case, and i am unable to think about such case (the person who told me that won't tell me).
Thank you for reading!

You solved looping through the digits by reversing the number, so a simple division by ten can be done to receive all digits in order.
The comma position is calculated from the right.
public static void printWithComma(int n, int d) {
if (n < 0) {
System.out.print('-');
n = -n;
}
if (n == 0) {
System.out.print('0');
return;
}
int length = numberLength(n);
int reversed = reverseNum(n);
for (int i = 0; i < length; ++i) {
int nextDigit = reversed % 10;
System.out.print(nextDigit);
reversed /= 10;
int fromRight = length - 1 - i;
if (fromRight != 0 && fromRight % d == 0) {
System.out.print(',');
}
}
}
This is basically the same code as yours. However I store the results of the help functions into variables.
A zero is a special case, an exception of the rule that leading zeros are dropped.
Every dth digit (from right) needs to print comma, but not entirely at the right. And not in front. Realized by printing the digit first and then possibly the comma.
The problems I see with your code are the two while loops, twice printing the comma, maybe? And the println with a newline when <= 0.
Test your code, for instance as:
public static void main(String[] args) {
for (int n : new int[] {0, 1, 8, 9, 10, 234,
1_234, 12_345, 123_456, 123_456_789, 1_234_567_890}) {
System.out.printf("%d : ", n);
printWithComma(n, 3);
System.out.println();
}
}

Your code seems overly complicated.
If you've learned about recursion, you can do it like this:
public static void printWithComma(int n, int d) {
printInternal(n, d, 1);
System.out.println();
}
private static void printInternal(int n, int d, int i) {
if (n > 9) {
printInternal(n / 10, d, i + 1);
if (i % d == 0)
System.out.print(',');
}
System.out.print(n % 10);
}
Without recursion:
public static void printWithComma(int n, int d) {
int rev = 0, i = d - 1;
for (int num = n; num > 0 ; num /= 10, i++)
rev = rev * 10 + num % 10;
for (; i > d; rev /= 10, i--) {
System.out.print(rev % 10);
if (i % d == 0)
System.out.print(',');
}
System.out.println(rev);
}

Are you allowed to use the whole Java API?
What about something as simple as using DecimalFormat
double in = 12345678;
DecimalFormat df = new DecimalFormat( ",##" );
System.out.println(df.format(in));
12,34,56,78
Using...
,# = 1 per group
,## = 2 per group
,### = 3 per group
etc...

It took me a bunch of minutes. The following code snippet does the job well (explanation below):
public static void printWithComma(int n, int d) { // n=number, d=commaIndex
final int length = (int) (Math.log10(n) + 1); // number of digits;
for (int i = 1; i < Math.pow(10, length); i*=10) { // loop by digits
double current = Math.log10(i); // current loop
double remains = length - current - 1; // loops remaining
int digit = (int) ((n / Math.pow(10, remains)) % 10); // nth digit
System.out.print(digit); // print it
if (remains % d == 0 && remains > 0) { // add comma if qualified
System.out.print(",");
}
}
}
Using (Math.log10(n) + 1) I find a number of digits in the integer (8 for 12345678).
The for-loop assures the exponents of n series (1, 10, 100, 1000...) needed for further calculations. Using logarithm of base 10 I get the current index of the loop.
To get nth digit is a bit tricky and this formula is based on this answer. Then it is printed out.
Finally, it remains to find a qualified position for the comma (,). If modulo of the current loop index is equal zero, the dth index is reached and the comma can be printed out. Finally the condition remains > 0 assures there will be no comma left at the end of the printed result.
Output:
For 4: 1234,5678
For 3: 12,345,678
For 2: 12,34,56,78
For 1: 1,2,3,4,5,6,7,8

Decrease execution time of the Java program

I wrote the following program for the second problem of project Euler, for the question: "Project Euler #3: Largest prime factor".It is supposed to print out all the highest prime factors of the provided inputs.
import java.util.Scanner;
public class euler_2 {
public static boolean isPrime(int n) {
if (n % 2 == 0) return false;
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return false;
}
return true;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
for (int i = 0; i < a; i++) {
int b = sc.nextInt();
for (int j = b; j >= 1; j--) {
boolean aa = isPrime(j);
if (aa == true && b % j == 0) {
b = j;
break;
}
}
System.out.println(b);
}
}
}
What changes can I make to the program to make it execute faster? What would be a better algorithm for this problem?

The problem with your approach is that for every number N, you try each number smaller or equal to N whether it is a prime and after that whether it is a divisor of N.
Obvious improvement is to check whether it is a divisor first and only then whether it is a prime. But most probably this will not help that much.
What you can do instead is just to start checking each number whether it is a divisor of a number. If it is a divisor, divide it. You continue this till sqrt(N).
I have not done anything with java in a long time, but here is Go implementation, which most probably any Java person will be able to transform to Java.
func biggestPrime(n uint64) uint64 {
p, i := uint64(1), uint64(0)
for i = 2; i < uint64(math.Sqrt(float64(n))) + uint64(1); i++ {
for n % i == 0 {
n /= i
p = i
}
}
if n > 1 {
p = n
}
return p
}
Using my algorithm it will take you O(sqrt(N)) to find the biggest prime of a number. In your case it was O(N * sqrt(N))

Attempt to factor the number into 2 factors. Repeat on the largest factor found so far until you find one that can't be factored -- that is the largest prime factor.
There are many different ways you might try to factor the numbers, but since they are only ints, then Fermat's method or even trial division (going down from sqrt(N)) will probably do. See http://mathworld.wolfram.com/FermatsFactorizationMethod.html

Binary search for square root [homework]

For an assignment I must create a method using a binary search to find the square root of an integer, and if it is not a square number, it should return an integer s such that s*s <= the number (so for 15 it would return 3). The code I have for it so far is
public class BinarySearch {
/**
* Integer square root Calculates the integer part of the square root of n,
* i.e. integer s such that s*s <= n and (s+1)*(s+1) > n
* requires n >= 0
*
* #param n number to find the square root of
* #return integer part of its square root
*/
private static int iSqrt(int n) {
int l = 0;
int r = n;
int m = ((l + r + 1) / 2);
// loop invariant
while (Math.abs(m * m - n) > 0) {
if ((m) * (m) > n) {
r = m;
m = ((l + r + 1) / 2);
} else {
l = m;
m = ((l + r + 1) / 2);
}
}
return m;
}
public static void main(String[] args) {
//gets stuck
System.out.println(iSqrt(15));
//calculates correctly
System.out.println(iSqrt(16));
}
}
And this returns the right number for square numbers, but gets stick in an endless loop for other integers. I know that the problem lies in the while condition, but I can't work out what to put due to the gap between square numbers getting much bigger as the numbers get bigger (so i can't just put that the gap must be below a threshold). The exercise is about invariants if that helps at all (hence why it is set up in this way). Thank you.

Think about it: Math.abs(m*m-n) > 0 is always true non-square numbers, because it is never zero, and .abs cannot be negative. It is your loop condition, that's why the loop never ends.
Does this give you enough info to get you going?

You need to change the while (Math.abs(m * m - n) > 0) to allow for a margin of error, instead of requiring it be exactly equal to zero as you do right now.
Try while((m+1)*(m+1) <= n || n < m * m)

#define EPSILON 0.0000001
double msqrt(double n){
assert(n >= 0);
if(n == 0 || n == 1){
return n;
}
double low = 1, high = n;
double mid = (low+high)/2.0;
while(abs(mid*mid - n) > EPSILON){
mid = (low+high)/2.0;
if(mid*mid < n){
low = mid+1;
}else{
high = mid-1;
}
}
return mid;}
As you can see above , you should simply apply binary search (bisection method)
and you can minimize Epsilon to get more accurate results but it will take more time to run.
Edit: I have written code in c++ (sorry)

As Ken Bloom said you have to have an error marge, 1. I've tested this code and it runs as expected for 15. Also you'll need to use float's, I think this algorithm is not possible for int's (although I have no mathematical proof)
private static int iSqrt(int n){
float l = 0;
float r = n;
float m = ((l + r)/2);
while (Math.abs(m*m-n) > 0.1) {
if ((m)*(m) > n) {
r=m;
System.out.println("r becomes: "+r);
} else {
l = m;
System.out.println("l becomes: "+l);
}
m = ((l + r)/2);
System.out.println("m becomes: "+m);
}
return (int)m;
}

Division By Zero Error For Project Euler?

I am trying to solve some project Euler problems and for whatever reason the following code gives me a division by zero error whenever I try to run it with large numbers. Can anyone tell me why?
import java.util.Scanner;
public class Problem3LargestPrimeFactor {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Please enter the number to find the largest prime factor for.");
long num = input.nextLong();
int largest = 1;
boolean isPrime = true;
for(int i = 2; i < num; i++) {
if(num % i == 0) {
for(int u = 2; u < i; u++){
if(i % u == 0)
isPrime = false;
}
if(isPrime)
largest = i;
}
}
}
}
Now I'm aware that this is not the most efficient way to design the algorithm but can anyone tell me what is going on here?

You're overflowing int. In the loop for(int i = 2; i < num; i++), num is a long and i is only an int. When i reaches it's capacity, it wraps around to -2,147,483,648 and keeps being incremented until it gets to 0, at which point you get your error.

Other responses have already pointed out your error. Let me give you a better algorithm for factoring a composite integer using trial division. The basic idea is to simply iterate through the possible factors, reducing n each time you find one. Here's pseudocode:
function factors(n)
f, fs := 2, {}
while f * f <= n
while n % f == 0
append f to fs
n := n / f
f := f + 1
if n > 1
append n to fs
return fs
If you're interested in programming with prime numbers, I modestly recommend this essay at my blog.

Find the largest palindrome made from the product of two 3-digit numbers

package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int value = 0;
for(int i = 100;i <=999;i++)
{
for(int j = i;j <=999;j++)
{
int value1 = i * j;
StringBuilder sb1 = new StringBuilder(""+value1);
String sb2 = ""+value1;
sb1.reverse();
if(sb2.equals(sb1.toString()) && value<value1) {
value = value1;
}
}
}
System.out.println(value);
}
}
This is the code that I wrote in Java... Is there any efficient way other than this.. And can we optimize this code more??

We suppose the largest such palindrome will have six digits rather than five, because 143*777 = 111111 is a palindrome.
As noted elsewhere, a 6-digit base-10 palindrome abccba is a multiple of 11. This is true because a*100001 + b*010010 + c*001100 is equal to 11*a*9091 + 11*b*910 + 11*c*100. So, in our inner loop we can decrease n by steps of 11 if m is not a multiple of 11.
We are trying to find the largest palindrome under a million that is a product of two 3-digit numbers. To find a large result, we try large divisors first:
We step m downwards from 999, by 1's;
Run n down from 999 by 1's (if 11 divides m, or 9% of the time) or from 990 by 11's (if 11 doesn't divide m, or 91% of the time).
We keep track of the largest palindrome found so far in variable q. Suppose q = r·s with r <= s. We usually have m < r <= s. We require m·n > q or n >= q/m. As larger palindromes are found, the range of n gets more restricted, for two reasons: q gets larger, m gets smaller.
The inner loop of attached program executes only 506 times, vs the ~ 810000 times the naive program used.
#include <stdlib.h>
#include <stdio.h>
int main(void) {
enum { A=100000, B=10000, C=1000, c=100, b=10, a=1, T=10 };
int m, n, p, q=111111, r=143, s=777;
int nDel, nLo, nHi, inner=0, n11=(999/11)*11;
for (m=999; m>99; --m) {
nHi = n11; nDel = 11;
if (m%11==0) {
nHi = 999; nDel = 1;
}
nLo = q/m-1;
if (nLo < m) nLo = m-1;
for (n=nHi; n>nLo; n -= nDel) {
++inner;
// Check if p = product is a palindrome
p = m * n;
if (p%T==p/A && (p/B)%T==(p/b)%T && (p/C)%T==(p/c)%T) {
q=p; r=m; s=n;
printf ("%d at %d * %d\n", q, r, s);
break; // We're done with this value of m
}
}
}
printf ("Final result: %d at %d * %d inner=%d\n", q, r, s, inner);
return 0;
}
Note, the program is in C but same techniques will work in Java.

What I would do:
Start at 999, working my way backwards to 998, 997, etc
Create the palindrome for my current number.
Determine the prime factorization of this number (not all that expensive if you have a pre-generated list of primes.
Work through this prime factorization list to determine if I can use a combination of the factors to make 2 3 digit numbers.
Some code:
int[] primes = new int[] {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,
73,79,83,89,97,101,103,107,109,113,,127,131,137,139,149,151,157,163,167,173,
179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,
283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,
419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,
547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,
661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,
811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,
947,953,967,971,977,983,991,997};
for(int i = 999; i >= 100; i--) {
String palstr = String.valueOf(i) + (new StringBuilder().append(i).reverse());
int pal = Integer.parseInt(pal);
int[] factors = new int[20]; // cannot have more than 20 factors
int remainder = pal;
int facpos = 0;
primeloop:
for(int p = 0; p < primes.length; i++) {
while(remainder % p == 0) {
factors[facpos++] = p;
remainder /= p;
if(remainder < p) break primeloop;
}
}
// now to do the combinations here
}

We can translate the task into the language of mathematics.
For a short start, we use characters as digits:
abc * xyz = n
abc is a 3-digit number, and we deconstruct it as 100*a+10*b+c
xyz is a 3-digit number, and we deconstruct it as 100*x+10*y+z
Now we have two mathematical expressions, and can define a,b,c,x,y,z as € of {0..9}.
It is more precise to define a and x as of element from {1..9}, not {0..9}, because 097 isn't really a 3-digit number, is it?
Ok.
If we want to produce a big number, we should try to reach a 9......-Number, and since it shall be palindromic, it has to be of the pattern 9....9. If the last digit is a 9, then from
(100*a + 10*b + c) * (100*x + 10*y + z)
follows that z*c has to lead to a number, ending in digit 9 - all other calculations don't infect the last digit.
So c and z have to be from (1,3,7,9) because (1*9=9, 9*1=9, 3*3=9, 7*7=49).
Now some code (Scala):
val n = (0 to 9)
val m = n.tail // 1 to 9
val niners = Seq (1, 3, 7, 9)
val highs = for (a <- m;
b <- n;
c <- niners;
x <- m;
y <- n;
z <- niners) yield ((100*a + 10*b + c) * (100*x + 10*y + z))
Then I would sort them by size, and starting with the biggest one, test them for being palindromic. So I would omit to test small numbers for being palindromic, because that might not be so cheap.
For aesthetic reasons, I wouldn't take a (toString.reverse == toString) approach, but a recursive divide and modulo solution, but on todays machines, it doesn't make much difference, does it?
// Make a list of digits from a number:
def digitize (z: Int, nums : List[Int] = Nil) : List[Int] =
if (z == 0) nums else digitize (z/10, z%10 :: nums)
/* for 342243, test 3...==...3 and then 4224.
Fails early for 123329 */
def palindromic (nums : List[Int]) : Boolean = nums match {
case Nil => true
case x :: Nil => true
case x :: y :: Nil => x == y
case x :: xs => x == xs.last && palindromic (xs.init) }
def palindrom (z: Int) = palindromic (digitize (z))
For serious performance considerations, I would test it against a toString/reverse/equals approach. Maybe it is worse. It shall fail early, but division and modulo aren't known to be the fastest operations, and I use them to make a List from the Int. It would work for BigInt or Long with few redeclarations, and works nice with Java; could be implemented in Java but look different there.
Okay, putting the things together:
highs.filter (_ > 900000) .sortWith (_ > _) find (palindrom)
res45: Option[Int] = Some(906609)
There where 835 numbers left > 900000, and it returns pretty fast, but I guess even more brute forcing isn't much slower.
Maybe there is a much more clever way to construct the highest palindrom, instead of searching for it.
One problem is: I didn't knew before, that there is a solution > 900000.
A very different approach would be, to produce big palindromes, and deconstruct their factors.

public class Pin
{
public static boolean isPalin(int num)
{
char[] val = (""+num).toCharArray();
for(int i=0;i<val.length;i++)
{
if(val[i] != val[val.length - i - 1])
{
return false;
}
}
return true;
}
public static void main(String[] args)
{
for(int i=999;i>100;i--)
for(int j=999;j>100;j--)
{
int mul = j*i;
if(isPalin(mul))
{
System.out.printf("%d * %d = %d",i,j,mul);
return;
}
}
}
}

package ex;
public class Main {
public static void main(String[] args) {
int i = 0, j = 0, k = 0, l = 0, m = 0, n = 0, flag = 0;
for (i = 999; i >= 100; i--) {
for (j = i; j >= 100; j--) {
k = i * j;
// System.out.println(k);
m = 0;
n = k;
while (n > 0) {
l = n % 10;
m = m * 10 + l;
n = n / 10;
}
if (m == k) {
System.out.println("pal " + k + " of " + i + " and" + j);
flag = 1;
break;
}
}
if (flag == 1) {
// System.out.println(k);
break;
}
}
}
}

A slightly different approach that can easily calculate the largest palindromic number made from the product of up to two 6-digit numbers.
The first part is to create a generator of palindrome numbers. So there is no need to check if a number is palindromic, the second part is a simple loop.
#include <memory>
#include <iostream>
#include <cmath>
using namespace std;
template <int N>
class PalindromeGenerator {
unique_ptr <int []> m_data;
bool m_hasnext;
public :
PalindromeGenerator():m_data(new int[N])
{
for(auto i=0;i<N;i++)
m_data[i]=9;
m_hasnext=true;
}
bool hasNext() const {return m_hasnext;}
long long int getnext()
{
long long int v=0;
long long int b=1;
for(int i=0;i<N;i++){
v+=m_data[i]*b;
b*=10;
}
for(int i=N-1;i>=0;i--){
v+=m_data[i]*b;
b*=10;
}
auto i=N-1;
while (i>=0)
{
if(m_data[i]>=1) {
m_data[i]--;
return v;
}
else
{
m_data[i]=9;
i--;
}
}
m_hasnext=false;
return v;
}
};
template<int N>
void findmaxPalindrome()
{
PalindromeGenerator<N> gen;
decltype(gen.getnext()) minv=static_cast<decltype(gen.getnext())> (pow(10,N-1));
decltype(gen.getnext()) maxv=static_cast<decltype(gen.getnext())> (pow(10,N)-1);
decltype(gen.getnext()) start=11*(maxv/11);
while(gen.hasNext())
{
auto v=gen.getnext();
for (decltype(gen.getnext()) i=start;i>minv;i-=11)
{
if (v%i==0)
{
auto r=v/i;
if (r>minv && r<maxv ){
cout<<"done:"<<v<<" "<<i<< "," <<r <<endl;
return ;
}
}
}
}
return ;
}
int main(int argc, char* argv[])
{
findmaxPalindrome<6>();
return 0;
}

You can use the fact that 11 is a multiple of the palindrome to cut down on the search space. We can get this since we can assume the palindrome will be 6 digits and >= 111111.
e.g. ( from projecteuler ;) )
P= xyzzyx = 100000x + 10000y + 1000z + 100z + 10y +x
P=100001x+10010y+1100z
P=11(9091x+910y+100z)
Check if i mod 11 != 0, then the j loop can be subtracted by 11 (starting at 990) since at least one of the two must be divisible by 11.

You can try the following which prints
999 * 979 * 989 = 967262769
largest palindrome= 967262769 took 0.015
public static void main(String... args) throws IOException, ParseException {
long start = System.nanoTime();
int largestPalindrome = 0;
for (int i = 999; i > 100; i--) {
LOOP:
for (int j = i; j > 100; j--) {
for (int k = j; k > 100; k++) {
int n = i * j * k;
if (n < largestPalindrome) continue LOOP;
if (isPalindrome(n)) {
System.out.println(i + " * " + j + " * " + k + " = " + n);
largestPalindrome = n;
}
}
}
}
long time = System.nanoTime() - start;
System.out.printf("largest palindrome= %d took %.3f seconds%n", largestPalindrome, time / 1e9);
}
private static boolean isPalindrome(int n) {
if (n >= 100 * 1000 * 1000) {
// 9 digits
return n % 10 == n / (100 * 1000 * 1000)
&& (n / 10 % 10) == (n / (10 * 1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (1000 * 1000) % 10)
&& (n / 1000 % 10) == (n / (100 * 1000) % 10);
} else if (n >= 10 * 1000 * 1000) {
// 8 digits
return n % 10 == n / (10 * 1000 * 1000)
&& (n / 10 % 10) == (n / (1000 * 1000) % 10)
&& (n / 100 % 10) == (n / (100 * 1000) % 10)
&& (n / 1000 % 10) == (n / (10 * 1000) % 10);
} else if (n >= 1000 * 1000) {
// 7 digits
return n % 10 == n / (1000 * 1000)
&& (n / 10 % 10) == (n / (100 * 1000) % 10)
&& (n / 100 % 10) == (n / (10 * 1000) % 10);
} else throw new AssertionError();
}

i did this my way , but m not sure if this is the most efficient way of doing this .
package problems;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class P_4 {
/**
* #param args
* #throws IOException
*/
static int[] arry = new int[6];
static int[] arry2 = new int[6];
public static boolean chk()
{
for(int a=0;a<arry.length;a++)
if(arry[a]!=arry2[a])
return false;
return true;
}
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int temp,z,i;
for(int x=999;x>100;x--)
for(int y=999;y>100;y--)
{
i=0;
z=x*y;
while(z>0)
{
temp=z%10;
z=z/10;
arry[i]=temp;
i++;
}
for(int k = arry.length;k>0;k--)
arry2[arry.length- k]=arry[k-1];
if(chk())
{
System.out.print("pelindrome = ");
for(int l=0;l<arry2.length;l++)
System.out.print(arry2[l]);
System.out.println(x);
System.out.println(y);
}
}
}
}

This is code in C, a little bit long, but gets the job done.:)
#include <stdio.h>
#include <stdlib.h>
/*
A palindromic number reads the same both ways. The largest palindrome made from the product of two
2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.*/
int palndr(int b)
{
int *x,*y,i=0,j=0,br=0;
int n;
n=b;
while(b!=0)
{
br++;
b/=10;
}
x=(int *)malloc(br*sizeof(int));
y=(int *)malloc(br*sizeof(int));
int br1=br;
while(n!=0)
{
x[i++]=y[--br]=n%10;
n/=10;
}
int ind = 1;
for(i=0;i<br1;i++)
if(x[i]!=y[i])
ind=0;
free(x);
free(y);
return ind;
}
int main()
{
int i,cek,cekmax=1;
int j;
for(i=100;i<=999;i++)
{
for(j=i;j<=999;j++)
{
cek=i*j;
if(palndr(cek))
{
if(pp>cekmax)
cekmax=cek;
}
}
}
printf("The largest palindrome is: %d\n\a",cekmax);
}

You can actually do it with Python, it's easy just take a look:
actualProduct = 0
highestPalindrome = 0
# Setting the numbers. In case it's two digit 10 and 99, in case is three digit 100 and 999, etc.
num1 = 100
num2 = 999
def isPalindrome(number):
number = str(number)
reversed = number[::-1]
if number==reversed:
return True
else:
return False
a = 0
b = 0
for i in range(num1,num2+1):
for j in range(num1,num2+1):
actualProduct = i * j
if (isPalindrome(actualProduct) and (highestPalindrome < actualProduct)):
highestPalindrome = actualProduct
a = i
b = j
print "Largest palindrome made from the product of two %d-digit numbers is [ %d ] made of %d * %d" % (len(str(num1)), highestPalindrome, a, b)

Since we are not cycling down both iterators (num1 and num2) at the same time, the first palindrome number we find will be the largest. We don’t need to test to see if the palindrome we found is the largest. This significantly reduces the time it takes to calculate.
package testing.project;
public class PalindromeThreeDigits {
public static void main(String[] args) {
int limit = 99;
int max = 999;
int num1 = max, num2, prod;
while(num1 > limit)
{
num2 = num1;
while(num2 > limit)
{
total = num1 * num2;
StringBuilder sb1 = new StringBuilder(""+prod);
String sb2 = ""+prod;
sb1.reverse();
if( sb2.equals(sb1.toString()) ) { //optimized here
//print and exit
}
num2--;
}
num1--;
}
}//end of main
}//end of class PalindromeThreeDigits

I tried the solution by Tobin joy and vickyhacks and both of them produce the result 580085 which is wrong here is my solution, though very clumsy:
import java.util.*;
class ProjEu4
{
public static void main(String [] args) throws Exception
{
int n=997;
ArrayList<Integer> al=new ArrayList<Integer>();
outerloop:
while(n>100){
int k=reverse(n);
int fin=n*1000+k;
al=findfactors(fin);
if(al.size()>=2)
{
for(int i=0;i<al.size();i++)
{
if(al.contains(fin/al.get(i))){
System.out.println(fin+" factors are:"+al.get(i)+","+fin/al.get(i));
break outerloop;}
}
}
n--;
}
}
private static ArrayList<Integer> findfactors(int fin)
{
ArrayList<Integer> al=new ArrayList<Integer>();
for(int i=100;i<=999;i++)
{
if(fin%i==0)
al.add(i);
}
return al;
}
private static int reverse(int number)
{
int reverse = 0;
while(number != 0){
reverse = (reverse*10)+(number%10);
number = number/10;
}
return reverse;
}
}

Most probably it is replication of one of the other solution but it looks simple owing to pythonified code ,even it is a bit brute-force.
def largest_palindrome():
largest_palindrome = 0;
for i in reversed(range(1,1000,1)):
for j in reversed(range(1, i+1, 1)):
num = i*j
if check_palindrome(str(num)) and num > largest_palindrome :
largest_palindrome = num
print "largest palindrome ", largest_palindrome
def check_palindrome(term):
rev_term = term[::-1]
return rev_term == term

What about : in python
>>> for i in range((999*999),(100*100), -1):
... if str(i) == str(i)[::-1]:
... print i
... break
...
997799
>>>

I believe there is a simpler approach: Examine palindromes descending from the largest product of two three digit numbers, selecting the first palindrome with two three digit factors.
Here is the Ruby code:
require './palindrome_range'
require './prime'
def get_3_digit_factors(n)
prime_factors = Prime.factors(n)
rf = [prime_factors.pop]
rf << prime_factors.shift while rf.inject(:*) < 100 || prime_factors.inject(:*) > 999
lf = prime_factors.inject(:*)
rf = rf.inject(:*)
lf < 100 || lf > 999 || rf < 100 || rf > 999 ? [] : [lf, rf]
end
def has_3_digit_factors(n)
return !get_3_digit_factors(n).empty?
end
pr = PalindromeRange.new(0, 999 * 999)
n = pr.downto.find {|n| has_3_digit_factors(n)}
puts "Found #{n} - Factors #{get_3_digit_factors(n).inspect}, #{Prime.factors(n).inspect}"
prime.rb:
class Prime
class<<self
# Collect all prime factors
# -- Primes greater than 3 follow the form of (6n +/- 1)
# Being of the form 6n +/- 1 does not mean it is prime, but all primes have that form
# See http://primes.utm.edu/notes/faq/six.html
# -- The algorithm works because, while it will attempt non-prime values (e.g., (6 *4) + 1 == 25),
# they will fail since the earlier repeated division (e.g., by 5) means the non-prime will fail.
# Put another way, after repeatedly dividing by a known prime, the remainder is itself a prime
# factor or a multiple of a prime factor not yet tried (e.g., greater than 5).
def factors(n)
square_root = Math.sqrt(n).ceil
factors = []
while n % 2 == 0
factors << 2
n /= 2
end
while n % 3 == 0
factors << 3
n /= 3
end
i = 6
while i < square_root
[(i - 1), (i + 1)].each do |f|
while n % f == 0
factors << f
n /= f
end
end
i += 6
end
factors << n unless n == 1
factors
end
end
end
palindrome_range.rb:
class PalindromeRange
FIXNUM_MAX = (2**(0.size * 8 -2) -1)
def initialize(min = 0, max = FIXNUM_MAX)
#min = min
#max = max
end
def downto
return enum_for(:downto) unless block_given?
n = #max
while n >= #min
yield n if is_palindrome(n)
n -= 1
end
nil
end
def each
return upto
end
def upto
return enum_for(:downto) unless block_given?
n = #min
while n <= #max
yield n if is_palindrome(n)
n += 1
end
nil
end
private
def is_palindrome(n)
s = n.to_s
i = 0
j = s.length - 1
while i <= j
break if s[i] != s[j]
i += 1
j -= 1
end
i > j
end
end

public class ProjectEuler4 {
public static void main(String[] args) {
int x = 999; // largest 3-digit number
int largestProduct = 0;
for(int y=x; y>99; y--){
int product = x*y;
if(isPalindormic(x*y)){
if(product>largestProduct){
largestProduct = product;
System.out.println("3-digit numbers product palindormic number : " + x + " * " + y + " : " + product);
}
}
if(y==100 || product < largestProduct){y=x;x--;}
}
}
public static boolean isPalindormic(int n){
int palindormic = n;
int reverse = 0;
while(n>9){
reverse = (reverse*10) + n%10;
n=n/10;
}
reverse = (reverse*10) + n;
return (reverse == palindormic);
}
}