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;

                }

            }
        }

        S开发者_开发百科ystem.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:

1. Start at 999, working my way backwards to 998, 997, etc
2. Create the palindrome for my current number.
3. Determine the prime factorization of this number (not all that expensive if you have a pre-generated list of primes.
4. 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);
}
}