﻿ Java - Find n in a range of seed for Lychrel numbers

# Java: Find the number of seed Lychrel number candidates and related numbers for n in the range 1..10000 inclusive.

## Java Numbers: Exercise-5 with Solution

Write a Java program to find the number of seed Lychrel number candidates and related numbers for n in the range 1..10000 inclusive. (With a 500-iteration limit).

A Lychrel number is a natural number that cannot form a palindrome through the iterative process of repeatedly reversing its digits and adding the resulting numbers. This process is sometimes called the 196-algorithm, after the most famous number associated with the process.
The first few Lychrel numbers are 196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, ... .

Pictorial Presentation: Sample Solution:

Java Code:

``````// https://rosettacode.org/
import java.math.BigInteger;
import java.util.*;
public class Example5 {
static Map<BigInteger, Tuple> cache = new HashMap<>();
static class Tuple {
final Boolean flag;
final BigInteger bi;

Tuple(boolean f, BigInteger b) {
flag = f;
bi = b;
}
}
static BigInteger rev(BigInteger bi) {
String s = new StringBuilder(bi.toString()).reverse().toString();
return new BigInteger(s);
}
static Tuple lychrel(BigInteger n) {
Tuple res;
if ((res = cache.get(n)) != null)
return res;

BigInteger r = rev(n);
res = new Tuple(true, n);
List<BigInteger> seen = new ArrayList<>();
for (int i = 0; i < 500; i++) {
r = rev(n);

if (n.equals(r)) {
res = new Tuple(false, BigInteger.ZERO);
break;
}
if (cache.containsKey(n)) {
res = cache.get(n);
break;
}
}
for (BigInteger bi : seen)
cache.put(bi, res);
return res;
}
public static void main(String[] args) {
List<BigInteger> seeds = new ArrayList<>();
List<BigInteger> related = new ArrayList<>();
List<BigInteger> palin = new ArrayList<>();

for (int i = 1; i <= 10_000; i++) {
BigInteger n = BigInteger.valueOf(i);
Tuple t = lychrel(n);
if (!t.flag)
continue;
if (n.equals(t.bi))
else

if (n.equals(t.bi))
}
System.out.printf("%d Lychrel seeds: %s%n", seeds.size(), seeds);
System.out.printf("%d Lychrel related%n", related.size());
System.out.printf("%d Lychrel palindromes: %s%n", palin.size(), palin);
}
}
```
```

Sample Output:

```5 Lychrel seeds: [196, 879, 1997, 7059, 9999]
244 Lychrel related
5 Lychrel palindromes: [196, 879, 1997, 7059, 9999]
```

Flowchart: Java Code Editor:

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﻿

## Java: Tips of the Day

IsPowerOfTwo

Checks if a value is positive power of two.

To understand how it works let's assume we made a call IsPowerOfTwo(4).

As value is greater than 0, so right side of the && operator will be evaluated.

The result of (~value + 1) is equal to value itself. ~100 + 001 => 011 + 001 => 100. This is equal to value.

The result of (value & value) is value. 100 & 100 => 100.

This will value the expression to true as value is equal to value.

```public static boolean isPowerOfTwo(final int value) {
return value > 0 && ((value & (~value + 1)) == value);
}
```

Ref: https://bit.ly/3sA5d4I

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