I'd like to create a program wherein a user will type a number and the program will tell if it is divisible by 3 or not. But %, /, +, * can't be used in the program. Anybody here got some ideas how to do that?
public static void main(String[] args) {
String number = "123456";
int sum = 0;
for (char c : number.toCharArray()) {
sum = sum - (0 - c) - '0';
while (sum >= 3) {
sum -= 3;
}
}
System.out.print("divisible by 3? ");
System.out.println(sum == 0);
}
Alternatively you can keep subtracting 3 until your number is either 0 (divisible by 3) or <0: not divisible by 3.
ps: it needs to be adapted if you want to deal with negative numbers
easy peasy...
boolean divisibleBy3(int n) {
return (""+n).matches("([0369]|[147]([0369]|[147][0369]*[258])*([" +
"258]|[147][0369]*[147])|[258]([0369]|[258]" +
"[0369]*[147])*([147]|[258][0369]*[258]))*");
}
A number is divisible by there if the sum of all the digits is also divisible by 3. You can iterate the process until you have a number smaller than 10 and compare it which known divisors (3,6 and 9)
Since it is most likely a game or homework and you can use + you can simple use minus two times: a - - b is equivalent to a + b
Assuming you can use the - operator then
bool divBy3(int n)
{
while (n >= 0)
{
n -= 3;
}
return n == 0;
}
This will return true if n is exactly divisible by 3, false otherwise. Note that this is really inefficient! Using the % operator would be far better.
Divisibility by 3 in binary representation is like divisibility by 11 in decimal (10+1): sum of digits in even places minus sum of digits on odd places is in turn divisible by 3 (maybe 0).
Related
I'm trying to count the digits of an integer with java, it supposed to give 4 but instead I get 1 .
I don't know if I did something wrong, if someone could help.
Here's the code :
public static void main(String[] args) {
int n = 4781, i = 0;
while (n != 0) {
n %= 10;
n /= 10;
i++;
}
System.out.println(i);
}
The n %= 10 is unnecessary and in fact the source of the problem here. If you remove it, the code will work.
Usually when you do these kinds of "do something for each digit" you use % 10 to find the current digit a little bit like this:
int digit = n % 10;
System.out.println("The current digit is " + digit);
But in your case you don't actually care what the digits are, you only care how many there are, so there's no need to do % 10 at all.
And the way you did it you overwrote n (which is supposed to hold the current state of the number) with the current digit and then divided by 10. So no matter what the first digit is, this will never return a number bigger than 1.
I am trying to write a Java method that checks whether a number is a perfect number or not.
A perfect number is a number that is equal to the sum of all its divisor (excluding itself).
For example, 6 is a perfect number because 1+2+3=6. Then, I have to write a Java program to use the method to display the first 5 perfect numbers.
I have no problem with this EXCEPT that it is taking forever to get the 5th perfect number which is 33550336.
I am aware that this is because of the for loop in my isPerfectNumber() method. However, I am very new to coding and I do not know how to come up with a better code.
public class Labreport2q1 {
public static void main(String[] args) {
//Display the 5 first perfect numbers
int counter = 0,
i = 0;
while (counter != 5) {
i++;
isPerfectNumber(i);
if (isPerfectNumber(i)) {
counter++;
System.out.println(i + " ");
}
}
}
public static boolean isPerfectNumber(int a) {
int divisor = 0;
int sum = 0;
for (int i = 1; i < a; i++) {
if (a % i == 0) {
divisor = i;
sum += divisor;
}
}
return sum == a;
}
}
This is the output that is missing the 5th perfect number
Let's check the properties of a perfect number. This Math Overflow question tells us two very interesting things:
A perfect number is never a perfect square.
A perfect number is of the form (2k-1)×(2k-1).
The 2nd point is very interesting because it reduces our search field to barely nothing. An int in Java is 32 bits. And here we see a direct correlation between powers and bit positions. Thanks to this, instead of making millions and millions of calls to isPerfectNumber, we will be making less than 32 to find the 5th perfect number.
So we can already change the search field, that's your main loop.
int count = 0;
for (int k = 1; count < 5; k++) {
// Compute candidates based on the formula.
int candidate = (1L << (k - 1)) * ((1L << k) - 1);
// Only test candidates, not all the numbers.
if (isPerfectNumber(candidate)) {
count++;
System.out.println(candidate);
}
}
This here is our big win. No other optimization will beat this: why test for 33 million numbers, when you can test less than 100?
But even though we have a tremendous improvement, your application as a whole can still be improved, namely your method isPerfectNumber(int).
Currently, you are still testing way too many numbers. A perfect number is the sum of all proper divisors. So if d divides n, n/d also divides n. And you can add both divisors at once. But the beauty is that you can stop at sqrt(n), because sqrt(n)*sqrt(n) = n, mathematically speaking. So instead of testing n divisors, you will only test sqrt(n) divisors.
Also, this means that you have to start thinking about corner cases. The corner cases are 1 and sqrt(n):
1 is a corner case because you if you divide n by 1, you get n but you don't add n to check if n is a perfect number. You only add 1. So we'll probably start our sum with 1 just to avoid too many ifs.
sqrt(n) is a corner case because we'd have to check whether sqrt(n) is an integer or not and it's tedious. BUT the Math Overflow question I referenced says that no perfect number is a perfect square, so that eases our loop condition.
Then, if at some point sum becomes greater than n, we can stop. The sum of proper divisors being greater than n indicates that n is abundant, and therefore not perfect. It's a small improvement, but a lot of candidates are actually abundant. So you'll probably save a few cycles if you keep it.
Finally, we have to take care of a slight issue: the number 1 as candidate. 1 is the first candidate, and will pass all our tests, so we have to make a special case for it. We'll add that test at the start of the method.
We can now write the method as follow:
static boolean isPerfectNumber(int n) {
// 1 would pass the rest because it has everything of a perfect number
// except that its only divisor is itself, and we need at least 2 divisors.
if (n < 2) return false;
// divisor 1 is such a corner case that it's very easy to handle:
// just start the sum with it already.
int sum = 1;
// We can stop the divisors at sqrt(n), but this is floored.
int sqrt = (int)Math.sqrt(n);
// A perfect number is never a square.
// It's useful to make this test here if we take the function
// without the context of the sparse candidates, because we
// might get some weird results if this method is simply
// copy-pasted and tested on all numbers.
// This condition can be removed in the final program because we
// know that no numbers of the form indicated above is a square.
if (sqrt * sqrt == n) {
return false;
}
// Since sqrt is floored, some values can still be interesting.
// For instance if you take n = 6, floor(sqrt(n)) = 2, and
// 2 is a proper divisor of 6, so we must keep it, we do it by
// using the <= operator.
// Also, sqrt * sqrt != n, so we can safely loop to sqrt
for (int div = 2; div <= sqrt; div++) {
if (n % div == 0) {
// Add both the divisor and n / divisor.
sum += div + n / div;
// Early fail if the number is abundant.
if (sum > n) return false;
}
}
return n == sum;
}
These are such optimizations that you can even find the 7th perfect number under a second, on the condition that you adapt the code for longs instead of ints. And you could still find the 8th within 30 seconds.
So here's that program (test it online). I removed the comments as the explanations are here above.
public class Main {
public static void main(String[] args) {
int count = 0;
for (int k = 1; count < 8; k++) {
long candidate = (1L << (k - 1)) * ((1L << k) - 1);
if (isPerfectNumber(candidate)) {
count++;
System.out.println(candidate);
}
}
}
static boolean isPerfectNumber(long n) {
if (n < 2) return false;
long sum = 1;
long sqrt = (long)Math.sqrt(n);
for (long div = 2; div <= sqrt; div++) {
if (n % div == 0) {
sum += div + n / div;
if (sum > n) return false;
}
}
return n == sum;
}
}
The result of the above program is the list of the first 8 perfect numbers:
6
28
496
8128
33550336
8589869056
137438691328
2305843008139952128
You can find further optimization, notably in the search if you check whether 2k-1 is prime or not as Eran says in their answer, but given that we have less than 100 candidates for longs, I don't find it useful to potentially gain a few milliseconds because computing primes can also be expensive in this program. If you want to check for bigger perfect primes, it makes sense, but here? No: it adds complexity and I tried to keep these optimization rather simple and straight to the point.
There are some heuristics to break early from the loops, but finding the 5th perfect number still took me several minutes (I tried similar heuristics to those suggested in the other answers).
However, you can rely on Euler's proof that all even perfect numbers (and it is still unknown if there are any odd perfect numbers) are of the form:
2i-1(2i-1)
where both i and 2i-1 must be prime.
Therefore, you can write the following loop to find the first 5 perfect numbers very quickly:
int counter = 0,
i = 0;
while (counter != 5) {
i++;
if (isPrime (i)) {
if (isPrime ((int) (Math.pow (2, i) - 1))) {
System.out.println ((int) (Math.pow (2, i -1) * (Math.pow (2, i) - 1)));
counter++;
}
}
}
Output:
6
28
496
8128
33550336
You can read more about it here.
If you switch from int to long, you can use this loop to find the first 7 perfect numbers very quickly:
6
28
496
8128
33550336
8589869056
137438691328
The isPrime method I'm using is:
public static boolean isPrime (int a)
{
if (a == 1)
return false;
else if (a < 3)
return true;
else {
for (int i = 2; i * i <= a; i++) {
if (a % i == 0)
return false;
}
}
return true;
}
I need to write a method which takes in an int and returns true if the number can be written as a sum of two or more consecutive positive integers and false otherwise.
boolean IsSumOfConsecutiveInts(int num)
I figured out that all odd numbers (except the number 1) can be written as the sum of 2 consecutive positive integers:
return (num > 1 && num % 2 == 1);
but this doesn't account for numbers that can be written as the sum of more than 2 consecutive positive integers (such as 6 == 1 + 2 + 3).
How can I determine whether a number can be written as a sum of two or more consecutive positive integers?
These numbers are called Polite Numbers.
And, conveniently, the only numbers that aren't polite are the powers of 2.
So, that gives us 2 options. We can either determine that a number is polite, OR we can determine that it is not a power of 2.
I did both; the latter is easier (and more efficient).
This determines whether or not a number is polite:
boolean IsSumOfConsecutiveInts(int num)
{
int sumOfFirstIIntegers = 3;
for (int i = 2; sumOfFirstIIntegers <= num; i++)
{
if (i%2 == 0 ? (num%i == i/2) : (num%i == 0))
{
return true;
}
sumOfFirstIIntegers += i + 1;
}
return false;
}
This one is pretty hard to understand. It took me a while to come up with.
Basically, i is the number of consecutive integers that we are checking;
sumOfFirstIIntegers is equal to the sum of the first i integers, so that means that all the numbers that can be expressed as a sum of i consecutive integers are greater than or equal to sumOfFirstIIntegers.
The last part that deserves discussing is the boolean statement i%2 == 0 ? (num%i == i/2) : (num%i == 0). Let's look at some examples:
i all sums of i consecutive positive integers
2 3, 5, 7, 9...
3 6, 9, 12, 15...
4 10, 14, 18, 22...
5 15, 20, 25, 30...
There are two cases, but in either case, we can express all possible numbers that are a sum of i consecutive integers pretty simply.
When i is even, num must be equal to (i * n) + (i / 2) where n is a non-negative integer. This can of course be written as num % i == i / 2.
When i is odd, num must be equal to i * n, where n is a non-negative integer. Which gives us our second condition num % i == 0.
In addition to these conditions, num can not be less than the sum of the first i positive integers. Hence, our for loop's conditional: sumOfFirstIIntegers <= num.
This determines whether a number is not a power of 2:
boolean IsSumOfConsecutiveInts(int num)
{
return (num & (num - 1)) != 0;
}
This answer does a good job of explaining why this works.
Note that both of the above solutions have the same result, they are just different ways of thinking about the problem.
public int sumOfDigits(int num){
return num == 0 ? 0 : num % 10 + sumOfDigits(num/10);
}
This is a recursive calculation of the sum of digits - as long as you haven't consumed all the digits, it adds the last one with the sum of all the previous ones.
sumOfDigits is a method which accepts int argument and the return type is also int.
If 0 is passed then it will return 0 other wise it will return num % 10 + sumOfDigits(num/10)
If a number other than 0 is passed this will continue as long as entered number becomes 0
return num == 0 ? 0 : num % 10 + sumOfDigits(num/10);
Is equivalent to:
if(num==0)
{
return 0;
}
else
{
return num % 10 + sumOfDigits(num/10); // this is recursive call. It gets called until num is reduce to 0
}
This basically calulates as following:
If num=768 result= 7+6+8.
That is the Java ternary operator. It is similar to ternary operators in other languages. In this case:
return <boolean expression> ? <if true> : <if false>;
You have posted a recursive algorithm to calculate the sum of digits in an integer.
Two equivalent methods are posted below:
public int sumOfDigitsIf(int num) {
if (num == 0) {
return 0;
} else {
return num % 10 + sumOfDigitsIf(num / 10);
}
}
public int sumOfDigitsLoop(int num) { // useful to examine if you aren't used to recursive algorithms
int value = 0;
while (num > 0) {
value += num % 10;
num /= 10;
}
return value;
}
This function calculates the sum of the digits of the number recursively. First, if the number is 0, it returns 0, as the sum of the digits of 0 is 0. Then, it calculates the digit in the 1s place with num % 10, and adds that to the sum of the digits of the remaining places, num / 10 is the remaining places, and it calls sumOfDigits() on that number.
This method calculates sum of the numbers representing base 10 integer numbers. The method is called recursively with % operator to separate digits in the number and then add them to form the result.
Eg - 234 into 2 + 3 + 4 = 9
However, if you pass a integer in base other than base 10, the method still gives the result for base 10
Eg sumOfDigits(0b1011010) -> answer is 9 for 90 in base 10
sumOfDigits(0x532) -> answer is 7 for 1330 in base 10
Following is the change for binary and octal values to get the sum of the representing numbers
return num == 0 ? 0 : num % 2 + sumOfDigits(num / 2);
return num == 0 ? 0 : num % 8 + sumOfDigits(num / 8);
That's a Java ternary operator, but in most cases it is not used since if-else statements are much more readable, which can also do the same function ternary operators can do. But if you would like to make shorter code lines, then this is preferred.
This question was asked in an interview (about prime numbers)
Russian Doll Primes
They are more commonly known as Truncatable Primes.
I found this code on wiki
public static void main(String[] args){
final int MAX = 1000000;
//Sieve of Eratosthenes (using BitSet only for odd numbers)
BitSet primeList = new BitSet(MAX>>1);
primeList.set(0,primeList.size(),true);
int sqroot = (int) Math.sqrt(MAX);
primeList.clear(0);
for(int num = 3; num <= sqroot; num+=2)
{
if( primeList.get(num >> 1) )
{
int inc = num << 1;
for(int factor = num * num; factor < MAX; factor += inc)
{
//if( ((factor) & 1) == 1)
//{
primeList.clear(factor >> 1);
//}
}
}
}
//Find Largest Truncatable Prime. (so we start from 1000000 - 1
int rightTrunc = -1, leftTrunc = -1;
for(int prime = (MAX - 1) | 1; prime >= 3; prime -= 2)
{
if(primeList.get(prime>>1))
{
//Already found Right Truncatable Prime?
if(rightTrunc == -1)
{
int right = prime;
while(right > 0 && primeList.get(right >> 1)) right /= 10;
if(right == 0) rightTrunc = prime;
}
//Already found Left Truncatable Prime?
if(leftTrunc == -1 )
{
//Left Truncation
String left = Integer.toString(prime);
if(!left.contains("0"))
{
while( left.length() > 0 ){
int iLeft = Integer.parseInt(left);
if(!primeList.get( iLeft >> 1)) break;
left = left.substring(1);
}
if(left.length() == 0) leftTrunc = prime;
}
}
if(leftTrunc != -1 && rightTrunc != -1) //Found both? then Stop loop
{
break;
}
}
}
System.out.println("Left Truncatable : " + leftTrunc);
System.out.println("Right Truncatable : " + rightTrunc);
}
This gives the output:
Left Truncatable : 998443
Right Truncatable : 796339
But I am not able to solve this particular Russian doll prime number problem like if you have a prime number and you remove either left or right digit of this prime number then if that resulting number is prime number or not?
I am new to this so please pardon any mistake.
Let's start from the beginning:
According to the link you supplied with your question:
"Russian Doll Primes are
prime numbers whose right digit can be repeatedly removed, and are
still prime."
I will assume that you have a function boolean isPrime(int) to find out if a number is prime.
Googling, we will find from Wikipedia that the number of right-truncatable prime numbers up to 73,939,133 is 83, which makes brute-force a viable option; but a few optimization techniques can be employed here:
Since we right-truncate, we can safely eliminate even numbers (since any even number won't be prime, and so any number generated upon it will never be a russian doll prime).
Since any number that starts with 5 is divisible by 5, then based on the same rule I mentioned in the previous point, we can eliminate 5.
That leaves us with numbers that contain 1, 3, 7, and 9.
Now if we wanted to generate all possible combinations of these 4 numbers that do not exceed the maximum you mentioned (1,000,000), it would take only 4,096 iterations.
The downside of this technique is that we now have 4,096 numbers that contain possible non-prime numbers, or prime numbers that are formed from non-prime numbers and hence are not russian doll primes. We can eliminate these numbers by looping through them and checking; or better yet, we can examine russian doll primes more closely.
Upon examining the rule I quoted from your link above, we find that a russian doll primes are prime numbers whose right digit can be repeatedly removed, and are still prime (and hence are still russian doll prime, given the word repeatedly)!
That means we can work from the smallest single-digit russian doll primes, work our generation magic that we used above, and since any prime number that is formed from russian doll prime numbers is a russian doll prime number, we can eliminate non-primes early on, resulting in a clean list of russian doll prime numbers, while reducing the running time of such a program dramatically.
Take a look at the generation code below:
void russianDollPrimesGeneration(int x) {
x *= 10;
if (x * 10 >= 1000000) return;
int j;
for (int i=1; i<=9; i+=2) {
if (i == 5) continue;
j = x + i;
if (isPrime(j)) {
addToRussianDollPrimesList(j);
russianDollPrimesGeneration(j);
}
}
}
Provided that void addToRussianDollPrimesList(int x) is a function that adds x to a list that we previously preserved to store the russian doll prime numbers.
UPDATED NOTE
Note that you can put the call to void russianDollPrimesGeneration(int x) that we made inside the if condition inside the void addToRussianDollPrimesList(int x) function, because whenever we call the former function, we will always call the latter function with the same arguments. I'm separating them here to emphasize the recursive nature of the generation function.
Also note that you must run this function with the integer 0.
A final note is that there are a number of cases that the generation function void russianDollPrimesGeneration(int x) above won't count, even though they are Russian Doll Primes.
Remember when we omitted 2 and 5, because even numbers and numbers divided by 5 cannot be primes and so cannot be Russian Doll Primes? and consequently cannot form Russian Doll Primes? Well, that case does not apply to 2 and 5, because they are prime, and since they are single digits, therefore they are Russian Doll Primes, and are eligible to form Russian Doll Primes, if placed in the left-side, like 23 and 53.
So how to correct our code to include these special cases?
We can make a wrapper function that adds these two numbers and checks for Russian Doll Primes that can be formed using them (which will be the same generation function we are using above).
void generationWrapperFunction(int x) {
addToRussianDollPrimesList(2);
russianDollPrimesGeneration(2);
addToRussianDollPrimesList(5);
russianDollPrimesGeneration(5);
russianDollPrimesGeneration(0);
}
END UPDATED NOTE
This little function will produce a list of russian doll prime numbers, which can then be searched for the number we are looking for.
An alternative, yet I believe will be more time-consuming, is the following recursive function:
boolean isRussianDollPrime(int n) {
if (!isPrime(n)) return false;
if (n < 10) return true;
return isRussianDollPrime(n / 10);
}
This function can be modified to work with left-truncatable primes. The generation-based solution, however, will be much difficult to implement for left-truncatable primes.
Your problem is to use this code or to solve the problem ?
if to solve it you can generate primes using Sieve algorithm then check if the element is prime or not (if it was prime then check if element/10 is also prime)
Let's start with a simple assumption that we know how to write code to detect if a value is a prime. In a coding interview, they won't likely won't expect you to pull out "Sieve of Eratosthenes". You should start with simple code that handles the special cases of x<=1 (false) and x==2(true). Then check for an even number !(x % 2)(false). Then loop on i from 3..sqrt(x) (incrementing by +2 each time) to see if there's an odd number divisor for x.
boolean isPrime(long x)
{
// your code goes here
}
And once we have a function to tell us if a value is prime, we can easily build the function to detect if a value is a Russian Prime. Therefore we just need to loop on our value, each time check for prime, and then chop off the right hand side. And the easiest way to remove the right-most digit from a number is to simply divide it by 10.
boolean isRussianPrime(long x)
{
boolean result = isPrime(x);
while ((x != 0) && result)
{
// chop off the right digit of x
x = x / 10;
if (x != 0)
{
result = isPrime(x);
}
}
return result;
}
And that's really all there is to it.
package com.example.tests;
public class RussianDollPrimeNumber {
public static void main(String[] args) {
int x= 373;
int k;
int n =x;
for ( k= String.valueOf(x).length()-1;k>0;k--){
System.out.println(n);
if (isPrime(n)){
String m=String.valueOf(n).substring(0, k);
n=Integer.parseInt(m);
continue;
}else {
break;
}
}
if( k==0){
System.out.println("Number is Russianl Doll Number "+x);
}else {
System.out.println("Number is not Russianl Doll Number "+x);
}
}
private static boolean isPrime(int x) {
boolean check=true;
for (int i=2;i<x/2;i++){
if( (x%i)==0){
check=false;
}
}
return check;
}
}