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Error in calculating sum of even place digits and odd place digits in java

In our directions, we have to get a 16 digit number, then sum all the digits in the odd place from right to left. After that, we have to sum all the even place digits from right to left, double the sum, then take module 9. When I try to run my code, I keep getting "Invalid", even if it is with a valid credit card number.
public static boolean validateCreditCard(long number) {
double cardSum = 0;
for (int i = 0; i < 16; i++) {
long cardnumber = (long) Math.pow(10, i);
double oddPlaceSum = 0;
double evenPlaceSum = 0;
if (i % 2 != 0) {
oddPlaceSum += ((int)(number % cardnumber / (Math.pow(10, i))));
} else { // so if i%2 ==0
evenPlaceSum += ((int)(number % cardnumber / (Math.pow(10, i)) * 2 % 9));
}
cardSum += evenPlaceSum + oddPlaceSum;
}
if (cardSum % 10 == 0) {
return true;
System.out.println("Valid");
} else {
return false;
System.out.println("Invalid");
}
}

Try this instead :
Convert the 16 digit number into a String using Long.toString(number).
Iterate through the String character by character and keep track of even and odd indexes.
Convert each char to an Integer using Integer.valueOf() thereby adding them incrementally.
Voila, you got your evenSum and oddSum. Next steps should be trivial.
public static boolean validateCreditCard(long number){
String x = Long.toString(number);
int evenSum = 0;
int oddSum = 0;
for(int i=0; i<x.length; i=i+2) {
oddSum += Integer.valueOf(s[i]);
evenSum += Integer.valueOf(s[i+1]);
}
//Do the next steps with odd and even sums.
Also, do handle IndexOutOfBoundsException as appropriate.

You can do it in a single while loop as digits are fixed, like this:
int digit,evensum,oddsum;
int i=16;
while(i > 0){
digit=number%10;
if(i%2 == 0)
evensum+=digit;
else
oddsum+=digit;
i--;
digit/=10;
}

Try this instead
using Recusion find sum of even placed of digit and sum of odd placed of digit.
class Recursion {
static int count = 0;
static int even =0;
static int odd =0;
public static int Digits(int num) {
if (num > 0) {
count++;
if(count%2 == 0){
even += num%10;
}
else{
odd += num%10;
}
Digits(num / 10);
}
return even;
// for odd
// return odd;
}
public static void main(String[] args) {
int num = 31593;
int res = Digits(num);
System.out.println("Total digits are: " + res);
}
}
Help me How to print both even and odd sum together?

how do i get even placed digits from a number in java

I want my program to get all the even digits from a number input. Then multiply those with digits with 2. If the result is a two digit number, add them. At the end i want it to give me the sum of all the even digits.
public class evenplaceadd {
public static void main(String[] args) {
System.out.println(sumOfevenPlace(5566));
}
public static int sumOfevenPlace(int number)
{
int maxDigitLength = 4;
int sum = 0;
for (int i = 0; i < maxDigitLength; i++)
{
if (i % 2 == 0)
{
int digita = number % 10;
int digitb =digita*2;
int digitc;
if(digita < 9)
{
sum = sum + digitb;
}
else if(digitb>9)
{
digitc =(digitb % 10)+ (digitb /10);
sum =sum + digitc;
}
}
else
{
number = number/10;
}
}
return sum;
}
}

Your code seems ok for the most part. There are some minor flaws in the code which I am sure you will be able to figure out after understanding the code provided below. I have changed it up a bit and made it easier to read. Please confirm it is working, and next time please provide the code when asking question. I know you are new to the community, and so am I. Its a learning experience for all of us. All the best in the future :)
public static void int sumOfEvenDigits(int num){
int sum = 0;
int lastDig = 0;
while(num/10 != 0)
{
lastDig = num % 10;
num = num / 10;
if(lastDig % 2 != 0)
{
continue;
}
if(lastDig > 10)
{
sum += lastDig / 10;
sum += lastDig % 10;
}
else
{
sum += lastDig;
}
}
return sum;
}

Programming a Luhn Algorithm in Java

I'm trying to program the Luhn Algorithm within Java.
My current code is :
import java.util.Scanner;
public class luhnAlgorithm {
public static void main(String[] args) {
System.out.println("main() : Entry Point");
Scanner input = new Scanner(System.in);
long num;
int digit;
int sum = 0;
System.out.println("Enter the digits of a credit card number : ");
num = input.nextLong();
while (num > 0) {
digit = (int) num % 10;
num = num / 10;
if (num % 2 != 0 ) {
digit *= 2;
}
if (digit > 9) {
digit -= 9;
}
sum += digit;
}
if(sum % 10 == 0) {
System.out.println("Credit card number is valid.");
}
else
System.out.println("Credit card number is invalid. Please try again.");
System.out.println("main() : Exit Point");
}
}
The problem I'm having is that when I enter in a valid credit card number, for example : 4012888888881881 (via PayPal Test Credit Card Accounts), it says it's invalid. But when I put in my own debit card number it says it's valid.
I know there's some code messed up in here, I just can't figure out what.
Any help would be appreciated and thanks in advance!

I think that I know where the problem is. You do the multiplication in a wrong way:
if (num % 2 != 0 ) {
digit *= 2;
}
You multiply digit when num is odd and in Luhn's algorithm you should multiply digit when this digit is on the even place in number moving from right to the left. Try to think about adding some index which will help you to know if a digit is on even place or not.
You can think about splitting your integer to array and then check if index of array is even or add to your while loop some index.
Carefully study this https://en.wikipedia.org/wiki/Luhn_algorithm
for example if you have 68 then you have: first iteration: digit = 8, num = 6 and sum =8 second iteration: digit = 6, num = 0 here you should multiply your digit by 2, because it is on the even place in number, but you don't do that and sum = 14 instead of 20

Okay I actually figured it out :).
import java.util.Scanner;
/*
* Author : Jonathan Patterson
* Date : 10/28/15
* Program : Luhn Algorithm; validates credit card numbers
*/
public class luhnAlgorithm {
public static void main(String[] args) {
System.out.println("main() : Entry Point");
Scanner input = new Scanner(System.in);
long num;
double digit = 0;
int sum = 0;
int n = 1;
int i = 0;
System.out.println("Enter the digits of a credit card number : ");
num = input.nextLong();
while (num > 0) {
digit = num % 10;
num = num / 10;
System.out.println(n + " digit is : " + digit);
if (i % 2 != 0 ) {
digit *= 2;
}
System.out.println(n + " digit is : " + digit);
if (digit > 9) {
digit = (digit % 10) + 1;
}
else
digit *= 1;
System.out.println(n + " digit is : " + digit);
sum += digit;
n++;
i++;
}
System.out.println("Sum of the digits is : " +sum);
if(sum % 10 == 0) {
System.out.println("Credit card number is valid.");
}
else
System.out.println("Credit card number is invalid. Please try again.");
System.out.println("main() : Exit Point");
}
}

Adding an additional answer to this in case anyone else finds this post.
There is a bitbucket project with a valid implementation:
https://bitbucket.org/javapda/npi-validator/src/master/
This if for verifying NPI provider numbers which is an implementation of the Luhn algorithm.
package org.bitbucket.javapda.npi;
import java.util.ArrayList;
import java.util.List;
public class NpiValidator {
private String npi;
public NpiValidator(String npi) {
this.npi = npi.trim();
}
public boolean isValid() {
return npi.length() == 10 && complies();
}
private boolean complies() {
if (!npi.matches("[0-9]{10}")) {
return false;
}
Character lastDigit = npi.charAt(9);
List<Integer> everyOther = listWithEveryOtherDoubled(npi.substring(0, 9));
int sum = 0;
for (Integer num : everyOther) {
sum += sumOfDigits(num);
}
int total = sum + 24; // 24 to account for 80840
int units = total % 10;
int checkDigit = (units != 0) ? (10 - units) : units;
return (Character.getNumericValue(lastDigit) == checkDigit);
}
private List<Integer> listWithEveryOtherDoubled(String str) {
List<Integer> nums = new ArrayList<Integer>();
for (int i = 0; i < str.length(); i++) {
if (i % 2 == 0) {
nums.add(2 * Character.getNumericValue(str.charAt(i)));
} else {
nums.add(Character.getNumericValue(str.charAt(i)));
}
}
return nums;
}
private static int sumOfDigits(int number) {
int num = number;
int sum = 0;
while (num > 0) {
sum += (num % 10);
num = num / 10;
}
return sum;
}
public static void main(String[] args) {
System.out.println("Hello, World!");
// System.out.println(sumOfDigits(16));
System.out.println("1234567890".matches("[0-9]{10}"));
System.out.println("123456789".matches("[0-9]{10}"));
}
}

how to get exponents without using the math.pow for java

This is my program
// ************************************************************
// PowersOf2.java
//
// Print out as many powers of 2 as the user requests
//
// ************************************************************
import java.util.Scanner;
public class PowersOf2 {
public static void main(String[] args)
{
int numPowersOf2; //How many powers of 2 to compute
int nextPowerOf2 = 1; //Current power of 2
int exponent= 1;
double x;
//Exponent for current power of 2 -- this
//also serves as a counter for the loop Scanner
Scanner scan = new Scanner(System.in);
System.out.println("How many powers of 2 would you like printed?");
numPowersOf2 = scan.nextInt();
System.out.println ("There will be " + numPowersOf2 + " powers of 2 printed");
//initialize exponent -- the first thing printed is 2 to the what?
while( exponent <= numPowersOf2)
{
double x1 = Math.pow(2, exponent);
System.out.println("2^" + exponent + " = " + x1);
exponent++;
}
//print out current power of 2
//find next power of 2 -- how do you get this from the last one?
//increment exponent
}
}
The thing is that I am not allowed to use the math.pow method, I need to find another way to get the correct answer in the while loop.

Powers of 2 can simply be computed by Bit Shift Operators
int exponent = ...
int powerOf2 = 1 << exponent;
Even for the more general form, you should not compute an exponent by "multiplying n times". Instead, you could do Exponentiation by squaring

Here is a post that allows both negative/positive power calculations.
https://stackoverflow.com/a/23003962/3538289
Function to handle +/- exponents with O(log(n)) complexity.
double power(double x, int n){
if(n==0)
return 1;
if(n<0){
x = 1.0/x;
n = -n;
}
double ret = power(x,n/2);
ret = ret * ret;
if(n%2!=0)
ret = ret * x;
return ret;
}

You could implement your own power function.
The complexity of the power function depends on your requirements and constraints.
For example, you may constraint exponents to be only positive integer.
Here's an example of power function:
public static double power(double base, int exponent) {
double ans = 1;
if (exponent != 0) {
int absExponent = exponent > 0 ? exponent : (-1) * exponent;
for (int i = 1; i <= absExponent; i++) {
ans *= base;
}
if (exponent < 0) {
// For negative exponent, must invert
ans = 1.0 / ans;
}
} else {
// exponent is 0
ans = 1;
}
return ans;
}

If there are no performance constraints you can do:
double x1=1;
for(int i=1;i<=numPowersOf2;i++){
x1 =* 2
}

You can try to do this based on this explanation:
public double myPow(double x, int n) {
if(n < 0) {
if(n == Integer.MIN_VALUE) {
n = (n+1)*(-1);
return 1.0/(myPow(x*x, n));
}
n = n*(-1);
return (double)1.0/myPow(x, n);
}
double y = 1;
while(n > 0) {
if(n%2 == 0) {
x = x*x;
}
else {
y = y*x;
x = x*x;
}
n = n/2;
}
return y;
}

It's unclear whether your comment about using a loop is a desire or a requirement. If it's just a desire there is a math identity you can use that doesn't rely on Math.Pow.
xy = ey∙ln(x)
In Java this would look like
public static double myPow(double x, double y){
return Math.exp(y*Math.log(x));
}
If you really need a loop, you can use something like the following
public static double myPow(double b, int e) {
if (e < 0) {
b = 1 / b;
e = -e;
}
double pow = 1.0;
double intermediate = b;
boolean fin = false;
while (e != 0) {
if (e % 2 == 0) {
intermediate *= intermediate;
fin = true;
} else {
pow *= intermediate;
intermediate = b;
fin = false;
}
e >>= 1;
}
return pow * (fin ? intermediate : 1.0);
}

// Set the variables
int numPowersOf2; //How many powers of 2 to compute
int nextPowerOf2 = 1; //Current power of 2
int exponent = 0;
/* User input here */
// Loop and print results
do
{
System.out.println ("2^" + exponent + " = " + nextPowerOf2);
nextPowerOf2 = nextPowerOf2*2;
exponent ++;
}
while (exponent < numPowersOf2);

here is how I managed without using "myPow(x,n)", but by making use of "while". (I've only been learning Java for 2 weeks so excuse, if the code is a bit lumpy :)
String base ="";
String exp ="";
BufferedReader value = new BufferedReader (new InputStreamReader(System.in));
try {System.out.print("enter the base number: ");
base = value.readLine();
System.out.print("enter the exponent: ");
exp = value.readLine(); }
catch(IOException e){System.out.print("error");}
int x = Integer.valueOf(base);
int n = Integer.valueOf(exp);
int y=x;
int m=1;
while(m<n+1) {
System.out.println(x+"^"+m+"= "+y);
y=y*x;
m++;
}

To implement pow function without using built-in Math.pow(), we can use the below recursive way to implement it. To optimize the runtime, we can store the result of power(a, b/2) and reuse it depending on the number of times is even or odd.
static float power(float a, int b)
{
float temp;
if( b == 0)
return 1;
temp = power(a, b/2);
// if even times
if (b%2 == 0)
return temp*temp;
else // if odd times
{
if(b > 0)
return a * temp * temp;
else // if negetive i.e. 3 ^ (-2)
return (temp * temp) / a;
}
}

I know this answer is very late, but there's a very simple solution you can use if you are allowed to have variables that store the base and the exponent.
public class trythis {
public static void main(String[] args) {
int b = 2;
int p = 5;
int r = 1;
for (int i = 1; i <= p; i++) {
r *= b;
}
System.out.println(r);
}
}
This will work with positive and negative bases, but not with negative powers.

To get the exponential value without using Math.pow() you can use a loop:
As long as the count is less than b (your power), your loop will have an
additional "* a" to it. Mathematically, it is the same as having a Math.pow()
while (count <=b){
a= a* a;
}

Try this simple code:
public static int exponent(int base, int power) {
int answer = 1;
for(int i = 0; i < power; i++) {
answer *= base;
}
return answer;
}

Better way to generate Armstrong Numbers recursively

Is there a more elegant way to write this recursively? Armstrong Numbers
PS: been out of school for 15 years this is not homework, just some code I am trying to convert from iterative to recursively.
import java.util.Scanner;
public class RecArmstrong {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter a number: ");
int number = keyboard.nextInt();
//Error checking
while(number < 0 || number > 100000){
System.out.print("Enter a number: ");
number = keyboard.nextInt();
}
if(arm(number) == number)
System.out.println(number + " is an armstrong number");
else
System.out.println(number + " is not an armstrong number.");
}
public static long arm(long n){
long temp, sum, digits = 0;
long remainder;
temp = n;
sum = 0;
if (temp == 0)
return 0; //base case
else{
while (temp != 0){
digits++; //number of digits for exponent
temp = temp / 10;
}
temp = n; //set temp back to original number
while (temp != 0){
remainder = temp % 10;
sum += Math.pow(remainder, digits);
temp = temp / 10;
}
return sum + arm(temp);
}
}
}

Maybe im mistaken (its 1AM here..), but as far as i can tell your implementation is actually not a recursion at all. With
while (temp != 0){
remainder = temp % 10;
sum += Math.pow(remainder, digits);
temp = temp / 10;
}
you do the whole calculation iteratively, until tmp is 0. Therefore in the next line
return sum + arm(temp);
arm(temp) will always return 0!
I hacked together a quick recursion of my own, which works digit by digit, starting from the last.
The function overload is necessary because every recursive call needs the total length of the original number.
public static long arm(long n){
return arm(n, Integer.toString(n).length());
}
public static long arm(long n, int num_digits){
if(n==0) //recursion finished
return;
// n%10 gives last digit
return java.lang.Math.pow(n%10,num_digits) + arm(n/10, num_digits);
}
I hope you like it ;-)

If you want to do it using java stream, you can do it like this.
public boolean isArmstrongNumber(String number) {
int exponent = number.length();
if (Integer.parseInt(number) == number.chars()
.map(n -> n - '0')
.map(n ->(int) Math.pow(Integer.parseInt("" + n),exponent))
.sum())
return true;
else return false;
}