public class Pow {
public double getAnswer(double a, double b) {
double b2 = 0;
if (b > 0) {
for (int i = 1; i < b; i++) {
a = a * a;
}
return a;
} else if (b < 0) {
int c = 0;
while (c > b) {
a = a * a;
c--;
}
b2 = 1 / a;
}
return b2;
}
}
I need the second part of my method to return the value of a negative power(i.e. 5^-2 = .04), but the output is always 0. The first part of the method works fine from what I have tested. I do have the last curly braces but they just wouldn't fit in the text box on here. Any help/suggestions would be much appreciated!
Running your code does not produce 0 as a result, but there is a bug.
a = a * a squares the number every iteration, so an will be calculated as: a2n-1.
Try accumulating the multiplication in a different variable:
double b2 = 0;
double result = 1;
if (b > 0) {
for (int i = 1; i <= b; i++) {
result *= a;
}
return result;
} else if (b < 0) {
int c = 0;
while (c > b) {
result *= a;
c--;
}
b2 = 1 / result;
}
return b2;
public class Pow {
public double getAnswer(double a, double b) {
double b2 = 0;
if (b > 0) {
for (int i = 1; i < b; i++) {
a = a * a;
}
return a;
} else if (b < 0) {
return getAnswer(a, -b);
}
return 1; // b is 0
}
Your algorithm for (b < 0) was wrong. If b < 0, the calculation is 1 / a^(-b).
Why your parameter b is of type double? What if b is 2.5? I suggest to change the type of b to int or you have to change your algorithm.
As Bohemian♦ said
a = a * a;
squares a each time.
What we want is for a (the base) to be multiplied by itself b (the power) times.
Since the only difference between a to the power of b and a to the power of (-b) is that the latter is the 1 / the former, we can use quite nearly the same code for both negative and positive exponents.
Code:
public double getAnswer(double a, double b) {//a is the base, b is the power
double a2 = 1;
double b2 = 0;
if (b > 0) {//for positive powers
for (int i = 0; i < b; i++) {//a needs to be multiplied by itself (b) times
a2 = a2 * a;
}
return a2;
}
else if (b < 0) {//for negative powers
for (int i = 0; i < -b; i++) {//a still needs to be multiplied by itself (-b) times, but since b is negative, we increment up to the opposite of b
a2 = a2 * a;
}
return 1 / a2;//finally, since the power (b) is negative, we need to return 1 / (a2)
}
}
Related
How I find among all pairs a and b with a "least common multiple" LCM(a,b) = 498960 and a "greatest common divisor" GDM(a, b) = 12 a pair with minimum sum a + b?
I solved this with O(n^2) time:
public class FindLcmAndGcdClass {
private int findGcd(int a, int b) {
if (a % b == 0) {
return b;
}
return findGcd(b, a % b);
}
private int findLcm(int a, int b, int gcd) {
return (a * b) / gcd;
}
private void run() {
int minSum = Integer.MAX_VALUE;
int foundNumberOne = 0;
int foundNumberTwo = 0;
for (int i = 12; i <= 498960; i += 12) {
for (int j = i; j <= 498960; j += 12) {
int gcd;
if (i < j) {
gcd = findGcd(j, i);
} else {
gcd = findGcd(i, j);
}
int lcm = findLcm(i, j, gcd);
if (gcd == 12 && lcm == 498960 && i + j < minSum) {
minSum = i + j;
foundNumberOne = i;
foundNumberTwo = j;
}
}
}
System.out.println(minSum);
System.out.println(foundNumberOne);
System.out.println(foundNumberTwo);
}
public static void main(String[] args) {
var o = new FindLcmAndGcdClass();
o.run();
}
}
And it executes quite slowly! I guess the problem can be solved with Dynamic Programming. Can anyone help with more fast solution?
I am not sure if this question can be solved with dynamic programming, but I think of a solution with time complexity O(sqrt(LCM * GCD)).
It is well known that for any two integers a and b, LCM(a, b) * GCD(a, b) = a * b. Therefore, you can first calculate the product of the gcd and lcm, (which is 5987520 in this question). Then for all its factors under sqrt(LCM * GCD), let a be one of the factors, then b = LCM * GCD / a. Test if gcd(a, b) = the required gcd, if so calculate the sum a + b, then find the minimum among the sums, and you are done.
class Main {
public static void main(String[] args) {
int abc;
int total = 1000;
for (abc = 220000000; abc < 240000000; abc++) {
String string1 = Integer.toString(abc);
int a, b, c;
a = 0;
b = 0;
c = 0;
int a2 = squared(a);
int b2 = squared(b);
int c2 = squared(c);
a = Integer.parseInt(string1.substring(0, 3));
b = Integer.parseInt(string1.substring(3, 6));
c = Integer.parseInt(string1.substring(6, 9));
if (a < b && a < c && b < c && a2 < b2 && b2 < c2 && a + b + c == total) {
System.out.println("The answer is " + abc);
} else {
System.exit(0); // testing
}
}
}
public static int squared(int x) {
return (x * x);
}
}
This is my code. I am trying to have the program print the value of abc so that a is less than b which is less than c. And where a squared is less than b squared which is less than c squared. Also, a+b+c must equal 1000. I tried making abc one whole value and using substrings to differ between a, b, and c. An example would be like 224,356,446. An with the substrings it would make a=224, b=356, and c=446. When I run the program, nothing is being printed. The else statement is being invoked because the system exits. Can anyone help?
You print nothing because you do not have a match on your first iteration. The lack of a match fires System.exit(0) so the program exits without another iteration.
You have appeared to make the assumption that that a, b, c are non-negative integers so I'll adopt that same assumption. As such, if a < b < c then a^2 < b^2 < c^2 so we do not have to test for the latter condition.
If a + b + c == 1000 and a < b < c then if a==0 and b==1 (their minimum values) then c == 999. So we know we do not have to iterate beyond 999 in any case. This bound could be further optimized a bit but doing so is pointless since he output operation far outweighs the iteration cost.
public static void main(String[] args) {
int count = 0;
for (int a = 0 ; a < 999 ; ++a) {
for (int b = a+1 ; b < 999 ; ++b) { // a < b
int c = 1000 - a - b;
if (c <= b) {
break;
}
System.out.printf("%d %d, %d, %d", ++count, a, b, c).println();
}
}
}
I've written this code that computes the sum of the positive divisors, and all the values have to be to the power of a.
For instance:
sigma(0,14) = 1^0 + 2^0 + 7^0 + 14^0 = 4;
sigma(2,12) = 1^2 + 2^2 + 3^2 + 4^2 + 6^2 + 12^2 = 210.
sigma(a, b).
I have tried different versions but I don't know how to add the power function.
try {
int a = Integer.parseInt(input1.getText());
int b = Integer.parseInt(input2.getText());
int result1 = 0;
for (int i = 2; i <= Math.sqrt(b); i++)
{
if (b % i == 0)
{
if (i == (b / i))
result1 += i;
else
result1 += (i + b / i);
}
}
result.setText(String.valueOf(result1 + b + 1));
}
}
In Java the ^ character means XOR.
The power function is provided by the Math.pow() method.
So 3^2 would be Math.pow(3, 2).
If you wanted to implement it yourself for integers, you could do it simply like this:
double power(int a, int b) {
int pow = (b < 0) ? -b : b;
double result = 1;
for (int i = 0; i < pow; i++) {
result *= a;
}
return (b < 0) ? 1 / result : result;
}
But I wouldn't do it myself. It gets a bit more complicated for floating points, and Java has a native underlying implementation which is much faster.
IntStream delivers beautiful concise calculation.
static int sigma(int exp, int num) {
IntStream.rangeClosed(1, num) // 1, ..., num
.filter(k -> num % k == 0) // Only divisors
.map(k -> pow(k, exp))
.sum();
}
static int pow(int k, int exp) {
if (exp == 0) {
return 1;
}
int squareRoot = pow(k, exp/2);
int n = squareRoot * squareRoot;
return (exp % 2) == 0 ? n : n*k;
}
The power calculation can be optimized by not using exp# multiplications of k but square roots.
For those interested in program transformation:
pow(k, exp) needs only to rely on exp with recursion to exp/2 (integer division). So you could turn the code inside out, have a vector of divisors,
and operate on that.
If you want to implement it without using Math.pow() you can simply follow the mathematical definition of the exponentiation for a positive exponent:
public static long exp(int a, int b){ //computes a^b
long result = 1;
for (int i = 0; i < b; i++) {
result *= a;
}
return result;
}
I would recommend that you use Java lambdas to accomplish what you're looking for.
Taking an input and returning a List of positive divisors seems useful on its own.
Raising every entry to a power could be done easily with a lambda.
Keep the two functions separate. Take a more functional approach.
Here is a simple code for you:
public static void main(String args[]) {
Scanner scanner = new Scanner(System.in);
List<Integer> listOfBs = new ArrayList<>();
System.out.println("Input your a");
int a = scanner.nextInt();
System.out.println("Input your b");
int b = scanner.nextInt();
int sqrt = (int) Math.sqrt(b);
for (int i = 1; i <= sqrt; i++) {
if (b % i == 0) {
listOfBs.add(i);
int d = b / i;
if (d != i) {
listOfBs.add(d);
}
}
}
int sigma = 0;
for(int e : listOfBs)
{
sigma += Math.pow(e,a);
}
System.out.println("Your sigma function is: "+sigma);
}
}
So i have a little problem with reducing a negative fraction
This is my reduce code
private void reduce() {
int g = Helper.gcd(this.num, this.den);
num /= g;
den /= g;
}
For example:
8/64 gives 1/8
But giving -8/64 let's the program crash
This is my gcd code
public static int gcd(int a, int b) {
while (a != b) {
if (a > b) {
a -= b;
} else {
b -= a;
}
}
return a;
}
You need to extract the sign first.
private void reduce() {
boolean neg = (num < 0) != (den < 0);
num = Math.abs(num);
den = Math.abs(den);
// obtain the GCD of the non-negative values.
int g = Helper.gcd(num, den);
num /= g;
den /= g;
if (neg) num *= -1;
}
Your gcd method only works for positive numbers. Negative numbers and zero need to be handled separately.
public static int gcd(int a, int b) {
a = Math.abs(a);
b = Math.abs(b);
if (a == 0) {
if (b == 0)
throw new IllegalArgumentException();
return b;
}
if (b == 0)
return a;
// The rest is just your code, unchanged.
while (a != b) {
if (a > b) {
a -= b;
} else {
b -= a;
}
}
return a;
}
I am currently attempting this question :
A Pythagorean triplet is a set of three natural numbers, a, b and c, for which
a2 + b2 = c2.
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
My code is as follows, I think it should be correct, but the site is telling me my answer is wrong? Can someone help me see the flaws in my logic please?
public class Pythagoras {
public static void main(String[] args) {
int sum = 1000;
int a;
int product=0;
for (a = 1; a <= sum/3; a++)
{
int b;
for (b = a + 1; b <= sum/2; b++)
{
int c = sum - a - b;
if ( c > 0 && (a*a + b*b == c*c) )
System.out.printf("a=%d, b=%d, c=%d\n",a,b,c);
product = a * b * c;
}
}
System.out.println(product);
}
}
Here are 5 solutions (from slow to fast):
1) Trivial implementation - 732857 microseconds (0.7 seconds)
private static void p1(int sum) {
for (int a = 0; a <= sum; a++) {
for (int b = 0; b <= sum; b++) {
for (int c = 0; c <= sum; c++) {
if (a < b && b < c && a + b + c == sum
&& (c * c == a * a + b * b)) {
System.out.print(a * b * c);
return;
}
}
}
}
}
2) Limit the lower bound for b & c (establish the order relation) - 251091 microseconds (0.2 seconds)
private static void p2(int sum) {
for (int a = 0; a <= sum; a++) {
for (int b = a + 1; b <= sum; b++) {
for (int c = b + 1; c <= sum; c++) {
if (a + b + c == sum && (c * c == a * a + b * b)) {
System.out.print(a * b * c);
return;
}
}
}
}
}
3) Limit the lower & upper bounds for b & c - 111220 microseconds (0.1 seconds)
private static void p3(int sum) {
for (int a = 0; a <= sum; a++) {
for (int b = a + 1; b <= sum - a; b++) {
for (int c = b + 1; c <= sum - a - b; c++) {
if (a + b + c == sum && (c * c == a * a + b * b)) {
System.out.print(a * b * c);
return;
}
}
}
}
}
4) Limit lower & upper bounds for b and fix value for c - 2625 microseconds
private static void p4(int sum) {
for (int a = 0; a <= sum; a++) {
for (int b = a + 1; b <= sum - a; b++) {
int c = sum - a - b;
if (c > b && c * c == a * a + b * b) {
System.out.print(a * b * c);
return;
}
}
}
}
5) Use Euclid's formula - 213 microseconds
private static void p5(int sum) {
// a = m^2 - n^2
// b = 2mn
// c = m^2 + n^2
int a, b, c;
int sqrt = (int)Math.sqrt(sum);
for (int n = 1; n <= sqrt; n++) {
for (int m = n+1; m <= sqrt; m++) {
a = m*m - n*n;
b = 2*m*n;
c = m*m + n*n;
if ( a + b + c == 1000 ) {
System.out.print(a * b * c);
return;
}
}
}
}
I think you're missing a set of braces. The indentation leads me to believe the two innermost statements go together but you need curly braces for that to be correct.
if ( c > 0 && (a*a + b*b == c*c) )
{
System.out.printf("a=%d, b=%d, c=%d\n",a,b,c);
product = a * b * c;
}
Without the braces product will always contain the product of the last values of a, b, and c. (333 * 500 * 167 == 27805500).
Though others have already given specific fixes for you code, here's a more general hint that will be useful on other problems as well. Test your code on a simpler version of the problem.
For example, see if your program can find 6,8,10 as a triplet with a sum of 24. With a smaller test you can actually step through the code to see where it's going wrong.
You may try it this way,
public class Pythagoras {
public static void main(String[] args) {
int m = 1, n = 0, a = 0, b = 0, c = 0, sum = 0;
int product = 0;
for (m = 2; m < 100; m++) {
for (n = 1; n < 100; n++) {
while (m > n) {
a = (m * m) - (n * n);
b = (2 * m) * n;
c = (m * m) + (n * n);
sum = a + b + c;
if (sum == 1000) {
product = a * b * c;
System.out.print("a :" + a + "b :" + b + "c : " + c);
System.out.println("Product is" + product);
break;
}
break;
}
}
}
}
}
This implements the Euclid's formula for generating Pythagorean triplet as explained here
Note that in this method we make only triplets hence unwanted repetitions are reduced.
and the output is a :375 b :200 c : 425 Product is 31875000
//
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
public javax.swingx.event.*;
public class Triplet extends JApplet implements ActionListener
{
JLabel l1, l2, l3;
JButton b1;
JTextFiel t1, t2;
public void init()
{
Container c = getContentPane();
c.setLayout(new FlowLayout());
l1=new JLabel("Enter the value of a: ");
l2=new JLabel("Enter the value of b: ");
t1 = new JTextField(20);
t2 = new JTextField(20);
b1=new JButton("Ok");
l2=new JLabel(" ");
add(l1);
add(t1);
add(l2);
add(t2);
add(b1);
add(l3);
b1.addActionListener(this);
public void ActionPerformed(ActionEvent e)
{
int a = Integer.parseInt(t1.getText());
int b = Integer.parseInt(t2.getText());
long c = Math.sqrt(a*a + b*b);
l3.setText(" " +c);
}
}
}
public class Pythagorean_Triplets
{
public static void main(long n)
{
long h=1,p=1,b1;
double b;
while(h<=n)
{
while(p<h)
{
b=Math.sqrt((h*h)-(p*p));
if(b%1==0)
{
b1=(long)b;
System.out.println(b1+","+p+","+h);
break;
}
p++;
}
h++;
p=1;
}
}
}