Questions over BigInteger? - java

import java.math.BigInteger;
public class ProjectEuler {
public static void main(String[] args) {
BigInteger bi = new BigInteger("600851475143");
int div = 7;
while (bi.compareTo(new BigInteger("1")) != 0) {
while (bi.mod(new BigInteger(div + "")).compareTo(new BigInteger("0")) == 0) {
bi = bi.divide(new BigInteger(div + ""));
}
div += 2;
}
System.out.println("" + div);
}
}
I was just looking over one of the basic but famous problems of "What is the largest prime factor of the number 600851475143". I found this solution different, i have a couple of questions on how this works.
The first condition checks whether the number equals 1 or not. From there i am not able to understand the rest of the code.
new BigInteger(div +""). why do we concatenate + "" here?

How is the div = 7 decided?
The author decided to "hard-code" his knowledge of the number itself to decide that the first three primes are not among the divisors
The first condition checks whether the number equals 1 or not. From there i am not able to understand the rest of the code.
The rest of the code looks like this in "regular" integers:
while (bi % div == 0) {
bi /= div;
}
div += 2;
new BigInteger(div +"") why do we concatenate + "" here
That is a short way of making an object a String. BigInteger has a parameter that takes String, so the alternative to this approach would be calling Integer.toString(div).
Note that this is not the most efficient solution: one could speed this up by observing that you could stop trying to divide when you reach the square root of the original number, because you can be sure that the next divisor will be the number itself.

How is the div = 7 decided?
Probably the author noticed that the number isn't divisible by 2 nor 3 nor 5. To know how the author did this, he/she should have known this rules: Divisibility Rules and Tests
The first condition checks whether the number equals 1 or not. From there i am not able to understand the rest of the code.
The author is making sure that the number is not BigInteger("1") since it's dividing the number and storing the results in bi in the loop iterations. Note this:
bi = bi.divide(new BigInteger(div + ""));
new BigInteger(div +""). why do we concatenate + "" here?
It uses the BigInteger(String) constructor. The author n̶a̶i̶v̶e̶l̶y̶ makes a new String by adding the int with an empty String.

Related

How can I correctly implement probability in Java?

I seem to have a problem I don't know why its happening its starting to look like witchcraft. This is the way I tried to implement probability:
int odd = (int)(100.0 * Math.random()); //Number between 1 and 100
if(odd<=50){ //50% chance
System.out.println("Lucky");
}
However, when I place it like this odd is ALWAYS inferior to 50, if I change the "<" to a ">" it ALWAYS generates number bigger than 50, which makes as the if statement happens every time.
I think you just had bad luck or whatever, so you got 10 or 100 times depending on how often you tested it, a larger or smaller number. Because your code works, I'm sure of it.
It is running well for me, and I do not find something wrong in your code. It may just generated many random numbers in a row that satisfied the statement.
I don't understand which problem did you have. I tried this code:
int b = 0;
for(int i = 0; i <100000;i++){
int odd = (int) (100.0 * Math.random()); //Number between 1 and 100
if (odd < 50) { //50% chance
System.out.println("Lucky");
b++;
} else {
System.out.println("Unlucky");
}
}
System.out.println("Lucky tries : " + b);
System.out.println("Unlucky tries : " + (100000 - b));
}
And it gave me the following result
Lucky tries : 50060
Unlucky tries : 49940

Euclid's algorithm shows wrong conclusion

Homework of mine is to create and Euclid's algorithm in java. The task binds me to use both while-loop and if statement. Futhermore - if statement has to be placed inside while-loop.
During this task i faced already infinity-loop problem, somehow manage to get pass it. Now my Euclid's algorithm is giving multiple answers (instead of one) and futhermore they are wrong...
I have searched a couple of topics over here, but none of answers shown in there gave me an answer. I tried to rewrite whole code, and also diffrent conditions for while-loop and if statement.
import java.lang.*;
class EuklidesAlgorithm {
public static void main (String[] args) throws java.lang.Exception{
int a = 25648;
int b = 15468;
while (a % b != 0 ){
int modulo = a % b;
if (modulo == 0){
break;
}else {
modulo = a % b;
System.out.println(" Checking for GCD");
a = b;
b = modulo;
}
System.out.println(" Number " + b + " is GDC of numbers(" + a + "," + b + ").");
}
}
}
I would like it to give a single answer what is GCD for a and b.
First of all the condition :
modulo==0
will alaways be false inside the loop...
and you dont have to change variable prices inside the loop and you also don't have to print answers in every loop so...
the if statement is probably goind to be used to check if any of those two numbers is 0 or if the result is 0 but you can do both

Add Two Binary Numbers Recursively, w/o Bitwise Operations; Java Homework

I think I've gotten mostly to a solution for a homework problem.
This is for a 201 CS class. Right now I just want to get the logic right. At present, it doesn't operate as intended, but it's close.
We don't want to use .toBinary, bitwise, or anything else. We also haven't been taught stringBuilder, so I'd like to avoid using it.
There's a System.out.println(); within the method which provides the correct answer if you read the console from bottom to top.
public static void main(String[] args) {
System.out.println(addBin(1100111011,1101110011));
}
public static String addBin(int num1,int num2){
String result = "";
if(num1 > 0 || num2 > 0){
int part1 = num1%10, part2 = num2%10;
int rem1 = num1/10, rem2 = num2/10;
result += Integer.toString((part1 + part2)%2);
//System.out.println(result);
int carry = (part1 + part2) /2;
addBin(rem1 + carry, rem2);
return result;
}
return result;
}
So, this example adds 1100111011 and 1101110011 with the output
0
1
1
1
0
1
0
1
0
1
1
0
when the correct answer is 11010101110.
I'm having trouble understanding how to properly "pop" the "result" part properly. Could you please help me understand this process, possibly within the context of this problem?
Thanks!
As you can see from your output, you are getting the correct result in the reverse order but you are not appending any of your older result to the ones that are being currently computed.
Inside your if condition, you are calling the addBin() function but you are not using the result that it gives anywhere. Just change that line to the following:
result = addBin(rem1 + carry, rem2)+result;
That should effective append all your results in front of the current answer so that you do not get the result in backwards direction. Hope this helps.

Calculate Dice Roll from Text Field

QUESTION:
How can I read the string "d6+2-d4" so that each d# will randomly generate a number within the parameter of the dice roll?
CLARIFIER:
I want to read a string and have it so when a d# appears, it will randomly generate a number such as to simulate a dice roll. Then, add up all the rolls and numbers to get a total. Much like how Roll20 does with their /roll command for an example. If !clarifying {lstThen.add("look at the Roll20 and play with the /roll command to understand it")} else if !understandStill {lstThen.add("I do not know what to say, someone else could try explaining it better...")}
Info:
I was making a Java program for Dungeons and Dragons, only to find that I have come across a problem in figuring out how to calculate the user input: I do not know how to evaluate a string such as this.
I theorize that I may need Java's eval at the end. I do know what I want to happen/have a theory on how to execute (this is more so PseudoCode than Java):
Random rand = new Random();
int i = 0;
String toEval;
String char;
String roll = txtField.getText();
while (i<roll.length) {
check if character at i position is a d, then highlight the numbers
after d until it comes to a special character/!aNumber
// so if d was found before 100, it will then highlight 100 and stop
// if the character is a symbol or the end of the string
if d appears {
char = rand.nextInt(#);
i + #'s of places;
// so when i++ occurs, it will move past whatever d# was in case
// d# was something like d100, d12, or d5291
} else {
char = roll.length[i];
}
toEval = toEval + char;
i++;
}
perform evaluation method on toEval to get a resulting number
list.add(roll + " = " + evaluated toEval);
EDIT:
With weston's help, I have honed in on what is likely needed, using a splitter with an array, it can detect certain symbols and add it into a list. However, it is my fault for not clarifying on what else was needed. The pseudocode above doesn't helpfully so this is what else I need to figure out.
roll.split("(+-/*^)");
As this part is what is also tripping me up. Should I make splits where there are numbers too? So an equation like:
String[] numbers = roll.split("(+-/*^)");
String[] symbols = roll.split("1234567890d")
// Rough idea for long way
loop statement {
loop to check for parentheses {
set operation to be done first
}
if symbol {
loop for symbol check {
perform operations
}}} // ending this since it looks like a bad way to do it...
// Better idea, originally thought up today (5/11/15)
int val[];
int re = 1;
loop {
if (list[i].containsIgnoreCase(d)) {
val[]=list[i].splitIgnoreCase("d");
list[i] = 0;
while (re <= val[0]) {
list[i] = list[i] + (rand.nextInt(val[1]) + 1);
re++;
}
}
}
// then create a string out of list[]/numbers[] and put together with
// symbols[] and use Java's evaluator for the String
wenton had it, it just seemed like it wasn't doing it for me (until I realised I wasn't specific on what I wanted) so basically to update, the string I want evaluated is (I know it's a little unorthodox, but it's to make a point; I also hope this clarifies even further of what is needed to make it work):
(3d12^d2-2)+d4(2*d4/d2)
From reading this, you may see the spots that I do not know how to perform very well... But that is why I am asking all you lovely, smart programmers out there! I hope I asked this clearly enough and thank you for your time :3
The trick with any programming problem is to break it up and write a method for each part, so below I have a method for rolling one dice, which is called by the one for rolling many.
private Random rand = new Random();
/**
* #param roll can be a multipart roll which is run and added up. e.g. d6+2-d4
*/
public int multiPartRoll(String roll) {
String[] parts = roll.split("(?=[+-])"); //split by +-, keeping them
int total = 0;
for (String partOfRoll : parts) { //roll each dice specified
total += singleRoll(partOfRoll);
}
return total;
}
/**
* #param roll can be fixed value, examples -1, +2, 15 or a dice to roll
* d6, +d20 -d100
*/
public int singleRoll(String roll) {
int di = roll.indexOf('d');
if (di == -1) //case where has no 'd'
return Integer.parseInt(roll);
int diceSize = Integer.parseInt(roll.substring(di + 1)); //value of string after 'd'
int result = rand.nextInt(diceSize) + 1; //roll the dice
if (roll.startsWith("-")) //negate if nessasary
result = -result;
return result;
}

Print out Yijing Hexagram Symbols

I encountered a problem while coding and I can't seem to find where I messed up or even why I get a wrong result.
First, let me explain the task.
It's about "Yijing Hexagram Symbols".
The left one is the original and the right one is the result that my code should give me.
Basically every "hexagram" contains 6 lines that can be either diveded or not.
So there are a total of
2^6 = 64 possible "hexagrams"
The task is to calculate and code a methode to print all possible combinations.
Thats what I have so far :
public class test {
public String toBin (int zahl) {
if(zahl ==0) return "0";
if (zahl ==1 ) return "1";
return ""+(toBin( zahl/2)+(zahl%2));
}
public void show (String s) {
for (char c : s.toCharArray()){
if (c == '1'){
System.out.println("--- ---");
}
if(c=='0'){
System.out.println("-------");
}
}
}
public void ausgeben (){
for(int i = 0 ; i < 64; i++) {
show (toBin(i));
}
}
}
The problem is, when I test the 'show'-methode with "10" I get 3 lines and not 2 as intended.
public class runner {
public static void main(String[] args){
test a = new test();
a.ausgeben();
a.show("10");
}
}
Another problem I've encoutered is, that since I'm converting to binary i sometimes have not enough lines because for example 10 in binary is 0001010 but the first "0" are missing. How can I implement them in an easy way without changing much ?
I am fairly new to all this so if I didn't explain anything enough or made any mistakes feel free to tell me.
You may find it easier if you use the Integer.toBinaryString method combined with the String.format and String.replace methods.
String binary = String.format("%6s", Integer.toBinaryString(zahl)).replace(' ', '0');
This converts the number to binary, formats it in a field six spaces wide (with leading spaces as necessary), and then replaces the spaces with '0'.
Well, there are many ways to pad a string with zeros, or create a binary string that is already padded with zeros.
For example, you could do something like:
public String padToSix( String binStr ) {
return "000000".substring( 0, 5 - binStr.length() ) + binStr;
}
This would check how long your string is, and take as many zeros are needed to fill it up to six from the "000000" string.
Or you could simply replace your conversion method (which is recursive, and that's not really necessary) with one that specializes in six-digit numbers:
public static String toBin (int zahl) {
char[] digits = { '0','0','0','0','0','0' };
int currDigitIndex = 5;
while ( currDigitIndex >= 0 && zahl > 0 ) {
digits[currDigitIndex] += (zahl % 2);
currDigitIndex--;
zahl /= 2;
}
return new String(digits);
}
This one modifies the character array ( which initially has only zeros ) from the right to the left. It adds the value of the current bit to the character at the given place. '0' + 0 is '0', and '0' + 1 is '1'. Because you know in advance that you have six digits, you can start from the right and go to the left. If your number has only four digits, well, the two digits we haven't touched will be '0' because that's how the character array was initialized.
There are really a lot of methods to achieve the same thing.
Your problem reduces to printing all binary strings of length 6. I would go with this code snippet:
String format = "%06d";
for(int i = 0; i < 64; i++)
{
show(String.format(format, Integer.valueOf(Integer.toBinaryString(i))));
System.out.println();
}
If you don't wish to print leading zeros, replace String.format(..) with Integer.toBinaryString(i).

Categories

Resources