Develop Reference

Getting the inverse of a function that uses summation in Java

I have a program with one class, which looks like this.
public class Functions {
public static void main(String[] args) {
System.out.println(summationFunction(1)); //Prints 13
System.out.println(summationFunction(2)); //Prints 29
System.out.println(summationFunction(3)); //Prints 48
System.out.println(summationFunction(4)); //Prints 70
}
public static int summationFunction(int input) {
int summedNumber = 0;
int i = input;
while (i > 0) {
summedNumber += i * 3;
i--;
}
return 10 * input + (summedNumber);
}
}
So, this program will take in a given number and apply this function to it:
And this all works well (I have run the class Functions and everything prints just as it's supposed to.) BUT, I need to find the inverse of this function, and I need to be able to translate it to code; I do not know how to do this.
I basically need a function that will return values like this:
public static void main(String[] args) {
System.out.println(summationFunction(13)); //Prints 1
System.out.println(summationFunction(29)); //Prints 2
System.out.println(summationFunction(48)); //Prints 3
System.out.println(summationFunction(70)); //Prints 4
}
which, (as you can tell) is the opposite of the original function.
So to sum everything up, I need a function that will return the inverse of my original function (summationFunction), and I would like to know how I would model this or if there is a quick solution, in code.
One more thing: I know that I can have the method take an input and search for the most similar output of the original method, but I would like to see if there is a simpler way to do this which does not involve searching, thus giving a quicker output speed. And if you wish you can safely assume that the input of the inversed function will always be a number which will give an integer output, like 13, 29, 48, 70, etc...
By the way, if you are going to downvote the question, will you at least give a reason somewhere? The comments perhaps? I can not see any reason that this question is eligible for being downvoted, and a reason would help.

Wolfram Alpha to the rescue !
It tells you that this function can be written as :
1/24*(6*x+23)^2-529/24
So if you want to solve f(x)=a, you have :
x = 1/6*(sqrt(24*a+529)-23)
a = 70
# => x = 4
Note : Using Wolfram shouldn't prevent you from finding the answer on your own.
sum(something*i) is equal to something*sum(i) because something (3 in this case ) doesn't depend on i.
sum(i,i=1..n) is equal to n*(n+1)/2, and it's easy to prove (see Wikipedia)
So your function becomes 10*x+3*x*(x+1)/2
Expanded, it is :
(3 x^2)/2+(23 x)/2
You need to solve (3 x^2)/2+(23 x)/2 = 70, in other words :
(3 x^2)/2+(23 x)/2 - 70 = 0
It is a quadratic equation, with a=3/2, b=23/2 and c=-70 or c=-29 or c=....

You sum can be written like this 3*x*(x+1)/2 so you have equation 10*x + 3*x*(x+1)/2 = y you need to solve it.
Wolfram alpha tells that result will be 1/6.0 * (-23.0+sqrt(529.0+24.0 * y))

Half String every second char

I an coding beginner.I have started practicing SPOJ basic problems.This was the one I was trying to solve , But the code is incorrect.
Please help me where I have coded this question wrong as I am unable to figure out:
public class Print2ndChar {
public static void main(String[] args) throws java.lang.Exception {
Print2ndChar mainObj = new Print2ndChar();
java.io.BufferedReader inputReader = new java.io.BufferedReader(new java.io.InputStreamReader(System.in));
String noOfTestCase;
if(((noOfTestCase = inputReader.readLine()) == null))
System.exit(0);
int noOfLines = 0;
try{
noOfLines = Integer.parseInt(noOfTestCase);
}catch(Exception e){
System.exit(0);
}
if(noOfLines<0 || noOfLines>100)
System.exit(0);
String [] randomWords = new String[noOfLines];
for(int i=0;i<noOfLines;i++){
randomWords[i] = inputReader.readLine();
if(randomWords[i] == null || randomWords[i].length()<2 || randomWords[i].length()%2!=0 || (randomWords[i].length()/2)>100)
System.exit(0);
}
for (String word : randomWords){
mainObj.letsBegin(word.substring(0, word.length() / 2));
System.out.println();
}
}
private void letsBegin(String data) {
if (data.length() <= 0) {
return;
} else {
System.out.print(data.charAt(0));
if (data.length() >= 3)
letsBegin(data.substring(2, data.length()));
}
}
}
EDIT :
I/P : 4
your
progress
is
noticeable
O/P
y
po
i
ntc

OK! So after a lot of hit and trials, I know what is wrong with your code. The code that you have written fails because of the condition randomWords[i].length()%2!=0 inside your if. There is nothing wrong with you putting this condition to check the input, but if you will select sample test case, inside the highlighted blue area you will notice an extra space after every string. Like this :
You can see that other than the last input all other input strings have a space character at the end. So, when you read the string from stdin the length of the string is 2*k + 1 (because of the space), and your program will exit without any output. Hence you get a wrong answer.
This problem exists with other test cases as well probably. And how do I know this? After spoj shows you wrong answer, if you click on the wrong answer, it will show you 2 failed test cases, something like this:
It shows your program's output is empty because your code exited because of the extra space at the end of strings.
So, I believe the person who wrote the test cases should be given a WT Error (Wrong Test Cases) :P :D
So, the possible correction is you remove the mentioned condition from the if and you will get AC. Because now you will be dividing 2*k + 1 by 2, which will not be an integer and which will get rounded to the nearest smallest integer, which will be same as dividing 2*k by 2 and the program will give the correct result.
A few things that you should take care while solving questions on spoj, you do not have to verify that every input lies within the range specified in the question, or if it is a valid data type. The range is given to tell you that Spoj will only test your program with cases which lie between those ranges and will not exceed them. So, even if you remove all the code where you check for exceptions and ranges of input data, you will get an AC. Moreover, writing such code only adds to the burden.
Hope this helps. :)

Java recursion class variable value is reset to 0

I was trying to implement the coin change problem using recursion. I have written the following code and am facing a problem with the static class variable. 'answer' is a class variable and i am trying to add the return value to it in the loop. This works fine within the while loop but after the while loop ends the answer is reset to 0;
while (i * currentCoin <= sum) {
System.out.println("inside while; answer is " + answer);
answer = answer
+ findCombinations(
sum - i * currentCoin,
new ArrayList<Integer>(denominations.subList(1,
denominations.size())));
i++;
}
Below is all the code that I have written. You can copy and run it to check.
import java.util.ArrayList;
import java.util.Collections;
public class CoinChangeHashMap {
static int answer = 0;
public static void main(String[] args) {
int[] array = new int[] { 7, 3, 2 };
ArrayList<Integer> input = new ArrayList<Integer>();
getList(array, input);
findCombinations(12, input);
System.out.println(answer);
}
private static void getList(int[] array, ArrayList<Integer> input) {
for (int i : array) {
input.add(i);
}
}
public static int findCombinations(int sum, ArrayList<Integer> denominations) {
if (denominations.size() == 1) {
if (sum % denominations.get(0) == 0) {
return 1;
}
return 0;
}
int i = 0;
int currentCoin = denominations.get(0);
while (i * currentCoin <= sum) {
System.out.println("inside while; answer is " + answer);
answer = answer
+ findCombinations(
sum - i * currentCoin,
new ArrayList<Integer>(denominations.subList(1,
denominations.size())));
i++;
}
return 0;
}}
**The output that I get is 0. but the expected output is 4. While debugging the output that I got is **
inside while; answer is 0
inside while; answer is 0
inside while; answer is 1
inside while; answer is 1
inside while; answer is 2
inside while; answer is 2
inside while; answer is 0
inside while; answer is 0
inside while; answer is 0
0
Any Help is appreciated.

The problem is related to your odd code structure, in which you convey the outcome of your recursive call sometimes by modifying static variable answer, and sometimes via the method's return value.
If you analyzed the problem more closely, you would discover that it is not upon exit from the loop that the partial results are lost, but rather some time after return from the method. Therefore, consider carefully the way you update the answer:
answer = answer + findCombinations( /* ... */ );
At the top-most level of your recursion, answer is initially 0. When Java evaluates the above expression, it evaluates first the left operand and then the right operand, then it adds them. That is, it evaluates answer, getting the result 0, before it performs the recursive call. The value of answer may be updated in the course of the recursive call, but those changes come too late. Only the bottom-most level of the recursion ever returns a value different from zero, so if the recursive call itself recurses at least one level deeper then it will return zero. In that case, the sum is computed as 0 + 0, and assigned to answer, clobbering any update the method performed.
You could resolve the problem by swapping the order of the operands in your sum, but it would be better, and not much harder, to get rid of the static variable altogether. Use a local variable within the method to accumulate results, and in all cases convey the total back to the caller via the method's return value.

Working of Return statement in a method in Java

I am trying to understand the working of return statement in JAVA.
My doubt is if inside a method with a Non void return type, I have a decision block which also has a return statement of its own, Still I have to return some value .
For understanding here is a sample code I have written :-
public int bunnyEars(int bunnies) {
//int count=0;
if (bunnies >=1) {
count = count + 2;
bunnyEars(bunnies -1);
return count1;
}
return count2 ;
}
In the mentioned code I just want to return the no. of bunnies which I am being able to do from inside the bunnyEars method count1. But still JAVA wont allow to have a non-void method without a return type which is totally understood and I have to add count2 return also. Now I am suspecting that I am having a conceptual understanding failure here. Kindly let me know if I am missing something? Kindly let me know If I am missing some more info here.
[Edited] Full code:
public class Test5 {
//public int ears=1;
public int count=0;
public int bunnyEars(int bunnies) {
//int count=0;
if (bunnies >=1) {
count = count + 2;
bunnyEars(bunnies -1);
return count;
}
return count ;
}
public static void main(String args[]){
Test5 test5= new Test5();
System.out.println(test5.bunnyEars(90));
}
}

Yes you need to return count2 which should be zero. Which means if there are no bunnies then there are no ears. So which returning you should be returning some value irrespective of the conditional block.
So in this case
return count1;
represents the number of ears if the bunnies are represent, while
return count2;
represents the number of ears when there are no bunnies, which should be 0.
I hope that gives you some clarification

I think your conceptual misunderstanding lies with understanding the flow of the program.
Supposed you were to use this method by calling:
bunnyEars(2)
Then, once you enter the method, the first thing the program does is check if 3 >= 1. Since this is true, you proceed into the code inside the {..} (called a 'block'). Inside this block, you increment count by 2. I am assuming count is defined elsewhere in the class, but suppose the current value for count is 10. Then, the new value of count will be 12.
After this, the program executes the line:
bunnyEars(bunnies - 1)
Which translates to:
bunnyEars(1)
Now, basically, you are calling the same method again, but passing in 1 instead of 2.
Once again, the program checks to see that 1 >= 1, which is true. So it goes into the
if-block which, again, increments count by 2. So now, count = 14. Then it calls the
same method again but this time passing in 0.
bunnyEars(0)
Since 0 >= 1 evaluates to false, you the program skips the if-block and continues
execution after the block. So know, you are in the method bunnyEars(), but you have
completely skipped over your "return" statement. But, alas, bunnyEars MUST return an int.
So this is why you must have a return after the block. In your case, bunnyEars(0) returns count2 and the program-execution returns to where you called bunnyEars(0).
Read up on recursive calls. The basic idea of a recursive method is that, inside the recursive method, you must have some case that terminates the recursion (otherwise you will loop forever).
For example, the following code will go on forever:
public int sum(int in)
{
return in + sum(in - 1);
}
This will keep going on forever, because sum(1) will call sum(0) which calls sum(-1).
So, I must have a condition that terminates the recursion:
public int sum(int in)
{
if(in == 0) return 0;
return in + sum(in - 1);
}
Now, I have a terminating-case. So if I call sum(1), it will call sum(0) which returns 0. So my result is 1 + 0 = 1.
Similarily,
sum(2) = 2 + sum(1) = 2 + 1 + sum(0) = 2 + 1 + 0
sum(3) = 3 + sum(2) = 3 + 2 + sum(1) = 3 + 2 + 1 + sum(0) = 3 + 2 + 1 + 0 = 6
Hope this helps!

So as I understand it, your question is why you still need to return count2 if you return count1. The answer is basically 'what happens if you don't enter the if block?'. In that case, without return count2, you wouldn't have a return value, which is what Java is complaining about. If you really don't want two return statements, you could probably do something like:
public int bunnyEars(int bunnies) {
int count=0;
if (bunnies >=1) {
count = count + 2;
bunnyEars(bunnies -1);
}
return count ;
}
On a side note, this and the code you posted in your question won't work for regression purposes, but the one in your comment does, and there it looks like you have a static variable for count, in which case you could set the return type to void and just print count.

Questions over BigInteger?

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.