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Java beginner: Dividing input value in half using a loop but cannot use arrays, built-in sorting routines or any other Java Collections classes

Java Beginner: I have most of the code complete to solve the problem below but having trouble with my loop section since it is currently dividing one value. It should continue dividing until it reaches 1. I'm not sure what's wrong so any assistance is greatly appreciated!! However, I cannot use arrays, built-in sorting routines or any other Java Collections classes
Problem:
Write a program that will prompt the user for a positive integer: N. The program will repeatedly divide the input in half using a loop, discarding any fractional part, until it becomes 1. The program should print on separate lines:
the sequence of 'halved' values, one per line
the number of iterations required
the value of log2(N)
My code and output when entering a value of 9:
import stdlib.StdIn;
import stdlib.StdOut;
public class DS1hw1b {
public static void main(String[] args) {
int countIteration = 0;
StdOut.println("enter a positive number: ");
int N = StdIn.readInt();
for (int i = 1; i <= 1; i++) {
countIteration++;
if ((N/2) != 1)
StdOut.println(N/2);
StdOut.println("number of iterations: " + countIteration);
//compute log formula
StdOut.println("log2 of input: " + (Math.log(N)/Math.log(2)));
}
}
}
Output:
enter a positive number:
9
4
number of iterations: 1
log2 of input: 3.1699250014423126
However, I should see is 9, 4, 2, and 1 on separate lines and iteration of 3.

Your loop will only loop once - you initialize i with 1, which adheres to the condition i<=1, and stop adhering to it once i is incremented.
You're asked to divide N by 2 until it becomes 1 - this calls for a while loop:
int counter = 0;
while (n > 1) {
System.out.println(n);
counter++;
n /= 2;
}
System.out.println("Number of iterations: " + counter);

Per my comment to #Murelink. Here is the updated code but now the log2 formula shows as 0.0 for any number entered. I thought it would be best to show it this way. Anyway, I don't understand why it doesn't work since I didn't modify the formula. Thank you all for your help!
public static void main(String[] args) {
int countIteration = 0;
StdOut.println("enter a positive number: ");
int N = StdIn.readInt();
while (N > 1) {
N /= 2;
countIteration++;
StdOut.println(N);
}
StdOut.println("number of iterations: " + countIteration);
//compute log formula
StdOut.println("log2 of input: " + (Math.log(N)/Math.log(2)));
}
}

I need to print the 215th Lucas Numbers using an efficient algorithm. Using recursion takes way too long.

The purpose of this class is to calculate the nth number of the Lucas Sequence. I am using data type long because the problems wants me to print the 215th number. The result of the 215th number in the Lucas Sequence is: 855741617674166096212819925691459689505708239. The problem I am getting is that at some points, the result is negative. I do not understand why I am getting a negative number when the calculation is always adding positive numbers. I also have two methods, since the question was to create an efficient algorithm. One of the methods uses recursion but the efficiency is O(2^n) and that is of no use to me when trying to get the 215th number. The other method is using a for loop, which the efficiency is significantly better. If someone can please help me find where the error is, I am not sure if it has anything to do with the data type or if it is something else.
Note: When trying to get the 91st number I get a negative number and when trying to get the 215th number I also get a negative number.
import java.util.Scanner;
public class Problem_3
{
static long lucasNum;
static long firstBefore;
static long secondBefore;
static void findLucasNumber(long n)
{
if(n == 0)
{
lucasNum = 2;
}
if(n == 1)
{
lucasNum = 1;
}
if(n > 1)
{
firstBefore = 1;
secondBefore = 2;
for(int i = 1; i < n; i++)
{
lucasNum = firstBefore + secondBefore;
secondBefore = firstBefore;
firstBefore = lucasNum;
}
}
}
static long recursiveLucasNumber(int n)
{
if(n == 0)
{
return 2;
}
if(n == 1)
{
return 1;
}
return recursiveLucasNumber(n - 1) + recursiveLucasNumber(n - 2);
}
public static void main(String[] args)
{
System.out.println("Which number would you like to know from "
+ "the Lucas Sequence?");
Scanner scan = new Scanner(System.in);
long num = scan.nextInt();
findLucasNumber(num);
System.out.println(lucasNum);
//System.out.println(recursiveLucasNumber(num));
}
}

Two observations:
The answer you are expecting (855741617674166096212819925691459689505708239) is way larger than you can represent using a long. So (obviously) if you attempt to calculate it using long arithmetic you are going to get integer overflow ... and a garbage answer.
Note: this observation applies for any algorithm in which you use a Java integer primitive value to represent the Lucas numbers. You would run into the same errors with recursion ... eventually.
Solution: use BigInteger.
You have implemented iterative and pure recursion approaches. There is a third approach: recursion with memoization. If you apply memorization correctly to the recursive solution, you can calculate LN in O(N) arithmetical operations.

Java data type long can contain only 64-bit numbers in range -9223372036854775808 .. 9223372036854775807. Negative numbers arise due to overflow.
Seems you need BigInteger class for arbitrary-precision integer numbers

I wasn't aware of the lucas numbers before this thread, but from wikipedia it looks like they are related to the fibonacci sequence with (n = nth number, F = fibonacci, L = lucas):
Ln = F_(n-1) + F_(n+1)
Thus, if your algorithm is too slow, you could use the closed form fibonacci and than compute the lucas number from it, alternative you could also use the closed form given in the wikipedia article directly (see https://en.wikipedia.org/wiki/Lucas_number).
Example code:
public static void main(String[] args) {
long n = 4;
double fibo = computeFibo(n);
double fiboAfter = computeFibo(n + 1);
double fiboBefore = computeFibo(n - 1);
System.out.println("fibonacci n:" + Math.round(fibo));
System.out.println("fibonacci: n+1:" + Math.round(fiboAfter));
System.out.println("fibonacci: n-1:" + Math.round(fiboBefore));
System.out.println("lucas:" + (Math.round(fiboAfter) + Math.round(fiboBefore)));
}
private static double computeFibo(long n) {
double phi = (1 + Math.sqrt(5)) / 2.0;
double psi = -1.0 / phi;
return (Math.pow(phi, n) - Math.pow(psi, n)) / Math.sqrt(5);
}
To work around the long size limit you could use java BigDecimal (https://docs.oracle.com/javase/7/docs/api/java/math/BigDecimal.html). This is needed earlier in this approach as the powers in the formula will grow very quickly.

Checking whether a number is in Fibonacci Sequence?

It was asked to find a way to check whether a number is in the Fibonacci Sequence or not.
The constraints are
1≤T≤10^5
1≤N≤10^10
where the T is the number of test cases,
and N is the given number, the Fibonacci candidate to be tested.
I wrote it the following using the fact a number is Fibonacci if and only if one or both of (5*n2 + 4) or (5*n2 – 4) is a perfect square :-
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
for(int i = 0 ; i < n; i++){
int cand = sc.nextInt();
if(cand < 0){System.out.println("IsNotFibo"); return; }
int aTest =(5 * (cand *cand)) + 4;
int bTest = (5 * (cand *cand)) - 4;
int sqrt1 = (int)Math.sqrt(aTest);// Taking square root of aTest, taking into account only the integer part.
int sqrt2 = (int)Math.sqrt(bTest);// Taking square root of bTest, taking into account only the integer part.
if((sqrt1 * sqrt1 == aTest)||(sqrt2 * sqrt2 == bTest)){
System.out.println("IsFibo");
}else{
System.out.println("IsNotFibo");
}
}
}
}
But its not clearing all the test cases? What bug fixes I can do ?

A much simpler solution is based on the fact that there are only 49 Fibonacci numbers below 10^10.
Precompute them and store them in an array or hash table for existency checks.
The runtime complexity will be O(log N + T):
Set<Long> nums = new HashSet<>();
long a = 1, b = 2;
while (a <= 10000000000L) {
nums.add(a);
long c = a + b;
a = b;
b = c;
}
// then for each query, use nums.contains() to check for Fibonacci-ness
If you want to go down the perfect square route, you might want to use arbitrary-precision arithmetics:
// find ceil(sqrt(n)) in O(log n) steps
BigInteger ceilSqrt(BigInteger n) {
// use binary search to find smallest x with x^2 >= n
BigInteger lo = BigInteger.valueOf(1),
hi = BigInteger.valueOf(n);
while (lo.compareTo(hi) < 0) {
BigInteger mid = lo.add(hi).divide(2);
if (mid.multiply(mid).compareTo(x) >= 0)
hi = mid;
else
lo = mid.add(BigInteger.ONE);
}
return lo;
}
// checks if n is a perfect square
boolean isPerfectSquare(BigInteger n) {
BigInteger x = ceilSqrt(n);
return x.multiply(x).equals(n);
}

Your tests for perfect squares involve floating point calculations. That is liable to give you incorrect answers because floating point calculations typically give you inaccurate results. (Floating point is at best an approximate to Real numbers.)
In this case sqrt(n*n) might give you n - epsilon for some small epsilon and (int) sqrt(n*n) would then be n - 1 instead of the expected n.
Restructure your code so that the tests are performed using integer arithmetic. But note that N < 1010 means that N2 < 1020. That is bigger than a long ... so you will need to use ...
UPDATE
There is more to it than this. First, Math.sqrt(double) is guaranteed to give you a double result that is rounded to the closest double value to the true square root. So you might think we are in the clear (as it were).
But the problem is that N multiplied by N has up to 20 significant digits ... which is more than can be represented when you widen the number to a double in order to make the sqrt call. (A double has 15.95 decimal digits of precision, according to Wikipedia.)
On top of that, the code as written does this:
int cand = sc.nextInt();
int aTest = (5 * (cand * cand)) + 4;
For large values of cand, that is liable to overflow. And it will even overflow if you use long instead of int ... given that the cand values may be up to 10^10. (A long can represent numbers up to +9,223,372,036,854,775,807 ... which is less than 1020.) And then we have to multiply N2 by 5.
In summary, while the code should work for small candidates, for really large ones it could either break when you attempt to read the candidate (as an int) or it could give the wrong answer due to integer overflow (as a long).
Fixing this requires a significant rethink. (Or deeper analysis than I have done to show that the computational hazards don't result in an incorrect answer for any large N in the range of possible inputs.)

According to this link a number is Fibonacci if and only if one or both of (5*n2 + 4) or (5*n2 – 4) is a perfect square so you can basically do this check.
Hope this helps :)

Use binary search and the Fibonacci Q-matrix for a O((log n)^2) solution per test case if you use exponentiation by squaring.
Your solution does not work because it involves rounding floating point square roots of large numbers (potentially large enough not to even fit in a long), which sometimes will not be exact.
The binary search will work like this: find Q^m: if the m-th Fibonacci number is larger than yours, set right = m, if it is equal return true, else set left = m + 1.

As it was correctly said, sqrt could be rounded down. So:
Even if you use long instead of int, it has 18 digits.
even if you use Math.round(), not simply (int) or (long). Notice, your function wouldn't work correctly even on small numbers because of that.
double have 14 digits, long has 18, so you can't work with squares, you need 20 digits.
BigInteger and BigDecimal have no sqrt() function.
So, you have three ways:
write your own sqrt for BigInteger.
check all numbers around the found unprecise double sqrt() for being a real sqrt. That means also working with numbers and their errors simultaneously. (it's horror!)
count all Fibonacci numbers under 10^10 and compare against them.
The last variant is by far the simplest one.

Looks like to me the for-loop doesn't make any sense ?
When you remove the for-loop for me the program works as advertised:
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int cand = sc.nextInt();
if(cand < 0){System.out.println("IsNotFibo"); return; }
int aTest = 5 * cand *cand + 4;
int bTest = 5 * cand *cand - 4;
int sqrt1 = (int)Math.sqrt(aTest);
int sqrt2 = (int)Math.sqrt(bTest);
if((sqrt1 * sqrt1 == aTest)||(sqrt2 * sqrt2 == bTest)){
System.out.println("IsFibo");
}else{
System.out.println("IsNotFibo");
}
}
}

You only need to test for a given candidate, yes? What is the for loop accomplishing? Could the results of the loop be throwing your testing program off?
Also, there is a missing } in the code. It will not run as posted without adding another } at the end, after which it runs fine for the following input:
10 1 2 3 4 5 6 7 8 9 10
IsFibo
IsFibo
IsFibo
IsNotFibo
IsFibo
IsNotFibo
IsNotFibo
IsFibo
IsNotFibo
IsNotFibo

Taking into account all the above suggestions I wrote the following which passed all test cases
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long[] fib = new long[52];
Set<Long> fibSet = new HashSet<>(52);
fib[0] = 0L;
fib[1] = 1L;
for(int i = 2; i < 52; i++){
fib[i] = fib[i-1] + fib[i - 2];
fibSet.add(fib[i]);
}
int n = sc.nextInt();
long cand;
for(int i = 0; i < n; i++){
cand = sc.nextLong();
if(cand < 0){System.out.println("IsNotFibo");continue;}
if(fibSet.contains(cand)){
System.out.println("IsFibo");
}else{
System.out.println("IsNotFibo");
}
}
}
}
I wanted to be on the safer side hence I choose 52 as the number of elements in the Fibonacci sequence under consideration.

Factorial loop results are incorrect after the 5th iteration

I am currently taking pre-calculus and thought that I would make a quick program that would give me the results of factorial 10. While testing it I noticed that I was getting incorrect results after the 5th iteration. However, the first 4 iterations are correct.
public class Factorial
{
public static void main(String[] args)
{
int x = 1;
int factorial;
for(int n = 10; n!=1; n--)
{
factorial = n*(n-1);
x = x * factorial;
System.out.printf("%d ", x);
}
}//end of class main
}//end of class factorial

That is an Integer Overflow issue. Use long or unsigned long instead of int. (And as #Dunes suggested, your best bet is really BigInteger when working with very large numbers, because it will never overflow, theoretically)
The basic idea is that signed int stores numbers between -2,147,483,648 to 2,147,483,647, which are stored as binary bits (all information in a computer are stored as 1's and 0's)
Positive numbers are stored with 0 in the most significant bit, and negative numbers are stored with 1 in the most significant bit. If your positive number gets too big in binary representation, digits will carry over to the signed bit and turn your positive number into the binary representation of a negative one.
Then when the factorial gets bigger than even what an unsigned int can store, it will "wrap around" and lose the carry-over from its most significant (signed) bit - that's why you are seeing the pattern of sometimes alternating positive and negative values in your output.

You're surpassing the capacity of the int type (2,147,483,647), so your result is wrapping back around to the minimum int value. Try using long instead.
Having said the that, the method you are currently employing will not result in the correct answer: actually, you are currently computing 10! ^ 2.
Why complicate things? You could easily do something like this:
long x = 1L;
for(int n = 1; n < 10; n++)
{
x *= n;
System.out.println(x);
}
1
2
6
24
120
720
5040
40320
362880
which shows successive factorials until 10! is reached.
Also, as others have mentioned, if you need values bigger than what long can support you should use BigInteger, which supports arbitrary precision.

Your formula for the factorial is incorrect. What you will have is this:
Step 1 : n*(n-1) = 10 * 9 = 90 => x = 1*90 = 90
Step 2 : n*(n-1) = 9 * 8 = 72 => x = 90*72 = 6480 or, it should be : 10 * 9 * 8 => 720
But the wrong results are coming from the fact that you reached the maximum value for the type int as pointed out by others
Your code should be
public class Factorial
{
public static void main(String[] args)
{
double factorial = 1;
for(int n = factorial; n>=1; n--)
{
factorial = factorial * n;
System.out.printf("%d ", factorial );
}
}
}

In addition to what the other answers mention about the overflow, your factorial algorithm is also incorrect. 10! should calculate 10*9*8*7*6*5*4*3*2*1, you are doing (10*9)*(9*8)*(8*7)*(7*6)*...
Try changing your loop to the following:
int x = 1;
for(int n = 10; n > 1 ; n--)
{
x = x * n;
System.out.printf("%d ", x);
}
You will eventually overflow if you try to calculate the factorial of higher numbers, but int is plenty large enough to calculate the factorial of 10.

Bounding this program to determine the sum of reciprocal integers not containing zero

Let A denote the set of positive integers whose decimal representation does not contain the digit 0. The sum of the reciprocals of the elements in A is known to be 23.10345.
Ex. 1,2,3,4,5,6,7,8,9,11-19,21-29,31-39,41-49,51-59,61-69,71-79,81-89,91-99,111-119, ...
Then take the reciprocal of each number, and sum the total.
How can this be verified numerically?
Write a computer program to verify this number.
Here is what I have written so far, I need help bounding this problem as this currently takes too long to complete:
Code in Java
import java.util.*;
public class recip
{
public static void main(String[] args)
{
int current = 0; double total = 0;
while(total < 23.10245)
{
if(Integer.toString(current).contains("0"))
{
current++;
}
else
{
total = total + (1/(double)current);
current++;
}
System.out.println("Total: " + total);
}
}
}

This is not that hard when approached properly.
Assume for example that you want to find the sum of reciprocals of all integers starting (i.e. the left-most digits) with 123 and ending with k non-zero digits. Obviously there are 9k such integers and the reciprocal of each of these integers is in the range 1/(124*10k) .. 1/(123*10k). Hence the sum of reciprocals of all these integers is bounded by (9/10)k/124 and (9/10)k/123.
To find bounds for sum of all reciprocals starting with 123 one has to add up the bounds above for every k>=0. This is a geometric serie, hence it can be derived that the sum of reciprocals of integers starting with 123 is bounded by 10*(9/10)k/124 and 10*(9/10)k/123.
The same method can of course be applied for any combination of left-most digits.
The more digits we examine on the left, the more accurate the result becomes.
Here is an implementation of this approach in python:
def approx(t,k):
"""Returns a lower bound and an upper bound on the sum of reciprocals of
positive integers starting with t not containing 0 in its decimal
representation.
k is the recursion depth of the search, i.e. we append k more digits
to t, before approximating the sum. A larger k gives more accurate
results, but takes longer."""
if k == 0:
return 10.0/(t+1), 10.0/t
else:
if t > 0:
low, up = 1.0/t, 1.0/t
else:
low, up = 0, 0
for i in range(10*t+1, 10*t+10):
l,u = approx(i, k-1)
low += l
up += u
return low, up
Calling approx(0, 8) for example gives the lower and upper bound:
23.103447707... and 23.103448107....
which is close to the claim 23.10345 given by the OP.
There are methods that converge faster to the sum in question, but they require more math.
A much better approximation of the sum can be found here. A generalization of the problem are the Kempner series.

For all values of current greater than some threshold N, 1.0/(double)current will be sufficiently small that total does not increase as a result of adding 1.0/(double)current. Thus, the termination criterion should be something like
while(total != total + (1.0/(double)current))
instead of testing against the limit that is known a priori. Your loop will stop when current reaches this special value of N.

I suspect that casting to string and then checking for the character '0' is the step that takes too long. If you want to avoid all zeroes, might help to increase current thus:
(Edited -- thanks to Aaron McSmooth)
current++;
for( int i = 10000000; i >= 10; i = i / 10 )
{
if ( current % i ) == 0
{
current = current + ( i / 10 );
}
}
This is untested, but the concept should be clear: whenever you hit a multiple of a power of ten (e.g. 300 or 20000), you add the next lower power of 10 (in our examples 10 + 1 and 1000 + 100 + 10 + 1, respectively) until there are no more zeroes in your number.
Change your while loop accordingly and see if this doesn't help performance to the point were your problem becomes manageable.
Oh, and you might want to restrict the System.out output a bit as well. Would every tenth, one hundreth or 10000th iteration be enough?
Edit the second:
After some sleep, I suspect my answer might be a little short-sighted (blame the late hour, if you will). I simply hoped that, oh, one million iterations of current would get you to the solution and left it at that, instead of calculating the correction cases using log( current ) etc.
On second thought, I see two problems with this whole problem. One is that your target number of 23.10345 is a leeeeettle to round for my tastes. After all, you are adding thousands of items like "1/17", "1/11111" and so on, with infinite decimal representations, and it is highly unlikely that they add up to exactly 23.10345. If some specialist for numerical mathematics says so, fine -- but then I'd like to see the algorithm by which they arrived at this conclusion.
The other problem is related to the first and concerns the limited in-memory binary representation of your rational numbers. You might get by using BigDecimals, but I have my doubts.
So, basically, I suggest you reprogram the numerical algorithm instead of going for the brute force solution. Sorry.
Edit the third:
Out of curiosity, I wrote this in C++ to test my theories. It's run for 6 minutes now and is at about 14.5 (roughly 550 mio. iterations). We'll see.
Current version is
double total = 0;
long long current = 0, currPowerCeiling = 10, iteration = 0;
while( total < 23.01245 )
{
current++;
iteration++;
if( current >= currPowerCeiling )
currPowerCeiling *= 10;
for( long long power = currPowerCeiling; power >= 10; power = power / 10 )
{
if( ( current % power ) == 0 )
{
current = current + ( power / 10 );
}
}
total += ( 1.0 / current );
if( ! ( iteration % 1000000 ) )
std::cout << iteration / 1000000 << " Mio iterations: " << current << "\t -> " << total << std::endl;
}
std::cout << current << "\t" << total << std::endl;
Calculating currPowerCeiling (or however one might call this) by hand saves some log10 and pow calculations each iteration. Every little bit helps -- but it still takes forever...
Edit the fourth:
Status is around 66,000 mio iterations, total is up to 16.2583, runtime is at around 13 hours. Not looking good, Bobby S. -- I suggest a more mathematical approach.

How about storing the current number as a byte array where each array element is a digit 0-9? That way, you can detect zeroes very quickly (comparing bytes using == instead of String.contains).
The downside would be that you'll need to implement the incrementing yourself instead of using ++. You'll also need to devise a way to mark "nonexistent" digits so that you don't detect them as zeroes. Storing -1 for nonexistent digits sounds like a reasonable solution.

For a signed 32-bit integer, this program will never stop. It will actually converge towards -2097156. Since the maximum harmonic number (the sum of integral reciprocals from 1 to N) of a signed 32-bit integer is ~14.66, this loop will never terminate, even when current wraps around from 2^31 - 1 to -2^31. Since the reciprocal of the largest negative 32-bit integer is ~-4.6566e-10, every time current returns to 0, the sum will be negative. Given that the largest number representable by a double such that number + + 1/2^31 == number is 2^52/2^31, you get roughly -2097156 as the converging value.
Having said that, and assuming you don't have a direct way of calculating the harmonic number of an arbitrary integer, there are a few things you can do to speed up your inner loop. First, the most expensive operation is going to be System.out.println; that has to interact with the console in which case your program will eventually have to flush the buffer to the console (if any). There are cases where that may not actually happen, but since you are using that for debugging they are not relevant to this question.
However, you also spend a lot of time determining whether a number has a zero. You can flip that test around to generate ranges of integers such that within that range you are guaranteed not to have an integer with a zero digit. That is really simple to do incrementally (in C++, but trivial enough to convert to Java):
class c_advance_to_next_non_zero_decimal
{
public:
c_advance_to_next_non_zero_decimal(): next(0), max_set_digit_index(0)
{
std::fill_n(digits, digit_count, 0);
return;
}
int advance_to_next_non_zero_decimal()
{
assert((next % 10) == 0);
int offset= 1;
digits[0]+= 1;
for (int digit_index= 1, digit_value= 10; digit_index<=max_set_digit_index; ++digit_index, digit_value*= 10)
{
if (digits[digit_index]==0)
{
digits[digit_index]= 1;
offset+= digit_value;
}
}
next+= offset;
return next;
}
int advance_to_next_zero_decimal()
{
assert((next % 10)!=0);
assert(digits[0]==(next % 10));
int offset= 10 - digits[0];
digits[0]+= offset;
assert(digits[0]==10);
// propagate carries forward
for (int digit_index= 0; digits[digit_index]==10 && digit_index<digit_count; ++digit_index)
{
digits[digit_index]= 0;
digits[digit_index + 1]+= 1;
max_set_digit_index= max(digit_index + 1, max_set_digit_index);
}
next+= offset;
return next;
}
private:
int next;
static const size_t digit_count= 10; // log10(2**31)
int max_set_digit_index;
int digits[digit_count];
};
What the code above does is to iterate over every range of numbers such that the range only contains numbers without zeroes. It works by determining how to go from N000... to N111... and from N111... to (N+1)000..., carrying (N+1) into 1(0)000... if necessary.
On my laptop, I can generate the harmonic number of 2^31 - 1 in 8.73226 seconds.

public class SumOfReciprocalWithoutZero {
public static void main(String[] args) {
int maxSize=Integer.MAX_VALUE/10;
long time=-System.currentTimeMillis();
BitSet b=new BitSet(maxSize);
setNumbersWithZeros(10,maxSize,b);
double sum=0.0;
for(int i=1;i<maxSize;i++)
{
if(!b.get(i))
{
sum+=1.0d/(double)i;
}
}
time+=System.currentTimeMillis();
System.out.println("Total: "+sum+"\nTimeTaken : "+time+" ms");
}
static void setNumbersWithZeros(int srt,int end,BitSet b)
{
for(int j=srt;j<end;j*=10)
{
for(int i=1;i<=10;i++)
{
int num=j*i;
b.set(num);
}
if(j>=100)
setInbetween(j, b);
}
}
static void setInbetween(int strt,BitSet b)
{
int bitToSet;
bitToSet=strt;
for(int i=1;i<=10;i++)
{
int nxtInt=-1;
while((nxtInt=b.nextSetBit(nxtInt+1))!=strt)
{
b.set(bitToSet+nxtInt);
}
nxtInt=-1;
int lim=strt/10;
while((nxtInt=b.nextClearBit(nxtInt+1))<lim)
{
b.set(bitToSet+nxtInt);
}
bitToSet=strt*i;
}
}
}
This is an implementation using BitSet.I calculated the sum of reciprocal's for all integer's in range (1-Integer.MAX_VALUE/10).The sum comes upto 13.722766931560747.This is the maximum I could calculate using BitSet since the maximum range for BitSet is Integer.MAX_VALUE.I need to divide it by 10 and limit the range to avoid overflow.But there is significant improvement in speed.I'm just posting this code in-case it might give you some new idea to improve your code.(Increase your memory using the VM argument -Xmx[Size>350]m)
Output:
Total: 13.722766931560747
TimeTaken : 60382 ms
UPDATE:
Java Porting of a previous , deleted answer :
public static void main(String[] args) {
long current =11;
double tot=1 + 1.0/2 + 1.0/3 + 1.0/4 + 1.0/5 + 1.0/6 + 1.0/7 + 1.0/8 + 1.0/9;
long i=0;
while(true)
{
current=next_current(current);
if(i%10000!=0)
System.out.println(i+" "+current+" "+tot);
for(int j=0;j<9;j++)
{
tot+=(1.0/current + 1.0/(current + 1) + 1.0/(current + 2) + 1.0/(current + 3) + 1.0/(current + 4) +
1.0/(current + 5) + 1.0/(current + 6) + 1.0/(current + 7) + 1.0/(current + 8));
current += 10;
}
i++;
}
}
static long next_current(long n){
long m=(long)Math.pow(10,(int)Math.log10(n));
boolean found_zero=false;
while(m>=1)
{
if(found_zero)
n+=m;
else if((n/m)%10==0)
{
n=n-(n%m)+m;
found_zero=true;
}
m=m/10;
}
return n;
}