I'm writing a program that is supposed to create chains of numbers and see witch one is the longest. The problem is that I run out of memory and I have no idea what eats all of the memory. Does anyone know were the problem is?
public static void main(String[] args) {
ArrayList<Integer> longestchain;
ArrayList<Integer> chain = new ArrayList<Integer>();
int size = 0;
int longestsize = 0;
int start;
int number = 0;
for(int i = 3; i < 1000000; i++)
{
start = i;
chain.clear();
chain.add(start);
size = 1;
while(true)
{
if(start == 1)
{
break;
}
if(iseven(start))
{
start = start / 2;
}
else
{
start = start*3 + 1;
}
chain.add(start);
size++;
}
if(size > longestsize)
{
longestsize = size;
longestchain = chain;
number = i;
}
//System.out.println(i);
}
System.out.println(number + ". " + longestsize);
}
public static boolean iseven(int n)
{
return (n % 2 == 0);
}
The problem is that when i=113383 you're running into an integer overflow. This results in an infinite loop that keeps adding to chain until you run out of heap space. Change chain and longestchain to ArrayList<Long>, start to long and iseven() to take long, and you'll solve that particular problem.
Another problem is that longestchain = chain assigns the reference, whereas you need to copy the contents. Otherwise the next iteration will wipe longestchain.
This will probably not be the solution, but a hint:
The default array size for an ArrayList is 10. Initialize your ArrayLists to a value closer to what you are expecting, or many array creations happen under the hood:
// e.g.
ArrayList<Integer> chain = new ArrayList<Integer>(50000);
Don't know why are you running out of memory, but I noticed a bug in your code - longestchain and chain point to the same object, so when you clear chain, you also clear longestchain. You can fix it by creating a new array with contents of chain instead of an assignment.
Related
I decided to reduce the number of comparisons required to find an element in an array. Here we replace the last element of the list with the search element itself and run a while loop to see if there exists any copy of the search element in the list and quit the loop as soon as we find the search element. See the code snippet for clarification.
import java.util.Random;
public class Search {
public static void main(String[] args) {
int n = 10000000;
int key = 10000;
int[] arr = generateRandomSize(n);
long start = System.nanoTime();
int find = sentinels(arr, key);
long end = System.nanoTime();
System.out.println(find);
System.out.println(end - start);
arr = generateRandomSize(n);
start = System.nanoTime();
find = linear(arr, key);
end = System.nanoTime();
System.out.println(find);
System.out.println(end - start);
}
public static int[] generateRandomSize(int n) {
int[] arr = new int[n];
Random rand = new Random();
for (int i = 0; i < n; ++i) {
arr[i] = rand.nextInt(5000);
}
return arr;
}
public static int linear(int[] a, int key) {
for(int i = 0; i < a.length; ++i) {
if (a[i] == key) {
return i;
}
}
return -1;
}
public static int sentinels(int[] a, int key) {
int n = a.length;
int last = a[n-1];
a[n-1] = key;
int i = 0;
while (a[i] != key) {
++i;
}
a[n-1] = last;
if ((i < n - 1) || a[n-1] == key ) {
return i;
}
return -1;
}
}
So using sentinel search we are not doing 10000000 comparisons like i < arr.length. But why linear search always shows up better performance?
You'd have to look at the byte code, and even deeper to see what hotspot is making from this. But I am quite sure that this statement is not true:
using sentinel search we are not doing 10000000 comparisons like i <
arr.length
Why? Because when you access a[i], i has to be bounds checked. In the linear case on the other hand, the optimiser can deduce that it can omit the bounds check since it "knows" that i>=0 (because of the loop structure) and also i<arr.length because it has already been tested in the loop condition.
So the sentinel approach just adds overhead.
This makes me think of a smart C++ optimisation (called "Template Meta Programming" and "Expression Templates") I did about 20 years ago that led to faster execution times (at cost of a much higher compilation time), and after the next compiler version was released, I discovered that the new version was able to optimise the original source to produce the exact same assembly - in short I should have rather used my time differently and stayed with the more readable (=easier to maintain) version of the code.
I've been trying to implement Euclid's algorithm in Java for 2 numbers or more.The problem with my code is that
a) It works fine for 2 numbers,but returns the correct value multiple times when more than 2 numbers are entered.My guess is that this is probably because of the return statements in my code.
b) I don't quite understand how it works.Though I coded it myself,I don't quite understand how the return statements are working.
import java.util.*;
public class GCDCalc {
static int big, small, remainder, gcd;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
// Remove duplicates from the arraylist containing the user input.
ArrayList<Integer> listofnum = new ArrayList();
System.out.println("GCD Calculator");
System.out.println("Enter the number of values you want to calculate the GCD of: ");
int counter = sc.nextInt();
for (int i = 0; i < counter; i++) {
System.out.println("Enter #" + (i + 1) + ": ");
int val = sc.nextInt();
listofnum.add(val);
}
// Sorting algorithm.
// This removed the need of conditional statements(we don't have to
// check if the 1st number is greater than the 2nd element
// before applying Euclid's algorithm.
// The outer loop ensures that the maximum number of swaps are occurred.
// It ensures the implementation of the swapping process as many times
// as there are numbers in the array.
for (int i = 0; i < listofnum.size(); i++) {
// The inner loop performs the swapping.
for (int j = 1; j < listofnum.size(); j++) {
if (listofnum.get(j - 1) > listofnum.get(j)) {
int dummyvar = listofnum.get(j);
int dummyvar2 = listofnum.get(j - 1);
listofnum.set(j - 1, dummyvar);
listofnum.set(j, dummyvar2);
}
}
}
// nodup contains the array containing the userinput,without any
// duplicates.
ArrayList<Integer> nodup = new ArrayList();
// Remove duplicates.
for (int i = 0; i < listofnum.size(); i++) {
if (!nodup.contains(listofnum.get(i))) {
nodup.add(listofnum.get(i));
}
}
// Since the array is sorted in ascending order,we can easily determine
// which of the indexes has the bigger and smaller values.
small = nodup.get(0);
big = nodup.get(1);
remainder = big % small;
if (nodup.size() == 2) {
recursion(big, small, remainder);
} else if (nodup.size() > 2) {
largerlist(nodup, big, small, 2);
} else // In the case,the array only consists of one value.
{
System.out.println("GCD: " + nodup.get(0));
}
}
// recursive method.
public static int recursion(int big, int small, int remainder) {
remainder = big % small;
if (remainder == 0) {
System.out.println(small);
} else {
int dummyvar = remainder;
big = small;
small = dummyvar;
recursion(big, small, remainder);
}
return small;
}
// Method to deal with more than 2 numbers.
public static void largerlist(ArrayList<Integer> list, int big, int small, int counter) {
remainder = big % small;
gcd = recursion(big, small, remainder);
if (counter == list.size()) {
} else if (counter != list.size()) {
big = gcd;
small = list.get(counter);
counter++;
largerlist(list, gcd, small, counter);
}
}
}
I apologize in advance for any formatting errors etc.
Any suggestions would be appreciated.Thanks!
I think these two assignments are the wrong way around
big = gcd;
small = list.get(counter);
and then big not used
largerlist(list, gcd, small, counter);
Also you've used static variables, which is usually a problem.
I suggest removing static/global variables and generally don't reuse variables.
Edit: Oh yes, return. You've ignored the return value of the recursion method when called from the recursion method. That shouldn't matter as you are printing out instead of returning the value, but such solutions break when, say, you want to use the function more than once.
I have developed a code for expressing the number in terms of the power of the 2 and I am attaching the same code below.
But the problem is that the expressed output should of minimum length.
I am getting output as 3^2+1^2+1^2+1^2 which is not minimum length.
I need to output in this format:
package com.algo;
import java.util.Scanner;
public class GetInputFromUser {
public static void main(String[] args) {
// TODO Auto-generated method stub
int n;
Scanner in = new Scanner(System.in);
System.out.println("Enter an integer");
n = in.nextInt();
System.out.println("The result is:");
algofunction(n);
}
public static int algofunction(int n1)
{
int r1 = 0;
int r2 = 0;
int r3 = 0;
//System.out.println("n1: "+n1);
r1 = (int) Math.sqrt(n1);
r2 = (int) Math.pow(r1, 2);
// System.out.println("r1: "+r1);
//System.out.println("r2: "+r2);
System.out.print(r1+"^2");
r3 = n1-r2;
//System.out.println("r3: "+r3);
if (r3 == 0)
return 1;
if(r3 == 1)
{
System.out.print("+1^2");
return 1;
}
else {
System.out.print("+");
algofunction(r3);
return 1;
}
}
}
Dynamic programming is all about defining the problem in such a way that if you knew the answer to a smaller version of the original, you could use that to answer the main problem more quickly/directly. It's like applied mathematical induction.
In your particular problem, we can define MinLen(n) as the minimum length representation of n. Next, say, since we want to solve MinLen(12), suppose we already knew the answer to MinLen(1), MinLen(2), MinLen(3), ..., MinLen(11). How could we use the answer to those smaller problems to figure out MinLen(12)? This is the other half of dynamic programming - figuring out how to use the smaller problems to solve the bigger one. It doesn't help you if you come up with some smaller problem, but have no way of combining them back together.
For this problem, we can make the simple statement, "For 12, it's minimum length representation DEFINITELY has either 1^2, 2^2, or 3^2 in it." And in general, the minimum length representation of n will have some square less than or equal to n as a part of it. There is probably a better statement you can make, which would improve the runtime, but I'll say that it is good enough for now.
This statement means that MinLen(12) = 1^2 + MinLen(11), OR 2^2 + MinLen(8), OR 3^2 + MinLen(3). You check all of them and select the best one, and now you save that as MinLen(12). Now, if you want to solve MinLen(13), you can do that too.
Advice when solo:
The way I would test this kind of program myself is to plug in 1, 2, 3, 4, 5, etc, and see the first time it goes wrong. Additionally, any assumptions I happen to have thought were a good idea, I question: "Is it really true that the largest square number less than n will be in the representation of MinLen(n)?"
Your code:
r1 = (int) Math.sqrt(n1);
r2 = (int) Math.pow(r1, 2);
embodies that assumption (a greedy assumption), but it is wrong, as you've clearly seen with the answer for MinLen(12).
Instead you want something more like this:
public ArrayList<Integer> minLen(int n)
{
// base case of recursion
if (n == 0)
return new ArrayList<Integer>();
ArrayList<Integer> best = null;
int bestInt = -1;
for (int i = 1; i*i <= n; ++i)
{
// Check what happens if we use i^2 as part of our representation
ArrayList<Integer> guess = minLen(n - i*i);
// If we haven't selected a 'best' yet (best == null)
// or if our new guess is better than the current choice (guess.size() < best.size())
// update our choice of best
if (best == null || guess.size() < best.size())
{
best = guess;
bestInt = i;
}
}
best.add(bestInt);
return best;
}
Then, once you have your list, you can sort it (no guarantees that it came in sorted order), and print it out the way you want.
Lastly, you may notice that for larger values of n (1000 may be too large) that you plug in to the above recursion, it will start going very slow. This is because we are constantly recalculating all the small subproblems - for example, we figure out MinLen(3) when we call MinLen(4), because 4 - 1^2 = 3. But we figure it out twice for MinLen(7) -> 3 = 7 - 2^2, but 3 also is 7 - 1^2 - 1^2 - 1^2 - 1^2. And it gets much worse the larger you go.
The solution to this, which lets you solve up to n = 1,000,000 or more, very quickly, is to use a technique called Memoization. This means that once we figure out MinLen(3), we save it somewhere, let's say a global location to make it easy. Then, whenever we would try to recalculate it, we check the global cache first to see if we already did it. If so, then we just use that, instead of redoing all the work.
import java.util.*;
class SquareRepresentation
{
private static HashMap<Integer, ArrayList<Integer>> cachedSolutions;
public static void main(String[] args)
{
cachedSolutions = new HashMap<Integer, ArrayList<Integer>>();
for (int j = 100000; j < 100001; ++j)
{
ArrayList<Integer> answer = minLen(j);
Collections.sort(answer);
Collections.reverse(answer);
for (int i = 0; i < answer.size(); ++i)
{
if (i != 0)
System.out.printf("+");
System.out.printf("%d^2", answer.get(i));
}
System.out.println();
}
}
public static ArrayList<Integer> minLen(int n)
{
// base case of recursion
if (n == 0)
return new ArrayList<Integer>();
// new base case: problem already solved once before
if (cachedSolutions.containsKey(n))
{
// It is a bit tricky though, because we need to be careful!
// See how below that we are modifying the 'guess' array we get in?
// That means we would modify our previous solutions! No good!
// So here we need to return a copy
ArrayList<Integer> ans = cachedSolutions.get(n);
ArrayList<Integer> copy = new ArrayList<Integer>();
for (int i: ans) copy.add(i);
return copy;
}
ArrayList<Integer> best = null;
int bestInt = -1;
// THIS IS WRONG, can you figure out why it doesn't work?:
// for (int i = 1; i*i <= n; ++i)
for (int i = (int)Math.sqrt(n); i >= 1; --i)
{
// Check what happens if we use i^2 as part of our representation
ArrayList<Integer> guess = minLen(n - i*i);
// If we haven't selected a 'best' yet (best == null)
// or if our new guess is better than the current choice (guess.size() < best.size())
// update our choice of best
if (best == null || guess.size() < best.size())
{
best = guess;
bestInt = i;
}
}
best.add(bestInt);
// check... not needed unless you coded wrong
int sum = 0;
for (int i = 0; i < best.size(); ++i)
{
sum += best.get(i) * best.get(i);
}
if (sum != n)
{
throw new RuntimeException(String.format("n = %d, sum=%d, arr=%s\n", n, sum, best));
}
// New step: Save the solution to the global cache
cachedSolutions.put(n, best);
// Same deal as before... if you don't return a copy, you end up modifying your previous solutions
//
ArrayList<Integer> copy = new ArrayList<Integer>();
for (int i: best) copy.add(i);
return copy;
}
}
It took my program around ~5s to run for n = 100,000. Clearly there is more to be done if we want it to be faster, and to solve for larger n. The main issue now is that in storing the entire list of results of previous answers, we use up a lot of memory. And all of that copying! There is more you can do, like storing only an integer and a pointer to the subproblem, but I'll let you do that.
And by the by, 1000 = 30^2 + 10^2.
I want to find the Nth number of the Recurrence Equation
T(n)=T(n-1)+3T(n-2)+3T(n-3)+(n-4),T(1)=T(4)=1,T(2)=T(3)=3
so if suppose you entered 2,5,9 as input, output should be T(2)=3,T(5)=20,T(9)=695
what I did is create an array of size equal to maximum of all input value and storing solution of T(i) at index i.Then look up into the array for specific index. eg array[3] for T(3),array[5] for T(5),etc
The code worked fine till maximum number is not greater than maximum integer value system can hold i.e
Integer.MAXValue.
Because the index of array can only be integer then
if number is n=1855656959555656 what should be the best way to find the solution of
T(1855656959555656)?
as clearly I cant create an array of size=1855656959555656..
I have even tried BigInteger from java.Math but with no success.
I have to find some other approach.please suggest some ideas..
Thanks
you do not need to store every T(i), you only need to store 3 values T(i-1), T(i-2), T(i-3). While looping over i, check if the current i should be part of your output, if so put it out immediately or save it to an "output"-array.
edit: this part is quite inefficient. You check in every iteation EVERY needed output.
for (int k = 0; k < arr.length; ++k) {
if (count == arr[k])
T[k] = temp[i];
else if (arr[k] == 1)
T[k] = 1;
else if (arr[k] == 2)
T[k] = 3;
else if (arr[k] == 3)
T[k] = 3;
else if (arr[k] == 4)
T[k] = 1;
}
so your code runs in time (max*arr.length) you can reduce it to only (max). Use a HashMap with key=neededPosition (=count) value=position in arr
Init the map like this:
Map<Long, Integer> map = new HashMap<Long, Integer>();
for (int i = 0; i < arr.length; i++) {
map.put(arr[i], i);
}
if (map.containsKey(count)) {
T[map.get(count)] = temp[i]
}
check the values 1-4 just once after the whole thing!
Not possible. The array size can be a maximum of Integer.MAX_VALUE (minus something usually 5 or 8, depending on the JVM capabilities). Why?. The index for an Array should be an integer thats a limitation.
It can't be done. So you need to solve the problem by introducing a sharding mechanism. The simplest way would be to just have arrays of arrays with a fixed length.
Edit: You really do not need this much storage for your problem at hand (as pointed out in another answer; this code fragment avoids arrays altogether to avoid bounds checks / indirection):
public void t(long n) {
if (n < 5) {
return (n == 2 || n == 3) ? 3 : 1;
}
long i = 5; // Initialize variables for n == 5;
long tn_1 = 1; // T(n-1) = T(4) = 1;
long tn_2 = 3; // T(n-2) = T(3) = 3;
long tn_3 = 1; // T(n-3) = T(2) = 1;
long tn_4 = 3; // T(n-4) = T(1) = 3;
while (true) {
long tn = tn_1 + 3*tn_2 + 3*tn_3 + tn_4;
if (i++ == n) {
return tn;
}
tn_4 = tn_3;
tn_3 = tn_2;
tn_2 = tn_1;
tn_1 = tn;
}
}
To answer the question in the title anyway:
If your array is sparse, use a map (TreeMap or HashMap) of Long or BigInteger:
Map<Long,Long> t = new TreeMap<Long,Long>()
The memory consumption of sparse arrays depends on the number of elements actually stored, so you may want to delete values from the map that are no longer needed.
If your array is not sparse, use a 2-level array (memory consumption will depend on the pre-allocated size only):
public class LongArray {
static final long BLOCK_SIZE = 0x40000000;
long[][] storage;
public LongArray(long size) {
long blockCount = (size + BLOCK_SIZE - 1) / BLOCK_SIZE;
storage = new long[][(int) blockCount];
for (long i = 0; i < blockCount; i++) {
if (i == blockCount - 1) {
storage[i] = new long[(int) size - BLOCK_SIZE * (blockCount - 1)];
} else {
storage[i] = new long[(int) BLOCK_SIZE];
}
}
}
public long get(long index) {
return storage[(int) (index / BLOCK_SIZE)][(int) (index % BLOCK_SIZE)];
}
public void put(long index, long value) {
storage[(int) (index / BLOCK_SIZE)][(int) (index % BLOCK_SIZE)] = value;
}
}
In both cases, use t.get(index) and t.put(index, value) instead of t[index] to access your array (if t is the name of the array).
You can do one thing. Check if the value of n is equal to 1855656959555656 in the beginning or if its multiple. Suppose, the value of n is twice of 1855656959555656. Then you can create two arrays and link them together virtually. This should solve your problem but it will involve a lot of overhead.
Use recursive call:
int T(int n){
if (n==1 || n==4){
return 1;
} else if (n==2 || n==3){
return 3;
} else {
return T(n-1)+3*T(n-2)+3T*(n-3)+T(n-4);
}
}
Edit: Time consumming. Won't work with large numbers
I got to know that memoization is better approach for playing with Fibonacci series. Today i wrote the program in this way--
private void printFibonacci(int lengthOfSeries) {
int lastNumb = 1;
int secondLastNumb = 1;
List<Integer> fabArray = new ArrayList<Integer>();
if (lengthOfSeries == 0) {
System.out.println("Please enter number to print series.");
} else {
fabArray.add(secondLastNumb);
fabArray.add(lastNumb);
while (fabArray.size() < lengthOfSeries) {
lastNumb = fabArray.get(fabArray.size() - 1);
secondLastNumb = fabArray.get(fabArray.size() - 2);
fabArray.add(lastNumb + secondLastNumb);
}
}
System.out.println(fabArray);
}
This is working fine. But when the input number is high (say 100,000,000), its throwing outOfMemoryerror. And its throwing error at line where printing List.
Can anybody please give some suggestion to make it powerful.
It's the growth of the ArrayList that is causing the OutOfMemoryError, and you don't need to either print the list or terminate your loop. The following code should have no problems with memory:
private void printFibonacci(int lengthOfSeries) {
int lastNumb = 1;
int secondLastNumb = 1;
int count = 2;
if (lengthOfSeries == 0) {
System.out.println("Please enter number to print series.");
} else {
System.out.print("1, 1");
while (count < lengthOfSeries) {
int next = lastNumb + secondLastNumb;
System.out.print(", " + next);
secondLastNumb = lastNumb;
lastNumb = next;
//Updated
count = count+1;
}
System.out.println();
}
}
That's the nature of this kind of algorithms, they're very memory-hungry. Also using int/Integer for it is a very bad idea as you'll quickly run out of Integer capacity (2*10^32 -1). Probably you want doubles, but still there is a limit of how far you can go.
100 million Integers takes about 400 million bytes.
This may well be larger than your JVMs default maximum heap size. Try adding command line switches to allocate more memory.
Some indications here: http://publib.boulder.ibm.com/infocenter/realtime/v2r0/index.jsp?topic=%2Fcom.ibm.rt.doc.20%2Fdiag%2Fappendixes%2Fdefaults.html
Following my comment, a trivial implementation would be:
int[] fabArray = new int[] {1, 1, 0};
for (int i = 0; i < lengthOfSeries; i++) {
fabArray[2] = fabArray[1] + fabArray[0];
System.out.print(fabArray[0] + ", ");
fabArray[0] = fabArray[1];
fabArray[1] = fabArray[2];
}
// End of line
System.out.println();