Results 1 to 11 of 11
Thread: binary search
 01102008, 09:12 PM #1Member
 Join Date
 Jan 2008
 Posts
 5
 Rep Power
 0
binary search
hey i got confused with this and maybe someone can help me...
i was experimenting about sorting a randomly generated array of 100 integers with binary search.. however i have some trouble..
heres d code:
class iamangry
{
public int binarySearch(int arr[], int key, int n)
{
int low = 0;
int ri = (arr[n]+ 1);
int mid;
do{ mid = ((low + ri)/2);
if(key < arr[mid]){
ri = (mid  1);
}
else low = mid + 1;
}
while (key == arr[mid]);
if(low < ri){
return 1;
}
else return mid;
}
public static void main(String[] args){
int x = 100;
int[]arr = binarySearch[x];
for(int n = 0; n < a.length; n++){
arr[n] = 1 + (int)(Math.random() *100);
}
for(int t=1; t<=100; t++){
System.out.print(""+arr[t]+" ");
}
}
}
the problem seems to be actually in: int[]arr = binarySearch[x]; ...
any help please? thanks
 01112008, 09:29 AM #2
One thing you need to understand about Binary Search, is that it is a search function for some value or object, thus your array is assumed to be already sorted. You're looking for some value within in the array and trying to discover its location.
I altered your code to form the below, although not thoroughly tested.. I hope you get the idea.
Java Code:public class BinarySearch { public int binarySearch(int arr[], int key, int size) { int low = 0; int mid = 0; int high = size; while (low <= high) { mid = ((low + high) / 2); if (key < arr[mid]) { high = (mid  1); } else { low = mid + 1; } } if (low < high) { return 1; } else { return mid; } } public static void main(String[] args){ int SIZE = 20; int[] arr = new int[SIZE]; int key = 5; for (int i = 0, k = 0; i < SIZE; i++, k++) { arr[k] = i; //(1 + (int)(Math.random() * 100)); } BinarySearch bs = new BinarySearch(); for (int j = 0; j <= arr.length  1; j++) { System.out.println(arr[j]); } System.out.println("Array size: " + arr.length); System.out.println("Value found at index " + bs.binarySearch(arr, key, SIZE)); } }
Vote for the new slogan to our beloved Java Forums! (closes on September 4, 2008)
Want to voice your opinion on your IDE/Editor of choice? Vote now!
Got a little Capt'n in you? (drink responsibly)
 01112008, 11:16 PM #3Member
 Join Date
 Jan 2008
 Posts
 5
 Rep Power
 0
ah ok ic..
so how can i sort an array of randomly generated numbers?
cheers
 01112008, 11:20 PM #4Vote for the new slogan to our beloved Java Forums! (closes on September 4, 2008)
Want to voice your opinion on your IDE/Editor of choice? Vote now!
Got a little Capt'n in you? (drink responsibly)
 01112008, 11:33 PM #5Member
 Join Date
 Jan 2008
 Posts
 24
 Rep Power
 0
i am assuming you are doing binary search code because you want to practice. Because obviously Arrays.binarySearch(...) does the thing.
 01122008, 10:32 PM #6Member
 Join Date
 Jan 2008
 Posts
 5
 Rep Power
 0
 01122008, 11:08 PM #7Member
 Join Date
 Jan 2008
 Posts
 24
 Rep Power
 0
well, did you read the java doc?
there is also a handy "sort" method there. first show me what do you have, then we can analyze the problem.
 01122008, 11:13 PM #8Member
 Join Date
 Jan 2008
 Posts
 24
 Rep Power
 0
By the way, leanring the internals of binary search algorithm is a very important thing, i strongly suggest understanding the way it works first (this is one of the favorite questions asked in job interviews even maybe by Google). But if you are developing an application that requires binary search, by all means use the built in Facilities in Java, like Arrays, Collections etc.
 01132008, 03:13 PM #9Member
 Join Date
 Jan 2008
 Posts
 5
 Rep Power
 0
well i thought that this method could sort the randomnumber array :(
public int binarySearch(int arr[], int key, int size) {
int low = 0;
int mid = 0;
int high = size;
while (low <= high) {
mid = ((low + high) / 2);
if (key < arr[mid]) {
high = (mid  1);
} else {
low = mid + 1;
}
}
if (low < high) {
return 1;
} else {
return mid;
}
}
actually i'm just starting to learn what sorting is and binary search seems one of the most efficient
 01132008, 05:10 PM #10Member
 Join Date
 Jan 2008
 Posts
 24
 Rep Power
 0
No, binary search is used for finding an element in an already sorted array quickly. For example if you have a sorted integer array like this:
a[]={1,2,4,5,6,7,9,10,11,12,14,16,19,21,22}
binary search can find the position of "19" in only three steps. However, as you see the array is already sorted.
there are several sorting algorithms, if you want to learn ebout them, you can start with easy ones, such as selection sort, bubble sort or insertion sort. there are more complicated sort algorithms also such as quick sort, merge sort, radix sort or heap sort.
Java platform uses a modified merge sort, and qick sort for its sort facilities. There are Arrays.sort() for arrays, and Collections.sort() for Lists available. But i got that you are in the learning phase, not sur eif you want to use them.
Make a search in wikipedia about search algorithms or binary search, you can see several different implementations in various languages in the links area. Or Google it. For binary search or sort, you can also check how java is doing it by checking the source.
 01142008, 08:13 PM #11Member
 Join Date
 Jan 2008
 Posts
 5
 Rep Power
 0
Similar Threads

Can anybody help with cuncurrent binary search tree guys)
By danylo in forum Threads and SynchronizationReplies: 1Last Post: 04232008, 07:22 PM 
Binary Search in Java
By Java Tip in forum AlgorithmsReplies: 0Last Post: 04152008, 08:43 PM 
Use recursion to convert binary to...
By coco in forum Advanced JavaReplies: 1Last Post: 08072007, 08:46 AM 
problem with recursive binary search program
By imran_khan in forum New To JavaReplies: 3Last Post: 08022007, 04:08 PM 
Binbot Binary Newsreader 1.1b3
By levent in forum Java SoftwareReplies: 0Last Post: 07272007, 02:18 PM
Bookmarks