### pllk's blog

By pllk, 6 years ago, ,

Binary search is a very familiar algorithm to most people here. A typical implementation looks like this:

// x = array of numbers
// n = length of the array
// k = search key
// returns "true" if the key is found, "false" otherwise
bool search(int x[], int n, int k) {
int l = 0, r = n-1;
while (l <= r) {
int m = (l+r)/2;
if (x[m] == k) return true;
if (x[m] < k) l = m+1; else r = m-1;
}
return false;
}


However, even if the idea is straightforward, there are often bugs in binary search implementations. How exactly to update the search bounds and what is the condition when the search should stop?

Here is an alternative binary search implementation that is shorter and should be easier to get correct:

bool search(int x[], int n, int k) {
int p = 0;
for (int a = n; a >= 1; a /= 2) {
while (p+a < n && x[p+a] <= k) p += a;
}
return x[p] == k;
}


The idea is to implement binary search like linear search but with larger steps. Variable p contains the current array index, and variable a contains the step length. During the search, the step length is n, n/2, n/4, ..., 1. Finally, p contains the largest index whose value is not larger than the search key.

A similar idea can be applied to binary search variations. For example, the following code counts how many times the key appears in the array. After the search, [p+1..q] is the range with value k.

int count(int x[], int n, int k) {
int p = -1, q = -1;
for (int a = n; a >= 1; a /= 2) {
while (p+a < n && x[p+a] < k) p += a;
while (q+a < n && x[q+a] <= k) q += a;
}
return q-p;
}


Do you know other tips how to implement binary search correctly?

• +29

 » 6 years ago, # |   +1 thanks for the second implementation it was a new and of course tricky implementation to me.:D
 » 6 years ago, # | ← Rev. 2 →   +29 Well, imagine you an array [0..n - 1] and let us have an invariant: some function F which returns True for every a[i] as an argument for any i from [0..k] and False for every a[j] for j from [k + 1..n - 1]. Then my binary search which actually finds the border between these two parts of the array looks the following way: l, r = -1, n while r-l > 1: m = (l+r)//2 if F(a[m]): l = m else: r = m You can easily prove that l = k and r = k + 1 by the end of the while loop. In this case no worries about whether to increase m by 1 or not.As an example this is the code which determines whether an element x exists in the sorted array [0..n-1]: def BinarySearch(x): l, r = -1, n while r-l > 1: m = (l+r)//2 if a[m] < x: l = m else: r = m return r != n and a[r] == x 
•  » » 2 years ago, # ^ |   0 http://codeforces.com/contest/812/problem/C I got many bugs earlier , but with your implementation , AC in one go :)BTW , do you always use this implementation , or adapt different version like h = mid + 1 etc , sometimes?
•  » » » 12 months ago, # ^ |   0 Hi I'm always use this binary search, you never need use r = mid + 1, but here are 2 tricks:1) To select the l, and r values you need:For l take a value that can never happen for left. (foribidden value left)For r take a value that can never happen for right. (forbidden value right)For example in an array we never can access to index -1, or the index N (assuming 0 to N-1 indexing) then l = -1, r = nOr on this problem: http://codeforces.com/gym/101908/problem/G(solution) https://pastebin.com/jqe0Zh5TFirst read the problem. A problem about binary search on a flow I pick 0 for left value, because I know that is impossible fill all stations in 0 minutes. and I picked 1000001 for right value, because I know that is unnecessary using more time of the max time (1000000 limit of T in Tthe problem) to travel from a point A to point B. (also I can use right = max(T[i], for i from 0 to n -1) + 1.So if your binary search ends and the value of b is the forbidden value means that no element meets the property.2) Nerver use this binary search for doubles, obviously u can code something like this: double EPS = 1e-9; double a = 0.0; //never gonna happend double r = 100000000000000000000.0; //never gonna happen while (b - a > EPS) { double mid = (a + b) / 2.0; if (propertie(mid))b = mid; else a = mid; } cout << b << end; But this implementation is prone to errors of precision, more than once it has happened to me. On the week 2 of: https://courses.edx.org/courses/course-v1:ITMOx+I2CPx+3T2016/course/ explains about the reason it gives problems with doubles. When u need find some propertie on doubles u can use: double a = 0.0; //never gonna happend double r = 100000000000000000000.0; //never gonna happen for (int it = 1; it <= 70; it++) { // some coders use 40, others 70, Same varies according to the problem, honestly I do not know how it is calculated double mid = (a + b) / 2.0; if (propertie(mid)) b = mid; else a = mid; } cout << b << endl; P.s I know the question was from months ago, but I hope someone finds it useful now.
 » 6 years ago, # |   +1 As a suggestion, As we all know that Binary Search runs in O(logn) so implementing it recursive will not only be that bad in case of running time or memory, but also it can decrease the number of bugs you might have :)And one other bug that anyone could have is stop condition of binary search. As a suggestion you can have a condition that when r-l<=2 check remaining items yourself. It can also decrease the number of bugs you might have.And as the final word, bounds are very important in implementing binary search. Searching for the first occurrence, last one, or the one after last one,or etc. will require different bounds of passing middle point and left or right to the function.Here is an example usage of Binary Search for Problem C,Round 218 : 5384392P.S. Sorry for my bad english :)
 » 6 years ago, # |   +1 Usually, when I have doubts about errors implementing binary search, which are common when I have to find things like "the largest element with this and that conditions..." (also lower_bound, upper_bound, etc), I do the following. I make the main loop of the binary search like this:while (hi-lo > 5) { // here the usual binary search code }and then, after that, a small linear loop for getting the correct result.while (lo != hi) // increment or decrement hi or lo, respectivelyThe number 5 has nothing of special. It only ensures that the remaining interval is small enough. This avoids the possibility of errors on termination of the main loop of binary search, or the wrong choose of lo or hi. Although this approach is nice for saving you during a contest, it is not that good for understanding and proving correctness, which is fair more valuable IMHO.I hope it was useful.
 » 6 years ago, # | ← Rev. 5 →   0 Let me try to generalize novaco's answer a bit:So far, every single time I used (integer) binary search, I could formulate the problem in one of two ways:(1) Given a range of integers R = {l, l + 1, ..., r - 1, r} and a monotonically increasing predicate P, find the smallest for which P(x) holds true. I then use the following code: while (l < r) { int mid = (l + r) / 2; if (P(mid)) r = mid; else l = mid + 1; } // after the loop, l = r = x This might not work if P(r) = 0 (in this case the algorithm will return x = r), but you can easily extend your search range to r + 1 and artificially set P(r + 1) = 1 or you can just precheck for that situation.(2) Given a range of integers R = {l, l + 1, ..., r - 1, r} and a monotonically decreasing predicate P, find the largest for which P(x) holds true. If we set Q(x) = !P(x), then Q is increasing and we can use (1) to find x + 1. We can also just use the following slightly modified variant: while (l < r) { int mid = (l + r + 1) / 2; if (P(mid)) l = mid; else r = mid - 1; } // after the loop, l = r = x I find this approach so intuitive that I haven't done a single mistake while implementing binary search since.In the case of finding the occurence of an element k in a sorted array, I would use variant (1) to find the first element i >= k and then check whether i = k. Note that in the case where k > x[n - 1] the algorithm still works because of the equality check at the end. bool search(int x[], int n, int k) { int l = 0, r = n - 1; while (l < r) { int mid = (l + r) / 2; if (x[mid] >= k) r = mid; else l = mid + 1; } return x[l] == k; } So, to summarize, you don't need to use this particular method, but you should have realized that there is really only one type of problem that can be solved with binary search and stick to only one particular implementation to solve this problem. You won't make a single mistake with binary search from there on.
•  » » 12 months ago, # ^ |   0 what is x in "if" condition, because search value is k??
•  » » » 12 months ago, # ^ |   0 Sorry, fixed
 » 6 years ago, # |   0 i found a good binary search problem. and in my experience when you wanna minimize the maximum of something or maximize the minimum of something binary search is a good idea.
 » 6 years ago, # |   0 I lost track of how many problems I couldn't solve because of bad binary search implementations. Even this year during IOI I couldn't manage to solve completely one of the problems because my binary search had a bug that I wasn't able to find :(
 » 5 years ago, # |   +5 if an array contains the following data: 1 2 3 4 4 4 4 4 4 6 7 8 Now I want to find the position of first occurance of data 4. How can I implement binary search for this problem? Here, the answer should be 3.
•  » » 5 years ago, # ^ |   0 Something like this: If the middle data is 4 or larger Set the right bound to be the middle Else Set the left bound to be the middle This guarantees that you won't miss the earliest occurrence of 4, since whenever you see a 4, it's the right bound that is modified, not the left bound.
 » 5 years ago, # |   0 Generalized implementation of lower bound for non-decreasing func : // result: min x of [firstX, lastX), when func[x] >= y template ArgType LowerBound(const ArgType& firstX, const ArgType& lastX, const ResultType& y, const Func& func) { ArgType l = firstX; ArgType r = lastX; while (l < r) { const ArgType m = l + (r - l) / 2; if (func(m) < y) l = m + 1; else r = m; } return l; } 
 » 5 years ago, # |   0 Actually you don't need while in your second implementation. You can use just if, can't you?
•  » » 5 years ago, # ^ |   +3 while is needed, because the steps are not always powers of 2.
 » 5 years ago, # |   +3 I used to write binary search like this: int binsearch(vector & x, int k) { int i = 0; for(int d = 1<
 » 4 years ago, # |   0 I think this is more like Exponential search. It always performs at least as efficient as binary search.
 » 4 months ago, # |   0 Thanks for the second implementation..