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Does a binary search beat an exponential search in any way, except in space complexity?
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However, in the average case scenario, hash lookup is significantly faster than binary search. In real applications, we mainly consider an average case scenario in order to test and compare the performance of different methods. Therefore, hash lookup is a better choice in such cases.
Exponential Search also known as finger search, searches for an element in a sorted array by jumping 2^i elements every iteration where i represents the value of loop control variable, and then verifying if the search element is present between last jump and the current jump.
Linear search is a search that finds an element in the list by searching the element sequentially until the element is found in the list. On the other hand, a binary search is a search that finds the middle element in the list recursively until the middle element is matched with a searched element.
The binary search algorithm works on the principle of divide and conquer and it is considered the best searching algorithm because it's faster to run.
Both these algorithms search for a value in an ordered list of elements, but they address different issues. Exponential search is explicitly designed for unbounded lists whereas binary search deals with bounded lists.
The idea behind exponential search is very simple: Search for a bound, and then perform a binary search.
Let's take an example. A = [1, 3, 7, 8, 10, 11, 12, 15, 19, 21, 22, 23, 29, 31, 37]. This list can be seen as a binary tree (although there is no need to build the tree):
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____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
A binary search for e = 27 (for example) will undergo the following steps
b0) Let T, R be the tree and its root respectively
15 (R)
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
b1) Compare e to R: e > 15. Let T, R be T right subtree and its root respectively
15
____/ \____
/ \
__8__ _23_(R)
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
b2) Compare e to R: e > 23. Let T, R be T right subtree and its root respectively
15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
b3) Compare e to R: e < 31. Let T, R be T left subtree and its root respectively
15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31___
/ \ / \ / \ / \
1 7 10 12 19 22 29 (R) 37
b4) Compare e to R: e <> 29: the element is not in the list, since T has no subtree.
An exponential search for e = 27 (for example) will undergo the following steps
Let T, R be the leftmost subtree (ie the leaf 1) and its root (1) respectively
15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31
/ \ / \ / \ / \
(R) 1 7 10 12 19 22 29 37
e1) Compare e to R: e > 1. Let R be the parent of R and T be the tree having R as root
15
____/ \____
/ \
__8__ _23__
/ \ / \
(R) 3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e2) Compare e to R: e > 3. Let R be the parent of R and T be the tree having R as root:
15
____/ \____
/ \
(R)_8__ _23__
/ \ / \
3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e3) Compare e to R: e > 8. Let R be the parent of R and T be the tree having R as root:
(R) 15
____/ \____
/ \
__8__ _23__
/ \ / \
3 11 21 31 (R)
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e4) Compare e to R: e > 15. R has no parent. Let T be the right subtree of T and R be its root:
15
____/ \____
/ \
__8__ _23_(R)
/ \ / \
3 11 21 31
/ \ / \ / \ / \
1 7 10 12 19 22 29 37
e5..7) See steps b2..4)
For the sake of demonstration, let N = 2^n be the size of A and let indices start from 1. If N is not a power of two, the results are almost the same.
Let 0 <= i <= n be the minimum so that A[2^(i-1)] < e <= A[2^i] (let A[2^-1] = -inf). Note that this kind of interval may not be unique if you have duplicate values, hence the "minimum".
You need i + 1 iterations to find i. (In the example, you are jumping from child to parent repeatedly until you find a parent greater than e or there is no more parent)
Then you use a binary search on the selected interval. The size of this interval is 2^i - 2^(i-1) = 2^(i-1).
The cost of a binary search in an array of size 2^k is variable: you might find the value in the first iteration, or after k iterations (There are sophisticated analysis depending on the distribution of the elements, but basically, it's between 1 and k iterations and you can't know it in advance)
Let j_i, 1 <= j_i <= i - 1 be the number of iterations needed for the binary search in our case (The size of this interval is 2^(i-1)).
Let i be the minimum so that A[2^(i-1)] < e <= A[2^i]. Because of the assumption that N = 2^n, the binary search will meet this interval:
We start with the root A[2^(n-1)]. If e > A[2^(n-1)], i = n because R = A[2^(n-1)] < e < A[2^n]. Else, we have e <= A[2^(n-1)]. If e > A[2^(n-2)], then i = n-1, else we continue until we find i.
You need n - i + 1 steps to find i using a binary search:
i = n, you know it at the first iteration (e > R) else, you select the left subtreei = n-1, you need two iterationsi = 0, you'll need n iterations.Then you'll need j_i iterations as shown above to complete the search.
As you see, the j_i iterations are common to both algorithms. The question is: Is i + 1 < n - i + 1? i.e. Is i < n - i or 2i < n? If yes, the exponential search will be faster than the binary search. If no, the binary search will be faster than the exponential search (or equally fast)
Let's get some distance: 2i < n is equivalent to (2^i)^2 < 2^n or 2^i < sqrt(2^n). While 2^i < sqrt(N), the exponential search is faster. As soon as 2^i > sqrt(N), the binary search is faster. Remember that the index of e is lower or equal than 2^i because e <= A[2^i].
In simple words, if you have N elements and if e is in the firstsqrt(N) elements, then exponential search will be faster, else binary search will be faster.
It depends on the distribution, but N - sqrt(N) > sqrt(N) if N > 4, and thus the binary search is likely to be faster than the exponential search unless you know that the element will be among the first ones or the list is ridiculously short.
2^n < N < 2^(n+1)
I won't go into details, but this does not change the general conclusion.
If the value is beyond the last power of two, the cost of exponential to find the bound is already n+2, more than the binary search (less than or equal to 2^(n+1)). Then you have a binary search to perform, maybe in a small interval, but binary search is already the winner.
Else you add the value A[N] to the list until you have 2^(n+1) value. This won't change anything for exponential search, and this will slow down the binary search. But this slow binary search remains faster if e is not in the firstsqrt(2^(n+1)) values.
That's an interesting question which I don't talk about, size of the pointer and things like that. If you are performing an exponential search and consuming elements as they arrive (imagine timestamps), you don't need to store the whole list at once. You just have to store one element (the first), then one element (the second), then two elements (the third and the fourth), then four elements, ... then 2^(i-1) elements. If i is small, then you won't need to store a large list as in a regular binary search.
Implementation is really not a problem here. See the Wikipedia pages for information: Binary search algorithm and Exponential search.
Use the exponential search only when the sequence is unbounded or when you know the value is likely to be among the first ones. Unbounded: I like the example of timestamps: they are strictly growing. You can imagine a server with stored timestamps. You can ask for n timestamps and you are looking for a specific timestamp. Ask 1, then 2, then 4, then 8,... timestamps and perform the binary search when one timestamps exceeds the value you are looking for.
In other cases, use the binary search.
Remark: the idea behind the first part of the exponential search has some applications:
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