how to calculate binary search complexity

Question

I heard somebody say that since binary search halves the input required to search hence it is log n  algorithm  Since I am not from a mathematics background I am not able to relate to it  Can somebody explain it in a little more detail  Does it have to do something with the logarithmic series

User · Answer

A binary search works by dividing the problem in half repeatedly, something like this (details omitted):

Example looking for 3 in [4,1,3,8,5]

Order your list of items [1,3,4,5,8]
Look at the middle item (4),
- If it is what you are looking for, stop
- If it is greater, look at the first half
- If it is less, look at the second half
Repeat step 2 with the new list [1, 3], find 3 and stop

It is a bi-nary search when you divide the problem in 2.

The search only requires log2(n) steps to find the correct value.

I would recommend Introduction to Algorithms if you want to learn about algorithmic complexity.

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The time complexity of the binary search algorithm belongs to the O log n  class  This is called big O notation  The way you should interpret this is that the asymptotic growth of the time the function takes to execute given an input set of size n will not exceed log n   This is just formal mathematical lingo in order to be able to prove statements  etc  It has a very straightforward explanation  When n grows very large  the log n function will out-grow the time it takes to execute the function  The size of the  input set   n  is just the length of the list   Simply put  the reason binary search is in O log n  is that it halves the input set in each iteration  It s easier to think about it in the reverse situation  On x iterations  how long list can the binary search algorithm at max examine  The answer is 2 x  From this we can see that the reverse is that on average the binary search algorithm needs log2 n iterations for a list of length n   If why it is O log n  and not O log2 n   it s because simply put again - Using the big O notation constants don t count

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Let s say the iteration in Binary Search terminates after k iterations  At each iteration  the array is divided by half  So let   s say the length of the array at any iteration is n At Iteration 1   Length of array   n   At Iteration 2   Length of array   n 2   At Iteration 3   Length of array    n 2  2   n 22   Therefore  after Iteration k   Length of array   n 2k   Also  we know that after After k divisions  the length of the array becomes 1 Therefore  Length of array   n 2k   1   gt  n   2k   Applying log function on both sides     gt  log2  n    log2  2k    gt  log2  n    k log2  2  As  loga  a    1    Therefore   As  loga  a    1  k   log2  n    Hence the time complexity of Binary Search is  log2  n

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Here is the solution using the master theorem  with readable LaTeX   For each recurrence in the recurrence relation for binary search  we convert the problem into one subproblem  with runtime T N 2   Therefore     Substituting into the master theorem  we get     Now  because  is 0 and f n  is 1  we can use the second case of the master theorem because     This means that

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Here a more mathematical way of seeing it  though not really complicated  IMO much clearer as informal ones   The question is  how many times can you divide N by 2 until you have 1  This is essentially saying  do a binary search  half the elements  until you found it  In a formula this would be this      1   N   2x   multiply by 2x      2x   N   now do the log2      log2 2x   nbsp  nbsp  nbsp   log2 N   x   log2 2    log2 N   x   1  nbsp  nbsp  nbsp  nbsp  nbsp  nbsp  nbsp  nbsp   log2 N   this means you can divide log N times until you have everything divided  Which means you have to divide log N   do the binary search step   until you found your element

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Let me make it easy for all of you with an example    For simplicity purpose  let s assume there are 32 elements in an array in the sorted order out of which we are searching for an element using binary search   1 2 3 4 5 6     32  Assume we are searching for 32   after the first iteration  we will be left with   17 18 19 20      32  after the second iteration  we will be left with   25 26 27 28      32  after the third iteration  we will be left with   29 30 31 32  after the fourth iteration  we will be left with   31 32  In the fifth iteration  we will find the value 32   So  If we convert this into a mathematical equation  we will get   32 X  1 25     1     n X  2-k    1      2k    n     k log22   log2n     k   log2n  Hence the proof

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Log2 n  is the maximum number of searches that are required to find something in a binary search  The average case involves log2 n -1 searches  Here s more info   http   en wikipedia org wiki Binary search Performance

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For Binary Search  T N    T N 2    O 1       the recurrence relation  Apply Masters Theorem for computing Run time complexity of recurrence relations   T N    aT N b    f N   Here  a   1  b   2    log  a base b    1  also  here f N    n c log k n        k   0  amp  c   log  a base b   So  T N    O N c log  k 1 N    O log N    Source   http   en wikipedia org wiki Master theorem

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Since we cut down a list in to half every time therefore we just need to know in how many steps we get 1 as we go on dividing a list by two  In the under given calculation x denotes the numbers of time we divide a list until we get one element In Worst Case    1   N 2x  2x   N  Taking log2  log2 2x    log2 N    x log2 2    log2 N   x   log2 N

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It doesn t half search time  that wouldn t make it log n    It decreases it logarithmicly   Think about this for a moment   If you had 128 entries in a table and had to search linearly for your value  it would probably take around 64 entries on average to find your value   That s n 2 or linear time   With a binary search  you eliminate 1 2 the possible entries each iteration  such that at most it would only take 7 compares to find your value  log base 2 of 128 is 7 or 2 to the 7 power is 128    This is the power of binary search

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ok see this for i 0 i lt n n n 2    i      1  Suppose at i k the loop terminate  i e  the loop execute k times   2  at each iteration n is divided by half   2 a n n 2                        AT I 1 2 b n  n 2  2 n  2 2  2 c n   n 2  2  2 n  2 3         aT I 3 2 d n    n 2  2  2  2 n  2 4   So at i k   n 1 which is obtain by dividing n  2 k times n 2 k 1 n 2 k  k log N     base 2

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Here is wikipedia entry  If you look at the simple iterative approach  You are just eliminating half of the elements to be searched for until you find the element you need    Here is the explanation of how we come up with the formula   Say initially you have N number of elements and then what you do is   N 2  as a first attempt  Where N is sum of lower bound and upper bound  The first time value of N would be equal to  L   H   where L is the first index  0  and H is the last index of the list you are searching for  If you are lucky  the element you try to find will be in the middle  eg  You are searching for 18 in the list     16  17  18  19  20  then you calculate   0 4  2    2 where 0 is lower bound  L - index of the first element of the array  and 4 is the higher bound  H - index of the last element of the array    In the above case L   0 and H   4   Now 2 is the index of the element 18 that you are searching found  Bingo  You found it   If the case was a different array 15 16 17 18 19  but you were still searching for 18 then you would not be lucky and you would be doing first N 2  which is   0 4  2    2 and then realize element 17 at the index 2 is not the number you are looking for  Now you know that you don   t have to look for at least half of the array in your next attempt to search iterative manner  Your effort of searching is halved  So basically  you do not search half the list of elements that you searched previously  every time you try to find the element that you were not able to find in your previous attempt    So the worst case would be      N  2     N 2   2      N 2  2   2       i e    N 21   N 22   N 23          N 2x           until    you have finished searching  where in the element you are trying to find is at the ends of the list   That shows the worst case is when you reach N 2x   where x is such that 2x   N  In other cases N 2x   where x is such that 2x  lt  N Minimum value of x can be 1  which is the best case   Now since mathematically worst case is when the value of 2x   N     log2 2x    log2 N      x   log2 2    log2 N      x   1   log2 N      More formally  log2 N  1

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T n  T n 2  1  T n 2   T n 4  1 1  Put the value of The n 2  in above so T n  T n 4  1 1         T n 2 k  1 1 1      1   T 2 k 2 k  1 1     1 up to k   T 1  k  As we taken 2 k n  K   log n  So Time complexity is O log n

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T N    T N 2    1    T N    T N 2    1    T N 4    1   1       T N    T N N     1   1   1        1    1   logN  base 2 log    1   logN  So the time complexity of binary search is O logN

[algorithm] how to calculate binary search complexity

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