complexity theory-sorting algorithm

Question

I'am taking a course in complexity theory,and so it's need some mathematical background which i have a problem,. so while i'am trying to do some practicing i stuck in the bellow example

1)  for (i = 1; i < n; i++) {
2)      SmallPos = i;
3)      Smallest = Array[SmallPos];
4)      for (j = i+1; j <= n; j++)   
5)          if (Array[j] < Smallest) {
6)              SmallPos = j;
7)              Smallest = Array[SmallPos]  
            }   
8)       Array[SmallPos] = Array[i];
9)       Array[i] = Smallest;
    }

Thus, the total computing time is:

T(n)    = (n) + 4(n-1) + n(n+1)/2 – 1 + 3[n(n-1) / 2]
        = n  + 4n - 4 + (n^2 + n)/2 – 1 + (3n^2 - 3n) / 2
        = 5n - 5 + (4n2 - 2n) / 2
        = 5n - 5 + 2n^2 - n
        = 2n^2 + 4n - 5 
        = O(n^2)

and what i don't understand or confused about line 4 analyzed to n(n+1)/2 – 1, and line 5 3[n(n-1) / 2]. i knew that the sum of positive series is =n(first+last)/2 ,but when i tried to calculate it as i understand it it gives me different result. i calculate for line no 4 so it shoulb be =n((n-1)+2)/2 according to n(first+last)/2 ,but here it's n(n+1)/2 – 1. and same for 3[n(n-1) / 2].....i don't understand this too

also here's what is written in the analysis it could help if anyone can explain to me,

Statement 1 is executed n times (n - 1 + 1); statements 2, 3, 8, and 9 (each representing O(1) time) are executed n - 1 times each, once on each pass through the outer loop. On the first pass through this loop with i = 1, statement 4 is executed n times; statement 5 is executed n - 1 times, and assuming a worst case where the elements of the array are in descending order, statements 6 and 7 (each O(1) time) are executed n - 1 times.

On the second pass through the outer loop with i = 2, statement 4 is executed n - 1 times and statements 5, 6, and 7 are executed n - 2 times, etc. Thus, statement 4 is executed (n) + (n-1) +... + 2 times and statements 5, 6, and 7 are executed (n-1) + (n-2) + ... + 2 + 1 times. The first sum is equal to n(n+1)/2 - 1, and the second is equal to n(n-1)/2.

Thus, the total computing time is:

T(n)    = (n) + 4(n-1) + n(n+1)/2 – 1 + 3[n(n-1) / 2]
        = n  + 4n - 4 + (n^2 + n)/2 – 1 + (3n^2 - 3n) / 2
        = 5n - 5 + (4n2 - 2n) / 2
        = 5n - 5 + 2n^2 - n
        = 2n^2 + 4n - 5 
        = O(n^2)

here's the link for the file containing this example: http://www.google.com.eg/url?sa=t&rct=j&q=Consider+the+sorting+algorithm+shown+below.++Find+the+number+of+instructions+executed+&source=web&cd=1&cad=rja&ved=0CB8QFjAA&url=http%3A%2F%2Fgcu.googlecode.com%2Ffiles%2FAnalysis%2520of%2520Algorithms%2520I.doc&ei=3H5wUNiOINDLswaO3ICYBQ&usg=AFQjCNEBqgrtQldfp6eqdfSY_EFKOe76yg

`T(n) = O(n^2)` is not syntatically correct, it should be: `T(n) is in O(n^2)`, since O(n^2) is a *set* of functions. — amit, Oct 06 '12 at 20:58

Origin · Accepted Answer · 2012-10-06T22:16:29.163

1

line 4: as the analysis says, it is executed n+(n-1)+...+2 times. This is a sum of (n-1) terms. In the formula you use, n(first+last)/2, n represents the number of terms. If you apply the formula to your sequence of n-1 terms, then it should be (n-1)((n)+(2))/2=(n²+n-2)/2=n(n+1)/2-1.

line 5: the same formula can be used. As the analysis says, you have to calculate (n-1)+...+1. This is a sum of n-1 terms, with the first and last being n-1 and 1. The sum is given by (n-1)(n-1+1)/2. The factor 3 is from the 3 lines (5,6,7) that are each being done (n-1)(n)/2 times

edited Oct 06 '12 at 22:16

answered Oct 06 '12 at 22:08

Origin

2,009
5
19
19

now i understand line 4 how it's calculated ,but I've question in line 5 why a sum of n-1 terms for (n-1)+...+1. why it's not n+(n-1)+...+2,since this is in the for loop starting at 2 to n. – HATEM EL-AZAB Oct 07 '12 at 06:48
also in line 4 i think something is missed there must be one additional check is required which will terminate the loop.where it is. – HATEM EL-AZAB Oct 07 '12 at 06:52
1

line 5: in your first inner loop, j will assume all values for 2 to n. For each value in that loop, you will execute line 5 1 time. In the formula `(n-1)+...+1` you are actually calculating how many times (in total) line 5 is executed by counting the number of times it is executed during each loop: n-1 times in the first loop (for i=1), n-2 times for i=2... If you do `n+n-1+...+2` you are summing all j-values, while you want to be summing the number of times line 5 is executed. – Origin Oct 07 '12 at 09:13
1

Concerning the for loop evaluation, it is indeed true the boolean expression will be checked each time the loop is performed (when it yields true) + 1 time at the end (when it will yield false). However, in analysing complexity, every statement is considered the same (a simple increment line adds 1 to the total count, the same as an exponentiation or even more complicated things) while they are clearly not the same. Thats why, usually, we are only interested in the O(n)-nature and dont care about that 1 extra check (because breaking down each statement would leads us too far) – Origin Oct 07 '12 at 09:26
what you mean is for example if i've a sequence of positive numbers 1,2,3,4,5 : for i=1->j=2,3,4,5, for i=2->j=3,4,5, for i=3->j=4,5,and for i=4-> j=5 so the line 5 will be exceuted 4+3+2+1=10 times and that's matched the sequence (n-1)+...+1,on the opposite side if i considered the sequence n+n-1+...+2 ,it will be the sum of 2+3+4+5=14 times which is not true....is that what u meant. – HATEM EL-AZAB Oct 07 '12 at 11:40
Yes, that exactly what i'm talking about – Origin Oct 07 '12 at 12:25

complexity theory-sorting algorithm

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