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Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Introduction • Sorting is the process of arranging the elements of a set in order. ALGORITHMS and COMPLEXITY THEORY • Sorting was one of the ﬁrst problem intensively studied by computer Lecture 6 scientists. Sorting Algorithms 1 • There are several methods of sorting , and most of them use the Divide and Conquer( to be studied later) techniques. • some sorting algorithms: Insertion sort, selection sort, Mergesort, Bubble sort, Radix sort , Quicksort. Lecture 6: Sorting Algorithms 1 1 Lecture 6: Sorting Algorithms 1 2 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Selection Sort If we want to sort a list of n elements, swap the smallest element with one at the ﬁrst position. Apply this Selection Sort Algorithm algorithm recursively to the n-1 remaining elements(placed from the 2nd to the nth positions). repeat the process until the list is reduce to a single element. SelectionSort(int[] t, int n) int i,j,min,q L I C E N C E begin We will have the following sequence of execution for i = 1 to n-1 min = i gives for j = i+1 to n L I C E N C E C I L E N C E if t[j] < t[min] then min = j C I L E N C E gives C C L E N I E endif endfor C C L E N I E gives C C E L N I E q = t[min] t[min]= t[i] C C E L N I E gives C C E E N I L t[i] = q end C C E E N I L gives C C E E I N L The maximum complexity of this algorithm is O(n2 ). C C E E I N L gives C C E E I L N Lecture 6: Sorting Algorithms 1 3 Lecture 6: Sorting Algorithms 1 4 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Bubble Sort If we want to sort a list of n elements, Place the biggest element at the end. Apply the algorithm recursively to the n-1 ﬁrst elements. Repeat the process until the list is reduce to a single element. For Bubble sort algorithm instance, sort the table: We want to sort the list C K G L B BUBBLESORT(int [] t, int n) int i,j,p step array operation Resulting array for i = n downto 2 do 1rst CKGLB No change CKGLB for j = 1 to i-1 do if t[j] > t[j + 1] then CKGLB K > G , swap K and G CGKLB p= t[j] CGKLB No change CGKLB t[j]= t[j+1] CGKLB L > B , swap L and B CGKBL t[j+1]= p 2nd CGKBL No change CGKBL end if CGKBL No Change CGKBL end for end for CGKBL K > B , swap K and B CGBKL 3rd CGBKL No change CGBKL Bubblesort is an O(n2 ) algorithm. CGBKL G > B , swap G and B CBGKL 4th CBGKL C > B , swap C and B BCGKL BCGKL end Lecture 6: Sorting Algorithms 1 5 Lecture 6: Sorting Algorithms 1 6 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 sequential Insertion sort We want to sort the following list: C K G A B step array Operation resulting array Insertion sort 1rst CKGAB Initialization, No Change CKGAB 2nd CKGAB Save the value(K) at the 2nd position Compare K to C, No change CKGAB • more natural way to sort. 3rd CKGAB Save the value(G) at the 3rd position Compare G to K, Move K to the right CKKAB • Insertion sort inserts each item into its proper place in the ﬁnal list. CKKAB compare G to C, No change CKKAB CKKAB insert G at the second position CGKAB Save the value(A) at the 4th position • The simplest implementation of this requires two list structures: the 4th CGKAB compare A to K, move K to the right CGKKB source list and the ﬁnal list, into which the sorted items are inserted. CGKKB compare A to G, move G to the right CGGKB CGGKB compare A to C, move C to the right CCGKB • But we can also work "in place" with one list. CCGKB insert A at the ﬁrst position ACGKB 5th ACGKB Save the value(B) at the 5th position compare B and K, move K to the right ACGKK ACGKK compare B to G, move G to the right ACGGK ACGGK compare B to C, move C to the right ACCGK ACCGK compare B to A, No change ACCGK ACCGK insert B at the second position ABCGK Lecture 6: Sorting Algorithms 1 7 Lecture 6: Sorting Algorithms 1 8 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Binary Insertion sort We want to use binary insertion sort to sort the following array C K G A B sequential Insertion sort(cont’d) (1) A single element C is already sorted (2) We will then try to insert K in the array which consists of a single element C, as K is greater than C , we will then have the array C K as SEQINSERTIONSORT(int [] t;int n) result, this will be done after one comparison , to know if we will insert K before or after C. int i,j,q (3) We want then to insert G in the array C K ; instead of trying to ﬁnd the place where to insert from C, we will BEGIN – split the array into two parts C and K , FOR i := 2 TO n DO – compare G to C , as G is greater to C , G can only be inserted in the second part of the list, which consists of a single element K which is greater than G , j := i-1; q = t[i]; – then G will be inserted between C and K, which results in the array C G K. WHILE((j >= 0) AND ( t[j] >q )) DO (4) We then have to insert A in the array C G K: t[j+1] = t[j] – split C G K into two parts C G and K, j := j - 1 – compare A to the last element (G) of C G, as G is greater than A, we will then recursively , using the same method, search for the END WHILE position where to insert A in C G. split the list into two parts C and G , compare A to C , as A is smaller than C , A will be inserted t[j+1] = q in the ﬁrst part of the array, which consists of a single element C ( and C is greater than A) , means A will be inserted at the END FOR beginning of the array C G, which gives to the array A C G. – We will then have at this step the list A C G K END (5) We then have to insert B in the array A C G K: – split A C G K into two parts A C and G K, This algorithm is an O(n2 ) – compare B to the last element (C) of A C, as C is greater than B, we will then recursively , search for the position where to insert B in A C. split the array into two parts A and C , compare B to A, as B is greater than A , B will be inserted between A and C, which gives the array A B C. – We will then have as result the array A B C G K Lecture 6: Sorting Algorithms 1 9 Lecture 6: Sorting Algorithms 1 10 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Binary search Binary Insertion sort int binsearch (int [] t, int Key ,int istart, int iend) SORTINSERTBIN (int [] t, int n) int j,k,position; int i,j,Key ,pos; BEGIN BEGIN position = istart; j=iend FOR i= 2 to n DO WHILE j > position DO Key = t[i] k = (position+j) div 2 pos = binsearch (t,Key,1,i-1) if Key <= t[k] then j:= k FOR j =i downto pos+ 1 DO t[j] := t[j-1] else position := k + 1 t[pos]= Key END WHILE END FOR if (Key > t[position]) then position := position + 1 END return(position) end Lecture 6: Sorting Algorithms 1 11 Lecture 6: Sorting Algorithms 1 12 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Complexity of binary sort complexity of Binary sort : worst case suppose we need m iterations to ﬁnd the right position for the element we α(i) = (i-1)(e+a) +a where a is time spent arithmetic operation and e for want to insert: store then n 2m−1 ≤ n − 1 < 2m then m = log2 (n-1) if n is the size of the array T(n) = (A1 i + Blog2 (i − 1) + C1 ) The running time of binary insertion sort is: i=2 n n n T(n) = (A + Blog2 (i − 1) + C + α(i)) but log2 i < i then T (n) ≤ (A1 i + B(i − 1) + C1 ) = (A2 i + B2 ) i=2 i=2 i=2 T(n) ≤ A2 (n − 1) − B2 + A2 n(n+1) 2 Were α(n) is the time spent to insert new item. thus T(n) = O(n2 ). Lecture 6: Sorting Algorithms 1 13 Lecture 6: Sorting Algorithms 1 14 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Algorithms and Complexity Theory UKZN - Computer Science - 2010 complexity of Binary sort : best case complexity of Binary sort : average case α(i) = 0 we have: n n−1 i(i−1) 0, 1, 2 , 3, . . . i-1 shifts and we have i type of shifts ( from 0 to i-1) T(n) = (A + Blog2 (i − 1)) = (A + Blog2 (i)) 2 i(i−1) i=2 i=1 then in average we have αm (i) = = i−1 2 2i n n−1 n i−1 T(n)=A(n − 1) + B log2 (i) = A(n − 1) + Blog2 ( i) T(n) = (A + Blog2 (i − 1) + C + ) i=2 2 i=1 i=1 n−1 √ as log2 n < n then but , i=1 i =(n − 1)! and n! = 2πn( n n [1 + O( n )] e 1 Thus T(n) = O(nlogn). T(n) = O(n2 ) Lecture 6: Sorting Algorithms 1 15 Lecture 6: Sorting Algorithms 1 16 Algorithms and Complexity Theory UKZN - Computer Science - 2010 Your To do List • Read Chapter 3 [ Sorting Algorithms] of your notes. • Prepare your tutorial 3 Lecture 6: Sorting Algorithms 1 17

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