Question

Problem 4: Sorting and Searching (55 points) a) (20 points) Scarch for the character K using the binary search algorithm on the following array of characters: KM 2. R For each iteration of binary search use the table below to list: (a) the left index and (b) the right index of the array that denote the region of the array that is still being searched, (c) the middle point index of the array, and (d) the number of character-to-character comparisons made uanine the search at lines 8 and 10 of the algorithm at the back of the exam iteration Hcompares left middle right b) (21 points) List the resulting array after each iteration of the outer loop of the insertion so algorithm at the back of the exam. Indicate the number of character-to-character compansc (line 6) made for each iteration order): GFTBRKMC sort the following array of letters (sort into alphabetical

Answer 1

Solution - Part a: العالم لعام سالم || || 7 0 1 2 3 4 5 6 Binary Search Algorithm : 1 e=0 11 the starting ind m = 8 ll last index mid (ltm)/2 while(l cim) te if (array (mid) = a scarich) { printf ("Element found" ); else if (array (mid] search) __lz mil +L; break; } else m? mid - mid (tm) 12; if (lim) printf ("Element not found Iteration 1: left index zo right inden 28 middle z ot 8 4 + 12 ansay (middle) z avay (4) 2M & . M 7K - mamid-1

Iteration 2 left index=0 inden 3 middle = 0+3 21:5=1 arvey (middle. 13 asicy C1] =E : FLK ez mid + 1 ez 1+1=2 Iteration 3 left index 2 night inden 3 middle z 273 32.5 = 2 array(middle) zarray (2) 2 G GLK 2 le mid til z 2 + 23 2 Iteration 4 left index=3 right index=3 middle - 3+3 23 arsay (msiddle) zavray (312K - The talele will be: Iteration left right middle 1 o 8 4 2 M o no 3 latalan 2 3 2 3 3 4 3 #compares 2 2 2 1 wN WWW

Part(e) - Insertion Sort Algorithm : for ( k = 1 king k++) do ja k-1 while ((charr Cil) && (>=0)) do arr ( ; + 1) = arr Ci j z j-1 enduchile are (j =c endfor ginen array: GET BIRIKMIC 0 1 2 3 4 5 6 7 compares Iteration 1: F G 1 T B 2 3 C R 4 K 5 M 6 0 7 Iteration 2: C F G I 1 2 B 3 R K 5 M 6 0 4 7 Iteration 3: B F G T R K m c 3 Iteration 4: B 0 F G R T K Me 2 3 4 5 6 7 1 I Iteration 5: B o F G K R T 2 3 4 5 MC 6 7 2 Iteration 6: - B o F G K 2 3 M R 4 5 TC 6 7 1 2 . Thration 7: b C F G K M R I