A. Take the Sub-String and Remember the First Character

Julian Corlaci CSC 255 - 702 Assignment # 5

Chapter 4

20. a. Take the sub-string and remember the first character. b. Compare the remembered character from sub-string with the next (of first) character from string 1 c. If characters match then remember the next character from sub-string 1 and compare it to the next character in string 1. d. If characters match then repeat step 3 until there are no character left in sub-string. Report True. e. If characters don’t match, stop and report False. f. If characters don’t match, compare remembered character to next character in string 1. g. If characters match, proceed in steps 3-5 h. If characters don’t match, proceed in step 6 until no character are left in sub-string or string1. If this occurs, report False. procedure SubStringSearch(FirstString, SecondString) P = 0; Success = false; while (P + length of FirstString) <= (length of SecondString) and Success = false) do { N = 1; while (P + Nth character in SecondString = character in FirstString) do (N = N + 1); if (N = length of FirstString) then (Success = true))) P = P + 1}

21.

The body of the loop is everything between "(" and ")". The initialization step is "assign Current the value 1". The modification step is "Current <- Last + Temp". The test step is "Current < 100" in the loop header.

Program prints the Fibonacci sequence < 100: 1 1 2 3 5 8 13 21 34 55 89 22.

1 1 2 3 4 5 8 13 21 34 55 89

23. procedure MysteryWrite (Last, Current) if (Current < 100) then (Temp = Current + Last; apply MysteryWrite to the values Current and Temp; print the value assign to Current)

For reverse order the print statement has been moved to the end of the then clause.

24.

A, B, C, D, E, F, G, H, I, J, K, L, M, N, O

First, the middle element is found. The index of the last element is 15, so the middle position is (1+15)/2 = 8. The 8th element is H. Since J is larger than H, we do a binary search in the upper half (from position 9 through position 15). The middle element in this part is (9+15)/2 = 12. The 12th element is L. But, J is smaller than L, so now we apply the binary search to the list between positions (9 through 11). The middle position here is (9+11)/2 = 10. The 10th position is J, so we stop.

In the case of "Z" (which is not in the list), we again find the middle element (at position 8), which is H. Since Z is bigger than H, we do a binary search in the upper half (from position 9 to position 15). In this case, the new middle is position (9+15)/2 = 12, which is L. Z is still bigger than L, so we again do a binary search, this time from position 13 through position 15. This leads to the interrogation of the element at location (13+15)/2 = 14, which is N. Z is still bigger than N, so we do another binary search from location 15 to location 15, the exact element O. Because O is less than Z and we have run out of elements to consider, we end in failure.

List of items examined by binary search for J and Z is:

Search for "J" ==> H L J (success) Search for "Z" ==> H L N O (fail)

25.

Sequential Search: least case: 1 worst case : 6000 add = 1+2+3+ ... +5999+6000 average = add/6000 = 18003000/6000 = 3000.5 Binary Search: least case: 1 worst case : 12

28. procedure Euclidean if (neither X nor Y is 0) then (divide the larger of X and Y by the smaller; X = the value of the remainder; Y = the value of the divisor; and apply Euclidean to X and Y)

33. procedure Factorial (Value) if (Value is 0) then (Return 1 as the answer) else (Apply Factorial to (Value - 1), multiply the result by Value, and X = the value of this product), return the number assigned to X as the answer)

43.

Binary Search: least case: 1 worst case: 11

Sequential Search: least case: 1 worst case: 4000 the largest number of entries interrogated: 4000

44.

The addition of two n-digit values would require 2n – 1 additions. Thus, addition would be in THETA(n). To multiply two n-digit values requires n2 multiplications. Thus, multiplication is in THETA(n2).

Chapter 7 5.

Using a contiguous block of memory cells in which the planes in the three-dimensional array are stored as consecutive two-dimensional planes in row major order. If each entry requires one memory cell and the first cell of the block is at address x, then the (i, j, k)th entry will be located at the address x + (R * C)(i - 1) + C(j - 1) + (k - 1), where R is the number of rows and C is the number of columns.

The list currently spells JANE. To spell JEAN, it should be changed to:

Address Contents Address Contents Address Contents 40 N 44 J 48 M 41 00 45 46 49 42 42 I 46 E 50 A 43 40 47 50 51 40

10.

Routine 1 is correct. The problem with Routine 2 is that it changes the value of the pointer field in PreviousEntry to be that of NewEntry first, and then copy this same value to the pointer field of NewEntry. Thus, the pointer field in NewEntry points to NewEntry or in other words New Entry pointing to itself.

12.

Combining two sorted lists into a single sorted list is called merging. set countS1, countS2 and countD all to zero while countS1 < lengthS1 and countS2 < lengthS2 if source1 [countS1] < source2 [countS2] then set dest [countD] to source1 [countS1] increment countS1 else set dest [countD] to source2 [countS2] increment countS2 end while increment countD if countS1 = lengthS1 then while countS2 < lengthS2 set dest [countD] to source2 [countS2] increment countS2 increment countD end while else while countS1 < lengthS1 set dest [countD] to source1 [countS1] increment countS1 increment countD end while end if

In the case of linked lists, the merge can be performed without producing a third list. The entries can remain in the same places in memory while their pointers are altered.

13.

Previous = NIL; Current = value of the head pointer; while (Current not NIL) do (Next = value pointed to by Current; pointer in the current entry = Previous; Previous = Current; Current = Next) head pointer = Previous