ALGORITHM 2-1 Solution for Exercise 4

Transcription

1 Chapter 2 Recursion Exercises 1. a. 3 * 4 = 12 b. (2 * (2 * fun1(0) + 7) + 7) = (2 * (2 * (3 * 0) + 7) + 7) = 21 c. (2 * (2 * fun1(2) + 7) + 7) = (2 * (2 * (3 * 2) + 7) + 7) = a. 3 b. (fun2(2, 6) + 3) = ( 3 + 3) = 0 c. fun2(12, 6) + 3) = ((fun2(6, 9) + 6) + 3) = ( ) = 6 3. a. 1 b. 1 c. (4 * fun3(5, 7)) = (4 * (5 * fun3(6, 7))) = 4 * (5 * (6 * fun3 (7, 7))) = 120 d See Algorithm 2.1. ALGORITHM 2-1 Solution for Exercise 4 algorithm squareroot (num, ans, tol) This algorithm computes the square root of a number using Newton's method and returns the computed square root. Pre num is number for square root ans is estimate of the square root initially the same as num tol is the tolerance (accuracy) desired Post square root calculated and returned Return square root 1 if (absolute value of (ans * ans num) <= tol) 1 return ans 2 else 1 return (squareroot (num, (ans * ans + num) / (2 * ans), tol) end squareroot 5. See Algorithm 2.2. ALGORITHM 2-2 Solution for Exercise 5 algorithm Combination (n, k) This algorithm calculates the combination of n objects taken k at a time 7

2 8 Chapter 2 Recursion ALGORITHM 2-2 Solution for Exercise 5 (continued) Pre n and k are positive integers and n >= k Post number of combination calculated and returned 1 if ((k is 0) OR (n equals k)) 1 return 1 2 else if (n > k AND k > 0) 1 return (C (n 1, k) + C (n 1, k 1)) end Combination 6. See Algorithm 2.3. ALGORITHM 2-3 Solution to Exercise 6 algorithm Ackerman (M, N) This algorithm calculates the Ackerman number Pre M an N are integers Post Ackerman number calculated and returned 1 if (M equal 0) 1 return (N + 1) 2 else if ((N equal 0) and (M > 0)) 1 return (Ackerman (M 1, 1)) 3 else 1 return (Ackerman (M 1, Ackerman (M, N 1))) end Ackerman Problems 7. See Algorithm 2.4. ALGORITHM 2-4 Solution for Problem 7 algorithm lengthoflist (list) This algorithm calculates and returns the length of a list Pre list is a valid list Post length is returned 1 if (empty list) 1 return 0 2 else 1 return (1 + lengthoflist (list minus first element)) end lengthoflist 8. See Algorithm 2.5. ALGORITHM 2-5 Solution for Problem 8 algorithm strtoint (string, num) This algorithm converts a number string to an integer. Pre string is an array of characters (all digits) num is address for converted number (initial call uses num of 0) Post the converted integer value in num

3 Problems 9 ALGORITHM 2-5 Solution for Problem 8 (continued) 1 return 0 3 set num to ((num * 10) + value of right character) 4 strtoint (string minus right character, num) end strtoint 9. See Algorithm 2.6. ALGORITHM 2-6 Solution to Problem 9 algorithm addn (n) This algorithm adds the first n elements of the series 1 + 1/2 + 1/3 + 1/4 + 1/ /n Pre n is an integer Post the sum is returned 1 if (n is 1) 1 return 1 2 else 1 return (1 / n + addn (n 1)) end addn 10. See ALGORITHM 2.7. ALGORITHM 2-7 Solution to Problem 10 algorithm palindrome (string) This algorithm will determine if a string is a palindrome Pre string is the string to be checked Post string has been checked Return true if palindrome false if not palindrome 1 return true 3 if (string only one character) 2 return true 5 if (first character not alphabetic) 1 return palindrom (string minus first character) 6 end if 7 if (last character not alphabetic) 1 return palindrom (string minus last character) 8 end if 9 if (lowercase first equals lowercase last) 1 return palindrom(string minus first and last characters) 10 end if 11 return false end palindrome

4 10 Chapter 2 Recursion 11. See Algorithm 2.8. ALGORITHM 2-8 Solution to Problem 11 algorithm checkcharacter (string, c) This algorithm checks if a character is in a string. Pre string is the string to search c is the character to look for Post returns true if found, false if not return false 3 if (first character equals c) 1 return true 5 return checkcharacter (string minus first character, c) end checkcharacter 12. See Algorithm 2.9. ALGORITHM 2-9 Solution to Problem 12 algorithm countchar (string, c) This algorithm counts all occurrences of a specified character in a string. Pre string is the string to search c is the character to count Post character count returned 1 return 0 3 if (first character equals c) 1 return (1 + countchar (string minus first character, c)) 5 return (countchar (string minus first character, c)) end countchar 13. See Algorithm ALGORITHM 2-10 Solution for Problem 13 algorithm removecharacter (string, newstring, c) This algorithm removes all occurrences of a specified character in a string and returns new string. Pre string is the string to searched newstring is empty string c is character to be deleted Post return new string with c-characters deleted 1 return newstring 3 if (first character in string not equal c) 1 append first character to newstring

5 Problems 11 ALGORITHM 2-10 Solution for Problem 13 (continued) 5 return removechar(string minus first character, newstring, c) 6 end if end removecharacter 14. See Algorithm ALGORITHM 2-11 Solution to Problem 14 algorithm findsubstring (string, substr, loclist) Find location of all occurrences of a specified substring in a string. Pre string is the string to search substr is substring to be located loclist is list to hold matching positions Post returns the count 1 return 3 if (substring matches beginning of string) 1 append location to loclist 2 findsubstring (string minus substring, substr, loclist) 4 else 1 findsubstring (string minus first character, substr, loclist) 5 end if end findsubstring 15. See Algorithm ALGORITHM 2-12 Solution to Problem 15 algorithm integertobinary (num, binary) This algorithm changes an integer to a binary number stored in a string. Pre num is an integer binary is a null string large enough to hold (size of integer * 8 + 1) characters must be zero on initial call Post binary number is stored in binary 1 if (num equals 0) 1 return 3 integertobinary(num / 2, binary) 4 if (num modulo 2 equals 0) 1 append (binary, "0") 5 else 1 append (binary, "1") 6 end if 7 return end integertobinary

6 12 Chapter 2 Recursion 16. See Program 2-1. PROGRAM 2-1 Solution to Problem 16 /* This program tests the calculation of a square root using Newton's method. Written by: Date: #include <stdio.h> #include <stdlib.h> #include <math.h> // Prototype Declarations float squareroot (float num, float ans, float tol); int main (void) // Local Definitions float result; result = squareroot (125, 11, 0.01); printf ("The square root of 125 is %f\n", result); result = squareroot (763, 27, 0.01); printf ("The square root of 763 is %f\n", result); result = squareroot (997, 31, 0.01); printf ("The square root of 997 is %f\n", result); return 0; } // main /* ==================== squareroot ==================== This function calculates the squareroot of a number using Newton's method Pre num is the number whose root is desired ans is an estimate of the square root tol is the tolerance desired Post the squareroot is returned float squareroot (float num, float ans, float tol) if (fabs (ans * ans - num) <= tol) return (ans); else return (squareroot (num, (num + ans * ans) / (2 * ans), tol)); } // squareroot 17. See Algorithm ALGORITHM 2-13 Solution to Problem 17 algorithm stringreverse This is a recursive algorithm that reads a string of characters from the keyboard and prints them reversed. Pre nothing

7 Problems 13 ALGORITHM 2-13 Solution to Problem 17 (continued) Post string is printed reversed 1 read (character) 2 if (end of input) 1 return 4 stringreverse 5 print (character) 6 return end stringreverse 18. See Program 2-2 PROGRAM 2-2 Solution to Problem 18 /* This program tests the combination function. Written by: Date: #include<stdio.h> #include<math.h> // Prototype Statements int combo (int n, int k); int main (void) printf ("The combination of 10 objects taken 3 " "at a time is: %d\n", combo(10, 3)); return 0; } // main /* ==================== combo ==================== This function calculates the combination of n objects taken k at a time Pre n and k are positive integers and n >= k Post the number of combination is returned int combo (int n, int k) if ((k == 0) (n == k)) return 1; else return (combo (n - 1, k) + combo (n - 1, k - 1)); } // combo 19. See Program 2-3. PROGRAM 2-3 Solution to Problem 19 /* This program tests the Ackerman number function. Written by: Date: #include <stdio.h> #include <math.h> // Prototype Statements int Ackerman (int M, int N);

8 14 Chapter 2 Recursion PROGRAM 2-3 Solution to Problem 19 (continued) int main (void) printf ("The Ackerman number for 2, 3 is %d\n", Ackerman (2, 3)); printf ("The Ackerman number for 2, 5 is %d\n", Ackerman (2, 5)); printf ("The Ackerman number for 0, 3 is %d\n", Ackerman (0, 3)); printf ("The Ackerman number for 3, 0 is %d\n", Ackerman (3, 0)); return 0; } // main /* ==================== Ackerman ==================== This function calculates Ackerman s number. Pre M and N are integers Post Ackerman's number is returned int Ackerman (int M, int N) if (M == 0) return (N + 1); else if ((N == 0) && (M > 0)) return (Ackerman (M - 1, 1)); else return (Ackerman (M - 1, Ackerman (M, N - 1))); } // Ackerman 20. See Program 2-4. PROGRAM 2-4 Solution to Problem 20 /* =================== printreverse =================== This algorithm prints the elements of a linked list in reverse order Pre plist is a node pointer Post list is printed in reverse order void printreverse (NODE *plist) if (plist == NULL) return; printreverse (plist >next); printf ("%d\n", plist >data); return; } // printreverse