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Understanding Recursive Functions and Palindrome Checks in Programming

This resource offers insights into recursive programming techniques used to solve problems, such as palindrome validation and summing arrays using divide-and-conquer methods. Various examples illustrate how recursion works, including tracing palindromes with sample inputs, summing array elements, and counting odd numbers in a list. The document also explains memory allocation for 2D arrays and how to initialize and manipulate data using recursive functions effectively. Perfect for beginners looking to deepen their understanding of recursion.

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Understanding Recursive Functions and Palindrome Checks in Programming

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  1. Exercise 2 CS 110 Sum 09

  2. what (104,16) X=104 y=16 X=88 y=16 X=72 y=16 X=56 y=16 X=40 y=16 X=24 y=16 X=8 y=16 X=8 y=8

  3. what (15,35) X=15 y=35 X=15 y=20 X=15 y=5 X=10 y=5 X=5 y=5

  4. what (63,18) X=63 y=18 X=45 y=18 X=27 y=18 X=9 y=18 X=9 y=9

  5. Palindrome bool palindrome(string s, int first, int last) { if(first>=last) return true; else if(s[first] != s[last]) return false; else return palindrome(s, first+1, last-1); }

  6. Tracing Palindrome “dad” and “daa” “dad” “dabd” true false F=0, L=2 F=0, L=3 true false F=1, L=2 F=1, L=1

  7. n = process (a , 12 , 10 ); 05 3 0 17 10 9 20 0 1 2 3 710 1+ n=12, x=10 A[10] = 0 n = 4 1+ 1+ X=9 A[9] = 0 + X=11 A[11] = 0 0 0 0 + 1+ + X=8 A[8] = 0 X=10 X=10 X=12 0 0 X=7 X=9

  8. long int z = Func( 5 , 3 ); z=10 n=5, m=3 6 4 n=4, m=3 + n=4, m=2 3 3 3 + 1 + n=3, m=3 n=3, m=2 n=3, m=2 n=3, m=1 1 1 1 2 2 + + 2 + n=2, m=2 n=2, m=1 n=2, m=2 n=2, m=1 n=2, m=1 n=2, m=0 1 1 1 + 1 1 1 + + 1,1 1,0 1,1 1,0 1,1 1,0

  9. sum (Divide & Conquer) int sum( int a[ ], int s, int e) { if ( s > e ) return 0; else if ( s == e ) return a [ s ]; else { int m = ( s + e ) / 2; return sum ( a, s, m ) + sum ( a, m+1, e ); } }

  10. sum (Divide & Conquer) 29 s=0, e=6 M=3 14 15 s=0, e=3 M=1 + s=4, e=6 M=5 9 12 A[6] + 6 + s=0, e=1 M=0 s=2, e=3 M=2 s=4, e=5 M=4 s=6, e=6 A[0] A[4] A[1] A[2] A[3] + + + A[5] s=0,e=0 s=1,e=1 s=2,e=2 s=3,e=3 s=4,e=4 s=5,e=5

  11. oddcount int oddcount(int A[ ], int s, int e) { if (s > e) return 0; else if ( A[s]%2 ) return 1 + oddcount (A, s+1, e); else return oddcount (A, s+1, e); }

  12. asciisum (Blanks included) int asciisum (string s, int first, int last) { if( first > last) return 0; return int ( s [first] ) + asciisum (s, first + 1, last); }

  13. asciisum (Blanksexcluded) int asciisum (string s, int first, int last) { if( first > last) return 0; if( s[first] == ‘ ‘) return asciisum (s, first + 1, last); return int ( s [first] ) + asciisum (s, first + 1, last); }

  14. Assignment 3 • int map ( int i, int j ) { return i * M + j; } • void zeros ( int * count ); • void zeros ( int * & count ); // if allocate memory inside zeros • void move ( int & ibug, int & jbug, int imove[ ], int jmove[ ] ) • bool inside ( int i, int j ); • void display ( int * count, int total ) • bool done ( int * count );

  15. Assignment 3 //in main function //Initialize int imove[9]={0,-1,0,1,1,1,0,-1,-1}; int jmove[9]={0,1,1,1,0,-1,-1,-1,0}; int sum=0;

  16. Assignment 3 //should be entered by user or randomly generated with validation int ibug=10; int jbug=12; M = 15; N = 15;

  17. Assignment 3 srand ((unsigned) time (NULL)); // Memory allocation int * count = new int [ M * N ]; //initial bug location count[map(ibug, jbug)] = 1;

  18. Assignment 3 Loop incrementing sum (total) call move function to move the bug increment array at that location until done call display function Free memory //delete [ ] count;

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