Recursion. Example R1
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1 Recursion Certain computer problems are solved by repeating the execution of one or more statements a certain number of times. So far, we have implemented the repetition of one or more statements by using iterations or loops: while loop, for loop, or do-while loop. Recursion, the process by which a function calls itself, is another way to implement the repetition of the execution of one or more statements. A function that calls itself is said to be defined recursively. You define a function recursively by providing the following two parts in its definition: 1. An anchor or base case or stopping case: in this part of the definition of the function, you specify the value of the function (output/returned) for one or more values of the arguments. 2. An inductive or recursive case: in this part of the definition of the function, you specify how the value of the function (output/returned) for the current value of the argument(s) can be obtained from the value(s) of the function for the previous value(s) of the arguments. Example R1 A function that receives as argument a positive integer value and outputs its digits in reverse order does the following: if the argument is 1234, the output will be: ; and if the argument is 3 the output will be 3. This function is written using a while loop as follows: void iwritereverse( unsigned num ) while( num!= 0 ) // condition: num cout << num % 10; num = num / 10; 2011 Gilbert Ndjatou Page 269
2 It is written using recursion as follows: void rwritereverse( unsigned num ) if ( num < 10 ) cout << num; else cout << num % 10; rwritereverse( num / 10); // stopping case // inductive/recursive case Example R2 A function that receives as argument a positive integer value and outputs its digits one per line does the following: if the argument is 1234, the output will be: if the argument is 3 the output will be: 3 The definition of this function using a while loop requires the use of two loops of the use of an array or a stack to hold the digits of a number. It is written using recursion as follows: void rwritedigits( unsigned num ) if ( num < 10 ) cout << num; else rwritedigits( num / 10); cout << num % 10 << endl; // stopping case // inductive/recursive case 2011 Gilbert Ndjatou Page 270
3 Example R3 Recursive definition of the function to compute the n th power of a double precision value: double rpower( double num, unsigned expo) if ( expo == 0 ) return( 1.0 ); else return( num * rpower( num, expo 1 ) ); // stopping case // inductive/recursive case Example R4 Recursive definition of the function to compute the n! (n factorial) for a positive integer value n: Note: 0! = 1 and n! = n * (n-1) * (n-2) * (n-3) *... 3 * 2 * 1 for n > 0 unsigned factorial( unsigned num) if ( num < 2 ) return( 1 ); else return( num * factorial( num 1 ) ); // stopping case // inductive/recursive case Example R5 Recursive definition of the function to compute the n th (n> 2 ) Fibonacci number. Fibonacci Numbers: 1, 1, 2, 3, 5, 8, 13, Note: Fibonacci( 1 ) = 1, Fibonacci(2 ) = 1, Fibonacci( n ) = Fibonacci(n-1) + Fibonacci(n-2) for n>2 unsigned fibonacci( unsigned num) if ( num < = 2 ) // stopping case return( 1 ); else return( Fibonacci(num 1) + fibonacci( num 2 ) ); // inductive/recursive case 2011 Gilbert Ndjatou Page 271
4 Exercise R1 1. Write a recursive definition of the function void writeasterisks( int num ) that receives as argument a positive integer value and outputs that number of asterisks * to the screen all on one line. 2. Write the recursive definition of the function void writersequence( int num ) that receives as argument a positive integer value n and outputs the sequence: n n-1 n Write the recursive definition of the function void writesequence( int num ) that receives as argument a positive integer value n and outputs the sequence: n. 4. Write a recursive definition of the function int countdigits( unsigned num ) that receives a positive integer value and returns its number of digits. Tracing the Execution of a Recursive Function Call We use the run-time stack to trace the execution of a recursive call as follows: For each function call, the following actions take place: A memory location is created in the run-time stack for each local variable and value parameter of the function. The address of the reentry point (the instruction to be executed after the function terminates) is also pushed on top of the stack. Note that in the examples that follow, we use the instruction itself instead of its address. After the execution of a function, the value returned by that function becomes available to the function that calls this function and the pop operation is performed on the stack. The execution of the function call terminates when the stack becomes empty. Example R6 The execution trace of the function call rwritereverse(123); in function main follows: Function Calls Value Parameters & Local Variables Returned Value Reentry point rwritereverse(1) num= 1 End of function rwritereverse(12) num= 12 End of function rwritereverse(123) num= 123 In function main Output Gilbert Ndjatou Page 272
5 Example R7 The execution trace of the function call rwritedigits(123); in function main follows: Function Calls Value Parameters & Local Variables Returned Value Reentry point rwritedigits(1) num= 1 cout << num % 10; rwritedigits(12) num= 12 cout << num % 10; rwritedigits(123) num= 123 In function main Output Example R8 The execution trace of the following two statements: in function main follows: result = rpower( 3.0, 4); cout << endl << Result=\t << result; Function Calls rpower( 3.0, 0) rpower( 3.0, 1) rpower( 3.0, 2) rpower( 3.0, 3) rpower( 3.0, 4) Value Parameters & Local Variables num= 3.0 expo= 0 num= 3.0 expo= 1 num= 3.0 expo= 2 num= 3.0 expo= 3 num= 3.0 expo= 4 Returned Value Reentry point 1.0 return( num * rpower( num, expo 1 ) ) 3.0 return( num * rpower( num, expo 1 ) ) 9.0 return( num * rpower( num, expo 1 ) ) 27.0 return( num * rpower( num, expo 1 ) ) 81.0 result = rpower( 3.0, 4 ) Output Result = 81.0 Exercise R2 1. Show the trace of the execution of the function call writersequence( 5 ); 2. Show the trace of the execution of the function call writesequence( 5 ); 3. Show the trace of the execution of the function call cout << countdigits( 12345); 2011 Gilbert Ndjatou Page 273
6 Exercise R3 1. Determine what is being computed by each of the following recursive functions: unsigned f1( unsigned num) if( num < 2 ) return 0; return ( 1 + f1( num / 2) ); unsigned f2( unsigned num) if( num == 0 ) return 0; return ( f2( num / 10) + num % 10 ); unsigned f3( unsigned num) if( num < 10 ) return num; return ( f3( num / 10) ); 2. Show the trace of the execution of the following function calls: a. cout << f1( 10 ); b. cout << f2( 1234); c. cout << f3( 3456 ); 3. Write a recursive definition of the function void reversearray( char list[ ], int first, int last ) that receives as arguments an array of characters, the index of the first element and the index of the last element in the array, and reverses the elements of the array. 4. Write a recursive definition of the function bool ispalindrome( char list[ ], int first, int last ) that receives as arguments an array of characters, the index of the first element and the index of the last element in the array, and returns true if the array is a palindrome, and false otherwise Gilbert Ndjatou Page 274
7 Algorithm Analysis An algorithm analysis is in general measured according to two criteria: The space utilization and the time efficiency. The space utilization refers to the amount of memory required to store the data used in the algorithm. The time efficiency refers to the amount of time required to process the data used in the algorithm. It is in general difficult to minimize both the space utilization and the time requirement of an algorithm simultaneously: A tradeoff between space utilization and time efficiency is therefore needed. This tradeoff should take into consideration the cost of the memory and the storage devices, and the importance of time in the execution of the algorithm. In addition to the size of the input, other factors that influence the execution time of an algorithm are: The quality of the source code that implements the algorithm. The quality of the machine code generated from this source code by the compiler. The kinds of instructions used in the algorithm. The speed with which the machine can execute the instructions used in the algorithm However, these factors are not considered in the analysis of the execution time of an algorithm for the following reasons: Some languages are better suited that others for certain algorithms. Some programmers write faster programs than others. Some compilers generate more efficient code that others. The kinds of instructions used in an algorithm and the speed with which the machine can execute them depend on the particular computer system being used, the processors speed, the amount and kind of memory and disk space, the number of users logged in and many other factors. So, the execution time of an algorithm is in general specified as a function T(n) of the size n of the input. T(n) is specified as an approximate count of the statements executed in the algorithm instead of real time units such as seconds Gilbert Ndjatou Page 275
8 Example AA1 /* Algorithm to find the mean of x[0], x[1],... x[n 1] */ /* Variables: sum to hold the sum of the elements of the array ct the index variable for the array */ 1. Initialize sum to Initialize the index variable ct to While ct < n do the following: a. Add x[ct] to sum. b. Increment ct by Calculate and return the mean: sum / n. Algorithm Analysis Statement # of Times Executed n+1 3.a n 3.b n 4 1 Total = TM(n) 3n + 4 Best, Average and Worst Cases In the example above, the computing time depends only on the size n of the input. However, in other problems, the computing time may also depend on the arrangement of the input items. For example, it will take less time to sort a list of items that is almost in order than to sort the same list of items which is in reverse order. With these type of problems, we can compute T(n) as follows: o T(n) for the best case - not very informative o T(n) for the average case - difficult to compute o T(n) for the worst case - is in general taken as a measure of the algorithm performance Example AA2 Selection Sort Algorithm /*- Algorithm to sort x[0], x[1],... x[n 1] in ascending order using the selection sort algorithm -*/ /* Variables: smallest to hold the smallest element in the part of the array being searched smallestpos to hold the position of the smallest element found pass to hold the number of passes made through the list ct the index variable for the array */ 2011 Gilbert Ndjatou Page 276
9 1. Initialize pass to While pass < n 2 do the following: a. Set smallestpos to pass. b. Set smallest to x[pass]. c. Set ct to pass + 1 d. While ct < n 1 do the following: i. If x[ct] < smallest then do the following: 1. Set smallest to x[ct]. 2. Set smallestpos to ct. ii. Increment ct by 1; e. Set x[smallpos] to x[pass]. f. Set x[pass] to smallest. g. Increment pass by 1. Algorithm Analysis Statement # of Times Executed n-1 2a n-2 2.b n-2 2.c n-2 2.d n-1 + n d.i n-2 + n d.i.1 n-2 + n d.i.2 n-2 + n d.ii n-2 + n e n-2 2.f n-2 2.g n (n-1) +6(n-2) + (n-1)n/2 + (n-2)(n-1)/2 Total = TSS(n) = 1 + (n-1) + 6(n-2) + (n-1)(n-1) = n(n-1) + 6n = n 2 + 5n - 11 Example AA3 Binary Search Algorithm /*- Algorithm to binary search x[0], x[1],... x[n 1] for item, where the elements are sorted in ascending order -*/ /* Variables: item to hold the item being searched in the array firstpos to hold the position of the first element of the list to be searched lastpos to hold the position of the last element of the list to be searched midpos to hold the position of the middle element of the list to be searched found a flag to indicate whether the item is found loc to hold the index of an element in the array */ 2011 Gilbert Ndjatou Page 277
10 1. Read the item to be found into variable item. 2. Set found to false. 3. Set firstpos to Set lastpos to n While firstpos < lastpos and not found do the following: a. Calculate midpos = ( firstpos + lastpos ) /2. b. If x[midpos] == item then do the following: i. Set found to true. c. Else if x[midpos] < item then do the following: i. Set firstpos to midpos + 1 d. Else do the following: i. Set lastpos to midpos 1. Algorithm Analysis Statement # of Times Executed If the number of repetitions of the loop is k, then n / 2 k = 1 because at each 5 iteration, the size of the list is divided by 2 and the loop stops when the size of the list is 1. Therefore k = log2(n ). 5.a k b k -1 5.c k -1 5.d 5.c.i k -1 5.d.i Total = 5log2(n ) TBS(n) Order of Magnitude f(n): O(f(n)) Given the functions of variable n f(n) and g(n), we said that g(n) has order of magnitude f(n), denoted by: Example g(n) = O(f(n)) if there exists the constant C and a value of n, n0 such that g(n) <= Cf(n) for every n > n0 1. 3n + 5 is O(n) 2. 5n2-4n 500 is O(n 2 ) 2011 Gilbert Ndjatou Page 278
11 The time efficiency of an algorithm is in general specified as an order of magnitude of a well-known simple function such as: O(1) constant computing time better O(log2(n )) O(n) O(nlog2(n )) O(n 2 ) O(n 3 ) Example logarithmic computing time linear computing time quadratic computing time cubic computing time O(2 n ) exponential computing time worst 1. TM(n) = 3n + 4 = O(n) linear algorithm 2. TSS(n) = n 2 + 5n 11 = O(n 2 ) quadratic algorithm 3. TBS(n) = 4log2(n ) = O(log2(n )) logarithmic algorithm Exercise AA1 1. What are the two criteria that are in general used to measure the efficiency of an algorithm? 2. The worst case computing time of the selection sort algorithm is O( )? 3. The worst case computing time of binary search is O( )? 4. Provide the order of magnitude that is the best O notation to describe each of the following computing times: a. T(n) = n + 7 b. T(n) = 5n 2 + n + 50 c. T(n) = n 3 + n + log2(n) d. T(n) = 2 n + n Provide the order of magnitude of each of the following algorithms: a. n = 0; sum = 0; cin >> num; while( num!= -99) sum += num; n++; cin >> num; cout << sum / n; 2011 Gilbert Ndjatou Page 279
12 b. /* matrix addition */ for( int i = 0; i < n ; i++ ) for(int j = 0 ; j < n; j++ ) list3[i][j] = list1[i][j] + list2[i][j]; c. /* matrix multiplication */ for( int i = 0; i < n ; i++ ) for(int j = 0 ; j < n; j++ ) c[i][j] = 0; for(int k = 0; k < n; k++) c[i][j] += a[i][k] + b[k][j]; d. while( n >= 1 ) n /= 2; e. x = 1; for( int i = 0; i < = n -1 ; i++ ) for(int j = 0 ; j <= x ; j++ ) cout << j << endl; x *= 2; Recursive Function to Solve the Towers of Hanoi Problem void move( unsigned numberofdisk, char source, char destination, char spare ) if( numberofdisk == 1 ) cout << endl << Move the top of the disk from\t << source << \tto\t << destination; else move( numberofdisk 1, source, spare, destination ); move( 1, source, destination, spare ); move( numberofdisk 1, spare, destination, source ); Example of execution of the program: move( 5, A, B, C ); 2011 Gilbert Ndjatou Page 280
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