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Explore the concepts of recursion in programming, including functions calling themselves and mathematical induction principles. Learn how to write recursive solutions, avoid infinite loops, and apply mathematical induction to prove statements. Gain insights into stack activation records and the dangers of excessive recursion. Includes examples and code snippets for a better understanding.
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Recursive Function Call • a recursion function is a function that either directly or indirectly makes a call to itself. • but we need to avoid making an infinite sequence of function calls (infinite recursion)
Finding a Recursive Solution • A recursive solution to a problem must be written carefully • The idea is for each successive recursive call to bring you one step closer to a situation in which the problem can easily be solved • The easily solved situation is called the base case • Each recursive algorithm must have at least one base case, as well as a general (recursive) case
Mathemetical Induction To prove • Let p(n) denote the statement involving the integer variable n. The Principle of Mathematical Induction states: • If p(1) is true and, for some integer K >=1 , p(k+1) is true whenever p(k) is true then p(n) is true for all n>=1 . • 4 steps in using Induction: • Base cases; --- p(1), p(2), … • Induction hypothesis (IH); --- assume p(k) is true • Statement to be proved in induction; --- it is true for p(k+1) • Induction step. --- prove p(k+1) is true based on IH
A recursive defination int s (int n) { if (n ==1) return 1; else return s(n-1) + n; } • A few of problems: • n 0 at the beginning; • return value might be too large to fit in an int.
Printing number in Any Base const string DIGIT_TABLE = "0123456789abcdef"; const int MAX_BASE = DIGIT_TABLE.length( ); void printIntRec( int n, int base ) { if( n >= base ) printIntRec( n / base, base ); cout << DIGIT_TABLE[ n % base ]; } • Potential problems in this code: • if base>16: --- out of bound; • if base = 0: --- division by 0; • if base = 1: --- infinite loop.
General format forMany Recursive Functions if (some easily-solved condition) // base case solution statement else// general case recursive function call
When a function is called... • A transfer of control occurs from the calling block to the code of the function--it is necessary that there be a return to the correct place in the calling block after the function code is executed; this correct place is called the return address • When any function is called, the run-time stack is used--on this stack is placed an activation record for the function call
Stack Activation Frames • The activation record contains the return address for this function call, and also the parameters, and local variables, and space for the function’s return value, if non-void etc. • The activation record for a particular function call is popped off the run-time stack when the final closing brace in the function code is reached, or when a return statement is reached in the function code. • At that time the function’s return value, if non-void, is brought back to the calling block return address for use there
A recursive function int Func ( /* in */ int a, /* in */ int b ) { int result; if ( b == 0 ) // base case result = 0; else if ( b > 0 ) // first general case result = a + Func ( a , b - 1 ) ) ; // instruction 50 return result; }
Run-Time Stack Activation Recordsx = Func(5, 2);// original call at instruction 100 FCTVAL ? result ? b 2 a 5 Return Address 100 original call at instruction 100 pushes on this record for Func(5,2)
record for Func(5,2) Run-Time Stack Activation Recordsx = Func(5, 2);// original call at instruction 100 FCTVAL ? result ? b 1 a 5 Return Address 50 FCTVAL ? result 5+Func(5,1) = ? b 2 a 5 Return Address 100 call in Func(5,2) code at instruction 50 pushes on this record for Func(5,1)
record for Func(5,1) record for Func(5,2) Run-Time Stack Activation Recordsx = Func(5, 2);// original call at instruction 100 call in Func(5,1) code at instruction 50 pushes on this record for Func(5,0) FCTVAL ? result ? b 0 a 5 Return Address 50 FCTVAL ? result 5+Func(5,0) = ? b 1 a 5 Return Address 50 FCTVAL ? result 5+Func(5,1) = ? b 2 a 5 Return Address 100
Run-Time Stack Activation Recordsx = Func(5, 2);// original call at instruction 100 FCTVAL 0 result 0 b 0 a 5 Return Address 50 FCTVAL ? result 5+Func(5,0) = ? b 1 a 5 Return Address 50 FCTVAL ? result 5+Func(5,1) = ? b 2 a 5 Return Address 100 record for Func(5,0) is popped first with its FCTVAL record for Func(5,1) record for Func(5,2)
Run-Time Stack Activation Recordsx = Func(5, 2);// original call at instruction 100 FCTVAL 5 result 5+Func(5,0) = 5+ 0 b 1 a 5 Return Address 50 FCTVAL ? result 5+Func(5,1) = ? b 2 a 5 Return Address 100 record for Func(5,1) is popped next with its FCTVAL record for Func(5,2)
Run-Time Stack Activation Recordsx = Func(5, 2);// original call at instruction 100 FCTVAL 10 result 5+Func(5,1) = 5+5 b 2 a 5 Return Address 100 record for Func(5,2) is popped last with its FCTVAL
Too much recursion Can Be Dangerous Fibonacci numbers. Long fib (int n) { If (n <=1) return n; Else return fib(n-1) + fib(n-2); }
Too much Recursion Can be Dangerous • This definition will lead to exponential running time. • Reason: -- too much redundant work. • Not necessary to use recursion.
Tree • Tree is a fundamental structure in computer science. • Recursive definition: A tree is a root and zero or more nonempty subtrees. • Nonrecursive definition: A tree consists of s set of nodes and a set of directed edges that connect pairs of nodes. That is, a connected graph without loop.
Some more examples: • Factorials • Binary Search template <class Comparable> int binarySearch( const vector<Comparable> & a, const Comparable & x, int low, int high ) { if( low > high ) return NOT_FOUND; int mid = ( low + high ) / 2; if( a[ mid ] < x ) return binarySearch( a, x, mid + 1, high ); else if( x < a[ mid ] ) return binarySearch( a, x, low, mid - 1 ); else return mid; }
Divide and Conquer • Given an instance of the problem to be solved, split that into several, smaller, sub-instances (of the same problem). • Independently solve each of the sub-instances and then combine the sub-instance solutions so as to yield a solution for the original instance.
Finding ‘Greatest Common Divisor’ • gcd(A,B) = gcd(A-B, B) • template <class HugeInt> HugeInt gcd (const HugeInt &a, Const HugeInt &b ) { if( b==0 ) return a; else return gcd(b, a%b); }
Recursion or Iteration? CLARITY EFFICIENCY