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COS201 Fundamentals of PL – Lecture 6
AUBG, COS dept
Lecture 6
Theory
Iteration and Recursion in Procedure Oriented Programming.
Basic notions. Syntax representation of iteration and recursion. Examples.
Comparison of iterative and recursive algorithms.
RS 6th ed., Section 8.3, Section 10.3.3
The essence of iteration and its syntax representation in the form of a variety of counter
controlled and logically controlled loop statements.
The essence of recursion and its syntax representation in the form of a subprogram call
statement. Recursive function: A function that calls itself is said to be recursive.
Recursion visualized: recursively-defined geometric figures, such as
the Koch curve,
the Sierpinski triangle,
the Cantor set
Recursion in Real Life, Recursion in Mathematics,
Recursion in Computing
Recursive algorithms:
Simple case: Problem case for which a straightforward solution is known.
The recursive algorithms generally consist of an if statement with the following form:
if ( <this is a simple case> ) <solve it>;
else <redefine the problem using recursion>;
Conditional expression after McCarthy:
Notation:
b1, b2, … , bn
Boolean expression(s)
e1, e2, … , en, en+1
Expression(s)
[ b1 > e1, b2 > e2, … , bn > en, en+1 ]
Classification of recursion:
 primitive recursion
Direct and indirect recursion.
 referenced recursion
Time Performance and Memory Consumption of iterative and recursive versions of a
certain set of algorithms: factorial, greatest common divisor, Fibonacci series:
C, C++, Perl, Python source texts for recursive and iterative versions of algorithms for
factorial fact(n)
greatest common divisor gcd(m,n)
sum of natural numbers 1 + 2 + 3 +… sum of the elements of 1dimensioned array
to calculate a square root
to calculate an element of Fibonacci series
Ackermann function A(m, n)
Hanoi towers game
addition, multiplication, division
Simulation of RTL itoa() function
Problem: Factorial
More on back page >>
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C/C++ Solution
int facti(int n) { if (n==0 || n==1) return 1;
Iteration
Recursion
C Driver
Program
C++ Driver
Program
int i, res=1;
for (i=2; i<=n; i++) res *= i;
return res; }
int factr(int n) {
if (n==0 || n==1) return 1;
return n * factr(n-1); }
#include <stdio.h>
int facti(int n);
int factr(int n);
void main()
{
int i;
printf("C Greeting Hello, World");
// Test Factoriel recursive and iterative versions
for(i=0; i<=10; i++)
{
printf("\n%d -> %d -> %d", i, facti(i), factr(i));
#include <iostream.h>
int facti(int n);
int factr(int n);
void main()
{
cout <<"CPP Greeting Hello, World";
for(int i=0;i<=10;i++) {cout<<"\n"<< i << " -> " << facti(i) << " -> " << factr(i); } }
Perl Solution
sub Facti {
Iteration
$Arg1 = shift(@_);
if($Arg1 == 0) {return(1);}
$Result = 1;
for ($i=1; $i <= $Arg1; $i++) { $Result = $Result * $i; }
return($Result);
}
my($Arg1) = shift(@_);
if($Arg1 == 0) {return(1);}
else {return($Arg1 * &Factr($Arg1-1)); }
}
Recursion
sub Factr {
Driver
Program
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
for($k=0; $k<=10; $k++)
{print "\n".$k." >> ".&Facti($k)." - ".&Factr($k);}
Python Solution
def facti(n):
Iteration
Recursion
Driver
Program
i, res = 1, 1
while i <= n:
res, i = res*i, i+1
return res
def factr(n):
if n==0: return 1
else:
return n * factr(n-1)
print "Python greeting Hello, world!“
listnum = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
print " "
print "Factorial recursive and iterative versions"
for k in listnum: print k, factr(k), facti(k)
} }
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Problem: GCD /Greatest Common Divisor/
C/C++ Solution
int gcdi(int m, int n)
Iteration
{
if(n==0) return m;
int r;
while ( (r=m%n)!=0 ) { m=n;
return n;
Recursion
C Driver
Program
C++ Driver
Program
}
int gcdr(int m, int n)
{
if(n==0) return m;
return gcdr(n, m%n);
}
#include <stdio.h>
int gcdi(int m, int n);
int gcdr(int m, int n);
void main()
{
int i;
printf("C Greeting Hello, World");
// Test GCD recursive and iterative versions
for(i=0; i<=10; i++)
{
printf("\n10,%d -> %d -> %d", i, gcdi(10,i), gcdr(10,i));
}
}
#include <iostream.h>
int gcdi(int m, int n);
int gcdr(int m, int n);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test GCD recursive and iterative versions
for(int i=0;i<=10;i++)
{
cout <<"\n10," << i << " -> " << gcdi(10,i) << " -> " << gcdr(10,i);
}
}
Perl Solution
sub GCDi {
Iteration
Recursion
Driver
Program
n=r; }
$Arg1 = shift(@_); $Arg2 = shift(@_);
if($Arg2 == 0) {return($Arg1);}
while( ($Rem=($Arg1%$Arg2)) != 0)
{
$Arg1 = $Arg2;
$Arg2 = $Rem;
}
return($Arg2);
}
sub GCDr {
my($Arg1) = shift(@_); my($Arg2) = shift(@_);
if($Arg2 == 0) {return($Arg1);}
else {return(&GCDr($Arg2, $Arg1%$Arg2)); }
}
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
print "\nGCD";
for($l=0; $l<=10; $l++)
{print "\n10,".$l." >> ".&GCDi(10, $l)." - ".&GCDr(10, $l);}
Python Solution
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Iteration
Recursion
Driver
Program
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def gcdi(m, n):
if n==0: return m
rem = m % n
while rem != 0:
m, n, rem = n, rem, m % n
return n
def gcdr(m, n):
if n==0: return m
else:
return gcdr(n, m%n)
print "Python greeting Hello, world!"
listnum = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
print " "
print "GCD recursive and iterative versions"
for k in listnum: print 10, k, gcdr(10, k), gcdi(10, k)
Problem: Sum of natural numbers Σi
C/C++ Solution
int sumi(int n)
Iteration
// Sum of natural numbers
{
if (n==0 || n==1) return n;
int i, sum = 0;
for (i=1; i<=n; i++) sum+= i;
return sum;
Recursion
C Driver
Program
C++ Driver
Program
}
int sumr(int n)
{
if (n==0 || n==1) return n;
return n + sumr(n-1);
}
#include <stdio.h>
int sumi(int n);
int sumr(int n);
void main()
{
int i;
printf("C Greeting Hello, World");
// Test Sum of natural numbers recursive and iterative versions
printf("\n\nSum of natural numbers");
for(i=0; i<=10; i++)
{ printf("\n%d -> %d -> %d", i, sumi(i), sumr(i)); }
}
#include <iostream.h>
int sumi(int n);
int sumr(int n);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test Sum of natural numbers recursive and iterative versions
cout <<"\n\nSum of natural numbers";
for(int i=0;i<=10;i++)
{ cout <<"\n" << i << " -> " << sumi(i) << " -> " << sumr(i); }
}
Perl Solution
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Iteration
sub sumi {
$Arg1 = shift(@_);
if($Arg1==0 || $Arg1==1) {return($Arg1);}
$Sum = 0;
for ($i=1; $i <= $Arg1 ; $i++) {$Sum += $i;}
return($Sum);
}
my($Arg1) = shift(@_);
if($Arg1==0 || $Arg1==1) {return($Arg1);}
return($Arg1 + &sumr($Arg1-1));
}
Recursion
sub sumr {
Driver
Program
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
print "\nSum of natural numbers";
for($l=0; $l<=10; $l++)
{print "\n".$l." >> ".&sumi($l)." - ".&sumr($l);}
Python Solution
# Sum of natural numbers
Iteration
Recursion
Driver
Program
def sumi(n):
if n==0: return n
i, sum = 1, 0
while i <= n:
sum, i = sum + i, i+1
return sum
def sumr(n):
if n==0: return n
else:
return n + sumr(n-1)
print "Python greeting Hello, world!"
listnum = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
print " "
print "Sum natural numbers recursive and iterative versions"
for k in listnum: print k, sumr(k), sumi(k)
Problem: Sum of 1-dimensioned array elements Σa[I]
C/C++ Solution
// Sum of 1-dimensioned array elements
Iteration
Recursion
int ar1sumi(int b[], int n)
{
int i, sum=0;
for (i=0; i<=n; i++) sum += b[i];
return sum;
}
int ar1sumr(int b[], int n)
{
if (n==0) return b[0];
return b[n] + ar1sumr(b, n-1);
}
#include <stdio.h>
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Program
C++ Driver
Program
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// Sum of 1-dmensioned array elements
int ar1sumi(int b[], int n);
int ar1sumr(int b[], int n);
void main()
{
int i;
printf("C Greeting Hello, World");
// Test Sum of 1-dimensioned array elements
printf("\n\nSum of 1-dimenisioned array elements\n");
int mas1[]={ 5, 10, 20, 30, 40, 50, 60};
for (i=0; i<=6; i++) printf("%d ", mas1[i]);
printf(" >> %d >> %d", ar1sumi(mas1, 6), ar1sumr(mas1, 6));
}
#include <iostream.h>
// Sum of 1-dmensioned array elements
int ar1sumi(int b[], int n);
int ar1sumr(int b[], int n);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test Sum of 1-dimensioned array elements
cout <<"\n\nSum of 1-dimenisioned array elements\n";
int mas1[] = { 5, 10, 20, 30, 40, 50, 60};
for (int i=0; i<=6; i++) cout << mas1[i] << " ";
cout << " > " << ar1sumi(mas1, 6) << " > " << ar1sumr(mas1, 6);
}
Perl Solution
sub ar1sumi{ @ArgList = @_; $sum = 0;
Iteration
Recursion
Driver
Program
for ($i=0;$i<=$#ArgList;$i++)
{$sum+=$ArgList[$i];}
return($sum);
}
sub ar1sumr{ my(@ArgList) = @_; my($Index)=$#ArgList;
if ($Index==0) {return ($ArgList[$Index]);}
return ($ArgList[0]+ar1sumr(@ArgList[1..$Index]));
}
#!/usr/bin/perl
print "\nSum of 1-dimensioned array elements ";
@mas = ( 5, 10, 20, 30, 40, 50, 60 );
for ($i=0; $i<=$#mas; $i++) { print $mas[$i]." ";}
print "\nIterative sum of an array/list = ".ar1sumi(@mas);
print "\nRecursive sum of an array/list = ".ar1sumr(@mas);
getc();
Python Solution
Iteration
Recursion
Driver
Program
Problem: Fibonacci series
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C/C++ Solution
int fibi(int n) {
Iteration
Recursion
C Driver
Program
C++ Driver
Program
}
int fibr(int n) {
int i, va=1, vb=1, vc=1;
if (n==0 || n==1) return 1;
for (i=2; i<=n; i++)
{ vc = va + vb; va = vb; vb = vc; }
return vc;
if (n==0 || n==1) return 1;
return fibr(n-1) + fibr(n-2);
}
#include <stdio.h>
int fibi(int n);
int fibr(int n);
void main()
{
int i;
printf("C Greeting Hello, World");
// Test Fibonacci recursive and iterative versions
printf("\n\nFibonacci");
for(i=0; i<=10; i++)
{printf("\n%d -> %d -> %d", i, fibi(i), fibr(i)); }
}
#include <iostream.h>
int fibi(int n);
int fibr(int n);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test Fibonacci recursive and iterative versions
cout <<"\n\nFibonacci";
for(int i=0;i<=10;i++)
{cout <<"\n" << i << " -> " << fibi(i) << " -> " << fibr(i);
}
Perl Solution
sub fibi {
Iteration
$Arg1 = shift(@_);
if($Arg1==0 || $Arg1==1) {return($Arg1);}
$va=0; $vb=1;
for ($i=2; $i <= $Arg1 ; $i++)
{$vc = $va+$vb; $va = $vb; $vb = $vc;}
return($vc);
}
my($Arg1) = shift(@_);
if($Arg1==0 || $Arg1==1) {return($Arg1);}
return( &fibr($Arg1-1) + &fibr($Arg1-2) );
}
Recursion
sub fibr {
Driver
Program
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
print "\nFibonacci";
for($l=0; $l<=10; $l++)
{print "\n".$l." >> ".&fibi($l)." - ".&fibr($l);}
getc();
Python Solution
}
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Iteration
Recursion
Driver
Program
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# Fibonacci series module
def fibi(n):
# Fibonacci series up to n
va, vb, i = 0, 1, 2
while i <= n:
va, vb, i = vb, va+vb, i+1
return vb
def fibr(n):
if n==0: return n
elif n==1:
return n
else:
return fibr(n-1) + fibr(n-2)
print "Python greeting Hello, world!"
listnum = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
print "Fibonacci recursive and iterative versions"
for k in listnum: print k, fibr(k), fibi(k)
#read (ch)
Problem: Square Root
C/C++ Solution
// Square Root simulations
Iteration
double sqrti(double x, double eps) {
Recursion
C Driver
Program
C++ Driver
Program
double yn=1, yn1=(x/yn+yn)/2;
while (abs(yn-yn1)>eps) {yn=yn1; yn1=(x/yn+yn)/2;}
return yn;
}
double sqrtr(double x, double start, double eps) {
if (abs(start*start-x) < eps) return start;
return sqrtr(x, (start*start +x)/(2*start), eps);
}
#include <stdio.h>
double sqrti(double x, double eps);
double sqrtr(double x, double start, double eps);
void main()
{
printf("C Greeting Hello, World");
// Test Square Root recursive and iterative versions
printf("\n\nSquare root simulations");
printf("\n 25 -> %lf -> %lf", sqrti( 25,0.001), sqrtr( 25,1,0.001));
printf("\n144 -> %lf -> %lf", sqrti(144,0.001), sqrtr(144,1,0.001));
printf("\n289 -> %lf -> %lf", sqrti(289,0.001), sqrtr(289,1,0.001));
}
#include <iostream.h>
double sqrti(double x, double eps);
double sqrtr(double x, double start, double eps);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test Square Root recursive and iterative versions
cout <<"\n\nSquare root simulations";
cout<<"\n"<< 25<<" -> "<<sqrti(25,0.001)<<" -> "<< sqrtr(25,1,0.001);
cout<<"\n"<<144<<" -> "<<sqrti(144,0.001)<<" -> "<< sqrtr(144,1,0.001);
cout <<"\n"<<289<<" -> "<<sqrti(289,0.001)<<" -> "<< sqrtr(289,1,0.001);
}
Perl Solution
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Iteration
Recursion
Driver
Program
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sub sqrti {
$Arg1 = shift(@_); $Arg2 = shift(@_);
$yn=1.; $yn1=($Arg1/$yn + $yn)/2.;
while( abs($yn-$yn1) > $Arg2)
{
$yn =$yn1;
$yn1=($Arg1/$yn + $yn)/2.
}
return($yn);
}
sub sqrtr {
my($Arg1) = shift(@_); my($Arg2) = shift(@_);
my($Arg3) = shift(@_);
if(abs($Arg2*$Arg2-$Arg1)<$Arg3) {return($Arg2);}
return(&sqrtr($Arg1,($Arg2*$Arg2+$Arg1)/(2.*$Arg2),$Arg3));
}
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
print "\nSquare root simulations";
print "\n 25 >> ".&sqrti( 25,0.001)." - ".&sqrtr( 25,1.,0.001);
print "\n144 >> ".&sqrti(144,0.001)." - ".&sqrtr(144,1.,0.001);
print "\n289 >> ".&sqrti(289,0.001)." - ".&sqrtr(289,1.,0.001);
getc();
Python Solution
# square root simulations
Iteration
Recursion
Driver
Program
def sqrti(x, eps):
yn = 1
yn1 = (x/yn + yn)/2
while abs(yn-yn1)>eps:
yn = yn1
yn1 = (x/yn+yn)/2
return yn
def sqrtr(x, yn, eps):
if abs(yn*yn-x)<eps:
return yn
else:
return sqrtr(x, (yn*yn + x)/(2*yn), eps)
print " "
print "SQRT recursive and iterative versions"
listx = [ 25, 49, 81, 100, 289 ]
for k in listx: print k, sqrtr(k, 1, 0.001), sqrti(k, 0.001)
Problem: Ackerman Function
C/C++ Solution
Iteration
// Ackerman function
Recursion
int acker(int m, int n)
{
if(m==0) return n+1;
if(n==0) return acker(m-1,1);
return acker(m-1, acker(m,n-1));
}
C Driver
#include <stdio.h>
int acker(int m, int n);
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C++ Driver
Program
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void main()
{
printf("C Greeting Hello, World");
// Test Ackerman recursive version
printf("\n\nAckerman function acker(m,n)");
printf("\nAckerman Arguments 0,0 > %d", acker(0,0));
printf("\nAckerman Arguments 1,1 > %d", acker(1,1));
printf("\nAckerman Arguments 2,1 > %d", acker(2,1));
printf("\nAckerman Arguments 0,51 > %d", acker(0,51));
}
#include <iostream.h>
int acker(int m, int n);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test Ackerman recursive version
cout <<"\n\nAckerman function acker(m,n)";
cout <<"\nAckerman Arguments 0,0 > " << acker(0,0);
cout <<"\nAckerman Arguments 1 1 > " << acker(1,1);
cout <<"\nAckerman Arguments 2 1 > " << acker(2,1);
cout <<"\nAckerman Arguments 0 51 > " << acker(0,51);
}
Perl Solution
Iteration
sub acker {
Recursion
Driver
Program
my($Arg1) = shift(@_); my($Arg2) = shift(@_);
if($Arg1==0) {return($Arg2 + 1);}
if($Arg2==0) {return(&acker($Arg1-1, 1));}
return(&acker($Arg1-1, &acker($Arg1, $Arg2-1)));
}
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
print "\nAckerman function acker(m,n)";
print "\nAckerman Arguments 0,0 > ".&acker(0,0);
print "\nAckerman Arguments 1,1 > ".&acker(1,1);
print "\nAckerman Arguments 2,1 > ".&acker(2,1);
print "\nAckerman Arguments 0,51 > ".&acker(0,51);
getc();
Python Solution
Iteration
# Ackerman function
Recursion
Driver
Program
def acker(m, n):
if m==0: return n+1
if n==0: return acker(m-1, 1)
else:
return acker(m-1, acker(m, n-1))
print "Python greeting Hello, world!"
print " "
print "Ackerman function recursive version"
print acker(0, 0), acker(1, 1), acker(2, 1), acker(0,51)
Problem: Hanoi towers
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C/C++ Solution
Iteration
// Hanoi Towers game
Recursion
C Driver
Program
C++ Driver
Program
void trantower(int ndsks, int fromndl, int tondl, int wrkndl)
{
if (ndsks > 0)
{
trantower(ndsks-1, fromndl, wrkndl, tondl);
printf("\n%2d >> %2d", fromndl, tondl);
trantower(ndsks-1, wrkndl, tondl, fromndl);
}
}
#include <stdio.h>
void trantower(int ndsks, int fromndl, int tondl, int wrkndl);
void main()
{
printf("C Greeting Hello, World");
// Test Hanoi Towers game recursive version
printf("\n\nHanoi towers game - 1 disk");
trantower(1, 1, 3, 2);
printf("\n\nHanoi towers game - 2 disks");
trantower(2, 1, 3, 2);
printf("\n\nHanoi towers game - 3 disks");
trantower(3, 1, 3, 2);
}
#include <iostream.h>
void trantower(int ndsks, int fromndl, int tondl, int wrkndl);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test Hanoi Towers game recursive version
cout <<"\n\nHanoi towers game - 1 disk";
trantower(1, 1, 3, 2);
cout <<"\n\nHanoi towers game - 2 disks";
trantower(2, 1, 3, 2);
cout <<"\n\nHanoi towers game - 3 disks";
trantower(3, 1, 3, 2);
}
Perl Solution
Iteration
sub trantower{ my($Arg1) = shift(@_); my($Arg2) = shift(@_);
Recursion
my($Arg3) = shift(@_); my($Arg4) = shift(@_);
if($Arg1 >= 1) {
trantower($Arg1-1,$Arg2,$Arg4,$Arg3);
print "\n".$Arg2." >> ".$Arg3;
trantower($Arg1-1,$Arg4,$Arg3,$Arg2);
}
}
Driver
Program
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
print "\n\nHanoi towers game - 1 disk "; &trantower(1, 1, 3, 2);
print "\n\nHanoi towers game - 2 disks"; &trantower(2, 1, 3, 2);
print "\n\nHanoi towers game - 3 disks"; &trantower(3, 1, 3, 2);
getc();
Python Solution
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Iteration
Recursion
Driver
Program
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# Hanoi towers game
def trantower(ndsks, fromndl, tondl, wrkndl):
if ndsks>=1:
trantower(ndsks-1, fromndl, wrkndl, tondl)
print fromndl, " >> ", tondl
trantower(ndsks-1, wrkndl, tondl, fromndl)
print "Python greeting Hello, world!"
print " "
print "Hanoi towers game - recursive function implemented"
trantower(1, 1, 3, 2)
print" "
trantower(2, 1, 3, 2)
print" "
trantower(3, 1, 3, 2)
Problem: RTL itoa() simulated
C/C++ Solution
Iteration
// C/C++ itoa run time function simulated
Recursion
void printd(int n) {
if (n<0) { putchar('-'); n = -n;}
if(n/10 != 0) printd(n/10);
putchar(n%10+'0');
C Driver
Program
C++ Driver
Program
}
#include <stdio.h>
void printd(int n);
void main()
{
printf("C Greeting Hello, World");
// Test C/C++ itoa run time function recursive version
printf("\n\nC/C++ itoa run time library function");
printf("\n"); printd(123);
printf("\n"); printd(-86);
}
// C/C++ itoa run time function simulated
void printd(int n) {
if (n<0) { cout << "-"; n = -n;}
if(n/10 != 0) printd(n/10);
cout << n%10;
}
//-----------------------------------------------------------------------#include <iostream.h>
void printd(int n);
int main()
{
cout <<"CPP Greeting Hello, World";
// Test C/C++ itoa run time function recursive version
cout <<"\n\nC/C++ itoa run time library function";
cout << "\n"; printd(123);
cout << "\n"; printd(-86);
}
Perl Solution
106761000
Iteration
Recursion
Driver
Program
Page 13 of 15
sub printd {
my($Arg1) = shift(@_);
if($Arg1<0){ print"-"; $Arg1 = -$Arg1; }
if( ($Arg1/10) != 0) { &printd($Arg1/10); }
if ($Arg1%10 != 0) {print chr($Arg1%10 + 48);}
}
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
print "\nC/C++ itoa run time library function";
print "\n"; &printd(123);
print "\n"; &printd(-86);
print "\n"; &printd(5);
getc();
Python Solution
Iteration
# C/C++ run-time function itoa recursively simulated
Recursion
Driver
Program
def printd(n):
if n<0:
print"-",
if n<0:
n = -n
if n/10!=0:
printd(n/10)
print n%10,
print "Python greeting Hello, world!"
print " "
print "The itoa run-time function recursively implemented as printd"
printd(123)
print " "
printd(-567890)
print " "
printd(86)
Problem: multiplication (*) operator
C/C++ Solution
int muli(int m, int n)
Iteration
{
if (n==0) return 0;
if (n==1) return m;
int i, res=0;
for (i=1; i<=n; i++) { res = res + m;}
return res;
Recursion
}
int mulr(int m, int n)
{
if (n==0) return 0;
if (n==1) return m;
return m + mulr(m, n-1);
}
106761000
C Driver
Program
C++ Driver
Program
Page 14 of 15
#include <stdio.h>
int muli(int m, int n);
int mulr(int m, int n);
void main()
{
int k, l;
printf("C Greeting Hello, World");
// Test multiplication mplemented as recursive and iterative rourtine
printf("\n\nMultiplication implemented as Recursion and Iteration");
for (k=0; k<=12; k++) {
printf("\n");
for (l=0; l<=12; l++)
printf("\n%d,%d -> %d -> %d", k, l, muli(k,l), mulr(k,l)); }
}
#include <iostream.h>
int muli(int m, int n);
int mulr(int m, int n);
void main()
{
cout <<"CPP Greeting Hello, World";
// Test multiplication mplemented as recursive and iterative rourtine
cout <<"\n\nMultiplication implemented as Recursion and Iteration";
for (int k=0; k<=12; k++) {
cout <<"\n";
for (int l=0; l<=12; l++)
cout <<"\n"<<k<<","<<l<<" -> "<<muli(k,l)<<" -> "<<mulr(k,l);}
}
Perl Solution
sub muli {
Iteration
$Arg1 = shift(@_); $Arg2 = shift(@_);
if($Arg2==0) { return 0;}
if($Arg2==1) { return $Arg1;}
$res = 0;
for ($i=1; $i<=$Arg2; $i++) { $res = $res + $Arg1; }
return $res;
}
$Arg1 = shift(@_); $Arg2 = shift(@_);
if($Arg2==0) { return 0;}
if($Arg2==1) { return $Arg1;}
return $Arg1 + &mulr($Arg1, $Arg2-1);
}
Recursion
sub mulr {
Driver
Program
#!/usr/bin/perl
# start main program untitled unstructured
print "\nPerl greeting: Hello, world!\n";
# Test for MULTIPLICATION based on Recursion and Iteration
print "\nMultiplication implemented using Recursion and Iteration";
for($k=0; $k<=12; $k++)
{
print "\n";
for($l=0; $l<=12; $l++)
{
print "\n".$k." ".$l." ".&muli($l, $k)." ".&mulr($l, $k);
}
}
getc();
Python Solution
106761000
Iteration
Recursion
Driver
Program
Page 15 of 15
# Recursive and iterative simulation of MULTIPLICATION
def muli(m, n):
if n==0: return 0
if n==1: return m
res, i = 0, 1
while i<=n:
res, i = res+m, i+1
return res
def mulr(m, n):
if n==0: return 0
if n==1: return m
return m + mulr(m,n-1)
print "Python greeting Hello, world!"
listnum = [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
print " "
print "Multiplication impl. using Recursion and Iteration"
for k in listnum:
print " "
for l in listx:
print k, l, mulr(k, l), muli(k, l)