recursion, memory layout · • recursion – stack grows as deepen the nested function call –...

CS101 Introduction to computing

Recursion Memory LayoutRecursion, Memory Layout and

Problem Solving

A. Sahu and S. V .RaoDept of Comp. Sc. & Engg.Dept of Comp. Sc. & Engg.

Indian Institute of Technology Guwahati

Outline• Memory layout of C program• Recursive Function

–ExampleExample

• Problem Solving Example –Using modular functionsUsing recursive functions–Using recursive functions

–Using Arrays : Max, Part, Sort

Memory layout of C programy y p g

Dynamic memory allocationDynamic memory allocation• Reduce wastage of memory • Useful when data size is unknown before hand• Array Declaration int A[100];y

– Easy, Not to use pointer, small size, known before

[ ];

Easy, Not to use pointer, small size, known before

• Array Creation: Not easy use of pointer typecast lager size– Not easy, use of pointer, typecast, lager size, necessary size

Memory management C: APIs• Application program interfaces (APIs)A il bl f ti /API t• Available function/APIs to manage memory

• Create/allocate/reserve space– malloc : memory allocation– calloc : memory allocation + initialization to 0

• Move a reserved space to another location– realloc: move the space to another locationp

• Destroy/de‐allocate/free space• free:free:

Memory Allocation• Memory can be allocated • Declaring a variableDeclaring a variable

int A[100];

• Explicitly requesting space

i t *Aint *A;A=(int*)malloc(sizeof(int)*100);

Example: Dynamic Array Allocation

• Given N persons (with their IQ level) in orderN may be dynamic variable– N may be dynamic, variable

• A person decide He/She is intelligent or dumb • Decides locally:

– If his/her IQ level is greater than equal to average of IQ level of both neighbors

– Left neighbor and right neighbor

Example: Dynamic Array Allocationmain(){main(){

int *IQScore, *Intelligent, i, N; printf(“Input N:”); scanf(“%d”, &N); p t ( put : ); sca ( d , & );IQScore=(int*)malloc(N*sizeof(int));Intelligent =(int*)calloc(N*sizeof(int));for(i=0;i<N;i++) scanf(“%d”,&IQScore[i]);for(i=1;i<N-1;i++){

if(IQScore[i]> (IQScore[i 1]+IQScore[i+1])/2)if(IQScore[i]>=(IQScore[i-1]+IQScore[i+1])/2)Intelligent[i]=1; else Intelligent[i]=0;

printf(“ I am %d person %s\n”, i, Intelligent[i]?”YES”:”NO”);

}

free(IQScore); free(Intelligent);free(IQScore); free(Intelligent);}

Memory layout of C program• Program: Input, Output, ProcessingC d (I t ti ) D t (St k H )• Code (Instruction), Data (Stack, Heap)

• To store: Require memory– Input data, output data, intermediate data

• Memory can be allocated – Declaring a variable– Explicitly requesting space

int A[100];

p y q g p

int *A;A=(int*)malloc(sizeof(int)*100);

Memory layout of C programMemory layout of C program• Stack

automatic (default) local– automatic (default), local– Initialized/uninitialized

• Data

Stack

• Data– Global, static, externBSS Bl k S d b S b l

Heap

– BSS: Block Started by Symbol– BBS: Uninitialized Data Seg.

Data

BSSBSS

• Code : program instructions• Heap

Data

– malloc, calloc Code

Memory layout of C program

int A;Stack

int B=10;

main(){

Stack

(){

int Alocal;

int *p

Heap

BSSint *p;

p=(int*)malloc(40); Data

}

Code$gcc test.c$size a out$size a.outtext data bss dec hex filename1200 544 8 1752 6d8 a.out

Examples ofExamples of Modular code using Functions

Modular Code for : Xn

int PowXtoN(int n, int x){int n, x, P=1, PS=x;while(n > 0) {

#include <stdio.h>while(n > 0) {

if ((n%2)==1) P=P*PS;n=n/2; PS = PS* PS;

int GetNum();int PowXtoN(int n,int x);void DispRes(int,int,int);

}return P;}

p ( , , );

int main(){int X, N, Res; }}

X=GetNum();N=GetNum();R P Xt N(X N)

int GetANum(){ int N; printf(“Enter a Number:");

Res= PowXtonN(X,N);DispRes(Res,X,N);

return 0;

scanf(“%d", &N);return N; }

void DispRes(int Res, return 0;}

p ( ,int X, int N){

printf(“%d to power %d=%d”,X, N, Res); }

Modular C Code : Square root of a Positive NumberPositive Number

#include <stdio.h>float GetAccuracy();fl t G tP N ()

float SqRoot(float m, float e){float GetPosNum();

float SqRoot(float,float);

void DispRes(float,float);

float e){

float r1, r2;r1=m/2;

int main(){float a, e, Res;

r1=m/2; r2=r1;while(abs(r1-r2)>e){

r1 r2;a=GetPosNum();e=GetAccuracy();Res S R t(a e);

r1=r2;r2=(r1+m/r1)/2;

}ret rn 2Res= SqRoot(a,e);

DispRes(a, Res);

return 0;

return r2;}

return 0;}

You need to write code of other functions

Modular C Code : Factorial of a Number#i l d tdi h#include <stdio.h>

int GetNum();i t F t i l(i t A)

int Factorial(int n) {int Factorial(int A);void DispRes(int,int); int Prod=1,i;

for (i=1;i<=N;i++){int main(){int a, Res;

a GetNum();

Prod=Prod*i; }

a=GetNum();Res= Factorial(a);DispRes(a, Res);

return prod;}

return 0;}


Modular C Code : Reverse a Number#i l d tdi h#include <stdio.h>

int GetNum();i t R (i t A)

int Reverse(int n) {int Reverse(int A);

void DispRes(int,int); int RevNum=0, Rem;RevNum=0;

int main(){int a, Res;

a GetNum();

while(n != 0) {Rem = n%10;RevNum=RevNum*10+ Rem;

a=GetNum();Res= Reverse(a);DispRes(a, Res);

n=n/10;}

return RevNum;return 0;

} }


Modular C Code : Binary Search#i l d tdi h#include <stdio.h>int GetMinOfRange();int GetMaxOfRange();int GetUnknown();

int BinSrch(int Rmin, intRmax, int X){

int GetUnknown();int BinSrch(int,

int, int);

void DispRes(int);

while (Rmin<Rmax){mid=(Rmin+Rmax)/2;void DispRes(int);

int main(){int Min, Max, X, Res;

mid (Rmin+Rmax)/2;if(X==mid)return mid;if (X>mid)

R i id+1Min=GetMinOfRange();Max=GetMaxOfRange();X=GetUnknown();

Rmin=mid+1; else Rmax=mid;}();

Res=BinSrch(Min,Max,X)DispRes(Res); return 0;

return -1;}

}


Modular C Code : Nth Fibonacci #i l d tdi h#include <stdio.h>

int GetNum();i t Fib(i t N)

int Fib(int N) {

int Fib(int N);

void DispRes(int,int);int fnm2=0; fnm1=1; n=2;if(N<=1) return 1; while(n<=N){int main(){

int N, Res;

N GetNum();

while(n<=N){fn = fnm2 + fnm1;fnm2=fnm1; f 1 fN=GetNum();

Res= Fib(N);DispRes(N, Res);

fnm1=fn;n = n + 1;}

return 0;}

return fn;}


Modular C Code : GCD#i l d tdi h#include <stdio.h>

int GetA();int GetB();

int GCD(int n1, int n2) {int GetB();int GCD(int A, int B);

Void DisRes(int,int,int);while(!(n1==0|| n2==0)){

if (n1>n2) n1=n1%n2;int main(){int a, b, Res;

a GetA();

( )else n2=n2%n1;

}if (n1==0) return n2;a=GetA();

b=GetB();Res= GCD(a,b);DisRes(a b Res);

if (n1 0) return n2; else return n1;}

DisRes(a, b, Res);

return 0;} }


Modular Code Sin(x)#i l d tdi h#include <stdio.h>float GetX();float GetAccuracy();fl t Si XC l(fl t

float SinXCal(float x,

float acc){float SinXCal(float x,

float acc);

void DispRes(float,float);

int i=1;

float SinXVal=0,term=x;

while (term < acc) {p ( )

int main(){float X, acc, Res;

( ) {

i = i+2;term *= - x*x/(i*(i-1));Sin Val Sin Val+ term

X=GetX();acc=GetAccuracy();R Si XC l(X )

SinxVal= SinxVal+ term;

}

return SinXVal;Res= SinXCal(X,acc);DispRes(X, Res);

return 0;

}

return 0;}


Modular C Code : Value of PI#include <stdio.h>

int GetABigNumer();

double ValPI(int N){int M=0,i;int GetABigNumer();

double ValPI(int N);

void Di R (i t fl t)

double x,y,z,pi; for(i=0;i<N;i++){x=(double)rand()/RAND MAX;DispRes(int,float);

int main(){int N;

x (double)rand()/RAND_MAX;y=(double)rand()/RAND_MAX;z = x*x+y*y;if ( < 1) M++int N;

double Res; N=GetABigNumber();Res= ValPI(N);

if (z<=1) M++;}pi=4.0*(double)M/N;Res= ValPI(N);

DispRes(a, b, Res);

return 0;

return pi;}

;}


Modular C Code : Bisection Method#include <stdio.h>float GetMinOfInterVal();float GetMaxOfInterVal();fl t G tA ()

float F(float a){return a*a*a-2*a-5;

}float GetAccuracy();float F(float a);

float Bisection(float a,

float Bisection(float a,

float a, float acc){float F Fb F

}

float a, float acc)

void DispRes(float,float,float);int main(){

float Fa, Fb, Fx;Fa=F(a); Fb=F(b);x=a; x1=b;while( abs(x-x1)>accuracy) {

float a, b, acc,Res;a=GetMinOfInterVal();b=GetMaxOfInterVal();

G tA ()

while( abs(x x1)>accuracy) {x1=x; x=(a+b)/2; Fx=F(x);if(Fa*Fx<0) b=x;else a=x;

acc=GetAccuracy();Res= Bisection(a,b,acc);DispRes(a,b,Res); return 0;

( ) ; ;}

return x;

}return 0;}

}


RecursionsRecursions and

Recursive functions

Recursions: Recursive functions • Functions that call themselves• Can only solve a base case• Divide a problem up into

– What it can do– What it cannot do

• What it cannot do resembles original problem• The function launches a new copy of itself (recursion step) to solve what it cannot do

• Eventually base case gets solvedEventually base case gets solved– Gets plugged in, works its way up and solves whole problemp

Recursions: Recursive functions • Many Problem , we define the problem it self using recursive p gdefinition

• Solving them using recursion• Solving them using recursion – Easier to think and implement

• Example• Example – Fibonacci, GCD, Binary Search, calculation of Xn, Reversing numberReversing number

• Recursive Functions– Functions that call themselves (directly/indirectly)Functions that call themselves (directly/indirectly)– Can only solve a base case

Recursion Example: factorials5! 5 * 4 * 3 * 2 * 1• 5! = 5 * 4 * 3 * 2 * 1

• Notice that– 5! = 5 * 4!– 4! = 4 * 3! ...

• Can compute factorials recursively S l b ( ) h l• Solve base case (1! = 0! = 1) then plug in–2! = 2 * 1! = 2 * 1 = 2;–3! = 3 * 2! = 3 * 2 = 6;3! 3 2! 3 2 6;

Recursion vs Iteration• Repetition

– Iteration: explicit loopp p–Recursion: repeated function calls

• Termination• Termination– Iteration: loop condition fails–Recursion: base case recognized

• Both can have infinite loopsBoth can have infinite loops• Balance

–Choice between performance (iteration) and good software engineering (recursion)

Recursion Example: factorials• From Definition: but no terminationint Fact(int N){

return N * Fact (N-1);}

• With proper base case (termination

}

With proper base case (termination guaranteed )

int Fact (int N){if(N<=1) return 1; t N * F t (N 1)return N * Fact (N-1);

}

Factorial: Recursive call Fact(1);

Fact(3);

Fact(2);1

int Fact (int N){if(N<=1) return 1; return N*Fact(N-1);

Fact(4);

Fact(3);2

6( );}int main(){

int X;

24

120int X;X=fact(5);printf(“Fact of 5=%d \n”,X);

120

return 0;}

Nested Function Call• Nested Function call uses

– Stack to store the return address, return result and any other information

• Recursion– Stack grows as deepen the nested function call – Recursive call: Stack grows when it call next grecursive function

– All the local variable need to be put into stackp– Stack contains grows like this: Fib Example

• Main, Fib(5), Fib(4), Fib(3), Fib(2), Fib(1)

Recursion vs Iteration• Any problem that can be solved recursively –Can also be solved iteratively (non‐recursively)

• A recursive approach is normally chosen in preference – To an iterative approach when the recursive approach more naturally mirrors the problem

d l h– And results in a program that is easier to understand and debug

• Another reason to choose a recursive solution• Another reason to choose a recursive solution – An iterative solution may not be apparent

Recursion vs Iteration• Avoid using recursion in performance situations

– Recursive calls take time – And consume additional memory

• Common Error by programmer– Accidentally a non‐recursive function may call itself either directly, or indirectly through another function

• Functionalizing programs in a neat, hierarchical manner promotes good practice – More easier to program, test, debug, maintain, and evolve.

– But it has a price, A heavily functionalized program: makes potentially large numbers of function calls

Solving Recursive ProblemsSolving Recursive Problems• See recursive solutions as two sections:

–CurrentR t–Rest

N! = N * (N‐1)!7! 7 * 6!7! = 7 * 6! 7! = 7 * (6 * 5 * 4 * 3 * 2 * 1 * 1)

Factorial FunctionFactorial Function

int Fact (int N){if(N<=1) return 1; return N*Fact(N-1);

}

Factorial Functionif (1 == 1)

Factorial Function

return 1;else

return 1 * Fact(0);

if (2 == 1) return 1;

elsereturn 2 * Fact(1);

if (3 == 1) return 1;


main {X = Fact(3)

}

Tracing Details

1 A t l t t d1. Actual parameters stored on the stack

3 C t

5. Return value and release stack frame

int Fact (Num =3)if (3 = 1)

return 1;else

3. Create a new Stack Frame


}

2. Recursive call to Fact

4. Unfinished Business

Activation Stack for FactorialActivation Stack for Factorial

C ll th f ti X F t(5)Call the function: X= Fact(5);

Main Function: Unfinished: X = Fact (5);


Fact. 1st: N=5, Unfinished: 5*Fact(4)



Fact 2nd: N 4 Unfinished: 4*Fact(3)


Fact. 2nd: N=4, Unfinished: 4*Fact(3)




Fact. 3rd: N=3, Unfinished: 3*Fact(2)





Fact. 4th: N=2, Unfinished: 2*Fact(1)



Fact. 4th: N 2, Unfinished: 2 Fact(1)





Fact. 4th: N=2, Unfinished: 2*Fact(1)

Fact. 5th: N=1, Finished: returns 1



Fact. 4th: N 2, Unfinished: 2 Fact(1)





Fact. 4th: N=2, Finished: returns 2*1



Fact. 4th: N 2, Finished: returns 2 1






Fact. 3rd: N=3, Finished: returns 3*2





Fact 2nd: N 4 Finished: returns 4*6


Fact. 2nd: N=4, Finished: returns 4*6



Fact. 1st: N=5, Finished: returns 5*24



Main Function: finished: X = 120;

Recursive Function: XnRecursive Function: X• Xn = X * X(n‐1)

• When n <=1; Xn=X

• Power(base, exp)= base * Power(base, exp‐1);

int Power (int base, int exp){

Activations Stack ExampleRecursive Function: Xn

int Power (int base, int exp){if(exp <=0) return 1;return X *Power(base, exp‐1);

}}Power base = 3 exp = 0 Finished: 11

11

Power base = 3 exp = 2 3 *Power(3,1)

Power base = 3 exp = 1 3 *Power(3,0)1

333

Power base = 3 exp = 3 3 *Power(3,2)

Power base 3 exp 2 3 Power(3,1)

999

2727Power base = 3 exp = 4 3 *Power(3,3)27

2727

8181main: total = Power(3,4)Total =81

The nth power of X• Is there any better approach?• From basic algebrag

– if n is even == > Xn = Xn/2.Xn/2

– If n is odd and n=2m+1 ==> Xn= X2m+1= xm . xm . x • From this above fact, can we calculate Xn in fewer steps

int Pow (int X, int N){if(N <=0) return 1;if(N < 0) return 1;int Y=Pow(X,N/2);if(N %2==0) return Y *Y;l t X*Y*Yelse return X*Y*Y;

}

A More Complex Recursive FunctionFibonacci Number SequenceA More Complex Recursive Function

if n = 1, then Fib(n) = 1if n = 2, then Fib(n) = 1if n > 2, then Fib(n) = Fib(n‐2) + Fib(n‐1)

Numbers in the series:Numbers in the series:1, 1, 2, 3, 5, 8, 13, 21, 34, ...

Fibonacci Sequence FunctionFibonacci Sequence Function

int Fib (int n){int Fib (int n){if (n == 1) ||(n == 2)) return 1;return 1;

elsereturn Fib(n-2)+Fib(n-1);return Fib(n 2)+Fib(n 1);

}

main(){int answer; answer=fib(5);

}printf(“5 th Fib Num is %d”, answer);}

ThanksThanks

recursion, memory layout · • recursion – stack grows as deepen the nested function call –...

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