Showing posts with label lispforc. Show all posts
Showing posts with label lispforc. Show all posts

Tuesday, August 25, 2015

Lisp for the C++ programmer: Numerical Integration

In the series of Lisp for the C++ programmer, we present some Common Lisp examples in a way similar to examples which are written in C++ or some other languages belong the same family tree of C++.

Here is the example of Riemann sum in Common Lisp. We first define a function f which takes a single parameter x and the function y = f(x) = x, for simplicity. We will change this function later. It is known that integration of this function from 0 to 1 is 0.50.

Common Lisp code for numerical integration (approximatly result) is:



 (defun f (x)  
     x  
 )  
 (defun integrate (f start stop)  
     (setq epsilon 0.0001)  
     (setq sum 0.0)  
     (loop for i from start to stop by epsilon do  
         (setq sum (+ sum (* (funcall f i) epsilon)))  
     )  
 sum  
 )  
 (print (integrate 'f 0 1))  

The result is 0.4999532. Now we can use a more complex function, for example a normal distribution function with zero mean and unit variance. This function can be defined in Common Lisp as


 (defun normal (x)  
     (setq a (* (/ 1 (sqrt (* 2 3.141592))) (exp (* -0.5 (* x x)))))  
 a  
 )  
 (defun integrate (f start stop)  
     (setq epsilon 0.0001)  
     (setq sum 0.0)  
     (loop for i from start to stop by epsilon do  
         (setq sum (+ sum (* (funcall f i) epsilon)))  
     )  
 sum  
 )  
 (print (integrate 'normal -1 1))  



In the code above, as it can clearly be seen, we integrate the standard normal distribution from -1 to 1 and the result is 0.6826645.

Lisp for the C++ programmer: Changing an element of a list

Here is the example of changing an element of a Common Lisp List and its equivalent code in C++ typed as List comment.






 ; #include <cstdlib>  
 ; #include <cstdio>  
 ;  
 ; int main(){  
 ;    double *d = (double*) malloc (sizeof(double) * 3);  
 ;    d[0] = 1;  
 ;    d[1] = 2;  
 ;    d[2] = 4;  
 ;  
 ;    puts("Current List:");  
 ;    printf("%f %f %f\n", d[0], d[1], d[2]);  
 ;  
 ;    puts("New List:");  
 ;    d[0] = 100;  
 ;    printf("%f %f %f\n", d[0], d[1], d[2]);  
 ;    return(0);  
 ;}  
   
   
 (setq aList (list 1 2 3))  
 (print "Current List:")  
 (print aList)  
   
 (setf (elt aList 0) 100)  
 (print "New List:")  
 (print aList)  
   
   

Lisp for the C++ programmer: for loop

Here is the example of for loop. Equivalent C++ code is commented on the top the Common Lisp code as comments.








 ; for (int i = 0; i <= 10; i++){  
 ;    printf("%f\n", i);  
 ;    printf("%f\n", i * 2);  
 ; }  
   
 (loop for i from 0 to 10  
     do  
     (progn  
         (print i)  
         (print (* i 2))  
     )  
 )  
   

Lisp for the C++ programmer: Function definitions and function calls

Here is the simple Common Lisp example, in which a sum and an ArithmaticMean functions defined. Both C++ and Common Lisp code here calculate the arithmetic mean of 1,2,3,4,5 and 6 which is 3.5. C++ code is commented as in the Common Lisp file.





 ; double sum (double *d, int len){  
 ;    double mysum = 0.0;  
 ;    for (int i = 0; i < len; i++){  
 ;        mysum += d[i];  
 ;    }  
 ;    return(mysum);  
 ; }  
 ;  
 ; double ArithmeticMean (double *d, int len){  
 ;    return ( sum(d) / len );  
 ; }  
   
 ; int main(){  
 ;    double *mylist = (double*) malloc(sizeof(double) * 6);  
 ;    for (int i = 0; i < 6; i++){  
 ;        d[i] = (double)i;  
 ;    }  
 ;    printf("%f\n", ArithmeticMean(d, 6);  
 ;    return(0); 
 ; }  
   
   
   
 (defun sum (aList)  
     (setq mysum 0.0)  
     (dotimes (i (length aList))  
         (setq mysum (+ mysum (nth i aList )))  
     )  
     mysum  
 )  
   
 (defun ArithmeticMean (aList)  
     (/ (sum aList) (length aList))  
 )  
   
 (print (ArithmeticMean '(1 2 3 4 5 6)))  
   

Lisp for the C++ programmer: cond expression

Here is the example for the cond expression of Common Lisp and its C++ equivalent.







 ; int x = 10;  
 ; if (x < 10) {  
 ;    puts("x is smaller than 10");  
 ; } else if (x > 10){  
 ;    puts("x is bigger than 10");  
 ; } else if (x == 10){  
 ;    puts("x equals to 10"));  
 ; }  
   
 (setq x 10)  
   
 (cond  
     ((< x 10) (print "x is smaller than 10"))  
     ((> x 10) (print "x is bigger than 10"))  
     ((= x 10) (print "x equals to 10"))  
 )