OCAML SAMPLE PROGRAMME

Exercise 3: Write a function area for computing the area of a circle whose diameter is given.
Solution1:
let pi = 22.0 /. 7.0 ;;(* Defining PI *)
let area : float->float =
function r-> pi *. r *. r;;
area 5.0;;(* Test Cases *)
area 3.0;;(* Test Cases *)
Solution2:
let pi = 22.0 /. 7.0 ;;(* Defining PI *)
let square : float->float =
function x -> x *. x ;;
let area : float->float =
function r-> pi *. square(r);;
* if PI is define locally(in side function), give more
credit
* With a square function will be efficient


Exercise 5 Write a Caml function to determine the roots of a quadratic equation. Figure out a way
to handle the case when the the roots are not real.
Solution1:
let desc : float*float*float->float =
function (a,b,c)->(b *. b) -. 4.0 *. a *. c ;;
let mdesc : float*float*float->float =
function (a,b,c)-> 4.0 *. a *. c -. (b *. b);;
let root: float*float*float->string =
function (a,b,c) ->
if ( desc(a,b,c) > 0.0 ) then
"root1=" ^ string_of_float(( -. b +. sqrt(desc(a,b,c) ) /. 2.0 *. a) )
^ "root2=" ^ string_of_float(( -. b -. sqrt(desc(a,b,c)) /. 2.0 *. a) )
else
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"root1=" ^ string_of_float((-. b)/. 2.0 *. a)^ " + "
^ string_of_float( sqrt( mdesc(a,b,c) ) /. 2.0 *. a) ^ " i "
^ "root2=" ^ string_of_float((-. b)/. 2.0 *. a)^ " - "
^ string_of_float( sqrt( mdesc(a,b,c) ) /. 2.0 *. a) ^ " i " ;;
Solution2:
a)mdesc is not required
b) In place of else part output, they may have output "No Real Root"
c) If root is float*float*float->float*float, they can not handle
string and output should be (0,0)
Solution3:
let desc : float*float*float->float =
function (a,b,c)->(b *. b) -. 4.0 *. a *. c ;;
let root: float*float*float->float*float =
function (a,b,c)= if ( desc(a,b,c) > 0.0 ) then
( ( -. b +. sqrt(desc(a,b,c) ) /. 2.0 *. a),
( -. b -. sqrt(desc(a,b,c)) /. 2.0 *. a))
else
(0.0,0.0);;



Exercise 6 Let d be an integer and m be a string. Write a Caml function that returns true iff d and
m form a valid date (assume non-leap year). For example 31 April is not a valid date.


Soution:
let validdate: int*string -> bool =
function (date, month)->
if( date <1>31 ) then
false
else
if (
month=="Jan"
(month=="Feb" && date<29) month="="" month="="" month="="" month="="" month="="" month="="" month="="" month="="" month="="" month="="" rel="nofollow" name="3">Page 3
Solution:
gcd(m, n) = gcd(m, n + m) is same as gcd(m, n + m) = gcd(m, n)
Adding -m to 2nd term
gcd(m, n+m-m ) = gcd(m,n-m)
Again it is same as gcd(m, n ) = gcd(m,n-m);
let rec gcd : int * int -> int =
function (m,n) ->
if (m==n)
then m
else if(m>n)
then gcd(n,m-n)
else
gcd(m,n-m) ;;
Solution1: for speeding up the convergence, mod operator can be used..
let rec gcd : int * int -> int =
function (m,n) ->
if (m==n)
then m
else if(m>n)
then gcd(n,m mod n)
else
gcd(m,n mod m) ;;
Exercise 9 Using the observation that ${x^k}^2= x^(2k)$, give an alternate recursive definition of
the function power(x, n) that computes $x^n$ and write a corresponding Caml program.
Hint : If n is odd then n-1 is even !
let rec power: int*int->int =
function (x,n)->
if (n==0) then 1
else if ((n mod 2)==0)
then power(x, n/2)*power(x,n/2)
else power(x, (n-1)/2)*power(x,(n-1)/2)*x;;
* Credit for Handling Base Case
* Extra Credit for Handling Negative and Negative Power Case
• Use of Square function may degrade the performance


Find out the value of max_int without using the the constant max_int
Solution1:


Assuming that the max_int number is a general number it may not be in the format
(2^n)-1.
Use of binary search ..to find max int and it have two phase, one is
multiplication phase and another is addition phase.
Page 4
example
Assume 123 is maximum ..
start with 1..
(Multiplication Phase)
search procedure start searching at 2 and go for 4,8,16,32,64,...
after then we encounter a negative..
(Addition Phase)
it will go for 96(64+32), 112(96+16), 120(112+8),
if we add 4 to 120 again we encounter a negative so
we will add 2 instead of 4..so go for 122(120+2) and then go
for (122+1)..and 123 is max_int
Start searching from zero
let rec findmax_addphase int*int ->int
function x,y -> if(y=0) then x
else
if ( x+ (y/2) > 0 )
then findmax_addphase(x+y,y/2)
else
findmax_addphase(x,y/2);;
let rec findmax_mulphase:int->int =
function x-> if( x*2 < max_int =" 1073741823" 1 ="2^(30)-1" 1 =" 2^(29)-1+2^(29)">int =
function x-> if( x*2 <>int =
function x->
if(power(2,x)>0) then findmax1(x+1)

else power(2,x-1)-1+power(2,x-1);;
findmax1(0);;
(* Wrting a function which make bound the input to 0 only*)
let rec findmax:int->int =
function x-> if(x<0>0) then findmax(0)
else findmax1(0);;