## Exercise 2.2 (Power)

```Ocaml
(*First version ; 6 multiplications*)
# let power28(x) =
	let x2 = x + x in
	let x4 = x2*x2 in
	let x8 = x4*x4 in
	let x16 = x8*x8 in
	x16*x8*x4 ;;

(*Second version ; 27 multiplications*)
# let power28(x) = x*x*x*x*x*x...*x ;;

(*Third version ; 11 multiplications*)
# let power28(x) = 
	let sq(x) = x*x in
	let pow4(x) = sq(sq(x)) in
	pow4(pow4(x))*sq(pow4(x))*pow4(x);;

(*Fourth version ; 7 multiplications*)
# let power28(x)=
	let sq(x) = x+x in
	let p4=sq(sq(x)) in
	sq(sq(p4))*sq(p4)*p4;;
```

## Exercise 2.3

```Ocaml
# (*my verison*)
# let mirror(n) = let diz = n/10 and uni = n mod 10 in uni*10 + diz;;
val mirror : int -> int = <fun>

# (*teatcher version*)
# let mirror n = 10 *(n mod 10)+n/10;;
val mirror : int -> int = <fun>
```

```Ocaml
# let abba(n) = n*100 + mirror(n) ;;
val abba: int -> int = <fun>
```

```Ocaml
# let stammer(n) = abba(mirror(n)) * 10 000 + abba(n) ;;
val stammer: int -> int = <fun>
```



## Exercice 2.6
```Ocaml
let sec_of_time h m s = 
	h*3600 + m*60 + s ;;

let time_of_sec s = 
	let hours = s/3600 in let minutes = (s - hours*3600)/60 in let seconds = s - hours *3600 - minutes * 60 in (hours, minutes, seconds);;

let add_times h1 m1 s1 h2 m2 s2 = 
	let sec1 = sec_of_time h1 m1 s1 and sec2 = sec_of_time h2 m2 s2 in let resultsec = sec1 + sec2 in time_of_sec resultsec ;;

```

## Exercise 3.1
```Ocaml
let f a b c =
	(if a > b && if b > c then
		a + b else c + a
	 else if a > c then a + b else b + c)*
	 (if a > b && b > c then
		 a - b else a - c
	 else if a > c then a-b else b-c);;
```

### Exercise 3.2
```OCaml
(*logical and*)
if a == b then 
	true
else false

(*logical or*)
if a && b then
	true
else if a && not b then
	true
else if not a && b then
	true
else 
	false

(*logical implication*)
if a && b then
	true
else if not a && b then
	true
else if not a && not b then
	true
else 
	false
```

### Exercise 3.3
```Ocaml
let max2 number1 number2 = if number1 > number2 then number1 else number2
let min2 number1 number2 = if number1 > number2 then number2 else number1

let max3 number1 number2 number3 = if number1 > number2 && number1> number2 then number1 else if number2 > number3 && number2>1 then number2 else number3

let min3 number1 number2 number3 = if number1 < number2 && number1< number2 then number1 else if number2 < number3 && number2<number1 then number2 else number3

let middle3 number1 number2 number3 = let min = min3(number1 number2 number2) and max = max3(number1 number2 number2) in if min = number1 && max = number3 then number2 else if min = number2 && max = number1 then number3 else number2

let max4 number1 number2 number3 number4 = let nb1 = max2(number1 number2) and nb2 = max2(number3 number4) in if nb1>nb2 then nb1 else nb2

let min4 number1 number2 number3 number4 = let nb1 = min2(number1 number2) and nb2 = min2(number3 number4) in if nb1<nb2 then nb1 else nb2
```
### Exercise 3.4
```Ocaml
let highest_square_sum x1 x2 x3 = let bigger = max3(x1 x2 x3) and middle = middle3(x1 x2 x3) in (bigger*bigger middle*middle)

```