# SICP Exercise 1-3

##### Define a procedure that takes three numbers as arguments and returns the sum of squares of the two larger numbers

I broke this problem into *3 steps*

Define a `square`

function which will take one argument and return its square.

```
(define (square x) (* x x))
; (square 6) 36
```

Define a `sum-of-squares`

function which will take two arguments and calculate sum of their squares.

```
(define (sum-of-squares x y) (+ (square x) (square y)))
; (sum-of-squares 3 4) 25
```

Define a `max`

function which will take two numbers and return maximum of both numbers.

```
(define (max x y) (if (> x y) x y))
; (max 3 4) 4
```

Now the solution to the problem is just combine all the 3 subproblems into one.

`EDIT:`

In my earlier logic, i had a flaw. I was just passing `(max x y)`

`(max y z)`

to the `sum-of-squares`

function. Thanks to Ankur for pointing out the mistake. The second argument actually depends on the result of `(max x y)`

.

```
(define (sum-of-squares-of-2-larger x y z)
(sum-of-squares (max x y)
(if (= (max x y) x)
(max y z)
(max x z)
)))
; (sum-of-squares-of-2-larger 1 2 3) 13
; (sum-of-squares-of-2-larger 5 4 3) 41
; (sum-of-squares-of-2-larger 1 3 2) 13
```

SICP chapter 1 tells about how problems can be solved by breaking them into smaller subproblems and then building from them and solving bigger problems.