CMPE260-Project 2 Solved

In this project, you will implement an automatic differentiation [1]. Automatic differentation is a computation technique to calculate partial gradients of an expression by a repetitive application of the chain rule of calculus [2]. This technique makes the gradient calculation of any expression very trivial. This is one of the very important methods (if not the most) for machine learning research. After completing this project, you will have a very minimal implementation of something like TensorFlow [3] or PyTorch [4] in which you can actually train an artificial neural network (albeit a bit slowly compared to them).

2        The Data Structure
We will first define a data structure called num which will have two fields: value and grad. value will hold the value of the number, and grad is for holding the partial gradient of a function with respect to this parameter. Don’t worry, you will not calculate the gradient by yourself, this will be automatically handled with some tricks.

;; plain, simplest version of this struct.

(struct num (value grad))

(define mynumber (num 12.3 1.0))

(num-value mynumber)

12.3

(num-grad mynumber)

1.0

mynumber #<num

I have added some additional descriptions to the num struct for printing so that the output will be (num 12.3 1.0) instead of #<num.

3      Helpers
You will implement the following two procedures which you can think as a warm-up exercise.

3.1 (get-value num-list)
This procedure will return the values of numbers in num-list as a list. If num-list is a num, then the return result should be a number.

Examples:

(get-value '((num 3 1) (num 4 2) (num 0.0001 0.0)))

'(3 4 0.0001))

(get-value (num 10 2)) 10

3.2 (get-grad num-list)
This procedure will return the gradient field of numbers in num-list as a list. If num-list is a num, then the return result should be a number.

Examples:

(get-grad '((num 3 1) (num 4 2) (num 0.0001 0.0)))

'(1 2 0.0))

(get-grad (num 10 2)) 2

4      Operators
You will extend +, -, and * operators so that they will also compute the gradients of their arguments. These operators will return a num struct. There will be normal return values of these operators in the value field of the number. In the grad field, there will be the derivative of the operator with respect to all of its arguments. So, given two numbers < v1,g1 , < v2,g2 , an operator f will return:

                                                                                                  (1)

Note that there may be more than two numbers for +, -, and * (see the examples below). If you feel like you cannot do it for n terms, just implement it for two terms for partial credit.

4.1 (add num1 num2 ...)
The derivative of add operator is the sum of gradients of its terms. This is rather easy, for the value, you will sum the values of numbers, and for the gradient, you will sum the gradients.

Examples:

(add (num 5 1) (num 2 0.2))

(num 7 1.2)

(add (num 5 1) (num 4 3) (num 6 2))

(num 15 6)

In these results, the first field is the value, and the second field is the gradient.

4.2 (mul num1 num2 ...)
For the value, you will multiply all the numbers’ values. For the gradient, the derivative of mul operator is the sum of multiplication of each term’s gradient with others’ values. For example:

 (2)

(3)

Hint: you can implement this recursively for n arguments. Again, if you have a hard time, implement it for only two arguments for partial credit. That would be rather easy.

Examples:

(mul (num 5 1) (num 2 1))

(num 10 7)

(mul (num 5 1) (num 4 3) (num 6 2)) (num 120 154)

4.3 (sub num1 num2)
For the value, you will subtract num2 from num1. For the gradient, the derivative of sub is the subtraction of num2’s gradients from num1’s gradients.

Examples:

(sub (num 5 1) (num 2 1)) (num 3 0)

(sub (num 5 0) (num 2 1))

(num 3 -1)

4.4 relu and mse
After you implement the primitive operations, you no longer need to implement the derivative for other stuff. You can basically calculate almost everything[1] using + and *. I have implemented rectified linear unit (ReLU, f(x) = max(x,0)) and mean square error function (mse, f(x,y) = (x − y)2) for you to use in the proceeding procedures.

(define relu (lambda (x) (if ( (num-value x) 0) x (num 0.0 0.0))))

(define mse (lambda (x y) (mul (sub x y) (sub x y))))

5       Higher-level procedures
In this part, you will implement some high-level helper functions that will help you parse the input and transform it to a valid expression, so that everything can be automated.

5.1 (create-hash names values var)
Given a list of symbols in names, respective values in values, and variable at interest in var, you will create a hash which contains numbers for each symbol with appropriate gradients. Here, initial gradients of each number will be 0 except for var, for which it will be 1.

Examples:

(create-hash '(a b c d) '(1 3 3 7) 'b)

'#hash((a . (num 1 0.0)) (b . (num 3 1.0)) (c . (num 3 0.0)) (d . (num 7 0.0))) (create-hash '(a b c d) '(1 3 3 7) 'c)

'#hash((a . (num 1 0.0)) (b . (num 3 0.0)) (c . (num 3 1.0)) (d . (num 7 0.0)))

This way, we will keep track of variables that we want to take the gradient. Numbers will 0 gradient will be automatically treated as constants.

5.2 (parse hash expr)
In this procedure, you will implement a simple parser which will translate the symbolic expression in expr into an appropriate expression by converting symbols in expr into their respective numbers in hash. For example, if we see x in expr, we will convert it into its number in hash. Operations in expr can be one of the following: +, *, -, mse, relu. Here, you have to convert * to mul, and so on. If you encounter a number in expr, you will construct the number with 0 gradient.

Examples:

(parse '#hash((x . (num 10 . 1.0)) (y . (num 20 0.0))) '(+ x y))

'(add (num 10 1.0) (num 20 0.0))

(parse (create-hash '(x y) '(10 20) 'x) '(+ x y))

'(add (num 10 1.0) (num 20 0.0))

(eval (parse '#hash((x . (num 10 . 1.0)) (y . (num 20 0.0))) '(+ x y)))

(num 30 1.0)

(parse (create-hash '(x y) '(10 20) 'x) '(+ (* (+ x y) x) (+ y x 5)))

'(add

(mul (add (num 10 1.0) (num 20 0.0)) (num 10 1.0))

(add (num 20 0.0) (num 10 1.0) (num 5 0.0)))

Here is a pseudocode to ease your work:

1.   If the expression is an empty list, return empty list.

2.   If the expression is a list, apply parse to the element of the list, apply parse to the rest of the list, and concatenate these two results.

3.   If the expression is is one of these: +, *, -, mse, relu, convert it to one of these: 'add, 'mul, 'sub, 'mse, 'relu.

4.   If the expression is a number, construct a number struct with 0 gradient.

5.   Else, it should be a symbol in our hash. Retrieve the number from the hash.

5.3 (grad names values var expr)
Given a list of symbols in names with respective values in values, take partial derivative of the expr with respect to var and return it. If you have implemented all the other parts, this procedure should be easy; you have all the tools you need.

Examples:

(grad '(x y) '(10 20) 'x '(+ (* (+ x y) x) (+ y x)))

41.0

(grad '(x) '(7) 'x '(* x x))

14.0

5.4 (partial-grad names values vars expr)
This is similar to grad except now we return a list of partial derivatives of expr with respect to each symbol in names. Here, if a symbol in names is not in vars, we treat it as a constant and its gradient should be 0. It should be easy to implement this procedure using grad (hint: for all elements in name, take grad if it is in vars, otherwise fill with 0).

Examples:

(partial-grad '(x y) '(10 20) '(x) '(+ (* (+ x y) x) (+ y x)))

'(41.0 0.0)

(partial-grad '(x y) '(10 20) '(y) '(+ (* (+ x y) x) (+ y x))) '(0.0 11.0)

(partial-grad '(x y) '(10 20) '(x y) '(+ (* (+ x y) x) (+ y x)))

'(41.0 11.0)

(partial-grad '(x y) '(10 20) '(x y) '(+ 1 2))

'(0.0 0.0)

5.5 (gradient-descent names values vars lr expr)
We are approaching the end. Now that we have everything in our toolbox, we can start optimizing our differentiable parameters (i.e. parameters that are in vars) to minimize an objective. In this procedure, you will implement a single iteration of the following equation for our parameter set:

                                                              values = values −lr ∗∇values(expr)                                                   (4)

Here, we change values of names in the direction that minimizes the value of expr [2]. If you have successfully implemented partial-grad, what you will return is just values - lr * partial-grad(names, values, vars, expr) (of course, in a functional programming style).

Examples:

(gradient-descent '(x y) '(10 20) '(x) 0.1 '(+ x y))

'(9.9 20.0) ;; x is updated to minimize x+y

(gradient-descent '(x y) '(10 20) '(x y) 0.1 '(+ x y))

'(9.9 19.9) ;; now both are updated to minimize x+y since both are in vars

(gradient-descent '(x y) '(10 20) '(x y) 0.001 '(+ x y))

'(9.999 19.999) ;; a smaller learning rate (lr)

(gradient-descent '(x y) '(10 20) '(x) 0.1 '(mse x y))

'(12.0 20.0) ;; now we try to bring x closer to y

5.6 (optimize names values vars lr k expr)
In this procedure, you will run gradient-descent iteratively k times. That is, the returning result of gradient-descent call should be the new values. In another words, you will call gradient-descent recursively k times and at each iteration you will its result for values in the next call. Here is an example for k=2:

(gradient-descent names (gradient-descent names vars lr expr) vars lr expr)

Examples:

(optimize '(x y) '(10 20) '(x) 0.1 100 '(+ x y))

'(1.8790524691780774e-14 20.0) ;; x is something closer to 0

(optimize '(x y) '(10 20) '(x y) 1 5 '(+ x y))

'(5.0 15.0)

;; 1000 iteration might take some time

(optimize '(x y) '(10 20) '(x) 0.01 1000 '(mse x y))

'(19.99999998317034 20.0) ;; don’t have to be exact

(optimize '(x y w1 w2 b) '(1 20 0.1 0.3 5) '(x) 0.001 1000

'(mse (+ (* x x w1) (* x w2) b) y))

'(10.83825591226222 20.0 0.1 0.3 5.0)

;; check for 0.1x2 + 0.3x + 5 = 20