Gradient Descent for SLE

Problem #115

Tags: mathematics

Who solved this?

We are already acquainted with the gradient calculation for arbitrary math functions. Let us learn how to apply this knowledge!

Gradient descent is a popular method of optimization - it works like following:

  1. We have a target function of the form f(x1, x2, ... , xn) - and want to optimize (e.g. minimize) its value by finding the best suit of xi values.
  2. Probe the gradient of this function at some random point (x1, x2, ... , xn).
  3. This gradient shows as the direction (as a vector (dx1, dx2, ... , dxn) in which the function grows most speedily.
  4. Let us make a step in the opposite direction and change the set of xi values accordingly.
  5. Calculate new value of the function and if we are not satisfied, repeate with the new point from the step 2.

We'll implement this algorithm for solving System of Linear Equations - extremely popular task which is encountered in many fields, like business predictions (and other machine learning), graphics, physics etc.

System of n equations with n variables is generally represented like:

a11 * x1  +  a12 * x2  +  ...  +  a1n * xn  =  b1
a21 * x1  +  a22 * x2  +  ...  +  a2n * xn  =  b2
   ...          ...                  ...       ...
an1 * x1  +  an2 * x2  +  ...  +  ann * xn  =  bn

The solution of it is such a vector of [x1, x2, ... , xn] for which equations become true - i.e. for which the left parts becomes equal to right ones.

Let us construct the target function in the following way:

f1  =  a11 * x1  +  a12 * x2  +  ...  +  a1n * xn  -  b1
f2  =  a21 * x1  +  a22 * x2  +  ...  +  a2n * xn  -  b2
   ...          ...                  ...       ...
fn  =  an1 * x1  +  an2 * x2  +  ...  +  ann * xn  -  bn

f(x1, x2, ... , xn)  =  f1 * f1  +  f2 * f2  +  ...  +  fn * fn

i.e. it is the sum of squares of the differences between left and right parts for each of equations. Obviously the minimum of this function is 0 and it is reached when the vector of x-s equals to solution of the system.


Algorithm

You will need to understand and implement the proposed algorithm for the gradient descent of the target function. Below is the description in pseudocode. Your goal is to count how many iteration algorithm performs before finishing.

X = [0, 0, ... , 0]        # solution vector [x1, x2, ... , xn] is initialized with zeroes

STEP = 0.01                # step of the descent - it will be adjusted automatically
ITER = 0                   # counter of iterations

WHILE (true)
    Y = F(X)               # calculate the target function at the current point
    IF (Y < 0.0001)        # condition to leave the loop
        BREAK
    END IF

    DX = STEP / 10         # mini-step for gradient calculation
    G = CALC_GRAD(X, DX)   # G(x1, x2, ... , xn) just as in "gradient calculation" problem

    XNEW = X               # copy the current X vector
    FOR (i = 1 .. n)       # and make the step in the direction specified by the gradient
        XNEW[i] -= G[i] * STEP
    END FOR

    YNEW = F(XNEW)         # calculate the function at the new point
    IF (YNEW < YN)         # if the new value is better
        X = XNEW           # shift to this new point and slightly increase step size for future
        STEP = MIN(0.1, STEP * 1.25)
    ELSE                   # if the function value grows it means step was too large
        STEP = STEP / 1.25
    END IF

    ITER += 1              # increment iteration counter
END WHILE

So we end the descent when the value of target function becomes small enough (less than 0.0001) and return the number of iterations.

To calculate gradient of the function at the given point x1, x2, ... , xn we need to calculate its values at N neighboring points, each of the "staying aside" by the value dx along one of coordinate axes:

y = f(x1, x2, ... , xn)

y1 = f(x1 + dx, x2, ... , xn)
y2 = f(x1, x2 + dx, ... , xn)
  ...
yn = f(x1, x2, ... , xn + dx)

After which the vector representing the gradient is composed like following:

g1 = (y1 - y) / dx
g2 = (y2 - y) / dx
  ...
gn = (yn - y) / dx

g = [g1, g2, ... , gn]

If you recollect "gradient calculation" problem - there we had letters f instead of y and x, y instead of x1, x2. (however for given task it is impractical to name many coordinates with separate letters, so we use x with indexes)

Example

Below is an example of first iterations for sample system with N = 3:

Matrix of coefficients "aij":
8 -1 -8
-4 2 9
-8 -9 2

Vector of right sides "bi":
9 -5 -7

"x1", "x2", "x3" and "step" on each iteration:
0:  0.00000 0.00000 0.00000   step: 0.01
1:  0.00000 0.00000 0.00000   step: 0.008
2:  0.00000 0.00000 0.00000   step: 0.0064
3:  0.00000 0.00000 0.00000   step: 0.00512
4:  0.00000 0.00000 0.00000   step: 0.004096
5:  0.00000 0.00000 0.00000   step: 0.0032768
6:  0.96977 0.28826 -0.85868  step: 0.004096
7:  0.96977 0.28826 -0.85868  step: 0.0032768
8:  0.26577 0.10317 -0.15674  step: 0.004096
9:  0.26577 0.10317 -0.15674  step: 0.0032768
10: 0.82771 0.24396 -0.66572  step: 0.004096
11: 0.82771 0.24396 -0.66572  step: 0.0032768

Input data will contain the number of systems to solve in the first line.
For each system there would be the description of several lines.
At the beginning the amount of equations / variables (N) is specified.
Then N lines with N coefficients aij follow.
The last line of the section contains the vector of N values of bi.
Answer should contain the amount of steps for the given algorithm to finish, for each case. We allow these answers have a small error of +/- 2 or 5% of the required value (which is greater).

Example:

input data:
4                # number of systems to solve
3                # the first system is of 3*3 size
-9 2 -1          # 3 lines with coefficients follow
1 -3 0
-3 -1 -7
-2 4 3           # the vector of right side values
4                # the second system is of 4*4 size
0 -4 -1 6        # and so on...
7 7 6 9
-6 3 -7 8
-7 -5 1 -4
1 9 -2 9
3
9 4 -8
-8 1 7
-6 -3 -8
6 4 6
5
6 5 -9 2 -1
-4 -6 -6 -6 8
7 6 4 2 -9
-2 -3 -9 -7 -5
-8 2 8 -7 -5
-2 -1 -2 1 3

answer:
101 114 178 279
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