Week 10 - Programming III

• So far:– Branches/condition testing– Loops

• Today:– Problem solving methodology– Examples

• Textbook chapter 7

Problem Solving Methodology

1. Define the problem and your goals2. Identify information (inputs) and desired

results (outputs)3. Select mathematical approach

(assumptions, scientific principles, solution method)

4. Write a program (use array operations) 5. Test your program (with a simple case)

Example 1: line fitting

Given N pairs of values (xi,yi), find the best straight line fit

y = m x + b

m ≡ slope of line, b ≡ y intercept

Problem solving methodology:

1. State the problem: find best line to fit data

2. Identify inputs and outputs:

(xi,yi) (m,b)

3. Describe algorithm: least squares

4. Write program

5. Test program

• method of least squares:

2

11

2

111

1

1

N

jj

N

jj

N

jj

N

jj

N

jjj

xN

x

yxN

yx

m

N

jj

N

jj x

N

my

Nb

11

1

Method of least squares

• Fit line, y = mx + b, to points

• Find m and b to Minimize the error,

• Solve for m and b

• Program flow:1. Read in data:

• How many data points• Get paired data

2. Compute best fit for m and b

3. Provide output to user:• Numerical values of both m and b• Graphical presentation

% Comments on the program…% Get number of pairs, n_pointsn_points = input('How many (x,y) pairs? ');

% Read in the input data x, yfor ii = 1:n_points temp = input('Enter [x y]: '); x(ii) = temp(1); y(ii) = temp(2);end

typical use of a loop,but could have user just enter vectors!

N

jj

N

jj x

N

my

Nb

11

12

11

2

111

1

1

N

jj

N

jj

N

jj

N

jj

N

jjj

xN

x

yxN

yx

m

% Calculate m and bm = ( sum(x.*y) – sum(x)*sum(y)/n_points )/( …

sum(x.^2) – sum(x)^2/n_points );b = ( sum(y) – m*sum(x) )/n_points;

% Plot the data

plot(x,y,'bo')

hold on

twopts = [ min(x), max(x) ];

plot( twopts, m*twopts+b ,'r-','Linewidth',2)

hold off

% Add labels, etc.

gtext([' y = ',num2str(m), ' x + ',num2str(b)])

xlabel('x data'), ylabel('y data'), title('line fit')

Typical result:

Simple check to

see if it works

Example 2: sorting

Given N numbers (x1,…xN), sort them into increasing order

3 1 4 5 2 7 1

0 1 2 3 4 5 7

Algorithm:

• Given N numbers, x1…xN:

– Search through x1…xN for the smallest, swap with x1

– Search through x2…xN for the smallest, swap with x2

– Continue a total of N-1 times

3145270

0145273

0145273

0125473

0123475

0123475

0123457

start

done

Program employs nested loops:

function array = simpsort(array)% function to sort from smallest to largest for k = 1:length(array)-1

loc = k; for k2 = k:length(array)

if array(k2)<array(loc) loc = k2;

end end if loc ~= k

array([k,loc ]) = array([loc,k]); endend

locate the index (loc) of the smallest value

swap values, if necessary

Another sort

function y = upsort(x)

% UPSORT sorts a vector x of any length.

% UPSORTs output y is the vector x with elements arranged

% in ascending order

z = x; % Put x in a temporary vector z.

for n=1:length(x)

[y(n),k]=min(z); % y(n) is the smallest element of z.

z(k)=[]; % Erase the smallest element of z.

end

Example 3: equation solving

• 3 x + 7 y = 0 and 2 x – 5 y = 4 (simultaneous linear equations)

• x2 + 4 x + 7 = 0 (rooting polynomials)

• x – sin x – 0.0236 = 0 (transcendental equations ?? )

Newton-Raphson method:• Assuming an equation of form f(x) = 0

• From the Taylor series expansion

f(x) = f(x1) + ( x – x1 ) f ’(x1) + …

keep only the first 2 terms

f(x) f(x1) + ( x – x1 ) f ’(x1) = 0

and solve for x

x = x1 – f(x1) / f ’(x1)

• Thinking of x1 as a first guess, then

x = x1 – f (x1) / f ’(x1)

yields a better estimate of the root

• Repeating is the NR iteration:

xk+1 = xk – f (xk) / f ’(xk)

Back to the example f(x) = x – sin x – 0.0236

• NR iteration is xk+1 = xk – f(xk) / f ’(xk) or

k

kkkk x

xxxx

cos1

0236.0sin1

implement this recursion; compare different values for first guess

x = -1

diff = 1;

while diff > 0.001

xnew = x - ( x - sin(x) - 0.0236 )/( 1 - cos(x) );

diff = abs(xnew-x);

x = xnew

end

compute untilit converges

abs to detect size of difference

Comparison of starting points: x – sin x – 0.0236 = 0

Start at x0= – 1.0

Start at x0= 0.1

iteration

value

Homework: tic-tac-toe analysis

• Find the winner in a 3-by-3 tic-tac-toe board. Assume representations:

– Empty cell = 0– X = +1– O = – 1

Problem solving methodology:

1. State the problem: look for occurrences of +1 or – 1 in 3 adjacent cells

2. Identify inputs and outputs:

board array winner or not

3. Describe algorithm:

4. Write program

5. Test program

011

110

111

111

111

111

win for X, column of 1’s sums to 3

draw – no empty cells,and no winner

Sum(board) and trace(board) are useful,

where board is the 3x3 array representing the game.