7.2: Flowcharts

By Carey A. Smith

A flowchart is a graphical representation of code logic. It is especially helpful to understand the logic of code with for loops and if-else logic.

An example is given here. More information can be found at:

https://en.wikipedia.org/wiki/Flowchart [en.Wikipedia.org]

Example \(\PageIndex{1}\) Odd-Even Flow Chart

Try to code this up
Solution

The corresponding code is:

for n = 1:12
    if(mod(n,2) == 0)
        disp([num2str(n),' is even'])
    else
        disp([num2str(n),' is odd'])
    end
end

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Example \(\PageIndex{2}\) flowchart for primes

This flowchart and code show an algorithm to determine which numbers between 1 and 48 are prime numbers.

We already know that 2, 3, and 5 are prime, so the for loop starts at k = 6.

 For each value it tests, it displays a message stating whether or not it is a prime number. It stores the prime numbers it finds in a vector called prime_list.

Try coding this logic.

Note that the for-loop automatically does the "k = k+1" shown at the bottom of the flowchart; do not code that line.

(There are more sophisticated and efficient prime-number algorithms; this one is an illustration of moderately complex logic.)

Solution

%% "for loop" example with “if-then-elseif” logic:
% Find the prime numbers between 1 to 48:
prime_list = [2,3,5] % Initial set of primes
for k = 6:48
    if( mod(k,2) == 0) % Tests if k is divisible by 2
        disp([num2str(k),' is not prime. It is divisible by 2'])
    elseif (mod(k,3) == 0) % Tests if k is divisible by 3
        disp([num2str(k),' is not prime. It is divisible by 3'])
    elseif( mod(k,5) == 0) % Tests if k is divisible by 5
        disp([num2str(k),' is not prime. It is divisible by 5'])
    else
        disp([num2str(k),' is prime.'])
        % Add k to the list of primes:
        prime_list = [prime_list, k];
    end % end if
end % for k = 6:48
% Display the primes found
prime_list

This code is the attached file for_loop_primes.m

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Example \(\PageIndex{3}\) Divergent Series1

This is an example of a divergent series. Instead of the terms getting smaller, these terms get larger in absolute value.

In this example the line continuation symbol of 3 dots is used to continue the code on the next line.

clear all; close all; format compact; clc;
n = 40; % maximum number of iterations
total = zeros(1,(n+1));
term_lim = 10; % Terminate the sequence if termk > term_lim
for k = 2:n
    termk = (-1)^k*(k^2)/(k+30);
    if( abs(termk) > term_lim)
        disp(['k= ',num2str(k),...
            ', termk = ',num2str(termk),...
            '>10, so break the loop'])
        break;
    else
        total(k) = total(k-1) + termk;
    end
end
figure;
plot(total)

Solution

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Homework:

1. Watch the following video from Robert Talbert

Then code the second example in the video about the temperature and rain. The corresponding flowchart is this:

 Figure \(\PageIndex{i}\): Flowchart for Temperature, Rain, Wearing a Coat.

Put parentheses around the conditional expressions, even though the video did not do this. Examples:

if (temp < 50)

elseif (raining == 1)

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Exercise \(\PageIndex{1}\) Divergent Series--Car Loss of Control

Read Matlab's description of the "break" keyword at this link: break keyword (Links to an external site.)(opens in new window)

Also watch this video: Break and Continue Statements in Matlab (Links to an external site.)(opens in new window)

% Divergent loop due to unstable feedback
% Car 1 (in freeway lane 1 next to the center divider) passes a truck in lane 2.
% Car 1 begins to enter lane 2.
% But car 1's driver sees car 3 in lane 3 (on the right) also start to change lanes into lane 2.
% So car 1's driver turns hard to the left.
% Due to the speed of the car and the delay to realize how far he has turned, each time he "over corrects".
% Set
r = 2.1 % Over correction factor
% Model this with the following, where x is the number of feet he has moved to the right.
% (Negative values are to the left.)

% Before the for loop, initial the vector of position vs. time with this code:

m = 1;
x(m) = 3; % position when he make his 1st correction

% Write a for loop with index m:

for m = 2:20

    In the for loop, include the following steps:
    % The change in position each iteration is
    dx = -r*x(m-1);
    % Add this to the current position:
    x(m) = x(m-1) + dx;
    % Use a compound if-statement such that if

    %   either the absolute value of dx is greater than 15,
    %   or the absolute value of x(m) is greater than 10,
    %   then use the "break" keyword to stop looping.

end

% After the for-loop, open a new figure and plot x.
figure;
plot(x)
% Turn on the grid
grid on
% Add this title:
title('Divergent car control loop')

% After running the code, check the value of the index, m. If m is >13, then the loop did not terminate when it should have. In this case, check the code to verify that the conditional statement used abs(dx) and abs(x(m)).

Answer

The solution is not shown here. The resulting figure should look like this:

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