Two problems regarding for loops

I'm working on a set of practice problems, which should be enforcing the use of for loops, but I can't seem to figure out how to do these two. I'm completely lost.

2) Create an x vector that has integers 1 through 10, and set a y vector equal to x.  Plot this straight line.  Now, add noise to the data points by creating a new y2 vector that stores the values of y plus or minus 0.25.  Plot the straight line and also these noisy points.

3) Write a script beautyofmath that produces the following output.  The script should iterate from 1 to 9 to produce the expressions on the left, perform the specified operation to get the results shown on the right, and print exactly in the format shown here.

>> beautyofmath

1 x 8 + 1 = 9

12 x 8 + 2 = 98

123 x 8 + 3 = 987

1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

123456 x 8 + 6 = 987654

1234567 x 8 + 7 = 9876543

12345678 x 8 + 8 = 98765432

123456789 x 8 + 9 = 987654321

[Matt Kindig]

Here's what can get you started on 2)

    x = 1:10;  y=x;
    plot(x,y);
    y2 = y+0.25;    
    %now how would you extend this to the minus 0.25 case?

By the way, this first problem doesn't actually use loops.

For 3)

for k=1:9  %outer loop 
   value = 0;
   for m=1:k   %inner loop
      power10 = 10^(??);   %what goes here?
      value = value + ?? % how to calculate? Think about how digits work!
   end
   answer = value*??   %how to calculate the answer?
   fprintf('%d x %d + %d = %d\n', value, 8, k, answer);
end

[Jan]

"I'm completely lost" is less useful than showing, what you have tried so far, even if it is not working. It is much easier to suggest an improvement than to write a program from scratch. So please post more details instead of simply repeating the text of the homework question.

Answer 1

[sixwwwwww]

Dear Paul,

Here is the solution for your task (2):

x = 1:10;
y = x;
y2(1:2:9) = y(1:2:9) + 0.25;
y2(2:2:10) = y(2:2:10) - 0.25;
plot(x, y, x, y2), legend('y', 'y2')

And following is solution for your task (3):

for i = 1:9
    if i == 1
        j(i, 1) = i;
    else
        j(i, 1) = j(i - 1, 1) * 10 + i;
    end
    j(i, 2) = i;
    j(i, 3) = j(i, 1) * 8 + j(i, 2);
    fprintf('%d x 8 + %d = %d\n', j(i, 1), j(i, 2), j(i, 3))
end

I hope it helps. Good luck!

[Samuel Sharon]

For those of you out there trying to solve Beauty of Math, this solution is terrible! I'm not going to post a solution here, but if you're using a nested for loop or a matrix, you're overthinking it! This problem should have no more than 5 lines of code, including one line for the for statement and one line for the word "end."

[Jian David Knight]

Wasn't able to get it to five lines of code, however I did get it down to 8.

for i = 1:9
    fprintf('%d', 1:i)
    fprintf(' x 8 + %d = ', i)
    for j = flip(i:-1:1)
        fprintf('%d', 10-j)
    end
    fprintf('\n')
end

The ouput should be this:

>> beautyofmath2
1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321