%Program to compute w and r
clear
num = input("Enter your value for n: ");
n=1:num;
w(1)=8000;
for i=2:num
w=[w(0.98*w(i-1)-250*(i-1)/((i-1)+1))];
end
r(1)=1.50*w(1);
for i=2:num
r=[r(1.50*w(i))];
end
format bank;
disp("N Weight Revenue");
disp([n', w', r']);
The output I am getting is:
A(I): Index out of bound; value 7715 out of bound 1
what mistake I am making here?
The error message also shows the line number where the error happened which you haven't posted. it's this line
w=[w(0.98*w(i-1)-250*(i-1)/((i-1)+1))];
Here you are trying to access element 7715 (=0.98*8000-250/2) but w only has size 1x1. This always happens for out of bounds read, for example:
octave:> a = [5 2 3]
a =
5 2 3
octave:> a(4)
error: A(I): index out of bounds; value 4 out of bound 3
Related
I hope you can help me with my request.
I have an integer input from a client. Lets say 250. The goal is to calculate the input to exactly 0 by only using 3 variables. x = 30, y = 50 and z = 100 and output how often i used each of them.
The given rules are:
I have to use as less of them as possible. means, first the z's then the y's then the x's. So i have to use as much as possible of the variables with greater values.
Im allowed to use a maximum of 3 of each of them.
I have to count how often i used each of them.
If a combination is not possible, lets say 70 ( 70 - y = 20, 70 - x - x = 10 ), give a message to the client, "Invalid input!".
Few examples:
Input = 250
Output:
z = 2
y = 1
####################
Input = 170
Output: Invalid input!
Because i can only calculate to zero by using 4 x's and 1 y. See rule 2.
Other solutions are not possible.
i.e. Input - z - y = 20 => error
Input - y - y - y = also 20 => error
####################
Input = 90
Output:
x = 3
####################
Input = 120
Output: Invalid input!
Input - z = 20 => error
Input - 50 - 50 = 20 => error
Input - x - x - x - x => See rule 2.
####################
and so on....
And how it could be extended if a further value is added to the 3 variables. i.e. 40
So if thats the case it should be somehowe dynamically.
I tried to explain that as much as possible. If you miss infos, just let me know.
Thank you.
dagogi
So I have this matrix:
E1 = [54 5 2 4;4 5 19 29;31 4 2 9; 1 3 99 34]
lets say I want to find the location of the value closest to 18.9. let A = 18.9
I would do
[r,c] = find(E1==min(min(abs(E1-A))))
This doesn't work. It returns r = "[](0x1)" and c = "[](0x1)"
however,
if I first do:
F = abs(E1-A) and then do
[r,c] = find(F==min(min(F)))
this gives r = 2 and c = 3 which is correct. 19 is the closest value and 19 lives in row 2 column 3.
Why doesnt this work then? F is simply abs(E1-A) so why can I not put abs(E1-A) in place of F in the find formula?
min(min(abs(E1-A)))
ans = 0.10000
This gives you the min over the absolute difference. Then you compare it to E1 which has absolute values. This is complete different from your second formular
[r,c] = find(F==min(min(F)))
where you comapre the minimum difference with the matrix containing the absolute of differences between E1 and A. If you replace in your second formula F with abs(E1-A) you would get
[r,c] = find(abs(E1-A)==min(min(abs(E1-A))))
Which would also work. Nevertheless I would suggest another approach:
E1 = [54 5 2 4;4 5 19 29;31 4 2 9; 1 3 99 34];
A = 18.9;
# get the index ( Column-major order) of the minimum
idx = nthargout (2, #min, abs (E1-A)(:));
# this returns 10
# convert it ro row, column
[r, c] = ind2sub (size (E1), idx)
r = 2
c = 3
I am new to programming and I would like to know how to solve questions like this. I was told to expect questions like this on the exam. Can someone please tell me how I would go about solving something like this? Thanks.
x = 0
for num in range(5):
if num % 2 == 0:
x = x + 2
else:
x = x + 1
print(x)
You need to work on a skill which is to "be the compiler", in the sense that you should be able to run code in your head. Step through line by line and make sure you know what is happening. In you code example, you have
for num in range(5) means you will be iterating with num being 0,1,2,3 and 4. Inside the for loop, the if statement num % 2 == 0 is true when num/2 does not have a remainder (how % mods work). So if the number is divisible by 2, x = x+2 will execute. The only numbers divisible by 2 from the for loop are 0,2 and 4. so x=x+2 will execute twice. The else statement x = x +1 runs for all other numbers (1,3) which will execute 2 times.
Stepping through the for loop:
num = 0 //x=x+2, x is now 2
num = 1 //x=x+1, x is now 3, print(x) prints 3
num = 2 //x=x+2, x is now 5
num = 3 //x=x+1, x is now 6, print(x) prints 6
num = 4 //x+x+2, x is now 8
Therefore the answer is that 3 and 6 will be printed
In my opinion,
Whatever language you are using, you need to learn some common elements of the modern programming languages, such as flow-control (if...else in your case), loop(for, in your case)
Some common used functions, in your case, you need to what does range do in Python,
docs.python.org is a good place for you.
As you are new to programming, you can go with the flow in you mind or draw it on the paper.
Using x to store our final result
loop through every item in [0, 1, 2, 3, 4] <- range(5)
a. if
the number is divisible by 2
then increase x by adding 2 to it.
b. else
increase x by adding 1 and print it out
So the result would be :
3
6
I am trying to understand if it's possible to use Octave more efficiently by removing the for loop I'm using to calculate a formula on each row of a matrix X:
myscalar = 0
for i = 1:size(X, 1),
myscalar += X(i, :) * y(i) % y is a vector of dimension size(X, 1)
...
The formula is more complicate than adding to a scalar. The question here is really how to iterate through X rows without an index, so that I can eliminate the for loop.
Yes, you can use broadcasting for this (you will need 3.6.0 or later). If you know python, this is the same (an explanation from python). Simply multiply the matrix by the column. Finnaly, cumsum does the addition but we only want the last row.
newx = X .* y;
myscalars = cumsum (newx, 1) (end,:);
or in one line without temp variables
myscalars = cumsum (X .* y, 1) (end,:);
If the sizes are right, broadcasting is automatically performed. For example:
octave> a = [ 1 2 3
1 2 3
1 2 3];
octave> b = [ 1 0 2];
octave> a .* b'
warning: product: automatic broadcasting operation applied
ans =
1 0 6
1 0 6
1 0 6
octave> a .* b
warning: product: automatic broadcasting operation applied
ans =
1 2 3
0 0 0
2 4 6
The reason for the warning is that it's a new feature that may confuse users and is not existent in Matlab. You can turn it off permanentely by adding warning ("off", "Octave:broadcast") to your .octaverc file
For anyone using an older version of Octave, the same can be accomplished by calling bsxfun directly.
myscalars = cumsum (bsxfun (#times, X, y), 1) (end,:);
I have a cell array, A. I would like to select all rows where the first column (for example) has the value 1234 (for example).
When A is not a cell array, I can accomplish this by:
B = A(A(:,1) == 1234,:);
But when A is a cell array, I get this error message:
error: binary operator `==' not implemented for `cell' by `scalar' operations
Does anyone know how to accomplish this, for a cell array?
The problem is the expression a(:,1) == 1234 (and also a{:,1} == 1234).
For example:
octave-3.4.0:48> a
a =
{
[1,1] = 10
[2,1] = 13
[3,1] = 15
[4,1] = 13
[1,2] = foo
[2,2] = 19
[3,2] = bar
[4,2] = 999
}
octave-3.4.0:49> a(:,1) == 13
error: binary operator `==' not implemented for `cell' by `scalar' operations
octave-3.4.0:49> a{:,1} == 13
error: binary operator `==' not implemented for `cs-list' by `scalar' operations
I don't know if this is the simplest or most efficient way to do it, but this works:
octave-3.4.0:49> cellfun(#(x) isequal(x, 13), a(:,1))
ans =
0
1
0
1
octave-3.4.0:50> a(cellfun(#(x) isequal(x, 13), a(:,1)), :)
ans =
{
[1,1] = 13
[2,1] = 13
[1,2] = 19
[2,2] = 999
}
I guess the Class of A is cell. (You can see in the Workspace box).
So you may need to convert A to the matrix by cell2mat(A).
Then, just like Matlab as you did: B = A(A(:,1) == 1234,:);
I don't have Octave available at the moment to try it out, but I believe that the following would do it:
B = A(A{:,1} == 1234,:);
When dealing with cells () returns the cell, {} returns the contents of the cell.