Save variable number in matrix for later iteration

Ok, this is the code I have:

clc, clearvars, home,

lim = input("Limit value: ");

V = zeros(1,lim);
V(1) = 5;
for i = 2:lim
    if mod(i,2) == 0
        V(i) = V(i-1) + 2;
    else
        V(i) = V(i-1) + 4;
    end
end

I = zeros(1,lim);
for i = 1:lim
    I(i) = (V(i)^2-1)/3;
end

J1 = zeros(1,lim);
J2 = zeros(1,lim);

J1(1) = 3;
J2(1) = 7;
for i = 2:lim
    if mod(i,2) == 0
        J1(i) = J1(i-1) + 2;
        J2(i) = J2(i-1) + 2;
    else
        J1(i) = J1(i-1) + 2;
        J2(i) = J2(i-1) + 6;
    end
end

J12 = [J1; J2];
J = sum(J12,1);

C = J-I;

K(1) = @(x) (x+C(1))/J(1);
for i = 2:lim
    K(i) = @(x) K{i-1} + (x+C(i))/J(i);
end

Val = zeros(1,max(I));
K = 0;
count = 0;
for i = 1:max(I)
    if ismember(i,I) == 1
        count = count + 1;
        K = K + K1(count)*i + K2(count);
        Val(i) = i/4 - (K);
    else
        Val(i) = i/4 - (K);
    end
end

It basicaly calculates three matrixes with values I, J and C with length "lim" and now I want to make a for loop where the value is for each i

Val(i) = i/4 - K

where I want K to be

K = i-C/J % count times

The thing is I want K to grow each time i appears in the matrix I, so for 2 occurences its desired value is

K = (i-C(1))/J(2) + (i-C(2))/J(2)

for 4 occurences is

K = (i-C(1))/J(2) + (i-C(2))/J(2) + (i-C(3))/J(3) + (i-C(4))/J(4)

and so.

The problem I'm facing is how to call a variable that contains a value that can change with the current index in the loop. I tried with functions, with the symbolic toolbox... And I'm currently stucked as all I make contains some kind of error.

If it's not clear what I want to make respond so I will clear it out. Thanks in advance for the help.

Answer 1

[Torsten]

Maybe like this ?

I'm not sure whether you want to take the elements of C and J according to the order of the elements in I or sort(I). In the case below, I and sort(I) are identical since I is strictly increasing.

lim = 4;
V = zeros(1,lim);
V(1) = 5;
for i = 2:lim
    if mod(i,2) == 0
        V(i) = V(i-1) + 2;
    else
        V(i) = V(i-1) + 4;
    end
end
I = zeros(1,lim);
for i = 1:lim
    I(i) = (V(i)^2-1)/3;
end
I

I = 1×4

8 16 40 56

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J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;
for i = 2:lim
    if mod(i,2) == 0
        J1(i) = J1(i-1) + 2;
        J2(i) = J2(i-1) + 2;
    else
        J1(i) = J1(i-1) + 2;
        J2(i) = J2(i-1) + 6;
    end
end
J12 = [J1; J2];
J = sum(J12,1)

J = 1×4

10 14 22 26

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C = J-I

C = 1×4

2 -2 -18 -30

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Val = (1:max(I))/lim;
count = 0;
for k = I
   count = count + 1;
   for j = k:max(I)
      Val(j) = Val(j) - (j + C(count))/J(count);
   end
end
Val(8)

ans = 1

8/4 - ((8+2)/10)

ans = 1

Val(9)

ans = 1.1500

9/4 - ((9+2)/10)

ans = 1.1500

Val(18)

ans = 1.3571

18/4 - ((18+2)/10 + (18-2)/14)

ans = 1.3571

[Joan]

This is exactly what I wanted, thank you so much!!! It only had to be changed the first definition of "Val" as it has to be divided by 4 and not by "lim", but the rest runs flawlessly. Again, thank you.

Answer 2

[Stephen23]

"The problem I'm facing is how to call a variable that contains a value that can change with the current index in the loop. I tried with functions, with the symbolic toolbox... "

All you are doing is adding numbers, so you do not need complicated things like function handles or the symbolic toolbox. Here is a much simpler approach using SUM:

lim = 4;
% Vectorized V calculation using alternating pattern
increments = 2*ones(1,lim-1);
increments(2:2:end) = 4;  % [2, 4, 2, 4, ...]
V = 5*ones(1,lim-1);
V(2:lim) = 5+cumsum(increments)

V = 1×4

5 7 11 13

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% Vectorized I calculation
I = (V.^2 - 1) / 3;
I = round(I)

I = 1×4

8 16 40 56

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% Vectorized J1 and J2 calculation
J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;

increments_J1 = 2*ones(1,lim-1);
J1(2:lim) = 3 + cumsum(increments_J1)

J1 = 1×4

3 5 7 9

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increments_J2 = 2*ones(1,lim-1);
increments_J2(2:2:end) = 6;  % [2, 6, 2, 6, ...]
J2(2:lim) = 7 + cumsum(increments_J2)

J2 = 1×4

7 9 15 17

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J = J1 + J2

J = 1×4

10 14 22 26

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C = J - I

C = 1×4

2 -2 -18 -30

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% Vectorized Val calculation with cumulative K
Val = zeros(1,max(I));
K = 0;

for i = 1:max(I)
    % Find all indices where I equals the current i
    indices = find(I==i);
    % Vectorized accumulation for all matching indices
    if ~isempty(indices)
        K = K + sum((i - C(indices)) ./ J(indices));
    end
    Val(i) = i/4 - K;
end
K

K = 7.8298

Val

Val = 1×56

0.2500 0.5000 0.7500 1.0000 1.2500 1.5000 1.7500 1.4000 1.6500 1.9000 2.1500 2.4000 2.6500 2.9000 3.1500 2.1143 2.3643 2.6143 2.8643 3.1143 3.3643 3.6143 3.8643 4.1143 4.3643 4.6143 4.8643 5.1143 5.3643 5.6143

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[Stephen23]

Compared to your original (non-vectorized) code:

lim = 4;

V = zeros(1,lim);
V(1) = 5;
for i = 2:lim
    if mod(i,2) == 0
        V(i) = V(i-1) + 2;
    else
        V(i) = V(i-1) + 4;
    end
end
V

V = 1×4

5 7 11 13

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I = zeros(1,lim);
for i = 1:lim
    I(i) = (V(i)^2-1)/3;
end
I

I = 1×4

8 16 40 56

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J1 = zeros(1,lim);
J2 = zeros(1,lim);

J1(1) = 3;
J2(1) = 7;
for i = 2:lim
    if mod(i,2) == 0
        J1(i) = J1(i-1) + 2;
        J2(i) = J2(i-1) + 2;
    else
        J1(i) = J1(i-1) + 2;
        J2(i) = J2(i-1) + 6;
    end
end
J1

J1 = 1×4

3 5 7 9

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J2

J2 = 1×4

7 9 15 17

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J12 = [J1; J2];
J = sum(J12,1)

J = 1×4

10 14 22 26

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C = J-I

C = 1×4

2 -2 -18 -30

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[Joan]

First of all, thanks a LOT for this response. I'm trying to learn to vectorize and this helps a ton.

Having said this, my main problem still persists as I want all values marked as "i" in "K" to change with every iteration. For exemple, for i=8 I want

Val(8) = 8/4 - ((8+2)/10) = 1

for i=9

Val(9) = 9/4 - ((9+2)/10) = 23/20

for i=18

Val(18) = 18/4 - ((18+2)/10 + (18-2)/14) = 19/14

and so.

The code as it is now assingns one "i" value to each set of "K" and I want it to change for each iteration.

PD: I have corrected the formula that I want. Before I posted K=(i-C)/J and I wanted to say K=(i+C)/J, sorry for the confusion that this may cause.

[Stephen23]

lim = 4;

% Vectorized V calculation using alternating pattern
increments = 2*ones(1,lim-1);
increments(2:2:end) = 4;  % [2, 4, 2, 4, ...]
V = 5*ones(1,lim);
V(2:lim) = 5 + cumsum(increments)

V = 1×4

5 7 11 13

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% Vectorized I calculation
I = round((V.^2 - 1) / 3)

I = 1×4

8 16 40 56

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% Vectorized J1 and J2 calculation
J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;

increments_J1 = 2*ones(1,lim-1); % [2, 2, 2, ...]
J1(2:lim) = 3 + cumsum(increments_J1)

J1 = 1×4

3 5 7 9

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increments_J2 = 2*ones(1,lim-1);
increments_J2(2:2:end) = 6;  % [2, 6, 2, 6, ...]
J2(2:lim) = 7 + cumsum(increments_J2)

J2 = 1×4

7 9 15 17

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J = J1 + J2

J = 1×4

10 14 22 26

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C = J - I

C = 1×4

2 -2 -18 -30

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% Initialize Val
Val = (1:max(I)) / 4;
% Apply corrections for each element in I
for count = 1:lim
    k = I(count);
    % Subtract (j + C(count))/J(count) from all Val(j) where j >= k
    Val(k:end) = Val(k:end) - ((k:max(I)) + C(count)) ./ J(count);
end

% Display results
Val(8)

ans = 1

8/4 - ((8+2)/10)

ans = 1

Val(9)

ans = 1.1500

9/4 - ((9+2)/10)

ans = 1.1500

23/20

ans = 1.1500

Val(18)

ans = 1.3571

18/4 - ((18+2)/10 + (18-2)/14)

ans = 1.3571

19/14

ans = 1.3571