# Need to run two counters in a foor loop

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Adil Ali on 1 May 2021
Answered: Steven Lord on 1 May 2021
Here is my problem.
I have two lists of the same length = A. The lists are P and T.
I anticipate to be running a for loop (for ls=1:A) embedded in another foor loop (for i = 1:ls).
Below is what I would like to get when the loops are running:
ls = 1
i = 1
P(1) + T(1);
ls = 2
i = 1
P(1) + T(2);
i = 2
P(2) + T(1);
ls = 3
i = 1
P(1) + T(3);
i = 2
P(2) + T(2);
i = 3
P(3) + T(1);
ls = 4
i = 1
P(1) + T(4);
i = 2
P(2) + T(3);
i = 3
P(3) + T(2);
i = 4
P(4) + T(1);
....... and so on until the lenght of A
I need help in making the counter with T go backwards from the value of ls. Create a dynamic vector in the loop is not a good idea
for ls=1:A
K = ls:-1:1
for i=1:ls
P(i)+T(K(i))
end
end
Id appreciate for someone's help as I have spent 3 hours trying to come up with a solution for this and got to no where. Maybe a formula instead of the "????" below.
for ls=1:A
for i=1:ls
P(i)+T(????)
end
end
Thank you.

dpb on 1 May 2021
Edited: dpb on 1 May 2021
If you describe the real problem to be solved instead, it'll lead to more efficient getting to the right answer, but if the result is to simply sum the two variables in obverse order, you don't need loop at all --
S=P+flipud(T);
Adil Ali on 1 May 2021
That is right. My apologies. Here is the acutal problem.
I am simulating emittion of laser pulses. I create pre allocated array of zeros
pulses=zeros(length(t),4); %% need other columns for other information but the first column needs to be distance travelled by each pulse at time = t.
Then i run my laser for time = LaserOnTime and the pulses are emitted (one after the other) at PRF = 500kHz. so I create an array
t = 1:1/PRF:LaserOnTime;
for ls=1:length(t)
k = ls:-1:1;
for i=1:(ls)
if(pulses(i,1)<=2*Altitude)%%increment the distance of pulse until 2*alt is %achieved
pulses(i,1) = pulses(i,1) + SpeedOfLight*t(k(i)); %%increment the %distance of each pulse
end
end
end
This works but k=ls:-1:1 changes its size at each counter and this is not the most idea solution.
Thank you.

David Fletcher on 1 May 2021
One way of recreating your original sequence via the loops would be:
A=ones(1,5);
for iterator=1:length(A)
sub=0;
fprintf('outer loop: %d\n',iterator)
for ls=1:iterator
fprintf('P(%d) + T(%d)\n',ls,iterator-sub)
sub=sub+1;
end
iterator=iterator+1;
end
This gives:
outer loop: 1
P(1) + T(1)
outer loop: 2
P(1) + T(2)
P(2) + T(1)
outer loop: 3
P(1) + T(3)
P(2) + T(2)
P(3) + T(1)
outer loop: 4
P(1) + T(4)
P(2) + T(3)
P(3) + T(2)
P(4) + T(1)
outer loop: 5
P(1) + T(5)
P(2) + T(4)
P(3) + T(3)
P(4) + T(2)
P(5) + T(1)

Steven Lord on 1 May 2021
F = 3;
S = 4;
for first = 1:F
fprintf("Counting up, first is %d\n", first)
for second = S:-1:1
fprintf("Counting down, second is %d\n", second)
fprintf("The equivalent counting up inner counter would be %d\n", S-second+1)
end
fprintf(" \n")
end
Counting up, first is 1
Counting down, second is 4
The equivalent counting up inner counter would be 1
Counting down, second is 3
The equivalent counting up inner counter would be 2
Counting down, second is 2
The equivalent counting up inner counter would be 3
Counting down, second is 1
The equivalent counting up inner counter would be 4
Counting up, first is 2
Counting down, second is 4
The equivalent counting up inner counter would be 1
Counting down, second is 3
The equivalent counting up inner counter would be 2
Counting down, second is 2
The equivalent counting up inner counter would be 3
Counting down, second is 1
The equivalent counting up inner counter would be 4
Counting up, first is 3
Counting down, second is 4
The equivalent counting up inner counter would be 1
Counting down, second is 3
The equivalent counting up inner counter would be 2
Counting down, second is 2
The equivalent counting up inner counter would be 3
Counting down, second is 1
The equivalent counting up inner counter would be 4

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