Stopping a for loop when certain values have been reached

I am running a for loop at want it to finish when a given number of values have all occured at least once. The ouputs of the for loop range from 1-9 (intagers) and I am saving the values into an array. I want to find how many iterations it takes for every number in this range to have occured at least once but am unsure how to code this. Any ideas would be greatly apprectiated.

 Accepted Answer

values = [];
n_tries = 1000;
for i = 1:n_tries
values(i) = randi(9);
if all(ismember(1:9,values))
break
end
end
display(i)
i = 27
display(values)
values = 1×27
2 1 9 5 9 3 1 4 4 2 7 3 9 2 5 6 1 2 1 2 7 3 5 1 7 6 8
It's probably better to use a while loop, so you don't have to set a maximum number of iterations (n_tries):
values = [];
i = 0;
while ~all(ismember(1:9,values))
i = i+1;
values(i) = randi(9);
end
display(i)
i = 16
display(values)
values = 1×16
7 8 2 6 6 5 7 4 5 9 3 4 3 7 2 1

4 Comments

This is perfect thank you so much. Lastly, how would i run this trail say 100 times and find the mean i value? i.e the average number of iterations needed to have all the values 1-9.
n_trials = 100;
n_iterations_needed = zeros(1,n_trials);
for j = 1:n_trials
values = [];
i = 0;
while ~all(ismember(1:9,values))
i = i+1;
values(i) = randi(9);
end
n_iterations_needed(j) = i;
end
n_avg = mean(n_iterations_needed);
display(n_iterations_needed)
n_iterations_needed = 1×100
13 31 26 21 43 39 56 37 18 13 22 27 24 13 25 25 39 21 22 30 29 22 21 19 18 15 25 36 28 26
display(n_avg)
n_avg = 25.9300
Theoretical result is
9*(1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9) = 25.46
Not so far apart.
@Ben Hatrick Answer moved here:
This is a massive help, thanks so much!

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Asked:

on 16 Jan 2022

Edited:

on 17 Jan 2022

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