Hi~
There is A=[1:1:37] in matlab, two numbers are randomly drawn from the list A simultaneously, and a total of 74 times are drawn. Requires are as fllows:
1. Each number is drawn 4 times in total.
2. And the same number will not appear twice every time you draw
3. The same number will not be drawn for two consecutive trials.
The final result is expressed as a 74×2 list, each row represents an extraction, the first column indicates the first number drawn, and the second column indicates the second number drawn.

6 Kommentare

Stephen23
Stephen23 am 2 Aug. 2023
It looks like you will need a WHILE loop, or something similar. What have you tried so far?
Jane
Jane am 2 Aug. 2023
I have tried:
"drawnNumbers = datasample(fullList, 2, 'Replace', false);"
I wonder know how to fufill the reqirement 2?
Stephen23
Stephen23 am 2 Aug. 2023
"I wonder know how to fufill the reqirement 2?"
By specifying "Replace"=false you have already fulfilled requirement 2.
Jane
Jane am 2 Aug. 2023
I misread it just now, it should be requirement three.
Stephen23
Stephen23 am 2 Aug. 2023
Bearbeitet: Stephen23 am 2 Aug. 2023
"I misread it just now, it should be requirement three."
You could make a sample, then test if its elements are different compared to the last sample. If the values are duplicated make another sample (e.g. a WHILE loop).
However you should be thinking about requirement 1, because how you approach that will determine the entire structure of your algorithm and code, probably more than requirements 2 and 3.
Jane
Jane am 3 Aug. 2023
Yes, very sensible~

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Bruno Luong
Bruno Luong am 2 Aug. 2023
Bearbeitet: Bruno Luong am 2 Aug. 2023

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succeed = false;
while ~succeed
c = repelem(4,37,1);
n = sum(c);
r = rand(1,n);
y = [];
for i=1:n
cn = c;
cn(y) = 0;
cc = cumsum(cn);
s = cc(end);
succeed = s > 0;
if ~succeed
break
end
x = discretize(r(i), [0; cc/s]);
r(i) = x;
c(x) = c(x)-1;
y = r(max(i-mod(i,2)-1,1):i);
end
end
r = reshape(r,2,[])';
r
r = 74×2
15 22 9 13 22 2 11 35 1 26 25 20 19 4 22 25 17 13 23 22

2 Kommentare

Jane
Jane am 3 Aug. 2023
Fabulous! Thanks a lot.
Bruno Luong
Bruno Luong am 3 Aug. 2023
You are welcome.

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