I want to make a recursive formula and execute two statements with the same variables at the same time
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clc
clear all
a=1
b=0
while irrelevant
b = [a] % The answer I want from this statement is b = [1]
a = [a b] % The answer I want from this statement is [1 0] , I dont want to use calculated b (from the statement before) but b=0 ('the first b').
end
% I dont want to really want to make b0,b1,b2,b3 because this needs to become recursive with an 'while' and 'for' statement.
% So I really want to execute b = [a] and a = [a b] at the same time; but I cannot figure out how to fix this.
ADDON:
Thank you for your suggestion.
Although it doesn't really work for our problem.
We'd like the following to be recursive:
... etc ...
clc
clear all
a1=1
b1=0
b2=[a1]
a2=[a1 b1]
b3=[a2]
a3=[a2 b2]
b4=[a3]
a4=[a3 b3]
... etc ...
3 件のコメント
Rik
2020 年 10 月 12 日
Please don't remove substantial parts of your question. It vastly reduces the usefulness of the answers to future readers.
採用された回答
Walter Roberson
2020 年 10 月 11 日
NO.
You cannot do what you want in MATLAB.
At the moment, I cannot think of any programming language that supports what you are asking for.
0 件のコメント
その他の回答 (4 件)
Alan Stevens
2020 年 10 月 11 日
What about a temporary variable for b:
...
bt = b;
b = [a];
a = [a bt];
...
Steven Lord
2020 年 10 月 12 日
Use a cell array (or two cell arrays, though b is a copy of most of a.)
a = {1};
b = {0};
cellToFill = 2;
while cellToFill < 20
b{cellToFill} = a{cellToFill-1};
a{cellToFill} = [a{cellToFill-1}, b{cellToFill-1}];
cellToFill = cellToFill + 1;
end
Instead of referring to numbered variables, refer to cells in a and b.
a{4}
b{5}
1 件のコメント
Steven Lord
2020 年 10 月 12 日
So you need all permutations of 8 ones and 4 zeros? That's a very different question than the one you asked.
If that's not the full question please state exactly what you're trying to do. In particular, what restrictions are there (if any) about where the 1's can be placed.
Alan Stevens
2020 年 10 月 12 日
I'm obviously not understanding something here (not unusual!),. Doesn't the following do what you want:
a = 1;
b= 0;
n = 5;
[a, b] = recur(a,b,n);
disp('Recursive')
disp(a)
disp(b)
% Compare with
a1 = 1; b1 = 0;
b2 = a1; a2 = [a1 b1];
b3 = a2; a3 = [a2 b2];
b4 = a3; a4 = [a3 b3];
b5 = a4; a5 = [a4 b4];
b6 = a5; a6 = [a5 b5];
disp('Incremental')
disp(a6)
disp(b6)
function [a, b] = recur(a,b,n)
if n>1
bt = b;
b = a;
a = [a bt];
n = n-1;
[a, b] = recur(a,b,n);
else
bt = b;
b = a;
a = [a bt];
end
end
This produces the following comparison:
Recursive
1 0 1 1 0 1 0 1 1 0 1 1 0
1 0 1 1 0 1 0 1
Incremental
1 0 1 1 0 1 0 1 1 0 1 1 0
1 0 1 1 0 1 0 1
0 件のコメント
Bruno Luong
2020 年 10 月 12 日
編集済み: Bruno Luong
2020 年 10 月 12 日
I'm surpised nobody proposes yet a very MATLABish solution
[a,b] = deal([a b],a]
0 件のコメント
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