# how can i have random numbers with criteria?

16 ビュー (過去 30 日間)
itay 2015 年 1 月 8 日
コメント済み: itay 2015 年 1 月 9 日
i need to have a 50 numbers matrice, while the numbers are between 1 to 5, and i need the neighbouring numbers to be different from each other. for example: 1 - 3 - 4 - 2 - 1 - 2 - 5 - 3 - 4 - 2...
i use:
r = [mod(randperm(50),5)+1];
but it gives me something like this: 1 - 2 - 4 - 4 - 3 - 4 - 5 - 5 - 5 - 3 - 1 - 2 - 2 - ....
how can i make the numbers to be different from the numbers before and after them?
thank you.

サインインしてコメントする。

### 回答 (3 件)

Guillaume 2015 年 1 月 8 日
Pick any of the 5 numbers for the first elements, for the subsequents, pick any of the four numbers making the set difference between 1:5 and the previous number:
r = [randi(5) randi(4, 1, 49)]; %number 2-50 are indices of which of the four numbers to pick in setdiff.
for idx = 2:numel(r)
rr = setdiff(1:5, r(idx-1)); %exclude previous number
r(idx) = rr(r(idx));
end
##### 3 件のコメント表示非表示 2 件の古いコメント
Image Analyst 2015 年 1 月 8 日
Uh yeah, dopey me (sound of hand slapping forehead).

サインインしてコメントする。

John D'Errico 2015 年 1 月 8 日

Easy peasy, and vectorized too. No setdiffs needed, no loops, no tests. So it will be quite fast. And essentially one line of code (if we ignore the fact that I defined n and k separately for clarity.)
For the numbers to be different from each other in sequence, this just means that each element must have a difference with its neighbors that is non-zero, modulo 5. I suppose you could also write it as a Markov process, where the state transition matrix is a simple one...
T = (ones(5) - eye(5))/4
T =
0 0.25 0.25 0.25 0.25
0.25 0 0.25 0.25 0.25
0.25 0.25 0 0.25 0.25
0.25 0.25 0.25 0 0.25
0.25 0.25 0.25 0.25 0
Anyway, the simple solution is to pick the first element randomly, then pick a set of random differences, doing arithmetic mod 5. I'll write it for a general set of n numbers, each of which must lie in the set of integers [1:k], but with no consecutive elements that are the same. For your problem, n=50, k=5.
n = 50;
k = 5;
r = 1 + mod(cumsum([randi(k),randi(k-1,[1,n-1])]),k);
r
r =
Columns 1 through 24
4 2 1 3 1 5 2 3 5 4 1 4 2 4 5 2 5 4 1 5 4 3 2 3
Columns 25 through 48
1 2 3 4 5 2 4 3 1 4 1 5 3 4 5 1 2 4 1 2 3 4 5 4
Columns 49 through 50
3 2
As you can see by looking at the difference vector, there are no zero differences.
diff(r)
ans =
Columns 1 through 24
-2 -1 2 -2 4 -3 1 2 -1 -3 3 -2 2 1 -3 3 -1 -3 4 -1 -1 -1 1 -2
Columns 25 through 48
1 1 1 1 -3 2 -1 -2 3 -3 4 -2 1 1 -4 1 2 -3 1 1 1 1 -1 -1
Column 49
-1
And the min and max elements are 1 and 5 respectively.
max(r)
ans =
5
min(r)
ans =
1
##### 1 件のコメント表示非表示 なし
itay 2015 年 1 月 9 日
thanks :]

サインインしてコメントする。

Isabella Osetinsky-Tzidaki 2015 年 1 月 8 日

r=nan(50,1); rng('shuffle') r(1)=randi([1,5]); n=1; while n<50, rng('shuffle') p = randi([1,5]); if p~=r(n), n=n+1; r(n)=p; end end
##### 1 件のコメント表示非表示 なし
John D'Errico 2015 年 1 月 8 日
Please use the code formatting button, else your answer is unreadable, just a long strung out mess. Just select your text, and click on the little button that says "{} Code".

サインインしてコメントする。

### カテゴリ

Find more on Logical in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by