Knuth Shuffle key-based

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Zeeshan Abbas
Zeeshan Abbas 2019 年 7 月 1 日
回答済み: Zeeshan Abbas 2019 年 7 月 4 日
I have this code of Knuth-Shuffle Algo in forward direction as below (Matlab):
X=[1 2 3 4 5 6];
n = numel(X);
for i = 2:n % Knuth shuffle in forward direction:
w = ceil(rand * i); % 1 <= w <= i
t = X(w);
X(w) = X(i);
X(i) = t;
end
I want to make it key dependent. i.e. whenever I change the key, the algorithm should give a different permutation.
  1 件のコメント
darova
darova 2019 年 7 月 1 日
waitforbuttonpress and waitfor?

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採用された回答

Zeeshan Abbas
Zeeshan Abbas 2019 年 7 月 4 日
X=[1:1:m];
n = numel(X);
key = 2538 %Just for example as it will be decided at run time
for i = 1:n % Knuth shuffle in forward direction: 1:n
w = mod(ceil((any fixed value) * key * i),m);
t = X(w+1);
X(w+1) = X(i);
X(i) = t;
end

その他の回答 (1 件)

Steven Lord
Steven Lord 2019 年 7 月 1 日
You want to control the random number generator, to allow you to generate random numbers repeatably? Use the rng function as shown on this documentation page.
  3 件のコメント
Steven Lord
Steven Lord 2019 年 7 月 1 日
What's the role of this KEY in permuting X? You want to control which elements get swapped? That requires controlling the value of w, which requires controlling the output of rand, which is exactly what rng does.
By the way, two ways in which your code can be made shorter:
  1. Rather than calling ceil(rand*i), consider using the randi function. It is designed specifically to return integer values.
  2. You don't need to use a temporary value to swap. You can use indexing.
sampledata = 1:10
valuesToSwap = [4 7];
sampledata(valuesToSwap) = flip(sampledata(valuesToSwap))
% or
sampledata(valuesToSwap) = sampledata(flip(valuesToSwap))
If you run both of the last two lines of code in that example, sampledata will be back to 1:10 since you will have swapped elements 4 and 7 twice.
Zeeshan Abbas
Zeeshan Abbas 2019 年 7 月 2 日
Thanks Steven. Actually I need not to use rand function that's why I have put a fixed value over there and taken mod. It's wroking fine now.
Here is the code below. Please, let me know if you find something wrong in it.
X=[1:1:202500];
n = numel(X);
key = 2538 %Just for example as it will be decided at run time
for i = 1:n % Knuth shuffle in forward direction: 1:n
w = mod(ceil(0.387 * key * i),202500);
t = X(w+1);
X(w+1) = X(i);
X(i) = t;
end

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