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Id
6953958
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0
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2
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CreationDate
2011-08-05T09:11:18.910
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2011-08-05T09:11:18.910
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0
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810312
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6947612
PostTypeId
2
Score
15
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0
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text
Body
Pure mathematics: Let's calculate the quantity of <code>rand()</code> function calls in both cases and compare the results: Case 1: let's see the mathematical expectation of calls on step <code>i = k</code>, when you already have k numbers chosen. The probability to get a number with one <code>rand()</code> call is equal to <code>p = (n-k)/n</code>. We need to know the mathematical expectation of such calls quantity which leads to obtaining a number we don't have yet. The probability to get it using <code>1</code> call is <code>p</code>. Using <code>2</code> calls - <code>q * p</code>, where <code>q = 1 - p</code>. In general case, the probability to get it exactly after <code>n</code> calls is <code>(q^(n-1))*p</code>. Thus, the mathematical expectation is <code>Sum[ n * q^(n-1) * p ], n = 1 --> INF</code>. This sum is equal to <code>1/p</code> (proved by wolfram alpha). So, on the step <code>i = k</code> you will perform <code>1/p = n/(n-k)</code> calls of the <code>rand()</code> function. Now let's sum it overall: <code>Sum[ n/(n - k) ], k = 0 --> m - 1 = n * T</code> - the number of <code>rand</code> calls in method 1. Here <code>T = Sum[ 1/(n - k) ], k = 0 --> m - 1</code> Case 2: Here <code>rand()</code> is called inside <code>random_shuffle</code> <code>n - 1</code> times (in most implementations). Now, to choose the method, we have to compare these two values: <code>n * T ? n - 1</code>. So, to choose the appropriate method, calculate <code>T</code> as described above. If <code>T < (n - 1)/n</code> it's better to use the first method. Use the second method otherwise. 
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1. POGenerating m distinct random numbers in the range [0..n-1]
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1. PTAnswer
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1. USGrigor Gevorgyan
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1. POGenerating m distinct random numbers in the range [0..n-1]
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