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This type of problem involves the use of a random permutation In general the problem is to generate a random permutation of 1 , 2, , N, with all permutations being equally likely The randomness of the random permutation is, of course, limited by the randomness of the pseudorandom number generator Thus all permutations being equally likely is contingent on all the random numbers being uniformly distributed and independent We demonstrate that random permutations can be generated in linear time, using one random number per item A routine, permute to generate a random permutation is shown in Figure 106 In the permute routine the first loop initializes the permutation with 1, 2, , N The second loop performs a random shuffling In each iteration of the loop we swap a [ j 1 with some array element in positions 0 to j (it is possible to perform no swap) Clearly, permute generates shuffled permutations But are all permutations equally likely The answer is both yes and no The answer based on the algorithm is yes There are N! possible permutations, and the number of different possible outcomes of the h'- 1 calls to randomInt at line 11 is also N! The reason is that the first call produces 0 or 1 so it has two outcomes The second call produces 0 1, or 2, so it has three outcomes The last call has N outcomes The total number of outcomes is the product of all these possibilities because each random number is independent of the previous random numbers All we have to show is that each sequence of random numbers corresponds to only one permutation We can do so by working backward (see Exercise 106) However the answer is actually no-all permutations are not equally likely There are only 2" - 2 initial states for the random number generator,.



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How To: OCR with Aspose in C# - YouTube
Duration: 10:59 Posted: Jun 21, 2016

The auto-selecting security level feature was designed to select the lowest possible security level It accomplishes this feat by analyzing speci c features used in your form s design The minimal security is chosen based on the highest level requested by any feature But the maximum security level that the automatic feature can select is domain Full trust will never be automatically selected, for three reasons First, InfoPath cannot detect many features that require full trust permission, such as speci c properties or methods used in custom form code Moreover, some reasons why a form could be full trust require the template designer to make subjective decisions (eg, to hide cross-domain security prompts) Second, even if the automatic feature could choose full trust, the form cannot be deployed as full trust without some additional preparatory work done by the form designer (We ll discuss details in the next section on full trust) Third, and most importantly, assigning a maximum level of full trust to a form template should be left as a nal decision to the form designer We ll see in the next section that there are more implications than meets the eye Having any of the following features in your form will bump up the security level to domain from restricted when Automatically determine security level (recommended) is checked: Custom task panes Form code (including script or managed code) Database, Web service, or SharePoint query or submit connections XML document data connections (when the XML le is not included in the form template) HTTP submit Custom ActiveX controls Roles Rules that open a new document





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so there can be only 3 1- 2 different permutations This condition could be a problem in some situations For instance, a program that generates 1,000,000 permutations (perhaps by splitting the work among many computers) to measure the performance of a sorting algorithm almost certainly generates some permutations twice-unfortunately Better random number generators are needed to help the practice meet the theory Note that rewriting the call to s w a p with the call to r randomInt ( 0, 1) does not work, even for three elements There are 3! = 6 possible npermutations, and the number of different sequences that could be computed by the three calls to randomInt is 3 3 = 27 Because 6 does not divide 27 exactly, some permutations must be more likely than others

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