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However, the precise optimal strategy depends on the implementation considered There is also the option to learn the levels, as is done in the xed-probability-of-success method of [8]QR-Code Recognizer In Visual Studio .NETUsing Barcode scanner for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications.Analysis in a simpli ed setting: a coin- ipping model Creating Barcode In VS .NETUsing Barcode drawer for .NET framework Control to generate, create barcode image in VS .NET applications.Suppose we have already selected an importance function and one of the splitting implementations discussed in the previous section For a given total computation budget, we would like to nd the number and the locations of the thresholds, or equivalently the numbers n, p0 , , pn , that minimize the variance of the estimator We are also interested in convergence results for the variance and the work-normalized variance, under various asymptotic regimes, such as when N while n and p0 , , pn are xed, or when 0 and n Here we study these questions and provide partial answers under a very simpli ed (but tractable) model, for the xed-effort and xed-splitting strategies The main focus is on the asymptotic behavior when N Our simpli ed setting is a coin- ipping model uniquely characterized by the initial probability p0 = P(A0 ) (ie, the occurrence of the event A0 depends only on the outcome of a {0, 1} Bernoulli trial with parameter p0 ), and by the transition probabilities pk = P(Ak | Ak 1 ) (ie, the occurrence of Ak , conditional on Ak 1 , depends only on the outcome of a {0, 1} Bernoulli trial with parameter pk ), for k = 1, , n This model is equivalent to assuming that there is only a single entrance state at each level For the work-normalized analysis, we need to make some assumptions on how much work it takes, on average, to run a trajectory from a given level k 1 until it reaches either the next level or the set A = \ B (ie, the stopping time T without reaching B) If there is a natural drift toward A, it appears reasonable to assume that the chains will reach A in O(1) expected time, independently of n, if A and B (and therefore ) are xed If we use truncation and/or Russian roulette, we still have O(1) expected time Then the total expected work for all stages is proportional to n E[Nk ] This is the assumption we will make k=0 everywhere in this section, unless stated otherwise If E[Nk ] = N for all k, then this sum is N (n + 1) For simplicity, we will further assume that the constant of proportionality (in the O(1) expected time mentioned above) is 1Barcode Decoder In .NETUsing Barcode reader for VS .NET Control to read, scan read, scan image in .NET applications.SPLITTING TECHNIQUES QR-Code Creation In Visual C#.NETUsing Barcode printer for .NET framework Control to generate, create QR Code image in .NET framework applications.In a different asymptotic regime, where 0 and n jointly, and if truncation and/or Russian roulette 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