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    <p>Assuming each node has a constant, known and independent availability rate, A divide and conquer approach come to mind.</p> <p>Say you have <code>N</code> nodes.</p> <ol> <li>Split them into two sets of <code>N/2</code> nodes.</li> <li>For each side, compute the probability that any number of nodes (<code>[0,N/2]</code>) are down.</li> <li>Multiply and sum these as needed to find the probability that any number (<code>[0,N]</code>) of the full set is down.</li> </ol> <p>Step 2 can be done recursively and at the top level you can sum as need to find how often more than some number are down.</p> <p>I don't know the complexity of this but if I has to guess, I'd say at or below <code>O(n^2 log n)</code></p> <hr> <p>The mechanics of this can be illustrated on a terminal case. Say we have 5 nodes with up times <img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=p_1...p_5" alt="p1...p5">. We can split this into segments <code>A</code> with<img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=p_1...p_2" alt="p1...p2"> and <code>B</code> with <img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=p_2...p_6" alt="p3...p5">. We then can process these to find the "N nodes up" times for each segment:</p> <p>For A:</p> <p><img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=a_2=p_1p_2" alt="a_2"></p> <p><img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=a_1=p_1(1-p_2)%2b(1-p_1)p_2" alt="a_1"></p> <p><img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=a_0=%281-p_1)%281-p_2)" alt="a_0"></p> <p>For B:</p> <p><img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=b_2=p_3p_4p_5" alt="b_3"></p> <p><img src="https://chart.apis.google.com/chart?cht=tx&amp;chl=b_2..b_0" alt="b_2"></p> <p>The final results for this stage can be found by multiplying each of the <code>a</code>'s with each of the <code>b</code>'s and summing appropriately.</p> <pre><code>v[0] = a[0]*b[0] v[1] = a[1]*b[0] + a[0]*b[1] v[2] = a[2]*b[0] + a[1]*b[1] + a[0]*b[2] v[3] = a[2]*b[1] + a[1]*b[2] + a[0]*b[3] v[4] = a[2]*b[2] + a[1]*b[3] v[5] = a[2]*b[3] </code></pre>
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    1. POCalculating the probability of system failure in a distributed network
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    1. CO1) How can you generalize this approach to the case of distinct failure probabilities? Example for `N=20`: If the probabilities that the three nodes N2, N3 and N7 are down is different from N1, N4 and N5 you still have complexity `O(2^N)` since you need to take into account all those distinct cases. 2) How can you generalize this approach to the case of node correlations? I.e., if a failure of node No. 2 results in a failure of nodes [N/2-1,...,N]? Such a non-locality cannot be handled efficiently in a recursive algorithm.
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    2. COYour example case for A contains four terms, corresponding to a combination of four different cases leading to three possible outcomes. Complexity is thus `2^2=4`. Case B corresponds to four possible outcomes; had you explicitly written it down you would have b0, b1, b2, b3 with altogether `2^3=8` individual terms, each representing one of those eight cases. "Multiplying each of the `a`'s with each of the `b`'s and summing appropriately" generates six possible outcomes with altogether `2^5=32` terms. Thus, the complexity of your proposal is identical to the one in the original question.
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    3. CO@user8472: Multiplying each of the `a`'s with each of the `b`'s and summing appropriately generates six possible outcomes with altogether '4*3=12' terms. `a0*b0 ... a0*b3, a1*b0 ... a1*b3, a2*b0 ... a2*b3` for a much reduced complexity, and the improvement gets better higher up.
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