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#34 | |
Jan 2020
22×7 Posts |
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According to your computations, after ten days the cities are contaminated ABOVE 100%. You noticed it, and you hid it with well placed if statements. But the main problem is that you are counting many infection events more than once. As an example, on day 2 the correct values should be (using your format): A = 10000, B = 1900, C = 1981, D = 1981, E = 199. Last fiddled with by 0scar on 2020-04-12 at 22:19 Reason: Adding correct values on day 2 |
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#35 | |
"Kebbaj Reda"
May 2018
Casablanca, Morocco
2·31 Posts |
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Day 0 infection B: 0 Day 1 infection B: 1000 (10000 * 0.1) Day 2 infection B: 2000 (1000(day 1) + 10000 * 0.1 = 2000) Where did your 1900 come from for city B. Please answer with calculations. Last fiddled with by Kebbaj on 2020-04-13 at 08:49 |
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#36 | |
Jan 2020
22·7 Posts |
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1000+9000*0.1= 1900. Otherwise, some people infected on previous days are counted over and over again, quickly exceeding 100%. |
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#37 |
"Kebbaj Reda"
May 2018
Casablanca, Morocco
2·31 Posts |
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Yes your reason is better than mine. I had seen that the source city A with a high viral load of 10000 infected cases can count 1000 healthy cases in city B.
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#38 | |
Jan 2017
8610 Posts |
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Anyone have an idea what the newly added bonus thing means? It says:
Quote:
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#39 | |
Jan 2020
348 Posts |
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My guess: somewhere before day 81. |
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#40 |
Oct 2017
101 Posts |
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Are there many equivalent adjacency matrices with 70,0074873 % after 10 days or is my code wrong?
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#41 |
Sep 2017
1348 Posts |
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Not sure, but there must be many equivalent matrices, at least the permutations of all the rest of the points except the originating city, minus the degenerate cases due to isomorphic connections. If the graph is dense, there will be many degenerate cases, but if the it is sparse, you will have little number of degenerate cases.
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#42 |
Jul 2015
23 Posts |
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Yes i am getting this result also not sure if it is good enough
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#43 | |
"Hugo"
Jul 2019
Germany
31 Posts |
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The optimum result of 70.007... for 10 days and 8 nodes corresponds to only one unlabeled connected graph. There are several ways to chose the root node in this graph all leading to the same probability of "all nodes infected". Since the updated description says
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#44 | |
Oct 2017
101 Posts |
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Could that be „Konjunktiv“? We also would accept... (wir würden auch...akzeptieren)? A question for a native speaker! Last fiddled with by Dieter on 2020-04-27 at 20:38 Reason: Complement |
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