20200402, 02:29  #12  
Nov 2003
16444_{8} Posts 
Quote:


20200402, 02:43  #13  
Romulan Interpreter
Jun 2011
Thailand
19·461 Posts 
Quote:
The puzzle is well formulated. I also get their number for the example, with a little Excel calculation. (edit: @virgo: you forgot the possibility of reverse infection, for example, B can also be infected by E, not only by A, assume he is lucky for 7 days, and meantime E gets the disease, by ADE or even ACDE route, then B gets it in the eighth day, from E). OTOH, the problem seems quite easy by brute force, 8 vertices, there are \(C_8^2=28\) possible edges, the matrix in fact is triangular (symmetric on the first diagonal, which is zero always), so there are just 28 bits of entropy, the search space is 2^28, less than a billion possible cases. This should be an extremely easy, few hours, pari "struggle" to calculate all probabilities and sort them in a list. (I didn't solve it yet, no time, but one idea would be to check if the probability of complete graph \(K_8\) is higher than 70%, first, to be sure there IS a solution. Then, start cutting edges in some fashion). Last fiddled with by LaurV on 20200402 at 03:05 

20200402, 03:06  #14  
Nov 2003
2^{2}·5·373 Posts 
Quote:
There are less than 2000000 such with 8 nodes. See http://oeis.org/A001187 

20200402, 03:09  #15 
Nov 2003
2^{2}·5·373 Posts 

20200402, 03:31  #16  
Romulan Interpreter
Jun 2011
Thailand
21067_{8} Posts 
Quote:
Last fiddled with by LaurV on 20200402 at 04:09 

20200403, 08:38  #17 
Nov 2003
1110100100100_{2} Posts 

20200403, 10:39  #18 
Romulan Interpreter
Jun 2011
Thailand
10001000110111_{2} Posts 
This is not what you initially said. I learned from you, to read it as it is, and not to interpret it. The two examples are the same, if I can label arbitrary the A node. Yes, the labeling is arbitrary. I always put A in the middle. Think about it. Arbitrary labeling means that, given a graph, I can label it the way I want. For 3 nodes, you have 4 cases (of connected, labeled graphs), but 2 are symmetric (ACB and ABC give the same probability), so you have to test 3 of them for probability, assuming you have a fast way to detect the symmetry. If not, you must test all 4. (Now, if you read the continuation of the discussion in the thread, you can see it was just nitpicking (or, better said, picking on you )).
Anyhow, this problem is extremely easy, once you come with the conditioned probability formula (which is actually the hardest part). Then, one way or the other, you have about 23 hundred millions cases to test. You can make a nice sorted list, in full, in memory, and pick the closest to 70 for the answer, but also provide the "full solution" to amaze the puzzle makers Edit: I have some vague idea about how I will spend my weekend tomorrow Last fiddled with by LaurV on 20200403 at 10:57 
20200403, 13:29  #19  
Nov 2003
2^{2}×5×373 Posts 
Quote:
The question is where A is placed on the graph. Quote:


20200403, 16:59  #20 
Apr 2020
4_{10} Posts 

20200405, 11:10  #21 
Romulan Interpreter
Jun 2011
Thailand
19·461 Posts 
Maybe for you...
Last fiddled with by LaurV on 20200405 at 11:15 
20200407, 10:31  #22  
Jul 2015
2×7 Posts 
what am i doing wrong?
Quote:
https://mersenneforum.org/images/smi...tra/picard.png 

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