First, I tried sketching it out, but that obviously was going to be too difficult (or far too time-consuming). Therefore, I made six columns, labeling them the six groups: Arachnid, Feline, Canine, Lupine, Bovine, and Porcine. Then I went through the paragraph with the distances and under each column I would write which connected to which and the distance. This sheet looked as:
Arachnid-Canine (15), Porcine (1)
Feline-Porcine (6), Lupine (11)
Canine-Arachnid (15), Lupine (3), Bovine (7)
Lupine-Bovine (12), Canine (3), Feline (11)
Bovine-Lupine (12), Porcine (9), Canine (7)
Porcine-Feline (6), Bovine (9), Arachnid (1)
From here, I started calculating the shortest difference. Basically, I took the two camps and wrote it out as stated. Then I took the possible paths from the first camp, including the distance to the next possible path. I proceeded to do this to get to the camp we were looking for. Basically, I did all possible combinations, and went with the shortest one. The asterisks are where I felt it unnecessary to continue the scenario. It is as follows, as I’m sure the work explains it better than I did;
Feline to Bovine
Shortest Distance is: Feline to Porcine to Bovine
Porcine (6) + Bovine (9) = 15
Lupine (11) + Bovine (12) = 23
Lupine to Porcine
Shortest Distance is: Lupine to Feline to Porcine
Bovine (12) + Porcine (9) = 21
Canine (3) + Arachnid (15) + Porcine (1) = 19
Canine (3) + Lupine (3) + Bovine (12) + Porcine (9) = 27
*Canine (3) + Lupine (3) + Canine = did not finish, goes back to Canine and definitely would not be the shortest route
Canine (3) + Lupine (3) + Feline (11) + Porcine (6) = 23
Canine (3) + Bovine (7) + Porcine (9) = 19
Feline (11) + Porcine (6) = 17
Canine to Feline
Shortest Distance is: Canine to Lupine to Feline
*Arachnid (15) + Canine (15) + () =already over the lupine/feline route distance
*Arachnid (15) + Porcine (1) + () =already over the lupine/feline route distance
Lupine (3) + Feline (11) = 14
*Bovine (7) + Lupine (12) + () = already over the lupine/feline route distance
*Bovine (7) + Porcine (9) + () = already over the lupine/feline route distance
*Bovine (7) + Canine (7) + () = already over the lupine/feline route distance
Arachnid to Lupine
Shortest Distance is: both the [Arachnid to Canine to Lupine] and the [Arachnid to Porcine to Feline to Lupine] are the same distance (18)
Canine (15) + Lupine (3) =18
Porcine (1) + Feline (6) + Lupine (11) =18
Porcine (1) + Bovine (9) + Lupine (12) =22
*Porcine (1) + Arachnid (1) + () =will be over the route distance
Canine to Porcine
Shortest Distance is: both the [Canine to Arachnid to Porcine] and the [Canine to Bovine to Porcine] are the same distance (16)
Arachnid (15) + Porcine (1) = 16
*Lupine (3) + Bovine (12) + () =will equal more than the Canine to Bovine to etc. route
*Lupine (3) + Canine (3) + Arachnid (15) + () =Equals more than the shortest distance
*Lupine (3) + Canine (3) + Lupine (3) + () =will equal more because it goes back to Lupine
Lupine (3) + Canine (3) + Bovine (7) + Porcine (9) =22
Lupine (3) + Feline (11) + Porcine (6) =20
Bovine (7) + Porcine (9) =16
Lupine to Bovine
Shortest Distance is: Lupine to Canine to Bovine
Bovine (12) = 12
Canine (3) + Bovine (7) =10
*Feline (11) + Porcine (6) =will equal more than the shortest distance
Arachnid to Feline
Shortest Distance is: Arachnid to Porcine to Feline
*Canine (15) + Arachnid (15) + () =will equal more than the shortest distance
*Canine (15) + Lupine (3) + () = will equal more than the shortest distance
*Canine (15) + Bovine (7) + () = will equal more than the shortest distance
Porcine (1) + Feline (6) =7
It took me a while to figure this out-probably half an hour. There is probably a much quicker way, but this worked for me and I am fairly certain that it is correct. I am not very good at math and writing it all out seems to help. I did cut a few corners by using asterisks when I knew that an equation was not going to be the shortest distance.