The answer is m=4 and p =2 , my question is why can't both m and p = 3 ?
#networks problem, please help
left sleet
vale lightBOT
gaunt nymph
At a glance it looks they could be, right? If you let m and b both be 3, then run Prims or Kruskals both edges m and b will be included. The only edge you’d have to consider being in competition with m and b is the right most 3, but even if you add that it looks like you’ll still add m=3 and p=3 because it doesn’t create a cycle