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You misunderstood the problem. That's not what it's saying at all.
by yongjik 2y ago
You misunderstood the problem. That's not what it's saying at all.
- bongodongobob 2y agoHow isn't it? There's no way to have 3 connected friends and 4 unconnected friends.
- Dylan16807 2y agoYou can have all 6 connected and still be in either category. For example, 6 friends in a loop can be divided into even and odd trios, where each trio has no direct friend connections within it. But six people in two separate triangles fall into the other category, with two trios of mutual friends. Notice how both of these cases give each person 2 friends. It's not nearly as simple as assigning everyone a color at the start. Another thing to think about: If you have a triangle of 3, and a line of 3, then you can find 3 mutual friends and three mutual non-friends. How would you use colors to differentiate this situation from the two-triangles situation?
- bongodongobob 2y agoNo, that's what I'm saying. You're either a member of the 3-group or you're not. It clearly requires 3 members, so...
- Dylan16807 2y agoIt's proving the 3 nodes with no direct connections to each other that is the complicated part. It's not "you have a group of 3 or you don't". You can have all six people tangled up in lots of different combinations, so how do you prove there's a way to pick three that only connect indirectly?
- drekipus 2y agoYou could have one green and five blue
- bongodongobob 2y agoUh, exactly. You've got 3 blue in that 5.
- drekipus 2y agoSorry I was thinking too fast