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>> No.11319809 [View]
File: 94 KB, 850x531, __patchouli_knowledge_and_remilia_scarlet_touhou_drawn_by_terimayo__sample-ad96284c916a2e5c6b26afd100f16342.jpg [View same] [iqdb] [saucenao] [google]
11319809

>>11318083
For a graph G with n vertices, if a vertex a has degree k, then the connected component it's in needs to have at least k+1 vertices. For another vertex b, if it has degree larger than n-k-1, then it's connected component needs to have more than n-k vertices.
Assuming a and b are on distinct connected components, you have n>k+1+n-k=n+1, and thus a and b are in the same component, by contradiction. If all vertices other than a have degree larger than n-k-1, then all vertices are on the same connected component as a, and the graph is connected.
>>11318674
Lewd.
>>11319744
Set [math]\Sigma_{k=1} ^n x_k = a[/math]. Then the new LHS for the induction becomes [math](a+x_{k+1})^2 = a^2 + 2ax_{k+1} + x_{k+1} ^2[/math].

>> No.11140661 [View]
File: 94 KB, 850x531, __patchouli_knowledge_and_remilia_scarlet_touhou_drawn_by_terimayo__sample-ad96284c916a2e5c6b26afd100f16342.jpg [View same] [iqdb] [saucenao] [google]
11140661

>>11140636
>wake up
>turn on light
>make alarm stop
>have two or three coffees in a row in a state of dizziness while checking your email, fiddling with random objects and thinking about stuff
>brush teeth while pissing
>dress up and apply deodorant
>check if you picked up your keys
>leave home

>> No.10241797 [View]
File: 94 KB, 850x531, __patchouli_knowledge_and_remilia_scarlet_touhou_drawn_by_terimayo__sample-ad96284c916a2e5c6b26afd100f16342.jpg [View same] [iqdb] [saucenao] [google]
10241797

>>10241790
Yeah, but if you put all the circles the image becomes a huge mess.

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