Ant cannot travel to opposite side of cube in even amount of steps Say an ant starts at vertex 1 of a cube and wants to travel to vertex 8. The ant cannot do this in 4 steps, nor any even amount of steps in general. How can I express this? It seems that after every first move of one, the number is guaranteed to be odd because there will be an even amount of moves made after the additional first move.

 A: The cube graph is an example of a bipartite graph, where the vertices can be  divided into two group such that the edges always connect a vertex from one group to a vertex in the other group, never to another vertex in the same group. 
The group can be seen in your diagram by moving vertex $1$ in amongst the other odd-numbered vertices and vertex  $8$ likewise into the even-numbered group. Then you will always traverse an odd number of edges to reach a vertex in the other group from the starting vertex, plus all cycles are even length, etc.
A: Indeed the graph is bipartite and 1 and 8 are on opposite sides (all edges go from one side to the other). Going from one to the other can only be done in an odd number of steps.

The graph can also be shown to be layered, so that the shortest path from 1 to 8 has length 3.

A: The 8 vertices of the graph can be divided into 2 disjoint sets $V_1=\{1,3,5,7\}$ and $V_2=\{2,4,6,8\}$. Each edge of the graph connects a vertex of $V_1$ to a vertex of $V_2$, therefore, moving from a point of one of the sets to another point of the same set requires an even number of steps, and moving from one set to another requires an odd number of steps. $1\in V_1$ and $8\in V_2$, so the trip from one to another cannot be made in an even number of steps.
A: Notice that every dot is connected to 3 other dots and these three dots are of different parity ( if first dot is even then other three will be odd and if first dot is odd then the other three will be even). So if you start with even you will go to odd in next step and similarly when you start from odd you will go to even in next step. 
Now just start from 1 and notice that:
step 1 -reached  even, 
step 2-reached  odd, 
step 3-reached even
.
.
.
.
Step odd-reached even.
Since you want to reach 8 which is even so you will have to take odd steps.
A: As you can see you have bipartite graph here and one of the things of bipartite graph is that for any u,v vertices all the paths from u to v has to odd or even (can't be 1 path odd while there is other even) so if we have bipartite graph and we show 1 odd path then all the paths are odd.
