Some friends of yours work on wireless networks, and they\'re currently studying

Question

Some friends of yours work on wireless networks, and they're currently studying the properties of a network of n mobile devices. As the devices move around (actually, as their human owners move around), they define a graph at any point in time as follows: there is a node representing each of the n devices, and there is an edge between device i and device j if the physical locations of i and j are no more than 500 meters apart. (lf so, we say that i and j are "in range" of each other.) They'd like it to be the case that the network of devices is connected at all times, and so they've constrained the motion of the devices to satisfy the following property: at all times, each device i is within 500 meters of at least n/2 of the other devices. (We'll assume n is an even number.) What they'd like to know is: Does this property by itself guarantee that the network will remain connected? Here's a concrete way to formulate the question as a claim about graphs Claim: Let G be a graph on n nodes, where n is an even number. If every node of G has degree at least n/2, then G is connected. Decide whether you think the claim is true or false, and give a proof of either the claim or its negation

Explanation / Answer

Claim : let G be a graph on n nodes, where n is an even number. If every node of G has degree at least n/2 then G is connected.
Let us prove this claim using Contradiction.

Let us assume that G is not connected and it has two components and every node of G has degree at least n/2.

Take vertices x from one component and vertices y from the other component,
Since they are not connected they have 0 common neighbours i.e Intersection of neighbours is 0 i.e
| N(x) N(y) | = 0.

Note : | N(x) | -->Total number of neigbours of node x

What about U (Union) of neighbours <= n -2 [because two nodes are excluded as they are not the neighbout of each other and are not connected ]
Since they are not connected they have 0 common neighbours i.e Intersection of neighbours is 0 i.e
| N(x) N(y) | = 0.



We know that | N(x) N(y) | = | N(x) | + | N(y) | - | N(x) U N(y) |

Now Since If every node of G has degree at least n/2
| N(x) | >= n/2 , | N(y) | >= n/2
| N(x) U N(y) | = (n-2) [calculated above ]
| N(x) N(y) | >= n/2 + n/2 - (n-2)
=> | N(x) N(y) | >= 2

Which means our assumption is wrong thats means x and y are connected and they have atleast 2 common neighbour.
Hence the graph is connected If every node of G has degree at least n/2 and n is even number.

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