394
395
396
397
398
399
400
401
Case 1: When a diagonal is joined between two consecutive points
Let A14
is joined
to
A12, so the point
A13 is now out of network. [Refer Fig.2].
Now since we have to cover each point of the network, A13 has to be joined to some other point. Let A13 is
joined to A3 and A4.These points are arbitrary
.The important point is not the
point but the property exhibited
by each point. If A13 is joined to any other point the property
exhibited by the point would be the same as with this point. Note we are talking
of topological
properties where only the relative position matters.
402\ Now new network distance is
403
N-A14A13-A13A12+A14A12+A3A13A4A13-A3A4
404
405
406
407
408
409
410
411
Now A14A12 >A14A13 (A14A12 is the adjacent diagonal of SCP and by the definition of SCP it
is
greater than the side forming it)
Further A3A13 >A13A12 (Since by the
definition
of
SCP the shortest distance from a point
on the
periphery
is next point to it on either side, all other branches from emerging from it are the
diagonals)
Finally A4A13>A3A4 (A4A13 is the adjacent diagonal of SCP and by the definition of SCP it is
greater than the side forming it)
412 c The Net
network distance increases
as sum
of the adding distances is
greater
than
the
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419
