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Divide the channels, remembering that... each channel actually occupies two neighboring channels on both sides, that is, 3 in total, => dividing the range of channels into groups, we get the following row:
1, 2 , 3 - 4, 5 , 6 - 7, 8 , 9 - 10, 11 , 12 - 13
therefore, it is possible to place in one place exactly "4 networks" that do not intersect at all, choosing the middle channels in each of the indicated groups.
Sometimes it is advised to start from channel 3, without touching the first at all (I don’t know why, I didn’t understand or understand, but I still don’t remember), that is, the following row is obtained:
1 - 2, 3 , 4 - 5, 6 , 7 - 8 , 9 , 10 - 11, 12, thirteen
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