The number of crosspoints needed for 10 lines in a crosspoint switch, which is…

A crosspoint (crossbar) switch that lets any of n lines connect to any other line is built from an n2 matrix of crosspoints. Two standard reductions apply here: first, since a line never connects to itself, the n diagonal (self-connection) positions of the matrix are dropped. Second, when the switch is full duplex, a SINGLE crosspoint connecting line i to line j already carries the call in both directions at once (the same physical pair of wires serves both directions) — so the matrix is symmetric about its diagonal, and only one of each mirrored pair, (i, j) or (j, i), needs to actually exist. Folding the matrix this way keeps just n(n − 1)/2 of the original positions.

1. Here n = 10, so the full n2 matrix has n2 = 102 = 100 positions.

2. Remove the 10 self-connection positions on the diagonal (a line does not connect to itself): 100 − 10 = 90.

3. Because the switch is full duplex, one crosspoint already carries both directions of the call between a pair of lines, so the matrix folds along its diagonal and only half of the remaining 90 positions are distinct: 90 ÷ 2 = 45.

This matches the direct formula n(n − 1)/2 = (10 × 9)/2 = 45. For contrast: keeping both the (i, j) and (j, i) positions — as if each direction of a call needed its own separate crosspoint rather than sharing one bidirectional connection — would give n(n − 1) = 90, and including the self-connections on top of that would give n2 = 100. Neither matches a switch that is both full duplex (one shared bidirectional crosspoint per pair) and free of self-connections, confirming 45 as the value consistent with both stem conditions.