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“mcs” — 2017/3/3 — 11:21 — page 433 — #441<br />

11.3. Network Designs 433<br />

À À<br />

2 n<br />

B n<br />

2 n 2 nC1<br />

B n<br />

new inputs<br />

B nC1<br />

new outputs<br />

Figure 11.6<br />

Beneš Net B nC1 with 2 nC1 inputs from two B n ’s.<br />

In the constructor step, we construct B nC1 out of two B n nets connected to a<br />

new set of 2 nC1 input switches and also a new set of 2 nC1 output switches. This is<br />

illustrated in Figure 11.6.<br />

The ith and 2 n C ith new input switches are each connected to the same two<br />

switches: the ith input switches of each of two B n components <strong>for</strong> i D 1; : : : ; 2 n ,<br />

exactly as in the Butterfly net. In addition, the ith and 2 n Cith new output switches<br />

are connected to the same two switches, namely, to the ith output switches of each<br />

of two B n components.<br />

Now, B nC1 is laid out in columns of height 2 nC1 by adding two more columns<br />

of switches to the columns in B n . So, the B nC1 switches are arrayed in 2.n C 1/<br />

columns. The total number of switches is the number of columns times the height<br />

2.n C 1/2 nC1 of the columns.<br />

All paths in B nC1 from an input switch to an output are length 2.n C 1/ 1, and<br />

the diameter of the Beneš net with 2 nC1 inputs is this length plus two because of<br />

the two edges connecting to the terminals.<br />

So Beneš has doubled the number of switches and the diameter, but by doing so<br />

he has completely eliminated congestion problems! The proof of this fact relies on<br />

a clever induction argument that we’ll come to in a moment. Let’s first see how the

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