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xi = 0 (on the right side of the gure) This completes the construction Take any solution of this instance of R UDRATA CYCLE WITH PAIRED EDGES As discussed before, it picks a value for each variable and a variable for every equation We claim that the values thus chosen are a solution to the original instance of ZOE If a variable x i has value 1, then the edge xi = 0 is not traversed, and thus all edges associated with x i on the equation side must be traversed (since they are paired in C with the x i = 0 edge) So, in each equation exactly one of the variables appearing in it has value 1 which is the same as saying that all equations are satis ed The other direction is straightforward as well: from a solution to the instance of ZOE one easily obtains an appropriate Rudrata cycle Getting Rid of the Edge Pairs So far we have a reduction from ZOE to R UDRATA CYCLE WITH PAIRED EDGES; but we are really interested in R UDRATA CYCLE, which is a special case of the problem with paired edges: the one in which the set of pairs C is empty To accomplish our goal, we need, as usual, to nd a way of getting rid of the unwanted feature in this case the edge pairs Consider the graph shown in Figure 812, and suppose that it is a part of a larger graph G in such a way that only the four endpoints a, b, c, d touch the rest of the graph We claim that this graph has the following important property: in any Rudrata cycle of G the subgraph shown must be traversed in one of the two ways shown in bold in Figure 812(b) and (c) Here is why Suppose that the cycle rst enters the subgraph from vertex a continuing to f Then it must continue to vertex g, because g has degree 2 and so it must be visited immediately after one of its adjacent nodes is visited otherwise there is no way to include it in the cycle Hence we must go on to node h, and here we seem to have a choice We could continue on to j, or return to c But if we take the second option, how are we going to visit the rest of the subgraph (A Rudrata cycle must leave no vertex unvisited) It is easy to see that this would be impossible, and so from h we have no choice but to continue to j and from there to visit the 252.
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Refer to the text in this chapter if necessary A good score is at least 18 correct Answers are in the back of the book 1 A radio wave has a frequency of 155 MHz The highest modulating frequency that can be used effectively is about (a) 155 kHz (b) 155 kHz (c) 155 kHz (d) 155 MHz 2 The reflected wave (a) arrives in phase with the direct wave (b) arrives out of phase with the direct wave (c) arrives in a variable phase compared with the direct wave (d) is always horizontally polarized
x 2 + x 5 = 2x + 1 We can subtract 1 from both sides to obtain x 2 + x 6 = 2x Then we can subtract 2x from both sides to get x2 x 6 = 0
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Next, we solve We ve derived a quadratic equation that can be solved using any of the methods from Chaps 22 and 23 We re lucky here, because this equation can be factored into
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rest of the graph as shown in Figure 812(b) By symmetry, if the Rudrata cycle enters this subgraph at c, it must traverse it as in Figure 812(c) And these are the only two ways But this property tells us something important: this gadget behaves just like two edges {a, b} and {c, d} that are paired up in the R UDRATA CYCLE WITH PAIRED EDGES problem (see Figure 812(d)) The rest of the reduction is now clear: to reduce R UDRATA CYCLE WITH PAIRED EDGES to R UDRATA CYCLE we go through the pairs in C one by one To get rid of each pair ({a, b}, {c, d}) we replace the two edges with the gadget in Figure 812(a) For any other pair in C that involves {a, b}, we replace the edge {a, b} with the new edge {a, f }, where f is from the gadget: the traversal of {a, f } is from now on an indication that edge {a, b} in the old graph would be traversed Similarly, {c, h} replaces {c, d} After |C| such replacements (performed in polynomial time, since each replacement adds only 12 vertices to the graph) we are done, and the Rudrata cycles in the resulting graph will be in one-to-one correspondence with the Rudrata cycles in the original graph that conform to the constraints in C
3 An advantage of FSK over on/off keying is the fact that FSK (a) offers better frequency stability (b) can provide faster data rates (c) reduces the number of receiving errors (d) Forget it! On-off keying is just as good as FSK 4 The highest layer of the ionosphere is (a) the D layer (b) the E layer (c) the F layer (d) none of the above 5 If an AM signal is modulated with audio having frequencies up to 5 kHz, then what is the complete signal bandwidth (a) 10 kHz (b) 20 kHz (c) 50 kHz (d) 500 kHz 6 An LSB, suppressed-carrier signal can be demodulated by (a) an envelope detector (b) a diode (c) a ratio detector (d) a product detector 7 Which of the following modes is used to send image data over telephone lines (a) On/off keying (b) Fax (c) AM (d) Product detection 8 The S/N ratio is a measure of (a) the sensitivity of a receiver (b) the selectivity of a receiver (c) the dynamic range of a receiver (d) the efficiency of a transmitter 9 Suppose an SSB suppressed carrier is at 14335 MHz, and audio data is contained in a band from 14335 to 14338 MHz What is this mode (a) DSB (b) LSB (c) USB (d) AFSK
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