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A factorial order of growth is even more extreme than an exponential order of growth For example, there are about 36 million permutations of 10 cities, but more than 2 trillion billion permutations of 20 If the computer can compute the distance for a million permutations a second, the TSP problem will take 18 seconds for 10 cities, but tens of thousands of years for 20 cities Figure 2-2 shows the rates of growth for lg n, n, n(lg n), n2, 2n, and n! Table 21 summarizes some different orders of growth, and the characteristics of associated algorithms ALGORITHMS AS TECHNOLOGY It s pretty exciting to buy a new computer with twice, four times, or even ten times the clock rate of the old computer Many people think of computer hardware speed as the measure of technological advance.

The sequential search is correct This means that it works The following argument is a proof of that fact If n = 0, then the sequence is empty and the loop does not execute at all Only step 4 executes, immediately returning 1 This satisfies the postconditions: x cannot equal any of the elements because there aren t any If n = 1, then the loop iterates only once, with i = 0 On that iteration, either s0 = x or s0 x If s0 = x, then 0 is returned and the postcondition is satisfied If s0 x, then the loop terminates, step 4 executes, and 1 is returned, and that satisfies the postcondition because the single element of the sequence is not equal to x Suppose n > 1.

Supply voltages and resistors bias a transistor; that is, they establish a speci c set of dc terminal voltages and currents, thus determining a point of active-mode operation (called the quiescent point or Q point). Usually, quiescent values are unchanged by the application of an ac signal to the circuit. With the universal bias arrangement of Fig. 3-8(a), only one dc power supply VCC ) is needed to establish active-mode operation. Use of the Thevenin equivalent of the circuit to the left of a; b leads to the circuit of Fig. 3-8(b), where R1 R2 R1 RB VBB V 3:5 R1 R2 R1 R2 CC If we neglect leakage current so that IEQ 1 IBQ and assume the emitter-to-base voltage VBEQ is constant % 0:7 V and % 0:3 V for Si and Ge, respectively), then KVL around the emitter loop of Fig. 3-8(b) yields IEQ VBB R VBEQ IEQ RE 3:6 1 B which can be represented by the emitter-loop equivalent bias circuit of Fig. 3-8(c). Solving (3.6) for IEQ and noting that ICQ IEQ % ICQ we obtain VBB VBEQ ICQ % IEQ 3:7 RB = 1 RE

CHAP. 3]

We want to apply the First Principle of Mathematical Induction to deduce that the loop invariant must be true on every iteration of the loop (See page 321) That requires the verification of the invariant on the first iteration and the deduction of the invariant on iteration i from the corresponding invariant on iteration i 1 On the first iteration of the loop, i = 0, and the loop invariant in step 2 is true vacuously because the subsequence {s0..si 1} is empty Then in step 3, either s0 = x or s0 x If s0 = x, then 0 is returned and the postcondition is satisfied If s0 x, then the loop continues on to a second iteration Then i = 1, and the loop invariant in step 2 is again true because the subsequence {s0..si 1} = {s0} and s0 x.

Having discussed algorithms and their performance, consider whether a better algorithm on a slower computer might be better than a slower algorithm on a faster computer As an example, consider a sorting task Suppose you need to sort a million numbers (social security numbers, for example) You have the choice of using your current computer with a merge sort program, or of buying a new computer, which is 10 times faster, but which uses an insertion sort The insertion sort on the new computer will require on the order of (10 6)2, or a million million cycles, while the merge sort will require on the order of 10 6(lg 10 6), or 10 6(20), or 20 million cycles Even when it runs on your old computer, the merge sort will still run four orders of magnitude faster than the insertion sort on the new machine.

a IBQ RB + VBB _ b IEQ RB +1 + VBB _