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HOWMANY BITS ARE
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The number of bits required represent numbers is logarithmic
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REQUIRED TO REPRESENT
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CONSECUTIVE INTEGERS
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A 16-bit s h o r t integer represents the 65,536 integers in the range -32,768 to 32,767 In general, B bits are sufficient to represent 2B different integers Thus the number of bits B required to represent N consecutive integers satisfies the equation 2B 2 N Hence we obtain B 2 log N, so the minimum num ber of bits is [log ~ 1(Here [XI is the ceiling function and represents the smallest integer that is at least as large as X The corresponding floor funcrepresents the largest integer that is at least as small as X) tion
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REPEATED DOUBLING STARTING X = 1 , HOW MANY FROM
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IS AT LEAST AS LARGE AS
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TIMES SHOULD X BE DOUBLED BEFORE IT
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The ~ o ~ a r i t h m - r n
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Suppose that we start with $1 and double it every year How long would it take to save a million dollars In this case, after 1 yr we would have $2; after 2 yr, $4; after 3 yr, $8, and so on In general, after K years we would have 2K dollars, so we want to find the smallest K satisfying 2 2 N This is the same K equation as before, so K = [log ~ 1After 20 yr, we would have more than a million dollars The repeated doubling principle holds that, starting from 1, we can repeatedly double only [log times until we reach N
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The repeated doubling principle holds that, starting at 1, we can repeatedly double only logarithmically many times until we reach N
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REPEATED HALVING STARTING FROM X = N , IF N IS REPEATEDLY HALVED, HOW M N AY ITERATIONS MUST BE APPLIED TO MAKE N SMALLER THAN O R EQUAL
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The repeated halving principle holds that, starting at N, we can halve only logarithmically many times This process is used to obtain logarithmic routines for searching The Nth harmonic number is the sum of the reciprocals of the first N positive integersThe growth rate of the harmonic number is logarithmic
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If the division rounds up to the nearest integer (or is real, not integer, division), we have the same problem as with repeated doubling, except that we are going in the opposite direction Once again the answer is [log N1 iterations If the division rounds down, the answer is Llog N ] We can show the difference by starting with X = 3 Two divisions are necessary, unless the division rounds down, in which case only one is needed Many of the algorithms examined in this text contain logarithms, introduced because of the repeated halving principle, which holds that, starting at N, we can halve only logarithmically many times In other words, an algo1)) rithm is O(1og N) if it takes constant (0( time to cut the problem size by a constant fraction (usually 112) This condition follows directly from the fact that there will be O(1og N) iterations of the loop Any constant fraction will do because the fraction is reflected in the base of the logarithm, and Theorem 64 tells us that the base does not matter All the remaining occurrences of logarithms are introduced (either directly or indirectly) by applying Theorem 65 This theorem concerns the Nth harmonic number, which is the sum of the reciprocals of the first N positive integers, and states that the Nth harmonic number, H,, satisfies HN = O(1og N) The proof uses calculus, but you do not need to understand the proof to use the theorem
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Let H N = I In N + 0577
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1 /i Then HN = @(logN ) A more precise estimate is
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Theorem 65
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-~Gthrn
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Analysis
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Proof
The intuition qf the proof is that a discrete ~ l r i 1r~ i t ~ l approximated by , l to show the (continuous) integral The proof uses n cor~st~~rctioi~ that the - 111sum H , can be bounded above and belo\\ 01 J - \tYrh uppropriate Y limits Details are left as Exercise 618