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Chapter 1 Introduction Mathematics Review

Chapter 1 Introduction Mathematics Review. Sections 1.1, 1.2. Outline. Why does program execution time for large inputs matter? Basic mathematical background. Selection Problem. Find the k th largest number from a group of N numbers How would you solve this? Algorithm 1:

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Chapter 1 Introduction Mathematics Review

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  1. Chapter 1IntroductionMathematics Review Sections 1.1, 1.2

  2. Outline • Why does program execution time for large inputs matter? • Basic mathematical background

  3. Selection Problem • Find the kth largest number from a group of N numbers • How would you solve this? • Algorithm 1: • Sort the N numbers and pick the kth one. • Easy to implement, but • Sorting requires many comparisons • Much work comparing elements having no chance to be at position K • To sort N elements, need N log2(N) comparisons/swaps in general • An unsorted data set with 10,000,000 elements and 1,000 swaps/sec will take around 2 hours

  4. Selection Problem (Cont’d) • Algorithm 2: (Better) • Sort first K elements in the array. • Then insert elements (K+1) to N, discarding the smallest element each time. • Then pick the kth element. • What if N=10 million and K=5,000,000? • How long do you think this would take? • Both algorithms are impractical. • A better algorithm can solve this in a second!

  5. Mathematics Review • Exponents • Logarithms • Series • Modular arithmetic • Proof techniques

  6. Exponents

  7. Logarithms • All logarithms are to the base 2, unless otherwise specified • Definition 1.1 • XA = B if and only if logxB = A

  8. Properties of logarithms • Theorem 1.1 (base-change) • ; A, B, C > 0, A  1 • Proof: Let X = logCB, Y = logCA, and Z = logAB • Then: CX = B, CY = A, and AZ = B • Implies: B = CX = (CY)Z = CYZ • Hence: X = YZ • And: Z = X/Y

  9. Logarithms (contd.) • Theorem 1.2 • log(AB) = log A + log B; A, B > 0 • Proof: (Assume base 2) • Let: X = log A, Y = log B, and Z = log AB • Hence: 2X = A, 2Y = B, and 2Z = AB • Combining the above: 2Z = AB = 2X2Y = 2(X+Y) • Implies: X + Y = Z

  10. Logarithms (contd.) • More results • log(A/B) = log A – log B • log(AB) = B log A • log X < X for all X > 0 • log 1 = 0 • log2 2 = 1 • log2 1024 = 10 • log2 1,048,576 = 20

  11. Geometric Series • If 0 < A < 1, • If N -> , we have • How??

  12. Arithmetic Series • How about 2+5+8+…+(3k-1) ?

  13. Modular Arithmetic • We say that A is congruent to B modulo N • Written as (mod N) • If N divides (A-B) • In other words, the remainder is the same if either A or B are divided by N • E.g. • (mod 10) • Similar to equality, if (mod N), then • (mod N), and (mod N)

  14. Proof techniques • Two common proof techniques in data structure and algorithm analysis (and in CS, in general) • Proof by induction • Proof by contradiction • Another common technique • Proof a statement false with a counterexample

  15. Proof by Induction • Given a theorem • First prove a base case • Show the theorem is true for some small degenerate values • Next assume an inductive hypothesis • Assume the theorem is true for all cases up to some limit k • Then prove that the theorem holds for the next value (k+1)

  16. Proof by Induction - example • Fibonacci Series • F0 = 1, F1 = 1, Fi = F(i-1) + F(i-2), for i>1 • Show that • Fi < (5/3)i,for i>0 • Base case: • F1 = 1 < 5/3 • F2 = 2 < (5/3)2=25/9 • Inductive Hypothesis • Assuming: Fi < (5/3)i , i = 1, 2, ..., k

  17. Proof by Induction - example (contd.) • Now prove that Fk+1 < (5/3)k+1 • From definition of Fibonacci Sequence • Fk+1 = Fk + Fk-1 • Using inductive hypothesis Fk+1 < (5/3)k+ (5/3)k-1 = (5/3)k+1[ 3/5 + (3/5)2]= (5/3)k+1[24/25]< (5/3)k+1

  18. Other types of proofs • Disprove with a counter-example • The statement is false in the Fibonacci series • Proof: F11 = 144 > 112 = 121 • Proof by contradiction • Initially assume that the theorem is false • Then show that some known property would be false as well. • Example: “There is an infinite number of prime numbers” • Proof: • Assume the theorem is false (so there are only finite prime) • Let P1, P2, ..., Pk be all the primes in increasing order. • Let N =P1P2 Pk + 1,N is > Pk , so it is not a prime • But it is also not divisible by any of the listed primes, contradicting the factorization of integers into primes. • We reach a contradiction

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