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Mathematics in OI

Mathematics in OI. Prepared by Ivan Li. Mathematics in OI. Greatest Common Divisor Finding Primes High Precision Arithmetic Partial Sum and Differencing. Greatest Common Divisor. Motivation Sometimes we want “k divides m” and “k divides n” occur simultaneously.

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Mathematics in OI

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  1. Mathematics in OI Prepared by Ivan Li

  2. Mathematics in OI • Greatest Common Divisor • Finding Primes • High Precision Arithmetic • Partial Sum and Differencing

  3. Greatest Common Divisor • Motivation • Sometimes we want “k divides m” and “k divides n” occur simultaneously. • And we want to merge the two statements into one equivalent statement: “k divides ?”

  4. Greatest Common Divisor • Definition • The greatest natural number dividing both n and m • A natural number k dividing both n and m, such that for each natural number h dividing both n and m, we have k divisible by h.

  5. Greatest Common Divisor • How to find it? • Check each natural number not greater than m and n if it divides both m and n. Then select the greatest one. • Euclidean Algorithm

  6. Euclidean Algorithm • Assume m > n GCD(m,n) While n > 0 m = m mod n swap m and n Return m

  7. Greatest Common Divisor • What if we want the greatest number which divides n1,n2, …, nm-1 and nm? • Apply GCD two-by-two • gcd(n1,n2, …,nm) = gcd(n1,gcd(n2,gcd(n3,…gcd(nm-1,nm)…))

  8. Applications • Solve mx + ny = a for integers x and y • Can be solved if and only if a is divisible by gcd(m,n)

  9. Applications • Simplifying a fraction m/n • If gcd(m,n) > 1, then the fraction can be simplified by dividing gcd(m,n) on the numerator and the denominator.

  10. Least Common Multiple • Definition • The least natural number divisible by both n and m • A natural number k divisible by both n and m, such that for each natural number h divisible by both n and m, we have k divides h. • Formula • lcm(m,n) = mn/gcd(m,n)

  11. Least Common Multiple • What if we want to find the LCM of more than two numbers? • Apply LCM two-by-two?

  12. Definition of Prime Numbers An integer p greater than 1 such that: • p has factors 1 and p only? • If p = ab, a b, then a = 1 and b = p ? • If p divides ab, then p divides a or p divides b ? • p divides (p - 1)! + 1 ?

  13. Test for a prime number • By Property 1 • For each integer greater than 1 and less than p, check if it divides p • Actually we need only to check integers not greater than sqrt(p) (Why?)

  14. Finding Prime Numbers • For each integer, check if it is a prime • Prime List • Sieve of Eratosthenes

  15. Prime List • Stores a list of prime numbers found • For each integer, check if it is divisible by any of the prime numbers found • If not, then it is a prime. Add it to the list.

  16. Sieve of Eratosthenes • Stores an array of Boolean values Comp[i] which indicates whether i is a known composite number

  17. Sieve of Eratosthenes for i = 2 … n If not Comp[i] output i j = 2*i while j n Comp[j] = true j = j + i

  18. Optimization • Consider odd numbers only • Do not forget to add 2, the only even prime

  19. High Precision Arithmetic • 32-bit signed integer:-2147483648 … 2147483647 • 64-bit signed integer:-9223372036854775808 … 9223372036854775807 • How to store a 100 digit number?

  20. High Precision Arithmetic • Use an array to store the digits of the number • Operations: • Comparison • Addition / Subtraction • Multiplication • Division and remainder

  21. High Precision Division • Locate the position of the first digit of the quotient • For each digit of the quotient (starting from the first digit), find its value by binary search.

  22. High Precision Arithmetic • How to select the base? • Power of 2 : Saves memory • Power of 10 : Easier input / output • 1000 or 10000 for 16-bit integer array • Beware of carry

  23. More on HPA • How to store • negative numbers? • fractions? • floating-point numbers?

  24. Partial Sum • Motivation • How to find the sum of the 3rd to the 6th element of an array a[i] ? • a[3] + a[4] + a[5] + a[6] • How to find the sum of the 1000th to the 10000th element? • A for-loop will take much time • In order to find the sum of a range in an array efficiently, we need to do some preprocessing.

  25. Partial Sum • Use an array s[i] to store the sum of the first i elements. • s[i] = a[1] + a[2] + … + a[i] • The sum of the j th element to the k th element = s[k] – s[j-1] • We usually set s[0] = 0

  26. Partial Sum • How to compute s[i] ? • During input s[0] = 0 for i = 1 to n input a[i] s[i] = s[i-1] + a[i]

  27. Differencing • Motivation • How to increment the 3rd to the 6th element of an array a[i] ? • a[3]++, a[4]++, a[5]++, a[6]++ • How to increment the 1000th to the 10000th element? • A for-loop will take much time • In order to increment(or add an arbitrary value to) a range of elements in an array efficiently, we will use a special method to store the array.

  28. Differencing • Use an array d[i] to store the difference between a[i] and a[i-1]. • d[i] = a[i] - a[i-1] • When the the j th element to the k th element is incremented, • d[j] ++, d[k+1] - - • We usually set d[1] = a[1]

  29. Differencing • Easy to compute d[i] • But how to convert it back to a[i]? • Before (or during) output a[0] = 0 for i = 1 to n a[i] = a[i-1] + d[i] output a[i] • Quite similar to partial sum, isn’t it?

  30. Relation between the two methods • They are “inverse” of each other • Denote the partial sum of a by a • Denote the difference of a by a • The difference operator • We have (a) = (a) = a

  31. Comparison • Partial sum - Fast sum of range query • Difference - Fast range incrementation • Ordinary array - Fast query and incrementation on single element

  32. Runtime Comparison

  33. Does there exist a method to perform range query and range update in constant time?

  34. Question?

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