1. Mathematics in OI Prepared by Ivan Li

2. Mathematics in OI A brief content Greatest Common Divisor Modular Arithmetic Finding Primes Floating Point Arithmetic High Precision Arithmetic Partial Sum and Difference Euler Phi function Fibonacci Sequence and Recurrence Twelvefold ways Combinations and Lucas Theorem Catalan Numbers Using Correspondence Josephus Problem

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 n 1,n 2, …, n m-1 and n m ? Apply GCD two-by-two gcd(n 1,n 2, …,n m ) = gcd(n 1,gcd(n 2,gcd(n 3,…gcd(n m-1,n m )…))

8. 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.

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

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. Extended Euclidean Algorithm The table method e.g. solve 93x + 27y = 6 General form: x = -4 + 9k, y = 14 - 31k, k integer Coeff of 93Coeff of 27 (1)9310 (2)2701 (3): (1)-3(2)121-3 (4): (2)-2(3)3 (gcd, 3*2=6) -2 (x = -4)7 (y = 14) (5): (3)-4(4)09-31

13. Modular Arithmetic Divide 7 by 3 Quotient = 2, Remainder = 1 7 ÷ 3 = 2...1 In modular arithmetic 7 ≡ 1 (mod 3) a ≡ b (mod m) if a = km + b for an integer k

14. Modular Arithmetic Like the equal sign Addition / subtraction on both sides 7 ≡ 1 (mod 3) => 7+2≡1+2 (mod 3) Multiplication on both side 7 ≡ 1 (mod 3) => 7*2≡1*2 (mod 3) We can multiply into m too 7≡1 (mod 3) 7*2≡1*2 (mod 3*2) Congruence in mod 6 is stronger than that in mod 3

15. Modular Arithmetic Division? Careful: 6≡4 (mod 2), but not 3≡2 (mod 2) ac≡bc (mod m) a≡b (mod m) when c, m coprime (gcd = 1) Not coprime? ac≡bc(mod cm) a≡b(mod m) ac ≡ bc(mod m) a ≡ b (mod m/gcd(c,m))

16. Modular Inverse Given a, find b such that ab ≡ 1 (mod m) Write b as a -1 We can use it to do “division” ax ≡ c (mod m)=> x ≡ a -1 c (mod m) Exist if and only if a and m are coprime (gcd = 1) When m is prime, inverse exists for a not congruent to 0

17. Modular Inverse ab ≡ 1 (mod m) ab + km = 1 Extended Euclidean algorithm

18. CAUTION!!! The mod operator “%” does not always give a non-negative integer below the divisor The answer is negative when the dividend is negative Use ((a % b)+b)%b

19. Definition of Prime Numbers An integer p greater than 1 such that: 1. p has factors 1 and p only? 2.If p = ab, a  b, then a = 1 and b = p ? 3.If p divides ab, then p divides a or p divides b ? 4. p divides (p - 1)! + 1 ?

20. 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?)

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

22. 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.

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

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

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

26. Other usages of sieve Prime factorization When we mark an integer as composite, store the current prime divisor For each integer, we can get a prime divisor instantly Get the factorization recursively

27. Floating point arithmetic Pascal: real, single, double C/C++: float, double Sign, exponent, mantissa Floating point error 0.2 * 5.0 == 1.0 ? 0.2 * 0.2 * 25.0 == 1.0?

28. Floating point arithmetic To tolerate some floating point Introduce epsilon EPS = 1e-8 to 1e-11 a a + EPS < b a a <= b + EPS a == b => abs(a - b) <= EPS

29. Floating point arithmetic Special values Positive / Negative infinity Not a number (NaN) Checked in C++ by x!=x Denormal number

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

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

32. 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.

33. 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

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

35. Partial Sum Motivation How to find the sum of the 3 rd to the 6 th element of an array a[i] ? a[3] + a[4] + a[5] + a[6] How to find the sum of the 1000 th to the 10000 th 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.

36. 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

37. 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]

38. Difference operation Motivation How to increment the 3 rd to the 6 th element of an array a[i] ? a[3]++, a[4]++, a[5]++, a[6]++ How to increment the 1000 th to the 10000 th 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.

39. Difference operation 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]

40. Difference operation 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?

41. 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

42. Comparison Partial sum - Fast sum of range query Difference - Fast range increment Ordinary array - Fast query and increment on single element

43. Runtime Comparison Range Query Single Query Single Update Range Update Partial sumConstant (treat a single element as a range) Linear (Have to update all sums involved) Linear Ordinary array LinearConstant Linear DifferenceLinear (Convert it back to the original array) LinearConstant

44. Does there exist a method to perform range query and range update in constant time? No, but there is a data structure that performs the two operations pretty fast.

45. Euler Phi Function  (n) = number of integers in 1...n that are relatively prime to n  (p) = p-1  (p n ) = (p-1)p n-1 = p n (1-1/p) Multiplicative property:  (mn)=  (m)  (n) for (m,n)=1  (n) = n  p|n (1-1/p)

46. Euler Phi Function Euler Theorem a  (n)  1 (mod n) for (a, n) = 1 Then we can find modular inverse aa  (n)-1  1 (mod n) a -1  a  (n)-1

47. Fibonacci Number F 0 = 0, F 1 = 1 F n = F n-1 + F n-2 for n > 1 The number of rabbits The number of ways to go upstairs How to calculate F n ?

48. What any half-wit can do F 0 = 0; F 1 = 1; for i = 2... n F i = F i-1 + F i-2 ; return F n ; Time Complexity: O(n) Memory Complexity: O(n)

49. What a normal person would do a = 0; b = 1; for i = 2... N t = b; b += a; a = t; return b; Time Complexity: O(n) Memory Complexity: O(1)

50. What a Math student would do Generating Function G(x) = F 0 + F 1 x + F 2 x 2 +... A generating function uniquely determines a sequence (if it exists) F n = d n G(x)/dx n (0) A powerful (but tedious) tool in solving recurrence

51. All the tedious works G(x) = F 0 + F 1 x + F 2 x 2 + F 3 x 3 +... xG(x) = F 0 x + F 1 x 2 + F 2 x 3 +... x 2 G(x) = F 0 x 2 + F 1 x 3 +... G(x) - xG(x) - x 2 G(x) = F 0 +F 1 x - F 0 x = x G(x) = x / (1 - x - x 2 ) Let a = (-1 - sqrt(5)) / 2, b = (-1 + sqrt(5)) / 2 By Partial Fraction: G(x) = ((5 + sqrt(5)) / 10) / (a-x)+((5 - sqrt(5)) / 10) / (b-x) = -(sqrt(5) / 5) / (1- x/a) + (sqrt(5) / 5) / (1- x/b) Note that 1 + rx + r 2 x 2 +... = 1 / (1 - rx) G(x) = (sqrt(5) / 5)(-1-x/a-x 2 /a 2 -...+1+x/b+x 2 /b 2 +...) By Uniqueness, F n = (sqrt(5) / 5)(-1/a n + 1/b n )

52. Shortcut Characteristic Equation F n - F n-1 - F n-2 = 0 f(x) = x 2 – x – 1 Then F n = Aa n + Bb n for some constants A, B where a, b are roots of f(x)

53. However How to compute ((-1-sqrt(5))/2) n ? The result must be a whole number, but the intermediate values may not Use floating point numbers Precision problem? If we are asked to find F n mod m?

54. What a programmer would do Note that ( )( ) = ( ) Then ( ) n ( ) = ( ) Matrix Exponential Just like fast exponential 0 1 1 F n F n+1 F n+2 0 1 1 F0F1F0F1 F n F n+1

55. Twelvefold ways Put n balls into m boxes How many ways? Balls identical / distinct? Boxes identical / distinct? Allow empty boxes? Allow two balls in one boxes?

56. Twelvefold ways

57. Combinations The number of ways to choose r objects among n different objects (without order) nCr = n! / r!(n-r)!

58. Combinations How to calculate nCr? Calculate n!, r!, (n-r)! ? Note nCr = n(n-1)...(n-r+1) / r! nCr = 1; for i = n-r+1... n nCr *= i; for i = 1... r nCr /= i;

59. Combinations Overflow problem? Note nCr = (n/r)(n-1)C(r-1) nCr = 1; //that is (n-r)C 0 for i = 1... r nCr *= (n - r + i); nCr /= i;

60. Combinations What if we are asked to find nCr mod p for very large n, r? Lucas Theorem Let n = n k n k-1...n 1 n 0 (base p) r = r k r k-1...r 1 r 0 Then nCr  (n k Cr k )(n k-1 Cr k-1 )...(n 0 Cr 0 ) (mod p) Works only for prime p

61. Combinations When p is large (larger than r), computing n i Cr i may be difficult (nCr)(r!) = n(n-1)...(n-r+1) nCr  n(n-1)...(n-r+1)(r!) -1 (mod p) where (r!)((r!) -1 )  1 (mod p) Fermat Little Theorem gives (r!) -1  (r!) p-2 (mod p)

62. Combinations When we are asked to mod a square-free number, factorize it into primes The situation becomes complicated when the number is not square-free Store the prime factorization of numerator and denominator

63. A drunken man A drunken man was standing beside a wall. On each second he moves left or right by 1m at random. What is the probability that he returns to the original position in k seconds without hitting the wall?

64. Observations If k is odd, then it is impossible Let k = 2n If the man returns to original position, then there are n left and n right moves Number of possible outcomes = 2 2n We have to find the number of moving sequence such that the man returns to original position without hitting the wall

65. Observations If the wall is removed, the number of ways is 2n C n Let A n be the number of ways If the first time the man returns to his original position is at the 2i th second Note that the first move must be rightward, and the 2i th move must be leftward

66. Observations Also in the 2i - 2 seconds after the first, he cannot return to his original position Think of an invisible wall on his original position A i-1 ways for the first 2i seconds A n-i ways for the remaining 2n-2i seconds (may return to original position again) A n =  i = 1...n A i-1 A n-i

67. Tedious works again Again, generating function g(x) =  n=0,1,... A n x n g(x) 2 =  n=0,1,...  i=0...n A i A n-i x n =  n=0,1,... A n+1 x n g(x) = A 0 + x  n=1,... A n x n-1 = 1 + xg(x) 2 xg(x) = (1-sqrt(1-4x))/2 or (1+sqrt(1-4x))/2 (rejected since xg(x)=0 when x = 0) Power series......

68. A much more elegant approach We now remove the wall and note when the man arrives at the position 1m on the left of the original position (If the man never arrives at it, he won’t hit the wall) When he arrives at the considered position on the first time, flip his remaining moves, i.e. left to right, right to left He will now end at the position 2m on the left of the original position

69. A much more elegant approach Wall removed: 2n C n ways End at 2m on the left of original position: 2n C n-1 ways Therefore the number of ways that the man never arrives at the considered position is (2nCn) – (2nCn-1) = (2nCn)/(n+1) This is called Catalan Number

70. Applications of Catalan Numbers The number of valid string consisting of n pairs of parenthesis ()()(), (())(), ()(()), (()()), ((())) The number of binary trees with n nodes

71. An evil game N people in a circle Kill the first person, skip the next k people, then kill the next, etc. Josephus Problem

72. N = 8, k = 2 0 6 2 4 5 7 1 3

73. 0 6 2 4 5 7 1 3

74. 0 6 2 4 5 7 1 3

75. 0 6 2 4 5 7 1 3

76. 0 6 2 4 5 7 1 3

77. 0 6 2 4 5 7 1 3

78. 0 6 2 4 5 7 1 3

79. 0 6 2 4 5 7 1 3

80. Josephus problem Closed form for k = 2

81. Josephus problem Naive simulation O(nk) With little technique in handling the index O(n) f(n, k) = (f(n-1, k) + k) mod n

82. Josephus problem With some more little technique in handling the index O(k log n)

83. Question?