mathematics in oi prepared by ivan li. mathematics in oi a brief content greatest common divisor...

Mathematics in OI

Prepared by Ivan Li

Mathematics in OIA 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

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 ?”


• 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.


• 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

Euclidean Algorithm

• Assume m > nGCD(m,n)While n > 0

m = m mod nswap m and n

Return m


• 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)

…))

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.

Applications

• Solve mx + ny = a for integers x and y

• Can be solved if and only if a is divisible by gcd(m,n)

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)

Least Common Multiple

• What if we want to find the LCM of more than two numbers?

• Apply LCM two-by-two?

Extended Euclidean Algorithm

• The table method• e.g. solve 93x + 27y = 6

• General form:• x = -4 + 9k, y = 14 - 31k, k

integer

Coeff of 93 Coeff of 27

(1) 93 1 0

(2) 27 0 1

(3): (1)-3(2) 12 1 -3

(4): (2)-2(3) 3 (gcd, 3*2=6)

-2 (x = -4) 7 (y = 14)

(5): (3)-4(4) 0 9 -31

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

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

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))

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-1c

(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

Modular Inverse

• ab ≡ 1 (mod m)• ab + km = 1• Extended Euclidean

algorithm

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

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 ?

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?)

Finding Prime Numbers

• For each integer, check if it is a prime

• Prime List• Sieve of Eratosthenes

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.

Sieve of Eratosthenes

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

Sieve of Eratosthenes

for i = 2 … nIf not Comp[i]

output ij = i *i //why

i*i?while j n

Comp[j] = truej = j + i

Optimization

• Consider odd numbers only• Do not forget to add 2, the

only even prime

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

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?


• To tolerate some floating point

• Introduce epsilon• EPS = 1e-8 to 1e-11

• a < b => a + EPS < b• a <= b => a <= b + EPS • a == b => abs(a - b) <=

EPS


• Special values• Positive / Negative infinity• Not a number (NaN)

• Checked in C++ by x!=x

• Denormal number

High Precision Arithmetic

• 32-bit signed integer:-2147483648 … 2147483647

• 64-bit signed integer:-9223372036854775808 … 9223372036854775807

• How to store a 100 digit number?


• Use an array to store the digits of the number

• Operations:• Comparison• Addition / Subtraction• Multiplication• Division and remainder

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.


• 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

More on HPA

• How to store• negative numbers?• fractions?• floating-point numbers?

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.

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

Partial Sum

• How to compute s[i] ?• During input

s[0] = 0for i = 1 to n

input a[i]s[i] = s[i-1] + a[i]

Difference operation

• 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.


• 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]


• Easy to compute d[i]• But how to convert it back to

a[i]?• Before (or during) output

a[0] = 0for i = 1 to n

a[i] = a[i-1] + d[i]output a[i]

• Quite similar to partial sum, isn’t it?

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

Comparison

• Partial sum - Fast sum of range query

• Difference - Fast range increment

• Ordinary array - Fast query and increment on single element

Runtime ComparisonRange Query

Single Query

Single Update

Range Update

Partial sum Constant

Constant(treat a single element as a range)

Linear(Have to update all sums involved)

Linear

Ordinary array

Linear Constant

Constant

Linear

Difference Linear(Convert it back to the original array)

Linear Constant

Constant

• 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.

Euler Phi Function

• (n) = number of integers in 1...n that are relatively prime to n

• (p) = p-1• (pn) = (p-1)pn-1 = pn(1-1/p)• Multiplicative property:

(mn)=(m)(n) for (m,n)=1

• (n) = np|n(1-1/p)

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

Fibonacci Number

• F0 = 0, F1 = 1

• Fn = Fn-1 + Fn-2 for n > 1

• The number of rabbits• The number of ways to go

upstairs

• How to calculate Fn?

What any half-wit can do

• F0 = 0; F1 = 1;for i = 2 . . . n

Fi = Fi-1 + Fi-2;return Fn;

• Time Complexity: O(n)• Memory Complexity: O(n)

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)

What a Math student would do

• Generating Function

• G(x) = F0 + F1 x + F2 x2 +. . .

• A generating function uniquely determines a sequence (if it exists)• Fn = dnG(x)/dxn (0)

• A powerful (but tedious) tool in solving recurrence

All the tedious works• G(x) = F0 + F1 x + F2 x2 + F3 x3 +. . .

• xG(x) = F0 x + F1 x2 + F2 x3 +. . .

• x2G(x) = F0 x2 + F1 x3 +. . .

• G(x) - xG(x) - x2G(x) = F0 +F1 x - F0 x = x

• G(x) = x / (1 - x - x2)• 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 + r2x2 +. . . = 1 / (1 - rx)

G(x) = (sqrt(5) / 5)(-1-x/a-x2/a2-...+1+x/b+x2/b2+...)

• By Uniqueness, Fn = (sqrt(5) / 5)(-1/an + 1/bn)

Shortcut

• Characteristic Equation

• Fn - Fn-1 - Fn-2 = 0

• f(x) = x2 – x – 1

• Then Fn = Aan + Bbn

for some constants A, Bwhere a, b are roots of

f(x)

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 Fn mod m?

What a programmer would do

• Note that

( )( ) = ( )• Then

( )n( ) = ( )• Matrix Exponential

Just like fast exponential

0 11 1

Fn

Fn+1

Fn+1

Fn+2

0 11 1

F0

F1

Fn

Fn+1

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?

Twelvefold ways

Combinations

• The number of ways to choose r objects among n different objects (without order)

• nCr = n!/r!(n-r)!

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;

Combinations

• Overflow problem?• Note nCr = (n/r)(n-1)C(r-1)

• nCr = 1; //that is (n-r)C0

for i = 1 . . . rnCr *= (n - r + i);nCr /= i;

Combinations

• What if we are asked to findnCr mod p for very large n, r?

• Lucas Theorem

• Let n = nknk-1...n1n0 (base p)r = rkrk-1...r1r0

• ThennCr (nkCrk)(nk-1Crk-1)...(n0Cr0) (mod p)

• Works only for prime p

Combinations

• When p is large (larger than r), computing niCri may be difficult

• (nCr)(r!) = n(n-1)...(n-r+1)• nCrn(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)

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

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?

Observations

• If k is odd, then it is impossibleLet k = 2n

• If the man returns to original position, then there are n left and n right moves

• Number of possible outcomes= 22n

• We have to find the number of moving sequence such that the man returns to original position without hitting the wall

Observations

• If the wall is removed, the number of ways is 2n C n

• Let An be the number of ways

• If the first time the man returns to his original position is at the 2ith second• Note that the first move must

be rightward, and the 2ith move must be leftward

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

• Ai-1 ways for the first 2i seconds• An-i ways for the remaining 2n-

2i seconds (may return to original position again)

• An = i = 1...nAi-1An-i

Tedious works again

• Again, generating function• g(x) = n=0,1,... Anxn

• g(x)2 = n=0,1,...i=0...n Ai An-i xn

= n=0,1,...An+1xn • g(x) = A0+ xn=1,... Anxn-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......

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

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

Applications of Catalan Numbers

• The number of valid string consisting of n pairs of parenthesis()()(), (())(), ()(()), (()()), ((()))

• The number of binary trees with n nodes

An evil game

• N people in a circle• Kill the first person, skip the

next k people, then kill the next, etc.

• Josephus Problem

N = 8, k = 2

0

62

4

5

71

3

Josephus problem

• Closed form for k = 2

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

Josephus problem

• With some more little technique in handling the index• O(k log n)

Question?

mathematics in oi prepared by ivan li. mathematics in oi a brief content greatest common divisor...

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