Mathematics in OI

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Mathematics in OI

• Prepared by Ivan Li

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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

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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 ?

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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.

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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

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Euclidean Algorithm

• Assume m gt n
• GCD(m,n)
• While n gt 0
• m m mod n
• swap m and n
• Return m

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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)))

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Applications

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

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Applications

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

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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)

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Least Common Multiple

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

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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, 326) -2 (x -4) 7 (y 14)
(5) (3)-4(4) 0 9 -31
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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

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Modular Arithmetic

• Like the equal sign
• Addition / subtraction on both sides
• 7 1 (mod 3) gt 7212 (mod 3)
• Multiplication on both side
• 7 1 (mod 3) gt 7212 (mod 3)
• We can multiply into m too
• 71 (mod 3)ltgt7212 (mod 32)
• Congruence in mod 6 is stronger than that in mod
  3

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Modular Arithmetic

• Division?
• Careful
• 64 (mod 2), but not 32 (mod 2)
• acbc (mod m)ltgtab (mod m) when c, m coprime
  (gcd 1)
• Not coprime?
• acbc(mod cm)ltgtab(mod m)
• ac bc(mod m) ltgt a b (mod m/gcd(c,m))

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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)gt 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

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Modular Inverse

• ab 1 (mod m)
• ab km 1
• Extended Euclidean algorithm

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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

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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 ?

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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?)

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Finding Prime Numbers

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

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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.

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Sieve of Eratosthenes

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

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Sieve of Eratosthenes

• for i 2 n
• If not Compi
• output i
• j i i //why ii?
• while j ? n
• Compj true
• j j i

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Optimization

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

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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

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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?

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Floating point arithmetic

• To tolerate some floating point
• Introduce epsilon
• EPS 1e-8 to 1e-11
• a lt b gt a EPS lt b
• a lt b gt a lt b EPS
• a b gt abs(a - b) lt EPS

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Floating point arithmetic

• Special values
• Positive / Negative infinity
• Not a number (NaN)
• Checked in C by x!x
• Denormal number

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High Precision Arithmetic

• 32-bit signed integer-2147483648 2147483647
• 64-bit signed integer-9223372036854775808
  9223372036854775807
• How to store a 100 digit number?

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High Precision Arithmetic

• Use an array to store the digits of the number
• Operations
• Comparison
• Addition / Subtraction
• Multiplication
• Division and remainder

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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.

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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

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More on HPA

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

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Partial Sum

• Motivation
• How to find the sum of the 3rd to the 6th element
  of an array ai ?
• a3 a4 a5 a6
• 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.

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Partial Sum

• Use an array si to store the sum of the first i
  elements.
• si a1 a2 ai
• The sum of the j th element to the k th element
  sk sj-1
• We usually set s0 0

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Partial Sum

• How to compute si ?
• During input
• s0 0
• for i 1 to n
• input ai
• si si-1 ai

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Difference operation

• Motivation
• How to increment the 3rd to the 6th element of an
  array ai ?
• a3, a4, a5, a6
• 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.

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Difference operation

• Use an array di to store the difference between
  ai and ai-1.
• di ai - ai-1
• When the the j th element to the k th element is
  incremented,
• dj , dk1 - -
• We usually set d1 a1

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Difference operation

• Easy to compute di
• But how to convert it back to ai?
• Before (or during) output
• a0 0
• for i 1 to n
• ai ai-1 di
• output ai
• Quite similar to partial sum, isnt it?

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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

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Comparison

• Partial sum - Fast sum of range query
• Difference - Fast range increment
• Ordinary array - Fast query and increment on
  single element

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Runtime Comparison
Range 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
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• 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.

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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) n?pn(1-1/p)

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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

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Fibonacci Number

• F0 0, F1 1
• Fn Fn-1 Fn-2 for n gt 1
• The number of rabbits
• The number of ways to go upstairs
• How to calculate Fn?

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What any half-wit can do

• F0 0 F1 1for i 2 . . . n Fi Fi-1
  Fi-2return Fn
• Time Complexity O(n)
• Memory Complexity O(n)

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What a normal person would do

• a 0 b 1for i 2 . . . N t b b
  a a t return b
• Time Complexity O(n)
• Memory Complexity O(1)

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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

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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 FractionG(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-...1x/bx2
  /b2...)
• By Uniqueness, Fn (sqrt(5) / 5)(-1/an 1/bn)

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Shortcut

• Characteristic Equation
• Fn - Fn-1 - Fn-2 0
• f(x) x2 x 1
• Then Fn Aan Bbn for some constants A,
  B where a, b are roots of f(x)

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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?

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What a programmer would do

• Note that( )( ) ( )
• Then( )n( ) ( )
• Matrix ExponentialJust like fast exponential

0 1 1 1
Fn Fn1
Fn1 Fn2
0 1 1 1
F0 F1
Fn Fn1
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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?

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Twelvefold ways
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Combinations

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

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Combinations

• How to calculate nCr?
• Calculate n!, r!, (n-r)! ?
• Note nCr n(n-1)...(n-r1)/r!
• nCr 1for i n-r1 . . . n nCr i for i
  1 . . . r nCr / i

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Combinations

• Overflow problem?
• Note nCr (n/r)(n-1)C(r-1)
• nCr 1 //that is (n-r)C0for i 1 . . . r nCr
  (n - r i) nCr / i

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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

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Combinations

• When p is large (larger than r), computing niCri
  may be difficult
• (nCr)(r!) n(n-1)...(n-r1)
• nCr?n(n-1)...(n-r1)(r!)-1 (mod p) where
  (r!)((r!)-1) ? 1 (mod p)
• Fermat Little Theorem gives(r!)-1 ? (r!)p-2 (mod
  p)

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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

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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?

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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 22n
• We have to find the number of moving sequence
  such that the man returns to original position
  without hitting the wall

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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

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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

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Tedious works again

• Again, generating function
• g(x) ?n0,1,... Anxn
• g(x)2 ?n0,1,...?i0...n Ai An-i xn
  ?n0,1,...An1xn
• g(x) A0 x?n1,... Anxn-1 1 xg(x)2
• xg(x) (1-sqrt(1-4x))/2 or (1sqrt(1-4x))/2
  (rejected since xg(x)0 when x 0)
• Power series......

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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 wont 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

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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)/(n1)
• This is called Catalan Number

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Applications of Catalan Numbers

• The number of valid string consisting of n pairs
  of parenthesis()()(), (())(), ()(()), (()()),
  ((()))
• The number of binary trees with n nodes

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An evil game

• N people in a circle
• Kill the first person, skip the next k people,
  then kill the next, etc.
• Josephus Problem

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N 8, k 2
0
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N 8, k 2
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N 8, k 2
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N 8, k 2
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N 8, k 2
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N 8, k 2
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N 8, k 2
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N 8, k 2
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Josephus problem

• Closed form for k 2

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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

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Josephus problem

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

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Question?