Mathematics in OI

Prepared by Ivan LiModified by LMH

Mathematics in OIA brief content

• Euler Phi function• Fibonacci Sequence and

Recurrence• Twelvefold ways• Combinations and Lucas

Theorem• Catalan Numbers• Josephus Problem

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 = 1/(n!) 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)) / 2By Partial

Fraction:• G(x) = A / (a – x) + B / (b – x)• Solve A, B by sub x = 0, x = 1 and form two equations

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 = Coefficient of x^k =

• (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 = pk nk + pk-1 nk-1...+ p n1 + n0

r = pk rk + pk-1 rk-1...+ r n1 + n0 ThennCr (nkCrk)(nk-1Crk-1)...(n0Cr0) (mod p)

• Works only for prime p• What if p is large such that storing

intermediate nCr wouldn’t be possible?

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)• Turning division in mod into

multiplication !

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• From 2 to 2i – 1 second, he

cannot return to original position due to definition of 2i

• In 2 and 2i – 1 second, he must be standing on 1m on the right of original position

• Problem in this time interval reduced to a sub-problem with 2i – 2 total moves• Think of an invisible wall on his

original position

Observations

• 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

• The number of monotonic paths along the edges of a grid with n × n square cells

An evil game

• N people in a circle• Labelled from 1 to n• Kill the k-th person, skip the

next k-1 people, then kill the next, etc.

• Josephus Problem

N = 8, k = 2

8

62

4

5

71

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N = 8, k = 2

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6

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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

Josephus problem

• Naive simulation• O(nk)

• With little technique in handling the index• O(n)• f(n, k) = (f(n-1, k) + k-1) mod n +

1• f(1, k) = 1

• Confused about the new index?

New indexing

• f(8, 2) = (f(7, 2) + 1) mod 8 + 1• 7 in f(7, 2) circle -> 1• 1 in f(7, 2) circle -> 3

8

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71

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Josephus problem

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

Question?