escola de verao - unicamp - aritmetica - fidel

7/29/2019 escola de verao - unicamp - aritmetica - fidel

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Review on arithmetics for ICPC

Escola de Verao

Maratona de Programacao

Fidel I. Schaposnik (UNLP) - [email protected] 28, 2013
http://find/


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Contents

Natural numbers

Finding prime numbersFactorization, Eulers (n) and the number of divisors of n

Escola de Verao Maratona de Programacao Review on arithmetics for ICPC
http://find/


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Contents

Natural numbers


Modular arithmetics

Basic operationsModExp

GCD and its extensionChinese Reminder Theorem

http://find/


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Contents

Natural numbers


Modular arithmetics



Matrices

Notation and basic operationsAdjacency matrix for a graph

Markov chains and other linear problemsSystems of equationsGauss-Jordan algorithmA particular case: bi-diagonal matrices

http://find/


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Contents

Natural numbers


Modular arithmetics



Matrices

Notation and basic operationsAdjacency matrix for a graph

Markov chains and other linear problemsSystems of equationsGauss-Jordan algorithmA particular case: bi-diagonal matrices

One additional problem

http://find/


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Natural numbers

Recall that

p N is prime 1 and p are ps only divisors in N

Given n N, we may uniquely factorize it as

n = pei1 . . . pekk

http://find/


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Natural numbers

Recall that



n = pei1 . . . pekk

Finding prime numbers and/or the factorization of a given numberwill be useful to solve problems involving:

(completely) multiplicative functions

http://find/


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Natural numbers

Recall that



n = pei1 . . . pekk


(completely) multiplicative functionsdivisors of a given number

http://find/


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Natural numbers

Recall that



n = pei1 . . . pekk


(completely) multiplicative functionsdivisors of a given number

prime numbers and factorizations in general :-)

http://find/http://goback/


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Algorithms to find prime numbers

We would like to find all prime numbers up to a given number N(e.g. to factorize m we need all prime numbers up to

m).

http://find/


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m).

A very naive algorithm: for each n [2,N), we check if n isdivisible by some prime number smaller than

n (we have

already found all of them). With some minor optimizations:

1 p [ 0 ] = 2 ; P = 1 ;

2 f o r ( i =3; i


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m).

A very naive algorithm: for each n [2,N), we check if n isdivisible by some prime number smaller than

n (we have

already found all of them). With some minor optimizations:

1 p [ 0 ] = 2 ; P = 1 ;

2 f o r ( i =3; i


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Algorithms to find prime numbers (cont.)

A very old algorithm: build a table with the numbers in therange [2,N), and iterate through it. Each un-crossed numberwe find is a prime, so we can cross all its multiples in the table:

2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 2122 23 24 25 26 27 28 29 30 3132 33 34 35 36 37 38 39 40 41

42 43 44 45 46 47 48 49 50 51

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2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 2122 23 24 25 26 27 28 29 30 31

32 33

34 35

36 37

38 39

40 41

42 43 44 45 46 47 48 49 50 51

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2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 2122 23 24 25 26 27 28 29 30 3132 33 34 35 36 37 38 39 40 4142 43 44 45 46 47 48 49 50 51


( )
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2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 2122 23 24 25 26 27 28 29 30 3132 33 34 35 36 37 38 39 40 4142 43 44 45 46 47 48 49 50 51


Al i h fi d i b ( )
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2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 2122 23 24 25 26 27 28 29 30 3132 33 34 35 36 37 38 39 40 4142 43 44 45 46 47 48 49 50 51


Al i h fi d i b ( )
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2 3 4 5 6 7 8 9 10 1112 13 14 15 16 17 18 19 20 2122 23 24 25 26 27 28 29 30 31

32

33

34

35

36 37

38

39

40 41

42 43 44 45 46 47 48 49 50 51


Al i h fi d i b ( )


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The corresponding code looks like

1 memset( is p , t r u e , s i z e o f ( i s p ) ) ;2 f o r ( i =2; i


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The corresponding code looks like

1 memset( is p , t r u e , s i z e o f ( i s p ) ) ;2 f o r ( i =2; i


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Factorization using the sieve

The sieve can actually hold more information

1 f o r ( i =4; i


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Factorization using the sieve

The sieve can actually hold more information

1 f o r ( i =4; i


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Divisors of a given number

We can now generate the divisors of a given number recursively

1 i n t d [MAXD] , D=0;23 v o i d d i v ( i n t cu r , i n t f [ ] , i n t s , i n t e ) {4 i f ( s == e ) d [ D++] = cu r ;5 e l s e {6 i n t m;7 f o r (m=s +1; m


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Some functions from number theory

If we can factorize a given number n, we can calculate somenumber theory related functions:


http://find/


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Eulers function: (n) is the number of positive integersless than n that are coprime with n. We have

(n) = (pe11 pe111 ) . . . (pekk pek1k )




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Eulers function: (n) is the number of positive integersless than n that are coprime with n. We have

(n) = (pe11 pe111 ) . . . (pekk pek1k )

The number of divisors of n is0(n) = (e1 + 1) . . . (ek + 1)

Larger numbers of divisors are achieved when there are manydistinct prime factors (n different prime factors

(n) = 2n):

6.469.693.230 = 2 3 5 7 11 13 17 19 23 29only has 1024 divisors.

There are similar formulas for m(n) =d|n

dm


Further reading
http://find/


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Further reading

There are other, faster algorithms to find prime numbers. Inparticular, the sieve of Eratosthenes can be further optimized:

in memory, to use just one bit per number;

using a wheel to achieve better running time:

1 i n t w [ 8 ] = {4 ,2 ,4 ,2 ,4 ,6 ,2 ,6 } ;23 f o r ( i =7 , c u r = 0; i i


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Further reading

There are other, faster algorithms to find prime numbers. Inparticular, the sieve of Eratosthenes can be further optimized:

in memory, to use just one bit per number;

using a wheel to achieve better running time:

1 i n t w [ 8 ] = {4 ,2 ,4 ,2 ,4 ,6 ,2 ,6 } ;23 f o r ( i =7 , c u r = 0; i i


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Modular arithmetic

Recall that given a, r Z and m N

a m r a = q.m + r with r = 0, 1, . . . , m 1

Summation, substraction and multiplication operations are triviallyextended and keep their well-known properties

a b = c = a bm c

a.b = c = a.bm c


Modular arithmetic


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Modular arithmetic

Recall that given a, r Z and m N

a m r a = q.m + r with r = 0, 1, . . . , m 1

Summation, substraction and multiplication operations are triviallyextended and keep their well-known properties

a b = c = a bm c

a.b = c = a.bm cDivision is defined as the inverse of multiplication, so

a/b = a.b1 with b.b1 = 1

Does the inverse mod m always exist? How do we calculate it?


ModExp
http://goforward/http://find/http://goback/


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ModExp

Sometimes we can avoid calculating inverses altogether: e.g.problem Last Digit (MCA07) asks us to calculate the last

non-zero digit of

=

N

m1 . . .mM

=

N!

m1! . . . mM!with

Mi=1

mi = N and N 106


ModExp


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p


non-zero digit of

=

N

m1 . . .mM

=

N!

m1! . . . mM!with

Mi=1

mi = N and N 106

We may factorize using the sieve, then evaluate it mod 10 aftereliminating any factors of 10 (i.e. the maximum possible numberof 5s and 2s). We also need to efficiently evaluate ab mod m:


ModExp


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p


non-zero digit of

=

N

m1 . . .mM

=

N!

m1! . . . mM!with

Mi=1

mi = N and N 106


Direct evaluation is O(b), which is (usually) too slow.


ModExp


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p


non-zero digit of

=

N

m1 . . .mM

=

N!

m1! . . . mM!with

Mi=1

mi = N and N 106


Direct evaluation is O(b), which is (usually) too slow.Write b in binary, then b = c0.2

0 +

+ clog b.2

log b and we

can evaluate ab in O(log b)

ab =

log bi=0,ci=0

a2i


ModExp (code)


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p ( )

1 l o n g l o n g modexp( l o n g l o n g a , i n t b ) {2 l o ng l o n g RES = 1LL;3 w h i l e ( b > 0 ) {4 i f ( b&1) RES = (RESa )% MOD;5 b >>= 1 ;

6 a = ( aa )% MOD;7 }8 r e t u r n RES;9 }

ModExp


Last Digit (MCA07)
http://find/


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( )

1 i n t c a l c ( i n t N, i n t m [ ] , i n t M) {2 i n t i , RES ;3

4 e [ N]++;5 f o r ( i =0; i 1) e [m[ i ] ];6 f o r ( i=N1; i >=2; i ) e [ i ] += e [ i +1] ;7 f o r ( i=N; i >=2; i ) i f ( p [ i ] ) {8 e [ i /p [ i ] ] += e [ i ] ;9 e [ p [ i ] ] += e [ i ] ;

10 e [ i ] = 0 ;11 }12 i n t tmp = min ( e [ 2 ] , e [ 5 ] ) ;13 e [ 2 ] = tmp ; e [ 5 ] = t m p ;1415 RES = 1 ;16 f o r ( i =2; i


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The greatest common divisor between a and b is the largest d suchthat d

|a and d

|b. Note that

a = q.b+ r = gcd(a, b) = gcd(b, r)

Which leads to

1 i n t g c d ( i n t a , i n t b ) {2 i f (b == 0 ) r e t u r n a ;3 r e t u r n gcd (b , a%b ) ;4 }

Euclids algorithm


GCD
http://find/


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The greatest common divisor between a and b is the largest d suchthat d

|a and d

|b. Note that

a = q.b+ r = gcd(a, b) = gcd(b, r)

Which leads to

1 i n t g c d ( i n t a , i n t b ) {2 i f (b == 0 ) r e t u r n a ;3 r e t u r n gcd (b , a%b ) ;4 }

Euclids algorithm

It can be shown that gcd(Fn+1,Fn) takes exactly n operations(here Fn are the Fibonacci numbers). Because Fn growsexponentially and is the worst-case input for this algorithm,running time is O(log n).


Extended GCD
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One may prove that

gcd(a,m) = 1

1 = a.x + m.y


Extended GCD


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One may prove that

gcd(a,m) = 1

1 = a.x + m.y

Then xm a1, so that a has an inverse mod m if and only ifgcd(a,m) = 1. [Corolary: Zp is a field.]


Extended GCD
http://find/


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One may prove that

gcd(a,m) = 1

1 = a.x + m.y

Then xm a1, so that a has an inverse mod m if and only ifgcd(a,m) = 1. [Corolary: Zp is a field.] To find x and y, we tracethem along Euclids algorithm:

1 p i i e g cd ( i n t a , i n t b ) {

2 i f (b == 0 ) r e t u r n m a k e p a i r ( 1 , 0 ) ;3 e l s e {4 p i i RES = e gcd ( b , a%b ) ;5 r e t u r n m a k e p a i r ( RES . s e c o n d , RES . f i r s t RES . sec ond ( a / b ) ) ;6 }7 }8

9 i n t i n v ( i n t n , i n t m) {10 p i i EGCD = e g c d ( n , m) ;11 r e t u r n ( (EGCD . f i r s t % m)+m)% m;12 }

Extension of Euclids algorithm; inverse mod m


Chinese Remainder Theorem


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Given a set of conditions

x ai mod ni for i = 1, . . . , k with gcd(ni, nj) = 1 i= j

there is a unique x mod N = n1 . . . nk satisfying all theseconditions simultaneously.


Chinese Remainder Theorem
http://find/


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Given a set of conditions

x ai mod ni for i = 1, . . . , k with gcd(ni, nj) = 1 i= j

there is a unique x mod N = n1 . . . nk satisfying all theseconditions simultaneously. We may construct it considering

mi =j=i

nj = gcd(ni,mi) = 1

Let mi = m1i mod ni, then

xk

i=1

mimiai mod N


Chinese reminder theorem (code)


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1 i n t c r t ( i n t n [ ] , i n t a [ ] , i n t k ) {2 i n t i , tmp , MOD, RES ;34 MOD = 1 ;5 f o r ( i =0; i


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An N M matrix is an array of N rows and M columns ofelements. A natural way to define the sum and the difference of

matrices is (O(N.M)):A B = C Cij = Aij Bij

The product of matrices is defined as (O(N.M.L))

ANM BML = CNL Cij =M

k=1

Aik.Bkj


Matrices
http://find/


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An N M matrix is an array of N rows and M columns ofelements. A natural way to define the sum and the difference of

matrices is (O(N.M)):A B = C Cij = Aij Bij

The product of matrices is defined as (O(N.M.L))

ANM BML = CNL Cij =M

k=1

Aik.Bkj

For square matrices (from now on, assume we are working with

N N), it makes sense to ask if a given matrix A has an inverse.One finds that if det A = 0, the inverse exists and satisfies

A A1 = 1 = A1 A


Matrices (cont.)


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We represent matrices using bi-dimensional arrays. In C++, to usea matrix as an argument of a function, it is convenient to define alist of pointers to avoid fixing one of the dimensions in thedefinition of the function

1 t yp e f u n c t i o n ( i n t A , i n t N, i n t M) {2 . . .3 }

45 i n t main ( ) {6 i n t a [MAXN] [MAXN] , pa [MAXN] ;78 f o r ( i n t i =0 ; i


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We may use a matrix to represent the edges of a graph.


Adjacency matrix
http://find/


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We may use a matrix to represent the edges of a graph. For agraph of N nodes, a matrix ANN can have Aij:

the weight of the edge going from node i to node j (or ifthis edge does not exist);


Adjacency matrix


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the number of edges that go from node i to node j (0 if thereare none).

In the latter case,

(A2)ij =

k

AikAkj = paths i j with exactly 2 edges

and in general An contains the number of paths between the pairsof nodes of the original graph with exactly n edges.


Adjacency matrix
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the number of edges that go from node i to node j (0 if thereare none).

In the latter case,

(A2)ij =

k

AikAkj = paths i j with exactly 2 edges

and in general An contains the number of paths between the pairsof nodes of the original graph with exactly n edges.We may calculate An using an adapted version of ModExp inO(N3 log n).


Linear recurrences
http://find/


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A linear recurrence of order K defines a sequence {an} through

an =

Kk=1

ckank

and a vector aK = (aK, aK1, . . . , a1) of initial values.


Linear recurrences
http://find/


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A linear recurrence of order K defines a sequence {an} through

an =

Kk=1

ckank

and a vector aK = (aK, aK1, . . . , a1) of initial values.

Let an be a vector containing an and its K

1 previous elements.Then the recurrence can be represented as

an = Ran1

anan1

..

.anK+2anK+1

=

c1 c2 cK1 0 00 1

0

......

0 1 0

an1an2

..

.anK+1

anK

So we can calculate the N-th term using aN = RNKaK


Markov chains
http://find/


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Let

{Si

}be a set of states, and pij the known probabilities to

observe a transition from state i to state j. Terminal states {Sk}are such that

i pki = 0. What is the expected number of

transitions Ei needed to reach a terminal state from state Si?


Markov chains
http://find/


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Let

{Si




transitions Ei needed to reach a terminal state from state Si?For terminal states, we clearly have

Ek = 0

For the rest,Ei = 1 +

j

pij.Ej


Markov chains
http://find/


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Let

{Si




transitions Ei needed to reach a terminal state from state Si?For terminal states, we clearly have

Ek = 0

For the rest,Ei = 1 +

j

pij.Ej

In other words, we need to solve a system of equations over theexpected number of transitions Ei.


Other linear problems


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Some linear systems of equations can be found in computational

geometry problems. As an example, something you may need latertoday: plane-plane intersection ;-)


Other linear problems


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Some linear systems of equations can be found in computational

geometry problems. As an example, something you may need latertoday: plane-plane intersection ;-)

A plane is defined by a point p0 on it and a vector n0 which isorthogonal to the plane. All other points on the plane satisfy

(p p0) n0 = 0

which leads to the familiar equation

ax + by + cz + d = 0 for n0 = (a, b, c) and d =

p0n0

In the general case, two planes defined by {p1,n1} and {p2,n2}intersect to give a line. How do we find this line?


Other linear problems (cont.)

A line is defined by a point p on it and a directing vector v


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A line is defined by a point p on it and a directing vector vpointing along its direction (either way).





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Our line should belong to both planes, so it is orthogonal to bothnormal vectors n1 and n2. Then

v = n1 n2

Ifv =0, the planes are parallel and there is no intersection. Buthow do we find p?





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Our line should belong to both planes, so it is orthogonal to bothnormal vectors n1 and n2. Then

v = n1 n2

Ifv =0, the planes are parallel and there is no intersection. Buthow do we find p?

Let p be that point on the line which is closest to some fixed,arbitrary point (e.g. the origin). Then p = (x, y, z) minimizes

D2 = x2 + y2 + z2

while being on both planes, i.e. satisfying

(p p1) n1 = 0 and (p p2) n2 = 0Escola de Verao Maratona de Programacao Review on arithmetics for ICPC


We can find the point p minimizing D2 under these constraints


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We can find the point p minimizing D under these constraintsusing Lagrange multipliers. Define

f(x, y, z, 1, 2) = x2 + y2 + z2 +1 (p p1) n1 +2 (p p2) n2Then we must ask

f

x= 2x+1n1x = 0

f

y= 2y+1n1y = 0

f

z= 2z+1n1z = 0

f

1= (p p1) n1 = 0 f

2= (p p2) n2 = 0



We can find the point p minimizing D2 under these constraints
http://find/


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We can find the point p minimizing D under these constraintsusing Lagrange multipliers. Define

f(x, y, z, 1, 2) = x2 + y2 + z2 +1 (p p1) n1 +2 (p p2) n2Then we must ask

f

x= 2x+1n1x = 0

f

y= 2y+1n1y = 0

f

z= 2z+1n1z = 0

f

1= (p p1) n1 = 0 f

2= (p p2) n2 = 0

Or in matrix notation

2 0 0 n1x n2x

0 2 0 n1y n2y0 0 2 n1z n2z

n1x n1y n1z 0 0n2x n2y n2z 0 0

x

yz12

=

0

00

p1 n1p2 n2


Systems of equations

A system of equations over N variables is
http://find/


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a11 x1 +

+ a1N xN = b1

...

aN1 x1 + + aN xN = bNIt can be represented by matrices as

Ax = b

a11 a1N...

. . ....

aN1 aNN

x1...

xN

=

b1...

bN


Systems of equations

http://find/


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a11 x1 +

+ a1N xN = b1

...

aN1 x1 + + aN xN = bNIt can be represented by matrices as

Ax = b

a11 a1N...

. . ....

aN1 aNN

x1...

xN

=

b1...

bN

Solving this system reduces to finding the inverse A1

, becausethen x = A1b. We can solve many systems of equations withdifferent independent terms bi by building a matrix B with thevectors bi as its columns. Then X = A

1B, and in particular ifB

1, X

A1.


Systems of equations (cont.)
http://find/


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To solve a system by hand, we solve for an arbitrary variable in one

of the equations, and then use this to eliminate that variable fromevery other equation, while working simultaneously with theindependent terms. In order to do this, we may:

Multiply or divide an equation (row) by any number.

Sum or subtract an equation (row) to another one.Swap two rows (leaves the equations unchanged)


Systems of equations (cont.)


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To solve a system by hand, we solve for an arbitrary variable in one

of the equations, and then use this to eliminate that variable fromevery other equation, while working simultaneously with theindependent terms. In order to do this, we may:

Multiply or divide an equation (row) by any number.

Sum or subtract an equation (row) to another one.Swap two rows (leaves the equations unchanged)

Gauss-Jordans algorithm is a formalization of this procedure, withone caveat: to reduce the numerical error, in each step the

corresponding variable is solved from the equation where it appearswith the largest coefficient in absolute value (this is calledpivoting).


Gauss-Jordan elimination

1 b o o l i n v e r t ( d o u b l e A , d o u b l e B , i n t N) {
http://find/


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( ) {2 i n t i , j , k , jmax ; d o u b l e tmp;3 f o r ( i =1; i= i5 f o r ( j=i +1; j a b s ( A [ j m ax ] [ i ] ) ) j m ax = j ;78 f o r ( j =1; j


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( )we can multiply two N N matrices in O (T(N)), then we caninvert a matrix or calculate its determinant in the same asymptotictime.


Gauss-Jordan elimination for bi-diagonal matrices

Gauss-Jordans algorithm is clearly O(N3). It can be seen that if
http://find/


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we can multiply two N N matrices in O (T(N)), then we caninvert a matrix or calculate its determinant in the same asymptotictime.

Usually, instead of optimizing the general algorithm it is a lot moreconvenient to take advantage of some special property of the

matrices we would like to invert: for instance, we can invert abi-diagonal or tri-diagonal matrix (with non-zero diagonalelements) in O(N).

a11 a12 0 0 0 . . . 0

a21 a22 a23 0 0 . . . 00 a32 a33 a34 0 . . . 0...

...0 . . . 0 aNN1 aNN


Further reading
http://find/


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There are more efficient algorithms that you can use to

multiply two matrices (Strassens is O(n2,807

),CoppersmithWinograds is O(n2,376)), but they are notnecessarily convenient for a competition...

There are many important decompositions of a matrix (i.e.

ways to write it as a sum or product of some other matriceswith special properties). These play an important role inmany algorithms, such as those that are used to calculatedeterminants.

There are many ways to generalize the treatment of linearrecurrences: one can work with sets of multiple dependentrecurrences, inhomogeneous recurrences, etc.


An additional problem
http://find/


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Bases (SARC08): Analyze in which bases an expression such as10000 + 3 5 334 = 3 5000 + 10 + 0 is true.


http://find/


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Things we have not covered:

Polynomials (evaluation, operations, properties, etc.).Evaluation of mathematical expressions (parsing).


http://find/


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Things we have not covered:

Polynomials (evaluation, operations, properties, etc.).Evaluation of mathematical expressions (parsing).

Maybe next time ;-)

http://find/

escola de verao - unicamp - aritmetica - fidel

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