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Programming Interest Group http://www.comp.hkbu.edu.hk/~chxw/pig/index.htm. Tutorial Five Combinatorics & Number Theory. Outline. Combinatorics ( 组合 ) Basic Counting Techniques Recurrence Relations Binomial Coefficients Number Theory Prime Number Congruences Fermat’s Theorem

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### Programming Interest Grouphttp://www.comp.hkbu.edu.hk/~chxw/pig/index.htm

Tutorial Five

Combinatorics & Number Theory

• Combinatorics (组合)

• Basic Counting Techniques

• Recurrence Relations

• Binomial Coefficients

• Number Theory

• Prime Number

• Congruences

• Fermat’s Theorem

• Euler’s Theorem

• Combinatorics is the mathematics of counting.

• It appears to be a collection of unrelated puzzles chosen at random.

• E.g. 1:

• Given n letters and n addressed envelopes, in how many ways can the letters be placed in the envelopes so that no letter is in the correct envelope?

• E.g. 2: (a Google interview problem)

• There are n stages. Every time you can go up either 1 stage or 2 stages. How many different ways can you reach the last stage?

• Product Rule

• If there are |A| possibilities from set A and |B| possibilities from set B, then there are |A|x|B| ways to combine one from A and one from B.

• E.g., suppose you own 5 shirts and 4 pants, then there are 5x4 =20 different ways you can get dressed.

• Sum Rule

• If there are |A| possibilities from set A and |B| possibilities from set B, then there are |A|+|B| ways for either A or B to occur (assuming the elements of A and B are distinct).

A

B

A

C

Basic Counting Techniques

• Inclusion-Exclusion Formula:

• |AUB| = |A| + |B| - |A∩B|

• |AUBUC| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B ∩C|

• Permutations

• A permutation is an arrangement of n items, where every item appears exactly once.

• There are n! different permutations.

• 2n-1≤ n! ≤ nn-1 =====> n! is a very large number.

• 232 = 4,294,967,296

• 10! = 3,628,800

• 11! = 39,916,800

• 12! = 479,001,600

• 13! = 6,227,020,800 > 232

• Number of subsets

• A subset is a selection of elements from n possible items. (including the empty set)

• There are 2n distinct subsets of n things.

• E.g., set {a, b, c} has 8 (=23) subsets:

• Ф, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b ,c}

• Recurrence relation is an equation which is defined in terms of itself.

• E.g. 1: an = an-1 + 1, a1 =1

•  an = n

• E.g. 2: an = 2an-1, a1 =2

•  an = 2n

• E.g. 3: an = nan-1, a1 =1

•  an = n!

• The interview question from Google

• To reach the nth stage, there are two possibilities:

• First reach the (n-2)th stage, then go to the nth stage directly

• First reach the (n-1)th stage, then go to the nth stage

• We can have the following recurrence relation:

• an = an-1 + an-2

• a1 = 1, a2 = 2

= -0.61803…

Fibonacci Numbers

• Fn = Fn-1 + Fn-2; F0 = 0; F1 = 1

• 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …

• Closed form:

• It can easily proved by mathematical induction.

• Golden ratio:

• Fibonacci numbers grow exponentially.

• How many ways are there to choose k things out of n possibilities?

• Coefficients of (a+b)n

• It’s difficult to calculate the value of binomial coefficients by using the previous equation directly: arithmetic overflow will happen for n > 12!

• Pascal’s triangle (1654), or杨辉三角 (1261)

• 1

• 1 1

• 1 2 1

• 3 3 1

• 1 4 6 4 1

• 5 10 10 5 1

• 1 6 15 20 15 6 1

#define MAXN 100

/* compute n choose m */

long binomial_coef(int n, int m) {

int i, j;

long bc[MAXN][MAXN}; /* table of binomial coef */

for(i = 0; i <= n; i++) bc[i][0] = 1;

for(j = 0; j <= n; j++) bc[j][j] = 1;

for(i = 1; i <= n; i++)

for(j = 1; j < i; j++)

bc[i][j] = bc[i-1][j-1] + bc[i-1][j];

return (bc[n][m]);

}

• How many ways are there to build a balanced formula from n pairs of parentheses?

• Divide the formula into two parts: (……)(……)

• The first part contains k pairs, the second part has n-1-k pairs.

• So the first part has Ck possibilities, while the second part has Cn-1-k possibilities.

Practice 1:How many pieces of land?

• http://acm.uva.es/p/v102/10213.html

• You are given an elliptical shaped land and you are asked to choose n arbitrary points on its boundary. Then you connect all these points with one another with straight lines (that’s n*(n-1)/2 connections for n points).

• What is the maximum number of pieces of land you will get by choosing the points on the boundary carefully?

• Input

• The first line of the input file contains one integer S (0 < S < 3500), which indicates how many sets of input are there. The next S lines contain S sets of input. Each input contains one integer N (0<=N<2^31).

• Output

• For each set of input you should output in a single line the maximum number pieces of land possible to get for the value of N.

Sample Input:

41234

Sample Output:

1248

• Denote the solution by f(n), so f(1) = 1, f(2) = 2, etc.

• What’s the relationship between f(n) and f(n-1)?

• Assume we have a maximum number of pieces for n-1 points, which is f(n-1).

• Now we add the nth point. We can connect this point to other n-1 points. So we have n-1 new lines.

• Each line will lead to more pieces.

k

i nodes on the left

Consider adding a new line between node n and k.

The new line will intersect with i(n-2-i) number of lines.

So, the new line will be divided into i(n-2-i)+1 segments.

Each segment means a new piece!

n-2-i nodes on the right

n

• http://acm.uva.es/p/v101/10198.html

• Gustavo knows how to count, but he is now learning how write numbers. As he is a very good student, he already learned 1, 2, 3 and 4. But he didn't realize yet that 4 is different than 1, so he thinks that 4 is another way to write 1. Besides that, he is having fun with a little game he created himself: he make numbers (with those four digits) and sum their values. For instance:

• 132 = 1 + 3 + 2 = 6

• 112314 = 1 + 1 + 2 + 3 + 1 + 1 = 9 (remember that Gustavo thinks that 4 = 1)

• After making a lot of numbers in this way, Gustavo now wants to know how much numbers he can create such that their sum is a number n. For instance, for n = 2 he noticed that he can make 5 numbers: 11, 14, 41, 44 and 2 (he knows how to count them up, but he doesn't know how to write five). However, he can't figure it out for n greater than 2. So, he asked you to help him.

• The Input

• Input will consist on an arbitrary number of sets. Each set will consist on an integer n such that 1 <= n <= 1000. You must read until you reach the end of file.

• The Output

• For each number read, you must output another number (on a line alone) stating how much numbers Gustavo can make such that the sum of their digits is equal to the given number.

• Sample Input

2

3

• Sample Output

5

13

• From the problem, we know f(1) = 2, f(2) = 5

• From the sample Input/Output, we know f(3) = 13

• A number can only start from 1, 2, 3, or 4

• If it starts from 1, then the remaining n-1 digits should have a summation of n-1

• If it starts from 2, then the remaining n-1 digits should have a summation of n-2

• If it starts from 3, then the remaining n-1 digits should have a summation of n-3

• If it starts from 4, then the remaining n-1 digits should have a summation of n-1 (because 4 is another way to write 1)

• So, f(n) = 2f(n-1) + f(n-2) + f(n-3)

• For this problem we can build a table for all f(n), 1 <= n <= 1000

• Remember to use big number library!

• Number theory is concerned with properties of integers.

• It is perhaps the most interesting and beautiful area of mathematics.

• Number theory plays an important role in modern cryptography.

• Some outstanding unsolved problems:

• Goldbach’s Conjecture: every even integer n > 2 is the sum of 2 primes.

• Twin Prime Conjecture: There are infinitely many twin primes. (If p and p+2 are primes, we say p and p+2 are twin primes.)

• Are there infinitely many primes of the form n2 + 1? How about the form 2n -1?

• Natural numbers: N = {1, 2, 3,…}

• Integers: Z = {…, -2, -1, 0, 1, 2, … }

• Rational numbers: Q={n/m | m≠0}

• Real numbers: R

• How to convert a decimal number to another base b?

#include <iostream>

#include <vector>

#include <algorithm>

using namespace std;

int main()

{

int d, b, t, r;

vector<int> v;

cin >> d >> b;

while (d > 0) {

r = d % b;

v.push_back(r);

d = d/b;

}

reverse(v.begin(), v.end());

for (int i = 0; i < v.size(); ++i)

cout << v[i];

cout << endl;

return 1;

}

• How to calculate A1A2…An mod p?

• The product may be very large and overflow!

• It can be calculated by:

• (A1 mod p) x (A2 mod p) x … (An mod p)

• Another example: how to calculate bp mod m?

long bigmod(long b, long p, long m) {

if (p == 0)

return 1;

else if (p%2 == 0)

return square(bigmod(b, p/2, m)) % m; // square(x) = x * x

else

return ((b % m) * bigmod(b, p-1, m)) % m;

}

• How to calculate bp?

• Assume no overflow

long square(long n) {

return n*n;

}

long fastexp(long base, long power) {

if (power == 0)

return 1;

else if (power%2 == 0)

return square(fastexp(base, power/2));

else

return base * (fastexp(base, power-1));

}

• Principle of Mathematical Induction:

• Let P(n) be a statement concerning the integer variable n. Let n0 be any fixed integer. P(n) is true for all integers n ≥ n0 if one can establish both of the following statements:

• (a) P(n) is true if n = n0

• (b) Whenever P(n) is true for n0 ≤ n ≤ k, then P(n) is true for n = k+1

• d | n means there is an integer k such that n = d·k.

• d | n has several readings:

• d divides n

• d is a divisor of n

• d is a factor of n

• n is a multiple of d

• n | n, and 1 | n

• d | n and n | m  d | m

• d | n and d | m  d | an + bm for all a and b

• d | n  ad | an

• d | 0

• If d and n are positive and d | n, then d ≤ n.

• If d | n, then n/d | n, and

• small(d, n/d) <= sqrt(n)

• If a and b are integers and b > 0, then there exist unique integers q and r satisfying the two conditions:

• a = bq + r, and

• 0 ≤ r < b

• For b > 0, define a mod b = r where r is the remainder given by the division algorithm when a is divided by b, that is, a = bq + r and 0 ≤ r < b.

• A prime number is an integer >1 and only has divisors of 1 and itself

• It cannot be written as a product of other numbers

• E.g. 2,3,5,7 are prime, 4,6,8,9,10 are not

• prime numbers are central to number theory

• list of prime number less than 200 is:

2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199

• We often need to find large prime numbers

• Traditionally we can usetrial division

• i.e. divide by all numbers (primes) in turn, less than the square root of the number

• only works for small numbers

• Alternatively, we can use statistical primality tests based on properties of primes

• for which all primes numbers satisfy property

• but some composite numbers, called pseudo-primes, also satisfy the property

int is_prime(int n) {

if (n <= 1) return 0;

if (n == 2) return 1; // 2 is a prime

if (n%2 == 0) return 0; // NO prime is EVEN, except 2

for (int i = 3; i * i <= n; i += 2) // start from 3, jump 2 numbers

if (n%i == 0) // no need to check even numbers

return 0;

return 1;

}

• number N is a prime if and only if no primes below sqrt(N) can divide N

• Generate a prime table

• Suppose max primes generated will not be greater than 2^31-1 (2,147,483,647).

• We need smaller primes below sqrt(N), so we need to store primes no greater that sqrt(2^31-1)

• sqrt(2^31) = 46340.95

• There will be at most 4792 primes in the range 1 to 46340.95

• We only need about array of size (roughly) 4800 elements

• check whether a big number N is a prime or not by dividing it with small primes up to sqrt(N)

• If you can find at least one small primes that can divide N, then N is not prime, otherwise N is prime

bool isPrime[MAXN];

int i, tot, prime[];

memset(isPrime, true, sizeof(isPrime));

tot = 0;

prime[tot++] = 2;

for ( i = 3; i < MAXN; i+=2) {

if (isPrime[i])

prime[tot++] = i;

for(j = i + i; j < MAXN; j+=i)

isPrime[j] = false;

}

}

bool isPrime[MAXN];

int i, tot, prime[];

memset(isPrime, true, sizeof(isPrime));

tot = 0;

for ( i = 2; i <= n; i++) {

if (isPrime[i])

prime[tot++] = i;

for (j = 0; j < tot && i * prime[j] <= n; j++) {

isPrime[i*prime[j]] = false;

if (i % prime[j] == 0)

break;

}

}

• There are an infinite number of primes.

• Can you prove it?

• Let π(x) denote the number of primes p such that p ≤ x.

• Prime number theorem states that primes occur roughly every ln(n) integers:

• π(x) ~ x/ln(x) for all x > 0.

• Since we can immediately ignore evens and multiples of 5, in practice we only need to test 0.4ln(n) numbers of size n before finding a prime

Prime Factorization

• to factor a number n is to write it as a product of other numbers: n=a × b × c

• note that factoring a number is relatively hard compared to multiplying the factors together to generate the number

• the prime factorisation of a number n is when its written as a product of primes

• E.g. 91=7×13 ; 3600=24×32×52

Prime Factorization

#include<iostream>

#include<vector>

using namespace std;

int is_prime(int n) { … }

vector<int> primefactors(int a)

{

vector<int> primefactors;

for (int i=1;i<=a;++i)

if(a%i==0&&is_prime(i))

{

primefactors.push_back(i);

a=a/i;

i=1;

}

return primefactors;

}

int main()

{

int b=2500;

if(!is_prime(b)) cout<<"not prime"<<endl;

vector <int> p = primefactors(b);

for(int i=0;i<p.size();++i)

cout<<p[i]<<endl;

return 0;

}

/* another C function for prime factorization */

primefactors(long x)

{

long I;

long c = x;

if ( c == 1) {

printf(“1\n”);

return;

}

while( (c%2) == 0) {

printf(“%ld\n”, 2);

c = c / 2;

}

i = 3;

while ( i <= (sqrt(c) + 1) ) {

if ( (c%i) == 0 ) {

printf(“%ld\n”, i);

c = c / I;

}

else

i = i + 2;

}

if (c > 1) printf(“%ld\n”, c);

}

• If the prime factorisation of a number n is

• Then the number of positive divisors of n is

• http://acm.uva.es/p/v101/10110.html

• The Problem: Light, More Light

• There is man named "mabu" for switching on-off light in our University. He switches on-off the lights in a corridor. Every bulb has its own toggle switch. That is, if it is pressed then the bulb turns on. Another press will turn it off.

• To save power consumption (or may be he is mad or something else) he does a peculiar thing. If in a corridor there is n bulbs, he walks along the corridor back and forth n times and in ith walk he toggles only the switches whose position is divisable by i. He does not press any switch when coming back to his initial position. Theith walk is defined as going down the corridor (while doing the peculiar thing) and coming back again. Now you have to determine what is the final state of the last bulb. Is it on or off?

1 2 3 4 …. n-1 n

• The question asks about the final state of the last bulb

• There are n rounds

• In the 1st round, because 1|n, press

• In the 2nd round, if 2|n, press;

• In the 3rd round, if 3|n, press;

• In the nth round, because n|n, press

• A better algorithm:

long i;

long press = 0;

for(i = 1; i <= n/2; i++) {

if(n % i == 0)

press++;

}

if( press % 2 == 0)

printf(“yes\n”);

else

printf(“no\n”);

• A simple algorithm:

long i;

long press = 0;

for(i = 1; i <= n; i++) {

if(n % i == 0)

press++;

}

if( press % 2 == 0)

printf(“no\n”);

else

printf(“yes\n”);

• But the previous algorithm is O(n)!

• Is there a better way?

• We can make some observations:

• Define the number of factors of n as f(n). If f(n) is even, then the final state of the nth bulb is “off”; if f(n) is odd, then the final state of the nth bulb is “on”.

• f(1) = 1; f(2) = 2; f(3) = 2; f(4) = 3;

• f(5) = 2; f(6) = 4; f(7) = 2; f(8) = 4;

• f(9) = 3; f(10) = 4; f(11) = 2; f(12) = 6;

• f(13) = 2; f(14) = 4; f(15) = 4; f(16) = 5

• It seems that, only if n is a square of another number, f(n) is odd !

• How to prove?

• Prime factorization:

• two numbers a, b are relatively prime if they have no common divisors apart from 1

• E.g. 8 & 15 are relatively prime since factors of 8 are 1,2,4,8 and of 15 are 1,3,5,15 and 1 is the only common factor

• We can determine the greatest common divisor by comparing their prime factorizations and using least powers

• E.g. 300=21×31×52 18=21×32hence GCD(18,300)=21×31×50=6

• Two observations:

• If b|a, then gcd(a,b) = b

• If a = bt + r, then gcd(a,b) = gcd(b, r)

• Euclid’s algorithm is recursive, repeated replacing the bigger integer by its remainder mod the smaller integer.

• E.g.: a = 34398, b = 2132

gcd(34398, 2132) = gcd(34398 mod 2132, 2132) = gcd(2132, 286)

= gcd(2132 mod 286, 286) = gcd(286, 130)

= gcd(286 mod 130, 130) = gcd(130, 26)

= gcd(130 mod 26, 26) = gcd(26, 0)

= 26

//Greatest Common Divisor

int gcd(int a, int b) {

while (b > 0) {

a = a % b;

a ^= b;

b ^= a;

a ^= b;

}

return a;

}

// Lowest Common Multiple

int lcm(int a, int b)

{

return a*b/gcd(a, b);

}

int t = b;

b = a;

a = t;

• gcd(a,b) can be expressed as ax+by

• Given a and b, find out not only gcd(a,b), but also x and y

• The previous function only gives gcd(a,b)

int ext_gcd(int a, int b, int& x, int& y) {

int t, ret;

if (!b){

x=1; y=0;

return a;

}

ret=ext_gcd(b, a%b, x, y);

t=x; x=y; y=t-a/b*y;

return ret;

}

• We say that a≡b (mod m) if m|(a-b).

• Congruences are an alternate notation for modular arithmetic.

• Addition: Suppose a≡b (mod n) and c≡d (mod n), then a+c≡b+d (mod n)

• Multiplication:

• If a≡b (mod n), then a·d≡b·d (mod n).

• If a≡b (mod n) and c≡d (mod n), then ac≡bd (mod n)

• Exponentiation: If a≡b (mod n), then am≡bm (mod n)

• Division??? 6·2≡6·1 (mod 3), but …

• Theorem 1: Let m ≥ 2. If a and m are relatively prime, there exists a unique integer a* such that aa* ≡ 1 (mod m) and 0 < a* < m.

• We call a* the inverse of a modulo m.

• Theorem 2: Let m > 0. If ab ≡ 1 (mod m), then both a and b are relatively prime to m.

• Theorem 3: Let m > 0 and assume gcd(c,m)=1. Then ca ≡ cb (mod m)  a ≡ b (mod m).

Fermat's Little Theorem

• If p is prime and gcd(a, p) = 1, then

• ap-1 mod p = 1

• E.g., how to calculate 12347865435 mod 11?

• 1234 ≡ 2 (mod 11)  12347865435≡ 2 7865435 (mod 11)

• 210≡ 1 (mod 11) (according to Fermat’s litter theorem)

• 2 7865435 ≡ 2786543x10+5 (mod 11)

≡ (2^10)786543 x25 (mod 11)

≡ 1786543x25 (mod 11)

≡ 25 (mod 11)

≡ 10 (mod 11)

• So 12347865435 mod 11 = 10.

• when doing arithmetic modulo n

• complete set of residues is: 0..n-1

• reduced set of residues is those numbers (residues) which are relatively prime to n

• E.g. for n=10,

• complete set of residues is {0,1,2,3,4,5,6,7,8,9}

• reduced set of residues is {1,3,7,9}

• number of elements in reduced set of residues is called the Euler Totient Function ø(n)

• To compute ø(n), we need to count number of elements to be excluded

• in general, it needs prime factorization, but

• for p (p prime) ø(p) = p-1

• for p.q (p,q prime) ø(p·q) = (p-1)(q-1)

• E.g.

• ø(37) = 36

• ø(21) = (3–1)×(7–1) = 2×6 = 12

• aø(n)mod n = 1

• where gcd(a, n)=1

• It is a generalisation of Fermat's Theorem

• E.g.

• a=3;n=10; ø(10)=4;

• hence 34 = 81 = 1 mod 10

• a=2;n=11; ø(11)=10;

• hence 210 = 1024 = 1 mod 11

• http://acm.uva.es/p/v101/10104.html

• http://acm.uva.es/p/v101/10139.html

• http://acm.uva.es/p/v101/10168.html

• http://acm.uva.es/p/v100/10042.html