Discrete Mathematics

Discrete Mathematics Chapter 3 The Fundamentals : Algorithms, the Integers, and Matrices 大葉大學資訊工程系黃鈴玲(Lingling Huang)

3.1 Algorithms Def 1. An algorithm is a finite sequence of precise instructions for performing a computation or for solving a problem. • Example 1. Describe an algorithm for finding the maximum value in a finite sequence of integers.(假設給定的sequence是a1,a2,…,an)

Solution : ( English language) • Set the temporary maximum equal to the first integer in the sequence. • Compare the next integer in the sequence to the temporary maximum, and if it is larger than the temporary maximum, set the temporary maximum equal to this integer. • Repeat the previous step if there are more integers in the sequence. • Stop when there are no integers left in the sequence. The temporary maximum at this point is the largest integer in the sequence.

Algorithm 1. Finding the Maximum Element procedure max(a1, a2, …, an : integers) max:=a1 for i:=2 to n if max<ai then max:= ai { maxis the largest element} Solution : (pseudo-code)

※ There are several properties that algorithms generally share : • Input • Output • Definiteness : The steps of an algorithm must be defined precisely. • Correctness : produce correct output values • Finiteness : produce the desired output after a finite number of step. • Effectiveness • Generality : The procedure should be applicable for all problems of the desired form, not just for a particular set of input values.

※ Searching Algorithms Problem : Locate an element x in a list of distinct elements a1,a2,…,an, or determine that it is not in the list. 做法 : linear search, binary search. Algorithm 2. The linear search algorithm procedure linear_search( x : integer, a1,a2,…,an: distinct integers) i := 1 While ( i ≤ n and x≠ai ) i := i + 1 if i≤n then location := i else location := 0 { location = jif x = aj; location = 0 if x≠ai, ∀i }

兩種search方式的概念 : Linear Search :從 a1開始，逐一比對 x 是否等於 ai，若找到則 location = i , 若到 an比完後還找不到，則 location = 0。 Binary Search :(必須具備 a1 < a2 < … < an的性質才能用)(1) 每次將 list 切成兩半，( ai , … , am )，(am+1 , … , aj )若 x > am表示 x應在右半，否則在左半。 (2) 重覆上一步驟至 list 只剩一個元素 ai，若 x = ai則 location = i，否則 location = 0。

Note : ai, ai+1, …, aj數列的切法 : 令 m= 則 am即切開紅線左邊那點。 Example 3. Search 19 from a1a2a3a4a5a6a7a8a9a10a11a12a13a14a15a16 1 2 3 5 6 7 8 10 12 13 15 16 18 19 20 22 12 13 15 16 18 19 20 22 18 19 20 22 18 19 19 1. ( 切兩半 ) ( 因 19 > 10，取右半 ) 2. ( 再切二半 ) ( 因 19 > 16，取右半 ) 3. ( 再切二半 ) ( 因 19 ≦19，取左半 ) 4. ( 再切二半 ) ( 因 19 > 18，取右半 ) 5 此時只剩一個元素 a14 = 19 因 19 = 19，故 location =14

Algorithm 3. The Binary Search Algorithm procedure binary_search( x : integer, a1,a2,…,an : increasing integers) i:=1 { i is left endpoint of search interval } j := n { j is right endpoint of search interval } while i < j begin m:= if x > am then i := m+1 else j := m end if x = ai then location := i else location :=0 { location = i if x = ai , location = 0 if x≠ai , ∀i }

※ Sorting Algorithms • Problem : Suppose that we have a list of elements, a sorting is putting these elements into a list in which the elements are in increasing order. • eg. 7, 2, 1, 4, 5, 9 => 1, 2, 4, 5, 7, 9 d, t, c, a, f => a, c, d, f, t • 解法有很多，此處僅介紹 : bubble sort (氣泡排序法)，及 insertion sort (插入排序法)。 • Bubble Sort 概念 : 設原 list 為 a1,…,an。 • 從a1,a2開始，向後兩兩比較，若ai > ai+1則交換，當檢查完 an時，an必定是最大數。 • 再從 a1,a2開始向後比較，若ai > ai+1則交換，此時只需檢查到 an-1 即可。 • 依此類推。

2 3 1 4 5 2 3 1 4 5 2 3 1 4 5 2 1 3 4 5 2 1 3 4 5 2 1 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 • Example 4. Use the bubble sort to put 3, 2, 4, 1, 5 into increasing order. • Sol : First pass (i=1) : Second pass (i=2) : 3 2 4 1 5 2 3 4 1 5 2 3 4 1 5 2 3 1 4 5 Third pass (i=3) : Fourth pass (i=4) : 1 2 3 4 5

Algorithm 4 The Bubble Sortprocedure bubble_sort (a1,…,an ) fori := 1 ton-1 forj := 1 ton-i ifaj > aj+1then interchange aj and aj+1 { a1,a2,…,an is in increasing order }

Insertion Sort 的概念 : • 從 j= 2開始，將 aj插入已排序好的 a1,…,aj-1間的位置，使得 a1,…,aj都由小 → 大排好。 • j 逐次遞增，重複上一步驟至做完。

a1a2a3a4a5 • Example 5. Use insertion sort to sort 3, 2, 4, 1, 5 • Sol : • (j=2時，a1=3可看成已經排序好的數列，此時要插入a2) : 3 < 2  2, 3 交換 2, 3, 4, 1, 5 • (j=3時，a1,a2已經排序好，此時要插入a3) : 4 > 2, 4 > 3  4的位置不變 2, 3, 4, 1, 5 • (j=4時，a1,a2 ,a3已經排序好，此時要插入a4) : 1 < 2 將 1 插在最前面 1, 2, 3, 4, 5 • (j=5時，a1,a2 ,a3 ,a4已經排序好，此時要插入a5) : 5 > 1, 5 > 2, 5 > 3, 5 > 4  5不變 1, 2, 3, 4, 5

Algorithm 5 The Insertion Sortprocedure insertion_sort ( a1,…,an : real numbers with n≥ 2 ) forj := 2 ton begin i := 1 whileaj > ai i := i + 1 m := aj fork := 0 toj – i – 1 aj-k := aj-k-1 ai := m end { a1,a2,…,an are sorted } 找出 aj應插入的位置最後ai-1 < aj <= ai 將 ai, ai+1, …, aj-1全部往右移一格 ( Exercise : 3, 9, 13, 23, 35, 39 )

3.2 The Growth of Functions • To analyze the practicality of the program, we need to understand how quickly the function (number of operations used by this algorithm) grows asn(number of input elements) grows. • eg. sort n objects • Alg. 1 : n2次計算 • Alg. 2 : 8n次計算 better!

Def 1. ( Big-O notation ) Let f and g be functions from the set of integers to the set of real numbers. We say that f (x) is O(g(x)) if there are constants C and k such that | f (x) | ≤ C | g(x) | whenever x > k . ( read as “f (x) is big-oh of g(x)” )

Example 1.Show that f (x) =x2+2x+1 is O(x2) Sol :Since x2+2x+1 ≤ x2+2x2+x2 = 4x2 whenever x> 1 , it follows that f (x) is O(x2) (take C = 4 and k =1 ) 另法： If x > 2, we see that x2+2x+1 ≤ x2+x2+x2 = 3x2 ( take C = 3 and k = 2 )

Cg(x) f (x) g(x) f (x) < Cg(x) for x > k k Figure 2. The function f (x) is O(g(x)) Example 1(補充).Show thatf (n)= n2+2n +2isO(n3) Sol :Since n2+2n+2 ≤ n3+n3+n3 = 3n3 whenever n> 1, we see that f (n) is O(n3) ( take C = 3 and k = 1 ) Note. The function g is chosen to be as small as possible.

Theorem 1. Let f (x) = anxn+an-1xn-1+…+a1x+a0 where a0, a1, …, an are real numbers. Then f (x) is O(xn). Example 5. How can big-O notation be used to estimate the sum of the first n positive integers? ( i.e.,  ) Sol : 1 + 2 + 3 + … + n≤ n + n + … + n = n2 ∴ is O(n2), taking C =1 and k =1.

Example 6. Give big-O estimates for f (n) = n! Sol : n! = 12  3  … n≤ nn… n = nn ∴ n! is O(nn) , taking C =1 and k =1. Example 7. (see Figure 3) 常見function的成長速度由小至大排列：1 < log n < n < n log n < n2 < 2n < n! Theorem 2,3 Suppose that f1(x) is O(g1(x)) and f2(x) is O(g2(x)), then (f1+f2)(x) is O(max(|g1(x)|, |g2(x)|)), (f1 f2)(x) is O(g1(x) g2(x)).

Exercise 7,11,19 Exercise 19(c) :f (n) = (n!+2n)(n3+log(n2+1))  (n!+n!)(n3+n3) = 4n3n! ∴ f (n) is O(n3n!)取 C = 4, k = 3

3.3 Complexity of Algorithms Q : How can the efficiency of an algorithm be analyzed ? Ans : (1) time (2) memory Def : • Time complexity : an analysis of the time required to solve a problem of a particular size. (評量方式 : 計算 # of operations，如 “comparison”次數，“加法” 或 “乘法” 次數等) • Space complexity : an analysis of the computer memory required to solve a problem of a particular size. (通常是資料結構探討的範圍)

Sol : (計算 # of comparisons) i值一開始 = 2 逐次加一，並比較是否>n. 當 i變成 n+1時因比 n大，故結束for迴圈。 ∴ 共有 n次 comparison 共有 n-1次 comparison 故整個演算法共做 2n-1次 comparison 其 time complexity 為 O(n). Example 1. Describe the time complexity of Algorithm 1. Algorithm 1. ( Find Max ) procedure max(a1,…,an: integers) max:= a1 for i:= 2 ton if max < ai then max := ai { max is the largest element }

Algorithm 2 ( Linear Search ) procedure ls ( x: integer , a1,…,an : distinct integers ) i := 1 While ( in and x ≠ai ) i := i +1 if i n then location := i else location := 0 location = ix = ai = 0 x ai i Example 2. Describe the time complexity of the linear search algorithm. Sol : ( 計算 # of comparisons ) (Case 1) 當 x = ai for some i  n時此行只執行 i次，故此行共2i次比較加上if，共計 2i +1次 comparisons. (Case 2) 當 x ≠ ai for all i時此行執行 n 次後第 n + 1次時i = n + 1 > n即跳出 ∴共計 2n+2 次 comparisons 由(1)、(2)取 worst-case演算法的 time complexity為 O(n)

Sol : ( 計算 “平均比較次數” ) 已知當 x = ai時，共需 2i + 1次比較. ( by Example 2 ) x = a1,a2, …,或 an的機率都是 1/n. ∴平均比較次數 (即期望值) = ( x = a1的比較次數 ) × ( x =a1的機率 ) + ( x = a2的比較次數 ) × ( x = a2的機率 ) + … + ( x = an的比較次數 ) × ( x = an的機率 ) = 3 × 1/n + 5 × 1/n + … + ( 2n+1) × 1/n = ( 3+5+…+(2n+1)) / n = / n = n + 2 Alg. 2 ( Linear Search ) procedure ls ( x,a1,…,an) i := 1 While ( in and x ≠ai ) i := i +1 if i n then location := i else location := 0 Example 4. Describe the average-case performance of the linear search algorithm, assuming that x is in the list. ∴average-case的time complexity為O(n)

Alg. 3 ( Binary Search ) procedurebs ( x : integer, a1,…,an : increasing integers ) i := 1 { left endpoint } j := n { right endpoint } whilei < j/* ( k+1 次) begin m :=  ( i + j ) / 2  ifx > am theni := m+1 /* ( k次) elsej := m end ifx = aithenlocation := i /* ( 1次) elselocation := 0 Example 3. Describe the time complexity of the binary search algorithm. Sol : 設 n = 2k以簡化計算 (若 n < 2k，其比較次數必小於等於 n = 2k的情況) 因while迴圈每次執行後整個 list 會切成兩半故最多只能切 k次就會因 i = j而跳出迴圈 ∴共比較 2k+2次 time complexity 為 O(k) = O(log n)

Sol : 共 n-1個 pass 第 i 個 pass 需 n – i次比較 ∴共計 (n-1)+(n-2)+…+1 = 次比較 ∴ O(n2) Example 5. What is the worst-case complexity of the bubble sort in terms of the number of comparisons made ? procedure bubble_sort( a1,…,an) for i := 1 to n -1 for j:= 1 to n – i if aj> aj+1 then interchange aj and ai+1 { a1,…,an is in increasing order } Note 1. 不管何種 case 都需做次比較。 Note 2. For 迴圈所需比較次數通常會省略，因此Example 5,6 不再考慮。

procedure insertion_sort ( a1,…,an ) for j := 2 to n begin i := 1 while aj > ai i := i +1 m := aj for k := 0 to j -i -1 aj-k := aj-k-1 ai := m end { a1,…,an are sorted } Sol : 做最多次比較的情況如下：在考慮 aj時 a1 < a2 < … < aj-1 < aj 此時共做 j 次比較故共計 2+3+…+n = -1 次比較 O(n2) (即 worst case 是 a1 < a2 < … < an) • Example 6. What is the worst-case complexity of the insertion sort in terms of the number of comparisons made ?

Table 1. Commonly Used Terminology Exercise : 7,8,13

3.4 The integers and division ※探討一些 Number Theory 的基本觀念 Def 1.a,b : integers, a≠0. a divides b (denote a | b) if cZ , b=ac . (a : a factor of b, b : a multiple of a) (ab if a does not divide b) Corollary 1. If a,b,cZ and a | b , a | c. then a | mb+nc whenever m,nZ Def 2. In the quality a = dq + rwith0r <d,d is called the divisor (除數), a is called the dividend (被除數), q is called the quotient (商數), and ris called the remainder (餘數). q = a div d, r = a mod d

Def 3. If a,bZ, mZ+, then a is congruent (同餘) to b modulo m if m | (a-b). (denote a≡b(mod m)). Thm 4. Let mZ+, a,bZ. a≡b (mod m) iff kZ, s.t. a=b+km. Thm 5. Let mZ+, a,bZ. If a≡b (mod m) andc≡d (mod m), then a+c≡b+d (mod m) and ac≡bd (mod m).

3.5 Primes and Greatest Common Divisors Def 1.pZ+ -{1} is called prime(質數)if ap, 1<a< p, aZ+. p is called composite (合成數) otherwise. Thm 1. (The fundamental theorem of arithmetic) Every positive integer greater than 1 can be written uniquely as a prime or as the product of two or more primes where the prime factors are written in order of nondecreasing size. Example 2. The prime factorization (因數分解) of 100 is 2252.

Thm 2. If n is a composite integer, then n has a prime divisor less than or equal to . Thm 3. There are infinitely many primes.Pf. 假設質數只有n個：p1, p2, …, 及 pn，令Q = p1p2…pn+1, 因 p1, …, pn都不整除Q，得證。 ※目前為止所知最大的質數是2p-1的形式, where p is prime. 稱為 Mersenne primes (梅森質數). Example 5.22-1=3, 23-1=7, 25-1=31 are primes, but 211-1=2047=2389 is not a prime. Def 2. gcd ( greatest common divisor ) Def 3. relatively prime (互質) Def 5. lcm ( least common multiple )

Exercise 14. How many zeros are there at the end of 100!? Sol :計算12  3  …  100=10k m, where 10 m ∵ 10=2  5，又2的次數必定比5多 ∴ 計算1  2  3  …  100=5k n, where 5 n ∵ 5,10,15,20,…,100才有因數5,而25,50,75,100有因數25 ∴ k=24  共有24個0 Homework : 試寫一alg. 求出  n 的所有質數

3.6 Integers and Algorithms ※The Euclidean Algorithm (輾轉相除法求 gcd ) Example : Find gcd(91,287) Sol: 287 = 91 3 + 14 91 = 14  6 + 7 14 = 7  2 ∴ gcd(91,287) = 7 Lemma 1 Let a = bq + r, where a, b, q, and rZ.Then gcd(a, b) = gcd (b, r). if x|91 & x|287  x|14 ∴gcd (91,287) = gcd(91,14) gcd (91,14) = gcd (14,7) gcd (14,7) = 7

eg. 求 gcd (6,12) x = 6 y = 12 while y≠0 r = 6 mod 12 =6 x = 12 y = 6 while y≠0 r = 12 mod 6 = 0 x = 6 y = 0 while y = 0 , end. ∴ gcd (6,12) = 6 Algorithm 6. ( The Euclidean Algorithm) procedure gcd ( a, b : positive integers) x := a y := b while y≠0 begin r := x mod y( if y > x then r = x) x:= y y := r end { gcd (a, b) = x } Exercise : 23,25

3.7 Applications of Number Theory gcd(a,b)can be expressed as a linear combination with integer coefficients of a and b. Theorem 1. If a and b are positive integers, then there exist integers s and t such that gcd(a,b) = sa+tb. ※將gcd(a,b)寫成a跟b的線性組合：The extended Euclidean Algorithm

Example 1 Express gcd(252, 198) =18 as a linear combination of 252 and 198. Sol: 252 = 1 198 + 54 198 = 3  54 + 36 54 = 1  36 + 18 36 = 2  18 ∴ gcd(252, 198) = 18 18 = 54 – 1  36 36 =198 – 3  54 54 =252 – 1  198  18 = 54 – 1  36 = 54– 1  (198 – 3  54 ) = 4  54 – 1  198 = 4  (252 – 1  198)– 1  198 = 4  252 – 5  198 Exercise : 1(g)

Lemma 1. If a, b and c are positive integers such that gcd(a,b) = 1and a | bc, then a | c. Lemma 2. If p is a prime and p | a1a2…an, where each ai is an integer, then p | aifor some i. Example 2 14  8 (mod 6), 但的左右兩邊同除以2後不成立 because 14/2=7, 8/2=4, but 7  4(mod 6). 另, 14  8 (mod 3), 同除以2後, 7  4 (mod 3)成立 Q: 何時可以讓  的左右同除以一數後還成立呢？

Theorem 2. Let m be a positive integer and let a, b, and c be integers. If ac  bc(mod m) and gcd(c, m) = 1, then a  b (mod m). ※ Linear Congruences A congruence (同餘式) of the form ax  b (mod m), wherem is a positive integer, a and b are integers, and x is a variable, is called a linear congruence. How can we solve the linear congruence ax  b (mod m)? Def: If ax  1(mod m), and let a be an answer of x,a is called an inverse (反元素) of a modulo m

Theorem 3. If a and m are relatively prime integers and m>1, then an inverse of a modulo m exists. Furthermore, this inverse is unique modulo m. Proof. (existence) (unique的部分是exercise) By Thm 1, because gcd(a, m) = 1, there exist integers s and t such that sa + tm =1.  sa + tm  1 (mod m). Because tm  0 (mod m), sa  1 (mod m),  s is an inverse of a modulo m.

Example 3 Find an inverse of 3 modulo 7. Sol. Because gcd(3, 7) = 1, find s, t such that 3s + 7t =1. 7 = 2  3 + 1 1 = -2  3 + 1  7 -2 is an inverse of 3 modulo 7. (Note that every integer congruent to-2 modulo 7 is also an inverse of 3, such as 5, -9, 12, and so on. ) Exercise : 5

Example 4 What are the solutions of the linear congruence 3x  4 (mod 7)? Sol. By Example 3  -2 is an inverse of 3 modulo 7  3  (-2) 1 (mod 7) 將 3x  4 (mod 7) 左右同乘-2 -2  3x  -2  4 (mod 7) -6x  -8 (mod 7) Because-6  1 (mod 7), and-8  6 (mod 7), If x is a solution, then x  -8  6 (mod 7). We need to determine whether every x with x  6 (mod 7)is a solution. Assume x  6 (mod 7), then3x  3  6 = 18  4 (mod 7). Therefore every such x is a solution: x = 6, 13, 20, …, and -1, -8, -15, …. Exercise : 11

The Chinese Remainder Theorem (中國餘數定理) Example 5.孫子算經 :「某物不知其數，三三數之餘二，五五數之餘三，七七數之餘二，問物幾何 ?」 (又稱為「韓信點兵」問題) i.e. x ≡ 2 (mod 3) x ≡ 3 (mod 5) x = ? x ≡ 2 (mod 7) Theorem 4.(The Chinese Remainder Theorem) Let m1,m2,…,mn be pairwise relatively prime positive integers and a1, a2, …, anarbitrary integers. Then the system x ≡ a1 (mod m1)x ≡ a2 (mod m2) :x ≡ an (mod mn) has a unique solution modulo m = m1m2…mn.(即有一解x, where 0 x < m , 且所有其他解 mod m都等於 x)

x ≡ a1 (mod m1)x ≡ a2 (mod m2) :x ≡ an (mod mn) m = m1m2…mn Proof of Thm 4: Let Mk = m / mk 1 k  n ∵ m1, m2,…, mn are pairwise relatively prime ∴ gcd (Mk, mk) = 1  integer yk s.t. Mk yk ≡ 1 (mod mk) ( by Thm. 3)  ak Mkyk ≡ ak (mod mk) ,  1  k  n Letx = a1 M1 y1+a2 M2 y2+…+an Mn yn ∵ mi | Mj , i ≠ j ∴ x ≡ ak Mkyk ≡ ak (mod mk)  1 k  n xis a solution. All other solution y satisfies y≡ x (mod mk).

M2 y2 M3 y3 x ≡ 2 (mod 3) x ≡ 3 (mod 5) x = ? x ≡ 2 (mod 7) Example 6. (Solve the system in Example 5) Let m = m1m2m3 = 357 = 105 M1 = m / m1 = 105 / 3 = 35 (也就是m2m3) M2 = m / m2 = 105 / 5 = 21 M3 = m / m3 = 105 / 7 = 15 35 ≡ 2 (mod 3)  35  2 ≡ 2  2 ≡ 1 (mod 3) 21 ≡ 1 (mod 5)  21  1 ≡ 1 (mod 5) 15 ≡ 1 (mod 7)  15  1 ≡ 1 (mod 7) ∴ x = a1M1y1 + a2M2y2 + a3M3y3 = 2  35  2 + 3  21  1 + 2  15  1 = 233 ≡ 23 (mod 105) ∴ 最小的解為23，其餘解都等於23+105t for some tZ+ 找 y1 使得M1y1 = 1 (mod 3) M1 y1

Exercise 18. Find all solutions to the system of congruences x ≡ 2 (mod 3) x ≡ 1 (mod 4) x ≡ 3 (mod 5) Sol : a1=2 , a2=1 , a3=3, m1=3 , m2=4 , m3=5 m=345=60 M1=20 , M2=15 , M3=12 20≡2 (mod 3)  202≡1 (mod 3) 15≡3 (mod 4)  153≡1 (mod 4) 12≡2 (mod 5)  123≡1 (mod 5) ∴ x = 2202+1153+3123 = 80+45+108=233≡53 (mod 60)

※ 補充：(when mi is not prime) Ex 20. Find all solutions, if any, to the system of congruences. x≡5 (mod 6) x≡3 (mod 10) x≡8 (mod 15) Sol. Rewrite the system as the following: x ≡ 1 (mod 2) x≡2 (mod 3) x ≡ 1 (mod 2) x≡3 (mod 5) x ≡ 2 (mod 3) x≡3 (mod 5) i.e., x≡1 (mod 2) x≡2 (mod 3) … x≡3 (mod 5) Exercise : 做完此題

※ 補充：(when mi is a prime power) Ex 21. Find all solutions, if any, to the system of congruences. x≡7 (mod 9) x≡4 (mod 12) x≡16 (mod 21) Sol. Rewrite the system as the following: x≡7 (mod 9) (不能拆！) x≡0 (mod 4) x≡1 (mod 3) x≡2 (mod 7) x≡1 (mod 3) i.e., x≡7 (mod 9) (此式取代 x≡1 (mod 3) 式子) x≡0 (mod 4) … x≡2 (mod 7)

Discrete Mathematics