이산수학(Discrete Mathematics)
귀납과 재귀
(Induction and Recursion)
2016년 봄학기
강원대학교 컴퓨터과학전공 문양세
강의 내용
Induction and Recursion
수학적 귀납법(Mathematical Inductions)
재귀(Recursion)
재귀 알고리즘(Recursive Algorithms)
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Discrete Mathematics
by Yang-Sae Moon
Introduction
Mathematical Induction
A powerful technique for proving that a predicate P(n) is
true for every natural number n, no matter how large.
(모든 자연수 n에 대해서 P(n)이 true임을 보일 수 있는 매우 유용한 방법임)
Essentially a “domino effect” principle.
(앞에 것이 성립하면(넘어지면), 바로 다음 것도 성립한다(넘어진다.)
Based on a predicate-logic inference rule:
P(0)
n0 (P(n)P(n+1))
n0 P(n)
P(0)가 true이고, P(n)이 true라 가정했을 때
P(n+1)이 true이면, 모든 n에 대해 P(n)이 true이다.
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Discrete Mathematics
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Validity of Induction
Mathematical Induction
Intuitively, we can prove that induction is correct.
P(1) = T since (P(0) = T)  (P(0)P(1) = T)
P(2) = T since (P(1) = T)  (P(1)P(2) = T)
P(3) = T since (P(2) = T)  (P(2)P(3) = T)
…
P(n) = T since (P(n-1) = T)  (P(n-1)P(n) = T)
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Discrete Mathematics
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Outline of an Inductive Proof
Mathematical Induction
Want to prove n P(n)…
• Induction basis (or base case): Prove P(0).
(기본 단계)
• Induction step: Prove n P(n)P(n+1).
(귀납적 단계)
- E.g. use a direct proof:
- Let nN, assume P(n). (induction hypothesis)
- Under this assumption, prove P(n+1).
• Inductive inference rule then gives n P(n).
Can also be used to prove nc P(n) for a given constant
cZ, where maybe c0. (Base로 0이 아닌 상수 c를 사용할 수 있다.)
• In this circumstance, the base case is to prove P(c) rather than P(0),
and the inductive step is to prove nc (P(n)P(n+1)).
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Discrete Mathematics
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Induction Examples (1/4)
Mathematical Induction
Prove that the sum of the first n odd positive integers is n2.
That is, prove: (처음 n개 홀수의 합은 n2와 동일하다.)
n
n  1 :  ( 2i  1)  n 2
i 1
P(n)
Proof by Induction
• Induction basis: Let n=1. Since 1 = 12, P(1) is true.
• Induction step: Let n1, assume P(n), and prove P(n+1).
 n

( 2i  1)    ( 2i  1)   ( 2( n  1)  1)

i 1
 i 1

By induction
 n 2  2n  1
hypothesis P(n)
2
 ( n  1)
n 1
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Discrete Mathematics
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Induction Examples (2/4)
Mathematical Induction
Prove that n>0, n<2n. Let P(n)=(n<2n) :
• Induction basis: P(1) = (1 < 21) = (1 < 2) = T.
• Induction step: For n>0, prove P(n)P(n+1).
- Assuming n<2n, prove n+1 < 2n+1.
- Note n + 1 < 2n + 1 (by inductive hypothesis)
< 2n + 2n (because 1 < 2=2∙20  2∙2n-1= 2n)
= 2n+1
- So n + 1 < 2n+1, and we’re done.
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Discrete Mathematics
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Induction Examples (3/4)
Mathematical Induction
Prove that the sum of the first n positive integers is
n(n+1)/2.
n( n  1) 
 n
i


Let P(n) =  
2
 i 1

1(1  1)  
2
 1
  1    T
• Induction basis: Let P(1) =   i 
2  
2
 i 1
• Induction step: Let n1, assume P(n), and prove P(n+1).
n( n  1)
n( n  1)  2n  2
i   i  ( n  1) 
 ( n  1) 

2
2
i 1
i 1
n 1
n
( n 2  3n  2) ( n  1)( n  2)


2
2
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Discrete Mathematics
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Induction Examples (4/4)
Mathematical Induction
Aj   Aj when n2. A1  A2  ...  An  A1  A2  ...  An 
Prove 
j 1
n
n
j 1

• Induction basis: n = 2, A1  A2  A1  A2

• Induction step: Assume P(n), and prove P(n+1).
n 1
n
j 1
j 1
 Aj   Aj  An 1
n
  Ak  An 1
j 1
n
  A j  An 1 (by induction hypothesis)
j 1
n 1
  Aj
j 1
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Discrete Mathematics
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Second Principle of Induction (강 귀납법)
Mathematical Induction
Characterized by another inference rule:
P(0)
n0: (0kn P(k))  P(n+1)
n0: P(n)
P(0)가 true이고, P(0),…,P(n)이 모두 true라 가정했을 때
P(n+1)이 true이면, 모든 n에 대해 P(n)이 true이다.
Difference with 1st principle is that the inductive step
uses the fact that P(k) is true for all smaller k<n+1, not
just for k=n.
(Induction step에서 k=n인 경우 대신에, k<n+1인 모든 k에 대해 P(k)가 true라 가정한다.)
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Discrete Mathematics
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Example of Second Principle
Mathematical Induction
Show that every n>1 can be written as a product p1p2…ps
of some series of s prime numbers. Let P(n)=“n has that
property”
(모든 양의 정수는 소수의 곱으로 나타낼 수 있다.)
• Induction basis: n=2, let s=1, p1=2. Thus, P(2) = T.
• Induction step: Let n2. Assume 2kn: P(k).
- Consider n+1. If prime, let s=1, p1=n+1. Done.
- Else n+1=ab, where 1<an and 1<bn.
- Then a=p1p2…pt and b=q1q2…qu.
- Then n+1= p1p2…pt q1q2…qu, a product of s=t+u primes.
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Discrete Mathematics
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강의 내용
Induction and Recursion
수학적 귀납법(Mathematical Inductions)
재귀(Recursion)
재귀 알고리즘(Recursive Algorithms)
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Discrete Mathematics
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Recursive Definitions (재귀의 정의)
Recursion
In induction, we prove all members of an infinite set have some
property P by proving the truth for larger members in terms of that of
smaller members.
(귀납적 정의에서는, “무한 집합의 모든 멤버가 어떠한 성질 P를 가짐”을 보이기
위하여, “작은 멤버들을 사용하여 큰 멤버들이 참(P의 성질을 가짐)임”을
증명하는 방법을 사용하였다.)
In recursive definitions, we similarly define a function, a predicate or
a set over an infinite number of elements by defining the function or
predicate value or set-membership of larger elements in terms of that
of smaller ones.
(재귀적 정의에서는, “작은 멤버들에 함수/술어/집합을 적용(정의)하여 모든 멤
버들을 정의”하는 방법을 사용한다.)
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Discrete Mathematics
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Recursion (재귀)
Recursion
Recursion is a general term for the practice of defining an object in
terms of itself (or of part of itself).
(재귀란 객체를 정의하는데 있어서 해당 객체 자신을 사용하는 것을 의미한다.)
An inductive proof establishes the truth of P(n+1) recursively in terms
of P(n).
(귀납적 증명은 P(n+1)이 참임을 증명하기 위하여 재귀적으로 P(n)을 사용하는
것으로 해석할 수 있다.)
There are also recursive algorithms, definitions, functions, sequences,
and sets.
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Discrete Mathematics
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Recursively Defined Functions
Recursion
Simplest case: One way to define a function f:NS (for
any set S) or series an=f(n) is to:
• Define f(0).
• For n>0, define f(n) in terms of f(0),…,f(n−1).
E.g.: Define the series an :≡ 2n recursively:
• Let a0 :≡ 1.
• For n>0, let an :≡ 2an-1.
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Another Example
Recursion
Suppose we define f(n) for all nN recursively by:
• Let f(0)=3
• For all nN, let f(n+1)=2f(n)+3
What are the values of the following?
• f(1)= 9
f(2)= 21
f(3)= 45 f(4)= 93
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Discrete Mathematics
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Recursive Definition of Factorial
Recursion
Given an inductive definition of the factorial function
F(n) :≡ n! :≡ 23…n.
• Base case: F(0) :≡ 1
• Recursive part: F(n) :≡ n  F(n-1).
− F(1)=1
− F(2)=2
− F(3)=6
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Discrete Mathematics
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The Fibonacci Series
Recursion
The Fibonacci series fn≥0 is a famous series defined
by:
f0 :≡ 0,
f1 :≡ 1,
fn≥2 :≡ fn−1 + fn−2
0
1 1
2 3
5 8
13
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Discrete Mathematics
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Inductive Proof about Fibonacci Series
Recursion
Theorem: fn < 2n.
Proof: By induction.
• Base cases: f0 = 0 < 20 = 1
f1 = 1 < 2 1 = 2
• Inductive step: Use 2nd principle of induction (strong induction).
Assume k<n, fk < 2k.
Then fn = fn−1 + fn−2 is < 2n−1 + 2n−2 < 2n−1 + 2n−1 = 2n. 
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Discrete Mathematics
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Recursively Defined Sets
Recursion
An infinite set S may be defined recursively, by giving:
• A small finite set of base elements of S.
(유한 개수의 기본 원소를 제시)
• A rule for constructing new elements of S from previouslyestablished elements. (새로운 원소를 만드는 방법을 제시)
Example: Let 3S, and let x+yS if x,yS.
What is S? (= {3, 6, 9, 12, 15, …})
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Discrete Mathematics
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강의 내용
Induction and Recursion
수학적 귀납법(Mathematical Inductions)
재귀(Recursion)
재귀 알고리즘(Recursive Algorithms)
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Discrete Mathematics
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Introduction
Recursive Algorithms
Recursive definitions can be used to describe algorithms as
well as functions and sets.
(재귀적 정의를 수행한 경우, 손쉽게 알고리즘의 함수/집합으로 기술할 수 있다.)
예제: A procedure to compute an.
procedure power(a≠0: real, nN)
if n = 0 then return 1
else return a · power(a, n−1)
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Discrete Mathematics
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Efficiency of Recursive Algorithms
Recursive Algorithms
The time complexity of a recursive algorithm may depend
critically on the number of recursive calls it makes.
(재귀 호출에서의 시간 복잡도는 재귀 호출 횟수에 크게 의존적이다.)
예제: Modular exponentiation to a power n can take log(n)
time if done right, but linear time if done slightly
differently.
(잘하면 O(log(n))이나, 조금만 잘못하면 O(n)이 된다.)
• Task: Compute bn mod m, where
m≥2, n≥0, and 1≤b<m.
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Discrete Mathematics
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Modular Exponentiation Algorithm #1
Recursive Algorithms
Uses the fact that
• bn = b·bn−1 and that
• x·y mod m = x·(y mod m) mod m.
procedure mpower(b≥1,n≥0,m>b N)
{Returns bn mod m.}
if n=0 then return 1;
else return (b·mpower(b,n−1,m)) mod m;
Note this algorithm takes (n) steps!
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Discrete Mathematics
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Modular Exponentiation Algorithm #2
Recursive Algorithms
Uses the fact that
• b2k = bk·2 = (bk)2,
• x·y mod m = x·(y mod m) mod m, and
• x·x mod m = (x mod m)2 mod m.
procedure mpower(b,n,m) {same signature}
if n=0 then return 1
else if 2|n then
return mpower(b,n/2,m)2 mod m
else return (mpower(b,n−1,m)·b) mod m
(첫 번째 mpower()는 한번 call되고, 그 값을 제곱하는 것임)
What is its time complexity?
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(log n) steps
Discrete Mathematics
by Yang-Sae Moon
A Slight Variation of Algorithm #2
Recursive Algorithms
Nearly identical but takes (n) time instead!
procedure mpower(b,n,m) {same signature}
if n=0 then return 1
else if 2|n then
return (mpower(b,n/2,m)·mpower(b,n/2,m)) mod m
else return (mpower(b,n−1,m)·b) mod m
The number of recursive calls made is critical.
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Discrete Mathematics
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Recursive Euclid’s Algorithm (예제)
procedure gcd(a, b: positive integer)
while b  0
begin
r := a mod b
a := b
b := r
end
return a
procedure gcd(a,bN)
if a = 0 then return b
Recursive Algorithms
{r = a % b}
else return gcd(b mod a, a)
Note recursive algorithms are often simpler to code than
iterative ones…
(Recursion이 코드를 보다 간단하게 하지만…)
However, they can consume more stack space, if your
compiler is not smart enough.
(일반적으로, recursion은 보다 많은 stack space를 차지한다.)
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Discrete Mathematics
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Recursion vs. Iteration (1/3)
Recursive Algorithms
Factorial – Recursion
procedure factorial(nN)
if n = 1 then return 1
else return n factorial(n – 1)
Factorial – Iteration
procedure factorial(nN)
x := 1
for i := 1 to n
x := i  x
return x
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Discrete Mathematics
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Recursion vs. Iteration (2/3)
Recursive Algorithms
Fibonacci – Recursion
procedure fibonacci(n: nonnegative integer)
if (n = 0 or n = 1) then return n
else return (fibonacci(n–1) + fibonacci(n–2))
Fibonacci – Iteration
procedure fibonacci(n: nonnegative integer)
if n = 0 then return 0
else
x := 0, y := 1
for i := 1 to n – 1
z := x + y, x := y, y := z
return z
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Discrete Mathematics
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Recursion vs. Iteration (3/3)
Recursive Algorithms
재귀(recursion)는 프로그램을 간단히 하고, 이해하기가 쉬
운 장점이 있는 반면에,
각 호출이 스택을 사용하므로, depth가 너무 깊어지지 않도
록 조심스럽게 프로그래밍해야 함
컴퓨터에게 보다 적합한 방법은 반복(iteration)을 사용한 프
로그래밍이나, 적당한 범위에서 재귀를 사용하는 것이 바람
직함
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Discrete Mathematics
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Merge Sort (예제) (1/2)
Recursive Algorithms
procedure sort(L = 1,…, n)
if n>1 then
m := n/2 {this is rough ½-way point}
L := merge(sort(1,…, m), sort(m+1,…, n))
return L
The merge takes (n) steps, and merge-sort takes
(n log n).
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Discrete Mathematics
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Merge Sort (예제) (2/2) – skip
Recursive Algorithms
The Merge Routine
procedure merge(A, B: sorted lists)
L = empty list
i:=0, j:=0, k:=0
while i<|A|  j<|B| {|A| is length of A}
if i=|A| then Lk := Bj; j := j + 1
else if j=|B| then Lk := Ai; i := i + 1
else if Ai < Bj then Lk := Ai; i := i + 1
else Lk := Bj; j := j + 1
k := k+1
return L
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Discrete Mathematics
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강의 내용
Induction and Recursion
수학적 귀납법(Mathematical Inductions)
재귀(Recursion)
재귀 알고리즘(Recursive Algorithms)
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Discrete Mathematics
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Homework #5
Recursive Algorithms
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Discrete Mathematics
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