Professor:%20Munehiro%20Fukuda

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CSS342 Induction

• Professor Munehiro Fukuda

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Todays Topics

• Review of sequence and summations
• Mathematical induction
• Strong form of mathematical induction

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Sequence
Review

• A function from an integer subset to a set S
• Geometric progression
• a, ar, ar2, , arn
• where a the initial term and r the common
  ratio
• Arithmetic progression
• a, a d, a 2d, , a nd
• where a the initial term and d the common
  difference

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Sequence
Review

• Finding a formula or a general rule for
  constructing the terms of a sequence
• Runs of the same value?
• 1, 1, 1, 1, 1
• Terms obtained from previous terms by adding the
  same amount?
• 1, 3, 5, 7, 9 2n 1 related to the arithmetic
  sequence
• Terms obtained from previous terms by multiplying
  a particular amount?
• 1 1, 1, 1, 1 (1)n1 related to the
  geometric sequence
• Terms obtained by combining previous terms?
• 1, 2, 2, 3, 3, 3, 4, 4, 4, 4
• Cycles among the terms?
• 1, 3, 5, 7, 9, 1, 3, 5, 7, 9

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Summation
Review

• ?jmnaj am am1 an
• ?jmnkaj k ?jmnaj
• ?i14?j13ij ?i14i (1 2 3) 6 ?i14i
  6 (1 2 3 4) 60
• ?j1nj n(n 1)/2
• ?j1nj2 n(n 1)(2n 1)/6
• ?jmnarj , (r ? 0) a(rn1 - 1)/(r - 1), (r?1)

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A Row of Dominos
Mathematical Induction

• If you push the first domino, all the dominoes
  will fall one after another.
• Formal representation
• The first domino falls.
• For any k 1, if the kth domino falls, the
  (k1)th domino will fall.
• Axiom the principle of mathematical induction
• A property P(n) that involves an integer n is
  true for all n 0 if
• the following are true
• P(0) is true.
• If P(k) is true for any k 0, then P(k1) is
  true.

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A Proof by Induction on N
Mathematical Induction

• Use the principle of mathematical induction
• Base case (or basis) prove P(0) is true.
• Inductive step
• Inductive hypothesis Assume P(k) is true for any
  k 0
• Inductive conclusion Prove P(k1) is true.
• Example 1
• Prove the correctness of a recursive
  solution.
• pow(int x, int n)
• if (n 0)
• return 1
• else
• return x pow(x, n-1)

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Example1 pow(x, n)
Mathematical Induction

• Basis
• When n 0, pow(x, 0) 1.
• x0 1.
• The recursive function is correct.
• Inductive hypothesis
• When n k, assume that pow(x, k) is correct,
  i.e., xk.
• Inductive step
• Show the pow(x, k1) is correct.
• pow(x, k1) x pow(x, k)
• By the inductive hypothesis, pow(x, k) returns
  the value xk.
• Thus, pow(x, k1) x xk
• xk1
• If pow(x, k) is correct, pow(x, k1) is correct.
  Therefore, by the principle of mathematical
  induction, pow(x, n) is correct for any n 1.

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Example2 1 2 . n n(n 1)/2
Mathematical Induction

• Basis
• When n 1, the S1(sum) is 1.
• S1 1(11)/2 1.
• The formula is true.
• Inductive hypothesis
• When n k, assume that Sk is k(k 1)/2.
• Inductive step
• Show the formula is true when n k 1.
• Sk1 1 2 k (k1) Sk (k1)
• k(k1)/2 (k1) k(k1)/2 2(k1)/2
• (k(k1) 2(k1))/2
• (k1)(k2)/2
• If Sk is true, Sk1 is true. Therefore, by the
  principle of mathematical induction, the formula
  is true when n 1.

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Example 3 Geometric Sum
Mathematical Induction

• Show that if r?1,
• a ar1 ar2 arn a(rn1 - 1)/(r - 1)
• Basis
• If n0, a(r1 1)/(r 1) a, which is true.
• Inductive hypothesis
• Assume that the formula is true for n k.
• Inductive step
• Show the formula is true for n k 1.
• a ar1 ar2 ark ark1 a(rk1 - 1)/(r
  - 1) ark1
• a(rk1 - 1)/(r - 1) ark1 (r - 1)/(r - 1)
• (a rk1 a a rk2 a rk1)/(r 1)
• (a rk2 a)/(r 1) a(rk2 1)/(r 1)
• If the formula is true for k, it is also true for
  k1. Therefore, by the principle of mathematical
  induction, the formula is true when n 0.

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Example 4 Triominoes
Mathematical Induction
A 4 4 deficient board with triominos
A n 8 8 deficient board with triominos
A triomino consisting of 3 squares joined at
the edges
A 2 2 deficient board with a triomino
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A Tiling Problem
Mathematical Induction
Can we tile a 2k 2k deficient board with
triominoes?

• Basis
• if k 1, the 2 x 2 deficient board is itself a
  tromino and can therefore be tiled with one
  tromino.
• Inductive hypothesis
• Assume that we can tile a deficient 2k 2k
  board.
• Inductive step
• Consider a 2k1 2k1 board and divide it into 4
  2k 2k boards.
• Rotate each divided board so that its missing
  square is placed at a corner.
• With three of those squares, make a new tromino.
• If a 2k 2k board is tiled with trominoes, a
  2k1 2k1 board can be tiled.

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Strong Form of Mathematical Induction
Strong Form of Mathematical Induction

• A property P(n) that involves an integer n is
  true for all
• n 0 if the following are true
• P(0) is true.
• If P(0), P(1), , P(k) are true for any k 0,
  then P(k1) is true.
• Difference from ordinary form of mathematical
  induction
• P(k1) is factorized in a combination of two or
  more of P(0) through to P(k).
• Proof by strong form of mathematical induction
• Base case or (basis) prove P(0) is true.
• Inductive step
• Inductive hypothesis Assume P(1),P(2),..,P(k) is
  true for any k 0
• Inductive conclusion Prove P(k1) is true.

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Example5 Every Integer gt 1 Can Be Written as A
Product of Prime Integers
Strong Form of Mathematical Induction

• Basis
• When n 2 ( gt 1), 2 is a prime number itself,
  and thus can be written as
• a product of prime integers.
• The proposition is true.
• Inductive hypothesis
• Assume that the proposition is true for each of
  the integers 2, 3, k
• Inductive step
• Show the proposition is true when n k 1.
• If k 1 is a prime number, there is nothing
  more to show.
• If k 1 is NOT a prime number, k 1 is
  divisible and can be written
• k 1 x y where 1 lt x lt k 1 and 1 lt y lt k
  1
• Notice that x and y can be written as a product
  of prime integers by inductive hypothesis.
• If k is a product of prime numbers, k1 is a
  product of prime numbers. Therefore, by the
  principle of mathematical induction, the
  proposition is true when n 1.

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Example6 Multiplying Rabbits
Strong Form of Mathematical Induction

• Our recursive solution was
• rabbit(1) 1
• rabbit(2) 1
• rabbit(n) rabbit(n-1) rabbit(n-2)
• The problem was its inherent efficiency.
• There is another solution that is much efficient
• rabbit(n) (an bn) / sqrt(5)
• where a 1 sqrt(5)/2 and b 1
  sqrt(5)/2
• Note
• a b 1/2 sqrt(5)/2 1/2 sqrt(5)/2 1
• b 1 a
• a2 1 2sqrt(5) 5/4 4 2
  2sqrt(5)/4 1 1 sqrt(5)/2 1 a
• b2 1 2sqrt(5) 5/4 4 2
  2sqrt(5)/4 1 1 sqrt(5)/2 1 b

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Example6 Multiplying Rabbits
Strong Form of Mathematical Induction

• Basis
• rabbit(1) (a1 b1)/sqrt(5) ((1 sqrt(5))/2
  (1 sqrt(5))/2)/sqrt(5)
• sqrt(5)/sqrt(5) 1
• rabbit(2) (a2 b2)/sqrt(5) (((1
  sqrt(5))/2)2 ((1 sqrt(5))/2)2)/sqrt(5)
• ((12sqrt(5)5)/4 (1 2sqrt(5)5)/4)/sqrt(
  5)
• (4sqrt(5)/4)/sqrt(5) sqrt(5)/sqrt(5) 1
• Thus, the formula is true.
• Inductive hypothesis
• Assume that the formula is true for all n such
  that 1n k, where k is at least 2.
• Inductive step
• Show the formula is true for n k 1.
• rabbit(k 1) rabbit(k) rabbit(k 1) (ak
  bk) (ak-1 bk-1)/sqrt(5)
• ak-1(a 1) bk-1(b1)/sqrt(5)
• ak-1(a2) bk-1(b2)/sqrt(5) // a 1
  a2, b 1 b2
• (ak1 bk1)/sqrt(5)
• If the formula is true for k, it is also true for
  k1. Therefore, by the principle of mathematical
  induction, the formula is true when n 1.

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In-Class Exercise 1
Lets form a group of three or four students.
Each group selects and solves one of the
following four math problems by adding up all
members student IDs and calculating its modulo
by 4, (i.e., ?id 4). Each group should write
their solution on a one-page scratch paper as
listing all its group members names, turn in the
solution to the professor, and choose its group
representative who will present the solution to
all the class audience in case if the group is
ask to do so. Group 0 Textbook p280 Exercise
11 Group 1 Textbook p280 Exercise 10 Group 2
Proof by Strong Induction (See http//courses.w
ashington.edu/css342/fukuda/exercise/ex1.pdf) Grou
p 3 Non-turn-in writing exercise Q15