24+Slides--Induction%2C+Trees

# 24+Slides--Induction%2C+Trees - CS103 HO#24...

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Unformatted text preview: CS103 HO#24 Slides--Induction, Trees 4/18/11 1 CS103 Mathematical Foundations of Computing 4/18/11 Tuesday: Special Review Session 7 9 pm Gates B12 Wednesday: Special Office Hours 7 9 pm Gates 104 Thursday: Office Hours 2:14 4:15 Gates 400 Midterm (open book, notes) 7 9 pm Cubberley Aud. Suppose b , b 1 , b 2 ,...is the sequence defined as follows: b = 1, b 1 = 2, b 2 = 3, b j = b j-3 + b j-2 + b j-1 for all integers j 3. Let P n be the assertion b n 3 n . Show that P n is true for n 0. BASE CASE: We will establish the three base cases: P , P 1 , P 2 . b = 1, which is 3 b 1 = 2, which is 3 1 b 2 = 3, which is 3 2 So P , P 1 , and P 2 are true. Suppose b , b 1 , b 2 ,...is the sequence defined as follows: b = 1, b 1 = 2, b 2 = 3, b j = b j-3 + b j-2 + b j-1 for all integers j 3. Let P n be the assertion b n 3 n . Show that P n is true for n 0. INDUCTION STEP: For an arbitrary integer k 2, assume P i for 0 i k; that is, b i 3 i Show P k+1 : b k+1 3 k+1 By the inductive hypothesis, b k-2 + b k-1 + b k 3 k-2 + 3 k-1 + 3 k 3 k + 3 k + 3 k 3(3 k ) = 3 k+1 b k+1 3 k+1 So P n is true for n 0 by the Principle of Strong Mathematical Induction. A jigsaw problem Block: a single piece or a number of pieces with matched boundaries that have been put together Blocks can be put together to make larger blocks. Putting two blocks together is called a move. Show that a puzzle of n pieces always takes n-1 moves to solve. BASE CASE: A puzzle of 1 piece takes no moves to solve. INDUCTION STEP: Assume P i for 1 i k: a puzzle of i pieces takes i-1 moves to solve. Show P k+1 : a puzzle of k+1 pieces takes k moves. n m Last move in k+1 piece puzzle total pieces n + m = k+1 total moves = n 1 + m 1 + 1 = n + m -1 = (k + 1) 1 = k The Pigeonhole Principle : If k+1 or more objects are placed in k boxes, then there is at least one box containing two or more of the objects. CS103 HO#24 Slides--Induction, Trees 4/18/11 2 Among any group of 367 people, there must be at least two with the same birthday since there are only 366 possible birthdays. In any group of 27 English words, there must be at least two that start with the same letter, since there are 26 letters in the alphabet. The Pigeonhole Principle : If k+1 or more objects are placed in k boxes, then there is at least one box containing two or more of the objects. The Generalized Pigeonhole Principle : If N objects are placed in k boxes, then there is at least one box containing at least ceil(N/k) objects. Among 100 people there are at least ceil(100/12) = 9 who were born in the same month. CI WOP Well Ordering Property: If A is a non-empty set of positive integers, then A has a least element....
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## 24+Slides--Induction%2C+Trees - CS103 HO#24...

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