



















































Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
This document, presented by konstantin busch from lsu, discusses the concepts of induction and recursion in the context of discrete structures. It includes proof techniques, theorems, and algorithms related to these concepts.
Typology: Study notes
1 / 59
This page cannot be seen from the preview
Don't miss anything!




















































Induction and Recursion
CSC-2259 Discrete Structures
Induction
Induction is a very useful proof technique
In computer science, induction is used
to prove properties of algorithms
Induction and recursion are closely related
for recursive algorithms
Inductive Step: Prove that P (^ k ^1 ) is true
Inductive Hypothesis:Assume P ( k ) is true
P ( k ) P ( k 1 )
In other words in inductive step we prove:
(for any positive integer k)
for every positive integer k
P ( k ) P ( k 1 ) P ( 1 )
True
True
Inductive basis Inductive Step
Proposition true for all positive integers
7
n n P n n Inductive Basis: 2 1 ( 1 1 ) ( 1 ): 1 P Inductive Hypothesis: 2 ( 1 ) ( ): 1 2 k k P k k Inductive Step: K. Busch - LSU Theorem: Proof: 2 ( 1 )(( 1 ) 1 ) ( 1 ): 1 2 ( 1 ) k k P k k k We will prove assume that it holds
2
( 1 )(( 1 ) 1 )
2
( 1 ) 2 ( 1 )
( 1 )
2
( 1 )
( 1 ): 1 2 ( 1 )
k k
k k k
k
k k
P k k k
Inductive Step:
(inductive
hypothesis)
End of Proof
Theorem:
2
n
n
Proof:
n 0
Inductive Basis:
1 2 2
0
n
n
n 0
Inductive Hypothesis: n^ k
2
k
k Suppose it holds:
Inductive Step: n k ^1
1
2
k
k We will show:
Theorem: (^) H n n
2
Proof:
n 0
Inductive Basis:
H H H n n
1 2 2
0
n 0
Inductive Hypothesis: n^ k
H k k
2
Suppose it holds:
Inductive Step: n k ^1
1 2
H k k
We will show:
n
2
We have shown:
It holds that: k k
H H H k
log log
2 2
H k
k
k
1 log
log
H (log k ) k
hole
hole
hole
2 2
3 3
Triominos
Inductive Hypothesis: n^ k
k k
Hole can be anywhere
Assume that a checkerboard
can be tiled with the hole anywhere
k k
Inductive Step: n k ^1
1 1
k k