5-Recursion -1

1
Strong Induction

• Normal Induction If we prove that 1)
  P(n0) 2) For any kn0, if P(k) then P(k1)
• Then P(n) is true for all nn0.
• Strong Induction If we prove that 1)
  Q(n0) 2) For any kn0, if Q(j) for all j from
  n0 to k then Q(k1) Then Q(k) is true for all
  kn0.
• These two are equivalent!
• Why? Set P(n) to be the predicate If n0 j
  n then Q(j)
• P(n0) Q(n0)
• P(k) Q(j) holds for all j between n0 and k

2
Proof by Strong Induction Jigsaw Puzzle

• Each step in assembling a jigsaw puzzle
  consists of putting together two already
  assembled blocks of pieces where each single
  piece is considered a block itself.
• P(n) It takes exactly n-1 steps to assemble a
  saw puzzle of n pieces.
• Basis Step P(1) is (trivially) true.
• Inductive Step We assume P(j) true for all jk
  and well argue P(k1)
• The last step in assembling a puzzle of k1
  pieces is to put together two blocks one of size
  jgt0 and one of size k1-j, for some j.
• Since 0ltj,k1-jk, P(j) and P(k1-j) are
  both assumed true.
• And so, the total number of steps to assemble a
  puzzle with n1 pieces is 1 (j-1)
  ((k1-j)-1) k (k1)-1.
• (This implies P(k1), and hence ends the
  inductive part, and thus also the proof.)

3
More Examples of Theorems with easy Proofs using
Strong Induction

• Thm1 The second player always wins the following
  game
• Starting with 2 piles each containing the same
  number of matches, players alternately remove any
  non-zero number of matches from one of the piles.
• The winner is the person who removes the last
  match.
• Thm2 Every ngt1 can be written as the product of
  primes.
• Thm3 Every postage amount of at least 18 cents
  can be formed using just 4-cent and 7-cent
  stamps
• - First 4 cases 18274, 19734, 2054,
  2137
• - Afterwards, using strong induction Take any
  k21. By (strong) inductive assumption, and by
  the 4 cases above, k-3i7j4 for some i,j.
  Therefore k1i7(j1)4, which ends the
  inductive step.
• (Btw, starting from 18 may seem weird, but
  you just cant do it for 17...)

4
Basis for Induction Integers are well ordered

• Induction is based on the fact that the integers
  are well ordered, i.e.
• Any non-empty set of integerswhich is bounded
  below i.e. there exists b (not necessarily in
  S) s.t. b x for all x in Scontains a smallest
  element i.e. e in S s.t. for all x in S we
  have e x.
• Why does well-ordering imply induction?
• Assume () P(0) and () P(k)?P(k1) for all
  k0.
• Well show that P(n)T for all n0 (thus showing
  that induction works)
• Define () S ngt0 P(n)F. Assume S
  non-empty.
• By well-ordering of N, there is a least element e
  of S.
• Consider element e1 Since P(e) F then by
  () also P(e1) F.
• Note that e1 cannot be equal to 0 because P(0)
  T by ().
• But then by (), e1 is in S, so e is not the
  least element of S
• Contradiction!
• Therefore S must be empty. Therefore P(n)T for
  all n 0.

5
Well Ordering can be used directly to prove
things (i.e. not necessarily via induction)

• Theorem If a and b are integers, not both 0,
  then ?s,t?Z (sa tb gcd( a,b )).
• Proof For any a,b define S ngt0 ?s,t?Z n
  sa tb.
• By the well-ordering of Z, S has a smallest
  element, call it d.
• Choose s and t so that d sa tb.
• Claim d is a common divisor of a and b.
• Proof that d a
• Writing d qa r where 0rltd. If r0 then d
  a.
• If rgt0 then since r d qa (sa tb)
  qa (s - q)a tb
• we would have r?S.
• But since rltd it would mean that d is not the
  smallest element of S
• gt contradiction gt and therefore r0 (and hence
  d a).
• Similarly, d b.
• d is the greatest common divisor since any common
  divisor of a and b must also divide sa tb d.

6
Recursive (Inductive) Definitions

• A function f N?R is defined recursively by
  specifying(1) f(0), its value at 0, and(2)
  f(n), for ngt0, in terms of f(1),.,f(n-1), i.e
  . in terms of f(k)s for kltn.
• Examples
• f(n)n! can be specified as f(0)1 and, for ngt0,
  f(n)nf(n-1).
• For any a, fa(n)an, can be specified as fa(0)1
  and,for ngt0, fa(n)afa(n-1)
• Note In many cases, we specify f(k) explicitly
  not only for f(0) but also for f(1), f(2), ,
  f(m) for some mgt0, and then use a recursive
  formula to define f(n) for all ngtm.

7
Fibonacci Numbers are Recursively Defined

• The Fibonacci numbers f0,f1,f2,,fn,, are
  defined by
• (1) f00, f11
• (2) fnfn-1fn-2, for n2
• (Btw, we can also use F(n) instead of fn to
  designate n-th Fibonacci number.)
• The first 18 Fibonacci numbers are
• f00, f11, f21, f32, f43, f55, f68,
  f713, f821, f934, f1055, f1089, f11144,
  f12233, f13377, f14610, f15987, f111597,
  f122584, f134181, f146765, f1510946,
  f1617711, f1728657, f1846368, ...
• Ref http//en.wikipedia.org/wiki/Fibonacci_number
  
• Question Are they growing exponentially, i.e.
  like fn an for some a?

8
Fibonacci numbers are related to a Golden
Sectioni.e. the splitting of an interval so that
(larger part) / whole (smaller p. / larger p.)
Ref http//en.wikipedia.org/wiki/Golden_ratio
9
Fibonacci Growth

• Theorem If n3 then fn gt f n-2
• where f is the solution to golden ratio, i.e. f
  (1?5) / 2
• Proof by Strong Induction
• Basis Step (for n3 and 4)
• f3 2 gt f 1.618
• f4 3 gt f2 (12?55)/4 (3?5)/2 f1
  2.168
• Fact you can check f2 f1
• Inductive Step Assume fkgtfk-2 for all k s.t.
  3kn.
• fn1 fnfn-1 gt fn-2fn-3 fn-3(f1) fn-3f2
  f(n1)-2

10
Recursively Defined Sets

• Always (1) Basis Step and (2) Recursive Step
• Set S of multiples of 3(1) 3?S (2) If x?S
  and y?S, then xy ?S
• Strings S over an alphabet S. Let ? be the
  empty string.(1) ?? S (2) If w?S and x?S,
  then wx?S
• Examples S0,1 S0,1,2,3,4,5,6,7,8,9
  Sa,b,c,d,e,,x,y,z
• Now recursively define length of a string, L
  S?N where (1) L(?)0 (2) L(wx)L(w)1
• String Catenation ()(1) If u?S, then
  u?u(2) If u,w?S and x?S, then u(wx)(uw)x

11
Recursively Defined Sets

• Always (1) Basis Step and (2) Recursive Step
• Well-Formed Boolean Formula
• T, F, and s, where s is a propositional variable
  are all well-formed formula (WFF).
• If E and F are WFFs, then (E), (E?F), (E?F),
  (E?F), and (E?F) are all WFFs.
• Well-Formed Arithmetic Expression
• x is a well-formed arithmetic expression (WFA) if
  x is either a numeral or a variable.
• If F and G are WFAs, then (FG), (F-G), (FG),
  (F/G), and (E?F) are all WFAs.

12
Recursively Defined StructuresRooted Trees

• Rooted Trees(1) A single vertex, r, is a rooted
  tree with root r.(2) Suppose that T1,T2,,Tn are
  rooted trees with roots r1,r2,,rn
  respectively. Then the graph formed by starting
  with a root, r, which is not in any of these
  trees and adding an edge from r to each of
  these roots is also a rooted tree, whose root is
  r.
• Basis Step 1
  Step 2

13
Recursively Defined StructuresExtended Binary
Trees

• Extended Binary Trees(1) The empty set is an
  Extended Binary Tree(2) If T1 and T2 are
  extended binary trees, then the following tree,
  denoted T1T2 , is also an extended binary
  tree pick a new root
• r, and attach T1 as the left subtree and
  T2 as the right subtree.
• Step 1 Step 2 Step 3

14
Full Binary Trees

• Full Binary Trees(1) A single vertex is a Full
  Binary Tree(2) If T1 and T2 are full binary
  trees, then the following tree, denoted T1T2 ,
  is also a full binary tree pick a new root, r,
  and attach T1 as the left subtree and attach T2
  as the right subtree.
• (Definition change only in the base case, but
  unlike Extended Binary
• Trees, Full Binary Tree has exactly 0 or 2
  child-nodes)
• Base Step 1 Step 2

15
(Recursive) Definitions of Functions on Full
Binary Trees

• Definition of Height, h(T), of a Full Binary Tree
  T(1) If T has a single (root) node/vertex then
  h(T)0.(2) O/w h(T1T2 )1max(h(T1),h(T2 ))
• The Number of vertices, n(T), of a Full Binary
  Tree, T, is given by(1) If T has a single (root)
  node/vertex, n(T)1.(2) n(T1T2 )1n(T1)n(T2 )

16
Structural Induction

• If a set is recursively defined, to show a
  predicate true for all elements in the set one
  needs to
• (1) Show the predicate true on all base cases
• (2) Show that if the predicate is true for each
  of the elements used to construct a new element,
  then the same predicate is also true for that new
  element.
• E.g. to prove P(T) holds for all T?FBT, these
  steps are
• (1) showing P(vertex) holds, because base case of
  FBT definition is just a vertex
• (2) Showing that if P(T1) and P(T2) then P(T1T2
  ), because the recursive clause of FBT definition
  defines a new FBT as T1T2 .
• Example Well show that if T is a full binary
  tree, then n(T) 2h(T)1-1
• (proof on next slide)

17
Structural Induction Example

• Thrm If T is a full binary tree, then
  n(T)2h(T)1-1
• Proof
• Basis Step If T is just the root vertex,
• n(T)1, h(T)0 and therefore n(T)1 201-1 1
• Inductive Step When T T1T2 , we compute
• n(T)1n(T1)n(T2) Definition of n(T)
  1(2h(T1)1-1)(2h(T2)1-1) Inductive
  hypothesis 2max(2h(T1)1,2h(T2)1)-1 Arithmeti
  c
• 221max(h(T1),h(T2)) -1 Arithmetic
• 22h(T)-1 Definition of h(T)

18
Recursive Algorithms

• An algorithm is recursive if it solves a
  problem by reducing it to an instance of the same
  problem with smaller input.
• Examples
• procedure factorial (n nonnegative integer)if
  n0 then factorial(n)1else factorial(n)n
  factorial(n-1)
• procedure power(a nonzero real, n nonnegative
  integer)if n0 then power(a,n)1else
  power(a,n)apower(a,n-1)
• procedure gcd(a,b nonnegative integers with
  altb)if a0 then gcd(a,b)belse gcd(a,b)gcd(b
  mod a, a)
• procedure fibonacci (n nonnegative integer)if
  n1 then fibonacci(n)1else fibonacci
  (n)fibonacci (n-1)fibonacci (n-2)