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01B - Inductive Reasoning - Great T
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Slides byVictor Adamchik
?
4
In order to share some candy,ܰ
students are assigned numbers from
ͳ
to
Then, the following rules are applied:1.
Student
ͳ
gets candy.
݅ : if student
݅ gets candy,
then student
݅ ͳ
also gets candy.
Question.
Will student
get candy?
Answer.
Of course!
6
A number of
dominos stand in a row:
They are close enough to each other, so:2. For all
݅ : if the
th݅
domino in the row
falls, then the
݅ ͳ
st
domino also falls.
Then, we push the
th
domino, so:
The
th
domino falls.
Question.
Does the
th
domino fall?
Answer.
Of course!
Question.
Does the 3
rd
domino fall?
Answer.
Of course not.
7
A number of
dominos stand in a row:
They are close enough to each other, so:2. For all
݅ : if the
th݅
domino in the row
falls, then the
݅ ͳ
st
domino also falls.
Then, we push the
th
domino, so:
The
th
domino falls.
Question.
What if
ܰ ൌ λ
Which of the infinitely many dominos will fall? Answer.
All dominos from the
th
one onwards.
[^ Why:
Pick any of them, say
th
, for some
Look at first
dominos. Then (as before) the
th
domino falls. So, they all do.]
Let P(n) be a statement about n.
(e.g., P(n):=“the n
th^
student gets candy”)
The value of it: Instead of attacking aproblem directly, we just explain:1.
why P(0) is true; and
Then the following holds: Parts of an Inductive Proof
Inductive step: Fix an arbitrary k
Assume that P(k) holds.
Use that to prove that P(k+1) holds too.
Base step(s): Show that P(0) holds.
Conclusion:
Therefore, P(n) holds for all n.
There is nothing sacred about the base case
n=0, we could just as well start at n = 11. We have chosen to describe the induction step
as moving from k to k+1, where k
There is the obvious alternative to change the
induction step from k-1 to k, where k
Parts of an Inductive Proof
Example
Prove that 2
n^ < n! for all n
Base step:
For n=4: We have 2
Inductive step: Fix arbitrary k
Assume 2
k^
< k!.
We will prove that 2
k+
< (k+1)!.
We have:
k+
k^ < 2·k! < (k+1)·k! = (k+1)!. Conclusion:
Therefore, 2n < n! for all n
A Template for Inductive Proofs
State that the proof is by induction. If there are
several variables, indicate which serves as n.
Clearly state your inductive hypothesis P(n).
Prove base case(s).
Fix arbitrary k. Assume P(k). Prove P(k+1).
State that the induction principle allows you to
conclude that P(n) is true for all n
Example
Induction is helpful for proving the
correctness of a statement, butnot helpful for discovering it.
Recall our recurrence
from last lecture for bounding P
:n
Prove T(n) = 2·n-.
T(n) = 2 + T(n-1)T(2) = 1
Example
Base case: n =2, T(2) = 2·2-3 = 1 Inductive step: Assume T(k) = 2·k-
Prove it for k+
T(k+1) = 2 + T(k) = 2 + 2·k-3 = 2·(k+1)-
T(n) = 2 + T(n-1)T(2) = 1Prove T(n) = 2·n-
Soundness of InductionHow do we know that INDUCTION
really works?
Soundness of Induction
Proof by contradiction
Assume that for statement P(n), wecan establish the base step P(0), andthe induction step, but nonetheless itis not true that P(n) holds for all n.
So, for some values of n,
P(n) is false.
Soundness of Induction
Certainly, n
0
cannot be 0.
Thus, it must be n
0
= n
+1, where 1
n
< n 1
Let n
0
be the least such n that P(n
) is false. 0
Now, by our choice of n
, this means that P(n^0
holds.
because of IH, since n
< n 1
0
Soundness of Induction
But then by Induction Step, P(n
+1) also holds. 1
Which is the same as P(n
), and we have a 0
contradiction.
0
to be the least n where P(n) fails.
Namely, we used the
LEAST ELEMENT PRINCIPLE:^ Every non-empty subset of thenatural numbers must contain a
least element.So, we proved that
LEP
ฺ
WI
Similarly, we can also prove that
WI
ฺ
LEP. So: LEP
WI.
Base case: 2 is primeInductive step:Fix any n
Assume all k
n can be factored into primes.
We must prove that n+1 can be factored too. Case 1: n+1 is primeCase 2: n+1 is composite, n+1 = p q
p, q > 1
(easy to finish from here…)
Theorem. Every natural number > 1
can be factored into primes
Strong Induction
How to justify the proof by strong
induction?
The proof is by contradiction.
Note that again we used theLEAST ELEMENT PRINCIPLE: Every non-empty subset of thenatural numbers must contain a
least element.So, we proved that
LEP
ฺ
SI
Similarly, we can also prove that
SI
ฺ
LEP. So: LEP
SI.
Strong Induction vs Weak InductionIt follows that there two forms are equivalent.The conversion from weak to strong form is trivialFrom strong in P to weak:Let Q(n) :
P(0) and P(1) and ... and P(n)
Base Step: Q(0) = P(0)Inductive Step: Q(n)
P(n+1)
Q(n)
Q(n) and P(n+1)
Q(n)
Q(n+1)
Therefore, the strong induction in P can bewritten as a weak induction in Q.
ATM Machine
ATM Machine: Proof
Base case: 54 = 2·7 + 4·10Induction step: assume k = 7·a + 10·b,for all k = 54, …, nHow do we proceed for k=n+1 dollars?
ATM Machine: Proof n + 1 = n
n
By IH:
n
6 = 7·a + 10·b
Hmm…,Therefore, we have to extend the base cases to55, 56, 57, 58 and 59
n
6 could be less than 54...
Chocolate Bar
Note, there are two variables n and k.We prove it by strong induction on thenumber of squares s = n k.If s = 1, no breaks are required andn k-1 = s -1 = 1-1 = 0
Chocolate Bar
IH: assume that this is true
s < n
Consider a chocolate bar with s = n totalpieces.After the first snap, there will be two smallerbars with k
and n-k 1
pieces. 1
Chocolate Bar
After the first snap, there will be two smallerbars with k
and n-k 1
pieces. 1
By IH, it requires k
-1 and n-k 1
-1 breaks. 1
Totally, (k
-1)+(n-k 1
-1)+1= n-1 1
In Programming.
A rule that
applies throughout the life of adata structure or procedure orloop. Each change to the datastructure or loop maintains the
15-122 again …
Loop invariants int fast_exp (int x, int y)//@requires y > 0;//@ensures result == pow(x,y);{
int r = 1; int b = x; int e = y;while (e > 1)//@loop_invariant r * pow(b,e) == pow(x,y);{
if (e % 2 == 1) r = b * r;b = b * b; e = e / 2; } return r * b; }
15-122 again …
Data structure invariants:
BST, Heaps, AVL Trees
h-
h-
Fractals
Fractals are geometric objects that
are self-similar, i.e. composed ofinfinitely many pieces, all of which
look the same.
The Koch Curve
Alphabet: { F, +, - }Rules:
Rule(-) = -
Rule(w
…w 1
) = Rule(wn
) … Rule(w 1
)n
Rule (F):
Rule(Rule (F)):F+F--F+F+F+F--F+F--F+F--F+F+F+F--F+F
Rule(F) = F+F--F+FRule(+) = +
Visual representation: F^
draw forward one unit
+^
turn 60 degree left
-^
turn 60 degrees right The Koch Curve Visual representation: F^
draw forward one unit
+^
turn 60 degree left
-^
turn 60 degrees right The Koch Curve
The Koch Curve a non-differentiable curveStructural Induction
Useful for proving properties ofinductively-defined sets.
Same steps as weak/strong induction.•^
Prove the property holds on each of the basecases of the definition.
-^
Prove the property holds on the result of anyinductive rule, assuming that it holds on eachof the parts on which the rule applies.
Base case:
3 is in S
Consider a set S defined byInductive rule:if x in S and y in S , then x+y in S
What does S contain?
