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CPSC 121: Models of Computation
2011 Winter Term 1
Proof Techniques
(Part B)
Steve Wolfman, based on notes by
Patrice Belleville and others
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Learning Goals: In-Class
By the end of this unit, you should be able to:
– Devise and attempt multiple different, appropriate
proof strategies—including all those listed in the
“pre-class” learning goals plus use of logical
equivalences, rules of inference, universal modus
ponens/tollens, and predicate logic premises—for
a given theorem.
– For theorems requiring only simple insights
beyond strategic choices or for which the insight
is given/hinted, additionally prove the theorem.
2
Outline
• More Proof Strategies
– Using Predicate Logic Premises
– Using Logical Equivalences
– Proof by contradiction
• Strategies Summary
• Problems and Discussion
• Next Lecture Notes
3
Using Predicate Logic
Premises: Universals
What can you say if you know (rather than needing
to prove) x  D, P(x)?
If you know x  D, P(x):
You can say P(d) is true for any particular d in D
of your choice, for an arbitrary d, or for every d.
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This is basically the opposite of how we go about proving a universal.
Using Predicate Logic
Premises: Existentials
What can you say if you know (rather than needing
to prove) y  D, Q(y)?
If you know y  D, Q(y):
Do you know Q(d) is true for every d in D?
Do you know Q(d) is true for a particular d of
your choice?
What do you know?
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This is basically the opposite of how we go about proving a existential.
Using Predicate Logic Premises
What can you say if you know (rather than needing
to prove) x  D, P(x) or y  D, Q(y)?
If you know x  D, P(x), you can say for any d in
D that P(d) is true. You can say P(d) is true for
any particular d in D or for an arbitrary one.
If you know y  D, Q(y), you can say that for
some d in D, Q(d) is true, but you don’t know
which one. So, assume nothing more about e
than that it’s from D.
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This is basically the opposite of how we go about proving the statements.
Problem: Anti-Symmetric
Generally Faster?
Let an algorithm be “generally faster” than another
algorithm exactly when it’s faster for all problem
sizes n greater than some minimum i.
Assume that if one algorithm is faster than another
algorithm for any particular n, then the other
algorithm is not faster than the first algorithm for
that n.
Problem: Prove that if one algorithm is generally
faster than another, then the other is not
generally faster than the first.
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How Shall We Start Our Strategy?
a.
b.
c.
d.
e.
Witness
Inequality proof
Antecedent assumption
WLOG
I have no idea
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Problem: Anti-Symmetric
Generally Faster?
Theorem: If one algorithm is generally faster than
another, then the other is not generally faster than the
first.
WLOG, let a and b be algorithms.
Assume a is generally faster than b.
Based on the definition of generally faster, then:
there is an i such that a is faster than b for all n > i.
We now need to prove that b is not generally faster than
a, that is that there is no i2 such that b is faster than a
for all n2 > i2.
Should we prove the negation or try something else?
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I renamed the variables so I wouldn’t get confused!
Problem: Anti-Symmetric
Generally Faster?
WLOG, let a and b be algorithms.
Assume a is generally faster than b.
Based on the definition of generally faster, then:
there is an i such that a is faster than b for all n > i.
We now need to prove that b is not generally faster than
a, that is that there is no i2 such that b is faster than a
for all n2 > i2.
Instead, we’ll prove the equivalent statement that for all
i2, there is an n2 > i2 such that b is not faster than a
for problem size n2.
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How Shall We Continue?
a.
b.
c.
d.
e.
Witness
Inequality proof
Antecedent assumption
WLOG
I have no idea
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Problem: Anti-Symmetric
Generally Faster?
Continuing:
WLOG, let i2 be a positive integer.
Let n2 = ??. (NOTE FOR LATER, better be > i2!)
We now need to prove that b is not faster than a for
problem size n2.
Does our assumption that a is generally faster than b
help? Under what conditions?
It’s common in scratch work to build up “NOTES
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FOR LATER” and handle them at the end.
Problem: Anti-Symmetric
Generally Faster?
Finishing the proof:
(NOTE FOR LATER, n2 better be > i as well!)
Based on our assumption and since n2 > i, a is faster
than b for problem size n2.
Based on our initial assumption about “faster”, we know
that since a is faster than b for problem size n2, b is
not faster than a for problem size n2.
QED!
(Note: we assumed nothing about i (the
existential) but that it’s a positive integer, but
we
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picked n (the universal) to be whatever we want!)
Crucial Steps and
Where They Came From??
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Outline
• More Proof Strategies
– Using Predicate Logic Premises
– Using Logical Equivalences
– Proof by contradiction
• Strategies Summary
• Problems and Discussion
• Next Lecture Notes
15
Using Logical Equivalences
Every logical equivalence that we’ve learned
applies to predicate logic statements.
For example, to prove ~x  D, P(x), you
can prove x  D, ~P(x) and then convert
it back with generalized De Morgan’s.
To prove x  D, P(x)  Q(x), you can
prove x  D, ~Q(x)  ~P(x) and convert
it back using the contrapositive rule.
In other words, Epp’s “proof by contrapositive” is
direct proof after applying a logical equivalence rule.
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Worked Problem:
Even Squares
Theorem: If the square of an integer n is even,
then n is even.
Problem: prove the theorem.
This is a tricky problem, unless you try some
different approaches.
An approach that may work with conditional
statements is to try the contrapositive (which is
logically equivalent to the original conditional).
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Worked Problem:
Even Squares
Theorem: If the square of an integer n is even,
then n is even.
Approach:
(1) Prove the contrapostive: If an integer n is
odd, then its square is also odd.
(2) Transform the result back into our
theorem.
We now focus on proving the contrapositive.
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Worked Problem:
Even Squares (part 1)
Theorem: If an integer n is odd, then its
square is also odd.
Proof: Without loss of generality, let n be an
integer. Assume n is odd.
We know (from Epp’s definition of “odd”) that
n = 2k + 1 for some integer k.
n2 = (2k + 1)2
= 4k2 + 4k + 1
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= 2(2k2 + 2k) + 1
Worked Problem:
Even Squares (part 2)
We know 2k2 + 2k is an integer (since k is
an integer and multiplication and addition
are “closed over the integers”).
n2 is 2q+1 for some integer q; so, n2 is odd.
Thus, if an integer n is odd, its square is
odd.
The contrapositive of this statement is also
true: if the square of an integer n is even,
then n is even.
QED
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Outline
• More Proof Strategies
– Using Predicate Logic Premises
– Using Logical Equivalences
– Proof by contradiction
• Strategies Summary
• Problems and Discussion
• Next Lecture Notes
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A New Proof Strategy
“Proof by Contradiction”
To prove p:
Assume ~p.
Derive a contradiction.
You have then shown that there was
something wrong (impossible) about
assuming ~p; so, p must be true.
Can you use this in predicate logic proofs?
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Of course you can, just like every other prop logic technique!
Example in Dialogue: Prove that
Achilles is not Omnipotent
Achilles: I am omnipotent you know.
Tortoise: Can you lift a mountain with your mind,
no matter how big?
Achilles: Of course!
Tortoise: Can you make a mountain out of nothing,
no matter how big?
Achilles: Certainly!
Tortoise: Can you make a mountain so big that
even you cannot lift it?
Achilles: See if I invite you over for dinner again.
I stole the characters from Gödel, Escher, Bach.
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I stole the story from some sci-fi novel.
Side Note:
Really a New Proof Strategy?
“Assume ~p and derive a contradiction” is
the same “assume ~p and prove F”.
That’s antecedent assumption!
What have we proven? ~p  F
What’s that logically equivalent to?
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Partly Worked Problem:
2 is Irrational
Note: a rational number can be expressed as a/b for
some a  Z, b  Z+ with no common factor except
1.
Theorem: The 2 is an irrational number.
Problem: prove the theorem.
This is a tricky problem to even start with. We know
from the definition that we want to show 2 cannot
be represented as a/b with the constraints given,
but where does that get us?
Let’s try contradiction, instead.
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Partly Worked Problem:
2 is Irrational
Theorem: The 2 is an irrational number.
Opening steps:
(1) Assume for contradiction that 2 is rational.
(2) Using our knowledge of rationals, we know
2 = a/b, where a  Z, b  Z+, and a and b
have no common factor except 1. [But, we
know nothing more about a and b!]
Next, play around with the formula 2 = a/b
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and see where it takes you!
Promising Strategies?
We know that 2 = a/b.
Which of these feels like the most promising
strategy?
a. Solve for a.
b. Solve for b.
c. Square both sides to ditch the radical.
d. Assume 2 is rational.
e. None of these is promising.
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Proof Scratchwork
a = b2
b = a/2
2 = a2/b2
a2 = 2b2
b2 = a2/2
b is a positive integer
a and b have no common factors
Plus, blast from the past:
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If the square of an integer is even, the integer is even.
Finishing the Proof
Assume for contradiction that 2 is rational.
Then, 2 = a/b for a  Z, b  Z+, where a and
b have no common factor except 1.
So, a2 = 2b2, and a2 is even.
Since a2 is even, a is even (prev. proof!!).
a = 2k for some integer k.
b2 = a2/2 = (2k)2/2 = 2k2. b2 is even and so is b.
Then, b and a share the factor 2.
CONTRADICTION! QED (2 is irrational.)
Notice: even with proofs, we can break them down into simpler pieces.
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When we use our previous proof this way, we call it a “lemma”.
Outline
• More Proof Strategies
– Using Predicate Logic Premises
– Using Logical Equivalences
– Proof by contradiction
• Strategies Summary
• Problems and Discussion
• Next Lecture Notes
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Strategies for
Predicate Logic Proofs (1 of 2)
Have lots of strategies on hand, and switch
strategies when you get stuck:
• Try using WLOG, exhaustion, or witness
approaches to strip the quantifiers
• Try antecedent assumption on
conditionals
• Try contradiction on the whole statement
or as part of other strategies
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Strategies for
Predicate Logic Proofs (2 of 2)
• Work forward, playing around with what
you can prove from the premises
• Work backward, considering what you’d
need to reach the conclusion
• Play with the form of both premises and
conclusions using logical equivalences
Finally, disproving something is just proving
its negation.
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Outline
• More Proof Strategies
– Using Predicate Logic Premises
– Using Logical Equivalences
– Proof by contradiction
• Strategies Summary
• Problems and Discussion
• Next Lecture Notes
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Problem: Aliens Attack
Aliens hold the Earth hostage and demand
that we help them get started proving:
x  D, y  E, z  F, P(x)  Q(y)  R(x,z).
Problem: Propose four different, promising
solution strategies that are each as complete as
possible (given the available information) and
could be used to address this theorem.
This is why a Computer Scientist is always kept on hand at Area
34 51.
That and the “use a computer virus to take out their shields” thing.
Problem:
Generally Faster than Itself?
Let an algorithm be “generally faster” than
another algorithm exactly when it’s faster
for all n greater than some minimum i.
Assume that no algorithm is faster than itself
for any particular n.
Problem: Prove that no algorithm is
generally faster than itself.
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Worked Problem:
Generally Faster than Itself?
Problem: Prove that no algorithm is
generally faster than itself.
Let’s roughly translate to predicate logic:
“~a that is generally faster than itself”.
We don’t have a technique for a ~ on the
outside, but we could: (1) disprove a that
is generally faster than itself, (2) use proof
by contradiction (and assume a that is
generally faster than itself), or (3) use De
Morgan’s to move the negation inward. 36
Worked Problem:
Generally Faster than Itself?
Let’s try: use De Morgan’s to move the
negation inward.
Using De Morgan’s, “~a that is generally
faster than itself” becomes “a ~(a is
generally faster than itself)”
Now, “consider an arbitrary algorithm a”.
We’re down to proving “~(a is generally
faster than itself)”.
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Worked Problem:
Generally Faster than Itself?
We’re down to proving “~(a is generally faster than
itself)”.
From the problem statement: Let an algorithm be
“generally faster” than another algorithm exactly
when it’s faster for all n greater than some
minimum i.
We’ll negate the definition of generally faster and
substitute a for both algorithms: “for any
minimum i, there is a larger number n such that
a is not faster than itself for problem size n.”
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Worked Problem:
Generally Faster than Itself?
We now need to prove: “for any minimum i,
there is a larger number n such that a is
not faster than itself for problem size n.”
Consider an arbitrary (positive integer) i. Let
n = i + 1. (We get to pick n, and i + 1
seems like a handy n to pick.)
So, we need to prove: “a is not faster than
itself for problem size i + 1 (where i is an
arbitrary positive integer)”
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Worked Problem:
Generally Faster than Itself?
From the problem statement: “Assume that no
algorithm is faster than itself for any particular n.”
That’s equivalent to: “for every algorithm and n, the
algorithm is not faster than itself for problem size
n”.
By universal instantiation on this premise, a is not
faster than itself for problem size i + 1 (where i is
an arbitrary positive integer).
QED!
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Worked Problem, Short Version:
Generally Faster than Itself?
The definition of “generally faster” requires us to
pick a “minimum i” after which the first algorithm
is always faster than the second for each
problem size.
Consider an algorithm a. We know from the
premise that a is not faster than itself for any
value of n. No matter what “minimum i” we use,
we can pick a larger n (such as i + 1) for which a
is not faster than itself. So, a is not generally
faster than itself.
QED
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Worked Problem, Short Version, Different Approach:
Generally Faster than Itself?
Assume for contradiction that some algorithm a is
generally faster than itself. Then, by the
definition of “generally faster”, for all n larger
than some minimum i, a is faster than itself.
Thus, for any number n1 larger than i (like,
n1 = i + 1). a is faster than itself for n1.
But, we also assumed that no algorithm (including
a) is faster than itself for any particular problem
size (including n1). This is a contradiction!
QED
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Outline
• Learning Goals, Quiz Notes, and Big Picture
• Proof Strategies
– Generalizing from the Generic Particular (WLOG) &
Proofs of Existence (Witness Proofs)
– Using Predicate Logic Premises &
Propositional Logic Rules
– Antecedent Assumption (generalization of “direct
proof”)
– Using Logical Equivalences
– Proof by contradiction
• Strategies Summary
• Problems and Discussion
• Next Lecture Notes
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Learning Goals: In-Class
By the start of class, you should be able to:
– Devise and attempt multiple different,
appropriate proof strategies—including all
those listed in the “pre-class” learning goals
plus use of logical equivalences, rules of
inference, universal modus ponens/tollens,
and predicate premises—for a given theorem.
– For theorems requiring only simple insights
beyond strategic choices or for which the
insight is given/hinted, additionally prove the
theorem.
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Next Learning Goals:
Pre-Class
We are mostly departing from the readings for
next class to talk about a new kind of circuit
on our way to a full computer: sequential
circuits.
The pre-class goals are to be able to:
– Trace the operation of a deterministic finite-state
automaton (represented as a diagram) on an
input, including indicating whether the DFA
accepts or rejects the input.
– Deduce the language accepted by a simple DFA
after working through multiple example inputs.
Next Lecture Prerequisites
See the “Sequential Circuits” readings at the
bottom of the “Textbook and References”
area of the website.
Complete the open-book, untimed quiz on
Vista that was due before class.
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More problems to solve...
(on your own or if we have time)
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Problem:
Generally, Transitively Faster?
Continue with our previous assumptions about
faster/generally faster.
Assume that our faster predicate is transitive
(Faster(a1, a2, n) and Faster(a2, a3, n) implies
Faster(a1, a3, n)).
Problem: Prove that if one algorithm is
generally faster than a second, and if the
second algorithm is generally faster than a
third, then the first algorithm is also generally
faster than the third.
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More More Practice:
A Representation for Every Number
Prove that any non-negative integer can be
represented using a finite number of bits
as an unsigned binary number.
(Don’t) Prove that with a finite number of
bits, we can represent any non-negative
integer as an unsigned binary number.
(What’s different between these two? Why
can’t we prove the second? How would
we disprove it?)
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