Office/cafe hours:
Tuesdays _and_ Thursdays, 10:50-11:50 in Van Vleck 813
Tuesdays, 2:45-4:00 at the State Street Steep and Brew
Wednesdays, 1:30-3:30 at the State Street Steep and Brew
READING
For next Tuesday: read second section of Chapter 3
For next Thursday: read third section of Chapter 3.
HOMEWORK
Assignment #2: due next Thursday (see web)
Remind students that I'm strict about deadlines.
On each assignment, you must indicate how many hours
you spent on each problem. This will help me to keep the
problem sets from being too hard or too easy. We'll keep
track of who isn't doing this and give a nominal (1%)
penalty on your homework grade.
If you didn’t work with anyone else (including me) you
should write “I worked alone”.
Reading the on-line solutions is important!
Notice that I acknowledge students.
One of them could be you next time around!
Note answers at back of book: feel free to use them
The problem set you've just done is pretty typical.
This is not a class where you can do the homework by
mimicking text illustrations; problems require thought, not
memorization. Some problems will require you to go back
to first principles. Sometimes the way to attack a problem
is to explore: make up some concrete examples, see if what
you're supposed to prove makes sense for those examples,
understand why it's true, and then come up with a proof
that holds in general. Or, if you can't solve a problem,
think about a related but simpler problem.
Thinking is good, but banging your head against a wall for
hours isn't. So, if you've given a problem some real
thought and haven't made any progress, ask me or a
classmate for help. For info on the rules that govern this
sort of cooperation, see the website (logistics).
If you're having trouble, start seeking help early in the
term.
Collect homework
Collect info sheets from new students.
PIGEONHOLE PRINCIPLE
Application 9: The Erdos-Szekeres Theorem
Suppose that we are given a sequence of 17 real numbers
a_1,...,a_17. Show that it contains an increasing
subsequence of length 5 or a decreasing subsequence of
length 5.
My proof (“take two”):
Suppose otherwise; that is, suppose that every increasing or
decreasing subsequence of a_1,…,a_17 has length at most
4.
For all k between 1 and 17, let {I(k) / D(k)} be the length
of the longest {increasing / decreasing} subsequence whose
first term is a_k.
By our supposition, I(k) and D(k) are between 1 and 4 for
all k between 1 and 17.
Let the ordered pairs (1,1),(1,2),…,(4,1) be the
PIGEONHOLES, and let the numbers 1,…,17 be the
OBJECTS.
Put the number k into the pigeonhole (I(k),D(k)).
Since there are 17 objects and only 4x4 = 16 pigeonholes,
two of the objects must be in the same pigeonhole.
That is, there must exist i and j with 1i<j17 such that
(I(i),D(i)) = (I(j),D(j)).
But: if a_i  a_j, then I(i) > I(j) (because we can always
take the longest increasing sequence that starts with a_j and
stick a_i at the front of it, obtaining an even longer
increasing subsequence),
whereas if a_i  a_j, then D(i) > D(j) (for the same reason).
Either way, we fail to have (I(i),D(i)) = (I(j),D(j)):
contradiction!
Alternatively: Use the strong form of the pigeonhole
principle (as in Brualdi).
Questions? ...
Is 17 best possible? ...
Note where the proof breaks down if 17 is replaced by 16.
But note also that for proving that 17 is best possible, it’s
not enough to show that some particular proof breaks down
when 17 is replaced by 16. Why not? …
So, is 17 best possible? …
The sequence 4,3,2,1,8,7,6,5,12,11,10,9,16,15,14,13 has no
increasing or decreasing subsequence of length greater than
4.
To prove this last assertion, divide the 16 positions in the
sequence into four quadruples, and note that (a) any
increasing subsequence can contain at most element of each
quadruplet, and (b) any decreasing subsequence must lie
entirely within a single quadruplet. Combining (a) and the
pigeonhole principle, we see that the sequence has no
increasing subsequence of length > 4; and combining (b)
and the pigeonhole principle, we see that the sequencehas
no decreasing subsequence of length > 4.
The Averaging Principle: Given a list of numbers, there is
always a number on the list whose value is at least as large
as the average value of all the numbers in the list. (Ditto
for "at least as small as".)
200 children sit in a circle of 200 chairs. Some of the
children are wearing blue, and the rest are wearing red.
100 of the chairs are blue, and the other 100 are red. Prove
that there exists some k between 0 and 199 such that if each
student moves k seats clockwise, at least half of the
students will be sitting in chairs that match their outfits.
Have you seen this problem before? … Application 8.
Proof: For all k between 0 and 199, let f(k) be the number
of children who will be in a matching chair if every student
moves k seats clockwise. We want to show that for some
k, f(k) ≥ 100. We will do this by proving that the average
value of f(k) as k varies is ≥ 100. To this end, we use the
“decomposition trick”, and write f(k) as
f(k) = a_{k,1}+a_{k,2}+…+a{k,200},
where a_{k,i} is 1 if the ith child will be in a color-matched
chair if he/she moves k seats clockwise, and a_{k,i} is 0
otherwise. What is the average value of a_{k,1} as k
varies? … 1/2. What about a_{k,2}? … So what is the
average value of f(k) as k varies? … 1/2 + 1/2 + … + 1/2 =
(200)(1/2) = 100.
Infinite combinatorics:
Show that at least one of the digits 0,1,...,9 must occur
infinitely often among the digits of pi. ...
What are the pigeonholes? … The digits 0 through 9.
What are the objects? … The numbers 1,2,3,…
When does the object n go into the pigeonhole k? … When
the nth digit of pi is k.
Mention that this can't be proved for any digit in particular,
with current methods!
“What are your questions on Chapter 2? … Any typos or
other mistakes?”
VOTE ON DATE OF MIDTERM!
The winner is: Tuesday, Mar. 9 (nobody voted by email)
COUNTING
Definitions:
A union B = {x: x in A or x in B},
A intersect B = {x: x in A and x in B},
A \ B = {x: x in A but x not in B},
A x B = {(x,y): x in A and y in B},
|A| = number of elements in A.
Addition Principle: For any two sets A and B, |A union B| =
|A| + |B| - |A intersect B|.
Consequently: |A union B| ≤ |A| + |B|, with equality
precisely when A and B are disjoint.
Subtraction Principle ("counting what you want by
counting what you don't want"): For any set A
contained in X, |X-A| = |X| - |A|.
Inclusion-exclusion principle (simple case): If A and B are
subsets of X,
|X-(A union B)| = |X| - |A| - |B| + |A intersect B|.