Math 22, Fall 2009, Home Page

Class 22, Th., Mar. 18, 2010
We reviewed. Here are some of the problems I posed for you:

Let A={x|x is congruent to 1, mod 4}, B = {a,b}.
Note A={1,5,9}.
Find the
(1) # of functions from A to B.
(2) # of functions from B to A.
(3) # of 1 to 1 functions from A to B.
(4) # of 1 to 1 functions from B to A.
(5) # of onto functions from A to B.
(6) # of onto functions from B to A.
(7) # of equivalence relations on A X A.
(8) # of subsets of AUB with 3 elements.
(9) # of ordered arrangements of AUB with 3 elements.
(10) # of functions f:AUB -> {1,2,3,...,10} which have f(1)+f(5)+f(9)+f(a)+f(b) = 5.
(11) Expand (3-4x^2)^5
(12) Expand (1-4x^2)^(-3) and show the first five terms.
(13) # of strings from AUB of length 5 using exactly 3 symbols of AUB.

Note solution to (13) which we briefly went over incorporates many of the ideas about counting:
There are C(5,3) ways to choose the 3 symbols from AUB = {1,5,9,a,b}.
There are two possible selections of numbers of symbols once those 3 are chosen, in order to have total of five:
(i) 1 of one of the 3 chosen symbols and 2 of each of the others.
(ii) 3 of one of the symbols and 1 of each of the others.
(Note that in general we could find the number of such selections using methods of counting selection with repetition, though we would still have to examine and count # of possible strings for each case.)
In both cases there are 3 ways, once the three symbols have been chosen to make these selections: in (i) there are three ways to choose the symbol that will appear in the string once, and in (ii) there are three ways to choose the symbol that will appear in the string three times.
In (i) there are 5!/(2!2!1!) ways to arrange the 5 symbols chosen.
In (ii) there are 5!/(3!1!1!) ways to arrange the 5 symbols chosen.
Thus total number is C(5,3) (3) [ 5!/(2!2!1!) + 5!/(3!1!1!)] = (10)(3)[30 + 20] = 1500.
(If this seems large, remember that the total number of strings of length 5 from 5 symbols is 5^5.)

Class 21, Tue., Mar. 16, 2010
We went over generating functions some more. Do homework through section 9.6.

Here's the extra homework problem:
We play the following game: start with a pile of n objects, and divide it into two piles or r and s objects. Calculate the sum 1/r + 1/s. Now take one of the two piles, separate it into two piles and calculate the same sum for these two numbers of objects. Repeat this process until you have n piles with 1 ojbect in each pile. Then add all the sums formed. What will that sum be, as a function of n? Prove this is always the case, using strong induction.

Class 20, Th., Mar. 11, 2010
We went over generating functions. Skip section 9.4.
Do homework through section 9.6.
Turn in homework for chapter 8 on Tuesday.

Class 19, Tue., Mar. 9, 2010
We went over sections 9.2 and 9.3. Do homework for sections 9.2 and 9.3.

Class 18, Th., March 4, 2010
We went over several areas: permutations and permutation notation, derangements, and several new problems. Here is the "100 prisoner problem:"

There are 100 prisoners in cells numbered 1 to 100. The keys, also numbered 1 to 100 are distributed among 100 boxes, also labeled 1 to 100. The prisoners are told that each will be given the chance to open any 50 boxes, looking for their own key. They may communicate - and strategize - before they start opening boxes, but no communication is allowed after the first one starts opening boxes. Once boxes are opened and checked by a prisoner, they are closed back again. If all of them find their own keys, then all will be released. What strategy do they adopt in order to maximize their probability of success?

I mentioned in class that there is a strategy that guarantees success (in the long run) 1-ln(2) = 31.2% of the time!
If you want to see a solution, look at this link to a paper by Peter Winkler.

I also mentioned the "Doodling Theorem:" draw as many doodles as you like on a sheet of paper. A doodle is a closed curve that is allowed to intersect itself, but must "close up" eventually. Several curves may also intersect at the same points. Can you always two-color the regions produced by the doodles so that two regions with the same color never share more than single isolated points along their boundaries? Can you prove this is true in a simple way?

We also played "Bulgarian Solitaire." Make several piles of objects, for example 3 piles of 2,3,and 5 objects. Remove one object from each pile and place them in a new pile. In this case you'll now have four piles of 1,2,3, and 4 objects. Continue in this fashion. What do "fixed points" look like? (These are piles of objects that are stable, for example, two piles with 1 and 2 objects in them, since the next set of objects will also be 1 and 2.) Do all such fixed points have the same structure? Are there any cyles of length 2? Consider the sets of numbers of objects to be vertices, with a directed edge to the vertex that represents the next set of numbers. Given, say, 9 objects, can you construct the directed graph that represents all possible states in the game?

Class 17, Tue., March 2, 2010
We went over the exam, and also the inclusion/exclusion principle (section 8.6), derangements, and introduced recursion relations (section 9.1). We will apply inclusion/exclusion to the problem of derangements, also known as the hat-check problem.We're skipping section 8.7 for now. Do homework thru section 9.1

In a previous class we went over the 6 people at the party problem, an example of Ramsey Theory.
Extra homework: Prove that R(3,4) = 9. That is, if the edges of the graph K9 are colored red and blue, then there is either a red K3 or a blue K4, or vice versa. Hint: see note at the very bottom of this page!

We'll also look at the birthday problem: what is the "break-even" point n for which the probability that n or more randomly chosen people include at least two with the same birthdate with probability more than 50%, and this is not true for fewer than n people?

Class 16, Th., Feb. 25, 2010
Had exam 2, also went over sections 8.4 and 8.5. Do homework thru section 8.5.

Class 15, Tue., Feb. 23, 2010
Went over 5.5, skipped 5.6 for now, also went over sections 8.1, 8.2 and 8.3. Exam will not cover chapter 8.

Class 14, Thu., Feb. 18, 2010
Exam 2 will be on Thursday, Feb. 21.
We went over more about number theory pertaining to the phi function.

Some properties of the phi function:
Phi(n) = the number of elements of {1,2,3,...,n–1} that are relatively prime to n.
Phi(p) = p–1, for p prime.
Phi(p^n) = p^n –p^(n–1), for p prime.
Phi(pq) = (p–1)(q–1), for p and q prime.
Phi(mn) = phi(m) phi(n), for m and n relatively prime.
a^phi(n) = 1, for a relatively prime to n.

We also went over sections 5.3, 5.4, and 5.5.
Do homework through 5.5.

Class 13, Tue., Feb. 16, 2010
We went over sections 5.1 and 5.2, do homework through 5.2.
We also went over solving linear congruences of the form ax == b (mod m).

The American Math Assoc. for Two Year Colleges contest exam is being given on Friday, Feb. 19, from 2:30-3:30 PM. Here is a site with old contest problems.

Class 12, Thu. Feb. 11, 2010
We went over material on codes and also some number theory related to "modular exponentiation."
We also went over sections 4.4 and 4.5 on coloring/scheduling problems and directed graphs. Do homework through section 4.5.
Be prepared to turn in assignments on this logic problem handout and this handout about quantifiers on Tuesday. Some of the logic puzzles are by Raymond Smullyan.

Here's a hat puzzle related to codes:

Three players enter a room and a red or blue hat is placed on each person's head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players' hats but not his own.
No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.

What strategy should the players adopt to insure their success rate is 3/4? How is this related to Hamming codes? What if there are 7 players?

Class 11, Tu., Feb. 9, 2010
We went over graph theory, sections 4.2 and 4.3. Also learned about "six degrees of separation."
Do homework through section 4.3.

Class 10, Th., Feb. 4, 2010
We went over homeowork, also codes and sections 3.6 and 4.1. Do homework for those sections.
Also find the number of (unlabelled) 4 vertex graphs with no multiple edges or loops.

Here are some puzzles by Raymond Smullyan, from his new book, Logical Labyrinths. (Recent quote by Smullyan: "Why should I be worried about dying? It's not going to happen in my lifetime!"):

An anthropologist visits an island where there are two kinds of people: knights, who always tell the truth; and knaves, who always lie.
(1) The anthropologist comes across three inhabitants, A, B, and C. He asks A, "Are you a knight or a knave?" A answers, but indistinctly so the anthropologist can not understand what he says. He then asks B, "What did A say?"B replies, "He said that he is a knave." At this point, C piped up and said, "Dont' believe that; it's a lie." Was C a knight or a knave?
(2) According to another version of the above story, the anthropologist instead asks A how many of the three are knaves. Again A answers indistictly, so he asks B what A has said. B then says that A has said that exactly two are knaves. Then, as before, C says that B is lying. Is it now possible to determine whether C is a knight or a knave?
(3) Next the anthropologist meets two inhabitants, D and E. D says, "Both of us are knaves." What is D and what is E?
Also do homework for sections 3.4 and 3.5.

Class 8, Th., Jan. 28, 2010
We went over exam 1, and also material from section 3.4.
Do homework through section 3.4.

Class 7, Tue., Jan. 26, 2010
We had exam 1, but also went over material from Appendices A.2 and A.3.
Do homework for these sections.

Class 6, Thursday, Jan. 21, 2010
We went over homework and also section 3.3 and (briefly) Appendix A.1.
Exam 1 is on Tuesday and will cover through section 3.3 and Appendix A.1.
Here is a study guide for Exam 1. Bring a Scantron sheet, part of the exam will be multiple choice, part will involve more extended answers (such as doing an induction proof.)

Class 5, Tue., Jan. 19, 2010
We worked on induction problems, and also went over sections 3.1 and 3.2 on congruence and the Euclidean Algorithm.
Do homework through the end of section 3.2.

We are investigatiing whether or when ac ≡ bc (mod m) implies a ≡ b (mod m).
Let d = GCD(c,m).
What conditions on a,b,c, and/or m allow us to say that ac ≡ bc (mod m) implies a ≡ b (mod m)?
What conditions on a,b,c, and m allow us to say that ac ≡ bc (mod m) implies a ≡ b (mod m/d)?
This is an important problem to solve, since we must be able to answer questions like this to know whether the simplest "linear congruence," ax ≡ b (mod m) has a solution. (A linear congruence is the equivalent, in a sense, of the linear equation ax=b for algebra of real numbers.)
Keep investigating!

Class 4, Th., Jan. 14, 2010
We went over sections 2.5, 2.6, and 3.1. Do homework through the end of chapter 3.1. Chapter 2 homework will be due to turn in next Thursday. Do the homework problem below listed under class 3 also.

Class 3, Tue., Jan. 12, 2010
We went over homework on equivalence relations, and introduced partial orders (section 2.3), including the lexicographic order and topological sorting. We also went over material from section 2.4 on functions, including one-to-one and onto functions.
Do homework through end of section 2.4. Also do the following problem:
Find a function from Z to Z which is onto and one-to-one.
Find a function from Z to Z which is onto but not one-to-one.
Find a function from Z to Z which is not onto but is one-to-one.
Find a function from Z to Z which is neither onto nor one-to-one.

Class 2, Th., Jan. 7, 2010
We went over sections 1.4, 2.1, and 2.2.
Do homework for sections 1.4-2.2.
We also went over the definition of "big oh" notation.
Here's more detail on Horner's method.
Here's more on Russell's Paradox.

Class 1, Tue., Jan. 5, 2010
We went over sections 1.1, 1.2, and 1.3.
Here's the list of all HW problems. Work all HW through section 1.3 for Thursday.

Homework List:
Ch. 1.1 # 7,15,19
Ch. 1.2 # 6,16,18,25,26,31,32
Ch. 1.3 # 1,9,13,14,19,20,25,29
Ch. 1.4 # 9,11,17,23-27,31,32
Ch. 2.1 #1,13,15,17,26,31,33,39
Ch. 2.2 Odd problems #1-19,18,20, 22, 23,29
Ch. 2.3 1,3,5,9,13,18,19,21,32,40,41,42
Ch. 2.4 #1,3,5,9,13,18,19,21,32,40,41,42
Ch. 2.5 # 6,8,12,17,19,26
Ch. 2.6 # 5,16,27,37,43,45,47,48
Ch. 3.1 # 4,5,9,10,20,32,33,39,41,42,47,,48,51
Ch. 3.2 # 1,5,11,19,23
Ch. 3.3 # 1,5,11,15,31-37
Appendix A.1 # 1,7,12,16-18,22,25,29,33,35Appendix A.2 # 1,9,15,21,30-34
Appendix A.3 # 3,11,15,16,18,20,23,26Ch. 3.4 # 1,9,11,23,33
Ch. 3.5 # 5,9,15,19,29,31,35,40
Ch. 3.6 # 1,5,9,13,17,29,31,33,35,40,41,44
Ch. 4.1 # 1,7,8,16,18,21,22,24,26,31,35,42,47,48,53,54
Ch. 4.2 #19,23,31-34,40,41,50,52,61,63
Ch. 4.3 # 7,11,13,17
Ch. 4.4 # 4,6,10,11,12,13,20,22,25,2932,34.35(a little difficult)
Ch. 4.5 # 5,11,17,19,24,27,45,63,67,73,77Ch. 5.1 # 13,18,25,27,32,37
Ch. 5.2 # 5,9 (we will definitely discuss this one in class!), 12,13,15,19,37,39
Ch. 5.3 # 3,25,28,32,33
Ch. 5.4 #13,16,21,30
Ch. 5.5 #5,16,22,28,35,41,45,53,56,64
Ch. 8.1 # 3,9,19,27,28
Ch. 8.2: # 1,3,6,7,15,19,23,34-36
Ch. 8.3: # 3,14,16,19-21,31,33-35
Ch. 8.4: # 3,9,10,15,30,34,36
Ch. 8.5: # 5,9,13,17,26,29-38
Ch. 8.6 #5,9,17,26,29-38
Ch. 9.1 #1,7,18,26,31
Ch. 9.2 # 1,2,5,11,27,34
Ch. 9.3 #1,14,23,25,
Skip Ch. 9.4 # 11,13,15,19,25
Ch. 9.5 # 3,8,13,27,31
Ch. 9.6 # 1,3,11,15,21,27
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Hint to R(3,4) problem: Consider two possibilities for a random vertex v: v has 5 blue edges from it to its neighbors, or v has 6 or more blue edges.