DLM 310- Quiz 2

Points to Ponder 7: 4. How many ways can you arrange the numbers {1, 2, 3, 4, 5} in a list if the first number has to be the 3?

1 x 4 x 3 x 2 x 1 = 4! = 24

Let's agree that we will always use the following rules to write a permutation in cycle notation.

1. Every individual cycle should list the smallest label in the cycle first. 2. No label should appear in more than one cycle. (We say the cycles are disjoint.) 3. The first label in a cycle should be smaller than the first label in any subsequent cycle. 4. We will not list any labels that stay fixed, unless they all stay fixed. We will represent the permutation with all labels remaining fixed as (1)

Points to Ponder 7: 5. How many ways can you arrange the numbers {1, 2, 3, 4, 5} if the first number must be 3 and the last number must be the 4?

1 x 3 x 1 x 2 x 1 = 3! = 6

Points to Ponder 7: Practice 15 puzzle permutations If the permutation is (1 4) (2 3 11 13 9 5 12 15 14 10) (6 8 7). There are three cycles with lengths 2, 10, and 3. We add the numbers 1+9+2 to get 12. What is your conjecture?

(1 8 3 11 5 9 6 2 14 12 10) (4 15 13) = 12 (1 8 2 5 15 3 4 13 12 9 11 10 6 14 7) = 15 (1 11 15 3 5 2 14) ( 4 10 12 6 9 13 8) = 12 there are almost always 12 gaps if it's not a 15 cycle

Points to Ponder 4: 8. Learn the rules for the puzzle Sudoku and solve at least three different puzzles.

- #s 1-9 appear exactly once in each row and columns of 9 - #s 1-9 appear once in each cage

Describe the permutation in S6 represented by (1 6 3) (2 5). Describe the permutation in S6 represented by (6 3 1) (5 2). Describe the permutation in S6 represented by (2 5) (3 1 6).

- 1 moves to 6; 6 moves to 3; 3 moves to 1; 2 and 5 swap places - they are all the same

Squarely

- constructed on a 5 x 5 grid with fewer clues than a Sudoku. - The digits 1-9 fill the grid so that no number is repeated in any single row or column or long diagonal but since there are only five cells per row (column, diagonal), you must determine which five digits appear. - One digit (the loner) appears only once in the grid. The other 8 digits appear three times each to fill the 25 cells. - In the easiest puzzles, the set of numbers is given for each row, column, diagonal. - You simply have to determine the location of each number. In more difficult puzzles you must deduce which numbers go in these sets.

Prufer's Algorithm

- converts an ordered list consisting of 7 positive integers (not necessarily distinct) less than 10 into a weighted tree. - provides a one-to-one matching between a tree with vertices labeled {1,2,... n} and an ordered list of length n-2 integers chosen from the set of labels. - numbers may be repeated - Changing the numbers in the list will create a different tree. - Keeping the same numbers but changing the order of the numbers in the list will create a different tree.

A permutation can also be thought of as describing a motion of labeled objects. What motion would the following permutation represent?

- objects 2, 4, 5, 6 stays fixed - objects 3 and 1 swap places

Points to Ponder 6: 5. 1. What is the prime factorization of 1680?

1680 = 8*210 = 8*7*6*5 = 2^4*3*5*7

Points to Ponder 6: 3. Which congruence class mod 9 will contain the integer 2024?

2024 = 9*224 +8 so 2024 is in the congruence class [8] mod 9. You can also know this since the sum of the digits is 8.

What are some rules for divisibility?

2: ends in 0, 2, 4, 6, 8, (even #s) 3: sum of digits divisible by 3 4: last two digits combined are divisible by 4 > 256924 -> (9+2+4) = 24/4 = 6 5: if and only if it ends in a 0 or 5 6: if and only divisible by 2 and 3 7: if he tells us it is, must be decided recursively 8: if and only if the last 3 digits form a number divisibly by 8 9: if the sum of the digits are divisible by 9

How many different permutations are there on the set {1, 2, 3, 4}?

4 x 3 x 2 x1 = 4! factorial = 24

Points to Ponder 7: 3. How many ways can you arrange the numbers {1, 2, 3, 4, 5} in a list?

5 x 4 x 3 x 2 x 1 = 5! = 120

Points to Ponder 5: Find the unique set of five distinct positive integers less than 10 with the given sum and product having the given prime factorization.

576- {1 , 3, 4, 6, 8} 720- {1, 3, 5, 6, 8} 2592- {2, 3, 5, 6, 9} 6048- {2, 6, 7, 8, 9} 9072- {3, 6, 7, 8, 9}

The Fundamental Theorem of Arithmetic says that every integer greater than 1 may be factored uniquely into a product of primes to powers.

6 = 2 * 3 31 = 31 100 = 4 * 25 = 2^2 * 5^2

You can also do some geometric motion called symmetries on the puzzle to make it look different. How many different symmetries are there on a square?

8 Rotation: R0, R90, R180, R270 Horizontal Flip: H Vertical Flip: V Diagonal Flip: D/ , D\

How can permutations be used to create new Sudoku or Squarely puzzles from existing ones? permute by ( 1 2 3 4 5) (8 9 )

9! = everywhere you see a 1 change it to a 2, etc...

Latin Square

A Latin square of order n is an n x n square with each of the numbers 1 - n appearing exactly once in each row and each column. - ex: sudoku > every sudoku is a Latin square but not every Latin square is a sudoku

Permutation

A permutation on the set {1, 2, 3, ... , n} is a one-to-one function from the set onto itself. - One way to describe such a function is in terms of a table of values with the top row being the domain of the function and the bottom row being the function values - each way you fill in the bottom row with the numbers 1, 2, 3, and 4 is a different permutation - # of permutations of n objects is n! > n x (n-1) x (n-2).... - can also be thought of as an arrangement of a set of labeled objects.

congruences classes

Definition: Let n be an integer greater than 1 and k be any integer in the set {0, 1, ... , n-1}. We define the congruence class of k mod n to be the set of all integers that leave a remainder of k when divided by the modulus n. We use the notation [k]n to represent this set of integers. - the difference between any two numbers in the same congruence class will always be a multiple of the modulus

Instant Insanity Puzzle

Four cubes: The color of each face of each cube is either red, white, blue, or green. Goal: Stack the four cubes so that you can see all four colors on each side of the stack.

Points to Ponder 7: 7. Analyze the two-player impartial game described as follows. Set-up: Markers are placed in an L shape in a rectangular grid. A move consists of removing all the markers in any single row or column. The player who clears the board is the winner. For example, player 1 has seven options available. She could take all the markers in the first column, or the marker in the second column, or the marker in the third column, or all the markers in the bottom row, or the marker in the second row, or the marker in the third row, or the marker in the fourth row. Would you want to be player 1 or player if faced with this position? Try this game with different numbers of markers and different arrangements. For example: How would this play out if you had a big plus sign with five markers in column 3 and five markers in row 3

I would want to be player 1. I would remove the very top row so that the two sides of the L have the same length.

Points to Ponder 5: 1. When solving a Squarely puzzle explain why the information provided in any single row or any single column is completely redundant. If only four rows are given, how do you know the numbers in the omitted row? Is the same true if you omit one of the diagonals?

If some digit does not appear at all in the 4 given rows(columns), it must be the loner and it must appear in the hidden row(column). If some digit appears exactly twice in the 4 given rows (columns), the third occurrence must be in the hidden row(column). This is not true if a diagonal is omitted.

Points to Ponder 6: 4. Explain how you know the number 12345678987654321 is divisible by 9.

The sum of the digits is 81 which is divisible by 9.

Points to Ponder 5: 3. Try to do Squarely # 1 from the notes if you hide the numbers in the bottom two rows.

There are two options (4 or 8) for the bottom two squares in column 3

Points to Ponder 5: 2. Try to do Squarely # 1 from the notes if you hide the numbers in the first diagonal. Is there a unique solution or more than one option?

There is still a unique solution

List elements in [0]5, [1]5, [2]5, [3]5, [4]5

[0]5 = all integers have remainder 0 when divided by 5 - (-5, 0, 5, 10, 15...) = 5x + 0 [1]5 = (-4, 1, 6, 11, 16...) = 5x +1

How many different congruence classes are there mod 6?

[0]6 [1]6 [2]6 [3]6 [4]6 [5]6

Points to Ponder 6: 1. List all the congruence classes mod 6. (Hint: What are the possible remainders when you divide any integer by 6?)

[0]6, [1]6, [2]6, [3]6, [4]6, [5]6

Points to Ponder 4: 5. Give a logical argument to explain why any tree with 9 vertices and exactly 5 leaves must have at least one vertex with degree 3 or more.

a tree with 9 vertices 8 edges so must have total degree = 16 - the 5 leaves have a total degree of 5, so the remaining 4 vertices have total degree = 11 - if some of these 4 had degree > 2, the total would not be 11

Points to Ponder 7: 1. Write each of the following permutations using standard cycle notation.

a. (3 9 4 7) (5 8 6) b. 3 9 4 7) (5 8 6) c. ( 1 2 4 6) (7 9) d. ( 1 5 8 6 7) (2 4 9)

Points to Ponder 4: 2. Construct the Prufer list for each given labeled tree.

a. [2 2 2 1 8 9 1] b. [2 6 8 9 7 9 ] c. [5 5 4 4 5 3 3]

Points to Ponder 7: 2. If α = (1 2 3 4 5 6), write all the powers of α in standard cycle form. α 2 = α 3 = α 4 = α 5 = α 6 = α 7 = Write a sentence connecting the powers of α and congruence classes.

aI = aj if and only if I and j are in the same congruence class mod 6.

Points to Ponder 4: 3. Do you see a connection between the degree of a vertex in a tree and a Prufer list?

degree = one more than the # of appearances in the Prufer list

Let α = (1 2 3 4)(5 6) in S6. What permutation is the result of doing α twice? We write α 2 = Let α = (1 2 3 4 5) in S5. This is a cycle of length 5. How many different permutations do you get when you raise α to powers? α 2 = α 3 = α 4 = α 5 = α 6 = α 7 = α 8 = How is this related to congruence classes?

it's the same congruence class mod 5

The set of all permutations on {1, 2, 3, ... , n} is denoted by Sn. The number of permutation in Sn is

n!

Points to Ponder 6: 2. Using the 5 congruence classes mod 5 to construct the mod 5 multiplication table shown below. If you ignore the 0 class, the table for the non-zero classes formed a 4 by 4 Latin square. Ignoring the 0 class in each case, make the multiplication table mod 6, mod 7, and mod 8. (This will go faster if you just drop off all the square brackets.) Which tables formed Latin squares? Make a guess about which moduli in general will have multiplication tables that form Latin squares and which will not. Write your guess as an if-then statement by filling in the blank. If the modulus is ___________________________, then the multiplication table for non-zero congruence classes will be a Latin Square.

prime

cycle notation

provides a more compact way to represent a permutation than having to use a table.

Points to Ponder 4: 4. What is the connection between the number of edges in any graph and the sum of the degrees of all its vertices? This should apply to graphs in general, not just trees.

total degree = twice the number of edges