Lecture 2 Summary: notation and modeling

Logistics

Definition: Set

Definition: a set is an unordered collection of elements. Two sets are considered the same if they contain the same elements; the order and duplication of elements is ignored. We write e ∈ S to indicate that e is an element of the set S (similarly e ∉ S indicates that e is not an element of S).

(Note: ("is an element of", \in in LaTeX) is different from e or E (the letter e) and ϵ (the greek letter epsilon, LaTeX \epsilon).)

There are many ways to write down sets.

You have written down a set if the reader can unambiguously determine whether a given element is or is not in the set.

Examples:

Definition: Function

A function f from a set A to a set B is an unambiguous rule, which gives an element of B for each element of A. A is called the domain of f, while B is called the codomain of f. We often write
f : A → B to represent the statement "f is a function from A to B".

As with sets, there are many ways to write down functions. The key rule is that everyone agrees (unambiguously!) what the output is for a given input (the input is the element of the domain, the output is the element of the codomain).

The domain and codomain are different: every element of the domain must have a corresponding element of the codomain, but an element of the codomain may have any number (including 0) of elements of the domain that map to it.

Some notation

Case study: Sudoku

We started modeling sudoku solutions. Here's where we are so far:

Let the set of numbers N = {1, 2, …, 9}. Let the set of cells C = N × N.

A sudoku solution f is a function f : C → N satisfying the following properties:

  1. For all i, j, and j, if j ≠ j then f(i, j)≠f(i, j′)
  2. For all i, i, and j, if i ≠ i then f(i, j)≠f(i′,j)
  3. For all c1 and c2 ∈ C, if c1 ≠ c2 and square(c1)=square(c2), then f(c1)≠f(c2).

It's often easier to work with the contrapositives (this is an equivalent definition):

  1. For all i, j, and j, if f(i, j)=f(i, j′) then j = j
  2. For all i, i, and j, if f(i, j)=f(i′,j) then i = i
  3. For all c1 and c2 ∈ C with square(c1)=square(c2), if f(c1)=f(c2) then c1 = c2.

Here square is a function that gives the number of the large box that contains the cell. Here are three styles of definition of square; decide which one you like best (my favorite is the first):

  1. square(i, j) is the number of the large box that contains the (i, j) cell, where we number the large 9x9 grids as follows:
1 4 7
2 5 8
3 6 9
  1. square(i, j) is the (i, j)th entry in the following table:
1 1 1 4 4 4 7 7 7
1 1 1 4 4 4 7 7 7
1 1 1 4 4 4 7 7 7
2 2 2 5 5 5 8 8 8
2 2 2 5 5 5 8 8 8
2 2 2 5 5 5 8 8 8
3 3 3 6 6 6 9 9 9
3 3 3 6 6 6 9 9 9
3 3 3 6 6 6 9 9 9
  1. (using integer division) square : C → N given by $square : (i,j) \mapsto 1 + \left\lfloor(i - 1)/3\right\rfloor + 3 \cdot \left(1 + \lfloor(j-1)/3\right\rfloor\right)$

A Proof

As you play sudoku, you are generating in your head lots of proofs. This is overkill, but let's try to write down one such proof as an example of proving something from definitions.

Suppose we are working on the following puzzle:

8
1 5 7
3 7 1 2 6 8
1 6 5 3
5 9
2 1
5 3 7 1 4
8 7
4 _ 9

It is clear that the last entry of the third column (cell (9,3)) must be 1. Let's prove it.

Claim: if f is a sudoku solution compatible with the above board (that is, if f(2, 2)=1, f(1, 5)=8, f(2, 5)=5 and so on), then f(9, 3)=1.

Proof: There must be a 1 somewhere in box 3 (in other words, there exists some (i, j) such that square(i, j)=3 and f(i, j)=1). By rule 2, j cannot be 1, because f(4, 1)=1. Similarly, since f(2, 2)=1, j cannot be 2. Thus j must be 3; and f(7, 3)≠1, and f(8, 3)≠1. The only cell left is (9, 3), so f(9, 3) must be 1.

Question: is this proof good? (Hint)