CS202 Fall 2011 Lecture (Lost Count) – 12/1 Relations

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Relations CS202 Fall 2011 Lecture (lost count) – 12/1 Recall the definition of the Cartesian (Cross) Product: The Cartesian Product of sets A and B, A x B, is the set Relations A x B = {(a,b) : a∈A and b∈B}.

Prof. Tanya Berger-Wolf aRb or (a,b) ∈ R means “a is related to b” A relation is just any subset of the Cartesian Product: R ⊆ AxB

 Ex1: A = {0,1,2}, B = {2,3} => AxB = {(0,2), (0,3), (1,2), (1,3), (2,2), (2,3)} R = {(a,b) | a < b}. So R = {(0,2), (0,3), (1,2), (1,3), (2,3)} = AxB-{(2,2)}

 Ex2: A = students at UIC; B = courses at UIC. R = {(a,b) | student a is enrolled in class b}

 Ex3: A = {3 letter strings}, B = {all English words} R = {(a,b) | a is a prefix of b}

Relations and Functions Properties of Relations

Recall the definition of a function: f = {(a,b) : b = f(a) , a∈A and b∈B}

Is every function a relation? Yes, a function is Reflexivity: a special kind of A relation R on AxA is reflexive if for all a∈A, (a,a) ∈R. relation.

Symmetry: Draw Venn diagram of cross products, relations, Cross product A relation R on AxA is symmetric if functions (a,b) ∈ R implies (b,a) ∈ R .

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Properties of Relations Properties of Relations - techniques…

How can we check for transitivity? Draw a picture of the relation (called a “graph”) [Epp p. 580]. Transitivity: Vertex for every element of A A relation on AxA is transitive if Edge for every element of R (a,b) ∈ R and (b,c) ∈ R imply (a,c) ∈ R.

R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)}

Anti-symmetry [Epp p.632]: A relation on AxA is anti-symmetric if (a,b) ∈ R implies (b,a) ∉ R. 1 A “short cut” 2 must be present for EVERY path of length 2. 3 4

Properties of Relations - techniques… Properties of Relations - techniques…

How can we check for the reflexive property? How can we check for the symmetric property? Draw a picture of the relation (called a “graph”). Draw a picture of the relation (called a “graph”). Vertex for every element of A Vertex for every element of A Edge for every element of R Edge for every element of R

R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)}

1 Loops must exist 1 EVERY edge 2 on EVERY vertex. 2 must have a return edge.

3 3 4 4

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Properties of Relations - techniques… Properties of Relations - techniques…

How can we check for the anti-symmetric property? Let R be a relation on People, Draw a picture of the relation (called a “graph”). R={(x,y): x and y have lived in the same country} Vertex for every element of A Edge for every element of R ?

1 2 R={(1,1),(1,2),(1,3),(1,4),(2,2),(2,3),(2,4),(3,3),(3,4),(4,4)} ? ? 3

Is R transitive? Is it symmetric? 1 NO edge can No Yes 2 have a return edge. Is it reflexive? Yes Is it anti-symmetric? No 3 4

Properties of Relations - techniques… Properties of Relations - techniques…

Let R be a relation on positive integers, Let R be a relation on positive integers, R={(x,y): 3|(x-y)} R={(x,y): 3|(x-y)}

Suppose (x,y) and (y,z) are in R. Is (x,x) in R, for all x? Definition of Definition of Then we can write 3j = (x-y) and 3k = (y-z) Does 3k = (x-x) for some k? “divides” “divides” Can we say 3m = (x-z)? Is (x,z) in R? Yes, for k=0.

Add prev eqn to get: 3j + 3k = (x-y) + (y-z)

3(j + k) = (x-z)

Is R transitive? Yes Is R transitive? Yes

Is it reflexive? Yes

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Properties of Relations - techniques… Properties of Relations - techniques…

Let R be a relation on positive integers, Let R be a relation on positive integers, R={(x,y): 3|(x-y)} R={(x,y): 3|(x-y)}

Suppose (x,y) is in R. Suppose (x,y) is in R. Definition of Definition of Then 3j = (x-y) for some j. “divides” Then 3j = (x-y) for some j. “divides” Does 3k = (y-x) for some k? Does 3k = (y-x) for some k?

Yes, for k=-j. Yes, for k=-j.

Is R transitive? Yes Is it symmetric? Yes Is R transitive? Yes Is it symmetric? Yes

Is it reflexive? Yes Is it reflexive? Yes Is it anti-symmetric? No

More than one relation More than one relation Let R be a relation from A to B (R ⊆ AxB), and let S be a relation from B to C (S ⊆ BxC). The composition of R and S is the relation from A to C (S°R ⊆ AxC): Suppose we have 2 relations, R1 and R2, and recall that relations are just sets! So we can take unions, intersections, complements, S°R = {(a,c): ∃ b∈B, (a,b) ∈ R, (b,c) ∈ S} symmetric differences, etc. There are other things we can do as well…

A B C R S 1 x s 2 y t

3 z u

4 v

S°R = {(1,u),(1,v),(2,t),(3,t),(4,u)}

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More than one relation More than one relation

Let R be a relation on A. Inductively define Let R be a relation on A. Inductively define R1 = R R1 = R Rn = Rn-1 ° R Rn = Rn-1 ° R

A A A A A A R R1 R R2 1 1 1 1 1 1 2 2 2 2 2 2

3 3 3 3 3 3

4 4 4 4 4 4 … = R4 = R5 R3 = R2°R = {(1,1),(1,2),(1,3),(2,3),(3,3),(4,1),(4,2),(4,3)} R2 = R1°R = {(1,1),(1,2),(1,3),(2,3),(3,3),(4,1), (4,2)} = R6…

Relations - A Theorem: Relations - A Theorem:

If R is a transitive relation, then Rn ⊆ R, ∀n.

Proof by induction on n. If R is a transitive relation, then Rn ⊆ R, ∀n. Base case (n=1): R1 ⊆ R because by definition, R1 = R. Inductive Step: IH: if R is transitive, then Rn-1 ⊆ R. Prove: if R is transitive, then Rn ⊆ R. Aside: notice that this theorem allows us to conclude that Typical the previous relation was NOT transitive. We are trying to prove that Rn ⊆ R. To do this, we select an way of Recall: “if p then q” ≡ “if not q then not p.” element of Rn and show that it is also an element of R. proving subset. We saw that Rn was not a subset of R (it was growing on Let (a,b) be an element of Rn. Since Rn = Rn-1 ° R, we know every iteration). there is an x so that (a,x) ∈ R and (x,b) ∈ Rn-1. Therefore, R is not transitive. By IH, since Rn-1 ⊆ R, (x,b) ∈ R.

But wait, if (a,x) ∈ R, and (x,b) ∈ R, and R is transitive, then (a,b) ∈ R.

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Relations - more techniques… Relations - more techniques…

Suppose we have our old relation R on AxB, where A={1,2,3,4}, and B={u,v,w}, R={(1,u),(1,v),(2,w),(3,w),(4,u)}. Some things to think about. Then we can represent R as: Let R be a relation on a set A, and let MR be the matrix representation of R. u v w The labels on Then R is reflexive if, ______. the outside are 1 1 1 0 for clarity. It’s really the

2 0 0 1 matrix in the u v w A. All entries in MR are 1. middle that’s 3 0 0 1 B. The \ diagonal of MR contains important. u 1 1 0 only 1s. 4 1 0 0 v 0 1 1 C. The first column of MR contains no 0s. This is a |A| x |B| matrix whose entries indicate membership w 0 0 1 D. None of the above. in R.

Relations - more techniques… Relations - more techniques…

Suppose we have R1 and R2 defined on A:

R1 u v w R2 u v w Some things to think about. u 1 0 1 u 1 1 0 Let R be a relation on a set A, and let MR be the matrix representation of R. Then R is symmetric if, ______. v 0 0 1 v 0 1 1 w 1 1 0 w 0 0 1 1 1 1 A. All entries above the \ are 1. Then R ∪ R is the bitwise “or” of the entries: 0 1 1 u v w 1 2 B. The first and last columns of 1 1 1 MR1∪R2 = MR1 v MR2 u 1 0 1 MR contain an equal # of 0s. C. MR is visually symmetric about Then R ∩ R is the bitwise “and” of the entries: 1 0 0 v 0 0 1 the \ diagonal. 1 2 M = M ∧ M 0 0 1 D. None of the above. R1∩R2 R1 R2 w 1 1 0 0 0 0

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Closure Closure

Consider relation R={(1,2),(2,2),(3,3)} on the set A = {1,2,3,4}. Definition: Is R reflexive? The closure of relation R on set A with respect to property P is the relation R’ with No 1. R ⊆ R’

2. R’ has property P What can we add to R to make it reflexive? 3. ∀S with R ⊆ S and S has property P, R’ ⊆ S. (1,1), (4,4)

R’ = R U {(1,1),(4,4)} is called the reflexive closure of R. P-Closure for a If relation R relation might has property P not exist! then R’ = R.

Reflexive Closure Symmetric Closure

Let s(R ) denote the symmetric closure of relation R. Let r(R ) denote the reflexive closure of relation R. Then s(R ) = R U { (b,a): (a,b) ∈ R } Then r(R ) = R U { ( a,a ): ∀ a ∈ A }

Fine, but does that satisfy the definition? Fine, but does that satisfy the definition?

1. R ⊆ s(R ) We added edges! 1. R ⊆ r(R ) We added edges!

2. s(R ) is symmetric By defn 2. r(R ) is reflexive By defn

3. Need to show that for any S with particular properties, 3. Need to show that for any S with particular properties, s(R ) ⊆ S. r(R ) ⊆ S. Let S be such that R ⊆ S and S is symmetric. Then Let S be such that R ⊆ S and S is reflexive. Then {(b,a): (a,b) ∈ R } ⊆ S (since S is symmetric) and R⊆S (given). {(a,a): ∀ a ∈ A } ⊆ S (since S is reflexive) and R⊆S (given). So, s(R ) ⊆ S. So, r(R ) ⊆ S.

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Transitive Closure (Epp p.587) Transitive Closure

Let c(R ) denote the transitive closure of relation R. So how DO we find the transitive closure? Then c(R ) = R U { (a,c): ∃ b (a,b),(b,c) ∈ R } Draw a graph. Example: A={1,2,3,4}, R={(1,2),(2,3),(3,4)}.

Example: A={1,2,3,4}, R={(1,2),(2,3),(3,4)}. Apply definition to get: 1 c(R ) = {(1,2),(2,3),(3,4), (1,3), (2,4) } 2

3 Which of the following is true: 4 a) This set is transitive, but we added too much. b) This set is the transitive closure of R. Define a path in a relation R, on A to be a sequence of elements from A: a,x ,…x ,…x ,b, with (a, x ) ∈ R, ∀i (x ,x ) ∈ R, (x ,b) ∈ R. c) This set is not transitive, one pair is missing. 1 i n-1 1 i i+1 n-1 d) This set is not transitive, more than 1 pair is missing. “Path from a to b.”

Transitive Closure Equivalence Relations

Example:

Let S = {people in this classroom}, and let Formally: R = {(a,b): a has same # of coins in his/her bag as b} If t(R) is the transitive closure of R, and if R contains a path from a to b, then (a,b) ∈ t(R) This is a special kind of relation, characterized Quiz time: by the properties it has. Note: Yes What’s special about it?  Later classes (401) will give you efficient algorithms for Is R reflexive? determining if there is a path between two vertices in a Is R symmetric? Yes Everyone graph (graph connectivity problem) with the same Is R transitive? Yes # of coins as you is just like you.

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Equivalence Relations Equivalence Relations

Formally: Example: Relation R on A is an equivalence relation if R is S = Z (integers), R = {(a,b) : a ≡ b mod 4} Reflexive (∀ a ∈ A, aRa) Is this relation an equivalence relation on S? Symmetric (aRb --> bRa) Transitive (aRb ^ bRc --> aRc) Start by thinking of R a different way: aRb iff there is an int k so that a = 4k + b. Your quest becomes one of finding ks.

Example: Let a be any integer. aRa since a = 4⋅0 + a. S = Z (integers), R = {(a,b) : a ≡ b mod 4} Consider aRb. Then a = 4k + b. But b = -4k + a. Is this relation an equivalence relation on S? a%4 = b? Consider aRb and bRc. Then a = 4k + b and b = 4j + c. So, a = 4k + 4j +c = 4(k+j) + c. Have to PROVE reflexive, symmetric, transitive.

Equivalence Relations Equivalence Classes

Example: Example: S = people in this room, Back to coins in bags. R = {(a,b) : total $ on a is within $1.00 of total $ on b} Is this relation an equivalence relation on S?

? Definition: Let R be an equivalence relation on S. The equivalence class of a ∈ S, [a]R, is

[a]R = {b: aRb}

Clearly reflexive and symmetric. Is it transitive? a is just a name for the equiv class. Any member of the class is a representative.

a) Yes, I can give a proof. b) Yes, I think so, but I can’t prove it. c) No, I can give a proof. d) No, I don’t think so, but I can’t prove it.

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Equivalence Classes Equivalence Classes

What equivalence relation we’ve seen recently has representatives Definition: Let R be an equivalence relation on S. The equivalence

[244], [7], [58], [1]? class of a ∈ S, [a]R, is

[a]R = {b: aRb} Notice this is just a subset of S.

What does the set of equivalence classes on S look like?

To answer, think about the relation from before: S = {people in this room} R = {(a,b) : a has the same # of coins in his/her bag as b} In how many different equivalence classes can each person fall?

Equivalence Classes Equivalence Classes

Lemma: Let R be an equivalence relation on S. Then Lemma: Let R be an equivalence relation on S. Then 1. If aRb, then [a]R = [b]R 1. If aRb, then [a]R = [b]R 2. If not aRb, then [a]R ∩ [b]R = ∅ 2. If not aRb, then [a]R ∩ [b]R = ∅

Proof: Proof:

1. Suppose aRb, and consider x ∈ S. 2. Suppose to the contrary that ∃ x ∈ [a]R ∩ [b]R.

x ∈ [a] ⇔ aRx x ∈ [a] x ∈ [b] ⇔ aRx and bRx R Defn of [a]R R R ⇔ xRa symmetry ⇔ aRx and xRb ⇔ xRb transitivity ⇔ aRb, contradicting “not aRb” ⇔ bRx symmetry ⇔ x ∈ [b] S = [a] Thus, [a]R and [b]R are either R Defn of [b]R  R identical or disjoint. a ∈S

€

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Equivalence Classes Equivalence Classes

So S is the union of disjoint equivalence classes of R. Give me a partition of the reals into 2 blocks:

A partition of a set S is a (perhaps infinite…or uncountably infinite) collection of sets {Ai} with

Each Ai non-empty Give me a partition of the reals into 5 blocks: Each Ai ⊆ S For all i, j, Ai ∩ Aj = ∅

S = ∪Ai Each Ai is called a block of the partition.

Equivalence Classes Equivalence Classes

Theorem: if R is a _____ S, then {[a]R : a ∈ S} is a _____ S. Theorem: if {Ai} is any partition of S, then there exists an equivalence relation R, whose equivalence classes are A. Partition of, equivalence relation on exactly the blocks Ai. B. Subset of, equivalence class of C. Relation on, partition of Proof If {Ai} partitions S then define relation R on S to be D. Equivalence relation on, partition of

E. I have no clue. R = {(a,b) : ∃ i, a ∈ Ai and b ∈ Ai} Next show that R is an equivalence relation. Reflexive and symmetric. Transitive? Theorem: if R is an equivalence relation on S, then {[a] : a ∈ R Suppose aRb and bRc. Then a and b are in A , and b S} is a partition of S. i and c are in Aj. Proof: we need to show that an equivalence relation R But b ∈ A ∩ A , so A = A . satisfies the definition of a partition. (we’ve spent the i j i j So, a, b, c ∈ A , thus aRc. whole day doing this!) i

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Partially Ordered Sets (POSets) Partially Ordered Sets (POSets)

Let R be a relation then R is a Partially Ordered Set (POSet) if it is Ex. (Z+, | ), the relation “divides” on positive integers.

Reflexive - aRa, ∀a Reflexive? Transitive - aRb ∧ bRc → aRc, ∀a,b,c Yes, x|x since x=1x (k=1)

Antisymmetric - aRb ∧ bRa → a=b, ∀a,b Transitive? a|b means b=ak, b|c means c=bj. Yes, or No? Does c=am for some m?

c = bj = akj (m=kj)

Ex. (R,≤), the relation “≤” on the real numbers, is a partial order. Antisymmetric? a|b means b=ak, b|a means a=bj. But b = bjk (subst) only if jk=1. Reflexive? = a ≤ a for any real How do you check? jk=1 means j=k=1, and we Transitive? order If a ≤ b, b ≤ c then a ≤ c have b=a1, or b=a Actually total Antisymmetric? order. < If a ≤ b, b ≤ a then a = b

Partially Ordered Sets (POSets) Partially Ordered Sets (POSets)

A poset. Ex. (Z, | ), the relation “divides” on integers. Not a poset. Ex. (2S, ⊆ ), the relation “subset” on set of all subsets of S.

Reflexive? Reflexive? Yes, x|x since x=1x (k=1) Yes, A ⊆ A, ∀A∈ 2S

Transitive? Transitive? a|b means b=ak, b|c means c=bj. A ⊆ B, B ⊆ C. Does that mean A ⊆ C? Does c=am for some m? A ⊆ B means x ∈A → x ∈B c = bj = akj (m=kj) B ⊆ C means x ∈B → x ∈C

Antisymmetric? Now take an x, and suppose 3|-3, and -3|3, but 3 ≠ -3. it’s in A. Must it also be in C? Yes,Yes, by… by MP Yes, or No? A. Modus Ponens B. Modus Tollens Antisymmetric?C. DeMorgan’s A ⊆ B, B ⊆ A → A=B D. Transitivity

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Partially Ordered Sets (POSets) Partially Ordered Sets (POSets)

n When we don’t have a special relation definition in mind, we use the Ex. A common partial order on bit strings of length n, {0,1} , is defined as: A. 0110 ≤ 1000 symbol “≤” to denote a partial order on a set. a1a2…an ≤ b1b2…bn B. 0110 ≤ 0000 If and only if ai ≤ bi, ∀ i. C. 0110 ≤ 1110 When we know we’re talking about a partial order, we’ll write “a ≤ b” instead of 0110 and 1000 are “incomparable” … We can’t tell “aRb” when discussing members of the relation. D. 0110 ≤ 10111 which is “bigger.”

Actually, this relation is exactly the same as the last example, (2S, ⊆ ).

We will also write “a < b” if a ≤ b and a ≠ b. In the string relation, we said 00≤01 because every bit in 00 is less than or = the corresp bit in 01. String on the right has at least all the 1 bits of the left, maybe more. If each 1 represents an element in S, then right side has all elts of the left, maybe more.

Partially Ordered Sets (POSets) Hasse Diagrams

Let (S, ≤ ) be a PO. If a ≤ b, or b ≤ a, then a and b are Hasse diagrams are a special kind of graphs used to describe posets. comparable. Otherwise, they are incomparable.

Ex. In poset (Z+, |), 3 and 6 are comparable, 6 and 3 are Ex. In poset ({1,2,3,4}, ≤), we can draw the following picture comparable, 3 and 5 are not, 8 and 12 are not. to describe the relation.

A total order is a partial order where every pair of elements 4 is comparable. Dictionary order, or 3 Ex. (Z+, ≤), is a total order, because alphabetic order, or 1. Draw edge (a,b) if a ≤b for every pair (a,b) in ZxZ, lexicographic order is a

either a ≤ b, or b ≤ a. partial order on words in 2 2. Don’t draw up arrows the english language. 3. Don’t draw self loops This idea can be generalized to strings 1 4. Don’t draw transitive edges over any alphabet.

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Hasse Diagrams Hasse Diagrams

Have you seen this one before? Consider this poset: Reds are maximal. Blues are minimal. {a,b,c} or 111

{a,b} or 110 {a,c} or 101 {b,c} or 011

{a} or 100 {b} or 010 {c} or 001

{} or 000

Hasse Diagrams Hasse Diagrams

Definition: In a poset S, an element z is a minimum element if ∀b∈S, z≤b. Theorem: In every poset, if the maximum element exists, it is unique. Similarly for minimum.

Intuition: If a is Write the defn of Did you get Proof: Suppose there are two maximum! maxiMAL, then no it right? maximum elements, a1 and one beats a. If a a2, with a1≠a2.

is maxiMUM, a Then a1 ≤ a2, and a2≤a1, by beats everything. defn of maximum.

Must minimum and maximum So a1=a2, a contradiction. exist? Thus, our supposition was A. Only if set is finite. incorrect, and the B. No. maximum element, if it C. Only if set is transitive. exists, is unique. D. Yes. Similar proof for minimum.

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Upper and Lower Bounds Upper and Lower Bounds

Defn: Let (S, ≤) be a partial order. If A⊆S, then an upper bound for A Defn: Let (S, ≤) be a partial order. If A⊆S, then an upper bound for A is any element x ∈ S (perhaps in A also) such that ∀ a ∈ A, a ≤ x. is any element x ∈ S (perhaps in A also) such that ∀ a ∈ A, a ≤ x.

A lower bound for A is any x ∈ S such that ∀ a ∈ A, a ≥ x. A lower bound for A is any x ∈ S such that ∀ a ∈ A, a ≥ x.

Ex. The upper bound of {g,j} Ex. The upper bound of {g,j} a b is a. Why not b? a b is a. Why not b? Ex. The upper bounds of {g,i} c d c d is/are… A. I have no clue. B. c and e e f j e f j C. a {a, b} has no D. a, c, and e UB. g h i g h i

Upper and Lower Bounds Upper and Lower Bounds

Defn: Let (S, ≤) be a partial order. If A⊆S, then an upper bound for A Defn: Given poset (S, ≤) and A⊆S, x ∈ S is a least upper bound (LUB) is any element x ∈ S (perhaps in A also) such that ∀ a ∈ A, a ≤ x. for A if x is an upper bound and for upper bound y of A, y ≥ x.

A lower bound for A is any x ∈ S such that ∀ a ∈ A, a ≥ x. x is a greatest lower bound for A if x is a lower bound and if x ≤ y for every UB y of A. Ex. The lower bounds of {a,b} a b are d, f, i, and j. a b Ex. LUB of {i,j} = d. Ex. The lower bounds of {c,d} Ex. GLB of {g,j} is… c d is/are… c d A. I have no clue. A. I have no clue. B. f, i B. a e f j e f j C. j, i, g, h {g, h, i, j} C. non-existent D. e, f, j has no LB. D. e, f, j g h i g h i

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Upper and Lower Bounds Total Orders (Epp p. 639)

This is because c and Ex. In the following poset, c and d are lower bounds for {a,b}, but d are Consider the problem of getting dressed. there is no GLB. incomparable. Precedence constraints are modeled by a poset in which a ≤ b iff you must put on a before b. Similarly, a and b are upper bounds for {c,d}, but there is no LUB.

In what order shoes belt jacket will you get dressed while a b respecting socks jeans swter jwlry constraints? c d

uwear shirt

Let (S, ≤) be a poset (S finite). We will extend ≤ to a total order on S, so we can decide for all incomparable pairs whether to make a ≤ b, or vice versa w/o violating T,R,A.

Total Orders Total Orders

Things we need: More things we need: Lemma: Every finite non-empty poset (S, ≤) has at least one Lemma: If (S, ≤) is a poset with a minimal, then (S-{a}, ≤) is minimal element. also a poset.

shoes belt jacket shoes belt jacket

socks jeans swter jwlry socks jeans swter jwlry NOT A REAL PROOF! Should use uwear shirt induction!! uwear shirt

Proof: choose a0 ∈ S. If a0 was not minimal, then there Proof: If you remove minimal a reflexivity and antisymmetry exists a1 ≤ a0, and so on until a minimal element is found. still hold. If x,y,z ∈ S-{a}, with x ≤ y and y ≤ z, then x ≤ z too, since (S, ≤) was transitive.

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Total Orders (Epp p. 641) Total Orders

alg Topological Sort Input: poset (S, ≤) Think about what this means: Out: elements of S in total Depending on which order 1. There is always a minimal element. min elt is chosen While S ≠ ∅ 2. If you remove it you still have a poset. each time, a Remove any min elt different total shoes belt jacket from S and output it. order is obtained, but all TOs will be shoes belt jacket consistent with the PO. socks jeans swter jwlry

socks jeans swter jwlry alg Topological Sort uwear shirt Input: poset (S, ≤) Out: elements of S in total uwear shirt order While S ≠ ∅ This suggests: Remove any min elt from S and output it.