::ICS 804::Theory of

Computation

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ICS 804ICS 804

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ICS 804 – Course TextICS 804 – Course Text

IntroductionIntroduction Computation from the point of view

of mathematics (discrete mathematics)

Introduction to computability and complexity theory

What kind of problems are inherently computable and which options are there?

Regular languages and finite-state automata

Very recent field started in the 1930s

Course OutlineCourse Outline

Mathematical PreliminariesTuring MachinesRecursion TheoryMarkov AlgorithmsRegister MachinesRegular Languages and finite-state

automataAspects of Computability

Course OutlineCourse Outline

Mathematical PreliminariesTuring MachinesRecursion TheoryMarkov AlgorithmsRegister MachinesRegular Languages and finite-state

automataAspects of Computability

Discrete MathematicsDiscrete Mathematics Theory of computability = new branch of

mathematics Presupposes background in discrete mathematics Topics:

◦ Set-theoretic concepts◦ Formal language theory◦ Functions◦ Big-O notation◦ Propositional logic◦ Proof Techniques◦ Number-theoretic predicates

SetsSets

SetsSets

Elements of a set: may be physical objects Set: never physical – is an abstract concept a S – a member of set S Members of sets may be sets

S = {0, 1, {0, 1}} Cardinality: card(S) = 3 Singleton: S = {a} – set containing single

element Sets lists (unordered, elements are unique) Set = family = class

Specifying (Naming) SetsSpecifying (Naming) Sets

Finite sets {a,e,i,o,u} – elements can be listed Infinite sets: come from mathematics

◦ enumerate elements in an extended sense

◦ Natural numbers N = { 0,1,2,…}◦ Integers Z = {…,-2,-1,0,1,2,…}

Set abstraction◦ {n|n is prime} or {n:n is prime}◦ {n|…n…} or {n:…n…}: abstraction operator: the

set of all n such that n…◦ {n|n N & 2|n}: all even natural numbers

Specifying (Naming) Sets Specifying (Naming) Sets ……

By convention:◦ Use variables n,m,k,j,…: natural

numbers / integers◦ Use variables x,y,z,…: rational/real

numbers◦ Set variables: uppercase letters (A,B,C,

…S)◦ Universal set: U – all elements in

current domain U by set abstraction {x|x=x}

◦ Empty Set: - with no element by set abstraction {x|xx}

SubsetsSubsets A = {0,1,2,3,4} A is subset of N or AN A is a subset of B

◦ If for any x, x A implies x B Consequences:

◦ Every set is a subset of itself i.e. A A, for any set A

◦ A, for any set A “proper subset”: non-identical subset

{1,2,3} {1,2,3,4} Power set – all subsets of a set

◦ For a set S: P(S) = { A|A S }◦ What is the power set of S = {1,2,3} ?◦ Card(P(S) = 2card(s) for a finite set S

TuplesTuples A tuple is an ordered list of elements Ordered pair: <a,b> = <c,d> iff a = c and b = d defined Ordered triple: <a,b,c> 0-tuple: A = {<>} 1-tuple: A = <a> For any two sets A and B:

◦ are said to be disjoint if AB = ◦ Card(AB) Card(A) + Card(B)◦ Card(AB) = Card(A) + Card(B) iff AB = ◦ Compliment of B relative to A denoted A\B is

defined as {x| x A and x B}◦ We write BC and speak of compliment of B for the

set U\B◦ Symmetric difference denoted by AB is as

{x| (x A and x B) or (x B and x A) }

Some set propertiesSome set propertiesA ∪ U = UA ∩ = C = UUC = Idempotent Law: For any set A,

◦ A ∪ A = A◦ A ∩ A = A

Identity Law: For any set A,◦ A ∪ = A◦ A ∩ U = A

Some set propertiesSome set propertiesCommutative Law: For any two sets A and B

◦ A ∪ B = B ∪ A◦ A ∩ B = B ∩ A

Associative Law: For any three sets A, B and C◦ (A ∪ B) ∪ C = A ∪ (B ∪ C)◦ A ∩ (B ∩ C) = (A ∩ B) ∩ C

  Distributive Law: For any three sets A, B and C◦ A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)◦ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

De Morgan’s Law: For any two sets A and B◦ (A ∪ B)' = A' ∩ B'◦ (A ∩ B)' = A' ∪ B'

Special SetsSpecial Sets N: set of natural numbers

closed under addition and multiplicationnot closed under subtraction nor division

Z: set of integersclosed under addition, subtraction and

multiplication Q: set of rational numbers

Q = {x|x = p/q for some p,q Z with q 0}closed under addition, subtraction and

multiplicationQ\{0} is closed under division

R: set of real numbersclosed under addition, subtraction and multiplicationR\{0} closed under division

We define a partition of a set A to be a family of mutually disjoint subsets such that A = A1 ∪ A2 ….An

ExercisesExercisesExercise 1If A and B are two sets, then A ∩ (A ∪ B) equals

◦ (A)   A      (B)      (C)        (D)       none of these

Exercise 2The set (A ∪ B ∪ C) ∩ (A ∩ B' ∩ C')' ∩ C' is

equal to◦  (A)  B ∩ C'   (B)   A ∩ C (C)    B' ∩C'    (D) None of

these

Formal Language TheoryFormal Language Theory

Formal LanguageFormal Language

Formal language: specific kind of set

Formal = not natural language, programming language, …but: NLP try to formalize natural language

Ignoring semantics:◦ possible to regard natural languages

like Spanish and programming languages such as PROLOG as formal

Formal LanguageFormal Language

Alphabet : finite set of symbols, e.g. = {a,b}

Word over alphabet : finite string of elements of allowing repetition

Length of word w = |w| = number of symbol occurrences in w

Card() = m then there exist mn distinct words over of length n◦ For alphabet = {a,b} we have 2n distinct words over

of length n

Formal LanguageFormal Language

Special superscript notation: anbm

word consisting of n times a and m times bExample: a3b4a2 aaabbbbaa

Distinct occurrences: na(aba)=2nb(aba)=1

Formal LanguageFormal Language

Null word: word of length 0◦ Zero occurrences of a and b◦ = a0

◦ na() = 0 and nb() = 0

Universal Language: * set of all words over Language over : any subset of *

◦ Empty language: - containing no word◦ Universal language: * - containing all words◦ Unit language L = {w} – containing word, w, only

Alphabet = {a,b}◦ {w *: |w| = 3} = aaa, aab, aba, abb, baa, bab, bba,

bbb}

Operations on wordsOperations on words

Concatenation: ww’ of words w and w’w is aba and w’ is bbaww’ = ababba

Reverse of a word: wR

abbR = bba

Palindromes over is the language {w*|where w=wR}

Palindrome ExamplesPalindrome Examples Examples:

Radar – level – mum – gig “Madam, in Eden, I’m Adam” “A man, a plan, a canal: Panama” “Doc, note. I dissent. A fast never prevents a

fatness. I diet on cod.” (Peter Hilton) Longest 1-word English palindrome:

detartrated - (to remove tartrates) Longest 1-word palindrome in the world (Finnish)

saippuakivikauppias - (lye dealer) Swahili palindromes: amesema, ataufuata,

akatutaka Longest palindrome: 31358 words

Operations on wordsOperations on words

A word w’ is a prefix of a word w if w can be written w’w’’ for some word w’’

◦ ,a,ab,abb,abba are prefixes of the word abba (first 4 are proper)

A word w’’ is a suffix of a word w if w can be written w’w’’ for some word w’

◦ ,a,ba,bba,abba are suffixes of the word abba (first 4 are proper)

Operations on wordsOperations on words

A word w’’ is a substring of word w if w can be written as w’w’’w’’’ for some words w’ and w’’

abba: a ab bb ba abb bba abba

ExercisesExercises

1 = {a,b,c} 2 = {a,b,ca} 3 = {a,b,Ab} Determine to which * each word belongs

ababAbcbacabcaabbaAb

SolutionSolution

1 = {a,b,c} 2 = {a,b,ca} 3 = {a,b,Ab} Determine to which * each word belongs

aba 1,2,3bAb 3cba1cab1,2caab 1,2baAb 3

ExercisesExercises

Calculate how many words there are in the following languages. List 3 elements in each of them. Which language contains

◦ *, where = {a,b,c}◦ {w*| |w| 3}, where = {a,b}◦ {w*| |w| =4}, where = {a,b}◦ {anb|n is prime}

SolutionSolution *, where = {a,b,c}

Infinite number of words {, abba, ccccba, aaccccccccc, …}

{w*| |w| 3}, where = {a,b} 15 words (23+22+21+20) = {, a, b, ab, ba, bb, aa, aaa, baa, aba, aab, bba, bab, abb, bbb}

{w*| |w| =4}, where = {a,b} 16 words (24) = {aaaa, aaab, aaba, abaa, baaa, aabb, abba, bbaa, baba, abab,

baab, abbb, babb, bbab, bbba, bbbb}

{anb|n is prime} Infinite number of words {aab, aaab, aaaaab, …}

ExercisesExercises

Let = {a,b,c,d}. List as many words as you can that are words in the language

{w *| |w| = 3 and w is a word in the English language}

SolutionSolution

Let = {a,b,c,d}. List as many words as you can that are words in the language

{w *| |w| = 3 and w is a word in the English language}

dad cab bad dab add cad baa

ExercisesExercises

Suppose w is ab and w’ is bab. Identify each of the following words:ww’wR

(ww’)R

ww’R

SolutionSolution

Suppose w is ab and w’ is bab. Identify each of the following words:ww’ abbabwR ba(ww’)R babbaww’R abbab

ExerciseExercise

Which of the following languages are palindrome languages◦ wwR for any w *◦ wR for any palindrome w *◦ ww for any palindrome w *◦ wawR for any w *◦ ww’wR for any w * and any palindrome w’ *◦

SolutionSolution Which of the following languages are palindrome languages

◦ wwR for any w *yes

◦ wR for any palindrome w *yes

◦ ww for any palindrome w *yes

◦ wawR for any w *yes

◦ ww’wR for any w * and any palindrome w’ *yes

◦ yes

Mapping and Mapping and FunctionsFunctions

Mapping and Mapping and FunctionsFunctions Mapping: association of members of one set with members of another (not necessarily different) set

First: domain Second: codomain or range Members of domain serve as: arguments Members of codomain serve as: values Mapping f between Dom(f)=A and

Cod(f)=Bf: A B

A mapping f is a function if each member of Dom(f) is mapped to one and only one member of Cod(f)

FunctionFunctionFunction = single valued mappingComplete description of a mapping f

consists of:a) specification of Dom(f) and Cod(f)b) description of values of f for any member of

Dom(f) taken as argumentsExamples

f: N Nf(n) = n + 1

Unary constant-7 function – returns 7 for all n

C17: N N

C17(n) = 7

f: * *f(w) = wR

FunctionsFunctionsExamples …continuedFloor function

f: R Zf(x) = [x] = greatest integer ≤ x

Ceiling functionf: R Zf(x) = [x] = least integer ≥ x

Multi-valued mapping◦ f: Z Z

f(n) = n ± 3

f(n) = n 3Not a function

FunctionsFunctionsImage of a function f defined as

Image(f) is subset of Cod(f) consisting precisely of the values of f for members of Dom(f)

Precisely in set language:

Image(f) = {yCod(f)|y=f(x) for some x Dom(f)}

Number-theoretic Number-theoretic FunctionsFunctionsNumber-theoretic functions: map natural numbers to natural numbers: f: N N

Unary number-theoretic functionsf(n) = n + 1

Binary number-theoretic functionsf: N 2 N f(n,m) = n + m

K-ary number-theoretic functions f: N k N0-ary number theoretic functions

C02: N 0 N

C02() = 2

Injective (1-to-1) Injective (1-to-1) FunctionsFunctionsA function f is 1-to-1 or injective if no two distinct elements of Dom(f) are mapped to one and the same member of the Cod(f)

Examples:Successor function

f: N Nf(n) = n + 1

Word reversal function f: * *

f(w) = wR

Ceiling, floor and constant-7 functions not injective

Surjective (onto) Surjective (onto) FunctionsFunctionsA function f is onto or surjective provided that every element of Cod(f) is a value of f for at least one element of the Dom(f)

Cod(f) = Image (f)

Examples:Word reversal function f: * *

f(w) = wR

Ceiling functionf: R Zf(x) = [x] = least integer ≥ x

Successor function not surjective – 0 not successor

A function that is both injective and surjective is termed as 1-1 correspondence or a bijective function

Functions forming operations

InverseInverseFor two functions g and f:

g: N Ng(n) = n/2

f: N Nf(n) = 2n

g is the inverse of fg = f-1

The inverse operation (though not always defined) -1is; Unary function forming operation

CompositionCompositionFor two functions g and f:

g: * N where ={a,b} g(w) = na(w)

f: N Nf(n) = 3n + 2

h: * Nh(w) = f(g(w))h: composition of g and f (h = g f)

The composition operation is:Binary function forming operation

Partial FunctionsPartial FunctionsFor a function g:

g: N 2 Ng(n,m) = n – m

The function g is undefined for pairs <n,m> where n < m

Number-theoretic functions that are undefined for some n N are Partial functions.

Partial FunctionsPartial FunctionsLet f(n) and g(n) be partial number-theoretic

functions We can define a new partial number-

theoretic function h(n) = max{ f(n), g(n) }

This means that h(n) is defined just in case both f(n) and g(n)

are defined

for any n Dom(h) = Dom(f) Dom(g), we have h(n) is the maximum of f(n) and g(n)

Polynomial FunctionsPolynomial FunctionsWe restrict ourselves to unary polynomial

functionsPolynomial in n of degree k is any expression

of form:

aknk + ak-1nk-1 + ak-2nk-2 + ….. + a1n1 + a0n0

where k is any natural constant and coefficients ak

, ak-1 , ak-2 , ….., a1 + a0 are

integer constants with ak ≠ 0Polynomials of degree 0 are numeralsPolynomials in n are not themselves natural

numbers or integers; they are expressions

Polynomial FunctionsPolynomial FunctionsFor example:

2n5 + 2n3 + 4n2 +16n + 7is a polynomial of degree 5

The unary polynomial function is defined by:f: N Nf(n) = 2n5 + 2n3 + 4n2 +16n + 7

f(2) = ?Polynomial functions with negative coefficients

may be partialSometimes we write p(n) for a unary polynomial

function of nFor p1(n) and p2(n) => p2[p1(n)]) – result of

replacing every occurrence of n in p2(n) with p1(n)

Big-O NotationBig-O Notation

Big-OBig-OBig-O (micron) notation provides standard

means of comparing growth of number-theoretic functions

Analyze time and space requirements of algorithms

Typical calculation: growth of a function as argument n goes to infinity

Monotone increasing: for all n1, n2 N, we have n1< n2 implies f(n1) f(n2)

Strictly monotone increasing: f(n1)<f(n2)

Big-OBig-OFour partial number-theoretic functions:

k(n) = 2n

h(n) = n2 + 1g(n) = 3nf(n) = 7[log2 n]

Big-OBig-O

Big-OBig-OWhich are (strictly) monotone increasing ?

Big-OBig-O

Four partial number-theoretic functionsk(n) = 2n

h(n) = n2 + 1 g(n) = 3nf(n) = 7[log2 n]

Value 2: k(2) = 4 h(2)=5 g(2) = 6 f(2)=7

Value 5: k(5) = 32 h(5)=26 g(2) = 15 f(2)=14

Big-OBig-O

Fastest growing function? Slowest?

Big-OBig-OFor small numbers f is higher, but it is k that grows

the fastest for large numbers.Big-O expresses how fast a function grows relative

to anotherFunction f(n) is O(g(n))

◦ g(n) is at least as great as f(n); ultimately for all sufficiently large n

Similarly; g(n) is O(h(n)) and h(n) is O(k(n))Transitivity holds:

◦ If f(n) is O(g(n)) and g(n) is O(h(n)) then f(n) is O(h(n))

Big-OBig-ODefinition:Let f(n) and g(n) be partial number-theoretic

functions. Then function f(n) is O(g(n)) if there exist natural number constants C and n0 such that for any nn0 we have f(n) C.g(n) whenever both f(n) and g(n) are defined.

e.g. Function f(n) is O(g(n)): with C=1 and n0=5

f(n) = 7[log2 n] C.g(n) = 1.3n

4 14 12

5=n0 14 15

7 14 21

16 28 48

Big-OBig-O • f1 is O((f2))• f3 is O((f4))• f5 is O((f6))• …..• …..• …..• f9 is O((f10))

• Growth rate of a function ftime is O(f72(n)) for instance

• In other words function ftime is O(n2)

Big-OBig-O

Big-OBig-O

Function f(0) is O(0)Function f(n) that is O(1) iff f(n) is bounded above by a

constant for sufficiently large arguments nFunction f(n) that is O(n) has linear growthFunction f(n) is O(n2) or O(n3): polynomial growthFunction f(n) is O(2n) or O(3n): exponential growth

ExercisesExercises

Characterize these functions as ◦ monotone increasing◦ strictly monotone increasing◦ neither

(a) (b)

(c) (d)

(e) (f)

SolutionSolution

• Strictly monotone• Neither• Strictly monotone• Monotone• Neither• Monotone

(a) (b)

(c) (d)

(e) (f)

ExercisesExercises

Which of the following functions are O(1)◦ f(n) = [log2 n]

◦ g(n) = [log10 n]

◦ h(n) = min(n,100)◦ j(n) = 4n2 + 7n + 3◦ k(n) = 5n5 + n3 + n + 68

SolutionSolution

Which of the following functions are O(1)◦ f(n) = [log2 n]

◦ g(n) = [log10 n]

◦ h(n) = min(n,100)

only this one is O(1)◦ j(n) = 4n2 + 7n + 3◦ k(n) = 5n5 + n3 + n + 68

Propositional LogicPropositional Logic

PropositionPropositionProposition: any statement with a truth value Any statement that is either true or falseExamples:

◦ For all n in P, n3 – n is divisible by 6◦ N2 is O(n3)◦ 6 is a prime number◦ Nairobi is located in Kenya◦ Goldbach’s Conjecture: every even integer strictly

greater than 4 is the sum of two primes

◦ Don’t forget to vote◦ Is Russian beer the best beer in the world?

◦ We define the degree of a vertex v within an undirected graph G to be the number of edges in G incident upon v.

NotationNotation

PL: infinite symbol set of

◦lower-case letters of the Roman alphabet, p,q,r,p’,q’,r’,p’’,q’’,r’’,… which stand for propositions: sentence letters

◦Propositional connectives: & ◦Aggregation signs - (), [], {}

Language of propositional logic- set of strings over alphabet PL

NegationNegation

Symbol: p pT FF T

p: John is tallp: John is not tall

Unary connective

ConjunctionConjunctionBinary connective: symbol - &

p q p&qT T TF T FT F FF F F

For conjuncts p and qp: John is tallq: Andrew is shortp&q: John is tall and Andrew is short

DisjunctionDisjunctionBinary connective: symbol -

p q p qT T T

F T T

T F T

F F F

Inclusive orFor disjuncts p and q

p: John is tall

q: Andrew is short

p q: Either John is tall or Andrew is short

ConditionalConditionalUsed to express conditional statementsBinary connective: symbol

p q p q

T T T

F T T

T F F

F F T

p: It is rainingq: The streets are wet

Antecedent p and consequent qFalse wherever antecedent is true and consequent is

false

ConditionalConditionalBinary connective: Not quite the same semantics as in natural languageIf Kenya is in Europe, then 2+2=4

= false

BiconditionalBiconditionalBinary connective: symbol

p q p qT T TF T FT F FF F T

If and only ifp: 2+2 = 4q: 1+1 = 2True if component sentence letters have same

truth values

Semantic IssuesSemantic IssuesSentences of propositional logic classified as:Tautologous

◦ True under all assignments of component sentence letters

◦ ppContradictory

◦ False under all assignments of component sentence letters

◦ p&pContingent

◦ Sentences that are neither tautologous or contradictory

◦ p & q and p q

Semantic IssuesSemantic IssuesSatisfiable

◦ A sentence is satisfiable if for at least one assignment of the component letters the sentence holds true

◦ Not contradictory◦ Truth tables: decision procedure for satisfiability◦ The satisfiability problem for propositional logic is

decidableEquivalence

Two sentences S1 and s2 are (logically) equivalent if for any given assignment of truth values to component sentence letters, the resulting truth value of S1 is identical with value of S2

Idempotence, Associativity, Commutativity,Distributivity and DeMorgans Laws also hold

PROOF TECHNIQUESPROOF TECHNIQUES

Proof TechniquesProof Techniques

Conditional Proofpq

assume truth of the antecedentreason to the truth of consequent q

Example 1If I drink too much alcohol, I can’t drive a car

• When I drink alcohol, the alcohol gets absorbed in my blood. My blood reaches my brain, where it will distort its functionality, reducing my coordination abilities. To drive a car I need coordination.

Proof TechniquesProof Techniques

Conditional Proof

Example 2

If Jones is elected, then the treasury will be bankrupted• Given that Jones is lavish and a spendthrift• He will spend vast amounts on annual banquets• Given his slackness in supervising his subordinates,

there will be widespread corruption

Therefore, if elected then the treasury will be bankrupted

Proof TechniquesProof TechniquesIndirect Proof

pqProof by contradiction

• 11 is not a prime number

if it is not a prime number, it must have a non-trivial factor. But it is not 2,3,4,5,6,7,8,9,10 and it can’t be larger than n. This is a contradiction, so 11 is a prime after all

• Appropriate when p is unstructured. • Derive a contradiction of the form p & p. • Having assumed p is true, continue to show that p &

p is true. • Conclude then that p is not true, thereby showing p

is true.

Number-theoretic Number-theoretic PredicatesPredicates

Number-theoretic Number-theoretic PredicatesPredicates• Properties of objects, relations, holding

between pairs of objects, relations, holding among triples of objects are expressed using predicates.•If the objects are members of N, then one speaks of number-theoretic predicates•e.g. prime(n) is a unary number-theoretic predicate that is satisfied by 2 and 3, but not by 4 or 6•e.g. predicate n|m hold just in case m is a multiple of n so that 2|6 and 2|0 hold

Number-theoretic PredicatesNumber-theoretic Predicates• The extension of a k-ary number-theoretic

predicate is the set of all and only those k-tuples of natural numbers satisfying that predicate

extension of the predicate prime(n) is the set {n N|prime(n)}

extension of cube_of(n,m) is the infinite set of ordered pairs {<0,0>,<1,1>,<8,2>,<27,3>,<64,4>,…}

Number-theoretic PredicatesNumber-theoretic Predicates

• Predicates can be combined using the five propositional connectives: &

C1(n1,n2,…nk) & C2(n1,n2,…nk)

is true if k-tuple <n1,n2,…,nk> satisfies both C1(n1,n2,…nk) and C2(n1,n2,…nk)

• A binary predicate C(n,n) is said to be reflexive if, for any and all n N, we have C(n,n)

• A binary predicate C(n1,n2) is said to be symmetric if, for any n, m N; C(n,m) implies C(m,n)

NotationNotation

Quantifier: may play a role in specification of predicates◦ Universal quantifier (n)(C1(n) C1(n))

C1(n): prime(n) Given any natural number n, either n is prime

or it is not

Given any natural number n is prime and n> 2 is odd ?

◦ Existential quantifier (n)(C1(n) & C2(n)) C1(n): prime(n)

C2(n): odd(n) There exists an n such that n is prime and n is

not odd

There is no prime greater than 2 that is even?

ExerciseExerciseGiven: the following predicates C1(n) C2(n,m,k) C3(n,m,k) with

n,m,k N C1(n1) n1 is prime C2(n1,n2,n3) n1 = n2 + n3

C3(n1,n2,n3) n1 = n2 . n3

For each of the predicates below, determine whether it is satisfied by the pair <4,12> n=4 m=12

(k)C2(n,m,k) (k)[C2(k,n,m) & C1(k)] (k)C3(n,m,k) (k)C3(k,n,m) (k)[C3(k,n,m)&C1(k)] (kn)[C3(m,k,n) & C1(k)]

ExerciseExerciseGiven: the following predicates C1(n) C2(n,m,k) C3(n,m,k) with

n,m,k N C1(n1) n1 is prime C2(n1,n2,n3) n1 = n2 + n3

C3(n1,n2,n3) n1 = n2 . n3

For each of the predicates below, determine whether it is satisfied by the pair <4,12> n=4 m=12

(k)C2(n,m,k) (k)[C2(k,n,m) & C1(k)] (k)C3(n,m,k) (k)C3(k,n,m) (k)[C3(k,n,m)&C1(k)] (kn)[C3(m,k,n) & C1(k)]

nonoyesyesnoyes