Lattice MCQ Quiz - Objective Question with Answer for Lattice - Download Free PDF

Given:

A = {x, y}

Calculation:

Subsets of A are ϕ, {x}, {y}, {x, y}

 ∴ P(A) = {ϕ, {x}, {y} {x, y}}

Correct answer is option 4.

Concept:

A partial order set, or poset, is a chain if for every a and b in the set, either a ≤ b or b ≤ a.

Explanation:

 

(3) is correct

Key Points

 The given lattice is,

I ) The upper bound is B

False, there is no upper bound of the above  lattice

II) minimal element is O

True, the Minimal element of the above lattice is 0 only. The meet of above lattice at only o.

III) maximal element s are A,B,C

True, The maximal elements of the above lattice is  A, B, C.

IV) Minimal elements are O, N

False, the Minimal element of the above lattice is 0 only. The meet of above lattice at only o.

Hence the correct answer is II, III are correct.

A bounded lattice is a lattice where both the upper bound and lower bound exists.

Every finite lattice is a bounded lattice because, for any finite lattice, there exists a  unique least as well as the unique greatest element.

Every finite lattice is a bounded lattice but the converse is not true. i.e. A bounded lattice may or may not be finite. For example, \([[0,1\left] {, \le } \right]\) is a bounded lattice but not finite. Here, the least element is 0 and the greatest element is 1.

Note that \(\left[ {\left( {0,1} \right), \le } \right]\) is not a bounded lattice because there is no least or greatest element.

So, the correct answer are option 1, option 2 and option 4

Concept:

A Hasse diagram becomes a lattice if for every pair of elements in the Hasse diagram there exist a LUB (Least upper bound) and GLB (Greatest lower bound).

Explanation:

Consider all the pairs one by one:

Pairs	LUB	GLB
(a, b)	b	a
(a, c)	c	a
(a, d)	d	a
(a, e)	e	a
(b, c)	c	b
(b, d)	e	a
(b, e)	e	b
(c, d)	e	a
(c, e)	e	c
(d, e)	e	d

 

For each pair of Hasse diagram, there exists a LUB and GLB, so there is no need to add any ordered pairs. It is already a lattice.

Concept:

A Hasse diagram becomes a lattice if for every pair of elements in the Hasse diagram there exist a LUB (Least upper bound) and GLB (Greatest lower bound).

Explanation:

Consider all the pairs one by one:

Pairs	LUB	GLB
(a, b)	b	a
(a, c)	c	a
(a, d)	d	a
(a, e)	e	a
(b, c)	c	b
(b, d)	e	a
(b, e)	e	b
(c, d)	e	a
(c, e)	e	c
(d, e)	e	d

 

For each pair of Hasse diagram, there exists a LUB and GLB, so there is no need to add any ordered pairs. It is already a lattice.

ℒ = {p, q, r, s, t}

|ℒ| = 5

∴ |ℒ ³| = 5 × 5 × 5 = 125

Example of Lattice

Hasse diagram of [{1,2,3,5,30};/]

x ⋁ (y ⋀ z) = (x ⋁  y) ⋀ (x ⋁  z) that satisifes

2 ⋁ (3 ⋀ 1 ) ≡ 2 ⋁ 3 ≡ 1

(2 ⋁ 3) ⋀ (2 ⋁ 1) ≡ 1 ∨ 1 ≡ 1

L.H.S = R.HS

Thefore pr ≠ 0

x ⋁ (y ⋀ z) = (x ⋁  y) ⋀ (x ⋁  z) that doesn't satisifes

2 ⋁ (3 ⋀ 5 ) ≡ 2 ⋁ 30 ≡ 2

(2 ⋁ 3) ⋀ (2 ⋁ 5) ≡ 1 ∨ 1 ≡ 1

L.H.S ≠ R.HS

Thefore pr ≠ 1

Number of ways to chose {2, 3, 5} = 3! =6

∴ only 6 combination will violate the distributive property

probability that satify the distributive property = \(1 - \frac{6}{125} = 0.952\)

Therefore \(\frac{1}{5} < {p_r} < 1\)is correct answer

Concept:

A partial order set, or poset, is a chain if for every a and b in the set, either a ≤ b or b ≤ a.

Explanation:

 

(3) is correct

• Hasse diagram for the given relation is:

 

It is not a lattice because it has two maximal elements 9 and 18 but lattice has unique maximal element called Least Upper Bound (LUB)

• If ordered pair (9,18) is added R then Hasse diagram will be

 

The above Hasse diagram represents is lattice

Hasse diagram of POSET [{1,2,3,5,30};/]

Since meet semi lattice and join semi lattice exists for the POSET, therefore, it is a lattice, but distributive property doesn’t holds for POSET [{1,2,3,5,30};/]

Explanation:

2 ∧ (3 ∨ 5 ) ≡ 2 ∧ 1 ≡ 2

2 ∧ (3 ∨ 5 ) ≡  (2 ∧ 3) ∨ (2 ∧ 5) ≡ 30 ∨ 30 ≡ 30

2 ∧ (3 ∨ 5 ) ≠ (2 ∧ 3) ∨ (2 ∧ 5)

Similarly,

2 ∧ (5 ∨ 3 ) ≠ (2 ∧ 5) ∨ (2 ∧ 3)

3 ∧ (2 ∨ 5 ) ≠ (3 ∧ 2) ∨ (3 ∧ 5)

3 ∧ (5 ∨ 2 ) ≠ (3 ∧ 5) ∨ (3 ∧ 2)

5 ∧ (2 ∨ 3 ) ≠ (5 ∧ 2) ∨ (5 ∧ 3)

5 ∧ (3 ∨ 2 ) ≠ (5 ∧ 3) ∨ (5 ∧ 2)

There are 6 combination which doesn't satisfy distributive property

Concept:

A Hasse diagram becomes a lattice if for every pair of elements in the Hasse diagram there exist a LUB (Least upper bound) and GLB (Greatest lower bound).

Explanation:

Consider all the pairs one by one:

Pairs	LUB	GLB
(a, b)	b	a
(a, c)	c	a
(a, d)	d	a
(a, e)	e	a
(b, c)	c	b
(b, d)	e	a
(b, e)	e	b
(c, d)	e	a
(c, e)	e	c
(d, e)	e	d

 

For each pair of Hasse diagram, there exists a LUB and GLB, so there is no need to add any ordered pairs. It is already a lattice.

Hasse diagram:

Greatest Lower Bound of lattice is { } or ϕ   

Sum of elements in GLB of lattice = 0 

A bounded lattice is a lattice where both the upper bound and lower bound exists.

Every finite lattice is a bounded lattice because, for any finite lattice, there exists a  unique least as well as the unique greatest element.

Every finite lattice is a bounded lattice but the converse is not true. i.e. A bounded lattice may or may not be finite. For example, \([[0,1\left] {, \le } \right]\) is a bounded lattice but not finite. Here, the least element is 0 and the greatest element is 1.

Note that \(\left[ {\left( {0,1} \right), \le } \right]\) is not a bounded lattice because there is no least or greatest element.

So, the correct answer are option 1, option 2 and option 4

ℒ = {p, q, r, s, t}

|ℒ| = 5

∴ |ℒ ³| = 5 × 5 × 5 = 125

Example of Lattice

Hasse diagram of [{1,2,3,5,30};/]

x ⋁ (y ⋀ z) = (x ⋁  y) ⋀ (x ⋁  z) that satisifes

2 ⋁ (3 ⋀ 1 ) ≡ 2 ⋁ 3 ≡ 1

(2 ⋁ 3) ⋀ (2 ⋁ 1) ≡ 1 ∨ 1 ≡ 1

L.H.S = R.HS

Thefore pr ≠ 0

x ⋁ (y ⋀ z) = (x ⋁  y) ⋀ (x ⋁  z) that doesn't satisifes

2 ⋁ (3 ⋀ 5 ) ≡ 2 ⋁ 30 ≡ 2

(2 ⋁ 3) ⋀ (2 ⋁ 5) ≡ 1 ∨ 1 ≡ 1

L.H.S ≠ R.HS

Thefore pr ≠ 1

Number of ways to chose {2, 3, 5} = 3! =6

∴ only 6 combination will violate the distributive property

probability that satify the distributive property = \(1 - \frac{6}{125} = 0.952\)

Therefore \(\frac{1}{5} < {p_r} < 1\)is correct answer

Concept:

A distributive lattice is the one having atmost one complement for each element.

Explanation:

1. A lattice with 4 or fewer elements is distributive.

A lattice with more than 4 elements can contain complement more than one for any element which is not the property of distributive lattice. Example :

In this for element a, we have two complement b and c. So, it is not distributive. So, given statement is correct.

2. Every totally ordered set is a distributive lattice.

This statement is correct. A POSET is a totally ordered set if all the elements are related to each other in the POSET. A totally ordered set is a chain and chain is always a distributive lattice as there is always at most one complement for each element.

3. Every sublattice of a distributive lattice is also distributive.

If diamond can be embedded in a lattice then that lattice has a non distributive sub lattice. Distributive lattices are closed sublattice property. Every sublattice of a distributive lattice is also distributive.

4. Every distributive lattice is a bounded lattice.

When both upper bound and lower bound exists for a lattice, then it is bounded lattice. Let N = {1,2,3,4,5……..}, the poset [N;<=] is totally ordered set and hence distributive but it is not bounded because the upper bound of lattice does not exist. So, given statement is incorrect.