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In the lattice defined by the Hasse diagram given in following figure, how many complements does the element ‘$e$’ have?

  1. $2$
  2. $3$
  3. $0$
  4. $1$
asked in Set Theory & Algebra by Veteran (59.6k points)
edited by | 1.4k views
+1

This might help ...

2 Answers

+15 votes
Best answer

Answer: B

Complement of an element a is a' if:

  • a ∧ a' = 0 (lowest vertex in the Hasse diagram)
  • a ∨ a' = 1 (highest vertex in the Hasse diagram)

g, c and d are the complements of e.

answered by Boss (34.1k points)
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0
0 and 1?

Also, 'd' is missed?
+1
0 means lowest vertex in the Hasse diagram (f in this case). 1 means highest vertex in the Hasse diagram (a in this case).
0
Okay. But a is not a complement of e rt? Answer should be 3 only.
0
Why not 'a'?
0
Because meet of a and e is e.
0
Oh, yes. You are right. Corrected.
0
@sir

i think there will be one more complement of e that would be f. am i right?
+1
Nopes. The join won't be the highest vertex for e, f.
+1
thanx. i'm considering only meet condition & missed join condition.
+1
To make things faster remember that compliment of an element is such an element which is not related to it and LUB of that goes to I and GLB to O.

So, the complement of e must be an element $e^{'}$ such that LUB(e,$e^{'}$)=I and GLB(e,$e^{'}$)=O, the topmost and bottom-most elements of hasse diagram respectively.

Now, candidates to be looked for the complement of e should be all those elements, to which there is no path in the hasse diagram and those are g,c,d and if you check them these 3 are complements of e.

Now why you only look for non-related elements to be the complement?

Say if I assume b to be a complement of e, from the diagram it is clear that eRb. So by lattice laws

LUB(e,b)=b(b $\not= I$ not okay!!) and GLB(e,b)=e(e=O okay!), so b can never be complement of e.
+2 votes

Option B

vertex 'e' have three complement like 'g','c' and 'd'.

When it will take LUB or GLB with g ,c or d get same answer.

answered by Loyal (9.1k points)

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