at 3:40, shouldn't the greatest lower bound (GLB) of (d,e) be a? Here is my working - 1. The lower bounds of d and e are b,c,a 2. As b and c are incomparable, they cannot be greatest lower bound 3. As a relates to both d and e transitively and also a relates to b and c (which fulfills the criteria for greatest lower bound), we can say a is the greatest lower bound?
No, _a_ is not the _greatest lower bound_ (GLB) of the set *{d,e}.* Recall the definition of a GLB: the GLB of a set is a lower bound of that set such that any other lower bound is either less than or equal to it. Elements _a, b_ and _c_ are indeed lower bounds of the set *{d,e},* but _a_ is less than _b,_ therefore _a_ cannot be the _greatest_ lower bound of that set, since there is another lower bound, namely _b,_ which is greater than _a._
You are very very very professinal teacher thank you very much you helped me a lot , but may sir upload more than one video a day, please. I need you to reach graph theory as i am a university student and my next exam include this topic, please.
I think even if transient state exists we may declare the Hassle diagram a lattice. As the transient state is not a hindrance in the LUB or GLB generation.
Solved both the questions on my own. Thanks to your teaching only.
at 3:40, shouldn't the greatest lower bound (GLB) of (d,e) be a?
Here is my working -
1. The lower bounds of d and e are b,c,a
2. As b and c are incomparable, they cannot be greatest lower bound
3. As a relates to both d and e transitively and also a relates to b and c (which fulfills the criteria for greatest lower bound), we can say a is the greatest lower bound?
No, _a_ is not the _greatest lower bound_ (GLB) of the set *{d,e}.* Recall the definition of a GLB: the GLB of a set is a lower bound of that set such that any other lower bound is either less than or equal to it. Elements _a, b_ and _c_ are indeed lower bounds of the set *{d,e},* but _a_ is less than _b,_ therefore _a_ cannot be the _greatest_ lower bound of that set, since there is another lower bound, namely _b,_ which is greater than _a._
@@anfauglitcan u explain why in the last diagram the glb of b and e is a but not d and both are related
You are very very very professinal teacher thank you very much you helped me a lot , but may sir upload more than one video a day, please. I need you to reach graph theory as i am a university student and my next exam include this topic, please.
Thank you 😃
I think even if transient state exists we may declare the Hassle diagram a lattice.
As the transient state is not a hindrance in the LUB or GLB generation.
So if we generate in the transient elements with GLB and LUB we may find a lattice.
Welcome to everyone
The last diagram there is intermediate has an element so there can be a chance of not being a lattice.
aoa sir, sir plzz ap ek apni vlog link banayein takhe hum wahan se apke discret maths notes download kar sakhey apne paas plzz?
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Hindi me smjha dijiye sir