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Describe, in pseudo-code, how to perform path compression on a path of length $h$ in $O(h)$ time in a tree-based partition union/find structure....

Question

Describe, in pseudo-code, how to perform path compression on a path of length $h$ in $O(h)$ time in a tree-based partition union/find structure.

Describe, in pseudo-code, how to perform path compression on a path of length $h$ in $O(h)$ time in a tree-based partition union/find structure.



Answers

Describe an algorithm based on the linear search for determining the correct position in which to insert a new element in an already sorted list.

All right. So here we are going to be using the binary search other than and that's other than three in the textbook. Let's call it buying search course you can name it, whatever makes sense to you. And we need for this elder than a list and into yours. And it should be a non empty list. So and not to be greater than or equal to one. And we also need in this case, we're not just searching for a moment in the list, but we're searching for where to put the elements. So that's the change that we're making. Rather than finding a preexisting one. We're placing one in, and the element will be placing his ex. And so this actually starts identical to the binary search. Other than where we want to set our left endpoint of the surgeon general I equal to one and then our right end point of our search interval J equal to end. So initially, we're setting it to the entire list here and then Well, say, while while I is less than J, we're going Thio, find the index of the point that is roughly halfway through. So we'll take I plus J Divide by two. But then we floor it in case it's not an even number of elements in the list. Once we've done that, you want to compare X to this value that's half flee through sorts. Check and see if X is greater than the value of the empath element of the list. And if it is, do we want to reset our surgeon point? And we want to reset the left endpoint to be one further than the of number. But if it's not greater than then, the assumption is that it's going to be less than or equal to. And so we would reset, are right. Most left point, then to be equal two. Em because this so far this is identical to the other than three. This was gonna change, then, is what we do after we've gone through this process. So rather than just finding the element once used or narrow it down to an element here in the previous case, the question was, Is this element equal to X or not? So that was us saying it's in the list at this location or it's not Melissa. All here were saying we've found the element that is greater than X, but the least greater than X. Okay, so then the question is, is ex less than or equal to the element that we've found? Okay. And if we have than the location is going thio equal, I okay. It's the only instance in which that wouldn't happen. Would be if nothing in the list is larger than X. And so if that's the case, well, then we just need to add X to the end of the list. So we'll set the location than equal to N plus one At the end of this. We want to return, and so is good. When you are running your return case to double check. What is it even that this algorithm is supposed to do? And it's supposed to determine the position in which to insert a new element. So we want to return the location, The position, not not the list

To this lesson in this lesson. Using algorithm to find the shortest distance between two two points. Okay. Yeah. Using the concept of Yeah, innovative. This is Okay, So the procedure Mhm. Yeah. Yeah. Okay. For an input M l Demetrius, That is 01 Uh huh. Squared mattress. Oh, Okay. Mm. For a court one. So a run from one to and the first rule to the end through if Mhm. Oh, if I've row and, uh, agave column she called to zero then yeah, has changed the one with a very high number. A very positive, higher number, like 10 billion or something else. Question that. I've rule and they give column of the mattress with just one. Okay, then for mhm. Okay, the variable key running from one to n. Yeah, for another variable running from 12 n Yeah. Mm. And for the last berry bogie running from Yeah. 12 And again we assigned that made Chris. Whoa. The the entry of a mattress. The I've rule and the gave column with the minimum of Thank you. Okay. The input. I'd the input I key and the input kg. Okay, then after that, we After all is done. We were 10. The maitresse w containing the shortest path. Okay, so thanks for your time. This is the policy idea. This is how the algorithm looks like. Thanks for your time. Good.

So we can assign of lead. You want two in charge, then? Is that just, uh, greet them?

So we need to extend the justice. Accurate maps follows. So first d'oh! Something? Something. So here, full. I could tow. Want to end. Wait I you called to And the key? How a you called to cereal? Here. We need to hide. Hey, culture zero us this empty while something Well, well. What? This is the knot in us. Yeah, out. You plus a few. You the last time l e then Healthy inn Seiko to out you last w you eat? Ah, here. We need to at even the this you we'll be here. This is, uh, okay, nation Then return. I don't see and


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