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Consider an open-address hash table with uniform hashing. Give upper bounds on the expected number of probes in an unsuccessful search and on the expected number of...

Question

Consider an open-address hash table with uniform hashing. Give upper bounds on the expected number of probes in an unsuccessful search and on the expected number of probes in a successful search when the load factor is $3 / 4$ and when it is $7 / 8$.

Consider an open-address hash table with uniform hashing. Give upper bounds on the expected number of probes in an unsuccessful search and on the expected number of probes in a successful search when the load factor is $3 / 4$ and when it is $7 / 8$.



Answers

[M] Construct a random $4 \times 4$ matrix $A$ with integer entries between $-9$ and $9,$ and compare det $A$ with $\operatorname{det} A^{T}, \operatorname{det}(-A)$ $\operatorname{det}(2 A),$ and $\operatorname{det}(10 A) .$ Repeat with two other random $4 \times 4$ integer matrices, and make conjectures about how these determinants are related. (Refer to Exercise 36 in Section $2.1 . )$ Then check your conjectures with several random $5 \times 5$ and $6 \times 6$ integer matrices. Modify your conjectures, if necessary, and report your results.

Okay, so we have an algorithm, Mikey, and want to determine the normal comparison. Jackie's being used. So this lion Kurt is the only line that does any comparisons. So this entire line here, he's one person, everyone to figure out. How many times does compassion is in May? So because we have a full appear from one to end, it repeats this loop in the amount of times. So then the comparisons is pain, so the water is just talking.

That's it for this problem. When it's gonna call this prove cedar turn Eri and turn ery takes in a list of integers. So some end number of integers, All right? And there's at least going to need to be a minimum of one of them. All right. And then we also have ah thes search algorithm, the Adam we're searching for, which is also an integer. OK, so first we wanted to find our search interval. So we'll set I equals one and j equal to end. So I is going to be the position of the first element of our search Interval and Jay will be the last element of our search internal. So at the moment, we've just initialized it to the entire list, OK? And so they will consider the case for while I is less than J minus one. Okay. And so this insurers, that builder them would stop when there's at most two elements in these search interval. We also to define a couple of new variables. Okay, so lots of definitions here, So let's let L Yeah, lower case l, uh, be equal to the floor. I plus J divided by three. Okay, we're gonna do essentially the same thing with you. This is gonna be the floor. Is there's gonna be two times I plus J divided by three. And so these are also setting positions in our list. Okay, so then we can determine in which Southern travel ex lies. Right? So if we I think about all these variables, feel lis a one and it helps you a, uh it's gonna go live to a U and then up to S O. J. Right. So were sort of crooning roughly evenly distributed intervals here. And so to find in which of these seven troubles exes in we. Well, then assign the boundaries of each search internal to the these values of the seven trolls. Right? So I say, if X is greater than or equal to hey, you then we'll set our our left interval R a. I equal to one plastic one past that, right? So they will say you plus one. It's basically an ex is larger than this element. Then we're gonna start our search at the next element cable. Then we'll do the same thing. Like what? If X is between l and you write so don't say. Well, if exes greater than or equal to a l Then in this case, we need to set. I am Jace. We need to say I is equal to l plus one. We also want our we want to stop it at Hey, you Right. So we also said j equal to you. It's that way. It's not, um, going to search from El forward, but rather just only search between l and you. No course. Lastly, it could be in this first interval, right? And so if that's the case, I would say else said j equal to L because our eyes already initialized at the 1st 1 Okay, so the thing is, this is all in this wild statement, right? So it's gonna once we narrow it down to 21 interval, it's gonna re split that interval up and do it again, and I'll split the interval. New intro found up and again and again up until we no longer have the property that eyes less than J minus one. Okay, so that will leave us with an interval containing at most two values. So continue, uh, just right side here. And so we just need to check if X is one of the values left in our search interval. Okay, so we'll say. Well, if X is equal to a I then begins at the location equal to ay right. And then we have else. If and this one we're here, we should also be in else F statement else. If X is equal to a J. The potential second item in this list then similarly will just set the location equal to J. And if neither of those cases occur, world is set the location equal to zero, which there is no zero location. So don't let us know that it does. X does not exist in in this list. And lastly, we need to return. The location is so little, Either give us a value or a zero, the location or an indication that it's not present.

All right. So here we are, reading an algorithm that will search for an element A list by splitting. Listen before sub list, but kind of over and over again. Right? So we'll just call this, uh, Quaternary. And so what we need in here is an integer that we are searching for, as well as the list that we are searching in. Okay, I'm the list to be non empty. So the 1st 1 to do is set ah, equal to the left. Most element of the list of the index of it. And jaegal toothy, right, Most index omen. And then, while eyes less than and not just Jay but Jay minus two. And this is because we need to be able to split the list into four parts. Uh, that is not possible. If, say, the distance between them was only to write, that would give us on the only a couple parts to split into. So we're making sure that we can split split it into four elements here. Okay. And we're gonna set some variable is equal to roughly the first quarter of the way through that. For instance, this is an ellen 11 and which will be the halfway point. Ruffles were using the floor in case it's not evenly divisible by fours that we get an integer. Since we're finding index values here, only you as the next one. This will be our 3/4 of the way through that list. Okay? And so essentially, we have our list. You we've put you know, l about 1/4 of the way through. And then, um, about halfway through, and then you would be 3/4 of the way through. And so now we want to say, Well, if the enemy are searching for is greater than equal to the element in the list at this 3/4 in next, then we're gonna reset our left surging in next to you plus one okay. And then the next one is going to compare it to a M. So if X is greater than or equal to a the EMP index, when the one you to set I two the left side of this interval and Jay to the right side, we're actually setting two elements here. And so then we get I being equal to N plus one, and Jay will set too. You okay. And then similarly, what is the same thing for these next interval, Right? So if and I should be clear, because it's gonna be in else if Okay, so else, if it's the same things that X is good ankle too. Is that l? Then we need to reset our search range so that I is going to be l plus one and J is equal to m then the final else case. We would not need to reset I for this left the first quarter of the list, but we would just need to set J and J would then be equal to l. Okay. And so then that would return, you know, 1/4 chunk of the list. And then if that chunk of the list is larger than for it will go through it again and split it up again and again. And eventually there should only be three elements remaining. And so then we're just gonna parse through and and compare our search element to all three of those. So So get out of the loop. Here. We leave myself enough room to sort of index left correctly, but, uh, so we would say if X There's going to be equal to case of eye. Then we can set our location too. I all right, And then the same thing for the next. Next. So if X And in this case, we could say either I plus one or J minus one. Really? It will be the same thing and should be in else. If so, that's else. If Elsa excess people that ace of my plus one then was that the location too? I plus one. And lastly, just for the third remaining element. If X is equal to Ace of J, then the location is equal to J. All right? And so the very last case would be if the element is not in the list and so that we would set our location 20 and finally see if I have room here. Finally, we would want to return. Okay. We want to return the location. Yes, we have some sort of output. All right. And that will

All right. So you were looking for the mode in a list of non decreasing into Jersey. We'll just call this bunker the mode. Of course, you can really call it whatever you want. To you, the input is a little less arbitrary. It needs to be our list of integers, which I will abbreviate as ence with their being at least one into during this list, and it's non decreasing. So that means that the first element is less than or equal to the second element and so on and so forth throughout the list. So first, we want to define a couple of values so well define I be equal to one and Jay also being equal to one. And these represent the frequency of a value in the list. For I case, this is the I's gonna be the frequency of the value, and Jay will represent the frequency of the most frequent value. Right? So So the current frequency than a placeholder for the most most frequent overall. Okay, we will not define one more element. So wants to find the mode, and we're just gonna initialize that to you. First element in the list and adjust it if need be. Okay, so then, for every element in the list, I want to check if the value is equal to the previous one. So we're really just checking if it's been repeated and if so, will increase our our account, which is checking the frequency by one, if not, will redefine it to one, uh, so that it only starts over the count with each new element. Okay, So to do this, we want to index starting at two. So at K equal to X, we've already said the mod equal to the first element are 31 to end, and then I want a chapel. Sable is if this element is equal to its previous element, then we will increase our frequency counter, which is I so well said, I equal to itself plus one more. And if not, we want a recent I, we'll tow one to start a counter over for the new element. Additionally, if at any point we obtained in our account larger than J and we want to adjust, J All right, So if I is greater than J, then we want to set J equal to ay and also set the MoD equal to a K. So they were keeping in memory the element which has the current largest number of repetitions. And once we've gone through all on elements, we should be able to return the item, which was repeated the most, and so that would be the mode.


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