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10) Calculate the bitwise AND, bitwise OR and bitwise XOR for the following bytes, 01111111 AND 1110100101111111 OR 111010L01111111 XOR 11101001...

Question

10) Calculate the bitwise AND, bitwise OR and bitwise XOR for the following bytes, 01111111 AND 1110100101111111 OR 111010L01111111 XOR 11101001

10) Calculate the bitwise AND, bitwise OR and bitwise XOR for the following bytes, 01111111 AND 11101001 01111111 OR 111010L 01111111 XOR 11101001



Answers

Find the bitwise $O R,$ bitwise $A N D,$ and bitwise $X O R$ of each of these pairs of bit strings.
a) $1011110,0100001$
b) $11110000,10101010$
c) $0001110001,1001001000$
d) $1111111111,0000000000$

For this problem, we're gonna be using bit wise operations on bit strings that are showing combinations of letters. Now, I've already written Ah, put together a grid. I think it would make it easier to see on the screen here and across the top of the grid. I've put all the 26 letters of the alphabet. Uh Z I was one column short so Z will be, um, just off of the right edge of the grid here. Now, let's take a look at what our sets are. We have set A and set A is made up of a, B, C, D and E. So if I'm going to make a string out of these, I'm gonna one in each of those positions because that's where I have elements. And everywhere else is going to be a zero because there are no other letters in this set. So they're set a How about set B? Well, set B is made up of letters B, C and D g p T N v. So those elements have ones. And since those were the Onley elements in set B to put together our bit string here, we're going to put zeros everywhere else. Okay, two more. We've got C c has elements C and e I owe you. Have I? We have Oh, we have you and then X y Z. Okay. And since those are the only elements in SETC, we're going to put zeros everywhere else. Okay? And let's look at D set d has elements. D e h I so d e h I en oh, and t you ex and why? And again, just like with the other sets anywhere that we did not put a one. That means that we don't have that element in this set. So we're gonna put zeros. Okay, Now, again, you don't need to use graph paper to do this, but I find it just it makes a little bit easier to see keeping everything straight. So those there are four sets. Now we have four things we're gonna find. So first the switch color so it stands out a little bit. I want to find the union of sets A and B, so that means I'm looking for any element that is in either a or B. So if I look up here, I'm comparing these two rows So that means I have a, B, C, D and E. I also have G p t envy. So all of those spots haven't won in either A or B. The other ones just have zeros. So they're gonna put zeros in all of those spots all the way across. This would be the string that shows what those elements are. And again, if you want to see the letters, you could look at the top and see that that first one corresponds to a the second one to be and so on. But that string would be, um, the bit wise operation that we did here. Okay, Next we want to find an intersection. What about the intersection of A and B? Well, again, we're going to be looking at those first two. But this time we want the intersection. Where do both of them have a one? Well, they both have a one in B, c and D, and those are the Onley elements where that is true. So all of the other 23 letters of the alphabet are going to get a zero in our strength all the way across to Z. Okay. Our next one Well, this is in two parts because we have the intersection of two unions. So what I'm going to do for this one is I'm going to break this up, okay? And I'll go back to blue for this one. So the first piece is the Union of Sets A and D and then the intersection of thinking myself a little bit of a break here. So the Union of A and D and then the Union of B and C. And then once I have those too, I'll take the intersection of those two. I'll take the intersection of the two unions. Okay, that's what I'll be doing here. So first a n d. So I am looking at the first and the last set of those red lines anywhere were either one has a one. Well, a has a one in a B, C, D and e. So we're gonna put ones there now. De also has a one in eight, and I and oh, t u ex And why so everywhere else will be zeros. Okay, now, the second piece of that, I'm not looking at a and e anymore, so I'm gonna raise those arrows instead of looking at B and C, So these two middle ones, or either one of those has a one that will be in the union. So that is going to be B C de e g I We move all the way over toe o and P T U V X y z. Okay, so that's got quite a few elements in it. So everywhere else is going to have a zero. Okay, now that that's done when ready to put them together and this will complete this portion of the question. So now, looking just that these two, what is their intersection? Where do they both have a one? So that will be B, C, D and E. They both have a one in I Oh, t u ex and why everywhere else. Either they're both zeros or only one of them has a one. So we're going to fill in the whole rest of the string with zeros. Okay, last piece. Okay, change color one more time. Now we're gonna be finding the union of a and be and see and D So any one of these four rows, if there's a one anywhere, it's gonna have a one in this because I'm putting all of them together. That's going to be a B C D e. Not F G H I. Now there's a string here. None of them had J K, L or M. But we have an element and oh and P, they get once now again, we have another stretch. Nobody had Q r or S, but we need to include T, u and V, not W X, Y and Z. So combining all of those together in the union that's what that would look like. So this is how you can use bit strings. Uh, you bit wise operations to do these intersections and unions of sets.

Mhm Okay, so in this problem we've got a bunch of binary representations that are going to use fractions And we want to convert that into base 10 so our answers are going to have some fractions in them. The first thing I want to point out this thing down here, The decimal point also called the red X point. Everything to the left of the ratings point works like normal. So this would be the twos place in the ones place, everything to the right of the readings point is going to be cut in half each time. So 1/2 place, 1 4th place. And then if we went any further it would be the one in its place. Forget about that one. Okay, So what would the inside of this one be? It would be three plus a half and 1/4. That is three and 3/4. Okay, Now for the next one again, the left side is just going to be like normal. That's the forest place so I can already go ahead and put forward down here. And then what are these? This is actually going to be the 4th place And the 16th place. So all we need to know to finish part B is what is 1/4 plus 1/16 is 5/16. Okay, go ahead and pass the video here and we will come back in a moment to see the rest. All right, here's what we've got uh Syrian city part C is going to be just a regular fraction because there's no numbers to the left of the red X point party has no numbers to the right of the red X point. So that's just one mm. One interesting thing I want to point out is that in part C, if all of these were the number one And the answer would be 15/16, because that's the closest you can get to the number one When you're using 16ths. Okay, that is the end of this problem.


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