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Drunk Knight. A knight starts at the upper left corner of achess board and moves uniformly at randomand independently according to how a knight moves in chess. Find...

Question

Drunk Knight. A knight starts at the upper left corner of achess board and moves uniformly at randomand independently according to how a knight moves in chess. Findthe expected return time (expected number of moves) off the knight back to the upper leftcorner.

Drunk Knight. A knight starts at the upper left corner of a chess board and moves uniformly at random and independently according to how a knight moves in chess. Find the expected return time (expected number of moves) off the knight back to the upper left corner.



Answers

Predict/Calculate Moving the Knight Two of the allowed chess
moves for a knight are shown in FlGURE $3-46$ . (a) Is the magnitude of
displacement 1 greater than, less than, or equal to the magnitude
of displacement 2$?$ Explain. (b) Find the magnitude and direction
of the knight's displacement for each of the two moves. Assume
that the checkerboard squares are 3.5 $\mathrm{cm}$ on a side.

Were given to possible chest moves for a night. One move is given. I will draw in this bread color we're in. The total displacement is this vector, and a second move is given by green. It is the displacement is given by the green vector. Oh, these have one component. One of the X and Y components has one and another as two moves. So we will say that the magnitude we will predict that the magnitude iss the same. Now let's actually find the magnitude and directions for each of these two possible moves, we're going to take the the Each of the squares of the chessboard is 3.5 centimeters in length. So let's go ahead with a move. Number one we have. And next, uh, this place that, uh, you negative seven centimeters because it moves to moves to the left and a wine displacement of 3.5 centimeters. Since it moves one move. So now, to find the SaiPan news, which will be the magnitude of a vector one, we take this. We're route of X squared plus y squared, and that gives us a magnitude of 7.38 centimeters. Similarly, for the green Vector X Here now is 3.5 centimeters to the right. And why is seven centimeters upwards? So the hi partners of this right triangle it is going to be again 7.38 centimeters because the value is going under the square root are the same. So now, to find the direction of Victor one, let's go ahead and use the inverse tangent we want to use. Why over eggs? Well, so this gives us inverse tangent of 3.5 over negative seven, and that gives us an angle of 20 NW six point six series. But notice 26.6 degrees points in negative 26.6 degrees which points in this direction. And we want to add 180 degrees to it, such that we get 153.4 degrees now for the other one for, uh, doctor, too. We have no two equals tangent members. I want to over x two, which we will if we plug in the numbers seven and 3.5, respectively, for why an ex we get 63.4 degrees

High in this question. We are asked to show that when we have an infinite chessboard with a night place at the lower left corner which we will position as Ciro Ciro, we have to show that the night can walk to any square on this infinite boss. So any squire, according it and we can arrive, they're using standout night move. We will prove this by induction on if wishes the some off the co ordinates. And then But before that, we should look at how to actually move a night. One said to the right and one step up because that will, so are our problem. And luckily is quite simple to fi. A way to move a night to the right for one step and up for one sip. So we can just do it like these. And once we have this, no matter wish co ordinate in everyone, right, we Kansas move to the right instead and move up instead and we are done. Okay, So the induction process I just for Molly D at this point because with this alone, we can actually move anywhere, regardless off the induction process. But they asked us to So let's do this. So, as I said, s is a some off in person basic step. When is one then it means us either. M o in is one right wishes what we just shown already. So this is clear that we can move it. So basic stick. We're clear for inductive steps, so supposes Sorry. Not this one. Oops. Sorry. I suppose it is true. For for some is we want to show that is true for his possible in SPL. But we don't have to do anything. We just say that. Okay, Wish, position him. And in such that impulse Candies. This number is plus one. We can move to the right bye in position and move up by m position using this. You're seeing these two step that we just introduce and and we're done. I hope this convince you. And so we have proven both step. And by Met induction we can move to any square on this infinite chessboard. And that is it. Thank you.

Okay, So here gonna let x b the number of balls that fall into the first been right. So from I was one toe em. We have M balls. And so x I x of I want to know the number of balls falling into that been. Okay, So what's happening is X I will be equal to one if the ice ball goes into Ben and zero otherwise, okay, so then we want to consider the expected value vexed by that's gonna be equal to the probability that equals one. So that probably that the ice ball has gone into her bin and that is equal to won over end. Okay, so then when we consider the expected value of X, I was gonna be this some toe em of the expected value. Lex. I right. The first n is tthe e number of bins. All right. So you have how many bins? There's a one over that number of Ben's chance that it's falling into the 1st 1 in Sedalia end balls all going into this one kind of about how many wolf go into this one been. Okay. Well, then we're getting me some from Michael's one tem of one over and, well, this is just equal to M over n and says he expected number of of balls.

So for this problem, no nets. Sure, you could say so. Because from exercise 56 we know the graph contains an isolated were tax and no, a Milton half. You could taste as graphs contains on dhe ass elevated war tax.


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