# exact motion quiz, moving to the left?

 0 may be I don't get it if the bot moves to the right, shouldn't then the second cell become the first one instead of the third one? the quiz takes 0.11 0.11 0.33 0.33 0.11 as the right answer, but shouldn't it be 0.33 0.33 0.11 0.11 0.11 instead? meaning if you have 1 - 2 - 3 - 4 - 5 * and you are in the first cell, and you move to the second: shouldn't the new world be: 2 - 3 - 4 - 5 - 1 * asked 26 Feb '12, 23:03 Andres Paglayan 47●1●1●5 accept rate: 0%

 2 Well, we assume that the robot moves, not the map. Map stays the same all the time, it's just cyclic. Imagine that the robot is moving around a round stadium with several entrances. To map it, we declare that one place is a start, but since the "building" itself is round, once you reach the far right end on map, you actually end up at the start. But still, it's you (or the robot!) that moves, not the building.  * 1 2 3 4 5 |_______| move right * 1 2 3 4 5 |_______| move right * 1 2 3 4 5 |_______| move right * 1 2 3 4 5 |_______| move right * 1 2 3 4 5 |_______| move right * 1 2 3 4 5 |_______| move right  answered 26 Feb '12, 23:10 Gundega ♦♦ 44.0k●70●170●315
 1 It's the robot that moves... If you remember the location vs probability graph that was drawn at the beginning of the localization lecture, that "location" is actually the "robot's location in the world" and the probability is the "probability of the robot's belief that it's at the correct location". The "bumps" on that graph will move as the robot moves. Ideally, as the robot moves the probability of it being correctly localized will increase and "travel" (shift cells) with the robot as it moves left/right or up/down. Eventually, it will find itself in the cell with the highest probability (biggest bump on the loc. vs prob. graph) therefore successfully localizing itself in the given "world". answered 26 Feb '12, 23:58 Ernur Karado... 680●5●8●23
 1 Moving the robot to the right is the same as moving the world to the left. And since you are trying to localize the robot you don't know exactly where it is, so you don't have a robot marker to move around. But you do have a world (believe probability of the world). A - B - C - D - E 1 - 2 - 3 - 4 - 5  Say the robot was at position C observing 3 and you do a right move. Then the robot should be at a position observing 4. After moving the believe world left (to simulate the right move). A - B - C - D - E 2 - 3 - 4 - 5 - 1  And this will work for when the robot was at A, B, D or E too. answered 27 Feb '12, 00:14 Marius 991●4●8●19 Nowhere in the lectures the robot is moved left. If you want to explain this with the notation of "Robot is the center of universe and the world moves while he stays in one place" then you have to say - robot moving to the right is the same as world moving to the left (27 Feb '12, 00:21) Gundega ♦♦ Sorry, I assumed the robot is in a 1D world and can move both left and right. (27 Feb '12, 00:31) Marius Well, it can. But this particular lecture deals with robot moving right, while the world stays as it was, and it is very confusing, when you change not only what moves, but also in which direction. And in the programming examples it is the robot that moves (when it moves, you change the belief - probability - in what coordinates it is now), not the world, so maybe it's better to try and use model of the world that is similar to that of lectures, not exactly the opposite, to help other students. (27 Feb '12, 00:47) Gundega ♦♦ Yes you are right. I should keep with the lecture in my examples. I changed the direction. (27 Feb '12, 01:52) Marius
 1 This is how I understand it. If it moves to the right exactly one, it still has the same probability that it knows where it is. In other words, it knows it moved exactly one space and so the probability of where it is at now is no better or worse than before. That is why the probability distribution shifts to the right with the robot. answered 27 Feb '12, 00:30 uofubry 17●1 Actually, it's going to be worse since movement only will decrease the probabilities. If "p_move" is equal to one, then the probabilities would remain the same (in a perfect world). (27 Feb '12, 11:07) Ernur Karado...
 0 No. The world should be 1 - 2 - 3 - 4 - 5 *  Whence, what you believed to be true when you were on 1 is now believed to be true in 2. The same follows for cell. answered 26 Feb '12, 23:09 Conrado Silv... 4.3k●7●27●68
 0 You're absolutely right if you view it as shifting the world instead of shifting the probability vector. Since you've shifted the world, you can look up the probabilities in the original vector. So the probabilities for 2 to 5 and then 1 are now: (0.11, 0.33,0.33,0.11,0.11). Since you're asked to give them for 1 to 5, just copy down the last followed by the first four:(0.11,0.11,0.33,0.33,0.11) which is the right answer. answered 26 Feb '12, 23:10 Adam Sherwin ♦♦ 17.9k●21●76●125 I just re-hear the video many times. At the beginning Prof. Thrun talks about the robot moving to the right. At around second 24 I now believe he says the "world" (not the bot) moves to the right, (Thus the bot moves to the left which makes sense). Can somebody clarify if the Professor refers to the bot or the world moving to the right? (26 Feb '12, 23:28) Andres Paglayan 1 Robot. The world does not move. I'm sure astronomers will disagree, but for the purpose of this class you may assume that the world does not move. P.S. I just rewatched - he said "robot" moves to the right, with a bit unclear pronuncation. (26 Feb '12, 23:34) Gundega ♦♦ It's simplest to think about the robot moving instead of the world but in tracking two entities, which one moves is irrelevant. You'll always get the same answer. It's only once you introduce a third object that selecting the right reference frame makes computations simpler. (27 Feb '12, 00:06) Adam Sherwin ♦♦ Which one moves is not irrelevant in the context of the question. If the bot moves right, the world shifts left. If the bot moves left, the world shifts right. I am still unsure if the question refers to the bot or the world moving right. (~ sec 24) (27 Feb '12, 11:13) Andres Paglayan
 0 I believe Prof Thrun mentions that the motion in the example is moving two steps to the right. In this case the robot moves from cell 1 to cell 3 (given exact motion). This was probably in order to be able to better show undershoot (which would be moving one step to the right and not staying in the same place). answered 27 Feb '12, 02:07 yasho 336●3●11
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