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I am trying to wrap my head around "Robot motion" concept in localization To begin with, we are given an initial "distribution over the cells" with values of [1/9, 1/3, 1/3, 1/9, 1/9] Now Robot "moves" to right. So what I am unable to grasp is why the probabilities shift to right by one cell? Thanks |
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If Robot was in cell-1 and then shifted to the right it will now be in cell-2. Therefore the probability that it is now in cell-2 == probability that it was previously in cell-1. Similarly, probability that it is now in cell-3 == probability that it was previously in cell-2. The net effect is that the probabilities for all cells shifts/rotates to the right. |
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The vector is the belief state of the robot.
The idea is that the vector represents the belief state of the robot and the values are the certainty of the robot about whether it is in that cell. |
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Lets take and simpler example. p = [0.5, 0.0, 0.5, 0.0] before moving, you have 50% confidence that you are in the first block and 50% confidence that you are in the third block. What happens if you move one block to the right (before any further measurement) ? p = [0.0, 0.5, 0.0, 0.5] Or you can take an extreme example where you know your location : p = [0.0. 0.0, 1.0, 0.0] if you move to the left, what is the new probability distribution ? |
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Search for question, "2nd video, why is the belief shifted at all?" Among the answers, look at the one that starts with "OK, I think I'm getting it now." It discusses shift and eliminates the confusion of advancing to being in front of a second door. Discussion there should help with deeper understanding. |
Thanks Leonardo, Conrado & icogs
Its clear now ..
I think the crucial point is to remember that if you believe that you are in a state-x with a prob X, then when you move one step, you are ONLY X-confident that you are in state x+1
Also videos 18-23 help reinforce this concept