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Knowledge of the Environment:
We must discriminate between knowledge an agent receives through its sensors and knowledge about the world from which the input comes. The world can be programmed at level of programming of complex data, and/or it can be learned through the sensor input. For example, a chess playing an agent would be programmed with the positions of the pieces at the start of a game, but would maintain a representation of the entire board by updating it with every move it is told about through the input it receives. Note that the sensor inputs are the opponent's moves and this is different to the knowledge of the world that the an agent maintains, which is the board situation
There are three main ways in which an agent can use knowledge of its world to update its actions. If an agent maintains a representation of the world, then it can use this information to decide how to act at any given time. Furthermore, if it stores its representations of the world, then it can also take out the information about previous world states in its operation. Finally, it can use knowledge about how its actions affect the challenge.
The RHINO an agent was provide with an accurate metric map of the museum and exhibits beforehand, carefully mapped out by the programmers. Having said this, the layout of the museum changed frequently as routes became blocked and chairs were moved. By updating its knowledge of the environment, however, RHINO consistently knew where it was, to accuracy better than 15cm. RHINO didn't move objects other than itself around the museum. However, as it moved around, people followed it, so its actions really were altering the environment. It was because of this (and other reasons) that the designers of RHINO made sure it updated its plan as it moved around.
Sample Program In this program we only display: Line Offset Numbers -----------------Source Code-----------------
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Simplify the given expressions using Boolean postulates Y = (A + B)(A‾ + C)(B + C) Ans. Y = (A + B)(A‾ + C)(B + C) = (A A‾ + AC + B A‾ + BC) (B + C) = (AC + B A‾ + BC) (B + C)
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