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Stack[StackPtr,3] = rGPSy() StackPtr = StackPtr+1 // assume two exits so take first one //(start with left, then CW) if dir & 4 rTurn -90 rForward 60 else // forward MUST be first option rForward 60 endif endif wend Return //============================================================= CheckBefore: yep = false if StackPtr = 0 then return dX = rGPSx()-Stack[StackPtr-1,2] dY = rGPSy()-Stack[StackPtr-1,3] if PolarR(dX,dY) < 10 then yep = true Return //============================================================= RetraceMaze: // now do it again - but no mistakes x=rGPSx() // save current position y=rGPSy() rLocate 108-60,90 // put robot back in the beginning rTurn 90 SetColor White // and erase old one circle x-20,y-20,x+20,y+20 rPen Down,LightGreen rForward 60 StackPtr = 0 while rLook()<>Red // determine if current cell has more than one exit numExits = 0 dir = 0 cnt = 0 if not(rFeel()& 16) cnt = cnt+1 dir = 4 // left side endif if not(rFeel() & 4) cnt = cnt+1 dir = dir+2 // straight ahead endif if not(rFeel() & 1) cnt = cnt+1 dir = dir+1 // right side endif // if there is only one exit, take it if cnt = 1 if dir=4 then rTurn -90 if dir=1 then rTurn 90 rForward 60 continue else
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FIGURE 14.5 (Continued )
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// take the exit on the stack dir = Stack[StackPtr,1] if dir & 4 rTurn -90 elseif dir&2 // do nothing else rTurn 90 endif rForward 60 StackPtr = StackPtr+1 endif wend Return //=============================================================
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FIGURE 14.5 (Continued )
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FIGURE 14.6
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Corridor maze solving.
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NEGOTIATING A MAZE
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dir (decimal) 0 1 2 3 4 5 6 7
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Binary equivalent 000 001 010 011 100 101 110 111
Meaning no exits exit only to right exit only forward exit forward or right exit only left exit left or right exit left or forward all three exits available
FIGURE 14.7 The variable dir holds a code for the possible exits from the current cell.
If the robot determines that it has been to this cell before, it will retrieve the saved information. There are four items that are saved:
1. The GPS (global positioning system) coordinates of the cell, which will be used to deter-
mine if the cell has been previously visited (two values the X and the Y coordinates).
2. The compass heading the robot had when it entered the cell. 3. The value dir discussed earlier.
Once it has retrieved the saved data, the robot orients itself so it has the same heading that it did when it originally entered the cell. It then looks at the value of dir to see what the possible exits are. The robot knows (since it is revisiting this cell) that the exit it took did not work (or it would not be back to this cell). The robot needs to mark this exit as not usable, and it can do so by simply removing the rst one bit (starting at the left most bit) in the binary number dir. Remember, the exits are tried starting from the left, which is why the rst one in dir represents the choice last made. After the bit is removed, all of the data is resaved to be used again if the cell is revisited. The robot then looks at the new value of dir and uses the rst one bit found to decide which path to try this time. If the value of dir is 0, then the robot knows that all the possible paths have been tried (and all have failed). When this happens, the robot turns around and goes back from where it came. It also needs to forget the information about this cell since it will not be coming back here again. It might not be obvious that the robot will not be back here again so let s examine this idea. The whole premise of this algorithm is that the robot methodically tries every possible option available to it when it visits a cell that has multiple exits. Each failed try eliminates that option from the future choices available to the robot. When all exits have failed, the robot returns to the last cell with multiple exits and tries the next exit available there. Either one of the choices in that cell will work, or the robot will go backward again until it eventually nds a cell that has an exit that does work. If the robot repeatedly executes the above logic, it will eventually get out of the maze. In our example a red line is placed at the end so the robot can use the camera to determine when it has completed its task. The data saved by the robot is placed in an array called Stack. To better understand the principle of a computer stack, think of a stack of dinner plates. When you want to add a new plate to the stack, you place it on the top of the stack. When you want to get a plate from the stack, you take the one at the top of the stack. This structure can be implemented in our program using an array. A variable is used to point to the array position currently being used. This variable (StackPtr) starts with a value
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