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Subject: 
Re: maze solving algorithm
Newsgroups: 
lugnet.robotics
Date: 
Sun, 8 Jun 2003 06:11:14 GMT
Original-From: 
scott davis <rcx2man@hotmail.*nospam*com>
Viewed: 
1325 times
  
I'm back to working with NQC.  I gave up on brickos.  I think I have a
program that works, I just need to find some time to test it.
Thank you for your help.
Scott

----- Original Message -----
From: "Bluey" <Wolf_and_eagle@spamblock.yahoo.com>
To: <lego-robotics@crynwr.com>
Sent: Saturday, June 07, 2003 10:32 PM
Subject: Re: maze solving algorithm


Ok, so did you get the idea about using the Binary system to identify the
"cell"?

Now, the next choice is what type of variable to use:

You can use a small 16 bit variable for each "cell" which will take up 400
bits=50 bytes. This was will conserve less memory, but it will make it • easier
for the robot to uncompress the data.

Or, you can use a larger 32 bit variable to hold more info efficiently at • 32 *
5=120 bits=20 bytes. This down side to this is, it will take more time to
uncompress this, and it won't fit in a an array as well (since each • variable
would hold 5 of the "cells"; therefore needing an extra subroutine to • extract
the number.)

I'm going to assume you chose the 16 bits per cell in an array option.
Instead of wrapping our cell info into a 2 digit number, we can leave the • raw
binary info in the 5 digit variable. So, if our "cell" has no walls, we • can just
enter the number 00000 into the variable.

Now for the next problem. The robot now has the ability to maneuver and • map a
5x5 maze, but one thing that might not come to mind is the robot doesn't • know
where it is in the map when you start it out. For all it knows, when it • enters
the maze it could be in the middle of the bottom of the maze, or to the • left or
the right. Unless it know where the entrance is, it won't know where to • place
the map info it gathers. Make sense?
The only two options I can think of are, 1. Have the entrance of the maze • at a
standard point in the maze (I.E. 3 "cells" from the right")
(provided you are going to start the robot on the outside of the maze • going in
through an entrance), or 2. Have the robot construct a temporary map that • is
[(2*5)-1]x[(2*5)-1]=9x9 so that no matter where you put it to enter the • maze, it
start its mapping in the middle of the maze in it memory (I.E. coordinates • 5,5),
and will have space to fit.

I don't know how you would create "temporary" variables in BrickOS unless • you
used disposable inline variables or classes (if those can be deconstructed • at an
earlier time). The problem is that the temporary map would need to be used • as
the main map until it figured out where it was......

Or here is another idea, have the robot use a 5x5 memory map, and have it • think
it starts at (3,3). Then as it finds that a passage is actually the edge • of the
map, have it move the known map over one "cell" to accommodate the new • info. In
other-words, if you start the robot actually at the bottom middle of the • maze,
and the robot starts at (3,3) (in its map) and then it moves "up/north" 2
"cells" (physically and in its map) then in its map, it will reach the • top,
middle "cell" of the "map in memory". Therefore, when it moves "north" one • more
space, it will be off the map, it could then see if there is an empty row • at the
bottom of the map. If there is an empty row, it could move all the cells • down
one space knowing that the map was oriented too far "north" compared to • the
actual maze.
Do you understand? if not, I'll try explaining it again.

I'm sorry if this is complicated or confusing. I'll try to elaborate on • anything
I didn't make clear.

Timothy



Message has 1 Reply:
  Re: maze solving algorithm
 
......... well, if you get a chance, send me a copy of the program. Timothy (...) (21 years ago, 8-Jun-03, to lugnet.robotics)

Message is in Reply To:
  Re: maze solving algorithm
 
Ok, so did you get the idea about using the Binary system to identify the “cell”? Now, the next choice is what type of variable to use: You can use a small 16 bit variable for each “cell” which will take up 400 bits=50 bytes. This was will conserve (...) (21 years ago, 8-Jun-03, to lugnet.robotics)

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