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image of a stop sign that has been edge detected, outlines of objects show in white on black background

We first tried to find the border of the sign.  The outside edge of the sign is a region most of the time, so this was a possibility.  However, there is not a good way of determining which region was the sign.  A first guess was to choose the biggest region, however this did not always work as the sign was not always the biggest object in the picture. We then noticed that the O of stop showed up as two regions, the inner ring and the outer ring.  Since these two regions have approximately the same center (fig. 7), it is possible to find regions that have centers close together and then analyze them to see if they are the sign in question.  For example, if the point is on a stop sign, then the point, the point a bit above it, and the point a bit below it will all be red.  If the point is on a do not enter sign, then the point will be white and the points above and below will be red.  

Both the inner and outer ring of the 'O' in a stop sign have the same center

The points above and below were chosen by using the height of the region found.  By choosing points the height of the region above and below the identified center, the points are on the outside of the letter.  This tells the background of the sign itself.  

This method of finding signs is not the most accurate.  It can be fooled by other signs. For example, a sign for a business that had an O on a red background would be considered a stop sign by this algorithm.  This could be helped by checking more points.  Points immediately around the center should also be red and some points in between the center and the currently checked outer points should be white.  This would improve the accuracy, but it could still be fooled. This algorithm also works best for signs with color.  Signs with white backgrounds such as speed limit signs are hard to find because values that are defined as ‘white’ cover a large range of pale colors and therefore many things not signs are identified as signs.  

Despite these drawbacks, this code is advantageous because it does not require large convolutions like the first method. This means that the method is much faster. To find the sign of the correct size with the first approach, it takes about 30 seconds. To find the sign including the size with the second approach, it takes about .5-1 second.

Sign Reference Percentage Above Threshold
Stop Stop 95%
Speed Limit Stop 0%
Do Not Enter Stop 15%
One Way Stop 0%

Approach iii: template matching and edge detection

For our final algorithm we used both template matching and edge detection in an attempt to combine the positive aspects of the two approaches.  The combined algorithm first utilizes edge detection to isolate regions within the image, throwing out all regions of a size below a certain threshold in order not to waste time with extraneous convolutions.  Once the significant regions, usually numbering between twenty and one hundred, have been identified, each is re-sized to a standard small template size and compared to each image in the template library.  Regions with high correlations with a template image are assumed to be the part of the sign that template was created from.

This combined algorithm performed quite well, although it was not error-free.  As was the case with pure template matching, attempting to match pictures of entire signs was prone to error because the background of any given sign can vary, occasionally leading to quite low correlations.  However, we discovered that using this algorithm to match only specific elements of a sign, such as the “S”, “T”, “O”, and “P” in stop (fig. 8), was much more reliable, as the conditions for such sub-sign features are much more consistent than for the overall sign.  In addition, searching for individual sign features and not just the sign itself means that partially obscured signs that would be beyond our first template matching algorithm would present no problems for this modified one. Our template library was expanded so as to contain a number of sub-sign features for both stop signs and speed limit signs, as both of these signs contain changing characteristics (background for the stop sign, speed limit for the speed limit sign) that can thwart conventional template matching.  Again, we found that matching individual letters yielded a high success rate (fig. 9).  Our basic algorithm was also fairly successful in identifying one way signs and do not enter signs, even though no sub-feature library was created for these signs and only the sign itself was used as a template.

pictures of individual letter 'S' 'T' 'O' 'P' and stop signs with white and black backgrounds

graph showing letters have higher correlation values than signs

This algorithm showed significant success in combining the positive aspects of both tablet matching and edge detection.  In comparison to pure template matching, this method performs correlation calculations on much smaller images and size matching is a non-issue because all regions are re-sized to match the templates.  In comparison to pure edge detection, this method is less prone to false positives and can be much more easily expanded to detect new signs simply by adding to the template library.  Overall, the combination of template matching and edge detection yields a more robust, flexible, and reliable method of sign detection than either algorithm used alone.

Sign Reference Percentage Above Threshold
Stop Stop 95%
Speed Limit Stop 15%
Do Not Enter Stop 66%
One Way Stop 33%

Questions & Answers

what is math number
Tric Reply
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
Kala Reply
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
Moses Reply
12, 17, 22.... 25th term
Alexandra Reply
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Shirleen Reply
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Adu
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
kinnecy Reply
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
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salma
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
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Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
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Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Kimberly Reply
Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
August Reply
What is the expressiin for seven less than four times the number of nickels
Leonardo Reply
How do i figure this problem out.
how do you translate this in Algebraic Expressions
linda Reply
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
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Source:  OpenStax, Street sign detection with template matching and edge detection methods. OpenStax CNX. Dec 19, 2011 Download for free at http://cnx.org/content/col11387/1.1
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