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public static bool ImagesEqual(image1, image2) {
if (wid1!=wid2) return false;
if (hei1!=hei2) return false;
if (crc1!=crc2) return false;
foreach(PixelPair p1,p2 in image1,image2) if (p1!=p2) return false;
return true;
}
Luc Pattyn [Forum Guidelines] [My Articles]
this months tips:
- use PRE tags to preserve formatting when showing multi-line code snippets
- before you ask a question here, search CodeProject, then Google
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Would use XOR functions ( blend with XOR , I think its possible with GDI images ... I used it some time ago to overlay two biztmaps to see where tehre is a differnce )
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Hi,
it depends what you like to match. If the pictures need to be exactly == binary identical, you could just use a md5 checksum, or something like this.
If you want to search for "similar" pictures e.g. showing the same thing but scaled (image for email, etc. ) then you could use a cooccourence matrix over the colors in the image.
If you want to check also if the contrast was changed then you need an even more elaborate algorithm, maybe doing a color histogram and then do a cross correllation. But these are only ideas...
Regards,
Tobias
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I have only 0...256 input in decimal. Some of readers they have knew my matter. I have explain it many time in this site. But still in the trouble. Could not decide it. I think about if the input one number is inside 256 then it would be search easy without any encryption. That is really trouble. Other's word on this matter is about think it much and search it. It would be had a method to survive this problem. I have thought this and researched on the matter then result it nothing then i have thought my cryptography algorithm without long number of the initial input, it would be easy work and can't keep as security better. If it is rose the work, which is that input number size would be extended, then it would be another huge work. this work request Not only me but also another people would be handle it then can decide the long input. Therefore i would like to survive my matter without extending input number. Extending input number means Initial number from other side would be extended there, not inside my algorithm. That is my good idea but i can't decide it easily. Would like to listen first other's important idea for this matter.
I think that professional people who are working on the cryptography and security for long time, they've got expirence and knew how to decide it.
I'm really glad if any good idea for this matter
Thank you
Professional people they almost be ready for any risk and know how to resolve when they got it.
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It is really hard to work out what you are trying to do. Sometimes you seem to have an 8 bit key - which suggests that you will have no hope of avoiding any sort of brute force attack, whereas at other times it seems that the message that you are encoding is only 8 bits.
If you are encoding only 8 bits, potentially you have only 256 possible messages, so an eavesdropper can determine, for example, if the same message is sent. The traditional way around this is to salt the message by appending a random number. As an example, to send 8 bits, add 56 random bits, then encode the resulting 64 bit message. Repeated messages with the same 8 message bits would be different and brute force searching becomes more complex.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Yes your right. I need to tell more about my algorithm. The algorithm has done on the input 8 bits and encrypted output 16-32 bytes, which are included bit Expansion, Inversion and Substitution operations.It is ready more complicated correlation between initial and converted data. Other words, it means that 8 bits input data became 128-256 bits ready. If you have another idea or i thought different idea from your idea, please could you make your idea more clearify? I need to know what your means. maybe it is important.
Thank you
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I have found Diffie-Hellman_key_exchange protocol ready and wanted to use it in my new algorithm but is that best one in the key exchange protocols? I need to make sure that protocol is suit for 32 bytes of the 8 bits key exchange before i use it in my algorithm. Most of the literature said that cryptographic algorithm could be good and perfect strength but using key exchange protocol should be enough strong security as the algorithm strong then whole algorithm can be said complete powerful one.
Please share me if you know about key exchange secret protocols. Which one is better to use it? My key generator is based one the 8 bits key. Algorithm input is 8 bits and output 32 bytes. Suggest me any thing else?
Thank you
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I don't understand how you can expect any sort of credible security using an 8 bit key - there are only 256 different keys to try so an exhaustive search would be trivial. Diffie-Hellman has a good reputation as a key exchange algorithm, and is used in many standard security protocols. You need to ensure that you use large numbers or else the algorithm can be inverted. I'd suggest you read Applied Cryptography[^]
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Yes. Peter That is good book. I'll do. So I think you need to know my problem first then you may say me what to do. I have only stable 0...256 number in decimal for input. That can be converted into the binary as one byte. Mission is keep security. I have gotten same idea with you since i saw it first time. This only 256 keys. Anyone can find it. In this case, how can i keep the security? Maybe you suggest me not only cryptography but also another thing to use it? If you have any idea please share me. I'm still in the trouble because of input 0...256 in decimal. thank you
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Good day for readers
This topic is related with patent for new creation
I have a question about new patent. If someone had creating new algorithm or equation then how the person can certify his or her algorithm or equation that is originality from him or her. There are many companies in internet for certification. But which one is really reliable and can certify the creation around the world. If any site or company for the best certification to approve new creation to the person, please share here your opinions. Your opinion is important for others, who are doing new something, and me. Certification purpose is that I'm the first person on the my new algorithm. I don't want others, who approves same thing with your creation. You know? Sometimes people are doing same thing with your creation mostly at same or at different time. Some of creators, who are doing discussing work, are cheating and stealing other's mind and technology to use it without authorization of the creator at another place from a original creator. The creator doesn't feel that. Even patent is also. Therefore i don't want others, who are doing new technical and technological novel,to repeat same thing in previous mistakes, which are about losing patent and intelligence knowledge. I hope you can share your expensive knowledge about patent and certification for new creation here if you have done something new creature before and certified that.
That your advice is not only important for me but also others who are reading this topic and doing new creation without no idea about certification.
Thank you for reading and sharing your knowledge.
Regards Bimbaa
-- modified at 11:25 Tuesday 30th October, 2007
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I have 2 patents for an internal combustion engine design so I'll try to answer your questions. The only place that can give you a patent ultimately is the united states government. Patents have to be prepared in a very specific way. For instance you can't just send in a description of your invention, it has to be written in a way that makes you patent application a legal document and is very spefic about what you are claiming as original and various variations of your idea. If you look in the phone book (assuming you are in the US) you should be able to find the names of patent attorneys in your local. Patents are typically made public unless I think there is some sort of national security reason not publish them.
I would recommend though that you do your own patent search first though at the US patent and trademark website.http://www.uspto.gov/patft/index.html[^]. This is the best place to check since that is what patent examiners look at. But I would also check other places to be safe.
One other thing, if you don't plan to apply for a patent right away it might not be a bad idea to write up a description of your idea and get it notorized at a notary. This will give you evidence of when you invented something if a dispute happens in the future about who invented something first.
good luck,
Mike
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MikeMarq wrote: But I would also check other places to be safe.
Such as this listing http://www.patentlawlinks.com/patoff.htm[^]
MikeMarq wrote: http://www.uspto.gov/patft/index.html[^]. This is the best place to check
One of the best no doubt, alongside those listed in my link above. But you must not restrict your searches to USA if you happen to live elsewhere.
modified 1-Aug-19 21:02pm.
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hai all,
is it possible to find the distance in meters between the two points if i have, their longitude, latitude and height.
iam aware that, the distance between two adjacent graticules is not standatd in all the places on the earth. but, based on the height, can we do any thing?
thanks for your views.
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Thinking out of the box, but I believe when you have 180° for the meridian and 360° for the equator (or vice versa)you have a full circle, right?
You could also find the circumference of the earth on google. Divide the circumference with 180° or 360° and that should give you (roundabout) the distance for 1° for your X axis and Y axis than you can use vector calculation to calculate the distance between 2 points.
I do think my idea is correct, BUT it is probably inaccurate.
V.
If I don't see you in this world, I'll see you in the next one... And don't be late. (Jimi Hendrix)
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we can do this,
but iam not sure weather this applies through out the globe.
any way, thanks for your response.
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I can, 100% sure. Just show me what and how GPS bases on, I'll find a way.
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Are you talking points that are close, or anywhere on Earth; that is, are you wanting to take into account the curvature of the Earth?
Another question: Are you talking about the great-circle distance (around the curvature of the Earth), or a straight line (which would have to tunnel thru the Earth if distances are very far)?
David
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Ha has a neutrino gun!
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
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CPallini wrote: Ha has a neutrino gun!
dagnabbit! I want one!!!!
_________________________
Asu no koto o ieba, tenjo de nezumi ga warau.
Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
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chandu004 wrote: is it possible to find the distance in meters between the two points if i have, their longitude, latitude and height.
The short answer: Yes.
The long answer: you need to deal with great-circle (geodesic solution) calculations. See: http://mathworld.wolfram.com/GreatCircle.html[^]
There is also an iterative solution that is often used for boat/air travel. Obviously continuously changing your heading to provide the absolute shortest route to the meter is difficult. So you create waypoints travel to them, change heading and travel to the next, etc., until you reach your destination. This has become easier with computers because a computer auto-pilot will auto-navigate to many more waypoints with micro changes in heading to keep a shorter path. But with ocean travel that is somewhat more difficult.
The question becomes how accurate do you need it, and what is the purpose of the answer. The piece-wise iterative approach would be:
<br />
surfacedistance(from: lat, lon, alt; to: lat, long, alt)<br />
{<br />
distance= ECEFdistance(ecef(to),ecef(from));<br />
if (distance<1000) return distance;<br />
ecef_mid=ecef_to*0.5+ecef_from*0.5;<br />
spher_mid=togeodetic(ecef_mid);<br />
spher_mid.alt=from.alt*0.5+to.alt*0.5;<br />
return surfacedistance(from,spher_mid)+surfacedistance(spher_mid,to);<br />
}<br />
or something similar (that is off the top of my head, google piecewise approximation of geodesic distances. The idea is that you can sum a series of vector distances in local space and the more accurate your division, the more accurate the answer. You can choose speed over accuracy at any point by changing the approximation distance change between mid-point division and vector-distance.
The bad news is, if you set the value too low, it is a heavy recursion level and you are better off calculating it directly. There are times in implimentation you may want to do either for other reasons related to use of the answer.
_________________________
Asu no koto o ieba, tenjo de nezumi ga warau.
Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
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chandu004 wrote: is it possible to find the distance in meters between the two points if i have, their longitude, latitude and height.
Like these:
http://mathforum.org/library/drmath/view/51720.html
http://jan.ucc.nau.edu/~cvm/latlongdist.html
http://mathforum.org/library/drmath/view/51711.html
http://www.movable-type.co.uk/scripts/latlong-vincenty.html
http://mathforum.org/library/drmath/view/54680.html
http://www.freevbcode.com/ShowCode.asp?ID=5532
http://mathforum.org/library/drmath/view/51717.html
http://www.meridianworlddata.com/Distance-Calculation.asp
http://mathforum.org/library/drmath/view/51722.html
http://www.csgnetwork.com/longlatdistance.html
http://mathforum.org/library/drmath/view/54941.html
http://www.csgnetwork.com/lldistcalc.html
http://mathforum.org/library/drmath/view/54680.html
"Normal is getting dressed in clothes that you buy for work and driving through traffic in a car that you are still paying for, in order to get to the job you need to pay for the clothes and the car and the house you leave vacant all day so you can afford to live in it." - Ellen Goodman
"To have a respect for ourselves guides our morals; to have deference for others governs our manners." - Laurence Sterne
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thanks a lot david,
one of the above links helped me to understand many interesting points.
and the logic worked for some of hte examples, so far i have tested.
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chandu004 wrote: and the logic worked for some of hte examples, so far i have tested.
the logic should work quite well. I haven't gone through most of the links, but I checked a few of them. I am not sure what you need the answer for, but it would be wise to clarify that your answer is for a spherical Earth not an ellipsoidal model. At least the ones I followed so far assume a spherical earth. This is a decent approximation, but again, it depends on what you are trying to do. No answer will be exactly right because even an ellipsoidal model is an approximate surface. The true surface is mapped in elevation points to the geoid.
_________________________
Asu no koto o ieba, tenjo de nezumi ga warau.
Talk about things of tomorrow and the mice in the ceiling laugh. (Japanese Proverb)
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You probably know that the "standard distance" for 1 nautical mile is 6076.115 feet. And that 1 degree of latitude is 60 Nautical miles (NM). There is no "standard" for longitude since it is wider at the equator and they all conjoin at the poles to zero distance. I think there is a formula for determining the length of 1 degree of longitude based on the latitude where the longitude is taken, but don't know it offhand. If you are somewhere in the range of 30 to 60 degrees latitude, the "standard distance" formula will get you close to the correct distance between two points.
John P.
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the easiest way to do it (assuming spherical earth)is to convert the lat-lon values into X-Y-Z values.
The center of the earth is (0,0,0),
the intersection of the prime meridian and the equator is (R,0,0)
lat 0, lon +90 is (0,R,0),
the north pole is (0,0,R)
the south pole is (0,0,-R)
where R = radius of earth
to go from lat lon to x y z is:
x = R * Cos(lon) * Cos(lat)
y = R * Sin(Lon) * Cos(lat)
z = R * Sin(lat)
You do that for point 1 and point 2, then determine the distance between them (straight line)
x' = x2-x1
y' = y2-y1
z' = z2-z1
d = sqrt((x'*x')+(y'*y')+(z'*z'))
One half of the angle from point 1 to point 2 is
a = ArcSin(d/2R)
to get the great circle distance take this angle (in radians), double it to get the actual angle, them multiply if by the radius of the earth
D = 2 * a * R
where D is the distance you are looking for.
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