|
the two points have rotation data which is how I consider them lines (infinitely long lines running thou the know point at the given angle)
I didn't have the line equations which turns out to be what I was actually trying to find.
I have worked it all out now anyway, but thanks for the reply
I may or may not be responsible for my own actions
|
|
|
|
|
In general, lines in 3D do not intersect. The closest thing is the line-line closest approach[^] which is too complex to explain here, but that article is good.
The Z axis is actually not important, I want an intersect point on only X and Y planes
This leads me to believe that you actually want 2D line intersection. There are a number of ways to formulate that but a nice one is using the line equations:
Line 1 = (a1,b1) + k1(dx1,dy1) =>
x1 = a1 + k1.dx1 and
y1 = b1 + k1.dy1
Line 2 = (a2,b2) + k2(dx2,dy2) =>
x2 = a2 + k2.dx2 and
y2 = b2 + k2.dy2
Target: L1 = L2 =>
a1 + k1.dx1 = a2 + k2.dx2
b1 + k1.dy1 = b2 + k2.dy2
There are only four unknowns (k1 and k2, and the meeting point (x1,x2)). The bottom two equations eliminate the coordinates leaving two equations and two unknowns. Now eliminate k2 (k2 = (k1.dy1 + b1 - b2)/dy2):
a1 + k1.dx1 = a2 + (dx2/dy2)(k1.dy1 + b1 - b2)
k1(dx1 - [dx2.dy1/dy2]) = a2 + (dx2/dy2)(b1 - b2)
k1 = [a2 + (dx2/dy2)(b1 - b2)] / (dx1 - [dx2.dy1/dy2])
Looks complex, but all those quantities are known at the start, so you can calculate k1 (unless dy2 is 0, in which case special case it, or the whole right hand side is 0, in which case the lines are parallel and there is no intersection).
Then you can use the k1 to calculate k2 from either of the last two equations, and then use k1 and k2 to find the intersection point. Note that if k1 or k2 is < 0 the intersection is 'before' the 'beginning' of the respective line; you may want to count such cases as failures, depending on the problem (and similarly for k > 1).
(NB I didn't check that maths, there may be a mistake, but the principle is sound.)
|
|
|
|
|
Thanks for the reply. I have solved this already now thou.
I did just want 2D intersection which is why I said the Z axis is not important, I can take a middle ground value for the resulting Z co-ordinate. The part I was really trying to find out (as it turns out) is to determine the line equations from the data I had.
For this I used cosine and sine methods to calculate a "second point" for each "line". From this I could then calculate the slope (m) and y-intersect (b) for my line equations. Then I solve the equations to get the intersect point. simple when you know how 
I may or may not be responsible for my own actions
|
|
|
|
|
NOTE: Please let me know if another section/forum on the site is more appropriate for this question.
I'm trying to solve an algorithm puzzle and I was hoping to get some help. Its not for any formal study, just a mind exercise/ puzzle but I've hit a road block:
The goal is to figure out what algorithm or formula can be used to decrypt and encrypted string of characters where you know the characters being encrypted and what their encrypted value is but not how the encrypted value is determined. Obviously this would be like looking for a needle in a haystack except that there are some parameters to this that help (I think) narrow the process down. I can't help but think there's got to be a pattermn or method for figuring out something like this when you know the values of both the unencrypted character and what it encrypts to.
The parameters of the puzzle are:
1) The encrypted characters are limited to the printiable set of characters within the standard 256 character ASCII set of characters (basically what's on a standard US keyboard). I suppose the thing woiuld work with non-visible charactres but I believe the intent is to use any of the keyboard characters a user could use and so that would include numbers, letters and some of the special characters too.
2) the string of characters to encrypt is limited to 12.
3) Each single character is encrypted to a 2 character value (I think it's always a number but I'm not certain ) so that a string of 12 characters will be encrypted to a 24 character long value.
4) Each character is encrypted to the exact same value based on its position in the string of characters to be encrypted.
For example if the text to encrypt is abcdefga1234 then the lower case a is encrypted to the value 57 for position 01 in this string of 12 characters and in the 8th position the lower case a is encrypted to a value of 0A.
No matter what combination of characters are in positions 1 thru 7, the lowercase a will always been encrypted as 0A when it is in the 8th position in the string of characters.
5) Each combination of characters plus position has a unique encrypted value and that value is constant when the position of the character is the same.
I hope the last 2 above make sense; it's hard to convey these in words.
I know that a large lookup table of characters along with their decrypted values and position could be sued to decrypt an encrypted string but unfortunately that’s not the answer to the puzzle. The goal is to decipher what formula or algorithm will decrypt an encrypted character using the above parameters.
If anyone has any links or suggested reading material on how to solve something like this I would most appreciate it.
|
|
|
|
|
Ok let's see if I can come up with something, sounds fun!
The hints we can gather from the given puzzle parameters are:
1) It seems we're talking ASCII characters, one byte per character and a well-known codification.
2) A maximum length of 12 chars, ok.
3) Putting together this with what I see in point 4, it seems the encrypted 2-character data seems to represent a byte value in hex (examples are "57" an "0A", both valid hex byte values).
4) So, the character position in the input string is the second variable to take into account (the first is the character ASCII code itself, see point 1).
5) This confirms what I said for points 1, 3 and 4.
So, we want to find an operation "op" which, applied to the two variables (character code and position) gives us the resulting hex code.
Now let's have a look at the examples we find in point 4.
We know that the ASCII code for "a" is decimal 97, hex 61h, so:
61h op 01h = 57h
61h op 08h = 0Ah
(where 01h and 08h are the character positions in the example, of course)
Let's have it in binary and decimal too, it will be useful:
0110 0001 op 0000 0001 = 0101 0111
0110 0001 op 0000 1000 = 0000 1010
97 op 1 = 87
97 op 8 = 10
Now, one interesting thing we can see is there seems to be a symmetrical relation between the three values:
97 - 10 = 87
This hints at a relation where the only variable is the character position (1 or 8).
So, let't try to find an operation which transforms our equation system in:
97 - 10 = 87
97 - 87 = 10
Looking at the binary representation of 87 and 10, I can seen that if I shift 87 three digits to the right I obtain 10 (0101 0111 shr 3 = 0000 1010).
So, one possibile solution for the puzzle is:
1) Take the charater's ASCII code ("a" = 97)
2) Subtract 10 (97 - 10 = 87)
3) Shift the resulting byte to the right by a number of digits equal to the character's position plus 2, where the first position in the input string is 1, but if the number of digits to shift is 10 or more, subtract 10. (so, 1 + 2 = 3 -> shr 3 ; 8 + 2 = 10 -> 10 - 10 = 0 -> shr 0).
3) Subtract the resulting byte from the character's ASCII code (so, 97 - 10 = 87 ; 97 - 87 = 10).
While this works for the two given examples, it's a very weak explanation. For example, what I say in point 3 can work in other ways too, and still give the correct results.
Maybe someone who is more experienced in cryptography than me will come up with a better solution.
Hope this can somehow help you, it sure was fun for me! 
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
In general, it is not doable: consider what happens when "a large lookup table" that you describe in your post is filled with randomly generated unique entries. Your target alphabet easily accommodates this, so the result satisfies each of your criteria 1 though 5. Yet the only way to "crack" this "pattern" would be to record the entire lookup table.
|
|
|
|
|
Clearly a very basic algorithm if each character is changed individually. I would say it is just a shift cipher of some sort.
First convert the characters to hex
a b c d e f g a 1 2 3 4 =
61 62 63 64 65 66 67 61 31 32 33 34
now find out what values you need to add to those to get your resulting encryption. As I only have two values to use, here is a sample
61 62 63 64 65 66 67 61 31 32 33 34
+ -10 -57
--------------------------------------------------
57 0A
fill in the rest and look for a pattern. Could be a repeating key, e.g. -10 +5 +23 - 57
interestingly (and maybe a coincidence) is that the second "a" is a reduction of the previously calculated "a"
So the algorithm could be...
1. start with a byte array of size [256] that contains all -10 values
2. loop each character
3. convert character to hex (iHex)
4. convert iHex to decimal (iDec)
5. encrypted hex (eHex) equals: iHex - bytearray[iDec]
6. set new bytearray value: bytearray[iDec] = eHex (ready for next use)
7. output eHex to result string
8. process next character
...of course that is just one possible to match the results you have given
I may or may not be responsible for my own actions
|
|
|
|
|
First off thanks to both Moreno Airoldi and musefan for providing such detailed and well thoughout replies, this is the kind of help internet forums/boards are all about.
I didn't realize how beenficial it would be to prvide additional sample values esle I would have given more then just the few I did. I am going to check out both your recomendations but it will be a day or so before I can get to them (thats why my own reply to each of you has been so slow coming) and so in the mean time here are more examples that might actually scream a pattern to you 2 who are clearly more versed in patterns.
I tried to provide a sampleing without going over board. The belwo restroicted to a,b & c (in upper & lower case) is for just the first 3 positions and the resulting number of rows is 18. To show all 12 positions for just a few characters is a lot.
Anyway, if anything about the pattern in teh eblow jumps out at you please let me know else I'll let you know what comes of testing your suggestions.
Thanks
Position, unencrypted character, encrytped value
1 a 57
2 a 24
3 a 2A
1 A 77
2 A 04
3 A 0A
1 b 54
2 b 27
3 b 29
1 B 74
2 B 07
3 B 09
1 c 55
2 c 26
3 c 28
1 C 75
2 C 06
3 C 08
|
|
|
|
|
Well obviously the same character in upper and lower case in the same position differs by (hex) 20. This strongly implies that the ANSI value of the character is being used in a relatively unsophisticated way (upper and lower case are separated by hex 20 in ANSI). The difference can be either way; to me this suggests an xor. However, it is not totally trivial as a->b->c is not just +1 each time:
(ASCII value a 61 b 62 c 63)
Slot 1 a 57 b 54 c 55
Slot 2 a 24 b 27 c 26
Slot 3 a 2A b 29 c 28
All numbers posted here are in the range 06-77. I would guess that it is a 7 bit algorithm (working on 00-7F) and operating with XORing the input character with some function of the input. Let's look at the XOR values for those slots:
Slot 1 a 36 b 36 c 36
Slot 2 a 45 b 45 c 45
Slot 3 a 4B b 4B c 4B
Looks like I'm onto something there. All that remains is to determine what the XOR parameter which is a function of position is. There isn't enough information in 3 slots and the pattern 36->45->4B does not immediately jump out at me as being meaningful. The encrypted value of 'aaaaaaaaaaaa' should provide all the XOR keys and hopefully that is enough to identify how they are calculated.
|
|
|
|
|
You beat me to it 
Your suggestion of encrypting "aaaaaaaaaaaa" should be enough to solve this you are right.
I think there is no pattern as it is just a random set of chars used to provide the XOR values, which should start with "6EK........."
I think the OP has been lucky that such a simple algorithm was decided on
If my jokes make me laugh, then I have already succeeded with 100% of my target audience
|
|
|
|
|
Well he said it was a mind exercise/puzzle, so I guess it's supposed to be solvable by a person in a reasonable time.
You may well be right that the XOR key is simply a 12 character pass code/encryption key.
Encrypting 'aaaaaaaaaaa' (or the equivalent with byte zeroes) is often useful for key analysis in simple encryption schemes. I had to modify the first version of my stream encryptor because by throwing a stream of zeroes at it, it would dispense the key. (Real life binary messages, particularly if you're using my socket library as well, often have quite a few zeroes in known places – for example if you request an EXE or DLL download you can rely on 16 or more consecutive zeroes in places in the header; and 4 byte length encoded messages will generally be possible to generate with three zeroes in known positions.)
|
|
|
|
|
Hello everybody
i am trying to disprove the claim:
the time complexity of merging K AVL trees is O(the sum of their nodes number)
i tried to calculate the complexity of meging K AVL trees but there could be more than one algorithm.
How do i disprove the claim.
Thanks
|
|
|
|
|
If you handle every node in every tree, the complexity will be O(the sum of their nodes number).
So one way to disprove this is to come up with a merging algorithm that avoids handling every node. One way to avoid handling every node is to reuse the existing structure of the AVL trees.
This can be done in trivial cases where the ranges of key values for the trees don't overlap. If they DO overlap (which is most likely), then you have a much more difficult problem, because you have to "weave" one tree into another, while maintaining the tree's balanced property.
You may be able to find subtrees whose key values don't overlap, and merge these. Using bigger building blocks like this avoids handling every node, and reduces the time complexity of your algorithm. After merging, you will probably have to do a little rearranging to restore the balanced property.
But it seems very hard, and in most cases, probably impossible.
|
|
|
|
|
If you diligently handle every node, I think the time will be O(N*Log2 N), where N is the sum of all nodes numbers.
|
|
|
|
|
To merge K trees in O(N), where N is the sum of node counts, you must never search on insertion: all your insertions must complete in O(1). In other words, on each step you must insert the new root node, and maybe do a rotation or two. However, the new root could come from any of the K trees that you are merging, so the time to search for that node would be O(K) (perhaps you could do O(Log2 K), but regardless of how you do it, K will be a factor to it). That's why I think the best you can do is O(N*Log2 K).
|
|
|
|
|
i liked that idea, lets say we've got k tree where each one got one node.
we know that there is no algorithm that can sort with better than teta(klogk) so, in order to sort the nodes in array and scan an empty tree to insert them, it will takes at least teta(klogk)
|
|
|
|
|
Who can tell me more about Eye-like Algorithm?
|
|
|
|
|
|
I don't want google I want to hear the opnion of our fulks here in CP what they have to say about this subject.
|
|
|
|
|
I never heard the term, but back in the 80s I worked at a place that was doing (among other things) research on computer vision.
One innovation (by Stan Sternberg) was the use of Lissajous curves (http://en.wikipedia.org/wiki/Lissajous_curve[^] ) to adaptively trace the outline of an object like a real eye.
The algorithm would sample a single point on the curve at regular intervals and slightly adjust the Lissajous parameters to adaptively fit the edge of the object. (E.g. if it hit the interior of the object, it would expand the curve accordingly, and if it hit the background it would contract.)
For some tasks it was orders of magnitude faster than traditional algorithms that looked at every pixel.
|
|
|
|
|
Finally some one with an answer.
|
|
|
|
|
This is a bit of a 'name that method' query - I'm sure there must be a way to do this!
Say I have a set of 2D points with data values attached to them. Let's say the data is 0 - 100. If I colour the points in the ranges 0-19, 20-30, 31-69, 70-80 and 81-100 I will get five or more regions (which may include "islands" of data, and there may be several of each colour, and they are not necessarily connected).
I want to display those regions as a set of congruent polygons - how can I find the polygons?
I could grid the data and calculate iso-lines, but I was wondering if I could search each point and "assign" it to a polygon somehow (like a flood fill on the points?).
My data is ordered in a grid-ish manner (by "column") if that helps any.
Suggested approaches welcome.
Thanks!
|
|
|
|
|
Kyudos,
your problem statement is not totally clear to me. A figure would be helpful.
You can form polygons around your point clouds by means of the convex hull construction (tightest convex polygon that includes all points); if you mention unconnected islands, this means that convexity is a too strong requirement. You could resort to Alpha-shapes, a generalization of the convex hull (see http://cgm.cs.mcgill.ca/~godfried/teaching/projects97/belair/alpha.html).
This suggestion only allows you to process each class independently, it will not guarantee that the polygons are nested. This requires a more general approach.
Another answer is provided by the Voronoi diagram, a tiling of the plane where every point is associated to all points that are closer to it than to another point. (http://en.wikipedia.org/wiki/Voronoi_diagram). Coloring the diagram will give you the desired polygon set (in reality some discrete form of iso-lines).
Maybe the specific arrangement of your data points can provide shortcuts to the solution...
|
|
|
|
|
Thanks for the reply. I created a couple of figures that may help.
Area[^]
Area Detail[^]
I'd want a polygon for each of the coloured areas, but bearing in mind that there may be no areas of any given colour, or several unconnected areas that may be islanded (e.g., there could be a "yellow" island in the middle of the "green").
Seems like Alpha-Shaped convex hulls might do the trick - will look into it 
|
|
|
|
|
Ah, the images sure help. I didn't have a clue what your description was about, but now I see.
Alas I don't know the official algorithm(s) that would do exactly this for you, however it resembles the flood fill problem, so I would read up on that, and study the A* algorithm.
Luc Pattyn [My Articles] Nil Volentibus Arduum
The quality and detail of your question reflects on the effectiveness of the help you are likely to get. Please use <PRE> tags for code snippets, they improve readability. CP Vanity has been updated to V2.3
|
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
