What is a mathematic diamond problem?
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In object-oriented programming languages with multiple inheritance, the diamond problem is an ambiguity that arises when two classes B and C inherit from A, and class D inherits from both B and C. If a method in D calls a method defined in A, which class does it inherit from, B or C?\n\n\n\nIt is called the 'diamond' problem because of the shape of the class inheritance diagram in this situation. Class A is at the top, both B and C separately beneath it, and D joins the two together at the bottom to form a diamond shape.\n\nDifferent programming languages have addressed this problem in different ways:\n\nC# / Java only allow single inheritance, but as a compromise allow the multiple inheritance of interfaces. Interfaces are essentially abstract base classes with all abstract methods and no data members. C# (and other .NET languages that allow interface definition) also allows property defintions to exist at the interface level. \nC++ by default follows each inheritance path separately, so a D object would actually contain two separate A objects, and uses of A's members have to be properly qualified. If the inheritance from A to B and the inheritance from A to C are both marked "virtual" ("class B : virtual A"), C++ takes special care to only create one A object, and uses of A's members work correctly. If virtual inheritance and nonvirtual inheritance are mixed, there is a single virtual A and a nonvirtual A for each nonvirtual inheritance path to A. \nEiffel handles this situation by a select and rename directives, where the ancestors' methods to use in a descendant are explicitly specified. This allows to share the methods of the base class between its descendants or even to give each of them a separate copy of the base class. \nPerl and Io handle this by specifying the inheritance classes as an ordered list. In the above ambiguity, class B and its ancestors would be checked before class C and its ancestors, so the method in A would be inherited through B. \nPython had to deal with this upon the introduction of new-style classes, all of which have a common ancestor, object. Python creates a list of the classes that would be searched in left-first depth-first order (D, B, A, C, A) and then removes all but the last occurrence of any repeated classes. Thus, the method resolution order is: D, B, C, A. \nThe diamond problem is not limited to inheritance. It also seems to arise when header files A, B, C, and D "#include" one another in a diamond as above and separate precompiled headers are created from B and C. If these two precompiled headers are combined, declarations in A are duplicated; the "#ifndef" convention (see the header file article) is ineffective
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The Ring of polynoms over the Field lets say G2 (just for denoting) is the right algebraic model to use if we want to study CRCs. \n\nA sequence of bit like 10011101 is modeled by 1x7 + 0x6 + 0x5 + 1x4 + 1x3 + 1x2 + 0x1 + 1x0 which we can write shorter as x7 + x4 + x3 + x2 + 1.\n\nI hope I helped...
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m1*m2=-1\n\nhold for the slope of two intersecting lines
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Diamond is one of the two best known forms (or allotropes) of carbon, whose hardness and high dispersion of light makes it useful for industrial applications and jewelry (the other equally well known allotrope is graphite). Diamonds are specifically renowned as a mineral with superlative physical qualities - they make excellent abrasives because they can only be scratched by other diamonds, which also means they hold a polish extremely well and retain luster. About 130 million carats (26,000 kg) are mined annually, with a total value of nearly USD $9 billion.\n\nThe name "diamond" derives from the ancient Greek adamas (αδάμας; "impossible to tame"). They have been treasured as gems since their use as religious icons in India at least 2,500 years ago—and usage in drill bits and engraving tools also dates to early human history. Popularity of diamonds has risen since the 19th century because of improved cutting and polishing techniques, and they are commonly judged by the "four Cs": carat, clarity, color, and cut. Although nearly four times the mass of natural diamonds are produced as synthetic diamond each year, the vast majority of synthetic diamond production remains small, imperfect diamonds suitable only for industrial-grade use, with gem-quality synthetic diamonds only recently becoming available.
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Lead: 11.35\nGraphite: 2.27\nDiamond: 3.51
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Dear friend,\n\nYou're proposing us the famous Basel Problem, solved by Leonhard Euler in 1735. You should try Euler's approach to the problem, you can see it in the following link:\n\nhttp://en.wikipedia.org/wiki/Basel_problem#Euler_attacks_the_problem\n\nHope I've helped on something, and wish you a nice day!
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Diamond is (under normal pressure) a chemically less stable form of elemental carbon. In order to synthesize diamond, you need to expose carbon to huge amounts of pressure and considerable heat, as can be found under geological conditions.\n\nSynthetical diamonds aren't fakes, btw.
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I'm pretty sure it is a Diamond. Here's my reasoning:\nDiamonds family name = carbon = black in appearance\n"Give a girl a buck" = Diamonds are a girls best friend\nMany nice colors = Refraction of light through a diamond displays many colors.....
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Yes, it is possible.\n\nDenote the number of black dots belonging to a red dot by N.\n\nThe present problem is formulated for N=10.\n\nLet us first solve a simpler problem with N=2.\n\nStep 1. \nTake a square, that is a four-sided regular polygon, with the side equal to R. Put the two red dots at the opposite [i.e., connected by a diagonal] vertices of the square. Put the two black dots at the remaining two vertices. If you draw a circle of radius R around each of the two red dots, you will find exactly 2 black dots lying precisely on the circle line. The problem is almost solved, but the number of red dots (2) is equal to the number of black dots (2).\n\nStep 2. \nTake another square and repeat the above procedure. \nNow we have the two square compounds. In total we have 4 red dots and 4 black dots.\n\nStep 3. \nPut the 5th red dot anywhere in the plane and draw a circle of radius R around it. \nDrag each of the above compounds as a whole closer to this circle. \nPlace one of the black dots belonging to the first compound and one of the black dots belonging to the second compound on the circle line surrounding the 5th red dot.\n\nWhile doing so, make sure that you move and rotate each compound as a whole, keeping all the sides of the squares equal to R. \n\nThe problem with N=2 has been solved: in total we have 5 red dots and 4 black dots satisfying the condition of the problem, 5>4. \n\nLet us solve the same problem for N=3.\n\nIn this case, one has to extend the Step 1.\n\nStep 1a. \nHaving constructed the square compound according to the Step 1 above, draw parallel vectors of length R starting at the vertices of the square. Make sure that the vectors do NOT coincide with the sides of the square. \nIf the vector starts at the red vertex, put the black dot at the end of the vector. If the vector starts at the black vertex, put the red dot at the end of the vector. \n\nNow we have one compound consisting of the two shifted squares containing 4 red dots and 4 black dots. If you draw a circle of R around each of the four red dots, you will find exactly 3 black dots lying precisely on the circle line.\n\nStep 2. Repeat the procedure 1+1a.\nStep 2a. Repeat the procedure 1+1a.\n\nNow that we have 3 compounds with equal numbers of red and black dots (in total 12 red dots and 12 black dots), put the 13th red dot anywhere in the plane and follow the instructions from Step 3 above.\n\nThe problem with N=3 has been solved: in total we have 13 red dots and 12 black dots satisfying the condition of the problem, 13>12. \n\nFor N=4, take the compound constructed for N=3 at Step 1a and translate ALL 8 vertices by a vector of length R.\nAgain, if the vector starts at the red vertex, put the black dot at the end of the vector and vice versa. You will have 8 red dots and 8 black dots, etc.\n\nThe above procedure applies to an arbitrary N, in particular to N=10.\n\n\nAn interesting question is: what is the minimal number of dots needed to solve the problem for a given N.
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umm none that im aware of, what kind of mathamatics leval are you looking at?
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Retail value will be around $25k. However, this is rough since there is not nearly enough information to give an accurate assessment of your diamond.
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http://www.maa.org/mathland/mathtrek_11_27_00.html
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A basic diamond-shaped kite satisfies the mathematical properties of a rhombus.
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have a look here, i think, you will find  all answers there.\n\nhttp://math-www.uni-paderborn.de/~axel/us-d.html
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It's the answer you get from a division problem.
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diamonds usually work
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wellz why not just help me with that matrix problem? :oP