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A general-purpose factoring algorithm, also known as a Category 2, Second Category, or Kraitchik family algorithm, [10] has a running time which depends solely on the size of the integer to be factored. This is the type of algorithm used to factor RSA numbers. Most general-purpose factoring algorithms are based on the congruence of squares method.
A decision problem is a computational problem where the answer for every instance is either yes or no. An example of a decision problem is primality testing: "Given a positive integer n, determine if n is prime." A decision problem is typically represented as the set of all instances for which the answer is yes. For example, primality testing ...
Integer factorization is the process of determining which prime numbers divide a given positive integer.Doing this quickly has applications in cryptography.The difficulty depends on both the size and form of the number and its prime factors; it is currently very difficult to factorize large semiprimes (and, indeed, most numbers that have no small factors).
A function problem is a computational problem where a single output (of a total function) is expected for every input, but the output is more complex than that of a decision problem—that is, the output is not just yes or no. Notable examples include the traveling salesman problem and the integer factorization problem.
The polynomial x 2 + cx + d, where a + b = c and ab = d, can be factorized into (x + a)(x + b).. In mathematics, factorization (or factorisation, see English spelling differences) or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.
The Kissing Number Problem. A broad category of problems in math are called the Sphere Packing Problems. They range from pure math to practical applications, generally putting math terminology to ...
An average-case assumption says that a specific problem is hard on most instances from some explicit distribution, whereas a worst-case assumption only says that the problem is hard on some instances. For a given problem, average-case hardness implies worst-case hardness, so an average-case hardness assumption is stronger than a worst-case ...
Since f is of degree d with integer coefficients, if a and b are integers, then so will be b d ·f(a/b), which we call r. Similarly, s = b e · g ( a / b ) is an integer. The goal is to find integer values of a and b that simultaneously make r and s smooth relative to the chosen basis of primes.