factorization | 易学教程

Factoring a quantum state

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问题 I'm looking for algorithms that take an arbitrary quantum state made up of a sum of weighted classical states made up of bits, like this: |0000>/2 - |0011>/2 + |0100>/2 - |0111>/2 and factor it into a more compact form using tensor products, like this: |0> x (|0> + |1>) x (|00> - |11>) / 2 I want to use the algorithm as a way of visualizing/simplifying the state of a (simulated) quantum circuit. For individual qubits I know I can just pair all the states with the state where the bit is

Calculating Eulers Totient Function for very large numbers JAVA

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I've managed to get a version of Eulers Totient Function working, albeit one that works for smaller numbers (smaller here being smaller compared to the 1024 bit numbers I need it to calculate) My version is here - public static BigInteger eulerTotientBigInt(BigInteger calculate) { BigInteger count = new BigInteger("0"); for(BigInteger i = new BigInteger("1"); i.compareTo(calculate) < 0; i = i.add(BigInteger.ONE)) { BigInteger check = GCD(calculate,i); if(check.compareTo(BigInteger.ONE)==0) {//coprime count = count.add(BigInteger.ONE); } } return count; } While this works for smaller numbers,

Implementing a factorisation method in Haskell

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I am doing question 266 at Project Euler and after a bit of searching, found this method of quickly finding the factors of a number. What you do is find all the permutations of the prime factors of a number: these are its factors. I already have a module to find the prime power factors of a number, eg: Main> primePowerFactors 196 [(2,2),(7,2)] This is basically showing that: 2^2 * 7^2 == 196 . From here I want to find all the permutations of those powers, to give the factors of 196 thusly: (2^0)(7^0) = 1 (2^1)(7^0) = 2 (2^2)(7^0) = 4 (2^0)(7^1) = 7 (2^1)(7^1) = 14 (2^2)(7^1) = 28 (2^0)(7^2) =

Calculating Eulers Totient Function for very large numbers JAVA

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问题 I've managed to get a version of Eulers Totient Function working, albeit one that works for smaller numbers (smaller here being smaller compared to the 1024 bit numbers I need it to calculate) My version is here - public static BigInteger eulerTotientBigInt(BigInteger calculate) { BigInteger count = new BigInteger("0"); for(BigInteger i = new BigInteger("1"); i.compareTo(calculate) < 0; i = i.add(BigInteger.ONE)) { BigInteger check = GCD(calculate,i); if(check.compareTo(BigInteger.ONE)==0) {/

Implementing a factorisation method in Haskell

阅读更多关于 Implementing a factorisation method in Haskell

问题 I am doing question 266 at Project Euler and after a bit of searching, found this method of quickly finding the factors of a number. What you do is find all the permutations of the prime factors of a number: these are its factors. I already have a module to find the prime power factors of a number, eg: Main> primePowerFactors 196 [(2,2),(7,2)] This is basically showing that: 2^2 * 7^2 == 196 . From here I want to find all the permutations of those powers, to give the factors of 196 thusly: (2

I have a new algorithm to find factors or primes in linear time - need verification for this

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问题 I have developed an algorithm to find factors of a given number. Thus it also helps in finding if the given number is a prime number. I feel this is the fastest algorithm for finding factors or prime numbers. This algorithm finds if a give number is prime in time frame of 5*N (where N is the input number). So I hope I can call this a linear time algorithm. How can I verify if this is the fastest algorithm available? Can anybody help in this matter? (faster than GNFS and others known)

Why is this Haskell code snippet not infinitely recursive?

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问题 To help me learn Haskell, I am working through the problems on Project Euler. After solving each problem, I check my solution against the Haskell wiki in an attempt to learn better coding practices. Here is the solution to problem 3: primes = 2 : filter ((==1) . length . primeFactors) [3,5..] primeFactors n = factor n primes where factor n (p:ps) | p*p > n = [n] | n `mod` p == 0 = p : factor (n `div` p) (p:ps) | otherwise = factor n ps problem_3 = last (primeFactors 317584931803) My naive

programmatically factorize a large number

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问题 Alright, so I have a huge number f . This number is just over 100 digits long, actually. I know that the factors are of approximately the same size. If I have limited resources and time, what language and algorithm should I use? I am including the length of time to code the algorithm in the restricted time. Thoughts? EDIT: By limited, I mean in the least amount of time possible. 回答1: The state-of-the-art prime factorization algorithm is the quadratic sieve and its variants. For numbers larger

I have a new algorithm to find factors or primes in linear time - need verification for this

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I have developed an algorithm to find factors of a given number. Thus it also helps in finding if the given number is a prime number. I feel this is the fastest algorithm for finding factors or prime numbers. This algorithm finds if a give number is prime in time frame of 5*N (where N is the input number). So I hope I can call this a linear time algorithm. How can I verify if this is the fastest algorithm available? Can anybody help in this matter? (faster than GNFS and others known) Algorithm is given below Input: A Number (whose factors is to be found) Output: The two factor of the Number.