U L?h@sdZddlmZmZddlmZddlmZddlmZddl m Z ddl m Z dd l mZdd lmZd d ZGd ddZeZddZeddddddZeedddZd-ddZddZd.dd Zd!d"Zd/d$d%Zd0d'd(Zd)d*Zd+d,ZdS)1z" Generating and counting primes. )bisect bisect_leftcount)array)randint)sqrt)isprime) deprecated)as_intcCsddlm}t||S)z Wrapping ceiling in as_int will raise an error if there was a problem determining whether the expression was exactly an integer or not.r)ceiling)Z#sympy.functions.elementary.integersr r )ar rH/var/www/html/venv/lib/python3.8/site-packages/sympy/ntheory/generate.py_as_int_ceilings rc@s~eZdZdZdddZddZddd Zd d Zd d ZddZ d ddZ ddZ ddZ ddZ ddZddZddZdS)!SieveaA list of prime numbers, implemented as a dynamically growing sieve of Eratosthenes. When a lookup is requested involving an odd number that has not been sieved, the sieve is automatically extended up to that number. Implementation details limit the number of primes to ``2^32-1``. Examples ======== >>> from sympy import sieve >>> sieve._reset() # this line for doctest only >>> 25 in sieve False >>> sieve._list array('L', [2, 3, 5, 7, 11, 13, 17, 19, 23]) @Bcsd_tdddddddg_tdd d d ddd g_td d d d d d d g_|d kr^td|_tfddjjjfDstdS)z Initial parameters for the Sieve class. Parameters ========== sieve_interval (int): Amount of memory to be used Raises ====== ValueError If ``sieve_interval`` is not positive. L rr iz+sieve_interval should be a positive integerc3s|]}t|jkVqdSN)len_n.0rselfrr Csz!Sieve.__init__..N) r!_array_list_tlist_mlist ValueErrorsieve_intervalallAssertionError)r%r,rr$r__init__-szSieve.__init__cCsddt|j|jd|jd|jd|jd|jddt|j|jd|jd|jd|jd|jdd t|j|jd|jd|jd|jd|jdfS) Nzs<%s sieve (%i): %i, %i, %i, ... %i, %i %s sieve (%i): %i, %i, %i, ... %i, %i %s sieve (%i): %i, %i, %i, ... %i, %i>primerr rrtotientmobius)r r(r)r*r$rrr__repr__Es0zSieve.__repr__NcCsjtdd|||fDr$d}}}|r:|jd|j|_|rP|jd|j|_|rf|jd|j|_dS)z]Reset all caches (default). To reset one or more set the desired keyword to True.css|]}|dkVqdSrrr"rrrr&VszSieve._reset..TN)r-r(r!r)r*)r%r0r2r3rrr_resetSs z Sieve._resetc Cst|}|jdd}||kr"dS|d}||kr^|jtd|||7_||d}}q*|jtd|||d7_dS)zGrow the sieve to cover all primes <= n. Examples ======== >>> from sympy import sieve >>> sieve._reset() # this line for doctest only >>> sieve.extend(30) >>> sieve[10] == 29 True rr Nrr)intr(r' _primerange)r%nnumZnum2rrrextend_s z Sieve.extendccs|dr|d8}||krt|j||d}dg|}|jdt|jt|d|dD]0}t|d| d|||D] }d||<q~q\t|D]\}}|r|d|dVq|d|7}qdS)a? Generate all prime numbers in the range (a, b). Parameters ========== a, b : positive integers assuming the following conditions * a is an even number * 2 < self._list[-1] < a < b < nextprime(self._list[-1])**2 Yields ====== p (int): prime numbers such that ``a < p < b`` Examples ======== >>> from sympy.ntheory.generate import Sieve >>> s = Sieve() >>> s._list[-1] 13 >>> list(s._primerange(18, 31)) [19, 23, 29] rr TFN)minr,r(rrrange enumerate)r%rb block_sizeblockptidxrrrr7xs *" zSieve._primerangecCs4t|}t|j|kr0|t|jddqdS)aExtend to include the ith prime number. Parameters ========== i : integer Examples ======== >>> from sympy import sieve >>> sieve._reset() # this line for doctest only >>> sieve.extend_to_no(9) >>> sieve._list array('L', [2, 3, 5, 7, 11, 13, 17, 19, 23]) Notes ===== The list is extended by 50% if it is too short, so it is likely that it will be longer than requested. rg?N)r r r(r:r6)r%rrrr extend_to_noszSieve.extend_to_noccsj|dkrt|}d}ntdt|}t|}||kr8dS|||jt|j|t|j|EdHdS)a(Generate all prime numbers in the range [2, a) or [a, b). Examples ======== >>> from sympy import sieve, prime All primes less than 19: >>> print([i for i in sieve.primerange(19)]) [2, 3, 5, 7, 11, 13, 17] All primes greater than or equal to 7 and less than 19: >>> print([i for i in sieve.primerange(7, 19)]) [7, 11, 13, 17] All primes through the 10th prime >>> list(sieve.primerange(prime(10) + 1)) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] Nr)rmaxr:r(r)r%rr>rrr primeranges  zSieve.primerangeccsNtdt|}t|}t|j}||kr,dS||krRt||D]}|j|Vq>n|jtdt||7_td|D]j}|j|}||dkr||d||}t|||D] }|j||j||8<q||krv|Vqvt||D]\}|j|}||kr4t|||D]"}|j||j||8<q||kr|j|VqdS)zGenerate all totient numbers for the range [a, b). Examples ======== >>> from sympy import sieve >>> print([i for i in sieve.totientrange(7, 18)]) [6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16] r Nr)rErr r)r<r')r%rr>r8rti startindexjrrr totientranges0       zSieve.totientrangeccs$tdt|}t|}t|j}||kr,dS||krRt||D]}|j|Vq>n|jtddg||7_td|D]T}|j|}||d||}t|||D]}|j||8<q||krz|Vqzt||D]D}|j|}td|||D]}|j||8<q||kr|VqdS)aGenerate all mobius numbers for the range [a, b). Parameters ========== a : integer First number in range b : integer First number outside of range Examples ======== >>> from sympy import sieve >>> print([i for i in sieve.mobiusrange(7, 18)]) [-1, 0, 0, 1, -1, 0, -1, 1, 1, 0, -1] r Nrrr)rErr r*r<r')r%rr>r8rmirHrIrrr mobiusranges,   zSieve.mobiusrangecCsrt|}t|}|dkr$td|||jdkr<||t|j|}|j|d|krb||fS||dfSdS)a~Return the indices i, j of the primes that bound n. If n is prime then i == j. Although n can be an expression, if ceiling cannot convert it to an integer then an n error will be raised. Examples ======== >>> from sympy import sieve >>> sieve.search(25) (9, 10) >>> sieve.search(23) (9, 9) rzn should be >= 2 but got: %srr N)rr r+r(r:r)r%r8testr>rrrsearch1s   z Sieve.searchc Cs^zt|}|dkstWnttfk r2YdSX|ddkrH|dkS||\}}||kS)NrFr)r r.r+rN)r%r8rr>rrr __contains__Ns zSieve.__contains__ccstdD]}||VqdS)Nr r)r%r8rrr__iter__Ys zSieve.__iter__cCst|trV||j|jdk r&|jnd}|dkr:td|j|d|jd|jS|dkrftdt|}|||j|dSdS)zReturn the nth prime numberNrr zSieve indices start at 1.) isinstanceslicerDstopstart IndexErrorr(stepr )r%r8rTrrr __getitem__]s   zSieve.__getitem__)r)NNN)N)__name__ __module__ __qualname____doc__r/r4r5r:r7rDrFrJrLrNrOrPrWrrrrrs  ) $%, rcCst|}|dkrtd|ttjkr.t|Sddlm}ddlm}d}t ||||||}||kr||d?}|||kr|}qf|d}qft |d}||krt |r|d7}|d7}q|dS)aK Return the nth prime, with the primes indexed as prime(1) = 2, prime(2) = 3, etc.... The nth prime is approximately $n\log(n)$. Logarithmic integral of $x$ is a pretty nice approximation for number of primes $\le x$, i.e. li(x) ~ pi(x) In fact, for the numbers we are concerned about( x<1e11 ), li(x) - pi(x) < 50000 Also, li(x) > pi(x) can be safely assumed for the numbers which can be evaluated by this function. Here, we find the least integer m such that li(m) > n using binary search. Now pi(m-1) < li(m-1) <= n, We find pi(m - 1) using primepi function. Starting from m, we have to find n - pi(m-1) more primes. For the inputs this implementation can handle, we will have to test primality for at max about 10**5 numbers, to get our answer. Examples ======== >>> from sympy import prime >>> prime(10) 29 >>> prime(1) 2 >>> prime(100000) 1299709 See Also ======== sympy.ntheory.primetest.isprime : Test if n is prime primerange : Generate all primes in a given range primepi : Return the number of primes less than or equal to n References ========== .. [1] https://en.wikipedia.org/wiki/Prime_number_theorem#Table_of_.CF.80.28x.29.2C_x_.2F_log_x.2C_and_li.28x.29 .. [2] https://en.wikipedia.org/wiki/Prime_number_theorem#Approximations_for_the_nth_prime_number .. [3] https://en.wikipedia.org/wiki/Skewes%27_number r z-nth must be a positive integer; prime(1) == 2rloglir) r r+r siever(&sympy.functions.elementary.exponentialr]'sympy.functions.special.error_functionsr_r6_primepir )nthr8r]r_rr>midZn_primesrrrr0vs(1       r0zgThe `sympy.ntheory.generate.primepi` has been moved to `sympy.functions.combinatorial.numbers.primepi`.z1.13z%deprecated-ntheory-symbolic-functions)Zdeprecated_since_versionZactive_deprecations_targetcCsddlm}||S)a Represents the prime counting function pi(n) = the number of prime numbers less than or equal to n. .. deprecated:: 1.13 The ``primepi`` function is deprecated. Use :class:`sympy.functions.combinatorial.numbers.primepi` instead. See its documentation for more information. See :ref:`deprecated-ntheory-symbolic-functions` for details. Algorithm Description: In sieve method, we remove all multiples of prime p except p itself. Let phi(i,j) be the number of integers 2 <= k <= i which remain after sieving from primes less than or equal to j. Clearly, pi(n) = phi(n, sqrt(n)) If j is not a prime, phi(i,j) = phi(i, j - 1) if j is a prime, We remove all numbers(except j) whose smallest prime factor is j. Let $x= j \times a$ be such a number, where $2 \le a \le i / j$ Now, after sieving from primes $\le j - 1$, a must remain (because x, and hence a has no prime factor $\le j - 1$) Clearly, there are phi(i / j, j - 1) such a which remain on sieving from primes $\le j - 1$ Now, if a is a prime less than equal to j - 1, $x= j \times a$ has smallest prime factor = a, and has already been removed(by sieving from a). So, we do not need to remove it again. (Note: there will be pi(j - 1) such x) Thus, number of x, that will be removed are: phi(i / j, j - 1) - phi(j - 1, j - 1) (Note that pi(j - 1) = phi(j - 1, j - 1)) $\Rightarrow$ phi(i,j) = phi(i, j - 1) - phi(i / j, j - 1) + phi(j - 1, j - 1) So,following recursion is used and implemented as dp: phi(a, b) = phi(a, b - 1), if b is not a prime phi(a, b) = phi(a, b-1)-phi(a / b, b-1) + phi(b-1, b-1), if b is prime Clearly a is always of the form floor(n / k), which can take at most $2\sqrt{n}$ values. Two arrays arr1,arr2 are maintained arr1[i] = phi(i, j), arr2[i] = phi(n // i, j) Finally the answer is arr2[1] Examples ======== >>> from sympy import primepi, prime, prevprime, isprime >>> primepi(25) 9 So there are 9 primes less than or equal to 25. Is 25 prime? >>> isprime(25) False It is not. So the first prime less than 25 must be the 9th prime: >>> prevprime(25) == prime(9) True See Also ======== sympy.ntheory.primetest.isprime : Test if n is prime primerange : Generate all primes in a given range prime : Return the nth prime r)primepi)Z%sympy.functions.combinatorial.numbersrf)r8Z func_primepirrrrfsX rf)r8returnc Csd|dkr dS|tjdkr(t|dSt|}dg|d}dg|d}td|dD] }|d||<||d||<qZtd|dD]}||||dkrq||d}tdt||||dD]J}||}||kr|||||8<q||||||8<qt|||d}t||dD]"}||||||8<q6q|dS)a Represents the prime counting function pi(n) = the number of prime numbers less than or equal to n. Explanation =========== In sieve method, we remove all multiples of prime p except p itself. Let phi(i,j) be the number of integers 2 <= k <= i which remain after sieving from primes less than or equal to j. Clearly, pi(n) = phi(n, sqrt(n)) If j is not a prime, phi(i,j) = phi(i, j - 1) if j is a prime, We remove all numbers(except j) whose smallest prime factor is j. Let $x= j \times a$ be such a number, where $2 \le a \le i / j$ Now, after sieving from primes $\le j - 1$, a must remain (because x, and hence a has no prime factor $\le j - 1$) Clearly, there are phi(i / j, j - 1) such a which remain on sieving from primes $\le j - 1$ Now, if a is a prime less than equal to j - 1, $x= j \times a$ has smallest prime factor = a, and has already been removed(by sieving from a). So, we do not need to remove it again. (Note: there will be pi(j - 1) such x) Thus, number of x, that will be removed are: phi(i / j, j - 1) - phi(j - 1, j - 1) (Note that pi(j - 1) = phi(j - 1, j - 1)) $\Rightarrow$ phi(i,j) = phi(i, j - 1) - phi(i / j, j - 1) + phi(j - 1, j - 1) So,following recursion is used and implemented as dp: phi(a, b) = phi(a, b - 1), if b is not a prime phi(a, b) = phi(a, b-1)-phi(a / b, b-1) + phi(b-1, b-1), if b is prime Clearly a is always of the form floor(n / k), which can take at most $2\sqrt{n}$ values. Two arrays arr1,arr2 are maintained arr1[i] = phi(i, j), arr2[i] = phi(n // i, j) Finally the answer is arr2[1] Parameters ========== n : int rrrr )r`r(rNrr<r;) r8ZlimZarr1Zarr2rrArIstZlim2rrrrcs.<   "rccCsTt|}t|}|dkr td|dkr4d}|d8}|tjdkrt|\}}||dttjkrxtj||dSttjd||ttjSd|krt|D] }t|}q|Sd|d}||kr|d7}t |r|S|d7}n6||d kr|d7}t |r|S|d7}n|d }t |r.|S|d7}t |rD|S|d7}q d S) aU Return the ith prime greater than n. Parameters ========== n : integer ith : positive integer Returns ======= int : Return the ith prime greater than n Raises ====== ValueError If ``ith <= 0``. If ``n`` or ``ith`` is not an integer. Notes ===== Potential primes are located at 6*j +/- 1. This property is used during searching. >>> from sympy import nextprime >>> [(i, nextprime(i)) for i in range(10, 15)] [(10, 11), (11, 13), (12, 13), (13, 17), (14, 17)] >>> nextprime(2, ith=2) # the 2nd prime after 2 5 See Also ======== prevprime : Return the largest prime smaller than n primerange : Generate all primes in a given range rzith should be positiverr r1rrrrN) r6r r+r`r(rNr nextprimer<r )r8Zithrl_nnrrrriusD(        ricCst|}|dkrtd|dkr4dddddd|S|tjdkrlt|\}}||krdt|dSt|Sd |d }||dkr|d}t|r|S|d 8}n|d}t|r|S|d8}t|r|S|d 8}qd S) a Return the largest prime smaller than n. Notes ===== Potential primes are located at 6*j +/- 1. This property is used during searching. >>> from sympy import prevprime >>> [(i, prevprime(i)) for i in range(10, 15)] [(10, 7), (11, 7), (12, 11), (13, 11), (14, 13)] See Also ======== nextprime : Return the ith prime greater than n primerange : Generates all primes in a given range rzno preceding primesrr)rrrrrrr rrN)rr+r`r(rNr )r8rjurlrrr prevprimes.    roNccs|dkrd|}}||krdStjd}||krFt||EdHdS||krttjttj|dEdH|d}n|dr|d8}t||d}||krt||EdH|}||krdSt|}||kr|VqdSqdS)a Generate a list of all prime numbers in the range [2, a), or [a, b). If the range exists in the default sieve, the values will be returned from there; otherwise values will be returned but will not modify the sieve. Examples ======== >>> from sympy import primerange, prime All primes less than 19: >>> list(primerange(19)) [2, 3, 5, 7, 11, 13, 17] All primes greater than or equal to 7 and less than 19: >>> list(primerange(7, 19)) [7, 11, 13, 17] All primes through the 10th prime >>> list(primerange(prime(10) + 1)) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] The Sieve method, primerange, is generally faster but it will occupy more memory as the sieve stores values. The default instance of Sieve, named sieve, can be used: >>> from sympy import sieve >>> list(sieve.primerange(1, 30)) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29] Notes ===== Some famous conjectures about the occurrence of primes in a given range are [1]: - Twin primes: though often not, the following will give 2 primes an infinite number of times: primerange(6*n - 1, 6*n + 2) - Legendre's: the following always yields at least one prime primerange(n**2, (n+1)**2+1) - Bertrand's (proven): there is always a prime in the range primerange(n, 2*n) - Brocard's: there are at least four primes in the range primerange(prime(n)**2, prime(n+1)**2) The average gap between primes is log(n) [2]; the gap between primes can be arbitrarily large since sequences of composite numbers are arbitrarily large, e.g. the numbers in the sequence n! + 2, n! + 3 ... n! + n are all composite. See Also ======== prime : Return the nth prime nextprime : Return the ith prime greater than n prevprime : Return the largest prime smaller than n randprime : Returns a random prime in a given range primorial : Returns the product of primes based on condition Sieve.primerange : return range from already computed primes or extend the sieve to contain the requested range. References ========== .. [1] https://en.wikipedia.org/wiki/Prime_number .. [2] https://primes.utm.edu/notes/gaps.html Nrrr )r`r(rFrr;r7ri)rr>Zlargest_known_primetailrrrrFs.K   rFcCsX||kr dStt||f\}}t|d|}t|}||krDt|}||krTtd|S)aG Return a random prime number in the range [a, b). Bertrand's postulate assures that randprime(a, 2*a) will always succeed for a > 1. Note that due to implementation difficulties, the prime numbers chosen are not uniformly random. For example, there are two primes in the range [112, 128), ``113`` and ``127``, but ``randprime(112, 128)`` returns ``127`` with a probability of 15/17. Examples ======== >>> from sympy import randprime, isprime >>> randprime(1, 30) #doctest: +SKIP 13 >>> isprime(randprime(1, 30)) True See Also ======== primerange : Generate all primes in a given range References ========== .. [1] https://en.wikipedia.org/wiki/Bertrand's_postulate Nr z&no primes exist in the specified range)mapr6rriror+)rr>r8rArrr randprime[s rrTcCsp|rt|}nt|}|dkr&tdd}|rPtd|dD]}|t|9}q>> from sympy.ntheory.generate import primorial, primerange >>> from sympy import factorint, Mul, primefactors, sqrt >>> primorial(4) # the first 4 primes are 2, 3, 5, 7 210 >>> primorial(4, nth=False) # primes <= 4 are 2 and 3 6 >>> primorial(1) 2 >>> primorial(1, nth=False) 1 >>> primorial(sqrt(101), nth=False) 210 One can argue that the primes are infinite since if you take a set of primes and multiply them together (e.g. the primorial) and then add or subtract 1, the result cannot be divided by any of the original factors, hence either 1 or more new primes must divide this product of primes. In this case, the number itself is a new prime: >>> factorint(primorial(4) + 1) {211: 1} In this case two new primes are the factors: >>> factorint(primorial(4) - 1) {11: 1, 19: 1} Here, some primes smaller and larger than the primes multiplied together are obtained: >>> p = list(primerange(10, 20)) >>> sorted(set(primefactors(Mul(*p) + 1)).difference(set(p))) [2, 5, 31, 149] See Also ======== primerange : Generate all primes in a given range r zprimorial argument must be >= 1r)r r6r+r<r0rF)r8rdrArrrr primorials2  rsFc cst|pd}d}}|||}}d}|r0|V||kr|rD||kr|d7}||krd|}|d9}d}|rn|V||}|d7}q0|r||kr|rdS|dfVdS|sd} |}}t|D] }||}q||kr||}||}| d7} q|| fVdS)awFor a given iterated sequence, return a generator that gives the length of the iterated cycle (lambda) and the length of terms before the cycle begins (mu); if ``values`` is True then the terms of the sequence will be returned instead. The sequence is started with value ``x0``. Note: more than the first lambda + mu terms may be returned and this is the cost of cycle detection with Brent's method; there are, however, generally less terms calculated than would have been calculated if the proper ending point were determined, e.g. by using Floyd's method. >>> from sympy.ntheory.generate import cycle_length This will yield successive values of i <-- func(i): >>> def gen(func, i): ... while 1: ... yield i ... i = func(i) ... A function is defined: >>> func = lambda i: (i**2 + 1) % 51 and given a seed of 4 and the mu and lambda terms calculated: >>> next(cycle_length(func, 4)) (6, 3) We can see what is meant by looking at the output: >>> iter = cycle_length(func, 4, values=True) >>> list(iter) [4, 17, 35, 2, 5, 26, 14, 44, 50, 2, 5, 26, 14] There are 6 repeating values after the first 3. If a sequence is suspected of being longer than you might wish, ``nmax`` can be used to exit early (and mu will be returned as None): >>> next(cycle_length(func, 4, nmax = 4)) (4, None) >>> list(cycle_length(func, 4, nmax = 4, values=True)) [4, 17, 35, 2] Code modified from: https://en.wikipedia.org/wiki/Cycle_detection. rr rN)r6r<) fZx0ZnmaxvaluespowerZlamZtortoiseZharermurrr cycle_lengths>3       rxc Csvt|}|dkrtdddddddd d d d g }|dkrD||dSdtjd }}||t|dkr||dkr||d?}|t|d|kr|}qh|}qht|r|d8}|Sddlm}ddlm }d}t ||||||}||kr(||d?}|||d|kr|}q|d}q|t|d}||kr`t|sT|d8}|d8}q8t|rr|d8}|S)a Return the nth composite number, with the composite numbers indexed as composite(1) = 4, composite(2) = 6, etc.... Examples ======== >>> from sympy import composite >>> composite(36) 52 >>> composite(1) 4 >>> composite(17737) 20000 See Also ======== sympy.ntheory.primetest.isprime : Test if n is prime primerange : Generate all primes in a given range primepi : Return the number of primes less than or equal to n prime : Return the nth prime compositepi : Return the number of positive composite numbers less than or equal to n r z1nth must be a positive integer; composite(1) == 4rrrm rrr\r^) r r+r`r(rcr rar]rbr_r6) rdr8Z composite_arrrr>rer]r_Z n_compositesrrr composite!sB            rcCs$t|}|dkrdS|t|dS)ak Return the number of positive composite numbers less than or equal to n. The first positive composite is 4, i.e. compositepi(4) = 1. Examples ======== >>> from sympy import compositepi >>> compositepi(25) 15 >>> compositepi(1000) 831 See Also ======== sympy.ntheory.primetest.isprime : Test if n is prime primerange : Generate all primes in a given range prime : Return the nth prime primepi : Return the number of primes less than or equal to n composite : Return the nth composite number rrr )r6rc)r8rrr compositepibsr)r )N)T)NF) r[rr itertoolsrrr'Zsympy.core.randomrZsympy.external.gmpyrZ primetestr Zsympy.utilities.decoratorr Zsympy.utilities.miscr rrr`r0rfr6rcrirorFrrrsrxrrrrrrs8       [K XX N/ i, B XA