def __new__(cls, partition, integer=None):
"""
Generates a new IntegerPartition object from a list or dictionary.
The partition can be given as a list of positive integers or a
dictionary of (integer, multiplicity) items. If the partition is
preceeded by an integer an error will be raised if the partition
does not sum to that given integer.
Examples
========
>>> from sympy.combinatorics.partitions import IntegerPartition
>>> a = IntegerPartition([5, 4, 3, 1, 1])
>>> a
IntegerPartition(14, (5, 4, 3, 1, 1))
>>> print a
[5, 4, 3, 1, 1]
>>> IntegerPartition({1:3, 2:1})
IntegerPartition(5, (2, 1, 1, 1))
If the value that the partion should sum to is given first, a check
will be made to see n error will be raised if there is a discrepancy:
>>> IntegerPartition(10, [5, 4, 3, 1])
Traceback (most recent call last):
...
ValueError: The partition is not valid
"""
from sympy.ntheory.residue_ntheory import int_tested
if integer is not None:
integer, partition = partition, integer
if isinstance(partition, (dict, Dict)):
_ = []
for k, v in sorted(partition.items(), reverse=True):
if not v:
continue
k, v = int_tested(k, v)
_.extend([k] * v)
partition = tuple(_)
else:
partition = tuple(sorted(int_tested(partition), reverse=True))
sum_ok = False
if integer is None:
integer = sum(partition)
sum_ok = True
else:
integer = int_tested(integer)
if not sum_ok and sum(partition) != integer:
raise ValueError("Partition did not add to %s" % integer)
if any(i < 1 for i in partition):
raise ValueError("The summands must all be positive.")
obj = Basic.__new__(cls, integer, partition)
obj.partition = list(partition)
obj.integer = integer
return obj
def __new__(cls, partition, integer=None):
"""
Generates a new IntegerPartition object from a list or dictionary.
The partition can be given as a list of positive integers or a
dictionary of (integer, multiplicity) items. If the partition is
preceeded by an integer an error will be raised if the partition
does not sum to that given integer.
Examples
========
>>> from sympy.combinatorics.partitions import IntegerPartition
>>> a = IntegerPartition([5, 4, 3, 1, 1])
>>> a
IntegerPartition(14, (5, 4, 3, 1, 1))
>>> print a
[5, 4, 3, 1, 1]
>>> IntegerPartition({1:3, 2:1})
IntegerPartition(5, (2, 1, 1, 1))
If the value that the partion should sum to is given first, a check
will be made to see n error will be raised if there is a discrepancy:
>>> IntegerPartition(10, [5, 4, 3, 1])
Traceback (most recent call last):
...
ValueError: The partition is not valid
"""
from sympy.ntheory.residue_ntheory import int_tested
if integer is not None:
integer, partition = partition, integer
if isinstance(partition, (dict, Dict)):
_ = []
for k, v in sorted(partition.items(), reverse=True):
if not v:
continue
k, v = int_tested(k, v)
_.extend([k]*v)
partition = tuple(_)
else:
partition = tuple(sorted(int_tested(partition), reverse=True))
sum_ok = False
if integer is None:
integer = sum(partition)
sum_ok = True
else:
integer = int_tested(integer)
if not sum_ok and sum(partition) != integer:
raise ValueError("Partition did not add to %s" % integer)
if any(i < 1 for i in partition):
raise ValueError("The summands must all be positive.")
obj = Basic.__new__(cls, integer, partition)
obj.partition = list(partition)
obj.integer = integer
return obj
def prevprime(n):
""" Return the largest prime smaller than n.
Potential primes are located at 6*j +/- 1.
>>> 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
"""
n = int_tested(n, strict=False)
if n < 3:
raise ValueError("no preceding primes")
if n < 8:
return {3: 2, 4: 3, 5: 3, 6: 5, 7: 5}[n]
nn = 6 * (n // 6)
if n - nn <= 1:
n = nn - 1
if isprime(n):
return n
n -= 4
else:
n = nn + 1
while 1:
if isprime(n):
return n
n -= 2
if isprime(n):
return n
n -= 4
def real_root(arg, n=None):
"""Return the real nth-root of arg if possible. If n is omitted then
all instances of -1**(1/odd) will be changed to -1.
Examples
========
>>> from sympy import root, real_root, Rational
>>> from sympy.abc import x, n
>>> real_root(-8, 3)
-2
>>> root(-8, 3)
2*(-1)**(1/3)
>>> real_root(_)
-2
See Also
========
sympy.polys.rootoftools.RootOf
sympy.core.power.integer_nthroot
root, sqrt
"""
if n is not None:
n = int_tested(n)
rv = C.Pow(arg, Rational(1, n))
if n % 2 == 0:
return rv
else:
rv = sympify(arg)
n1pow = Transform(
lambda x: S.NegativeOne, lambda x: x.is_Pow and x.base is S.NegativeOne
and x.exp.is_Rational and x.exp.p == 1 and x.exp.q % 2)
return rv.xreplace(n1pow)
def primepi(n):
""" Return the value of the prime counting function pi(n) = the number
of prime numbers less than or equal to n. The number n need not
necessarily be an integer.
Examples
========
>>> from sympy import primepi
>>> primepi(25)
9
See Also
========
sympy.ntheory.primetest.isprime : Test if n is prime
primerange : Generate all primes in a given range
prime : Return the nth prime
"""
n = int_tested(n, strict=False)
if n < 2:
return 0
else:
n = int(n)
return sieve.search(n)[0]
def prime(nth):
""" Return the nth prime, with the primes indexed as prime(1) = 2,
prime(2) = 3, etc.... The nth prime is approximately n*log(n) and
can never be larger than 2**n.
References
==========
- http://primes.utm.edu/glossary/xpage/BertrandsPostulate.html
Examples
========
>>> from sympy import prime
>>> prime(10)
29
>>> prime(1)
2
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
"""
n = int_tested(nth)
if n < 1:
raise ValueError("nth must be a positive integer; prime(1) == 2")
return sieve[n]
def primepi(n):
""" Return the value of the prime counting function pi(n) = the number
of prime numbers less than or equal to n. The number n need not
necessarily be an integer.
Examples
========
>>> from sympy import primepi
>>> primepi(25)
9
See Also
========
sympy.ntheory.primetest.isprime : Test if n is prime
primerange : Generate all primes in a given range
prime : Return the nth prime
"""
n = int_tested(n, strict=False)
if n < 2:
return 0
else:
n = int(n)
return sieve.search(n)[0]
def prevprime(n):
""" Return the largest prime smaller than n.
Potential primes are located at 6*j +/- 1.
>>> 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
"""
n = int_tested(n, strict=False)
if n < 3:
raise ValueError("no preceding primes")
if n < 8:
return {3: 2, 4: 3, 5: 3, 6: 5, 7: 5}[n]
nn = 6*(n//6)
if n - nn <= 1:
n = nn - 1
if isprime(n):
return n
n -= 4
else:
n = nn + 1
while 1:
if isprime(n):
return n
n -= 2
if isprime(n):
return n
n -= 4
def real_root(arg, n=None):
"""Return the real nth-root of arg if possible. If n is omitted then
all instances of -1**(1/odd) will be changed to -1.
Examples
========
>>> from sympy import root, real_root, Rational
>>> from sympy.abc import x, n
>>> real_root(-8, 3)
-2
>>> root(-8, 3)
2*(-1)**(1/3)
>>> real_root(_)
-2
See Also
========
L{sqrt}, L{RootOf}, L{root}, L{integer_nthroot}
"""
if n is not None:
n = int_tested(n)
rv = C.Pow(arg, Rational(1, n))
if n % 2 == 0:
return rv
else:
rv = sympify(arg)
n1pow = Transform(
lambda x: S.NegativeOne,
lambda x: x.is_Pow and x.base is S.NegativeOne and x.exp.is_Rational and x.exp.p == 1 and x.exp.q % 2
)
return rv.xreplace(n1pow)
def nextprime(n, ith=1):
""" Return the ith prime greater than n.
i must be 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
"""
n = int(n)
i = int_tested(ith)
if i > 1:
pr = n
j = 1
while 1:
pr = nextprime(pr)
j += 1
if j > i:
break
return pr
if n < 2:
return 2
if n < 7:
return {2: 3, 3: 5, 4: 5, 5: 7, 6: 7}[n]
nn = 6 * (n // 6)
if nn == n:
n += 1
if isprime(n):
return n
n += 4
elif n - nn == 5:
n += 2
if isprime(n):
return n
n += 4
else:
n = nn + 5
while 1:
if isprime(n):
return n
n += 2
if isprime(n):
return n
n += 4
def nextprime(n, ith=1):
""" Return the ith prime greater than n.
i must be 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
"""
n = int(n)
i = int_tested(ith)
if i > 1:
pr = n
j = 1
while 1:
pr = nextprime(pr)
j += 1
if j > i:
break
return pr
if n < 2:
return 2
if n < 7:
return {2: 3, 3: 5, 4: 5, 5: 7, 6: 7}[n]
nn = 6 * (n // 6)
if nn == n:
n += 1
if isprime(n):
return n
n += 4
elif n - nn == 5:
n += 2
if isprime(n):
return n
n += 4
else:
n = nn + 5
while 1:
if isprime(n):
return n
n += 2
if isprime(n):
return n
n += 4
def prime(nth):
""" Return the nth prime, with the primes indexed as prime(1) = 2,
prime(2) = 3, etc.... The nth prime is approximately n*log(n) and
can never be larger than 2**n.
References
==========
- http://primes.utm.edu/glossary/xpage/BertrandsPostulate.html
Examples
========
>>> from sympy import prime
>>> prime(10)
29
>>> prime(1)
2
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
"""
n = int_tested(nth)
if n < 1:
raise ValueError("nth must be a positive integer; prime(1) == 2")
return sieve[n]
def unrank(self, rank, n):
"""Finds the unranked Prufer sequence.
Examples
========
>>> from sympy.combinatorics.prufer import Prufer
>>> Prufer.unrank(0, 4)
Prufer([0, 0])
"""
n = int_tested(n)
rank = int_tested(rank)
L = defaultdict(int)
for i in xrange(n - 3, -1, -1):
L[i] = rank % n
rank = (rank - L[i]) // n
return Prufer([L[i] for i in xrange(len(L))])
def unrank(self, rank, n):
"""Finds the unranked Prufer sequence.
Examples
========
>>> from sympy.combinatorics.prufer import Prufer
>>> Prufer.unrank(0, 4)
Prufer([0, 0])
"""
n = int_tested(n)
rank = int_tested(rank)
L = defaultdict(int)
for i in xrange(n - 3, -1, -1):
L[i] = rank % n
rank = (rank - L[i])//n
return Prufer([L[i] for i in xrange(len(L))])
def __contains__(self, n):
try:
n = int_tested(n)
assert n >= 2
except (ValueError, AssertionError):
return False
if n % 2 == 0:
return n == 2
a, b = self.search(n)
return a == b
def __contains__(self, n):
try:
n = int_tested(n)
assert n >= 2
except (ValueError, AssertionError):
return False
if n % 2 == 0:
return n == 2
a, b = self.search(n)
return a == b
def extend_to_no(self, n):
"""Extend to include (at least) the nth prime numbers
Examples
========
>>> from sympy import sieve
>>> sieve.extend_to_no(9)
>>> sieve[10] == 29
True
"""
n = int_tested(n, strict=False)
while len(self._list) < n:
self.extend(int(self._list[-1] * 1.5))
def extend_to_no(self, n):
"""Extend to include (at least) the nth prime numbers
Examples
========
>>> from sympy import sieve
>>> sieve.extend_to_no(9)
>>> sieve[10] == 29
True
"""
n = int_tested(n, strict=False)
while len(self._list) < n:
self.extend(int(self._list[-1] * 1.5))
def give(a, b, seq=seed):
from sympy.ntheory.residue_ntheory import int_tested
a, b = int_tested(a, b)
w = b - a
if w < 0:
raise ValueError('_randint got empty range')
try:
x = seq.pop()
except AttributeError:
raise ValueError('_randint expects a list-like sequence')
except IndexError:
raise ValueError('_randint sequence was too short')
if a <= x <= b:
return x
else:
return give(a, b, seq)
def give(a, b, seq=seed):
from sympy.ntheory.residue_ntheory import int_tested
a, b = int_tested(a, b)
w = b - a
if w < 0:
raise ValueError('_randint got empty range')
try:
x = seq.pop()
except AttributeError:
raise ValueError('_randint expects a list-like sequence')
except IndexError:
raise ValueError('_randint sequence was too short')
if a <= x <= b:
return x
else:
return give(a, b, seq)
def primorial(n, nth=True):
"""
Returns the product of either 1. the first n primes (default) or
2. the primes less than or equal to n (when ``nth=False``).
>>> from sympy.ntheory.generate import primorial, randprime, primerange
>>> from sympy import factorint, Mul, primefactors
>>> primorial(4) # the first 4 primes are 2, 3, 5, 7
210
>>> primorial(4, nth=0) # primes <= 4 are 2 and 3
6
>>> primorial(1)
2
>>> primorial(1, nth=0)
1
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 primes must divide this
product of primes.
>>> factorint(primorial(4) + 1)
{211: 1}
>>> factorint(primorial(4) - 1)
{11: 1, 19: 1}
>>> 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
"""
n = int_tested(n, strict=False)
if n < 1:
raise ValueError("primorial argument must be >= 1")
p = 1
if nth:
for i in range(1, n + 1):
p *= prime(i)
else:
for i in primerange(2, n + 1):
p *= i
return p
def primorial(n, nth=True):
"""
Returns the product of either 1. the first n primes (default) or
2. the primes less than or equal to n (when ``nth=False``).
>>> from sympy.ntheory.generate import primorial, randprime, primerange
>>> from sympy import factorint, Mul, primefactors
>>> primorial(4) # the first 4 primes are 2, 3, 5, 7
210
>>> primorial(4, nth=0) # primes <= 4 are 2 and 3
6
>>> primorial(1)
2
>>> primorial(1, nth=0)
1
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 primes must divide this
product of primes.
>>> factorint(primorial(4) + 1)
{211: 1}
>>> factorint(primorial(4) - 1)
{11: 1, 19: 1}
>>> 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
"""
n = int_tested(n, strict=False)
if n < 1:
raise ValueError("primorial argument must be >= 1")
p = 1
if nth:
for i in range(1, n + 1):
p *= prime(i)
else:
for i in primerange(2, n + 1):
p *= i
return p
def __new__(cls, *args):
# expand range
slc = slice(*args)
start, stop, step = slc.start or 0, slc.stop, slc.step or 1
try:
start, stop, step = [S(int_tested(w)) for w in (start, stop, step)]
except ValueError:
raise ValueError("Inputs to Range must be Integer Valued\n"+
"Use TransformationSets of Ranges for other cases")
n = ceiling((stop - start)/step)
if n <= 0:
return S.EmptySet
# normalize args: regardless of how they are entered they will show
# canonically as Range(inf, sup, step) with step > 0
start, stop = sorted((start, start + (n - 1)*step))
step = abs(step)
return Basic.__new__(cls, start, stop + step, step)
def __new__(cls, *args):
# expand range
slc = slice(*args)
start, stop, step = slc.start or 0, slc.stop, slc.step or 1
try:
start, stop, step = [S(int_tested(w)) for w in (start, stop, step)]
except ValueError:
raise ValueError(
"Inputs to Range must be Integer Valued\n" +
"Use TransformationSets of Ranges for other cases")
n = ceiling((stop - start) / step)
if n <= 0:
return S.EmptySet
# normalize args: regardless of how they are entered they will show
# canonically as Range(inf, sup, step) with step > 0
start, stop = sorted((start, start + (n - 1) * step))
step = abs(step)
return Basic.__new__(cls, start, stop + step, step)
def __add__(self, other):
"""
Return permutation whose rank is ``other`` greater than current rank,
(mod the maximum rank for the set).
Examples
========
>>> from sympy.combinatorics.partitions import Partition
>>> a = Partition([[1, 2], [3]])
>>> a.rank
1
>>> (a + 1).rank
2
>>> (a + 100).rank
1
"""
other = int_tested(other)
offset = self.rank + other
result = RGS_unrank((offset) % RGS_enum(self.size), self.size)
return Partition.from_rgs(result, self.members)
def search(self, n):
"""For n >= 2, return the tightest a, b such that
self[a] <= n <= self[b]
Examples
========
>>> from sympy import sieve
>>> sieve.search(25)
(9, 10)
"""
n = int_tested(n, strict=False)
if n < 2:
raise ValueError("n must be greater than 1")
if n > self._list[-1]:
self.extend(n)
b = bisect(self._list, n)
if self._list[b-1] == n:
return b, b
else:
return b, b+1
def search(self, n):
"""For n >= 2, return the tightest a, b such that
self[a] <= n <= self[b]
Examples
========
>>> from sympy import sieve
>>> sieve.search(25)
(9, 10)
"""
n = int_tested(n, strict=False)
if n < 2:
raise ValueError("n must be greater than 1")
if n > self._list[-1]:
self.extend(n)
b = bisect(self._list, n)
if self._list[b - 1] == n:
return b, b
else:
return b, b + 1
def extend_to_no(self, i):
"""Extend to include the ith prime number.
i must be an integer.
The list is extended by 50% if it is too short, so it is
likely that it will be longer than requested.
Examples
========
>>> from sympy import sieve
>>> from array import array # this line and next for doctest only
>>> sieve._list = array('l', [2, 3, 5, 7, 11, 13])
>>> sieve.extend_to_no(9)
>>> sieve._list
array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23])
"""
i = int_tested(i)
while len(self._list) < i:
self.extend(int(self._list[-1] * 1.5))
def __add__(self, other):
"""
Return permutation whose rank is ``other`` greater than current rank,
(mod the maximum rank for the set).
Examples
========
>>> from sympy.combinatorics.partitions import Partition
>>> a = Partition([[1, 2], [3]])
>>> a.rank
1
>>> (a + 1).rank
2
>>> (a + 100).rank
1
"""
other = int_tested(other)
offset = self.rank + other
result = RGS_unrank((offset) %
RGS_enum(self.size),
self.size)
return Partition.from_rgs(result, self.members)
def extend_to_no(self, i):
"""Extend to include the ith prime number.
i must be an integer.
The list is extended by 50% if it is too short, so it is
likely that it will be longer than requested.
Examples
========
>>> from sympy import sieve
>>> from array import array # this line and next for doctest only
>>> sieve._list = array('l', [2, 3, 5, 7, 11, 13])
>>> sieve.extend_to_no(9)
>>> sieve._list
array('l', [2, 3, 5, 7, 11, 13, 17, 19, 23])
"""
i = int_tested(i)
while len(self._list) < i:
self.extend(int(self._list[-1] * 1.5))
def random_integer_partition(n, seed=None):
"""
Generates a random integer partition summing to ``n`` as a list
of reverse-sorted integers.
Examples
========
>>> from sympy.combinatorics.partitions import random_integer_partition
For the following, a seed is given so a known value can be shown; in
practice, the seed would not be given.
>>> random_integer_partition(100, seed=[1, 1, 12, 1, 2, 1, 85, 1])
[85, 12, 2, 1]
>>> random_integer_partition(10, seed=[1, 2, 3, 1, 5, 1])
[5, 3, 1, 1]
>>> random_integer_partition(1)
[1]
"""
from sympy.utilities.randtest import _randint
n = int_tested(n)
if n < 1:
raise ValueError('n must be a positive integer')
randint = _randint(seed)
partition = []
while (n > 0):
k = randint(1, n)
mult = randint(1, n // k)
partition.append((k, mult))
n -= k * mult
partition.sort(reverse=True)
partition = flatten([[k] * m for k, m in partition])
return partition
def random_integer_partition(n, seed=None):
"""
Generates a random integer partition summing to ``n`` as a list
of reverse-sorted integers.
Examples
========
>>> from sympy.combinatorics.partitions import random_integer_partition
For the following, a seed is given so a known value can be shown; in
practice, the seed would not be given.
>>> random_integer_partition(100, seed=[1, 1, 12, 1, 2, 1, 85, 1])
[85, 12, 2, 1]
>>> random_integer_partition(10, seed=[1, 2, 3, 1, 5, 1])
[5, 3, 1, 1]
>>> random_integer_partition(1)
[1]
"""
from sympy.utilities.randtest import _randint
n = int_tested(n)
if n < 1:
raise ValueError('n must be a positive integer')
randint = _randint(seed)
partition = []
while (n > 0):
k = randint(1, n)
mult = randint(1, n//k)
partition.append((k, mult))
n -= k*mult
partition.sort(reverse=True)
partition = flatten([[k]*m for k, m in partition])
return partition
def partitions(n, m=None, k=None):
"""Generate all partitions of integer n (>= 0).
'm' limits the number of parts in the partition, e.g. if m=2 then
partitions will contain no more than 2 numbers, while
'k' limits the numbers which may appear in the partition, e.g. k=2 will
return partitions with no element greater than 2.
Each partition is represented as a dictionary, mapping an integer
to the number of copies of that integer in the partition. For example,
the first partition of 4 returned is {4: 1}: a single 4.
>>> from sympy.utilities.iterables import partitions
Maximum key (number in partition) limited with k (in this case, 2):
>>> for p in partitions(6, k=2):
... print p
{2: 3}
{1: 2, 2: 2}
{1: 4, 2: 1}
{1: 6}
Maximum number of parts in partion limited with m (in this case, 2):
>>> for p in partitions(6, m=2):
... print p
...
{6: 1}
{1: 1, 5: 1}
{2: 1, 4: 1}
{3: 2}
Note that the _same_ dictionary object is returned each time.
This is for speed: generating each partition goes quickly,
taking constant time independent of n.
>>> [p for p in partitions(6, k=2)]
[{1: 6}, {1: 6}, {1: 6}, {1: 6}]
If you want to build a list of the returned dictionaries then
make a copy of them:
>>> [p.copy() for p in partitions(6, k=2)]
[{2: 3}, {1: 2, 2: 2}, {1: 4, 2: 1}, {1: 6}]
Reference:
modified from Tim Peter's version to allow for k and m values:
code.activestate.com/recipes/218332-generator-for-integer-partitions/
See Also
========
sympy.combinatorics.partitions.Partition
sympy.combinatorics.partitions.IntegerPartition
"""
from sympy.ntheory.residue_ntheory import int_tested
if n < 0:
raise ValueError("n must be >= 0")
if m == 0:
raise ValueError("m must be > 0")
m = min(m or n, n)
if m < 1:
raise ValueError("maximum numbers in partition, m, must be > 0")
k = min(k or n, n)
if k < 1:
raise ValueError("maximum value in partition, k, must be > 0")
if m * k < n:
return
n, m, k = int_tested(n, m, k)
q, r = divmod(n, k)
ms = {k: q}
keys = [k] # ms.keys(), from largest to smallest
if r:
ms[r] = 1
keys.append(r)
room = m - q - bool(r)
yield ms
while keys != [1]:
# Reuse any 1's.
if keys[-1] == 1:
del keys[-1]
reuse = ms.pop(1)
room += reuse
else:
reuse = 0
while 1:
# Let i be the smallest key larger than 1. Reuse one
# instance of i.
i = keys[-1]
newcount = ms[i] = ms[i] - 1
reuse += i
if newcount == 0:
del keys[-1], ms[i]
room += 1
# Break the remainder into pieces of size i-1.
i -= 1
q, r = divmod(reuse, i)
need = q + bool(r)
if need > room:
if not keys:
return
continue
ms[i] = q
keys.append(i)
if r:
ms[r] = 1
keys.append(r)
break
room -= need
yield ms
def __getitem__(self, n):
"""Return the nth prime number"""
n = int_tested(n)
self.extend_to_no(n)
return self._list[n - 1]
def partitions(n, m=None, k=None):
"""Generate all partitions of integer n (>= 0).
'm' limits the number of parts in the partition, e.g. if m=2 then
partitions will contain no more than 2 numbers, while
'k' limits the numbers which may appear in the partition, e.g. k=2 will
return partitions with no element greater than 2.
Each partition is represented as a dictionary, mapping an integer
to the number of copies of that integer in the partition. For example,
the first partition of 4 returned is {4: 1}: a single 4.
>>> from sympy.utilities.iterables import partitions
Maximum key (number in partition) limited with k (in this case, 2):
>>> for p in partitions(6, k=2):
... print p
{2: 3}
{1: 2, 2: 2}
{1: 4, 2: 1}
{1: 6}
Maximum number of parts in partion limited with m (in this case, 2):
>>> for p in partitions(6, m=2):
... print p
...
{6: 1}
{1: 1, 5: 1}
{2: 1, 4: 1}
{3: 2}
Note that the _same_ dictionary object is returned each time.
This is for speed: generating each partition goes quickly,
taking constant time independent of n.
>>> [p for p in partitions(6, k=2)]
[{1: 6}, {1: 6}, {1: 6}, {1: 6}]
If you want to build a list of the returned dictionaries then
make a copy of them:
>>> [p.copy() for p in partitions(6, k=2)]
[{2: 3}, {1: 2, 2: 2}, {1: 4, 2: 1}, {1: 6}]
Reference:
modified from Tim Peter's version to allow for k and m values:
code.activestate.com/recipes/218332-generator-for-integer-partitions/
See Also
========
sympy.combinatorics.partitions.Partition
sympy.combinatorics.partitions.IntegerPartition
"""
from sympy.ntheory.residue_ntheory import int_tested
if n < 0:
raise ValueError("n must be >= 0")
if m == 0:
raise ValueError("m must be > 0")
m = min(m or n, n)
if m < 1:
raise ValueError("maximum numbers in partition, m, must be > 0")
k = min(k or n, n)
if k < 1:
raise ValueError("maximum value in partition, k, must be > 0")
if m*k < n:
return
n, m, k = int_tested(n, m, k)
q, r = divmod(n, k)
ms = {k: q}
keys = [k] # ms.keys(), from largest to smallest
if r:
ms[r] = 1
keys.append(r)
room = m - q - bool(r)
yield ms
while keys != [1]:
# Reuse any 1's.
if keys[-1] == 1:
del keys[-1]
reuse = ms.pop(1)
room += reuse
else:
reuse = 0
while 1:
# Let i be the smallest key larger than 1. Reuse one
# instance of i.
i = keys[-1]
newcount = ms[i] = ms[i] - 1
reuse += i
if newcount == 0:
del keys[-1], ms[i]
room += 1
# Break the remainder into pieces of size i-1.
i -= 1
q, r = divmod(reuse, i)
need = q + bool(r)
if need > room:
if not keys:
return
continue
ms[i] = q
keys.append(i)
if r:
ms[r] = 1
keys.append(r)
break
room -= need
yield ms
def __getitem__(self, n):
"""Return the nth prime number"""
n = int_tested(n)
self.extend_to_no(n)
return self._list[n - 1]
def solve_congruence(*remainder_modulus_pairs, **hint):
"""Compute the integer ``n`` that has the residual ``ai`` when it is
divided by ``mi`` where the ``ai`` and ``mi`` are given as pairs to
this function: ((a1, m1), (a2, m2), ...). If there is no solution,
return None. Otherwise return ``n`` and its modulus.
The ``mi`` values need not be co-prime. If it is known that the moduli are
not co-prime then the hint ``check`` can be set to False (default=True) and
the check for a quicker solution via crt() (valid when the moduli are
co-prime) will be skipped.
If the hint ``symmetric`` is True (default is False), the value of ``n``
will be within 1/2 of the modulus, possibly negative.
Examples
========
>>> from sympy.ntheory.modular import solve_congruence
What number is 2 mod 3, 3 mod 5 and 2 mod 7?
>>> solve_congruence((2, 3), (3, 5), (2, 7))
(23, 105)
>>> [23 % m for m in [3, 5, 7]]
[2, 3, 2]
If you prefer to work with all remainder in one list and
all moduli in another, send the arguments like this:
>>> solve_congruence(*zip((2, 3, 2), (3, 5, 7)))
(23, 105)
The moduli need not be co-prime; in this case there may or
may not be a solution:
>>> solve_congruence((2, 3), (4, 6)) is None
True
>>> solve_congruence((2, 3), (5, 6))
(5, 6)
The symmetric flag will make the result be within 1/2 of the modulus:
>>> solve_congruence((2, 3), (5, 6), symmetric=True)
(-1, 6)
See also: crt and sympy.polys.galoistools.gf_crt
"""
def combine(c1, c2):
"""Return the tuple (a, m) which satisfies the requirement
that n = a + i*m satisfy n = a1 + j*m1 and n = a2 = k*m2.
References
==========
- http://en.wikipedia.org/wiki/Method_of_successive_substitution
"""
from sympy.core.numbers import igcdex
a1, m1 = c1
a2, m2 = c2
a, b, c = m1, a2 - a1, m2
g = reduce(igcd, [a, b, c])
a, b, c = [i // g for i in [a, b, c]]
if a != 1:
inv_a, _, g = igcdex(a, c)
if g != 1:
return None
b *= inv_a
a, m = a1 + m1 * b, m1 * c
return a, m
rm = remainder_modulus_pairs
symmetric = hint.get('symmetric', False)
if hint.get('check', True):
rm = [int_tested(*pair) for pair in rm]
# ignore redundant pairs but raise an error otherwise; also
# make sure that a unique set of bases is sent to gf_crt if
# they are all prime.
#
# The routine will work out less-trivial violations and
# return None, e.g. for the pairs (1,3) and (14,42) there
# is no answer because 14 mod 42 (having a gcd of 14) implies
# (14/2) mod (42/2), (14/7) mod (42/7) and (14/14) mod (42/14)
# which, being 0 mod 3, is inconsistent with 1 mod 3. But to
# preprocess the input beyond checking of another pair with 42
# or 3 as the modulus (for this example) is not necessary.
uniq = {}
for r, m in rm:
r %= m
if m in uniq:
if r != uniq[m]:
return None
continue
uniq[m] = r
rm = [(r, m) for m, r in uniq.iteritems()]
del uniq
# if the moduli are co-prime, the crt will be significantly faster;
# checking all pairs for being co-prime gets to be slow but a prime
# test is a good trade-off
if all(isprime(m) for r, m in rm):
r, m = zip(*rm)
return crt(m, r, symmetric=symmetric, check=False)
rv = (0, 1)
for rmi in rm:
rv = combine(rv, rmi)
if rv is None:
break
n, m = rv
n = n % m
else:
if symmetric:
return symmetric_residue(n, m), m
return n, m
def crt(m, v, symmetric=False, check=True):
r"""Chinese Remainder Theorem.
The moduli in m are assumed to be pairwise coprime. The output
is then an integer f, such that f = v_i mod m_i for each pair out
of v and m. If ``symmetric`` is False a positive integer will be
returned, else \|f\| will be less than or equal to the LCM of the
moduli, and thus f may be negative.
If the moduli are not co-prime the correct result will be returned
if/when the test of the result is found to be incorrect. This result
will be None if there is no solution.
The keyword ``check`` can be set to False if it is known that the moduli
are coprime.
As an example consider a set of residues ``U = [49, 76, 65]``
and a set of moduli ``M = [99, 97, 95]``. Then we have::
>>> from sympy.ntheory.modular import crt, solve_congruence
>>> crt([99, 97, 95], [49, 76, 65])
(639985, 912285)
This is the correct result because::
>>> [639985 % m for m in [99, 97, 95]]
[49, 76, 65]
If the moduli are not co-prime, you may receive an incorrect result
if you use ``check=False``:
>>> crt([12, 6, 17], [3, 4, 2], check=False)
(954, 1224)
>>> [954 % m for m in [12, 6, 17]]
[6, 0, 2]
>>> crt([12, 6, 17], [3, 4, 2]) is None
True
>>> crt([3, 6], [2, 5])
(5, 6)
Note: the order of gf_crt's arguments is reversed relative to crt,
and that solve_congruence takes residue, modulus pairs.
Programmer's note: rather than checking that all pairs of moduli share
no GCD (an O(n**2) test) and rather than factoring all moduli and seeing
that there is no factor in common, a check that the result gives the
indicated residuals is performed -- an O(n) operation.
"""
if check:
m = int_tested(*m)
v = int_tested(*v)
result = gf_crt(v, m, ZZ)
mm = prod(m)
if check:
if not all(v % m == result % m for v, m in zip(v, m)):
result = solve_congruence(*zip(v, m),
**dict(check=False, symmetric=symmetric))
if result is None:
return result
result, mm = result
if symmetric:
return symmetric_residue(result, mm), mm
return result, mm
def crt(m, v, symmetric=False, check=True):
"""Chinese Remainder Theorem.
The moduli in m are assumed to be pairwise coprime. The output
is then an integer f, such that f = v_i mod m_i for each pair out
of v and m. If ``symmetric`` is False a positive integer will be
returned, else |f| will be less than or equal to the LCM of the
moduli, and thus f may be negative.
If the moduli are not co-prime the correct result will be returned
if/when the test of the result is found to be incorrect. This result
will be None if there is no solution.
The keyword ``check`` can be set to False if it is known that the moduli
are coprime.
As an example consider a set of residues ``U = [49, 76, 65]``
and a set of moduli ``M = [99, 97, 95]``. Then we have::
>>> from sympy.ntheory.modular import crt, solve_congruence
>>> crt([99, 97, 95], [49, 76, 65])
(639985, 912285)
This is the correct result because::
>>> [639985 % m for m in [99, 97, 95]]
[49, 76, 65]
If the moduli are not co-prime, you may receive an incorrect result
if you use ``check=False``:
>>> crt([12, 6, 17], [3, 4, 2], check=False)
(954, 1224)
>>> [954 % m for m in [12, 6, 17]]
[6, 0, 2]
>>> crt([12, 6, 17], [3, 4, 2]) is None
True
>>> crt([3, 6], [2, 5])
(5, 6)
Note: the order of gf_crt's arguments is reversed relative to crt,
and that solve_congruence takes residue, modulus pairs.
Programmer's note: rather than checking that all pairs of moduli share
no GCD (an O(n**2) test) and rather than factoring all moduli and seeing
that there is no factor in common, a check that the result gives the
indicated residuals is performed -- an O(n) operation.
"""
if check:
m = int_tested(*m)
v = int_tested(*v)
result = gf_crt(v, m, ZZ)
mm = prod(m)
if check:
if not all(v % m == result % m for v, m in zip(v, m)):
result = solve_congruence(*zip(v, m),
**dict(check=False,
symmetric=symmetric))
if result is None:
return result
result, mm = result
if symmetric:
return symmetric_residue(result, mm), mm
return result, mm
def solve_congruence(*remainder_modulus_pairs, **hint):
"""Compute the integer ``n`` that has the residual ``ai`` when it is
divided by ``mi`` where the ``ai`` and ``mi`` are given as pairs to
this function: ((a1, m1), (a2, m2), ...). If there is no solution,
return None. Otherwise return ``n`` and its modulus.
The ``mi`` values need not be co-prime. If it is known that the moduli are
not co-prime then the hint ``check`` can be set to False (default=True) and
the check for a quicker solution via crt() (valid when the moduli are
co-prime) will be skipped.
If the hint ``symmetric`` is True (default is False), the value of ``n``
will be within 1/2 of the modulus, possibly negative.
Examples::
>>> from sympy.ntheory.modular import solve_congruence
What number is 2 mod 3, 3 mod 5 and 2 mod 7?
>>> solve_congruence((2, 3), (3, 5), (2, 7))
(23, 105)
>>> [23 % m for m in [3, 5, 7]]
[2, 3, 2]
If you prefer to work with all remainder in one list and
all moduli in another, send the arguments like this:
>>> solve_congruence(*zip((2, 3, 2), (3, 5, 7)))
(23, 105)
The moduli need not be co-prime; in this case there may or
may not be a solution:
>>> solve_congruence((2, 3), (4, 6)) is None
True
>>> solve_congruence((2, 3), (5, 6))
(5, 6)
The symmetric flag will make the result be within 1/2 of the modulus:
>>> solve_congruence((2, 3), (5, 6), symmetric=True)
(-1, 6)
See also: crt and sympy.polys.galoistools.gf_crt
"""
def combine(c1, c2):
"""Return the tuple (a, m) which satisfies the requirement
that n = a + i*m satisfy n = a1 + j*m1 and n = a2 = k*m2.
Reference:
http://en.wikipedia.org/wiki/Method_of_successive_substitution
"""
from sympy.core.numbers import igcdex
a1, m1 = c1
a2, m2 = c2
a, b, c = m1, a2 - a1, m2
g = reduce(igcd, [a, b, c])
a, b, c = [i//g for i in [a, b, c]]
if a != 1:
inv_a, _, g = igcdex(a, c)
if g != 1:
return None
b *= inv_a
a, m = a1 + m1*b, m1*c
return a, m
rm = remainder_modulus_pairs
symmetric = hint.get('symmetric', False)
if hint.get('check', True):
rm = [int_tested(*pair) for pair in rm]
# ignore redundant pairs but raise an error otherwise; also
# make sure that a unique set of bases is sent to gf_crt if
# they are all prime.
#
# The routine will work out less-trivial violations and
# return None, e.g. for the pairs (1,3) and (14,42) there
# is no answer because 14 mod 42 (having a gcd of 14) implies
# (14/2) mod (42/2), (14/7) mod (42/7) and (14/14) mod (42/14)
# which, being 0 mod 3, is inconsistent with 1 mod 3. But to
# preprocess the input beyond checking of another pair with 42
# or 3 as the modulus (for this example) is not necessary.
uniq = {}
for r, m in rm:
r %= m
if m in uniq:
if r != uniq[m]:
return None
continue
uniq[m] = r
rm = [(r, m) for m, r in uniq.iteritems()]
del uniq
# if the moduli are co-prime, the crt will be significantly faster;
# checking all pairs for being co-prime gets to be slow but a prime
# test is a good trade-off
if all(isprime(m) for r, m in rm):
r, m = zip(*rm)
return crt(m, r, symmetric=symmetric, check=False)
rv = (0, 1)
for rmi in rm:
rv = combine(rv, rmi)
if rv is None:
break
n, m = rv
n = n % m
else:
if symmetric:
return symmetric_residue(n, m), m
return n, m