# Generic data structures and algorithms for rings#

AUTHORS:

class sage.rings.generic.ProductTree(leaves)#

Bases: object

A simple binary product tree, i.e., a tree of ring elements in which every node equals the product of its children. (In particular, the root equals the product of all leaves.)

Product trees are a very useful building block for fast computer algebra. For example, a quasilinear-time Discrete Fourier Transform (the famous Fast Fourier Transform) can be implemented as follows using the remainders() method of this class:

sage: from sage.rings.generic import ProductTree
sage: F = GF(65537)
sage: a = F(1111)
sage: assert a.multiplicative_order() == 1024
sage: R.<x> = F[]
sage: ms = [x - a^i for i in range(1024)]               # roots of unity
sage: ys = [F.random_element() for _ in range(1024)]    # input vector
sage: zs = ProductTree(ms).remainders(R(ys))            # compute FFT!
sage: zs == [R(ys) % m for m in ms]
True


This class encodes the tree as layers: Layer $$0$$ is just a tuple of the leaves. Layer $$i+1$$ is obtained from layer $$i$$ by replacing each pair of two adjacent elements by their product, starting from the left. (If the length is odd, the unpaired element at the end is simply copied as is.) This iteration stops as soon as it yields a layer containing only a single element (the root).

Note

Use this class if you need the remainders() method. To compute just the product, prod() is likely faster.

INPUT:

• leaves – an iterable of elements in a common ring

EXAMPLES:

sage: from sage.rings.generic import ProductTree
sage: R.<x> = GF(101)[]
sage: vs = [x - i for i in range(1,10)]
sage: tree = ProductTree(vs)
sage: tree.root()
x^9 + 56*x^8 + 62*x^7 + 44*x^6 + 47*x^5 + 42*x^4 + 15*x^3 + 11*x^2 + 12*x + 13
sage: tree.remainders(x^7 + x + 1)
[3, 30, 70, 27, 58, 72, 98, 98, 23]
sage: tree.remainders(x^100)
[1, 1, 1, 1, 1, 1, 1, 1, 1]

sage: vs = prime_range(100)
sage: tree = ProductTree(vs)
sage: tree.root().factor()
2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 * 29 * 31 * 37 * 41 * 43 * 47 * 53 * 59 * 61 * 67 * 71 * 73 * 79 * 83 * 89 * 97
sage: tree.remainders(3599)
[1, 2, 4, 1, 2, 11, 12, 8, 11, 3, 3, 10, 32, 30, 27, 48, 0, 0, 48, 49, 22, 44, 30, 39, 10]


We can access the individual layers of the tree:

sage: tree.layers
[(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97),
(6, 35, 143, 323, 667, 1147, 1763, 2491, 3599, 4757, 5767, 7387, 97),
(210, 46189, 765049, 4391633, 17120443, 42600829, 97),
(9699690, 3359814435017, 729345064647247, 97),
(32589158477190044730, 70746471270782959),
(2305567963945518424753102147331756070,)]

leaves()#

Return a tuple containing the leaves of this product tree.

EXAMPLES:

sage: from sage.rings.generic import ProductTree
sage: R.<x> = GF(101)[]
sage: vs = [x - i for i in range(1,10)]
sage: tree = ProductTree(vs)
sage: tree.leaves()
(x + 100, x + 99, x + 98, ..., x + 93, x + 92)
sage: tree.leaves() == tuple(vs)
True

remainders(x)#

Given a value $$x$$, return a list of all remainders of $$x$$ modulo the leaves of this product tree.

The base ring must support the % operator for this method to work.

INPUT:

• x – an element of the base ring of this product tree

EXAMPLES:

sage: from sage.rings.generic import ProductTree
sage: vs = prime_range(100)
sage: tree = ProductTree(vs)
sage: n = 1085749272377676749812331719267
sage: tree.remainders(n)
[1, 1, 2, 1, 9, 1, 7, 15, 8, 20, 15, 6, 27, 11, 2, 6, 0, 25, 49, 5, 51, 4, 19, 74, 13]
sage: [n % v for v in vs]
[1, 1, 2, 1, 9, 1, 7, 15, 8, 20, 15, 6, 27, 11, 2, 6, 0, 25, 49, 5, 51, 4, 19, 74, 13]

root()#

Return the value at the root of this product tree (i.e., the product of all leaves).

EXAMPLES:

sage: from sage.rings.generic import ProductTree
sage: R.<x> = GF(101)[]
sage: vs = [x - i for i in range(1,10)]
sage: tree = ProductTree(vs)
sage: tree.root()
x^9 + 56*x^8 + 62*x^7 + 44*x^6 + 47*x^5 + 42*x^4 + 15*x^3 + 11*x^2 + 12*x + 13
sage: tree.root() == prod(vs)
True

sage.rings.generic.prod_with_derivative(pairs)#

Given an iterable of pairs $$(f, \partial f)$$ of ring elements, return the pair $$(\prod f, \partial \prod f)$$, assuming $$\partial$$ is an operator obeying the standard product rule.

This function is entirely algebraic, hence still works when the elements $$f$$ and $$\partial f$$ are all passed through some ring homomorphism first. One particularly useful instance of this is evaluating the derivative of a product of polynomials at a point without fully expanding the product; see the second example below.

INPUT:

• pairs – an iterable of tuples $$(f, \partial f)$$ of elements of a common ring

ALGORITHM: Repeated application of the product rule.

EXAMPLES:

sage: from sage.rings.generic import prod_with_derivative
sage: R.<x> = ZZ[]
sage: fs = [x^2 + 2*x + 3, 4*x + 5, 6*x^7 + 8*x + 9]
sage: prod(fs)
24*x^10 + 78*x^9 + 132*x^8 + 90*x^7 + 32*x^4 + 140*x^3 + 293*x^2 + 318*x + 135
sage: prod(fs).derivative()
240*x^9 + 702*x^8 + 1056*x^7 + 630*x^6 + 128*x^3 + 420*x^2 + 586*x + 318
sage: F, dF = prod_with_derivative((f, f.derivative()) for f in fs)
sage: F
24*x^10 + 78*x^9 + 132*x^8 + 90*x^7 + 32*x^4 + 140*x^3 + 293*x^2 + 318*x + 135
sage: dF
240*x^9 + 702*x^8 + 1056*x^7 + 630*x^6 + 128*x^3 + 420*x^2 + 586*x + 318


The main reason for this function to exist is that it allows us to evaluate the derivative of a product of polynomials at a point $$\alpha$$ without ever fully expanding the product as a polynomial:

sage: alpha = 42
sage: F(alpha)
442943981574522759
sage: dF(alpha)
104645261461514994
sage: us = [f(alpha) for f in fs]
sage: vs = [f.derivative()(alpha) for f in fs]
sage: prod_with_derivative(zip(us, vs))
(442943981574522759, 104645261461514994)