mò ógÈFc @s€dZd„Zd„Zd„Zd„ZdZdZd„Zd„Zd„Z d„Z de fd„ƒYZ e ƒZ [ dS(sCSimple tools for manipulating types linear over the positive integers. Example linear spaces over the positive integers: the natural numbers the integers, whether modulo some value or not the polynomials with integer coefficients the rational, real or complex numbers any vector or linear space over any of the above c CsŒd}gd}}|||||}} }yÐxÉ|oÁt || ƒ\}}d|jo"|jo| jo djnpt ‚|o|i || |fƒn||jo|| | jo | |jpt ‚||| | }} }q5WWn)t j otd|||f‚nXxV|oN|iƒ\}} }t ||| || ƒ\}}|djpt ‚q.W||S(s„Division modulo a base. Returns a q for which num - q*den is a multiple of base, if possible: raises ValueError on failure.sG First reduce num and den mod base. If num is 0, q is 0; if den is 0, we have ValueError. Thereafter, to solve for n-q*d = s*b: wlog, q = q%b and s = s%d rearrange to n%d - s * (b%d) = q*d i.e. s = (n%d) / (b%d) mod d use this in n - s*b = q*d to discover q = n / d mod b This may be implemented recursively as q, r = divmod(n - dividemod(n%d, b%d, d) * b, d) assert r is 0 return q However, we can unroll the recursion to a pair of while loops. Compare and contrast with Euclid's algorithm (below). is%d / %d mod %dN(t __algorithmtstacktqtnumtbasetdentntdtbtdivmodtrtAssertionErrortappendtZeroDivisionErrort ValueErrortpop( RRRRRRRR RR((t)/home/eddy/.sys/py/study/maths/natural.pyt dividemod s* =2!!cCso|djo | }n|djo | }n|djo|Snx#|djo|||}}qHW|S(sPair-wise highest common factor. The value returned is, strictly, that value whose set of factors is the intersection of the sets of factors of the two arguments, ignoring all universal factors (e.g. 1, -1: values which are factors of everything). iN(Rta(RR((Rtgcd:s    cGs*d}x|D]}t||ƒ}q W|S(s¯The highest natural common factor of its arguments. All arguments should be members of a linear space over the natural numbers, eg (optionally long) integers or polynomials with such coefficients; there may be arbitrarily many arguments. Formally, the % operator, as defined for the given arguments, must be guaranteed to yield a positive or zero value whenever its (two) arguments are positive. This is true for integers. Formally, at least one argument (to hcf) must be non-zero: if (there are no arguments, or) the arguments are all zero, of which every value is a factor, there is no highest common factor - all integers are factors of 0. In this case, the value zero is returned. To justify not raising an error when (there are no inputs, or) all inputs are zero, we can re-cast the definition as: `that non-negative value which has, as its factors, exactly those naturals which are factors of all the arguments' (reading 'is n a factor of all arguments' as the negation of 'no argument does not have n as a factor' for the case of no arguments). So the result must be a multiple of every common factor of the arguments, but of nothing else. This coincides with `highest common factor' when at least one argument is non-zero: and gives zero when all arguments are zero. Note that -1 is a factor of every value (just as is 1), hence the need to specify `non-negative'. With this (undeniably less catchy) re-definition, we also get: a concatenation of lists of values has, as its hcf, the hcf of the values obtained by taking the hcfs of the lists seperately; i.e. hcf is a transitive binary operator. iN(tthistotherstotherR(RRR((RthcfLs cGs]d}x:|D]2}|p|Snt||ƒ}|||}q W|djo | Sn|S(sThe smallest common multiple of its arguments. All arguments should be members of a linear space over the natural numbers, e.g. (optionally long) integers or polynomials with such coefficients. There may be arbitrarily many arguments. If any entry is zero, so is the result: zero is a multiple of everything, and is smaller than any other value; furthermore, nothing else is a multiple of zero, so it is the only candidate. The return when no arguments are supplied is 1, to ensure that lcm is transitive. iiN(RRRRtc(RRRR((Rtlcmps  s»Any rational whose square is an integer is, itself, an integer. As a special case, this tells us that the square root of 2 is irrational. Proof: Suppose, for positive integers p, q, that the square of p/q is an integer, n. Thus p.p = q.q.n and n is positive. There are positive integers m, r for which m.r.r = n and m has no perfect square as a factor, yielding p.p = (q.r).(q.r).m. Expressing p, q.r and m in terms of their prime factors we now find that every prime factor of m has odd multiplicity as a factor of p.p, all of whose factors have even multiplicity; thus m cannot have any prime factors, so m is 1 and p = q.r has q as a factor so p/q = r is a positive integer. iiiii i iiiicCs€x5tD]-}t||ƒ\}}|djoPqqWd}x;|djo-||djo||7}n|d8}qAW|S(sgReturns the sum of the proper factors of N. Thus N is perfect precisely if N == factorsum(N). iiN(t _early_primestpR tNtiR tS(RRRRR ((Rt factorsum—s  ccs.d}x!tox5tD]-}t||ƒ\}}|djoPqqWd}xH|djo:||jo-||djo||7}n|d8}qQW||jou|Vd}x|d|>@p|d7}q³W||?dd|>djo)dGt |ƒG|Gt ||d?ƒGHqn|d7}q WdS(sÕReturns an iterator over the perfect numbers. All known perfect numbers are 2**n * (2**(1+n) -1) for some positive natural n (for which the second factor, 2**(1+n) -1, is prime), so iterating over these (which is *much* quicker, even pausing to check primality of the second factor) shall (probably) yield the same result. The iterator yielded by this implementation checks for other perfect numbers and prints a message if it ever finds one. iilsUnusually perfectN( RtTrueRRR tjR RRthex(RR!RRRR ((Rtperfectªs0  -ccsN|Vx=|djo/|dod|d}n |d}|VqWt‚dS(sqIterator for the Collatz conjecture's sequence for n. It is conjectured that, whatever positive integer n you give to this function, the resulting sequence shall ultimately terminate (by yielding 1). It is known (by experiment) that the conjecture is good up to n = 10 * 2**58 (and, hence, also good for any n which is just a power of 2 times some positive integer less than this limit). The function iterated is, with Z+ = {positive integers}, the union of (: n ← 2.n :Z+) and (: 6.j+4 ←2.j+1 :Z+). The conjecture effectively says that its transitive closure subsumes ({1}:|Z+). iiiN(Rt StopIteration(R((RtCollatzÊs     cCs¼|djotd|ƒ‚n|d}}x |o|dL}|d7}q0Wdd|>}xW|oO|dL}||>|}|d8}||jo|d|>O}||8}qaqaW|S(s<Returns the highest natural whose square does not exceed valis!Negative value has no square rootiiN(tvalRtvtbittbbtup(R&R)R*R'R(((Rtsqrtßs$      tNaturalscBsNtZdefd„ƒYZeie_eƒd„Z[eiZ d„ZRS(NtSuccBstZdd„Zd„ZRS(NcCs6|iƒ|dj o|i|ƒ|||