I’m trying to find ‘highly composite’ pythagorean triples - numbers (c) that have more than one unique a,b (in the naturals) that satisfy a² + b² = c².
I’ve written a short python script to find these - it cycles through c in the range (0,1000), and for each c, finds all possible (a,b) such that b < a < c. This is a more brute force method, and I know if I did some reading on number theory I could find some more methods for different cases of a and b.
I have a feeling that my script isn’t particularly efficient, especially for large c. I don’t really know what to change or how to make it more efficient.
I’d be really grateful for any help or pointers!
a = 0
b = 0
l=[]
for i in range (0,1000):
#i is our c.
while a<i:
while b<a:
#for each a, we cycle through b = 1, b = 2, … until b = a.
#Then we make b = 0 and a = a+1, and start the iterative process again.
if a*a + b*b == i*i:
l.append(a)
l.append(b)
#I tried adding a break here - my thought process was that we can’t find any
#other b^2 that satisfies a^2 + b^2 = i^2 without changing our a^2. This
#actually made the runtime longer, and I don’t know why.
b = b+1
a = a+1
b = 0
if len(l) > 4:
#all our pairs of pythagorean triples, with the c at the end.
print(l, i)
#reset, and find pairs again for i = i+1.
l = []
b = 0
a = 0
Your code seems quite inefficient, because you are doing many times the same computations. You could make it more efficient by not calculating things that are not useful. The most important detail is the computation of a and b. You are looping through all possible values for a and b and checking if it's a pythagorean triplet. But once you give yourself a value for a, there is only one possible choice for b, so the b loop is useless.
By removing that loop, you're basically lowering the degree of the polynomial complexity by one, which will make it increasingly faster (compared to your current script) when c grows
Also, your code seems to be wrong, as it misses some triplets. I ran it and the first triplets found were with 65 and 85, but 25, 50 and 75 are also highly composite pythagoren triplets. That's because you're checking len(l)>4, while you should check len(l)>=4 instead because you're missing numbers that have two decompositions.
As a comparison, I programmed a similar python script as yours (except I did it myself and tried to make it as efficient as possible). On my computer, your script ran in 66 seconds, while mine ran in 4 seconds, so you have a lot of room for improvement.
EDIT : I added my code for the sake of sharing. Here is a list of what differs from yours :
I stored all squares of numbers from 1 to N in a list called squares so I can check efficiently if a number is a square
I store the results in a dictionary where the value at key c is a list of tuples corresponding to (a, b)
The loop for a goes from 1 to floor(c/sqrt(2))
Instead of looping for b, I check whether c²-a² is a square
On a general note, I pre-compute every value that has to be used several times (invsqrt2, csqr)
from math import floor, sqrt
invsqrt2 = 1/sqrt(2)
N=1000
highly_composite_triplets = {}
squares = list(map(lambda x: x**2, range(0,N+1)))
for c in range(2,N+1):
if c%50==0: print(c) # Just to keep track of the thing
csqr = c**2
listpairs = []
for a in range(1,floor(c*invsqrt2)+1):
sqrdiff = csqr-a**2
if sqrdiff in squares:
listpairs.append((a, squares.index(sqrdiff)))
if len(listpairs)>1:
highly_composite_triplets[c] = listpairs
print(highly_composite_triplets)
First of all, and as already mentioned, you should fix that > 4 by >= 4.
For performance, I would suggest using the Tree of primitive Pythagorean triples. It allows to generate all possible primitive triples, such that three "children" of a given triple have a c-value that is at least as great as the one of the "parent".
The non-primitive triples can be easily generated from a primitive one, by multiplying all three values with a coefficient (until the maximum value of c is reached). This has to only be done for the initial triplet, as the others will follow from it.
That is the part where most efficiency gain is made.
Then in a second phase: group those triples by their c value. You can use itertools.groupby for that.
In a third phase: only select the groups that have at least 2 members (i.e. 4 values).
Here is an implementation:
import itertools
import operator
def pythagorian(end):
# DFS traversal through the pythagorian tree:
def recur(a, b, c):
if c < end:
yield c, max(a, b), min(a, b)
yield from recur( a - 2*b + 2*c, 2*a - b + 2*c, 2*a - 2*b + 3*c)
yield from recur( a + 2*b + 2*c, 2*a + b + 2*c, 2*a + 2*b + 3*c)
yield from recur(-a + 2*b + 2*c, -2*a + b + 2*c, -2*a + 2*b + 3*c)
# Start traversal from basic triplet, and its multiples
for i in range(1, end // 5):
yield from recur(4*i, 3*i, 5*i)
def grouped_pythagorian(end):
# Group by value of c, and flatten the a, b pairs into a list
return [
(c, [a for _, *ab in group for a in ab])
for c, group in itertools.groupby(sorted(pythagorian(end)),
operator.itemgetter(0))
]
def highly_pythagorian(end):
# Select the groups of triples that have at least 2 members (i.e. 4 values)
return [(group, c) for c, group in grouped_pythagorian(end) if len(group) >= 4]
Run the function as follows:
for result in highly_pythagorian(1000):
print(*result)
This produces the triples within a fraction of a second, and is thousands of times faster than your version and the one in #Mateo's answer.
Simplified
As discussed in comments, I provide here code that uses the same algorithm, but without imports, list comprehensions, generators (yield), and unpacking operators (*):
def highly_pythagorian(end):
triples = []
# DFS traversal through the pythagorian tree:
def dfs(a, b, c):
if c < end:
triples.append((c, max(a, b), min(a, b)))
dfs( a - 2*b + 2*c, 2*a - b + 2*c, 2*a - 2*b + 3*c)
dfs( a + 2*b + 2*c, 2*a + b + 2*c, 2*a + 2*b + 3*c)
dfs(-a + 2*b + 2*c, -2*a + b + 2*c, -2*a + 2*b + 3*c)
# Start traversal from basic triplet, and its multiples
for i in range(1, end // 5):
dfs(4*i, 3*i, 5*i)
# Sort the triples by their c-component (first one),
# ...and then their a-component
triples.sort()
# Group the triples in a dict, keyed by c values
groups = {}
for c, a, b in triples:
if not c in groups:
groups[c] = []
groups[c].append(a)
groups[c].append(b)
# Select the groups of triples that have at least 2 members (i.e. 4 values)
results = []
for c, ab_pairs in sorted(groups.items()):
if len(ab_pairs) >= 4:
results.append((ab_pairs, c))
return results
Call as:
for ab_pairs, c in highly_pythagorian(1000):
print(ab_pairs, c)
Here is a solution based on the mathematical intuition behind Gaussian integers. We are working in the "ring" R of all numbers of the form
a + ib
where a, b are integers. This is the ring of Gaussian integers.
Here, i is the square root of -1. So i² = -1.
Such numbers lead to a similar arithmetic as in the case of the (usual) integers. Each such number has a unique decomposition in gaussian primes. (Up to the order of the factors.) Such a domain is called a unique factorization domain, UFD.
Which are the primes in R? (Those elements that cannot be split multiplicatively in more than two non-invertible pieces.) There is a concrete characterization for them.
The classical primes of the shapes 4k + 3 remain primes in R, are inert. So we cannot split primes like 3, 7, 11, 19, 23, 31, ... in R. But we can always split uniquely (up to unit conjugation, a unit being one among 1, -1, i, -i) the (classical) primes of the shape 4k + 1 in R. For instance:
(*)
5 = (2 + i)(2 - i)
13 = (3 + 2i)(3 - 2i)
17 = (4 + i)(4 - i)
29 = (5 + 2i)(5 - 2i)
37 = (6 + i)(6 - i)
41 = (5 + 4i)(5 - 4i)
53 = (7 + 2i)(7 - 2i)
61 = (6 + 5i)(6 - 5i)
and so on, i hope the scheme is clear. For our purpose, the remained prime two is the oddest prime. Since we have its decomposition
2 = (1 + i)(1 -i), where the two Gaussian primes (1 + i) and (1 - i) are associated, multiplying with a unit bring one in the other one. I will avoid this prime below.
Now consider the product of some of the numbers on the L.H.S. in (*). For instance 5.5.13.17 = 5525 - and let us pick from each of the four (classical) prime factors one of the Gaussian primes inside.
We may thus pick (2 + i) twice from the two 5-factors, (3 - 2i) from 13 and (4 + i) from the 17. We multiply and get:
sage: (2 + i)^2 * (3 - 2*i) * (4 + i)
41*I + 62
And indeed, a = 41 and b = 62 is a solution of 41² + 62² = 5525. Unfortunately 5525 is not a square. OK, let us start with a square, one like
1105² = 5².13².17² = (2+i)²(2-i)² . (3+2i)²(3-2i)² . (4+i)²(4-i)²
and now separate the factors in "two parts", so that in one part we have some factors, and in the other part the conjugates. Here are the possibilities for 25 = 5²:
(2+i)² and (2-i)²
5 and 5
(2-i)² and (2+i)²
There are three possibilities. Do the same for the other two squares, then combine. For instance:
sage: (2 + i)^2 * (3 - 2*i)^2 * 17
-272*I + 1071
And indeed, 272² + 1071² = 1105² . This solution is not "primitive", in the sense that 17 is a divisor of the three involved numbers, 272, 1071, 1105. Well, this happens because we took the factor 17 from the separation of 17² in two (equal) parts. To get some other solutions, we may take
each possible first part from 5² with...
each possible first part from 13² with...
each possible first part from 17²
and thus get "many solutions". Here are they:
sage: [ (m, n) for m in range(1, 1105) for n in range(1, 1105)
....: if m <= n and m2 + n2 == 1105**2 ]
[(47, 1104),
(105, 1100),
(169, 1092),
(264, 1073),
(272, 1071),
(425, 1020),
(468, 1001),
(520, 975),
(561, 952),
(576, 943),
(663, 884),
(700, 855),
(744, 817)]
We expect 3.3.3 solutions. One of them is the trivial one, 1105² = 1105² + 0².
The other solutions of 1105² = a² + b² may be arranged to have a < b. (No chance to get equality.) So we expect (27 - 1)/2 = 13 solutions, yes, the ones above.
Which solution is produced by taking the "first parts" as follows: (2 + i)^2 * (3 - 2*i)^2 * (4 + i)^2 ?!
sage: (2 + i)^2 * (3 - 2*i)^2 * (4 + i)^2
264*I + 1073
And indeed, (264, 1073) is among the solutions above.
So if getting "highly composite" numbers is the issue, with an accent on highly, then just pick for c such a product of primes of the shape 4k + 1.
For instance c = 5³.13.17 or c = 5.13.17.29. Then compute all representations c² = (a + ib)(a - ib) = a² + b² best by using the UFD property of the Gaussian integers.
For instance, in a python3 dialog with the interpreter...
In [16]: L25 = [complex(2, 1)**4, complex(2, 1)**2 * 5, 25, complex(2, -1)**2 * 5, complex(2, -1)**4]
In [17]: L13 = [complex(3, 2)**2, 13, complex(3, -2)**2]
In [18]: L17 = [complex(4, 1)**2, 17, complex(4, -1)**2]
In [19]: solutions = []
In [20]: for z1 in L25:
...: for z2 in L13:
...: for z3 in L17:
...: z = z1 * z2 * z3
...: a, b = int(abs(z.real)), int(abs(z.imag))
...: if a > b:
...: a, b = b, a
...: solutions.append((a, b))
...:
In [21]: solutions = list(set(solutions))
In [22]: solutions.sort()
In [23]: len(solutions)
Out[23]: 23
In [24]: solutions
Out[24]:
[(0, 5525),
(235, 5520),
(525, 5500),
(612, 5491),
(845, 5460),
(1036, 5427),
(1131, 5408),
(1320, 5365),
(1360, 5355),
(1547, 5304),
(2044, 5133),
(2125, 5100),
(2163, 5084),
(2340, 5005),
(2600, 4875),
(2805, 4760),
(2880, 4715),
(3124, 4557),
(3315, 4420),
(3468, 4301),
(3500, 4275),
(3720, 4085),
(3861, 3952)]
We have 23 = 22 + 1 solutions. The last one is the trivial one. All other solutions (a, b) listed have a < b, so there are totally 1 + 22*2 = 45 = 5 * 3 * 3, as expected from the triple for loop above. A similar code can be written for c = 5 * 13 * 17 * 29 = 32045 leading to (3^4 - 1)/2 = 40 non-trivial solutions.
In [26]: L5 = [complex(2, 1)**2, 5, complex(2, -1)**2]
In [27]: L13 = [complex(3, 2)**2, 13, complex(3, -2)**2]
In [28]: L17 = [complex(4, 1)**2, 17, complex(4, -1)**2]
In [29]: L29 = [complex(5, 2)**2, 29, complex(5, -2)**2]
In [30]: z_list = [z1*z2*z3*z4
...: for z1 in L5 for z2 in L13
...: for z3 in L17 for z4 in L29]
In [31]: ab_list = [(int(abs(z.real)), int(abs(z.imag))) for z in z_list]
In [32]: len(ab_list)
Out[32]: 81
In [33]: ab_list = list(set([(min(a, b), max(a, b)) for (a, b) in ab_list]))
In [34]: ab_list.sort()
In [35]: len(ab_list)
Out[35]: 41
In [36]: ab_list[:10]
Out[36]:
[(0, 32045),
(716, 32037),
(1363, 32016),
(2277, 31964),
(2400, 31955),
(3045, 31900),
(3757, 31824),
(3955, 31800),
(4901, 31668),
(5304, 31603)]
(Feel free to also use powers of two in c.)
#There is a general formula for pythagoran triples
take 2 numbers, m & n where m > n
a = (m^2) - (n^2)
b = 2mn
c = (m^2) + (n^2)
That will always give you a pythagoran triple. Its more efficient but it might not be what you're looking for.
This is for an assignment I'm doing through school. I am having trouble generating a private key. My main problem is understanding the relation of my equations to each other. To set everything up, we have:
p = 61
q = 53
n = p * q (which equals 3233)
From here we have the totient of n (phi(n)) which equals 3120, now we can choose prime e; where 1 < e < 3120
e = 17
Okay easy enough.
For my assignment we've been made aware that d = 2753, however I still need to be able to arbitrarily generate this value.
Now here is where I am having trouble. I've been perusing wikipedia to understand and somewhere something isn't connecting. I know that I need to find the modular multiplicative inverse of e (mod phi(n)) which will be d, our private exponent.
Reading though wikipedia tells us to find the mmi we need to use the Extended Euclidian Algorithm. I've implemented the algorithm in python as follows:
def egcd(a, b):
x, lastX = 0, 1
y, lastY = 1, 0
while (b != 0):
q = a // b
a, b = b, a % b
x, lastX = lastX - q * x, x
y, lastY = lastY - q * y, y
return (lastX, lastY)
This is where I am lost. To my understanding now, the equation ax + bx = gcd(a, b) = 1 is the same e*x + phi(n)*y = gcd(e, phi(n)) = 1.
So we call egcd(e, phi(n)), and now I get [-367, 2] for my x and y.
From here I honestly don't know where to go. I've read this similar question and I see that there are some substitutions that happen, but I don't understand how those number relate to the answer that I got or the values I have started out with. Can someone explain to me pragmatically what I need to do from here? (When I say pragmatically, I mean without actual code. Pseudo code is fine, but if I get actual code I won't be able to learn without plagiarism on my assignment which is a big no-no).
As always, any help is genuinely appreciated. Thanks everyone!
(And yes, I have seen these:RSA: Private key calculation with Extended Euclidean Algorithm and In RSA encryption, how do I find d, given p, q, e and c?)
The implementation of the Extended Euclidean algorithm you have is not complete, since it is generating a negative number for the private key. Use this code instead:
https://en.wikibooks.org/wiki/Algorithm_Implementation/Mathematics/Extended_Euclidean_algorithm
For your example the private key, d, is 2753.
p=61
q=53
n = 3233
phi(n)=3120
e=17
d=modinv(17,3120)=2753
Try it out:
message m m=65
encryption: m^e mod n = c (65**17) % 3120 = 65
decryption: c^d mod n = m (65**2753) % 3120 = 65
Its all explained here:
http://southernpacificreview.com/2014/01/06/rsa-key-generation-example/
def egcd(a,b):
s1, s2 = 1, 0
t1, t2 = 0, 1
while b!=0:
q = a//b
r = a%b
a, b = b, r
s = s1-q*s2
s1, s2 = s2, s
t = t1-q*t2
t1, t2 = t2, t
return (s1, t1)
try comparing above.
i will tell you where was your mistake:
a, b = b, a % b
a has the value of b now
(b=a%b)==(b=b%b)
and similar reason for proceeding two lines
I have got two strings in Python,
A m * B s / (A m + C m)
and
C m * B s / (C m + A m)
that are both equivalent functions of the unordered set (A, C) and the unordered set (B). m and s indicate units that can be swapped among the same but not with another unit.
So far, I'm doing permutations of A, B, and C and testing them using eval and SymPy's == operator. This has multiple drawbacks:
for more complicated expressions, I have to generate a large number of permutations (in my case 8 nested for loops)
I need to define A, B, C as symbols, which is not optimal when I don't know which parameters I will have (so I have to generate all of them -> terribly inefficient and messing up my variable namespace)
Is there a pythonian way to test for this kind of equivalence? It should work an arbitrary expressions.
Here is a simplified approach based on my previous answer.
The idea is that if two expressions are equivalent under permutations, the permutation carrying one to the other must map the ith symbol in the first string (ordered by index of first occurrence) to the ith symbol in the second string (again ordered by index of first occurrence). This principle can be used to construct a permutation, apply it to the first string and then check for equality with the second string - if they are equal they are equivalent, otherwise they are not.
Here is one possible implementation:
import re
# Unique-ify list, preserving order
def uniquify(l):
return reduce(lambda s, e: s + ([] if e in s else [e]), l, [])
# Replace all keys in replacements with corresponding values in str
def replace_all(str, replacements):
for old, new in replacements.iteritems():
str = str.replace(old, new)
return str
class Expression:
units = ["m", "s"]
def __init__(self, exp):
self.exp = exp
# Returns a list of symbols in the expression that are preceded
# by the given unit, ordered by first appearance. Assumes the
# symbol and unit are separated by a space. For example:
# Expression("A m * B s / (A m + C m)").symbols_for_unit("m")
# returns ['A', 'C']
def symbols_for_unit(self, unit):
sym_re = re.compile("(.) %s" % unit)
symbols = sym_re.findall(self.exp)
return uniquify(symbols)
# Returns a string with all symbols that have units other than
# unit "muted", that is replaced with the empty string. Example:
# Expression("A m * B s / (A m + C m)").mute_symbols_for_other_units("m")
# returns "A m * s / (A m + C m)"
def mute_symbols_for_other_units(self, unit):
other_units = "".join(set(self.units) - set(unit))
return re.sub("(.) ([%s])" % "".join(other_units), " \g<2>", self.exp)
# Returns a string with all symbols that have the given unit
# replaced with tokens of the form $0, $1, ..., by order of their
# first appearance in the string, and all other symbols muted.
# For example:
# Expression("A m * B s / (A m + C m)").canonical_form("m")
# returns "$0 m * s / ($0 m + $1 m)"
def canonical_form(self, unit):
symbols = self.symbols_for_unit(unit)
muted_self = self.mute_symbols_for_other_units(unit)
for i, sym in enumerate(symbols):
muted_self = muted_self.replace("%s %s" % (sym, unit), "$%s %s" % (i, unit))
return muted_self
# Define a permutation, represented as a dictionary, according to
# the following rule: replace $i with the ith distinct symbol
# occurring in the expression with the given unit. For example:
# Expression("C m * B s / (C m + A m)").permutation("m")
# returns {'$0':'C', '$1':'A'}
def permutation(self, unit):
enum = enumerate(self.symbols_for_unit(unit))
return dict(("$%s" % i, sym) for i, sym in enum)
# Return a string produced from the expression by first converting it
# into canonical form, and then performing the replacements defined
# by the given permutation. For example:
# Expression("A m * B s / (A m + C m)").permute("m", {"$0":"C", "$1":"A"})
# returns "C m * s / (C m + A m)"
def permute(self, unit, permutation):
new_exp = self.canonical_form(unit)
return replace_all(new_exp, permutation)
# Test for equality under permutation and muting of all other symbols
# than the unit provided.
def eq_under_permutation(self, unit, other_exp):
muted_self = self.mute_symbols_for_other_units(unit)
other_permuted_str = other_exp.permute(unit, self.permutation(unit))
return muted_self == other_permuted_str
# Test for equality under permutation. This is done for each of
# the possible units using eq_under_permutation
def __eq__(self, other):
return all([self.eq_under_permutation(unit, other) for unit in self.units])
e1 = Expression("A m * B s / (A m + C m)")
e2 = Expression("C m * B s / (C m + A m)")
e3 = Expression("A s * B s / (A m + C m)")
f1 = Expression("A s * (B s + D s) / (A m + C m)")
f2 = Expression("A s * (D s + B s) / (C m + A m)")
f3 = Expression("D s")
print "e1 == e2: ", e1 == e2 # True
print "e1 == e3: ", e1 == e3 # False
print "e2 == e3: ", e2 == e3 # False
print "f1 == f2: ", f1 == f2 # True
print "f1 == f3: ", f1 == f3 # False
As you pointed out, this checks for string equivalence under permutations without any regard to mathematical equivalence, but it is half the battle. If you had a canonical form for mathematical expressions, you could use this approach on two expressions in canonical form. Perhaps one of sympy's Simplify could do the trick.
Instead of iterating over all possible permutations, assume one exists and attempt to construct it. I believe that done in the right way, failure of the algorithm would imply inexistence of the permutation.
Here is the outline of the idea applied to the expressions above:
let:
str1 = "A m * B s / (A m + C m)"
str2 = "C m * B s / (C m + A m)"
We're looking for a permutation of the set (A, C) that would render the expressions identical. Relabel A and C as X1 and X2 according to the order of their first appearance in str2, so:
X1 = C
X2 = A
because C appears before A in str2. Next, create the array Y such that y[i] is the ith symbol A or C in order of first appearance in str1. So:
Y[1] = A
Y[2] = C
Because A appears before C in str1.
Now construct str3 from str2 by replacing A and C with X1 and X2:
str3 = "X1 m * B s / (X1 m + X2 m)"
And then start substituting Xi for Y[i]. First, X1 becomes Y[1]=A:
str3_1 = "A m * Bs / (A m + X2 m)"
At this stage, compare str3_1 and str1 up to the first occurrence of any of the Xi's, in this case X2, so because these two strings are equal:
str3_1[:18] = "A m * B s / (A m + "
str1[:18] = "A m * B s / (A m + "
You have a chance of constructing the permutation. If they were unequal, you'd have proven no suitable permutation exists (because any permutation would have had to make at least that substitution) and could deduce inequivalence. But they are equal, so you proceed to the next step, substituting X2 for Y[2]=C:
str3_2 = "A m * B s / (A m + C m)"
And this is equal to str1, so you have your permutation (A->C, C->A) and have shown the equivalence of the expressions.
This is only a demonstration of the algorithm to a particular case, but it should generalize. Not sure what the lowest order you could get it down to is, but it should be quicker than the n! of generating all permutations on n variables.
If I understand the significance of the units correctly, they limit which variables may be swapped for which others by the permutations. So that A can be substituted with C in the above expressions because both have 'm' units, but not with B which has 's' units. You can handle this in the following way:
construct expressions str1_m and str2_m from str1 and str2 by removing all symbols that don't have m units, and then carry out the above algorithm for str1_m and str2_m. If construction fails, no permutation exists. If construction succeeds, keep that permutation (call it the m-permutation) and construct str1_s and str2_s from str1 and str2 by removing all symbols that don't have s units, then carry out the algorithm again for str1_s and str2_s. If construction fails, they are not equivalent. If it succeeds, the final permutation will be a combination of the m-permutation and the s-permutation (although you probably don't even need to construct it, you just care that it exists).
If you pass a string to SymPy's sympify() function, it will automatically create the Symbols for you (no need to define them all).
>>> from sympy import *
>>> x
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'x' is not defined
>>> sympify("x**2 + cos(x)")
x**2 + cos(x)
>>> sympify("diff(x**2 + cos(x), x)")
2*x - sin(x)
I did it once, in one simulater of mathemathics estudies..
Well, in my case, i knew what were the variables that will be used.
So, i tested the result putting values inside the vars.
A = 10
B = 20
C = 30
m = Math.e
s = Math.pi
And so, we solve:
s1 = 'A m * B s / (A m + C m)'
s2 = 'C m * B s / (C m + A m)'
If s1 != s2, was proved there isn't equivalence
With this method is impossible say that two expressions are equivalent,
But you can say that both isn't equivalent
if s1 != s2:
print "Not equivalent"
else:
print "Try with another sample"
Well.. I hope that this can help you.
This, like all other answers to date is not a robust solution to the problem, but instead contains more helpful information for our future meticulous friend to solve it.
I provide a difficult example using Euler's Formula https://en.wikipedia.org/wiki/Euler%27s_formula
I am certain all other overflow answers to date will not succeed in my example.
I show that all the suggestions on sympy's website also fail on my example. (https://github.com/sympy/sympy/wiki/Faq)
#SOURCE FOR HELPERS: https://github.com/sympy/sympy/wiki/Faq
import sympy
import sympy.parsing.sympy_parser
ExampleExpressionString1 = 'exp( i*( (x0 - 1)*(x0 + 2) ) )'
ExampleExpressionSympy1 = sympy.parsing.sympy_parser.parse_expr(ExampleExpressionString1)
ExampleExpressionString2 = 'i*sin( (x0 - 1)*(x0 + 2) ) + cos( (x0 - 1)*(x0 + 2) )'
ExampleExpressionSympy2 = sympy.parsing.sympy_parser.parse_expr(ExampleExpressionString2)
print '(ExampleExpressionSympy1 == ExampleExpressionSympy2):'
print ' ', (ExampleExpressionSympy1 == ExampleExpressionSympy2)
print '(ExampleExpressionSympy1.simplify() == ExampleExpressionSympy2.simplify()):'
print ' ', (ExampleExpressionSympy1.simplify() == ExampleExpressionSympy2.simplify())
print '(ExampleExpressionSympy1.expand() == ExampleExpressionSympy2.expand()):'
print ' ', (ExampleExpressionSympy1.trigsimp() == ExampleExpressionSympy2.trigsimp())
print '(ExampleExpressionSympy1.trigsimp() == ExampleExpressionSympy2.trigsimp()):'
print ' ', (ExampleExpressionSympy1.trigsimp() == ExampleExpressionSympy2.trigsimp())
print '(ExampleExpressionSympy1.simplify().expand().trigsimp() == ExampleExpressionSympy2.simplify().expand().trigsimp()):'
print ' ', (ExampleExpressionSympy1.simplify().expand().trigsimp() == ExampleExpressionSympy2.simplify().expand().trigsimp())
MORE NOTES:
I suspect this is a difficult problem to solve generically, and robustly. To properly check mathematical equivalence, you not only have to try order permutations, but you also have to have a library of mathematical equivalent transformations and try all those permutations as well.
I do however believe this might be a solvable problem, because Wolfram Alpha seems to have 'alternate expression' section, which seems to do the trick of providing all permutations most of the time on arbitrary expressions using these kinds of equivalences.
IN SUMMATION:
I suggest the following with the expectation that it will break:
import sympy
import sympy.parsing.sympy_parser
Expression.simplify().expand().trigsimp()