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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.
I'm trying to understand how complex numbers get multiplied. when I multiply two numbers it always seems to give me an arbitrary amount. for example,
complex(10,9)*complex(11,13) equals complex(-7,229)and complex(10,1)*complex(10,2) equals complex(98,30). is there a mathematical way to figure out 2 complex numbers multiplied and if so, what is is it?
The result is not arbitrary, it is following the definition of complex multiplication:
For example if you have
x = a + j * b
y = c + j * d
then the expression for multiplication is
x * y = (a * c - b * d) + j (a * d + b * c)
For your example complex(10,9) * complex(11,13) that would evaluate to
(10 * 11 - 9 * 13) + j * (10 * 13 + 9 * 11)
(-7 + 229j)
which is exactly what Python shows
>>> complex(10,9) * complex(11,13)
(-7+229j)
Complex number multiplication operates in this way:
(a + ib) * (c + id) = a * c + a * id + ib * c + ib * id
= a * c - b * d + i(a * d + b * c)
So, in Python, the result is like this:
complex(a, b) * complex(c, d) = complex(a * c - b * d, a * d + b * c)
Example:
complex(10, 9) * complex(11, 13) = complex(10 * 11 - 9 * 13, 10 * 13 + 9 * 11)
= complex(-7, 227)
If you have 2 complex numbers, the first a + bj, and the second c + dj, then the product (a + bj) * (c + dj) = a*c - b*d + (a*d + b*c)j. The way to think about it is that j is equal to the square root of -1, so j*j = -1, and then just multiply out the brackets as normal. See below:
a, b = 10, 9
c, d = 11, 13
print(complex(a, b)*complex(c, d))
print(a*c - b*d, a*d + b*c)
Output:
(-7+229j)
-7 229
Another way to understand complex number multiplication is geometrically. We can think of complex numbers as two dimensional vectors, things with a length and direction. Then when multiplying a complex number with length r and direction a by another with length s and direction b, you get a complex number with length r*s and direction a+b, i.e. a complex number with length r and direction a acts on others by multiplication by scaling them by r and rotating them through a.
If you work out the lengths and directions of your (10,9) and (11,13) and combine them as above, you will get the length and direction of (-7,229)
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I'm a beginner in programming and I'm looking for a nice idea how to generate three integers that satisfy a condition.
Example:
We are given n = 30, and we've been asked to generate three integers a, b and c, so that 7*a + 5*b + 3*c = n.
I tried to use for loops, but it takes too much time and I have a maximum testing time of 1000 ms.
I'm using Python 3.
My attempt:
x = int(input())
c = []
k = []
w = []
for i in range(x):
for j in range(x):
for h in range(x):
if 7*i + 5*j + 3*h = x:
c.append(i)
k.append(j)
w.append(h)
if len(c) == len(k) == len(w)
print(-1)
else:
print(str(k[0]) + ' ' + str(c[0]) + ' ' + str(w[0]))
First, let me note that your task is underspecified in at least two respects:
The allowed range of the generated values is not specified. In particular, you don't specify whether the results may include negative integers.
The desired distribution of the generated values is not specified.
Normally, if not specified, one might assume that a uniform distribution on the set of possible solutions to the equation was expected (since it is, in a certain sense, the most random possible distribution on a given set). But a (discrete) uniform distribution is only possible if the solution set is finite, which it won't be if the range of results is unrestricted. (In particular, if (a, b, c) is a solution, then so is (a, b + 3k, c − 5k) for any integer k.) So if we interpret the task as asking for a uniform distribution with unlimited range, it's actually impossible!
On the other hand, if we're allowed to choose any distribution and range, the task becomes trivial: just make the generator always return a = −n, b = n, c = n. Clearly this is a solution to the equation (since −7n + 5n + 3n = (−7 + 5 + 3)n = 1n), and a degenerate distribution that assigns all probability mass to single point is still a valid probability distribution!
If you wanted a slightly less degenerate solution, you could pick a random integer k (using any distribution of your choice) and return a = −n, b = n + 3k, c = n − 5k. As noted above, this is also a solution to the equation for any k. Of course, this distribution is still somewhat degenerate, since the value of a is fixed.
If you want to let all return values be at least somewhat random, you could also pick a random h and return a = −n + h, b = n − 2h + 3k and c = n + h − 5k. Again, this is guaranteed to be a valid solution for any h and k, since it clearly satisfies the equation for h = k = 0, and it's also easy to see that increasing or decreasing either h or k will leave the value of the left-hand side of the equation unchanged.
In fact, it can be proved that this method can generate all possible solutions to the equation, and that each solution will correspond to a unique (h, k) pair! (One fairly intuitive way to see this is to plot the solutions in 3D space and observe that they form a regular lattice of points on a 2D plane, and that the vectors (+1, −2, +1) and (0, +3, −5) span this lattice.) If we pick h and k from some distribution that (at least in theory) assigns a non-zero probability to every integer, then we'll have a non-zero probability of returning any valid solution. So, at least for one somewhat reasonable interpretation of the task (unbounded range, any distribution with full support) the following code should solve the task efficiently:
from random import gauss
def random_solution(n):
h = int(gauss(0, 1000)) # any distribution with full support on the integers will do
k = int(gauss(0, 1000))
return (-n + h, n - 2*h + 3*k, n + h - 5*k)
If the range of possible values is restricted, the problem becomes a bit trickier. On the positive side, if all values are bounded below (or above), then the set of possible solutions is finite, and so a uniform distribution exists on it. On the flip side, efficiently sampling this uniform distribution is not trivial.
One possible approach, which you've used yourself, is to first generate all possible solutions (assuming there's a finite number of them) and then sample from the list of solutions. We can do the solution generation fairly efficiently like this:
find all possible values of a for which the equation might have a solution,
for each such a, find all possible values of b for which there still have a solution,
for each such (a, b) pair, solve the equation for c and check if it's valid (i.e. an integer within the specified range), and
if yes, add (a, b, c) to the set of solutions.
The tricky part is step 2, where we want to calculate the range of possible b values. For this, we can make use of the observation that, for a given a, setting c to its smallest allowed value and solving the equation gives an upper bound for b (and vice versa).
In particular, solving the equation for a, b and c respectively, we get:
a = (n − 5b − 3c) / 7
b = (n − 7a − 3c) / 5
c = (n − 7a − 5b) / 3
Given lower bounds on some of the values, we can use these solutions to compute corresponding upper bounds on the others. For example, the following code will generate all non-negative solutions efficiently (and can be easily modified to use a lower bound other than 0, if needed):
def all_nonnegative_solutions(n):
a_min = b_min = c_min = 0
a_max = (n - 5*b_min - 3*c_min) // 7
for a in range(a_min, a_max + 1):
b_max = (n - 7*a - 3*c_min) // 5
for b in range(b_min, b_max + 1):
if (n - 7*a - 5*b) % 3 == 0:
c = (n - 7*a - 5*b) // 3
yield (a, b, c)
We can then store the solutions in a list or a tuple and sample from that list:
from random import choice
solutions = tuple(all_nonnegative_solutions(30))
a, b, c = choice(solutions)
Ps. Apparently Python's random.choice is not smart enough to use reservoir sampling to sample from an arbitrary iterable, so we do need to store the full list of solutions even if we only want to sample from it once. Or, of course, we could always implement our own sampler:
def reservoir_choice(iterable):
r = None
n = 0
for x in iterable:
n += 1
if randrange(n) == 0:
r = x
return r
a, b, c = reservoir_choice(all_nonnegative_solutions(30))
BTW, we could make the all_nonnegative_solutions function above a bit more efficient by observing that the (n - 7*a - 5*b) % 3 == 0 condition (which checks whether c = (n − 7a − 5b) / 3 is an integer, and thus a valid solution) is true for every third value of b. Thus, if we first calculated the smallest value of b that satisfies the condition for a given a (which can be done with a bit of modular arithmetic), we could iterate over b with a step size of 3 starting from that minimum value and skip the divisibility check entirely. I'll leave implementing that optimization as an exercise.
import numpy as np
def generate_answer(n: int, low_limit:int, high_limit: int):
while True:
a = np.random.randint(low_limit, high_limit + 1, 1)[0]
b = np.random.randint(low_limit, high_limit + 1, 1)[0]
c = (n - 7 * a - 5 * b) / 3.0
if int(c) == c and low_limit <= c <= high_limit:
break
return a, b, int(c)
if __name__ == "__main__":
n = 30
ans = generate_answer(low_limit=-5, high_limit=50, n=n)
assert ans[0] * 7 + ans[1] * 5 + ans[2] * 3 == n
print(ans)
If you select two of the numbers a, b, c, you know the third. In this case, I randomize ints for a, b, and I find c by c = (n - 7 * a - 5 * b) / 3.0.
Make sure c is an integer, and in the allowed limits, and we are done.
If it is not, randomize again.
If you want to generate all possibilities,
def generate_all_answers(n: int, low_limit:int, high_limit: int):
results = []
for a in range(low_limit, high_limit + 1):
for b in range(low_limit, high_limit + 1):
c = (n - 7 * a - 5 * b) / 3.0
if int(c) == c and low_limit <= c <= high_limit:
results.append((a, b, int(c)))
return results
If third-party libraries are allowed, you can use SymPy's diophantine.diop_linear linear Diophantine equations solver:
from sympy.solvers.diophantine.diophantine import diop_linear
from sympy import symbols
from numpy.random import randint
n = 30
N = 8 # Number of solutions needed
# Unknowns
a, b, c = symbols('a, b, c', integer=True)
# Coefficients
x, y, z = 7, 5, 3
# Parameters of parametric equation of solution
t_0, t_1 = symbols('t_0, t_1', integer=True)
solution = diop_linear(x * a + y * b + z * c - n)
if not (None in solution):
for s in range(N):
# -10000 and 10000 (max and min for t_0 and t_1)
t_sub = [(t_0, randint(-10000, 10000)), (t_1, randint(-10000, 10000))]
a_val, b_val, c_val = map(lambda t : t.subs(t_sub), solution)
print('Solution #%d' % (s + 1))
print('a =', a_val, ', b =', b_val, ', c =', c_val)
else:
print('no solutions')
Output (random):
Solution #1
a = -141 , b = -29187 , c = 48984
Solution #2
a = -8532 , b = -68757 , c = 134513
Solution #3
a = 5034 , b = 30729 , c = -62951
Solution #4
a = 7107 , b = 76638 , c = -144303
Solution #5
a = 4587 , b = 23721 , c = -50228
Solution #6
a = -9294 , b = -106269 , c = 198811
Solution #7
a = -1572 , b = -43224 , c = 75718
Solution #8
a = 4956 , b = 68097 , c = -125049
Why your solution can't cope with large values of n
You may understand that everything in a for loop with a range of i, will run i times. So it will multiply the time taken by i.
For example, let's pretend (to keep things simple) that this runs in 4 milliseconds:
if 7*a + 5*b + 3*c = n:
c.append(a)
k.append(b)
w.append(c)
then this will run in 4×n milliseconds:
for c in range(n):
if 7*a + 5*b + 3*c = n:
c.append(a)
k.append(b)
w.append(c)
Approximately:
n = 100 would take 0.4 seconds
n = 250 would take 1 second
n = 15000 would take 60 seconds
If you put that inside a for loop over a range of n then the whole thing will be repeated n times. I.e.
for b in range(n):
for c in range(n):
if 7*a + 5*b + 3*c = n:
c.append(a)
k.append(b)
w.append(c)
will take 4n² milliseconds.
n = 30 would take 4 seconds
n = 50 would take 10 seconds
n = 120 would take 60 seconds
Putting it in a third for-loop will take 4n³ milliseconds.
n = 10 would take 4 seconds
n = 14 would take 10 seconds.
n = 24 would take 60 seconds.
Now, what if you halved the original if to 2 milliseconds? n would be able to increase by 15000 in the first case... and 23 in the last case. The lesson here is that fewer for-loops is usually much more important than speeding up what's inside them. As you can see in Gulzar's answer part 2, there are only two for loops which makes a big difference. (This only applies if the loops are inside each other; if they are just one after another you don't have the multiplication problem.)
from my perspective, the last number of the three is never a random number. let say you generate a and b first then c is never a random because it should be calculated from the equation
n = 7*a + 5*b + 3*c
c = (7*a + 5*b - n) / -3
this means that we need to generate two random values (a,b)
that 7*a + 5*b - n is divisible by 3
import random
n = 30;
max = 1000000;
min = -1000000;
while True:
a = random.randint(min , max);
b = random.randint(min , max);
t = (7*a) + (5*b) - n;
if (t % 3 == 0) :
break;
c = (t/-3);
print("A = " + str(a));
print("B = " + str(b));
print("C = " + str(c));
print("7A + 5B + 3C =>")
print("(7 * " + str(a) + ") + (5 * " + str(b) + ") + (3 * " + str(c) + ") = ")
print((7*a) + (5*b) + (3*c));
REPL
I have a case where a variable (a, in this case) is calculated at each loop iteration and stops where the increment of value between two iterations is small enough.
I would like to know of a general way to find the value for that variable in this kind of case, without having to do that "convergence" work using loops.
There I would like to know if the solution is to put everything in equations, or if some tools exist to tackle that.
a = 10
b = 10
diff = 1
while diff > .1:
old_a = a
a += b
diff = 1 - (old_a/a)
print(diff)
The present code produces:
0.5
0.33333333333333337
0.25
0.19999999999999996
0.16666666666666663
0.1428571428571429
0.125
0.11111111111111116
0.09999999999999998
Therefore, it takes 9 iterations to find a relative difference of the value of a between two iterations inferior to 10%.
You have
a_n = a_0 + n * b
and try to find where
1 - (a_(n-1) / a_n)
= 1 - (a_0 + (n--1)*b) / ( a_0 + n * b)
= 1 - (a_0 + n*b -b) / (a_0 + n*b)
= 1 - 1 + b / (a_0 + n*b)
= b / (a_0 + n * b)
< 0.1
That is the same as
(a_0 / b) + n * b / b
= (a_0 / b) + n
> 10
(because 0.1 = 1 / 10 and 1/x > 1/y <=> y > x if x,y != 0)
Since you metion in the comments that your actual problem is more complex: If finding a closed form solution like above is not feasible, look at this wikipedia page about fixed point iteration, which is exactly the kind of problem you try to solve.
I am looking to compute a^b mod m where a & b are floating point numbers and m is an non-negative integer. The trivial solution is to do b multiplications which takes O(n) time, however my numbers a & b can be largish (~10 digits before the decimal point) and I would like to do this efficiently. When a,b and m are integers we can compute the modpow quickly in log(n) time via: Exponentiation_by_squaring.
How would I use this method (or another) for floating point numbers? I am using Python to do this computation and the pow function only allows integers. Here is my attempt at doing exponentiation by squaring with Decimal numbers, but the answer is not coming out right:
from decimal import Decimal
EPS = Decimal("0.0001")
# a, b are Decimals and m is an integer
def deci_pow(a, b, m):
if abs(b) < EPS:
return Decimal(1)
tmp = deci_pow(a, b / 2, m) % m # Should this be // ?
if abs(b % 2) < EPS:
return (tmp * tmp) % m
else:
if b > 0:
return (a * tmp * tmp) % m
else:
return ((tmp * tmp)/a) % m
print(deci_pow(Decimal(2.4), Decimal(3.5), 5)) # != 1.416
When a,b,m are all integers this is what the method looks like:
# a, b, m are Integers
def integer_pow(a, b, m):
if b == 0: return 1
tmp = integer_pow(a, b // 2, m) % m
if b % 2 == 0:
return (tmp * tmp) % m
else:
if b > 0:
return (a * tmp * tmp) % m
else:
return ((tmp * tmp) / a) % m
I don't think there is an easy way to do this in general, if a and b can be 10 digits (assuming before the decimal point). The problem is that for floats x and y, you don't necessarily have the property
((x % m) * (y % m)) % m == (x * y) % m
If you tell us your specific context and why you want to do this, there might be other approaches possible.