问题
Lets N be a number (10<=N<=10^5).
I have to break it into 3 numbers (x,y,z) such that it validates the following conditions.
1. x<=y<=z
2. x^2+y^2=z^2-1;
3. x+y+z<=N
I have to find how many combinations I can get from the given numbers in a method.
I have tried as follows but it's taking so much time for a higher number and resulting in a timeout..
int N= Int32.Parse(Console.ReadLine());
List<String> res = new List<string>();
//x<=y<=z
int mxSqrt = N - 2;
int a = 0, b = 0;
for (int z = 1; z <= mxSqrt; z++)
{
a = z * z;
for (int y = 1; y <= z; y++)
{
b = y * y;
for (int x = 1; x <= y; x++)
{
int x1 = b + x * x;
int y1 = a - 1;
if (x1 == y1 && ((x + y + z) <= N))
{
res.Add(x + "," + y + "," + z);
}
}
}
}
Console.WriteLine(res.Count());
My question:
My solution is taking time for a bigger number (I think it's the for loops), how can I improve it?
Is there any better approach for the same?
回答1:
Here's a method that enumerates the triples, rather than exhaustively testing for them, using number theory as described here: https://mathoverflow.net/questions/29644/enumerating-ways-to-decompose-an-integer-into-the-sum-of-two-squares
Since the math took me a while to comprehend and a while to implement (gathering some code that's credited above it), and since I don't feel much of an authority on the subject, I'll leave it for the reader to research. This is based on expressing numbers as Gaussian integer conjugates. (a + bi)*(a - bi) = a^2 + b^2. We first factor the number, z^2 - 1, into primes, decompose the primes into Gaussian conjugates and find different expressions that we expand and simplify to get a + bi, which can be then raised, a^2 + b^2.
A perk of reading about the Sum of Squares Function is discovering that we can rule out any candidate z^2 - 1 that contains a prime of form 4k + 3 with an odd power. Using that check alone, I was able to reduce Prune's loop on 10^5 from 214 seconds to 19 seconds (on repl.it) using the Rosetta prime factoring code below.
The implementation here is just a demonstration. It does not have handling or optimisation for limiting x and y. Rather, it just enumerates as it goes. Play with it here.
Python code:
# https://math.stackexchange.com/questions/5877/efficiently-finding-two-squares-which-sum-to-a-prime
def mods(a, n):
if n <= 0:
return "negative modulus"
a = a % n
if (2 * a > n):
a -= n
return a
def powmods(a, r, n):
out = 1
while r > 0:
if (r % 2) == 1:
r -= 1
out = mods(out * a, n)
r /= 2
a = mods(a * a, n)
return out
def quos(a, n):
if n <= 0:
return "negative modulus"
return (a - mods(a, n))/n
def grem(w, z):
# remainder in Gaussian integers when dividing w by z
(w0, w1) = w
(z0, z1) = z
n = z0 * z0 + z1 * z1
if n == 0:
return "division by zero"
u0 = quos(w0 * z0 + w1 * z1, n)
u1 = quos(w1 * z0 - w0 * z1, n)
return(w0 - z0 * u0 + z1 * u1,
w1 - z0 * u1 - z1 * u0)
def ggcd(w, z):
while z != (0,0):
w, z = z, grem(w, z)
return w
def root4(p):
# 4th root of 1 modulo p
if p <= 1:
return "too small"
if (p % 4) != 1:
return "not congruent to 1"
k = p/4
j = 2
while True:
a = powmods(j, k, p)
b = mods(a * a, p)
if b == -1:
return a
if b != 1:
return "not prime"
j += 1
def sq2(p):
if p % 4 != 1:
return "not congruent to 1 modulo 4"
a = root4(p)
return ggcd((p,0),(a,1))
# https://rosettacode.org/wiki/Prime_decomposition#Python:_Using_floating_point
from math import floor, sqrt
def fac(n):
step = lambda x: 1 + (x<<2) - ((x>>1)<<1)
maxq = long(floor(sqrt(n)))
d = 1
q = n % 2 == 0 and 2 or 3
while q <= maxq and n % q != 0:
q = step(d)
d += 1
return q <= maxq and [q] + fac(n//q) or [n]
# My code...
# An answer for https://stackoverflow.com/questions/54110614/
from collections import Counter
from itertools import product
from sympy import I, expand, Add
def valid(ps):
for (p, e) in ps.items():
if (p % 4 == 3) and (e & 1):
return False
return True
def get_sq2(p, e):
if p == 2:
if e & 1:
return [2**(e / 2), 2**(e / 2)]
else:
return [2**(e / 2), 0]
elif p % 4 == 3:
return [p, 0]
else:
a,b = sq2(p)
return [abs(a), abs(b)]
def get_terms(cs, e):
if e == 1:
return [Add(cs[0], cs[1] * I)]
res = [Add(cs[0], cs[1] * I)**e]
for t in xrange(1, e / 2 + 1):
res.append(
Add(cs[0] + cs[1]*I)**(e-t) * Add(cs[0] - cs[1]*I)**t)
return res
def get_lists(ps):
items = ps.items()
lists = []
for (p, e) in items:
if p == 2:
a,b = get_sq2(2, e)
lists.append([Add(a, b*I)])
elif p % 4 == 3:
a,b = get_sq2(p, e)
lists.append([Add(a, b*I)**(e / 2)])
else:
lists.append(get_terms(get_sq2(p, e), e))
return lists
def f(n):
for z in xrange(2, n / 2):
zz = (z + 1) * (z - 1)
ps = Counter(fac(zz))
is_valid = valid(ps)
if is_valid:
print "valid (does not contain a prime of form\n4k + 3 with an odd power)"
print "z: %s, primes: %s" % (z, dict(ps))
lists = get_lists(ps)
cartesian = product(*lists)
for element in cartesian:
print "prime square decomposition: %s" % list(element)
p = 1
for item in element:
p *= item
print "complex conjugates: %s" % p
vals = p.expand(complex=True, evaluate=True).as_coefficients_dict().values()
x, y = vals[0], vals[1] if len(vals) > 1 else 0
print "x, y, z: %s, %s, %s" % (x, y, z)
print "x^2 + y^2, z^2-1: %s, %s" % (x**2 + y**2, z**2 - 1)
print ''
if __name__ == "__main__":
print f(100)
Output:
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 3, primes: {2: 3}
prime square decomposition: [2 + 2*I]
complex conjugates: 2 + 2*I
x, y, z: 2, 2, 3
x^2 + y^2, z^2-1: 8, 8
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 9, primes: {2: 4, 5: 1}
prime square decomposition: [4, 2 + I]
complex conjugates: 8 + 4*I
x, y, z: 8, 4, 9
x^2 + y^2, z^2-1: 80, 80
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 17, primes: {2: 5, 3: 2}
prime square decomposition: [4 + 4*I, 3]
complex conjugates: 12 + 12*I
x, y, z: 12, 12, 17
x^2 + y^2, z^2-1: 288, 288
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 19, primes: {2: 3, 3: 2, 5: 1}
prime square decomposition: [2 + 2*I, 3, 2 + I]
complex conjugates: (2 + I)*(6 + 6*I)
x, y, z: 6, 18, 19
x^2 + y^2, z^2-1: 360, 360
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 33, primes: {17: 1, 2: 6}
prime square decomposition: [4 + I, 8]
complex conjugates: 32 + 8*I
x, y, z: 32, 8, 33
x^2 + y^2, z^2-1: 1088, 1088
valid (does not contain a prime of form
4k + 3 with an odd power)
z: 35, primes: {17: 1, 2: 3, 3: 2}
prime square decomposition: [4 + I, 2 + 2*I, 3]
complex conjugates: 3*(2 + 2*I)*(4 + I)
x, y, z: 18, 30, 35
x^2 + y^2, z^2-1: 1224, 1224
回答2:
Here is a simple improvement in Python (converting to the faster equivalent in C-based code is left as an exercise for the reader). To get accurate timing for the computation, I removed printing the solutions themselves (after validating them in a previous run).
- Use an outer loop for one free variable (I chose
z), constrained only by its relation toN. - Use an inner loop (I chose
y) constrained by the outer loop index. - The third variable is directly computed per requirement 2.
Timing results:
-------------------- 10
1 solutions found in 2.3365020751953125e-05 sec.
-------------------- 100
6 solutions found in 0.00040078163146972656 sec.
-------------------- 1000
55 solutions found in 0.030081748962402344 sec.
-------------------- 10000
543 solutions found in 2.2078349590301514 sec.
-------------------- 100000
5512 solutions found in 214.93411707878113 sec.
That's 3:35 for the large case, plus your time to collate and/or print the results.
If you need faster code (this is still pretty brute-force), look into Diophantine equations and parameterizations to generate (y, x) pairs, given the target value of z^2 - 1.
import math
import time
def break3(N):
"""
10 <= N <= 10^5
return x, y, z triples such that:
x <= y <= z
x^2 + y^2 = z^2 - 1
x + y + z <= N
"""
"""
Observations:
z <= x + y
z < N/2
"""
count = 0
z_limit = N // 2
for z in range(3, z_limit):
# Since y >= x, there's a lower bound on y
target = z*z - 1
ymin = int(math.sqrt(target/2))
for y in range(ymin, z):
# Given y and z, compute x.
# That's a solution iff x is integer.
x_target = target - y*y
x = int(math.sqrt(x_target))
if x*x == x_target and x+y+z <= N:
# print("solution", x, y, z)
count += 1
return count
test = [10, 100, 1000, 10**4, 10**5]
border = "-"*20
for case in test:
print(border, case)
start = time.time()
print(break3(case), "solutions found in", time.time() - start, "sec.")
回答3:
The bounds of x and y are an important part of the problem. I personally went with this Wolfram Alpha query and checked the exact forms of the variables.
Thanks to @Bleep-Bloop and comments, a very elegant bound optimization was found, which is x < n and x <= y < n - x. The results are the same and the times are nearly identical.
Also, since the only possible values for x and y are positive even integers, we can reduce the amount of loop iterations by half.
To optimize even further, since we compute the upper bound of x, we build a list of all possible values for x and make the computation parallel. That saves a massive amount of time on higher values of N but it's a bit slower for smaller values because of the overhead of the parallelization.
Here's the final code:
Non-parallel version, with int values:
List<string> res = new List<string>();
int n2 = n * n;
double maxX = 0.5 * (2.0 * n - Math.Sqrt(2) * Math.Sqrt(n2 + 1));
for (int x = 2; x < maxX; x += 2)
{
int maxY = (int)Math.Floor((n2 - 2.0 * n * x - 1.0) / (2.0 * n - 2.0 * x));
for (int y = x; y <= maxY; y += 2)
{
int z2 = x * x + y * y + 1;
int z = (int)Math.Sqrt(z2);
if (z * z == z2 && x + y + z <= n)
res.Add(x + "," + y + "," + z);
}
}
Parallel version, with long values:
using System.Linq;
...
// Use ConcurrentBag for thread safety
ConcurrentBag<string> res = new ConcurrentBag<string>();
long n2 = n * n;
double maxX = 0.5 * (2.0 * n - Math.Sqrt(2) * Math.Sqrt(n2 + 1L));
// Build list to parallelize
int nbX = Convert.ToInt32(maxX);
List<int> xList = new List<int>();
for (int x = 2; x < maxX; x += 2)
xList.Add(x);
Parallel.ForEach(xList, x =>
{
int maxY = (int)Math.Floor((n2 - 2.0 * n * x - 1.0) / (2.0 * n - 2.0 * x));
for (long y = x; y <= maxY; y += 2)
{
long z2 = x * x + y * y + 1L;
long z = (long)Math.Sqrt(z2);
if (z * z == z2 && x + y + z <= n)
res.Add(x + "," + y + "," + z);
}
});
When ran individually on a i5-8400 CPU, I get these results:
N: 10; Solutions: 1; Time elapsed: 0.03 ms (Not parallel,
int)N: 100; Solutions: 6; Time elapsed: 0.05 ms (Not parallel,
int)N: 1000; Solutions: 55; Time elapsed: 0.3 ms (Not parallel,
int)N: 10000; Solutions: 543; Time elapsed: 13.1 ms (Not parallel,
int)N: 100000; Solutions: 5512; Time elapsed: 849.4 ms (Parallel,
long)
You must use long when N is greater than 36340, because when it's squared, it overflows an int's max value. Finally, the parallel version starts to get better than the simple one when N is around 23000, with ints.
回答4:
No time to properly test it, but seemed to yield the same results as your code (at 100 -> 6 results and at 1000 -> 55 results).
With N=1000 a time of 2ms vs your 144ms also without List
and N=10000 a time of 28ms
var N = 1000;
var c = 0;
for (int x = 2; x < N; x+=2)
{
for (int y = x; y < (N - x); y+=2)
{
long z2 = x * x + y * y + 1;
int z = (int) Math.Sqrt(z2);
if (x + y + z > N)
break;
if (z * z == z2)
c++;
}
}
Console.WriteLine(c);
回答5:
#include<iostream>
#include<math.h>
int main()
{
int N = 10000;
int c = 0;
for (int x = 2; x < N; x+=2)
{
for (int y = x; y < (N - x); y+=2)
{
auto z = sqrt(x * x + y * y + 1);
if(x+y+z>N){
break;
}
if (z - (int) z == 0)
{
c++;
}
}
}
std::cout<<c;
}
This is my solution. On testing the previous solutions for this problem I found that x,y are always even and z is odd. I dont know the mathematical nature behind this, I am currently trying to figure that out.
回答6:
I want to get it done in C# and it should be covering all the test cases based on condition provided in the question.
The basic code, converted to long to process the N <= 100000 upper limit, with every optimizaion thrown in I could. I used alternate forms from @Mat's (+1) Wolfram Alpha query to precompute as much as possible. I also did a minimal perfect square test to avoid millions of sqrt() calls at the upper limit:
public static void Main()
{
int c = 0;
long N = long.Parse(Console.ReadLine());
long N_squared = N * N;
double half_N_squared = N_squared / 2.0 - 0.5;
double x_limit = N - Math.Sqrt(2) / 2.0 * Math.Sqrt(N_squared + 1);
for (long x = 2; x < x_limit; x += 2)
{
long x_squared = x * x + 1;
double y_limit = (half_N_squared - N * x) / (N - x);
for (long y = x; y < y_limit; y += 2)
{
long z_squared = x_squared + y * y;
int digit = (int) z_squared % 10;
if (digit == 3 || digit == 7)
{
continue; // minimalist non-perfect square elimination
}
long z = (long) Math.Sqrt(z_squared);
if (z * z == z_squared)
{
c++;
}
}
}
Console.WriteLine(c);
}
I followed the trend and left out "the degenerate solution" as implied by the OP's code but not explicitly stated.
来源：https://stackoverflow.com/questions/54110614/how-to-enumerate-x2-y2-z2-1-with-additional-constraints