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I'm doing this for my school library where I need to print asterisk pattern like below with *'s
* *
* *
***
*******
***
* *
* *
Below is what I tried,
using System;
using System.Text;
namespace Patterns
{
class Asterisks
{
static void Main(string[] args)
{
int n = 7;
for(int i = 1; i<=n; i++){
for(int j =1; j<=n; j++){
if(j==i || j==n-i+1 || i==n/2+1){
Console.Write("*");
}else{
Console.Write(" ");
}
}
Console.WriteLine();
}
}
}
}
Outputs below :
* *
* *
* *
*******
* *
* *
* *
My logic above is faulty, I tried couple of edits on the loops and conditions but somewhere I'm going wrong. Need help to fix this. Thanks
Break the pattern into parts, and write checks for them individually. If any of checks pass, print a *, otherwise a space. This seems to be what you are already doing. You just seem to be missing some disjunctions in your if statement.
Right now, your condition to print a * is:
j==i || // the "\" part
j==n-i+1 || // the "/" part
i==n/2+1 // the "-" part
You really just need one more condition that those two missing positions would match. One way to check for this is to check that the column (j) is the middle column, and that the row (i) is one away from the middle row.
j == n / 2 + 1 && (i == n / 2 + 2 || i == n / 2)
There's quite a lot of + 1s in your conditions. You can remove them if you start your loop at 0. Rather than repeating n / 2 all the time, you can extract to a variable.
Your code ends up being like this:
int n = 7;
int midRow = n / 2;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (j == i ||
j == n - i - 1 ||
i == midRow ||
j == midRow && (i == midRow + 1 || i == midRow - 1))
{
Console.Write("*");
}
else
{
Console.Write(" ");
}
}
Console.WriteLine();
}
Output:
* *
* *
***
*******
***
* *
* *
Think of the problem as painting a picture line by line and pixel by pixel, starting from the top left (TL) corner and moving to the bottom right (BR).
Depending on the brush/cursor's current position, the program should either "paint" a * or a . This is what the if inside the for does.
If you only wanted to paint a diagonal going from TL to BR then the if only needed one check:
if (j == i) {
Console.Write("*");
} else {
Console.Write(" ");
}
Try it out for yourself and you'll see this shape:
*
*
*
*
*
*
*
Also try and understand how i and j are changed by the two for loops over time and why this works.
For the opposite diagonal we'll have to get the "inverted" value of i which is n - i + 1. This inverted i is what the value of i would have been if we were drawing the picture going right to left, but still top to bottom (i.e. from TR to BL). Knowing this inverted i we can now add an additional check to the if that makes the if more inclusive and draw on more positions:
if (j == i // diag from TL to BR
|| j == n - i + 1 // diag from TR to BL
) {
Console.Write("*");
} else {
Console.Write(" ");
}
This will give us the following shape:
* *
* *
* *
*
* *
* *
* *
Again, try it out for yourself.
The horizontal center line is easier, because it only depends on i, but not on j. We just have to check if we are currently in the center line:
if (j == i // diag from TL to BR
|| j == n - i + 1 // diag from TR to BL
|| i == n / 2 + 1 // center horizontal
) {
Console.Write("*");
} else {
Console.Write(" ");
}
Make sure you understand why the + 1 in n / 2 + 1 is necessary.
Similarly we can also include the center vertical line like so:
if (j == i // diag from TL to BR
|| j == n - i + 1 // diag from TR to BL
|| i == n / 2 + 1 // center horizontal
|| j == n / 2 + 1 // center vertical
) {
Console.Write("*");
} else {
Console.Write(" ");
}
...and we'll end up with this shape:
* * *
* * *
***
*******
***
* * *
* * *
The only problem now is that we don't want to draw the whole vertical line, only it's middle part. So the last check we added to the if is too inclusive, it needs to be more strict. Only the parts of the vertical line that are right next to the center point should be drawn. Or, more precisely, only the point above and the point below the center point should be drawn. (It doesn't matter if we include or exclude the actual center point in the check, because all other checks will cause it to be drawn anyway).
The cursor is right on the center point when i == n / 2 + 1 and j == n / 2 + 1. The point above the center has an i value of n / 2 and the point below an i value of n / 2 + 2.
So when j == n / 2 + 1 is true (we're currently somewhere on the vertical center line) we must also ensure that we're either directly above or below the center point like so:
if (j == i // diag from TL to BR
|| j == n - i + 1 // diag from TR to BL
|| i == n / 2 + 1 // center horizontal
|| (j == n / 2 + 1 && (i == n / 2 || i == n / 2 + 2))
) {
Console.Write("*");
} else {
Console.Write(" ");
}
* *
* *
***
*******
***
* *
* *
Here is an idea - you can map your shape using a jagged array, where 1 means a sign in this case'*' and 0 is a space.
So our pattern will be like this:
int[][] myPattern = new int[][]
{
new int[] {1,0,0,0,0,0,0,0,0,0,1 },
new int[] {0,0,0,0,0,0,0,0,0,0,0 },
new int[] {0,1,0,0,0,0,0,0,0,1,0 },
new int[] {0,0,0,0,0,0,0,0,0,0,0 },
new int[] {0,0,0,0,1,1,1,0,0,0,0 },
new int[] {0,0,0,0,0,0,0,0,0,0,0 },
new int[] {1,1,1,1,1,1,1,1,1,1,1 },
new int[] {0,0,0,0,0,0,0,0,0,0,0 },
new int[] {0,0,0,0,1,1,1,0,0,0,0 },
new int[] {0,0,0,0,0,0,0,0,0,0,0 },
new int[] {0,1,0,0,0,0,0,0,0,1,0 },
new int[] {0,0,0,0,0,0,0,0,0,0,0 },
new int[] {1,0,0,0,0,0,0,0,0,0,1 }
};
If you look at it - represent our asterisk sign.
What you then have to is loop per each item in the jagged array.
for (int j = 0; j < myPattern.Length; j++)
{
for (int i = 0; i < myPattern[j].Length; i++)
{
if (myPattern[j][i] > 0)
{
Console.Write('*');
}
else
{
Console.Write(' ');
}
if (i+1 == myPattern[j].Length)
{
Console.WriteLine();
}
}
}
if you want you can remove altogether the patter with all 0's and replace this
if (i+1 == myPattern[j].Length)
{
Console.WriteLine();
}
with
Console.WriteLine();
I have left it in for demonstration purpose.
I'm going through "sololearn" and udemy courses to try to learn C#. I am doing the challenges but could not figure out how the below code resulted with 32 (as in 32 is the correct answer here and I am trying to find out why). Can someone explain this process to me, the method calling itself is throwing me I think.
static double Pow(double x, int y)
{
if (y == 0)
{
return 1.0;
}
else
{
return x * Pow(x, y - 1);
}
}
static void Main()
{
Console.Write(Pow(2, 5));
}
Please excuse my bad coding. I am trying to do it on mobile, which is difficult, the answer was 32. Can someone explain why?
Edit: Aplogies here is how I work through it. Pass 2 and 5 to Pow, check if y == 0 which is false, it is now y == 5 so x * pow(x, y-1) formula will be active. X is still 2, y is now 4 which means it fails the check again on whether it equals 0, this loop continues until it returns the 1.0, x has remained at 2 so 2 * 1.0 = 2 and not 32?
First thing to note is that this is NOT how you would normally implement a power function. It's done this way to demonstrate recursion.
With that out the way, let's look at what happens when you call Pow(2, 5):
Pow(x = 2, y = 5)
-> return 2 * Pow(2, 4)
<- 2 * 16 = 32
Pow(x = 2, y = 4)
-> return 2 * Pow(2, 3)
<- 2 * 8 = 16
Pow(x = 2, y = 3)
-> return 2 * Pow(2, 2)
<- 2 * 4 = 8
Pow(x = 2, y = 2)
-> return 2 * Pow(2, 1)
<- 2 * 2 = 4
Pow(x = 2, y = 1)
-> return 2 * Pow(2, 0)
<- 2 * 1 = 2
Pow(x = 2, y = 0)
-> return 1 (because y == 0)
<- 1
To read this representation of the recursive call stack, work your way from the top to the bottom looking at how the arguments change; then work your way back up from the bottom to the top looking at the return values (which I indicate with <-).
Ok so let's go through the whole thing.
First of all, a static function is one that can be called without need to instantiate an object. There is one signature that all objects of the same class share. The double is a type within C# and its appearing here to show what the final output type of the function will be. Pow is the name of the function, double x, int y are parameters described by their type (not very well named but we'll leave that for another day)
So x is a number and y is the power of that number. There is a conditional here to check for two outcomes. If y is 0 then the answer is always 1, simple maths. Otherwise, the function performs the arithmetic using recursion (it calls itself again until it meets a terminating condition). The reason we get 32 is because 2x2x2x2x2 = 32. It is 2 to the power of 5.
I'm presuming you know what main and console.write are.
That method basically computes "x raised to the power of y". It does this in a recursive manner.
First, it defines a base case: anything raised to the power of 0 is 1.
Then, it defines what to do in all other cases: x * Pow(x, y - 1). Assuming y is big, what's x * Pow(x, y - 1)? It's x * x * Pow(x, y - 2), which in turn is x * x * x * Pow(x, y - 3). See the pattern here? Eventually, you will reach the point where the second argument, y - N, is 0, which as we have established, is 1. At that point, how many x * have we got? Exactly y.
Let's see this in action for Pow(2, 5):
Pow(2, 5)
2 * Pow(2, 4)
2 * 2 * Pow(2, 3)
2 * 2 * 2 * Pow(2, 2)
2 * 2 * 2 * 2 * Pow(2, 1)
2 * 2 * 2 * 2 * 2 * Pow(2, 0)
2 * 2 * 2 * 2 * 2 * 1
Hence the result 32.
Hello its recursion and it repeat until y=1, then return 2,then return 4, 8, 16, 32 at then end. 2^5=32
To be able to understand each action in this recursive behavior log all the details to see what is actually going on. Such as :
using System;
namespace Tester
{
class test
{
// What Pow actually does:
static double logPow(double x, int y) {
var old = x; // Hold the x
for (var i = 0; i < y; i++){ // do it y times
x = old * x; // Multiply with it's first self
}
return x;
}
static int counter = 0;
static double Pow(double x, int y) {
counter++;
Console.Write("Recursive action[" + counter + "] Y status ["+ y +"] : ");
if (y == 0)
{
Console.Write("return 1.0 = " + logPow(x, y) + " \n");
return 1.0;
}
else
{
Console.Write("return " + x + " * Pow(" + x + ", " + y + " - 1) = " + logPow(x,y-1) + " \n");
return x * Pow(x, y - 1);
}
}
static void Main() {
Console.Write("Last Result : " + Pow(2, 5));
}
}
}
Which gives the result :
Recursive action[1] Y status [5] : return 2 * Pow(2, 5 - 1) = 32
Recursive action[2] Y status [4] : return 2 * Pow(2, 4 - 1) = 16
Recursive action[3] Y status [3] : return 2 * Pow(2, 3 - 1) = 8
Recursive action[4] Y status [2] : return 2 * Pow(2, 2 - 1) = 4
Recursive action[5] Y status [1] : return 2 * Pow(2, 1 - 1) = 2
Recursive action[6] Y status [0] : return 1.0 = 2
Last Result : 32
You can debug your code by looking at these details.
Also you can have fun with it using this link : https://onlinegdb.com/Bysbxat9H
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?
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
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 to N.
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.")
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.
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);
#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.
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.
Trying to find functions that will assist us to draw a 3D line through a series of points.
For each point we know: Date&Time, Latitude, Longitude, Altitude, Speed and Heading.
Data might be recorded every 10 seconds and we would like to be able to guestimate the points in between and increase granularity to 1 second. Thus creating a virtual flight path in 3D space.
I have found a number of curve fitting algorithms that will approximate a line through a series of points but they do not guarantee that the points are intersected. They also do not take into account speed and heading to determine the most likely path taken by the object to reach the next point.
From a physics viewpoint:
You have to assume something about the acceleration in your intermediate points to get the interpolation.
If your physical system is relatively well-behaved (as a car or a plane), as opposed to for example a bouncing ball, you may go ahead supposing an acceleration varying linearly with time between your points.
The vector equation for a constant varying accelerated movement is:
x''[t] = a t + b
where all magnitudes except t are vectors.
For each segment you already know v(t=t0) x(t=t0) tfinal and x(tfinal) v(tfinal)
By solving the differential equation you get:
Eq 1:
x[t_] := (3 b t^2 Tf + a t^3 Tf - 3 b t Tf^2 - a t Tf^3 - 6 t X0 + 6 Tf X0 + 6 t Xf)/(6 Tf)
And imposing the initial and final contraints for position and velocity you get:
Eqs 2:
a -> (6 (Tf^2 V0 - 2 T0 Tf Vf + Tf^2 Vf - 2 T0 X0 + 2 Tf X0 +
2 T0 Xf - 2 Tf Xf))/(Tf^2 (3 T0^2 - 4 T0 Tf + Tf^2))
b -> (2 (-2 Tf^3 V0 + 3 T0^2 Tf Vf - Tf^3 Vf + 3 T0^2 X0 -
3 Tf^2 X0 - 3 T0^2 Xf + 3 Tf^2 Xf))/(Tf^2 (3 T0^2 - 4 T0 Tf + Tf^2))}}
So inserting the values for eqs 2 into eq 1 you get the temporal interpolation for your points, based on the initial and final position and velocities.
HTH!
Edit
A few examples with abrupt velocity change in two dimensions (in 3D is exactly the same). If the initial and final speeds are similar, you'll get "straighter" paths.
Suppose:
X0 = {0, 0}; Xf = {1, 1};
T0 = 0; Tf = 1;
If
V0 = {0, 1}; Vf = {-1, 3};
V0 = {0, 1}; Vf = {-1, 5};
V0 = {0, 1}; Vf = {1, 3};
Here is an animation where you may see the speed changing from V0 = {0, 1} to Vf = {1, 5}:
Here you may see an accelerating body in 3D with positions taken at equal intervals:
Edit
A full problem:
For convenience, I'll work in Cartesian coordinates. If you want to convert from lat/log/alt to Cartesian just do:
x = rho sin(theta) cos(phi)
y = rho sin(theta) sin(phi)
z = rho cos(theta)
Where phi is the longitude, theta is the latitude, and rho is your altitude plus the radius of the Earth.
So suppose we start our segment at:
t=0 with coordinates (0,0,0) and velocity (1,0,0)
and end at
t=10 with coordinates (10,10,10) and velocity (0,0,1)
I clearly made a change in the origin of coordinates to set the origin at my start point. That is just for getting nice round numbers ...
So we replace those numbers in the formulas for a and b and get:
a = {-(3/50), -(3/25), -(3/50)} b = {1/5, 3/5, 2/5}
With those we go to eq 1, and the position of the object is given by:
p[t] = {1/60 (60 t + 6 t^2 - (3 t^3)/5),
1/60 (18 t^2 - (6 t^3)/5),
1/60 (12 t^2 - (3 t^3)/5)}
And that is it. You get the position from 1 to 10 secs replacing t by its valus in the equation above.
The animation runs:
Edit 2
If you don't want to mess with the vertical acceleration (perhaps because your "speedometer" doesn't read it), you could just assign a constant speed to the z axis (there is a very minor error for considering it parallel to the Rho axis), equal to (Zfinal - Zinit)/(Tf-T0), and then solve the problem in the plane forgetting the altitude.
What you're asking is a general interpolation problem. My guess is your actual problem isn't due to the curve-fitting algorithm being used, but rather your application of it to all discrete values recorded by the system instead of the relevant set of values.
Let's decompose your problem. You're currently drawing a point in spherically-mapped 3D space, adjusting for linear and curved paths. If we discretize the operations performed by an object with six degrees of freedom (roll, pitch, and yaw), the only operations you're particularly interested in are linear paths and curved paths accounting for pitch and yaw in any direction. Accounting for acceleration and deceleration also possible given understanding of basic physics.
Dealing with the spherical mapping is easy. Simply unwrap your points relative to their position on a plane, adjusting for latitude, longitude, and altitude. This should allow you to flatten data that would otherwise exist along a curved path, though this may not strictly be necessary for the solutions to your problem (see below).
Linear interpolation is easy. Given an arbitrary number of points backwards in time that fit a line within n error as determined by your system,* construct the line and compute the distance in time between each point. From here, attempt to fit the time points to one of two cases: constant velocity or constant acceleration.
Curve interpolation is a little more difficult, but still plausible. For cases of pitch, yaw, or combined pitch+yaw, construct a plane containing an arbitrary number of points backwards in time, within m error for curved readouts from your system.* From these data, construct a planar curve and once again account for constant velocity or acceleration along the curve.
You can do better than this by attempting to predict the expected operations of a plane in flight as part of a decision tree or neural network relative to the flight path. I'll leave that as an exercise for the reader.
Best of luck designing your system.
--
* Both error readouts are expected to be from GPS data, given the description of the problem. Accounting and adjusting for errors in these data is a separate interesting problem.
What you need (instead of modeling the physics) is to fit a spline through the data. I used a numerical recipies book (http://www.nrbook.com/a has free C and FORTRAN algorithms. Look into F77 section 3.3 to get the math needed). If you want to be simple then just fit lines through the points, but that will not result in a smooth flight path at all. Time will be your x value, and each parameter loged will have it's own cublic spline parameters.
Since we like long postings for this question here is the full code:
//driver program
static void Main(string[] args)
{
double[][] flight_data = new double[][] {
new double[] { 0, 10, 20, 30 }, // time in seconds
new double[] { 14500, 14750, 15000, 15125 }, //altitude in ft
new double[] { 440, 425, 415, 410 }, // speed in knots
};
CubicSpline altitude = new CubicSpline(flight_data[0], flight_data[1]);
CubicSpline speed = new CubicSpline(flight_data[0], flight_data[2]);
double t = 22; //Find values at t
double h = altitude.GetY(t);
double v = speed.GetY(t);
double ascent = altitude.GetYp(t); // ascent rate in ft/s
}
// Cubic spline definition
using System.Linq;
/// <summary>
/// Cubic spline interpolation for tabular data
/// </summary>
/// <remarks>
/// Adapted from numerical recipies for FORTRAN 77
/// (ISBN 0-521-43064-X), page 110, section 3.3.
/// Function spline(x,y,yp1,ypn,y2) converted to
/// C# by jalexiou, 27 November 2007.
/// Spline integration added also Nov 2007.
/// </remarks>
public class CubicSpline
{
double[] xi;
double[] yi;
double[] yp;
double[] ypp;
double[] yppp;
double[] iy;
#region Constructors
public CubicSpline(double x_min, double x_max, double[] y)
: this(Sequence(x_min, x_max, y.Length), y)
{ }
public CubicSpline(double x_min, double x_max, double[] y, double yp1, double ypn)
: this(Sequence(x_min, x_max, y.Length), y, yp1, ypn)
{ }
public CubicSpline(double[] x, double[] y)
: this(x, y, double.NaN, double.NaN)
{ }
public CubicSpline(double[] x, double[] y, double yp1, double ypn)
{
if( x.Length == y.Length )
{
int N = x.Length;
xi = new double[N];
yi = new double[N];
x.CopyTo(xi, 0);
y.CopyTo(yi, 0);
if( N > 0 )
{
double p, qn, sig, un;
ypp = new double[N];
double[] u = new double[N];
if( double.IsNaN(yp1) )
{
ypp[0] = 0;
u[0] = 0;
}
else
{
ypp[0] = -0.5;
u[0] = (3 / (xi[1] - xi[0])) *
((yi[1] - yi[0]) / (x[1] - x[0]) - yp1);
}
for (int i = 1; i < N-1; i++)
{
double hp = x[i] - x[i - 1];
double hn = x[i + 1] - x[i];
sig = hp / hn;
p = sig * ypp[i - 1] + 2.0;
ypp[i] = (sig - 1.0) / p;
u[i] = (6 * ((y[i + 1] - y[i]) / hn) - (y[i] - y[i - 1]) / hp)
/ (hp + hn) - sig * u[i - 1] / p;
}
if( double.IsNaN(ypn) )
{
qn = 0;
un = 0;
}
else
{
qn = 0.5;
un = (3 / (x[N - 1] - x[N - 2])) *
(ypn - (y[N - 1] - y[N - 2]) / (x[N - 1] - x[N - 2]));
}
ypp[N - 1] = (un - qn * u[N - 2]) / (qn * ypp[N - 2] + 1.0);
for (int k = N-2; k > 0; k--)
{
ypp[k] = ypp[k] * ypp[k + 1] + u[k];
}
// Calculate 1st derivatives
yp = new double[N];
double h;
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
yp[i] = (yi[i + 1] - yi[i]) / h
- h / 6 * (ypp[i + 1] + 2 * ypp[i]);
}
h = xi[N - 1] - xi[N - 2];
yp[N - 1] = (yi[N - 1] - yi[N - 2]) / h
+ h / 6 * (2 * ypp[N - 1] + ypp[N - 2]);
// Calculate 3rd derivatives as average of dYpp/dx
yppp = new double[N];
double[] jerk_ij = new double[N - 1];
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
jerk_ij[i] = (ypp[i + 1] - ypp[i]) / h;
}
Yppp = new double[N];
yppp[0] = jerk_ij[0];
for( int i = 1; i < N - 1; i++ )
{
yppp[i] = 0.5 * (jerk_ij[i - 1] + jerk_ij[i]);
}
yppp[N - 1] = jerk_ij[N - 2];
// Calculate Integral over areas
iy = new double[N];
yi[0] = 0; //Integration constant
for( int i = 0; i < N - 1; i++ )
{
h = xi[i + 1] - xi[i];
iy[i + 1] = h * (yi[i + 1] + yi[i]) / 2
- h * h * h / 24 * (ypp[i + 1] + ypp[i]);
}
}
else
{
yp = new double[0];
ypp = new double[0];
yppp = new double[0];
iy = new double[0];
}
}
else
throw new IndexOutOfRangeException();
}
#endregion
#region Actions/Functions
public int IndexOf(double x)
{
//Use bisection to find index
int i1 = -1;
int i2 = Xi.Length;
int im;
double x1 = Xi[0];
double xn = Xi[Xi.Length - 1];
bool ascending = (xn >= x1);
while( i2 - i1 > 1 )
{
im = (i1 + i2) / 2;
double xm = Xi[im];
if( ascending & (x >= Xi[im]) ) { i1 = im; } else { i2 = im; }
}
if( (ascending && (x <= x1)) || (!ascending & (x >= x1)) )
{
return 0;
}
else if( (ascending && (x >= xn)) || (!ascending && (x <= xn)) )
{
return Xi.Length - 1;
}
else
{
return i1;
}
}
public double GetIntY(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double h2 = h * h;
double a = (x-x1)/h;
double a2 = a*a;
return h / 6 * (3 * a * (2 - a) * y1
+ 3 * a2 * y2 - a2 * h2 * (a2 - 4 * a + 4) / 4 * y1pp
+ a2 * h2 * (a2 - 2) / 4 * y2pp);
}
public double GetY(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double h2 = h * h;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return A * y1 + B * y2 + h2 / 6 * (A * (A * A - 1) * y1pp
+ B * (B * B - 1) * y2pp);
}
public double GetYp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1 = yi[i];
double y2 = yi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return (y2 - y1) / h + h / 6 * (y2pp * (3 * B * B - 1)
- y1pp * (3 * A * A - 1));
}
public double GetYpp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
double A = 1 - (x - x1) / (x2 - x1);
double B = 1 - A;
return A * y1pp + B * y2pp;
}
public double GetYppp(double x)
{
int i = IndexOf(x);
double x1 = xi[i];
double x2 = xi[i + 1];
double y1pp = ypp[i];
double y2pp = ypp[i + 1];
double h = x2 - x1;
return (y2pp - y1pp) / h;
}
public double Integrate(double from_x, double to_x)
{
if( to_x < from_x ) { return -Integrate(to_x, from_x); }
int i = IndexOf(from_x);
int j = IndexOf(to_x);
double x1 = xi[i];
double xn = xi[j];
double z = GetIntY(to_x) - GetIntY(from_x); // go to nearest nodes (k) (j)
for( int k = i + 1; k <= j; k++ )
{
z += iy[k]; // fill-in areas in-between
}
return z;
}
#endregion
#region Properties
public bool IsEmpty { get { return xi.Length == 0; } }
public double[] Xi { get { return xi; } set { xi = value; } }
public double[] Yi { get { return yi; } set { yi = value; } }
public double[] Yp { get { return yp; } set { yp = value; } }
public double[] Ypp { get { return ypp; } set { ypp = value; } }
public double[] Yppp { get { return yppp; } set { yppp = value; } }
public double[] IntY { get { return yp; } set { iy = value; } }
public int Count { get { return xi.Length; } }
public double X_min { get { return xi.Min(); } }
public double X_max { get { return xi.Max(); } }
public double Y_min { get { return yi.Min(); } }
public double Y_max { get { return yi.Max(); } }
#endregion
#region Helpers
static double[] Sequence(double x_min, double x_max, int double_of_points)
{
double[] res = new double[double_of_points];
for (int i = 0; i < double_of_points; i++)
{
res[i] = x_min + (double)i / (double)(double_of_points - 1) * (x_max - x_min);
}
return res;
}
#endregion
}
You can find an approximation of a line that intersects points in 3d and 2d space using a Hough Transformation algorithm. I am only familiar with it's uses in 2d however but it will still work for 3d spaces given that you know what kind of line you are looking for. There is a basic implementation description linked. You can Google for pre-mades and here is a link to a 2d C implementation CImg.
The algorithm process (roughly)... First you find equation of a line that you think will best approximate the shape of the line (in 2d parabolic, logarithmic, exponential, etc). You take that formula and solve for one of the parameters.
y = ax + b
becomes
b = y - ax
Next, for each point you are attempting to match, you plugin the points to the y and x values. With 3 points, you would have 3 separate functions of b with respect to a.
(2, 3) : b = 3 - 2a
(4, 1) : b = 1 - 4a
(10, -5): b = -5 - 10a
Next, the theory is that you find all possible lines which pass through each of the points, which is infinitely many for each individual point however when combined in an accumulator space only a few possible parameters best fit. In practice this is done by choosing a range space for the parameters (I chose -2 <= a <= 1, 1 <= b <= 6) and begin plugging in values for the variant parameter(s) and solving for the other. You tally up the number of intersections from each function in an accumulator. The points with the highest values give you your parameters.
Accumulator after processing b = 3 - 2a
a: -2 -1 0 1
b: 1
2
3 1
4
5 1
6
Accumulator after processing b = 1 - 4a
a: -2 -1 0 1
b: 1 1
2
3 1
4
4
5 2
6
Accumulator after processing b = -5 - 10a
a: -2 -1 0 1
b: 1 1
2
3 1
4
5 3
6
The parameter set with the highest accumulated value is (b a) = (5 -1) and the function best fit to the points given is y = 5 - x.
Best of luck.
My guess is that a serious application of this would use a http://en.wikipedia.org/wiki/Kalman_filter. By the way, that probably wouldn't guarantee that the reported points were intersected either, unless you fiddled with the parameters a bit. It would expect some degree of error in each data point given to it, so where it thinks the object is at time T would not necessarily be where it was at time T. Of course, you could set the error distribution to say that you were absolutely sure you knew where it was at time T.
Short of using a Kalman filter, I would try and turn it into an optimisation problem. Work at the 1s granularity and write down equations like
x_t' = x_t + (Vx_t + Vx_t')/2 + e,
Vx_t_reported = Vx_t + f,
Vx_t' = Vx_t + g
where e, f, and g represent the noise. Then create a penalty function such as e^2 + f^2 + g^2 +...
or some weighted version such as 1.5e^2 + 3f^2 + 2.6g^2 +... according to your idea of what the errors really are and how smooth you wnat the answer to be, and find the values that make the penalty function as small as possible - with least squares if the equations turn out nicely.
screenshot of debug: http://img1.uploadscreenshot.com/images/orig/12/36121481470-orig.jpg
note how x, y have values (i have no idea why x and y stopped on 69 in the for loop - x should've went up to 86 and y to 183) yet r has no value at all. (the variable doesn't exist? what?) how should I fix this?
code if you want to read:
public float[] cht(int[,] matrix)
{
float[] xyrd = new float[4];
int xthreshold, ythreshold;
float slope;
double dir;
float zone;
int[] limitsStorage = new int[3] { matrix.GetLength(0), matrix.GetLength(1), matrix.GetLength(0) / 2 - 10 };
short[,,] accumulator = new short[limitsStorage[0]+1, limitsStorage[1]+1,limitsStorage[2]+1];
for (int x = 0; x < limitsStorage[0]; x++)
{ //first dimension loop of matrix 100
for (int y = 0; y < limitsStorage[1]; y++)
{ //second dimension loop of matrix 120
if (matrix[x, y] == 225)
{//the data at the for loop location is a 1 and not 0 hit.
xthreshold = x - limitsStorage[0] / 2;
ythreshold = y - limitsStorage[1] / 2;
//forget angle, search slope: float angle = xthreshold > 0 ? ((float)Math.Atan2(xthreshold, ythreshold)) : ((float)Math.Atan2(xthreshold, ythreshold) + 180);
slope = xthreshold / ythreshold;
//initiate if loops.
dir = 180 + Math.Round(Math.Atan2(ythreshold, xthreshold) * 57.2957 / 45, 0) * 45 + 45 * (Math.Round(((Math.Atan2(ythreshold, xthreshold) * 57.2957) % 45) / 45));
if (slope > .404 || slope < -.404)
{
if (slope < 2.3558 || slope > -2.3558)
{
if (xthreshold > 0)
{
if (ythreshold > 0)
{
//+x+y zone
zone = 45 + 180;
}
else
{
//+x-y zone
zone = 180 - 45;
}
}
else
{
if (ythreshold > 0)
{
//-x+y zone
zone = 360 - 45;
}
else
{
//-x-y zone
zone = 45;
}
}
}
else if (ythreshold > 0)
{
//+y zone
zone = 360 - 90;
}
else
{
//-y zone
zone = 90;
}
}
else if (xthreshold > 0)
{
//+x zone
zone = 180;
}
else
{
//-x zone
zone = 0;
}
for (int R = 6; R < limitsStorage[2]; R++)
{ //Radius loop for scan 44
float delta = (float)((1 / R) * 57.2957);
for (float Theta = zone - 25; Theta < zone + 25; Theta += delta)
{
accumulator[(int)(((R * Math.Cos(Theta / 57.2957)) < 0 || (R * Math.Cos(Theta / 57.2957)) > limitsStorage[0]) ? 0 : R * Math.Cos(Theta / 57.2957)), (int)(((R * Math.Sin(Theta / 57.2957)) < 0 || (R * Math.Sin(Theta / 57.2957)) > limitsStorage[1]) ? 0 : R * Math.Sin(Theta / 57.2957)),R]++;
//btw, 0,0,R is always the non-center area.
}
}
}
}
}
for (int x = 1; x < limitsStorage[0]; x++)
{
for (int y = 1; y < limitsStorage[1]; y++)
{
for (int r = 6; r < limitsStorage[2]; r++)
{
if (xyrd[3] > accumulator[x, y, r])
{
xyrd[0] = x;
xyrd[1] = y;
xyrd[2] = r;
xyrd[3] = accumulator[x, y, r];
}
}
}
}
if (accPrint)
{
//do something for debugging?
accPrint = false;
}
return xyrd;
}
I just noticed that the x and y have the little lock symbol under them indicating that you have private variables named x and y in the class in which this method is executing. Those are the x and y that you are seeing in the debugger.
r is appropriately out of scope as you've exited the loop in which it is declared.
By the way, x and y are ridiculously bad member variable names, and are ridiculously bad names for for loop variables of type int, especially if they are contained in a class with member variables named x and y.
The only place you declare r is in the for statement, right? That means r goes out of scope as soon as the loop ends. So naturally if you inspect variables at he end of the function, r won't be there.
Confessing I don't know why x and y are in scope based on the comments. They could be class variables, but the asker says not. That's the only explanation I can think of, though.
The behaviour is not weird -- you actually get exactly what you expect.
Please note that the watch window can only accurately show you values that are in scope at the breakpoint.
At the highlighted breakpoint, only accumulator[x, y, r] is in scope, and you see exactly the values you expected.