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suppose i want to find any value of x and y such that they satisfy x . W + y . D = P
this can be done by the following using extended euclidean algorithm
int exgcd(int a, int b, int &x, int &y)
{
if (b == 0)
{
x = 1;
y = 0;
return a;
}
int g, xi, yi;
g = exgcd(b, a % b, xi, yi);
x = yi;
y = xi - (a / b * yi);
return g;
}
but this will be just some random x and y satisfying the equation
suppose i add an additional constraint that i want any x y z such that
x>=0 y>=0 z>=0 and x + y + z = n
how can i effectively (please share code/pseudocode if possible) find all such x y and z??
my question boils down to find any x and y (using extended euclidean algorithm) which
1) satisfy a linear equation
2) fall in a given range
here is the link to the question if you want
Ok so we have 2 equations and 3 unknowns, so doing some simple mathematics we can find the equations we need to solve
x * W + y * D + z * 0 = p
And
x + y + z = n
given
x,y,z >=0
So firstly we will reduce one unknown by looping through any one of the unknowns lets say z
We iterate through 0 - n for z and our new equations will be
x * w + y * d = p
And
x + y = m { m is n - z for value of z in current iteration
Now we have 2 equations and 2 unknowns
So our solution for x and y can now be reduced by substituting x in first equation
Which makes it
(m - y) * w + y * d = p
This can be reduced to
y = (p - m * w) / (d - w)
and
x = m - y
You can stop at first value where x and y are integers
Related
I want to calculate the sum of the series as below
Lim
X->1 (2/3 - x/3 -(x^2)/3 +(x^3)*2/3 -..). I am not sure whether we have a formula for finding the sum of this kind of series. Tried a lot but couldn't find any. Any help is appreciated.
This seems to be more maths than computing.
It factorises as (1 + x^3 + x^6 + ...)(2 - x - x^2)/3
If x = 1-d (where d is small), then to first order in d, the (2 - x - x^2) term becomes (2 - (1-d) - (1-2d)) = 3d
And the (1 + x^3 + x^6 + ...) term is a geometric progression, with sum 1/(1-x^3), or here 1/(1-(1-d)^3), and the denominator to first order in d is (1 - (1-3d)) = 3d
Hence the whole thing is (1/3d) (3d) / 3 = 1/3
But we can also verify computationally with a value close to 1 (Python code here):
x = 0.999999
s = 0
f = (2 - x - x*x) / 3.
x3 = x ** 3
s_prev = None
while s != s_prev:
s_prev = s
s += f
f *= x3
print(s)
gives:
0.33333355556918565
This question already has answers here:
Calculating (a^b)%MOD
(7 answers)
Closed 8 years ago.
I am wondering how to calculate x^y mod z. x and y are very large (can't fit in 64 bit integer) and z will fit 64 bit integer.
And one thing don't give answers like x^y mod z is same as (x mod z)^y mod z.
Here is the standard method, in pseudo-code:
function powerMod(b, e, m)
x := 1
while e > 0
if e % 2 == 1
x := (b * x) % m
e := e - 1
else
b := (b * b) % m
e := e / 2
return x
The algorithm is known as exponentiation by squaring or the square and multiply method.
If you are using Java, then turn x, y, and z into BigInteger.
BigInteger xBI = new BigInteger(x);
BigInteger yBI = new BigInteger(y);
BigInteger zBI = new BigInteger(z);
BigInteger answer = xBI.modPow(yBI,zBI);
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I have 2 3D points A and B I now assume the parametric equation like:
x = Ax + (Bx*t)
y = Ay + (By*t)
z = Az + (Bz*t)
So this can be refined to:
x = Ax + (Bx * ((y - Ay)/By));
is correct.
Given this I want to know what the coordinates of a point(on A-B) at height 0 are.
so I now do this:
float y = 0;
float t = ((y - Ay) / By)
float x = Ax + (Bx * t);
float z = Az + (Bz * t);
Is there anything mathematically wrong with this?
(my code is not doing what should be happening with this)
Thanks!
PS: The relevance of this question to programming:
In a game engine when projecting points(in this case corners of my view frustum) onto a plane such as the xz-plane with y = 0 this mathematical problem coincides with my game-programming
I would write them this way:
x = A*(1-t) + B*t
So x(0) = A and x(1) = B. Writing it this way assumes 0 <= t <= +1.
Another way to think about it is to assume -1 <= t <= +1. If you go that way the shape functions look like this:
x = A*(1-t)/2.0 + B*(1+t)/2.0
Once again x(-1) = A and x(+1) = B.
And it's easy to generalize to higher-order functions:
x(t) = A*t(t-1)/2 + C*(1-t^2) + B*t(t+1)/2
So x(0) = A, x(0.5) = C, x(1) = B.
I think your parametrization is wrong:
x = Ax + (Bx*t)
...
But it should be:
x = Ax + ((Bx-Ax)*t)
...
This is a fairly simple question. I need need an equation to determine whether two 2 dimensional lines collide with each other. If they do I also need to know the X and Y position of the collision.
Put them both in general form. If A and B are the same then they're parallel. Otherwise, create two simultaneous equations and solve for x and y.
Let A and B represented by this parametric form : y = mx + b
Where m is the slope of the line
Now in the case of parallelism of A and B their slope should be equal
Else they will collide with each other at point T(x,y)
For finding the coordinates of point T you have to solve an easy equation:
A: y = mx + b
B: y = Mx + B
y(A) = y(B) means : mx + b = Mx + B which yields to x = (B - b)/(m - M) and by putting
the x to the line A we find y = ((m*(B - b))/(m - M)) + b
so : T : ((B - b)/(m - M) , ((m*(B - b))/(m - M)) + b)
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Well it is not a program problem. Is there any hint for such quiz?
I am thinking about focusing on two random R1, R2, both of which is in range (0, 1). and supposing R2 > R1
and then fulfill two equation:
R1 + (1 - R2) > R2 - R1 // two sticks sum longer then the rest one
|R1 - (1 - R2)| < R2 - R1 // the difference of these two should be shorter the rest one
but I cannot move further...
Think of (r1, r2) as a point in the unit square.
Which part of the unit square is allowed for r2 > r1?
Which part of that leads to three lengths that can form a triangle?
The answer is 1/4.
Here is the explanation.
Let x is the length of the leftmost stick and y is the length of the rightmost stick.
Then the middle stick has length n-x-y, if the original stick's length was n.
The possible values for x,y are those for which:
x > 0
y > 0
x + y < n
In the plane Oxy this is equivalent to say that the point (x,y) lies within the triangle with vertices (0, 0), (n, 0), (0, n).
Now these three numbers (x, y, n-x-y) form a triangle if all of the three are satisfied:
x + y > n - x - y <=> x + y < n/2
x + (n - x - y) > y <=> y < n/2
y + (n - x - y) > x <=> x < n/2
Again in the Oxy plane these are satisfied when the point (x,y) lies within the triangle with vertices (0, n/2), (n/2, n/2), (n/2, 0).
The area of this triangle is a quarter of the area of the (0, 0), (n, 0), (0, n) triangle, since it's the 'middle' triangle (whose vertices are the midpoints) of the bigger one.
Here is a simple C# program to verify the answer:
Random r = new Random();
int count = 0, total = 0, tries = 1000000;
double x, y;
for (int i = 0; i < tries; i++)
{
x = r.NextDouble();
y = r.NextDouble();
if (x + y > 0.5 && x < 0.5 && y < 0.5) ++count;
if (x + y < 1.0) ++total;
}
Console.WriteLine((double)count / total);
I have just made a program to verify it, and I found it is 1/4:
class Program
{
static void Main(string[] args)
{
int nIsTriangle = 0;
Random ran = new Random(0);
int nTry = 1000000;
for (int i = 0; i < nTry; i++)
{
double r1 = ran.NextDouble();
double r2 = ran.NextDouble();
if (Check(r1, r2)) nIsTriangle++;
}
Console.WriteLine((double)nIsTriangle / (double)nTry);
Console.ReadKey();
}
static bool Check(double r1, double r2)
{
double first = Math.Min(r1, r2);
double second = Math.Abs(r1 - r2);
double third = 1 - Math.Max(r1, r2);
bool conditionA = (first + second) > third;
bool conditionB = Math.Abs(first - second) < third;
return conditionA && conditionB;
}
}