Given a arbitrary integers p, g, and r and given y such that y = gx mod p where x is an unknown integer quantity, how would one solve for C where C = gr• (gx)-1 mod p?
My math is below, but when I input it in a verifier function, it says the answer is incorrect.
y•u = 1 mod p
y•u = 1 + mp
uy - mp = 1
where u is the inverse of y and m is the set of natural numbers (as inverse of mod requires this)
If I see it right you are looking for Inverse modpow. The math is like this:
ab = a^b % p
ab + c*p = a^b
log(ab+c*p)/log(a) = b
(ab+c*p)^(1/b) = a
where c is integer c={ 0,1,2,3,4... } converting between normal and modular arithmetics. So in your case you want to compute b. The problem is that log(ab+c*p)/log(a) grows very slow with increasing c if p is not much bigger than a. So in such case its faster to use all combinations of b instead until a fit is found something like this in C++:
//---------------------------------------------------------------------------
ALU32 alu;
DWORD modmul(DWORD a,DWORD b,DWORD p) // ans = a*b % p
{
DWORD ch,cl,c,d;
alu.mul(ch,cl,a,b);
alu.div(c,d,ch,cl,p);
return d;
}
//---------------------------------------------------------------------------
DWORD modinv(DWORD a,DWORD p) // a * ans % p = 1
{
DWORD b,c,db,dc,i=0;
db=p/a;
dc=db*a;
for (b=1,c=a;b<p;i++)
{
if (c==1) return b;
b+=db; c+=dc;
while (c<p){ b++; c+=a; }
c-=p;
}
return 0;
}
//---------------------------------------------------------------------------
DWORD modpow(DWORD a,DWORD b,DWORD p) // ans = a^b % p
{ // b is not mod(p) !
DWORD i,d=1;
for (a%=p,i=0;i<32;i++,b<<=1)
{
d=modmul(d,d,p);
if (DWORD(b&0x80000000)) d=modmul(d,a,p);
}
return d;
}
//---------------------------------------------------------------------------
DWORD imodpow(DWORD ab,DWORD a,DWORD p) // ab = a^ans % p
{ // ans is not mod(p) !
DWORD b,AB;
for (AB=1,b=0;;)
{
if (AB==ab) return b;
b++; if (!b) return 0;
AB=modmul(AB,a,p);
}
}
//---------------------------------------------------------------------------
of coarse this is SLOOOOW that is why is this used for cryptography...
Also beware there are multiple valid solutions and the first one found might not be the one you're seeking so you need to add additional conditions ...
The ALU32.h can be found in here Cant make value propagate through carry
And the modular arithmetics is based on this: Modular arithmetics and NTT (finite field DFT) optimizations
Here a sample for comparison (ignore VCL and tbeg/tend/tstr functions):
DWORD a=87654321,b=12345678,p=0xC0000001,ab,bb;
tbeg(); ab=modpow(a,b,p); tend(); mm_log->Lines->Add(AnsiString().sprintf("%8u^%8u mod %u = %u ",a,b ,p,ab)+tstr(1));
tbeg(); bb=imodpow(ab,a,p); tend(); mm_log->Lines->Add(AnsiString().sprintf("%8u^%8u mod %u = %u ",a,bb,p,ab)+tstr(1));
and output:
87654321^12345678 mod 3221225473 = 3038293251 [ 0.002 ms]
87654321^12345678 mod 3221225473 = 3038293251 [ 421.910 ms]
PS.
There might be some more advanced approaches from number theory if the p is special like prime, composite of two primes or even n-th root of unity ... but that is in galaxy far far away from mine reach of expertise.
[edit1]
from your newly posted question its finally more clear that you really just wanted modular inverse and has nothing to do with imodpow. So what you want is this:
a*b % p = 1
where b is unknown so simply try all b in increasing manner where a*b % p is just truncated by p towards zero and if the result is 1 you found your answer. I updated the code above with modinv function doing exactly that + some optimization. However I think there are even faster approaches for this using GCD or something.
Here another test sample:
DWORD a=87654321,b=12345678,p=0xC0000001,ab,bb;
ab=modmul(a,b,p);
tbeg(); bb=modinv(b,p); tend(); mm_log->Lines->Add(AnsiString().sprintf(" 1/%8u mod %u = %u ",b,p,bb)+tstr(1));
tbeg(); a =modmul(b,bb,p); tend(); mm_log->Lines->Add(AnsiString().sprintf("%8u*%8u mod %u = %u ",b,bb,p,a)+tstr(1));
tbeg(); a =modmul(ab,bb,p); tend(); mm_log->Lines->Add(AnsiString().sprintf("%8u*%8u mod %u = %u ",ab,bb,p,a)+tstr(1));
And output:
1/12345678 mod 3221225473 = 165081805 [ 4.999 ms]
12345678*165081805 mod 3221225473 = 1 [ 0.000 ms]
652073126*165081805 mod 3221225473 = 87654321 [ 0.000 ms]
Last week I was in an interview and there was a test like this:
Calculate N/9 (given that N is a positive integer), using only
SHIFT LEFT, SHIFT RIGHT, ADD, SUBSTRACT instructions.
first, find the representation of 1/9 in binary
0,0001110001110001
means it's (1/16) + (1/32) + (1/64) + (1/1024) + (1/2048) + (1/4096) + (1/65536)
so (x/9) equals (x>>4) + (x>>5) + (x>>6) + (x>>10) + (x>>11)+ (x>>12)+ (x>>16)
Possible optimization (if loops are allowed):
if you loop over 0001110001110001b right shifting it each loop,
add "x" to your result register whenever the carry was set on this shift
and shift your result right each time afterwards,
your result is x/9
mov cx, 16 ; assuming 16 bit registers
mov bx, 7281 ; bit mask of 2^16 * (1/9)
mov ax, 8166 ; sample value, (1/9 of it is 907)
mov dx, 0 ; dx holds the result
div9:
inc ax ; or "add ax,1" if inc's not allowed :)
; workaround for the fact that 7/64
; are a bit less than 1/9
shr bx,1
jnc no_add
add dx,ax
no_add:
shr dx,1
dec cx
jnz div9
( currently cannot test this, may be wrong)
you can use fixed point math trick.
so you just scale up so the significant fraction part goes to integer range, do the fractional math operation you need and scale back.
a/9 = ((a*10000)/9)/10000
as you can see I scaled by 10000. Now the integer part of 10000/9=1111 is big enough so I can write:
a/9 = ~a*1111/10000
power of 2 scale
If you use power of 2 scale then you need just to use bit-shift instead of division. You need to compromise between precision and input value range. I empirically found that on 32 bit arithmetics the best scale for this is 1<<18 so:
(((a+1)<<18)/9)>>18 = ~a/9;
The (a+1) corrects the rounding errors back to the right range.
Hardcoded multiplication
Rewrite the multiplication constant to binary
q = (1<<18)/9 = 29127 = 0111 0001 1100 0111 bin
Now if you need to compute c=(a*q) use hard-coded binary multiplication: for each 1 of the q you can add a<<(position_of_1) to the c. If you see something like 111 you can rewrite it to 1000-1 minimizing the number of operations.
If you put all of this together you should got something like this C++ code of mine:
DWORD div9(DWORD a)
{
// ((a+1)*q)>>18 = (((a+1)<<18)/9)>>18 = ~a/9;
// q = (1<<18)/9 = 29127 = 0111 0001 1100 0111 bin
// valid for a = < 0 , 147455 >
DWORD c;
c =(a<< 3)-(a ); // c= a*29127
c+=(a<< 9)-(a<< 6);
c+=(a<<15)-(a<<12);
c+=29127; // c= (a+1)*29127
c>>=18; // c= ((a+1)*29127)>>18
return c;
}
Now if you see the binary form the pattern 111000 is repeating so yu can further improve the code a bit:
DWORD div9(DWORD a)
{
DWORD c;
c =(a<<3)-a; // first pattern
c+=(c<<6)+(c<<12); // and the other 2...
c+=29127;
c>>=18;
return c;
}
In Go's crytography library, I found this function ConstantTimeByteEq. What does it do, how does it work?
// ConstantTimeByteEq returns 1 if x == y and 0 otherwise.
func ConstantTimeByteEq(x, y uint8) int {
z := ^(x ^ y)
z &= z >> 4
z &= z >> 2
z &= z >> 1
return int(z)
}
x ^ y is x XOR y, where the result is 1 when the arguments are different and 0 when the arguments are the same:
x = 01010011
y = 00010011
x ^ y = 01000000
^(x ^ y) negates this, i.e., you get 0 when the arguments are different and 1 otherwise:
^(x ^ y) = 10111111 => z
Then we start shifting z to the right for masking its bits by itself. A shift pads the left side of the number with zero bits:
z >> 4 = 00001011
With the goal of propagating any zeros in z to the result, start ANDing:
z = 10111111
z >> 4 = 00001011
z & (z >> 4) = 00001011
also fold the new value to move any zero to the right:
z = 00001011
z >> 2 = 00000010
z & (z >> 2) = 00000010
further fold to the last bit:
z = 00000010
z >> 1 = 00000001
z & (z >> 1) = 00000000
On the other hand, if you have x == y initially, it goes like this:
z = 11111111
z (& z >> 4) = 00001111
z (& z >> 2) = 00000011
z (& z >> 1) = 00000001
So it really returns 1 when x == y, 0 otherwise.
Generally, if both x and y are zero the comparison can take less time than other cases. This function tries to make it so that all calls take the same time regardless of the values of its inputs. This way, an attacker can't use timing based attacks.
It does exactly what the documentation says: It checks if x and y are equal. From a functional point it is just x == y, dead simple.
Doing x == y in this cryptic bit-fiddling-way prevent timing side attacks to algorithms: A x == y may get compiled to code which performs faster if x = y and slower if x != y (or the other way around) due to branch prediction in CPUs. This can be used by an attacker to learn something about the data handled by the cryptographic routines and thus compromise security.
Given an array of x,y points, how do I sort the points of this array in clockwise order (around their overall average center point)? My goal is to pass the points to a line-creation function to end up with something looking rather "solid", as convex as possible with no lines intersecting.
For what it's worth, I'm using Lua, but any pseudocode would be appreciated.
Update: For reference, this is the Lua code based on Ciamej's excellent answer (ignore my "app" prefix):
function appSortPointsClockwise(points)
local centerPoint = appGetCenterPointOfPoints(points)
app.pointsCenterPoint = centerPoint
table.sort(points, appGetIsLess)
return points
end
function appGetIsLess(a, b)
local center = app.pointsCenterPoint
if a.x >= 0 and b.x < 0 then return true
elseif a.x == 0 and b.x == 0 then return a.y > b.y
end
local det = (a.x - center.x) * (b.y - center.y) - (b.x - center.x) * (a.y - center.y)
if det < 0 then return true
elseif det > 0 then return false
end
local d1 = (a.x - center.x) * (a.x - center.x) + (a.y - center.y) * (a.y - center.y)
local d2 = (b.x - center.x) * (b.x - center.x) + (b.y - center.y) * (b.y - center.y)
return d1 > d2
end
function appGetCenterPointOfPoints(points)
local pointsSum = {x = 0, y = 0}
for i = 1, #points do pointsSum.x = pointsSum.x + points[i].x; pointsSum.y = pointsSum.y + points[i].y end
return {x = pointsSum.x / #points, y = pointsSum.y / #points}
end
First, compute the center point.
Then sort the points using whatever sorting algorithm you like, but use special comparison routine to determine whether one point is less than the other.
You can check whether one point (a) is to the left or to the right of the other (b) in relation to the center by this simple calculation:
det = (a.x - center.x) * (b.y - center.y) - (b.x - center.x) * (a.y - center.y)
if the result is zero, then they are on the same line from the center, if it's positive or negative, then it is on one side or the other, so one point will precede the other.
Using it you can construct a less-than relation to compare points and determine the order in which they should appear in the sorted array. But you have to define where is the beginning of that order, I mean what angle will be the starting one (e.g. the positive half of x-axis).
The code for the comparison function can look like this:
bool less(point a, point b)
{
if (a.x - center.x >= 0 && b.x - center.x < 0)
return true;
if (a.x - center.x < 0 && b.x - center.x >= 0)
return false;
if (a.x - center.x == 0 && b.x - center.x == 0) {
if (a.y - center.y >= 0 || b.y - center.y >= 0)
return a.y > b.y;
return b.y > a.y;
}
// compute the cross product of vectors (center -> a) x (center -> b)
int det = (a.x - center.x) * (b.y - center.y) - (b.x - center.x) * (a.y - center.y);
if (det < 0)
return true;
if (det > 0)
return false;
// points a and b are on the same line from the center
// check which point is closer to the center
int d1 = (a.x - center.x) * (a.x - center.x) + (a.y - center.y) * (a.y - center.y);
int d2 = (b.x - center.x) * (b.x - center.x) + (b.y - center.y) * (b.y - center.y);
return d1 > d2;
}
This will order the points clockwise starting from the 12 o'clock. Points on the same "hour" will be ordered starting from the ones that are further from the center.
If using integer types (which are not really present in Lua) you'd have to assure that det, d1 and d2 variables are of a type that will be able to hold the result of performed calculations.
If you want to achieve something looking solid, as convex as possible, then I guess you're looking for a Convex Hull. You can compute it using the Graham Scan.
In this algorithm, you also have to sort the points clockwise (or counter-clockwise) starting from a special pivot point. Then you repeat simple loop steps each time checking if you turn left or right adding new points to the convex hull, this check is based on a cross product just like in the above comparison function.
Edit:
Added one more if statement if (a.y - center.y >= 0 || b.y - center.y >=0) to make sure that points that have x=0 and negative y are sorted starting from the ones that are further from the center. If you don't care about the order of points on the same 'hour' you can omit this if statement and always return a.y > b.y.
Corrected the first if statements with adding -center.x and -center.y.
Added the second if statement (a.x - center.x < 0 && b.x - center.x >= 0). It was an obvious oversight that it was missing. The if statements could be reorganized now because some checks are redundant. For example, if the first condition in the first if statement is false, then the first condition of the second if must be true. I decided, however, to leave the code as it is for the sake of simplicity. It's quite possible that the compiler will optimize the code and produce the same result anyway.
What you're asking for is a system known as polar coordinates. Conversion from Cartesian to polar coordinates is easily done in any language. The formulas can be found in this section.
After converting to polar coordinates, just sort by the angle, theta.
An interesting alternative approach to your problem would be to find the approximate minimum to the Traveling Salesman Problem (TSP), ie. the shortest route linking all your points. If your points form a convex shape, it should be the right solution, otherwise, it should still look good (a "solid" shape can be defined as one that has a low perimeter/area ratio, which is what we are optimizing here).
You can use any implementation of an optimizer for the TSP, of which I am pretty sure you can find a ton in your language of choice.
Another version (return true if a comes before b in counterclockwise direction):
bool lessCcw(const Vector2D ¢er, const Vector2D &a, const Vector2D &b) const
{
// Computes the quadrant for a and b (0-3):
// ^
// 1 | 0
// ---+-->
// 2 | 3
const int dax = ((a.x() - center.x()) > 0) ? 1 : 0;
const int day = ((a.y() - center.y()) > 0) ? 1 : 0;
const int qa = (1 - dax) + (1 - day) + ((dax & (1 - day)) << 1);
/* The previous computes the following:
const int qa =
( (a.x() > center.x())
? ((a.y() > center.y())
? 0 : 3)
: ((a.y() > center.y())
? 1 : 2)); */
const int dbx = ((b.x() - center.x()) > 0) ? 1 : 0;
const int dby = ((b.y() - center.y()) > 0) ? 1 : 0;
const int qb = (1 - dbx) + (1 - dby) + ((dbx & (1 - dby)) << 1);
if (qa == qb) {
return (b.x() - center.x()) * (a.y() - center.y()) < (b.y() - center.y()) * (a.x() - center.x());
} else {
return qa < qb;
}
}
This is faster, because the compiler (tested on Visual C++ 2015) doesn't generate jump to compute dax, day, dbx, dby. Here the output assembly from the compiler:
; 28 : const int dax = ((a.x() - center.x()) > 0) ? 1 : 0;
vmovss xmm2, DWORD PTR [ecx]
vmovss xmm0, DWORD PTR [edx]
; 29 : const int day = ((a.y() - center.y()) > 0) ? 1 : 0;
vmovss xmm1, DWORD PTR [ecx+4]
vsubss xmm4, xmm0, xmm2
vmovss xmm0, DWORD PTR [edx+4]
push ebx
xor ebx, ebx
vxorps xmm3, xmm3, xmm3
vcomiss xmm4, xmm3
vsubss xmm5, xmm0, xmm1
seta bl
xor ecx, ecx
vcomiss xmm5, xmm3
push esi
seta cl
; 30 : const int qa = (1 - dax) + (1 - day) + ((dax & (1 - day)) << 1);
mov esi, 2
push edi
mov edi, esi
; 31 :
; 32 : /* The previous computes the following:
; 33 :
; 34 : const int qa =
; 35 : ( (a.x() > center.x())
; 36 : ? ((a.y() > center.y()) ? 0 : 3)
; 37 : : ((a.y() > center.y()) ? 1 : 2));
; 38 : */
; 39 :
; 40 : const int dbx = ((b.x() - center.x()) > 0) ? 1 : 0;
xor edx, edx
lea eax, DWORD PTR [ecx+ecx]
sub edi, eax
lea eax, DWORD PTR [ebx+ebx]
and edi, eax
mov eax, DWORD PTR _b$[esp+8]
sub edi, ecx
sub edi, ebx
add edi, esi
vmovss xmm0, DWORD PTR [eax]
vsubss xmm2, xmm0, xmm2
; 41 : const int dby = ((b.y() - center.y()) > 0) ? 1 : 0;
vmovss xmm0, DWORD PTR [eax+4]
vcomiss xmm2, xmm3
vsubss xmm0, xmm0, xmm1
seta dl
xor ecx, ecx
vcomiss xmm0, xmm3
seta cl
; 42 : const int qb = (1 - dbx) + (1 - dby) + ((dbx & (1 - dby)) << 1);
lea eax, DWORD PTR [ecx+ecx]
sub esi, eax
lea eax, DWORD PTR [edx+edx]
and esi, eax
sub esi, ecx
sub esi, edx
add esi, 2
; 43 :
; 44 : if (qa == qb) {
cmp edi, esi
jne SHORT $LN37#lessCcw
; 45 : return (b.x() - center.x()) * (a.y() - center.y()) < (b.y() - center.y()) * (a.x() - center.x());
vmulss xmm1, xmm2, xmm5
vmulss xmm0, xmm0, xmm4
xor eax, eax
pop edi
vcomiss xmm0, xmm1
pop esi
seta al
pop ebx
; 46 : } else {
; 47 : return qa < qb;
; 48 : }
; 49 : }
ret 0
$LN37#lessCcw:
pop edi
pop esi
setl al
pop ebx
ret 0
?lessCcw##YA_NABVVector2D##00#Z ENDP ; lessCcw
Enjoy.
vector3 a = new vector3(1 , 0 , 0)..............w.r.t X_axis
vector3 b = any_point - Center;
- y = |a * b| , x = a . b
- Atan2(y , x)...............................gives angle between -PI to + PI in radians
- (Input % 360 + 360) % 360................to convert it from 0 to 2PI in radians
- sort by adding_points to list_of_polygon_verts by angle we got 0 to 360
Finally you get Anticlockwize sorted verts
list.Reverse()..................Clockwise_order
I know this is somewhat of an old post with an excellent accepted answer, but I feel like I can still contribute something useful. All the answers so far essentially use a comparison function to compare two points and determine their order, but what if you want to use only one point at a time and a key function?
Not only is this possible, but the resulting code is also extremely compact. Here is the complete solution using Python's built-in sorted function:
# Create some random points
num = 7
points = np.random.random((num, 2))
# Compute their center
center = np.mean(points, axis=0)
# Make arctan2 function that returns a value from [0, 2 pi) instead of [-pi, pi)
arctan2 = lambda s, c: angle if (angle := np.arctan2(s, c)) >= 0 else 2 * np.pi + angle
# Define the key function
def clockwise_around_center(point):
diff = point - center
rcos = np.dot(diff, center)
rsin = np.cross(diff, center)
return arctan2(rsin, rcos)
# Sort our points using the key function
sorted_points = sorted(points, key=clockwise_around_center)
This answer would also work in 3D, if the points are on a 2D plane embedded in 3D. We would only have to modify the calculation of rsin by dotting it with the normal vector of the plane. E.g.
rsin = np.dot([0,0,1], np.cross(diff, center))
if that plane has e_z as its normal vector.
The advantage of this code is that it works on only one point at the time using a key function. The quantity rsin, if you work it out on a coefficient level, is exactly the same as what is called det in the accepter answer, except that I compute it between point - center and center, not between point1 - center and point2 - center. But the geometrical meaning of this quantity is the radius times the sin of the angle, hence I call this variable rsin. Similarly for the dot product, which is the radius times the cosine of the angle and hence called rcos.
One could argue that this solution uses arctan2, and is therefore less clean. However, I personally think that the clearity of using a key function outweighs the need for one call to a trig function. Note that I prefer to have arctan2 return a value from [0, 2 pi), because then we get the angle 0 when point happens to be identical to center, and thus it will be the first point in our sorted list. This is an optional choice.
In order to understand why this code works, the crucial insight is that all our points are defined as arrows with respect to the origin, including the center point itself. So if we calculate point - center, this is equivalent to placing the arrow from the tip of center to the tip of point, at the origin. Hence we can sort the arrow point - center with respect to the angle it makes with the arrow pointing to center.
Here's a way to sort the vertices of a rectangle in clock-wise order. I modified the original solution provided by pyimagesearch and got rid of the scipy dependency.
import numpy as np
def pointwise_distance(pts1, pts2):
"""Calculates the distance between pairs of points
Args:
pts1 (np.ndarray): array of form [[x1, y1], [x2, y2], ...]
pts2 (np.ndarray): array of form [[x1, y1], [x2, y2], ...]
Returns:
np.array: distances between corresponding points
"""
dist = np.sqrt(np.sum((pts1 - pts2)**2, axis=1))
return dist
def order_points(pts):
"""Orders points in form [top left, top right, bottom right, bottom left].
Source: https://www.pyimagesearch.com/2016/03/21/ordering-coordinates-clockwise-with-python-and-opencv/
Args:
pts (np.ndarray): list of points of form [[x1, y1], [x2, y2], [x3, y3], [x4, y4]]
Returns:
[type]: [description]
"""
# sort the points based on their x-coordinates
x_sorted = pts[np.argsort(pts[:, 0]), :]
# grab the left-most and right-most points from the sorted
# x-roodinate points
left_most = x_sorted[:2, :]
right_most = x_sorted[2:, :]
# now, sort the left-most coordinates according to their
# y-coordinates so we can grab the top-left and bottom-left
# points, respectively
left_most = left_most[np.argsort(left_most[:, 1]), :]
tl, bl = left_most
# now that we have the top-left coordinate, use it as an
# anchor to calculate the Euclidean distance between the
# top-left and right-most points; by the Pythagorean
# theorem, the point with the largest distance will be
# our bottom-right point. Note: this is a valid assumption because
# we are dealing with rectangles only.
# We need to use this instead of just using min/max to handle the case where
# there are points that have the same x or y value.
D = pointwise_distance(np.vstack([tl, tl]), right_most)
br, tr = right_most[np.argsort(D)[::-1], :]
# return the coordinates in top-left, top-right,
# bottom-right, and bottom-left order
return np.array([tl, tr, br, bl], dtype="float32")
With numpy:
import matplotlib.pyplot as plt
import numpy as np
# List of coords
coords = np.array([7,7, 5, 0, 0, 0, 5, 10, 10, 0, 0, 5, 10, 5, 0, 10, 10, 10]).reshape(-1, 2)
centroid = np.mean(coords, axis=0)
sorted_coords = coords[np.argsort(np.arctan2(coords[:, 1] - centroid[1], coords[:, 0] - centroid[0])), :]
plt.scatter(coords[:,0],coords[:,1])
plt.plot(coords[:,0],coords[:,1])
plt.plot(sorted_coords[:,0],sorted_coords[:,1])
plt.show()