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Given is a bitstream (continuous string of bits too long to be processed at once) and the result should be a matching stream of base20 numbers.
The process is simple for a small number of bits:
Assuming most significant bit right:
110010011 = decimal 403 (1 * 1 + 1 * 2 + 1 * 16 + 1 * 128 + 1 * 256)
403 / 20 = 20 R 3
20 / 20 = 1 R 0
1 / 20 = 0 R 1
Result is [3, 0, 1] = 3 * 1 + 0 * 20 + 1 * 400
But what if the bits are too much to be converted to a decimal number in one step?
My approach was doing both processes in a loop: Convert the bits to decimal and converting the decimal down to base20 numbers. This process requires the multipliers (position values) to be lowered while walking through the bits, because otherwise, they'll quickly increase too much to be calculated probably. The 64th bit would have been multiplied by 2^64 and so on.
note: I understood the question that a bitstream is arriving of unknown length and during an unknown duration and a live conversion from base 2 to base 20 should be made.
I do not believe this can be done in a single go. The problem is that base 20 and base 2 have no common ground and the rules of modular arithmetic do not allow to solve the problem cleanly.
(a+b) mod n = ( (a mod n) + (b mod n) ) mod n
(a*b) mod n = ( (a mod n) * (b mod n) ) mod n
(a^m) mod n = ( (a mod n)^m ) mod n
Now if you have a number A written in base p and q (p < q) as
A = Sum[a[i] p^i, i=0->n] = Sum[b[i] q^i, i=0->n]
Then we know that b[0] = A mod q. However, we do not know A and hence, the above tells us that
b[0] = A mod q = Sum[a[i] p^i, i=0->n] mod q
= Sum[ (a[i] p^i) mod q, i=0->n] mod q
= Sum[ ( (a[i] mod q) (p^i mod q) ) mod q, i=0->n] mod q
This implies that:
If you want to know the lowest digit b0 of a number in base q, you need to have the knowledge of the full number.
This can only be simplified if q = pm as
b[0] = A mod q = Sum[a[i] p^i, i=0->n] mod q
= Sum[ (a[i] p^i) mod q, i=0->n] mod q
= Sum[ a[i] p^i, i=0->m-1]
So in short, since q = 20 and p = 2. I have to say, no, it can not be done in a single pass. Furthermore, remind yourself that I only spoke about the first digit in base q and not yet the ith digit.
As an example, imagine a bit stream of 1000 times 0 followed by a single 1. This resembles the number 21000. The first digit is easy, but to get any other digit ... you are essentially in a rather tough spot.
I am trying to Implement gradient Descent algorithm from scratch to find the slope and intercept value for my linear fit line.
Using the package and calculating slope and intercept, I get slope = 0.04 and intercept = 7.2 but when I use my gradient descent algorithm for the same problem, I get slope and intercept both values = (-infinity,-infinity)
Here is my code
x= [1,2,3,4,5,6,7,8,9,10,11,12,13,141,5,16,17,18,19,20]
y=[2,3,4,5,6,7,8,9,10,11,12,13,141,5,16,17,18,19,20,21]
function GradientDescent()
m=0
c=0
for i=1:10000
for k=1:length(x)
Yp = m*x[k] + c
E = y[k]-Yp #error in predicted value
dm = 2*E*(-x[k]) # partial derivation of cost function w.r.t slope(m)
dc = 2*E*(-1) # partial derivate of cost function w.r.t. Intercept(c)
m = m + (dm * 0.001)
c = c + (dc * 0.001)
end
end
return m,c
end
Values = GradientDescent() # after running values = (-inf,-inf)
I have not done the math, but instead wrote the tests. It seems you got a sign error when assigning m and c.
Also, writing the tests really helps, and Julia makes it simple :)
function GradientDescent(x, y)
m=0.0
c=0.0
for i=1:10000
for k=1:length(x)
Yp = m*x[k] + c
E = y[k]-Yp
dm = 2*E*(-x[k])
dc = 2*E*(-1)
m = m - (dm * 0.001)
c = c - (dc * 0.001)
end
end
return m,c
end
using Base.Test
#testset "gradient descent" begin
#testset "slope $slope" for slope in [0, 1, 2]
#testset "intercept for $intercept" for intercept in [0, 1, 2]
x = 1:20
y = broadcast(x -> slope * x + intercept, x)
computed_slope, computed_intercept = GradientDescent(x, y)
#test slope ≈ computed_slope atol=1e-8
#test intercept ≈ computed_intercept atol=1e-8
end
end
end
I can't get your exact numbers, but this is close. Perhaps it helps?
# 141 ?
datax = [1,2,3,4,5,6,7,8,9,10,11,12,13,141,5,16,17,18,19,20]
datay = [2,3,4,5,6,7,8,9,10,11,12,13,141,5,16,17,18,19,20,21]
function gradientdescent()
m = 0
b = 0
learning_rate = 0.00001
for n in 1:10000
for i in 1:length(datay)
x = datax[i]
y = datay[i]
guess = m * x + b
error = y - guess
dm = 2error * x
dc = 2error
m += dm * learning_rate
b += dc * learning_rate
end
end
return m, b
end
gradientdescent()
(-0.04, 17.35)
It seems that adjusting the learning rate is critical...
i have a setup like this:
2 coordinate systems. (x,y) is the main coordinate system and (x',y') is a coordinate system that lives inside (x,y). The system (x',y') is defined by the points x1 or x2 and if i move these 2 points around then (x',y') moves accordingly. The origin of (x',y') is defined as the middle of the vector going from x1 to x2, and the y' axis is the normal vector on x1->x2 going through the origin. If i have a point x3 defined in (x',y') and i move either of x1 or x2 to make the origin shift place, how do i then move x3 accordingly such that it maintains its position in the new (x',y') ?
And how do i make a transformation which always converts a point in (x,y) to a point in (x',y') nomatter how x1 and x2 have been set?
I was thinking that if i had more points than just the one i am moving (x1 or x2) i guess i could try to estimate theta, tx, ty of the transformation
[x2'] [cos(theta) , sin(theta), tx][x2]
[y2'] = [-sin(theta), cos(theta), ty][y2]
[ 1 ] [ 0 , 0 , 1 ][1 ]
and just apply that estimated transformation to x3 and i would be good...mmm but i think i would need 3 points in order to estimate theta, tx and ty right?
I mean i could estimate using some least squares approach...but 3 unknowns requires 3 coordinate sets right?
I tried to implement this and calculate an example. I hope you understand the syntax. Its not really giving me what i expect:
import math
import numpy as np
x1=[ 0,10]
x2=[10,20]
rx = x2[0] - x1[0]
ry = x2[1] - x1[1]
rlen = math.sqrt(rx*rx+ry*ry)
c = rx / rlen
s = ry / rlen
dx = - ( x1[0] + x2[0] )/2 # changing the sign to be negative seems to
dy = - ( x1[1] + x2[1] )/2 # rectify translation. Rotation still is wrong
M = np.array([[c, -s, 0],[s, c, 0],[dx, dy, 1]])
print( np.dot(x2 + [1],M) )
# Yields -> [ 15.92031022 -8.63603897 1. ] and should yield [5,0,1]
Since I am trying to transform the x2 coordinate, should the result then not have the value 0 in the y-component since its located in the x-axis?
Ok, I tried doing the implementation for x3 from dynamic1 to dynamic2 which the check is that x3 should end up with the same coordinate in both d1 and d2. I did that as you suggested, but I do not get the same coordinate in both d1 and d2. Did i misunderstand something?
import math
import numpy as np
x1=[ 1,1]
x2=[ 7,9]
x3=[4,3]
rx = (x2[0] - x1[0])
ry = (x2[1] - x1[1])
rlen = math.sqrt( rx*rx + ry*ry )
c = rx / rlen
s = ry / rlen
dx = ( x1[0] + x2[0] )/2
dy = ( x1[1] + x2[1] )/2
M = np.array([[c, -s, 0],[s, c, 0],[-dx*c-dy*s, dx*s-dy*c, 1]])
Minv = np.array([[c, s, 0],[-s, c, 0],[dx, dy, 1]])
x1new=[ 1,1]
x2new=[ 17,4]
rxnew = (x2new[0] - x1new[0])
rynew = (x2new[1] - x1new[1])
rlennew = math.sqrt( rxnew*rxnew + rynew*rynew )
cnew = rxnew / rlennew
snew = rynew / rlennew
dxnew = ( x1new[0] + x2new[0] )/2
dynew = ( x1new[1] + x2new[1] )/2
Mnew = np.array([[cnew, -snew, 0],[snew, cnew, 0],[-dxnew*cnew-dynew*snew, dxnew*snew-dynew*cnew, 1]])
Mnewinv = np.array([[cnew, snew, 0],[-snew, cnew, 0],[dxnew, dynew, 1]])
M_dyn1_to_dyn2 = np.dot(Minv,Mnew)
print( np.dot(x3 + [1], M) )
print( np.dot(x3 + [1], M_dyn1_to_dyn2))
#yields these 2 outputs which should be the same:
[-1.6 -1.2 1. ]
[-3.53219692 8.29298408 1. ]
Edit. Matrix correction.
To translate coordinates from static system to (x1,x2) defined one, you have to apply affine transformation.
Matrix of this transformation M consists of shift matrix S and rotation about origin R.
Matrix M is combination of S and R:
c -s 0
M = s c 0
-dx*c-dy*s dx*s-dy*c 1
Here c and s are cosine and sine of rotation angle, their values are respectively x- and y- components of unit (normalized) vector x1x2.
rx = x2.x - x1.x
ry = x2.y - x1.y
len = Sqrt(rx*rx+ry*ry)
c = rx / Len
s = ry / Len
And shift components:
dx = (x1.x + x2.x)/2
dy = (x1.y + x2.y)/2
To translate (xx,yy) coordinates from static system to rotate one, we have to find
xx' = xx*c+yy*s-dx*c-dy*s = c*(xx-dx) + s*(yy-dy)
yy' = -xx*s+yy*c+dx*s-dy*c = -s*(xx-dx) + c*(yy-dy)
Quick check:
X1 = (1,1)
X2 = (7,9)
dx = 4
dy = 5
rx = 6
ry = 8
Len = 10
c = 0.6
s = 0.8
for point (4,5):
xx-dx = 0
yy-dy = 0
xx',yy' = (0, 0) - right
for point X2 =(7,9):
xx-dx = 3
yy-dy = 4
xx' = 0.6*3 + 0.8*4 = 5 -right
yy' = -0.8*3 + 0.6*4 = 0 -right
P.S. Note that matrix to transform dyn.coordinates to static ones is inverse of M and it is simpler:
c s 0
M' = -s c 0
dx dy 1
P.P.S. You need three pairs of corresponding points to define general affine transformations. It seems here you don't need scaling and sheer, so you may determine needed transform with your x1,x2 points
I think you need double dimension array to save and set your value in that
the structure gonna be like this
=============|========|========|
index number |x |y |
=============|========|========|
first point | [0][0] | [0][1] |
second point | [1][0] | [1][1] |
third point | [2][0] | [2][1] |
=============|========|========|
I will use java in my answer
//declare the double dimension array
double matrix[][] = new double[3][2];
//setting location first point, x
matrix[0][0] = 1;
//setting location first point, y
matrix[0][1] = 1;
//fill with your formula, i only give example
//fill second point with first point and plus 1
//setting location second point, x
matrix[1][0] = matrix[0][0] + 1;
//setting location second point, y
matrix[1][1] = matrix[0][1] + 1;
//fill with your formula, i only give example
//fill third point with second point and plus 1
//setting location third point, x
matrix[2][0] = matrix[1][0] + 1;
//setting location third point, y
matrix[2][1] = matrix[1][1] + 1;
I have this recurrence formula:
P(n) = ( P(n-1) + 2^(n/2) ) % (X)
s.t. P(1) = 2;
where n/2 is computer integer division i.e. floor of x/2
Since i am taking mod X, this relation should repeat at least with in X outputs.
but it can start repeating before that.
How to find this value?
It needn't repeat within x terms, consider x = 3:
P(1) = 2
P(2) = (P(1) + 2^(2/2)) % 3 = 4 % 3 = 1
P(3) = (P(2) + 2^(3/2)) % 3 = (1 + 2) % 3 = 0
P(4) = (P(3) + 2^(4/2)) % 3 = 4 % 3 = 1
P(5) = (P(4) + 2^(5/2)) % 3 = (1 + 4) % 3 = 2
P(6) = (P(5) + 2^(6/2)) % 3 = (2 + 8) % 3 = 1
P(7) = (P(6) + 2^(7/2)) % 3 = (1 + 8) % 3 = 0
P(8) = (P(7) + 2^(8/2)) % 3 = 16 % 3 = 1
P(9) = (P(8) + 2^(9/2)) % 3 = (1 + 16) % 3 = 2
P(10) = (P(9) + 2^(10/2)) % 3 = (2 + 32) % 3 = 1
P(11) = (P(10) + 2^(11/2)) % 3 = (1 + 32) % 3 = 0
P(12) = (P(11) + 2^(12/2)) % 3 = (0 + 64) % 3 = 1
and you see that the period is 4.
Generally (suppose X is odd, it's a bit more involved for even X), let k be the period of 2 modulo X, i.e. k > 0, 2^k % X = 1, and k is minimal with these properties (see below).
Consider all arithmetic modulo X. Then
n
P(n) = 2 + ∑ 2^(j/2)
j=2
It is easier to see when we separately consider odd and even n:
m m
P(2*m+1) = 2 + 2 * ∑ 2^i = 2 * ∑ 2^i = 2*(2^(m+1) - 1) = 2^((n+2)/2) + 2^((n+1)/2) - 2
i=1 i=0
since each 2^j appears twice, for j = 2*i and j = 2*i+1. For even n = 2*m, there's one summand 2^m missing, so
P(2*m) = 2^(m+1) + 2^m - 2 = 2^((n+2)/2) + 2^((n+1)/2) - 2
and we see that the length of the period is 2*k, since the changing parts 2^((n+1)/2) and 2^((n+2)/2) have that period. The period immediately begins, there is no pre-period part (there can be a pre-period for even X).
Now k <= φ(X) by Euler's generalisation of Fermat's theorem, so the period is at most 2 * φ(X).
(φ is Euler's totient function, i.e. φ(n) is the number of integers 1 <= k <= n with gcd(n,k) = 1.)
What makes it possible that the period is longer than X is that P(n+1) is not completely determined by P(n), the value of n also plays a role in determining P(n+1), in this case the dependence is simple, each power of 2 being used twice in succession doubles the period of the pure powers of 2.
Consider the sequence a[k] = (2^k) % X for odd X > 1. It has the simple recurrence
a[0] = 1
a[k+1] = (2 * a[k]) % X
so each value completely determines the next, thus the entire following part of the sequence. (Since X is assumed odd, it also determines the previous value [if k > 0] and thus the entire previous part of the sequence. With H = (X+1)/2, we have a[k-1] = (H * a[k]) % X.)
Hence if the sequence assumes one value twice (and since there are only X possible values, that must happen within the first X+1 values), at indices i and j = i+p > i, say, the sequence repeats and we have a[k+p] = a[k] for all k >= i. For odd X, we can go back in the sequence, therefore a[k+p] = a[k] also holds for 0 <= k < i. Thus the first value that occurs twice in the sequence is a[0] = 1.
Let p be the smallest positive integer with a[p] = 1. Then p is the length of the smallest period of the sequence a, and a[k] = 1 if and only if k is a multiple of p, thus the set of periods of a is the set of multiples of p. Euler's theorem says that a[φ(X)] = 1, from that we can conclude that p is a divisor of φ(X), in particular p <= φ(X) < X.
Now back to the original sequence.
P(n) = 2 + a[1] + a[1] + a[2] + a[2] + ... + a[n/2]
= a[0] + a[0] + a[1] + a[1] + a[2] + a[2] + ... + a[n/2]
Since each a[k] is used twice in succession, it is natural to examine the subsequences for even and odd indices separately,
E[m] = P(2*m)
O[m] = P(2*m+1)
then the transition from one value to the next is more regular. For the even indices we find
E[m+1] = E[m] + a[m] + a[m+1] = E[m] + 3*a[m]
and for the odd indices
O[m+1] = O[m] + a[m+1] + a[m+1] = O[m] + 2*a[m+1]
Now if we ignore the modulus for the moment, both E and O are geometric sums, so there's an easy closed formula for the terms. They have been given above (in slightly different form),
E[m] = 3 * 2^m - 2 = 3 * a[m] - 2
O[m] = 2 * 2^(m+1) - 2 = 2 * a[m+1] - 2 = a[m+2] - 2
So we see that O has the same (minimal) period as a, namely p, and E also has that period. Unless maybe if X is divisible by 3, that is also the minimal (positive) period of E (if X is divisible by 3, the minimal positive period of E could be a proper divisor of p, for X = 3 e.g., E is constant).
Thus we see that 2*p is a period of the sequence P obtained by interlacing E and O.
It remains to be seen that 2*p is the minimal positive period of P. Let m be the minimal positive period. Then m is a divisor of 2*p.
Suppose m were odd, m = 2*j+1. Then
P(1) = P(m+1) = P(2*m+1)
P(2) = P(m+2) = P(2*m+2)
and consequently
P(2) - P(1) = P(m+2) - P(m+1) = P(2*m+2) - P(2*m+1)
But P(2) - P(1) = a[1] and
P(m+2) - P(m+1) = a[(m+2)/2] = a[j+1]
P(2*m+2) - P(2*m+1) = a[(2*m+2)/2] = a[m+1] = a[2*j+2]
So we must have a[1] = a[j+1], hence j is a period of a, and a[j+1] = a[2*j+2], hence j+1 is a period of a too. But that means that 1 is a period of a, which implies X = 1, a contradiction.
Therefore m is even, m = 2*j. But then j is a period of O (and of E), thus a multiple of p. On the other hand, m <= 2*p implies j <= p, and the only (positive) multiple of p satisfying that inequality is p itself, hence j = p, m = 2*p.
This may be quite a basic question for someone who knows linear programming.
In most of the problems that I saw on LP has somewhat similar to following format
max 3x+4y
subject to 4x-5y = -34
3x-5y = 10 (and similar other constraints)
So in other words, we have same number of unknown in objective and constraint functions.
My problem is that I have one unknown variable in objective function and 3 unknowns in constraint functions.
The problem is like this
Objective function: min w1
subject to:
w1 + 0.1676x + 0.1692y >= 0.1666
w1 - 0.1676x - 0.1692y >= -0.1666
w1 + 0.3039x + 0.3058y >= 0.3
w1 - 0.3039x - 0.3058y >= -0.3
x + y = 1
x >= 0
y >= 0
As can be seen, the objective function has only one unknown i.e. w1 and constraint functions have 3 (or lets say 2) unknown i.e w1, x and y.
Can somebody please guide me how to solve this problem, especially using R or MATLAB linear programming toolbox.
Your objective only involves w1 but you can still view it as a function of w1,x,y, where the coefficient of w1 is 1, and the coeffs of x,y are zero:
min w1*1 + x*0 + y*0
Once you see this you can formulate it in the usual way as a "standard" LP.
Prasad is correct. The number of unknowns in the objective function does not matter. You can view unknowns that are not present as having a zero coefficient.
This LP is easily solved using Matlab's linprog function. For more
details on linprog see the documentation here.
% We lay out the variables as X = [w1; x; y]
c = [1; 0; 0]; % The objective is w1 = c'*X
% Construct the constraint matrix
% Inequality constraints will be written as Ain*X <= bin
% w1 x y
Ain = [ -1 -0.1676 -0.1692;
-1 0.1676 0.1692;
-1 -0.3039 -0.3058;
-1 0.3039 0.3058;
];
bin = [ -0.166; 0.166; -0.3; 0.3];
% Construct equality constraints Aeq*X == beq
Aeq = [ 0 1 1];
beq = 1;
%Construct lower and upper bounds l <= X <= u
l = [ -inf; 0; 0];
u = inf(3,1);
% Solve the LP using linprog
[X, optval] = linprog(c,Ain,bin,Aeq,beq,l,u);
% Extract the solution
w1 = X(1);
x = X(2);
y = X(3);