text:
Check if a point is inside a polygon

To check if a given point is inside a polygon or not is a very useful piece of code.

Here is my implementation in JavaScript of an algorithm counting the number of times a ray crosses the perimeter of the polygon, and subsequently checking the parity. I have included an initial test doing a fast comparison with the rectangular bounding box of the polygon, and also support for the possibility that the bounding box and/or if the polygon is convex is precalculated.

Here is my function:

pointInsidePolygon = function(x,y,polygon){//returns true/false
//also consider to use the'pointInsiderectangle' function for simpler tasks
/* sturcture of the polygon object:
{
corners:[
{
x:number,
y:number
}//repeat for any number of corners > 2
]
//optional, if it has a precomputed bounding box, it has these properties as well:
xmin:number,
xmax:number,
ymin:number,
ymax:number,
convex:boolean
}
*/
//fast "false/maybe" test: check if the point is inside the outer bounding box of the polygon.
if(polygon.xmin === undefined){//if the bounding box of the polygon is not precomputed, then compute it.
var xmin = xmax = polygon.corners[0].x;
var ymin = ymax = polygon.corners[0].y;
for(var i = polygon.corners.length;--i;){
if(polygon.corners[i].x < xmin){
xmin = polygon.corners[i].x;
}
else if(polygon.corners[i].x > xmax){
xmax = polygon.corners[i].x;
};
if(polygon.corners[i].y < ymin){
ymin = polygon.corners[i].y;
}
else if(polygon.corners[i].y > ymax){
ymax = polygon.corners[i].y;
};
};
polygon.xmin = xmin;
polygon.xmax = xmax;
polygon.ymin = ymin;
polygon.ymax = ymax;
};
if(x < polygon.xmin || x > polygon.xmax){
return false;
}
else if(y < polygon.xmin || x > polygon.ymax){
return false;
};
/* complete "true/false" test: check the number of times a ray from the point intersects the perimeter of the polygon
*  odd number  == inside
*  even number == outside
* I cast the ray along the x axis
*/
var intersections = 0;
for(var i = polygon.corners.length;i--;){
var xpos = polygon.corners[i].x - x;//the polygon is translated with the test point in origo
var ypos = polygon.corners[i].y - y;
if(i){
var xpos2 = polygon.corners[i - 1].x - x;
var ypos2 = polygon.corners[i - 1].y - y;
}
else{//the last side of the polygon wraps around to the first point of the corner list, it requires special handling.
var xpos2 = polygon.corners[polygon.corners.length - 1].x - x;
var ypos2 = polygon.corners[polygon.corners.length - 1].y - y;
};
//each side of the polygon does potentially intersect the ray. I skip the iteration for those that do not.
if(xpos < 0 && xpos2 < 0){//the ray can not cross a side starting and ending at negative x, so no intersection then
continue;
}
else if(ypos * ypos2 > 0){//still no intersection if both y's have the same sign
continue;
}
else if(xpos + (xpos2 - xpos) * ypos/(ypos - ypos2) < 0){//intersections of the x-axis before x=0 do not count
continue;
}
else if(ypos === 0){//a corner may be exactly at the ray
continue;
};
intersections++;
if(polygon.convex && intersections === 2){//a ray through a convex polygon can intersect at most 2 times
return false;
};
};
if(intersections % 2 === 0){
return false;
};
return true;
};

It seems like somebody has already posted something similar in Python, but it was implemented quite differently.

• 'A corner is exactly at the ray' is the trickiest point. Consider 3 consecutive corners, all with positive x and (case 1) negative, zero and negative y or (case 2) negative, zero and positive y. You should get a single intersection in the case 1 and no (or two) intersections in case 2. – CiaPan Sep 19 '16 at 12:29
• @CiaPan Good catch. That is certainly a weakness with the current algorithm. – Hohmannfan Sep 19 '16 at 12:47
• Whoops, I have mismatched example data and results: case 1 (minus, zero, minus) should give no intersection (the polygon's edge turns back at the ray) while case 2 (minus, zero, plus) should give a single intersection (the edge actually crosses the ray). I'm sorry for that. – CiaPan Sep 19 '16 at 13:45
• There are even more complicated cases when several consecutive vertices lay at the ray... However, you can easily solve them all at once: treat each vertex at the ray as if it lays slightly above the ray. :) Simply replace y==0 with some constant epsilon. – CiaPan Sep 19 '16 at 13:47
• The real problem would be a polygon's side coincident with the ray with one vertex at negative and the other one at positive x (that is your point belonging to a 'horizontal' line segment being the polygon's side). That however appears a special case of a general decision whether the point on the border is inside or outside the polygon. – CiaPan Sep 19 '16 at 13:50


text:
• https://me.yahoo.com
COST (GBP)
0.37
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0

# Rank

Computes the rank of the given permutation, as generated by Permutation Class 2.
Controller: CodeCogs
Contents

C++

## Rank

 intrank( int n int* p )
This function calculates at which call the Permutation Class 2 component would generate the given permutation. The value of the rank will be between 1 and $\inline&space;n!$. This function may be used with the Maths/Combinatorics/Permutations/Unrank component in order to give a method of recognizing certain permutations only by their rank number.

## Example:

#include <iostream>
int main()
{
int sigma[3] = {2, 3, 1};
std::cout << "The rank of the Sigma permutation is: ";
std::cout << Maths::Combinatorics::Permutations::rank(3, sigma) << std::endl;
return 0;
}

## Output:

The rank of the Sigma permutation is: 5

## References:

SUBSET, a C++ library of combinatorial routines, http://www.csit.fsu.edu/~burkardt/cpp_src/subset/subset.html

### Parameters

 n the size of the permutation p the actual permutation stored as an array

### Returns

the rank of the given permutation, as generated by the <em> Permutation Class 2 </em> component

### Authors

Lucian Bentea (August 2005)
##### Source Code

Source code is available when you agree to a GP Licence or buy a Commercial Licence.

Not a member, then Register with CodeCogs. Already a Member, then Login.

format: plain


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# Turbulence scales

Although a huge number of theoretical, experimental and computational investigations on turbulence have been performed, the structure and description of turbulence still remain open issues. The motivation behind the study of turbulent flows is due to the fact that the vast majority of flows is turbulent and turbulence significantly increases the rates of transport and mixing of matter, momentum and heat in the flows (PopeSB:2000). Turbulence itself is a very complex phenomenon, involving different time and length scales. In the well-known energy-cascade theory, the kinetic energy enters the turbulence at the largest scales of motion through the production mechanism. This energy is transferred from the largest scale (integral scale) to smaller and smaller scales until it is dissipated by viscous action at the smallest scale, the so-called Kolmogorov length scale. The wide range of length-scales and time-scales, the production, the dissipation and the energy transfer between different length-scales make the modelling of turbulence one of the greatest scientific challenge. The level of difficulty in investigating premixed flames is greatly enhanced when also turbulence comes into play, because of the complex interactions between the turbulence itself and the combustion.

The transfer rate of kinetic energy per mass unit of fluid over any particular scale between the macro structure and the micro structure obeys the relationship (this is known as Kolmogorov’s law), e.g., (BatchelorGK:1993):

\varepsilon =-\frac{d}{dt} U_{}^{2} =A\frac{U_{}^{3} }{L} \ \ \ \ \ \ (1)

where A is a constant in the order of unity. Together with the root-mean-square of the instantaneous velocity fluctuations, this expression can be used to characterise the macro structure in terms of a macro velocity scale, a macro length scale and a macro time scale:

The microstructure can be characterised by the Kolmogorov velocity, length and time scale:

where \nu denotes the kinematic viscosity. It should be noted that equation (1) describes the kinetic energy transfer on the larger scales as well as the viscous dissipation on the micro structure despite the apparent absence of the molecular viscosity.

A combustion zone may also be characterised in terms of velocity, length and time scales. If the laminar burning velocity, the laminar flame thickness, and the chemical time scale:

are taken as the fundamental scales of the combustion ((PetersN:1991), (BorghiR:1988)), the structure of a turbulent flame may be considered by relating these combustion scales to those of the turbulent flow field. The occurrence of a particular combustion regime is determined by the value of the Damkohler number:

Da=\frac{L/U}{\tau _{c} } =\frac{\ell _{t} /u_{rms}^{'} }{\delta _{L} /S_{u L} }

When the scales of the micro structure of the turbulent flow field are substituted for L and U, it constitutes the reciprocal Karlovitz number:

Ka^{-1} =\frac{L/U}{\tau _{c} } =\frac{\ell _{K} /u_{K} }{\delta _{L} /S_{u L} }

The Damkohler and Karlovitz number can be used to identify different combustion regimes [14, 15] by employing two dimensionless parameters,

\left(\frac{u_{rms}^{,} }{S_{u L} } \right) and \left(\frac{\ell _{t}^{} }{\delta _{L}^{} } \right)

and by rewriting their definitions into

\frac{u_{rms}^{,} }{S_{u L} } = Da_{}^{-1} \left(\frac{\ell _{t}^{} }{\delta _{L}^{} } \right) (2)

and

\frac{u_{rms}^{,} }{S_{u L} } = Ka_{}^{2/3} \left(\frac{\ell _{t}^{} }{\delta _{L}^{} } \right)_{}^{1/3} (3)

Since the expressions for the laminar burning velocity and the laminar flame thickness (DahoeAE:2003),

\frac{S_{u L} }{S_{u L}^{o}} =\sqrt{\frac{\lambda (T_{u} )}{\lambda (T_{u 0} )} } \left(\frac{T_{u} }{T_{u 0}} \right)\left(\frac{P_{}^{} }{P_{0}^{} } \right)_{}^{\frac{n-2}{2} } \left(\frac{T_{f} }{T_{f}^{o} } \right)^{-\frac{n}{2} } \left(\frac{T_{f} }{T_{f}^{o} } \right)^{\frac{m}{2} } \exp \left[-\frac{Ea}{R} \left(\frac{1}{T_{f}} -\frac{1}{T_{f}^{o} } \right)\right]

\frac{\delta _{L}^{} }{\delta _{L}^{o} } =\sqrt{\frac{\lambda (T_{u})}{\lambda (T_{u 0})} } \left(\frac{P }{P_{0}} \right)^{-\frac{n}{2} } \left(\frac{T_{f}}{T_{f}^{o} } \right)_{}^{\frac{n}{2} } \left(\frac{T_{f}^{} }{T_{f}^{o} } \right)_{}^{-\frac{m}{2} } \exp \left[+\frac{Ea}{R} \left(\frac{1}{T_{f}^{} } -\frac{1}{T_{f}^{o} } \right)\right]

in conjunction with the definition of the thermal conductivity and the viscosity in terms of molecular properties imply that

S_{u L} \delta _{L}^{} =\frac{\lambda }{\hat{C}_{P}^{} } =\frac{\mu }{\rho }

a third relationship may be derived as

\frac{u_{rms}^{,} }{S_{u L} } = Re\left(\frac{\ell _{t}^{} }{\delta _{L}^{} } \right)_{}^{-1} (4)

Equations (2), (3) and (4) may then be used to construct a figure which is known as the Borghi diagram.

Pope S.B. (2000) Turbulent flows. Cambridge University Press, Cambridge, United Kingdom.(BibTeX)
Batchelor G.K. (1993) The theory of homogeneous turbulence. Cambridge University Press.(BibTeX)
Peters N. (1991) Length scales in laminar and turbulent flames. Progress in Astronautics and Aeronautics, 135:155-182.(BibTeX)
Borghi R. (1988) Turbulent Combustion Modelling. Progress in Energy and Combustion Science, 14:245-292.(BibTeX)


text:
Welcome to the broken dawn

# machinery to extract gold from the lithosphere

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text:
# Tag Info

10

Pyth, $(ω^ω, 6)$ >_PE_P Try it online! Compares the reversed prime factorizations of the two inputs. Pyth, $(ε_0, 17)$ L_SmylP#U_dPb>yEy Try it online! L def y(b): Pb prime factors of b m map over d: _d -d U range [-d, …, -1] P# ...

9

///, 80 bytes /!1/1!//!/S'SS'.'//1/S.B'S.B|B'SS.B|S|.SS'B|S'SS|B.SxBxSSxBxS|B..S//B/\\//S/\//! Takes input in unary as 1s appended to the program. Produces output in unary as .s. Try it online! The program sends the ! to the end of the input, then makes several more substitutions. After completing the substitutions that exist in the initial program, each 1 ...

9

K (ngn/k), 43 41 bytes {{+/'y*/:(u*|+\|+/x*u)-x*++\u:=#x}\x:\:x} Try it online! Brutally golfed by ngn. -2 bytes because it is allowed to output starting with A itself. input: matrix A with size n * n {{...}\x:\:x} scan over the list of n copies of A {...} inner fn: x is running matrix, y is A u:=#x u: identity matrix of size n x*++...

9

Haskell, $(\omega, 3)$ This solution is dead simple, really can't be beat in Haskell. (<) Try it online! Haskell, $(\omega+n,19+2\left\lfloor\log_{10}(n+1)\right\rfloor)$ Here's a scheme to get $\omega + n$ for any $n\in\mathbb{N}$. Just replace the 9s below with one more than $n$. This will make the first $n$ numbers bigger than all other ...

7

R, 113 109 103 96 95 bytes Or R>=4.1, 88 bytes by replacing the word function with \. -4 bytes and another -7 thanks to @Dominic van Essen. function(A,B=A)for(i in 2:nrow(A))B=print((diag(sum(b<-diag(B))-cumsum(b))-B*upper.tri(B))%*%A) Try it online! Straightforward approach. The terms on the diagonal are calculated as cumsum-sum. Solution shorter ...

7

JavaScript (V8), 21 bytes b=>g=n=>n&&1+g(n/b|0) Try it online!

7

Python 2, 28 bytes f=lambda x,b:x and-~f(x/b,b) Try it online! thanks @Kevin Cruijssen for -1 byte by using python 2 instead of python 3

7

Pari/GP, 29 bytes n->!sumdiv(n,d,Str(d)>Str(n)) Try it online! Using the hint.

7

05AB1E, 5 bytes Using the hint, suggested by Kevin Cruijssen Ñ§{θQ Try it online! Cast the divisors Ñ to string §, sort {, is the last one θ equal to the input Q? Or 9 bytes without the hint, -1 byte thanks to Kevin Cruijssen: ÑÂøεS∍‹}P Try it online! Ñ # divisors of the input Âø # zip with its reverse ε } # map over the divisor ...

6

Jelly, 13 12 bytes Ø.,æ*ÆḊN,ÆṭƲ Try it online! Takes input as [v, u] on the left, and a on the right. Outputs as [β, α] -1 byte thanks to ovs! Uses xnor's formula that \alpha = \operatorname{tr} \left( \begin{bmatrix} 0 & 1 \\ v & u \end{bmatrix}^a \right) \\ \beta = -(-v)^a = - \det \left( \begin{bmatrix} 0 & 1 \\ v & u \end{bmatrix}^a \...

6

Python 2, 182 bytes def E(a,b): if b:x,y,d=E(b,a%b);return y,x-a/b*y,d return 1,0,a def f(a,b,c,d):x,y,g=E(a,b);z,w,h=E(c,d);j=E(g,h)[2];return(b/g,-a/g,0,0),(0,0,d/h,-c/h),(-x*h/j,-y*h/j,z*g/j,w*g/j) Try it online! Yes, it is possible to solve it without fancy built-ins. And this gives relatively good results for a non-LLL one. How it works I derived a ...

6

Python 3 + numpy, 117, 99, 97, 91, 89(@Dingus), 88 bytes from numpy import* def f(M): N=M;T=triu(M==M) for _ in M:N=T@N[T&T.T,print(N)]*M-T*N@M Try it online! Old version Old version Old version Old version Old version Detailed explanation (refers to the latest and greatest versison (88 bytes)) We start with a sizeable upfront investment to create ...

6

APL (Dyalog Extended), 2 bytes (SBCS) Anonymous tacit infix function taking $b$ as left argument and $n$ as right argument. ≢⊤ Try it online! ≢ tally the digits of ⊤ the anti-base (i.e. the representation in the given base) Note that this is wasteful in that it actually does the base conversion. APL (Dyalog Unicode), 4 bytes Anonymous tacit infix ...

5

Husk, 4 bytes Σṁdḣ Try it online! ḣ makes a range from 1 to input ṁd flatmap digits Σ sum

5

Zsh, 20 bytes My first Zsh answer, cobbled together using these very helpful posts. <<<${#v=$[[##$2]$1]} Try it online or run the test suite. Full program taking $n$ and $b$ as arguments. Commented <<< # print ${# } # length in characters of v= # set dummy variable v to$[[##$2]... 5 Proton, 67 bytes f=(n,k)=>k?f("".join(str(n.count(x))+x for x:sorted(set(n))),k-1):n Try it online! Python 3, 78 bytes f=lambda n,k:k and f("".join(str(n.count(x))+x for x in sorted({*n})),k-1)or n Try it online! -7 bytes thanks to Mukundan314 5 Jelly, 8 bytes ṢŒrUFVµ¡ Try it online! Credit to hyper-neutrino for the µ trick, and for telling me I was wrong :P Full program. Takes n then k on the command line How it works ṢŒrUFVµ¡ - Main link. Takes n on the left, k on the right µ - Last links as a monadic chain f(n): Ṣ - Sort the digits of n Œr - Run length encode UF - ... 5 Wolfram Language (Mathematica), 55 bytes ToString@#2<>#&@@@Sort@Tally@Characters@#<>""&~Nest~##& Try it online! Input ["n", k]. 5 JavaScript (Node.js), 40 bytes x=>b=>b.flatMap(v=>x?x-(x=~-x/v|0)*v:[]) Try it online! 5 Python 2, $\phi_{\Omega^\omega}(0)\cdot\omega$, 266 bytes l=len T=lambda i,s='(',d=1:T(i/2,s+['),','('][i%2],d+i*2%4-1)if d else(eval(s)[0],i) L=lambda A,B:any(A==b or L(A,b)for b in B)or all(L(a,B)for a in A)and(l(A)-l(B),0)<(0,next((L(a,b)for a,b in zip(A,B)if a!=b),0)) def f(a,b):A,a=T(a);B,b=T(b);return(a,0)<(b,L(A,B)) Try it online! At least I ... 5 R, 45 40 39 38 bytes Or R>=4.1, 31 bytes by replacing the word function with \. function(n,k=1:n)any(c("",k[!n%%k])>n) Try it online! Using the hint (proof in @alephalpha's answer). Outputs FALSE for digit small number and TRUE otherwise. Converts list of divisors k[!n%%k] to character by prepending empty string. Then compares the vector ... 5 Python 2, 44 bytes lambda n:any(i>n>n%i<1for i in range(n)) Try it online! Outputs True/False negated. Looks for an i that's a divisor of n and is lexicographically smaller than n. The two conditions are combined into an inequality chain, connected by Python 2's willingness to compare strings and numbers (strings are bigger). This chain also ... 4 Charcoal, 5 bytes ＩＬ↨ＮＮ Try it online! Link is to verbose version of code. Works for any b>1. Unfortunately Base doesn't accept implicit arguments, so I have to explicitly input the n and b. Explanation: Ｎ Input n as an integer Ｎ Input b as an integer ↨ Convert n to base b as an array Ｌ Take the length Ｉ Cast to ... 4 Japt, 4 bytes sV l Try it here 4 Retina 0.8.2, 35 bytes \d+$* +\b(1+) (\1)+1* $1$#2$* Try it online! Link includes test cases. Takes input in the order b, n. Explanation: \d+$* Convert b and n to unary. +\b(1+) (\1)+1* $1$#2\$* Repeatedly integer divide n by b until n is less than b. With each division, prefix a space to b. Count the total number of spaces, including the ...

4

05AB1E, 2 bytes вg First input is the base $b$, second input is $n$. Try it online or verify all test cases. Using logarithm like most other answers is 4 bytes: .n>ï Try it online or verify all test cases. Explanation: # First (implicit) input-integer = b # Second (implicit) input-integer = n в # Convert n to a base-b list g ...

4

Pyth, 4 3 bytes ljF Try it online! # implicitly output l # len of jF # convert first element in input to base second element of input -1 byte thanks to Mr. Xcoder

4

Vyxal, 7 bytes (søeRf∑ Try it Online! ( # k times... s # Sort øe # Run length encode R # Reverse each f # Flatten ∑ # Sum

4

Python 3, $(\omega^2,28)$ lambda a,b:(a&-a,a)<(b&-b,b) Try it online! Old version This orders by number of trailing zeros (of binary rep) first and then by the number itself.

4

Haskell, $\omega \cdot 2$, 22 bytes (.q).(<).q q=odd>>=(,) Try it online! Based off Grain Ghost's Haskell answers. The function q is pointfree for q x=(odd x,x). That is, we sort numbers by whether they're odd, then the number itself, which gives the evens followed by the odds. We can generalize this idea as: Haskell, $\omega \cdot 4$, 24 ...

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# Raspberry Turk Robotic Arm¶

## Introduction¶

The Raspberry Turk uses a robotic arm (SCARA) to pick up and move chess pieces. The arm consists of two links, whose rotation is each controlled by a separate servo. This allows it to move freely in the X-Y direction while remaining rigid in the Z direction. There is a third servo that controls the vertical motion at the end of the arm.

In order to move pieces, the robot needs to be able to adjust the servos rotation $(s_1, s_2)$ to accurately move the arm to any given $x, y$ coordinate on the board.

$s_1, s_2 = g(x, y)$

Our goal in this notebook is to create this function $g(x, y)$.

## Working Backwards¶

We'll eventualy get to $g(x, y)$, but to get there we need to work backwards. To begin, let's first formalize what we know.

The arm consists of two links of length $l_i$. Since we know the lengths of the links, we can come up with an equation to determine $x, y$ given their angles $(\theta_i)$.

$$f_x(\theta_1, \theta_2) = x_{offset} + cos(\theta_1) \cdot l_1 + cos(\theta_1 + \theta_2) \cdot l_2$$$$f_y(\theta_1, \theta_2) = y_{offset} + sin(\theta_1) \cdot l_1 + sin(\theta_1 + \theta_2) \cdot l_2$$

Now, in order to derive $\theta_i$ we need to learn more about the servo and how it's attached to the arm.

Diagram from Robotis Dynamixel AX-12A support manual

You can adjust the rotation of the servo by setting the goal position address in the RAM of the servo. The goal position address is a number between 0 and 1023 (10 bit value) that maps to a 0° to 300° rotation. This equates to an accuracy of 0.29° per unit change in value (not bad!).

The servo is attached to a pinion gear which rotates a larger gear that the link of the arm is directly attached to. The pinion gear has 32 teeth and the larger gear has 76 teeth. This gives us a gear ratio of $\frac{76}{32}$ or $2.375 : 1$. From now on we'll refer to this gear ratio as $\phi$. Additionally, we must account of the rotational offset of the link when the servo goal position has a value of $0$. We'll call this rotational offset $\theta'_i$. Given this, we can how calculate $\theta_i$ given the goal position ($s_i$) of the servo.

$$\theta_i(s_i) = \theta'_i + \frac{1023-s_i}{1023}\cdot 300°\cdot \phi$$

Now, we have everything we need to convert any $s_1, s_2$ to an $x, y$.

$$f_x(s_1, s_2) = x_{offset} + \cos(\frac{1023-s_1}{1023}\cdot 300°\cdot \phi + \theta'_1)\cdot l_1 + \cos((\frac{1023-s_1}{1023}\cdot 300°\cdot \phi + \theta'_1) + (\frac{1023-s_2}{1023}\cdot 300°\cdot \phi + \theta'_2)) \cdot l_2 \\[10pt]$$$$f_y(s_1, s_2) = y_{offset} + \sin(\frac{1023-s_1}{1023}\cdot 300°\cdot \phi + \theta'_{s_1})\cdot l_1 + \sin((\frac{1023-s_1}{1023}\cdot 300°\cdot \phi + \theta'_1) + (\frac{1023-s_2}{1023}\cdot 300°\cdot \phi + \theta'_2)) \cdot l_2$$

However, both $f_x$ and $f_y$ are assuming ideal conditions. In reality there is error introduced by inaccuracies in measurements, and unaccounted for real world effects on the system. In order to achieve better accuracy in reality we need to better understand this error component and account for it.

$$f'_x(s_1, s_2) = f_x(s_1, s_2) + \epsilon_x(s_1, s_2)$$$$f'_y(s_1, s_2) = f_y(s_1, s_2) + \epsilon_y(s_1, s_2)$$

$f'_x$ and $f'_y$ represent the actual real world $x, y$ given any $s_1, s_2$, where $\epsilon_x$ and $\epsilon_y$ is the respective error. In a more simple form:

$$f'(s) = f(s) + \epsilon(s)$$

To achieve a $f'(s)$ that is suffienct for our purposes we need to, as accurately as possible, model $\epsilon$.

## Data¶

In order to model $\epsilon$ we first need to collect a sample of data points. To do this I manually moved the arm to to 81 points and recorded the real $x, y$ and the respective $s_1, s_2$.

In [1]:
import project
import numpy as np
import pandas as pd

In [2]:
data.head()

Out[2]:
s1 s2 x y
0 92 482 0.125 0.0000
1 177 1023 0.000 18.0000
2 497 386 9.000 9.0000
3 950 0 18.000 0.1875
4 674 660 18.000 18.0000

We need two new columns for $x$ and $y$ error. We can calculate this by first calulating the theoretical coordinate and then subtracting the observed coordinate.

In [3]:
dynamixel_range = 300.0
gear_ratio = 76.0 / 32.0
l_1 = 15.25
l_2 = 13.75
x_offset = 17.0
y_offset = -2.97
gripper_offset = 0.5

def f_with_theta(s, theta_prime_1, theta_prime_2):
theta_1 = ((1023.0-s[:,0])/1023.0) * np.deg2rad(dynamixel_range / gear_ratio) + theta_prime_1
theta_2 = ((1023.0-s[:,1])/1023.0) * np.deg2rad(dynamixel_range / gear_ratio) + theta_prime_2
x = x_offset + np.cos(theta_1) * l_1 + np.cos(theta_1 + theta_2) * l_2 + np.cos(theta_1 + theta_2 + np.pi/2.0) * gripper_offset
y = y_offset + np.sin(theta_1) * l_1 + np.sin(theta_1 + theta_2) * l_2 + np.sin(theta_1 + theta_2 + np.pi/2.0) * gripper_offset
return np.array([x, y]).T

## Optimization of Theta Values¶

$\theta'_1$ and $\theta'_2$ were measured by hand to be 4° and 40° respectively. To find a more accurate measurement, we will brute force values close to those and find the two values that results in the smallest mean squared error when comparing the theoretical results to the known values in the dataset. This will provide a better starting point when solving for $\epsilon$.

In [4]:
from sklearn.metrics import mean_squared_error
from scipy.optimize import brute

def mse(theta):
xy = f_with_theta(data.values[:,0:2], theta[0], theta[1])
return np.array([mean_squared_error(data.x, xy[:,0]), mean_squared_error(data.y, xy[:,1])]).mean()

def slice_for_value(value):
inc = value * 0.2
return slice(value - inc, value + inc)

t = brute(mse, [slice_for_value(theta_prime_1_estimate),
slice_for_value(theta_prime_2_estimate)])

θ′: [  3.82714566  39.9262376 ]

In [5]:
def f(s):
return f_with_theta(s, t[0], t[1])

In [6]:
theoretical_xy = f(data.values[:,0:2])
data["x_theory"] = theoretical_xy[:,0]
data["y_theory"] = theoretical_xy[:,1]
data["x_error"] = data.x - data.x_theory
data["y_error"] = data.y - data.y_theory

In [7]:
data.head()

Out[7]:
s1 s2 x y x_theory y_theory x_error y_error
0 92 482 0.125 0.0000 0.378050 0.236596 -0.253050 -0.236596
1 177 1023 0.000 18.0000 0.263558 18.327552 -0.263558 -0.327552
2 497 386 9.000 9.0000 8.948043 8.989270 0.051957 0.010730
3 950 0 18.000 0.1875 18.112372 0.139365 -0.112372 0.048135
4 674 660 18.000 18.0000 17.901277 18.107595 0.098723 -0.107595

## Visualizing Error¶

Below is a visualization of the error. The x and y axis on the graph represent inches from the bottom left corner of the chessboard. The actual $x,y$ is in blue, and the theoretical values are red. The equation is pretty accurate, but in the far corners, there is noticable error.

In [8]:
import matplotlib.pyplot as plt

def scatter(x, y):
plt.scatter(data.x, data.y, color='blue')
plt.scatter(x, y, color='red', marker=',')
plt.show()

scatter(data.x_theory, data.y_theory)

Below is another way to visualize the error. This time the x and y axes represent servo values.

In [9]:
from matplotlib import cm
from matplotlib.colors import Normalize

plt.scatter(data.s1, data.s2, c=data.x_error, cmap=cm.coolwarm, norm=Normalize(-0.5, 0.5), linewidth=0.4, s=40.0)
plt.xlim(0,1023)
plt.ylim(0,1023)
plt.show()
plt.scatter(data.s1, data.s2, c=data.y_error, cmap=cm.coolwarm, norm=Normalize(-0.5, 0.5), linewidth=0.4, s=40.0)
plt.xlim(0,1023)
plt.ylim(0,1023)
plt.show()

## Model¶

The model is a simple polynomial regression.

In [10]:
from sklearn.model_selection import cross_val_score
from sklearn.model_selection import ShuffleSplit
from sklearn.linear_model import LinearRegression
from sklearn.preprocessing import PolynomialFeatures
from sklearn.pipeline import Pipeline

S_train = data.values[:,0:2]
z_train = data.values[:,6:8]
poly = PolynomialFeatures(degree=3)
x_error_model = Pipeline([('poly', poly), ('linear', LinearRegression(normalize=True))])
y_error_model = Pipeline([('poly', poly), ('linear', LinearRegression(normalize=True))])

cv = ShuffleSplit(n_splits=3, test_size=0.3, random_state=0)
cvs_x = cross_val_score(x_error_model, S_train, y=z_train[:, 0], scoring='neg_mean_squared_error', cv=cv)
cvs_y = cross_val_score(y_error_model, S_train, y=z_train[:, 1], scoring='neg_mean_squared_error', cv=cv)
print "Cross validation error for x: {}".format(cvs_x)
print "Cross validation error for y: {}".format(cvs_y)

x_error_model.fit(S_train, z_train[:, 0])
y_error_model.fit(S_train, z_train[:, 1])

def predict_error(s):
return np.array([x_error_model.predict(s), y_error_model.predict(s)]).T

Cross validation error for x: [-0.01337804 -0.01384453 -0.00966184]
Cross validation error for y: [-0.00928625 -0.0084753  -0.0051186 ]

Below is a plot of the predict_error function with the real error plotted in circles on top.

In [11]:
from itertools import product

s_plot = np.array(list(product(np.linspace(0.0, 1023.0, 32), np.linspace(0.0, 1023.0, 32))))
predicted_error = predict_error(s_plot)
x_error_predicted = predicted_error[:,0]
y_error_predicted = predicted_error[:,1]

plt.scatter(s_plot[:,0], s_plot[:,1], c=x_error_predicted, marker=',', cmap=cm.coolwarm, norm=Normalize(-0.5, 0.5), linewidth=0.0, s=120.0)
plt.scatter(data.s1, data.s2, c=data.x_error, cmap=cm.coolwarm, norm=Normalize(-0.5, 0.5), linewidth=0.4, s=40.0)
plt.xlim(-20,1043)
plt.ylim(-20,1043)
plt.show()

plt.scatter(s_plot[:,0], s_plot[:,1], c=y_error_predicted, marker=',',cmap=cm.coolwarm, norm=Normalize(-0.5, 0.5), linewidth=0.0, s=120.0)
plt.scatter(data.s1, data.s2, c=data.y_error, cmap=cm.coolwarm, norm=Normalize(-0.5, 0.5), linewidth=0.4, s=40.0)
plt.xlim(-20,1043)
plt.ylim(-20,1043)
plt.show()

## Building the Final Equation¶

Using the predict_error function $f'(s)$ can now be created,.

In [12]:
def f_prime(s):
return f(s) + predict_error(s)

To visualize the difference $f'(s)$ makes, we can, once again, plot the predicted $x,y$ (red) on top of the real $x,y$ (blue) for each given $s_1,s_2$.

In [13]:
xy_theoretical_with_error = f_prime(data.values[:,0:2])
data['x_theory_with_error'] = xy_theoretical_with_error[:,0]
data['y_theory_with_error'] = xy_theoretical_with_error[:,1]

scatter(data.x_theory_with_error, data.y_theory_with_error)

Now that we have a working $f'(s)$, we can set out to accomplish our original goal, to create $g(x, y)$. The function $f'$ takes a $s_1, s_2$ and returns $x, y$. The function $g$ takes $x, y$ and returns $s_1, s_2$. In order to find $g$, we just need to invert $f'$! But $f'$ doesn't invert easily and we actually don't need $g$ to work over all values of $s_1,s_2$–only when $s_1,s_2 \in \{0, 1, 2, ..., 1023\}$. So instead, let's solve for every possible $s_1,s_2$ combination, and then create a lookup tree with the results. We can the query the lookup tree to find the closest $s_1,s_2$ for any given point $x,y$.

In [14]:
srange = np.array(list(product(range(1024), range(1024))))
pts = f_prime(srange)

In [15]:
from sklearn.neighbors import KDTree
tree = KDTree(pts, metric='euclidean')

In [16]:
def g(pts):
return np.array(srange[tree.query(pts, return_distance=False)]).reshape(-1,2)

## Visualizing the Final Equation¶

Below are two plots visualizing the function $g(x,y)$ over the entire 18"x18" chessboard. The first plot is $s_1$ and the second is $s_2$.

In [17]:
pts_grid = np.array(list(product(np.linspace(0.0, 18.0, 32), np.linspace(0.0, 18.0, 32))))
projected_s = g(pts_grid)

plt.scatter(pts_grid[:,0], pts_grid[:,1], c=projected_s[:,0], marker=',', cmap=cm.coolwarm, norm=Normalize(0, 1023), linewidth=0.0, s=120.0)
plt.xlim(0,18)
plt.ylim(0,18)
plt.colorbar()
plt.title("s1")
plt.show()

plt.scatter(pts_grid[:,0], pts_grid[:,1], c=projected_s[:,1], marker=',', cmap=cm.coolwarm, norm=Normalize(0, 1023), linewidth=0.0, s=120.0)
plt.xlim(0,18)
plt.ylim(0,18)
plt.colorbar()
plt.title("s2")
plt.show()

## Saving the Model¶

The $x,y$ associated with the each $s_1,s_2$ from 0 to 1023 is saved to disk so the lookup tree can be created on demand for the Raspberry Turk to use.

In [18]:
path = project.path('data', 'processed', 'arm_movement_engine_pts.npy')
with open(path, 'w') as f:
np.save(f, pts)

The full source of this notebook can be found on GitHub.


text:
Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > DETAIL:

Paper:

TR22-068 | 5th May 2022 00:56

Sketching Approximability of (Weak) Monarchy Predicates

TR22-068
We analyze the sketching approximability of constraint satisfaction problems on Boolean domains, where the constraints are balanced linear threshold functions applied to literals. In particular, we explore the approximability of monarchy-like functions where the value of the function is determined by a weighted combination of the vote of the first variable (the president) and the sum of the votes of all remaining variables. The pure version of this function is when the president can only be overruled by when all remaining variables agree. For every $k \geq 5$, we show that CSPs where the underlying predicate is a pure monarchy function on $k$ variables have no non-trivial sketching approximation algorithm in $o(\sqrt{n})$ space. We also show infinitely many weaker monarchy functions for which CSPs using such constraints are non-trivially approximable by $O(\log(n))$ space sketching algorithms. Moreover, we give the first example of sketching approximable asymmetric Boolean CSPs. Our results work within the framework of Chou, Golovnev, Sudan, and Velusamy (FOCS 2021) that characterizes the sketching approximability of all CSPs. Their framework can be applied naturally to get a computer-aided analysis of the approximability of any specific constraint satisfaction problem. The novelty of our work is in using their work to get an analysis that applies to infinitely many problems simultaneously.


text:
## Elementary Algebra

Published by Cengage Learning

# Chapter 3 - Equations and Problem Solving - 3.2 - Equations and Problem Solving - Problem Set 3.2 - Page 107: 55

The number is 6

#### Work Step by Step

Let X represent the number we need to find: Three times a certain number means 3X. If 6 is added to three times a certain number, we write 6 + 3X. Thus: $6 + 3X = 24$ Subtract 6 from both sides: $3X = 18$ Divide both sides by 3: $X = 6$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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 " tommy quaternion " tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/18/2021, 11:03 PM (This post was last modified: 01/21/2021, 11:40 PM by tommy1729.) I always like stuff like bicomplex numbers , hypercomplex numbers etc And some of you here do to. So I thought I share my " tommy quaternion " with you. Do not confuse with the normal quaternion or hyperbolic quaternion. The nonreal units are $A,B,C$ And the rules for the units are : $A*B=B*A=C,A*C=C*A=-B$ $B*C=C*B=A$ $A*A=B*B=-1$ $C*C=1$ The distributive property holds. And clearly this is a commutative number system. But not associative. A pair of zero divisors are $(A+B)(A-B)=A^2 - B^2 = -1 - -1 = 0$ The word " anti-associative " comes to mind. Im considering doing tetration with them. Feel free to comment. Regards tommy1729 Tom Marcel Raes marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/21/2021, 12:49 PM Can you post the matrix representation? I have the result, but I do not yet know how to get it. tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/21/2021, 01:17 PM (This post was last modified: 01/21/2021, 11:40 PM by tommy1729.) (01/21/2021, 12:49 PM)marraco Wrote: Can you post the matrix representation? Hi marraco. These numbers have NO matrix representation because multiplication is NOT associative. So doing simple linear algebra probably will not work. This is one of the simplest algebra's that have no matrix representation. Or should I say smallest. Dimension 3 does not have this property if we require it to have some group like structure. So dimension 4 is the smallest. HOWEVER it is not the only one in dimension 4. There exists at least one other nonisomorphic one : unfortunately neither of them are closed under square root. ( $1 + 2 A + 3 B + 4 C$  has no square root in either one ) Another one is  $A*B=B*A=C,A*C=C*A=B$ $B*C=C*B=A$ $A*A=B*B=-1$ $C*C=1$ Thank you for your interest. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 01/21/2021, 09:27 PM (This post was last modified: 01/21/2021, 10:59 PM by MphLee.) Hi, why you don't specify that $C*B=A$ in your list of identities? You forgot to add it or it is possible to derive it from the others? Also, you see it as a non-associative algebra over the reals or the complex? Trivial question (sorry but I'm new to linear algebra): being an algebra means that $\mathbb T$ is a $k$-vector space of dimension d=4 on which is defined a bilinear application $*:{\mathbb T} \times {\mathbb T}\to {\mathbb T}$ that is not associative... but it must be bilinear because we want it distributive: An element is of the form $T=x1+yA+zB+wC$. Let  $T,R,S,U\in{\mathbb T}$, the application being distributive means that $(T+R)*(S+U)=T*S+R*S+T*U+R*U$ i.e. every translation by a Tommy quaternion is a endomorphism of  the addition group; and we also want that for every scalar $\lambda$ $\lambda R*S= R*\lambda S=\lambda (R*S)$ i.e we want it to contain a copy of the base field, i.e multiplication by scalar  is multiplication by $\lambda1$ where $1\in \mathbb T$. So the previous means that the Tommy quaternions of the form $T=x1+0A+0B+0C$ commute and "associate" with everything. But this means that the multiplication is bilinear: every element should it should have a representation.. e.g. multiplication by $T$ has a 4x4 matrix $M_T$. Obviously if it is not associative we "should not" have $M_{T*S}=M_{T}M_{S}$ in general... Probably my understanding of non-associative alg. is so poor that I'm missing something obvious. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/21/2021, 11:47 PM (01/21/2021, 09:27 PM)MphLee Wrote: Hi, why you don't specify that $C*B=A$ in your list of identities? You forgot to add it or it is possible to derive it from the others? Also, you see it as a non-associative algebra over the reals or the complex? Trivial question (sorry but I'm new to linear algebra): being an algebra means that $\mathbb T$ is a $k$-vector space of dimension d=4 on which is defined a bilinear application $*:{\mathbb T} \times {\mathbb T}\to {\mathbb T}$ that is not associative... but it must be bilinear because we want it distributive: An element is of the form $T=x1+yA+zB+wC$. Let  $T,R,S,U\in{\mathbb T}$, the application being distributive means that $(T+R)*(S+U)=T*S+R*S+T*U+R*U$ i.e. every translation by a Tommy quaternion is a endomorphism of  the addition group; and we also want that for every scalar $\lambda$ $\lambda R*S= R*\lambda S=\lambda (R*S)$ i.e we want it to contain a copy of the base field, i.e multiplication by scalar  is multiplication by $\lambda1$ where $1\in \mathbb T$. So the previous means that the Tommy quaternions of the form $T=x1+0A+0B+0C$ commute and "associate" with everything. But this means that the multiplication is bilinear: every element should it should have a representation.. e.g. multiplication by $T$ has a 4x4 matrix $M_T$. Obviously if it is not associative we "should not" have $M_{T*S}=M_{T}M_{S}$ in general... Probably my understanding of non-associative alg. is so poor that I'm missing something obvious. yeah ... I forgot to add C*B sorry. And thanks for pointing it out. It is corrected now. Well matrix representation is based on representing the multiplication. But matrix multiplication is associative , however these numbers are not associative by multiplication. Functional composition is also associative btw. So easy representations might not exist. In my opinion this makes these commutative algebras interesting. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 01/22/2021, 12:03 AM Ah I see... you didn't mean that multiplication is not a bilinear map, we can represent only the translations... you mean representable in the sense of a morphism from the Algebra multiplication to a $M_{n\times n}(k)$ or to a GL(n), e.g. like a group representation, where we study characters and so on... In fact how can we "represent" non associative operations? Sorry but I'm not very fluent with this yet.. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/22/2021, 12:10 AM (This post was last modified: 01/23/2021, 01:23 PM by tommy1729.) Then again maybe dimension 4 is too small and simple to be interesting. It does not even have a square root if we use real coefficients. ( they have 8 solutions over the complex coefficients though ) Now it turns out commutative nonassociative hypercomplex numbers that have a square root must be of dimension 4n. Powerassociative is also a thing. Im not an expert in this though. Informally speaking , the main issue is constructing such cayley tables such that it does not have any " subcayleys " (the analogue of subgroup), yet do have a structure that makes the square root possible. In dimension 8 we have for instance this  (see picture) But Im not sure about the properties yet. 8 = 4*2 , so the complex might be a subgroup and then the other subgroup would be one of those 4 dim ones. ( edit : it has no subalgebra , this has been checked see follow up replies ) sqrt has not been investigated. I think it is power-associative. regards tommy1729 Attached Files Image(s)     tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/22/2021, 12:15 AM (01/22/2021, 12:03 AM)MphLee Wrote: Ah I see... you didn't mean that multiplication is not a bilinear map, we can represent only the translations... you mean representable in the sense of a morphism from the Algebra multiplication to a $M_{n\times n}(k)$ or to a GL(n), e.g. like a group representation, where we study characters and so on... In fact how can we "represent" non associative operations? Sorry but I'm not very fluent with this yet.. Powerassociativity together with newton's binomium and tensor and nonlinear theory are useful. However Im not an expert and your questions are good. The complexity is the reason Im into this. I feel hypercomplex numbers deserve more attention. Maybe it is hidden or maybe it is not investigated sufficiently. Or both. regards tommy1729 MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 01/22/2021, 12:41 AM (This post was last modified: 01/22/2021, 12:47 AM by MphLee.) I've to look deeper on this power associativity... sounds familiar... with unitBtw I agree. Basically if we regard as hypercomplex everything that is a finite dimensional algebra, not necessarily commutative or associative, over the real, e.g. (C, H, O, S) , in the last years, doing some readings for hobby... I grow more convinced that those structures conceal a great deal of fundamentally natural properties that has to do with the very essence of the universe in some way. For example I tried, not very successfully, to follow John Baez on his blog posts about the octonions and their relevance to the standard model, or the dimensions in which some kind of hypercomplex algebras has some properties that have to do with the dimensions were spacetime appears in some obscure and esoteric string theory black magic, or still about dual numbers were we can rebuild analysis algebraically by the use of idempotent elements acting as infinitesimals. EDIT: I'm not sure that my brain can process that 8-dim caley table. I might try too look at it with more calm. I'm way less expert at it than you btw. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 01/23/2021, 01:17 PM The 8 dimensional " tommy octonion " - or whatever we will call it - is not constructed at random nor by trial and error. It has to satisfy properties and those properties help in the construction. For instance commutative implies (x*y)^2 = x^2 * y^2. So that is " respected " in the construction. That also implies how many -1 and +1 we have in the table. Another thing is (x*x)*y = x*(x*y). This also helps alot. ( alternative algebra , jordan algebra , power associative ) I also did a subalgebra test ( the analogue of a subgroup test ). To be more specific we have dimension 8. If dimension 8 has a subalgebra then it must be one of order 4 and of order 2 ( since 2*4 = 8 analogue to subgroups ). So we only need to check for subgroups of order 2. So I checked all the A,B,C,D for being the complex imaginary i^2 = -1 and all the E,F,G for being the split complex number j^2 = 1. None of them matched so the split-complex or complex numbers are NOT a subalgebra. Hence the " tommy quaternion " is not an extension. For the subalgebra test see the 2 added pics. regards tommy1729 Attached Files Image(s) « Next Oldest | Next Newest »

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# mathr

## Interior coordinates in the Mandelbrot set

There is an algorithm to find points on the boundary of the Mandelbrot set, given a particular hyperbolic component and the desired internal angle. It involves Newton's method in two complex variables to solve

$F^p(z,c) = z \\ \frac{\partial}{\partial z} F^p(z,c) = b$

where $$F^0(z,c) = z$$ and $$F^{q+1}(z,c) = F^q(F(z,c)^2+c)$$, $$p$$ is the period of the target component, and $$b=e^{2 \pi i \theta}$$ with the $$\theta$$ the desired internal angle. The resulting $$c$$ is the coordinates of the point on the boundary. It can also be modified to find points in the interior, simply set $$b = r e^{2 \pi i \theta}$$ with $$\left|r\right| \le 1$$.

After seeing this fractal Mandelbrot in an IRC channel, I wondered if it was possible to invert the algorithm, namely find $$b$$ given $$c$$, and use $$b$$ as polar texture coordinates to place copies of an image inside the Mandelbrot fractal. The answer is yes, and in some sense it's actually easier than the other direction.

The algorithm I developed works like this:

1. When $$c$$ is outside the Mandelbrot set, give up now;
2. For each period $$p$$, starting from $$1$$ and increasing:
1. Find $$z_0$$ such that $$F^p(z_0,c) = z_0$$ using Newton's method in one complex variable;
2. Find $$b$$ by evaluating $$\frac{\partial}{\partial z} F^p(z,c)$$ at $$z_0$$;
3. If $$\left|b\right| \le 1$$ then return $$b$$, otherwise continue with the next $$p$$.

I wrote this implementation.


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# How to get circuit run time in Q#

What is the best way to get the circuit run time in Q# ?

More precisely, I run a circuit using dotnet -run command and I would like to know how many time it took for running my circuit. Actually I compute the delay between the moment I start the command and the return, but I suppose it includes building time, I really just need the circuit running time.

Thank you.

Best regards.

Here is what I did when I was exploring optimizing the code to run as fast as possible in simulation (here is the blog post describing the things I looked into).

1. Add a classical driver to your project that calls your Q# code.
I used C# (you can also use Python if you prefer it) and the code you'll use is going to be something like this:
using QuantumSimulator qsim = new QuantumSimulator();
bool[] result = GroversSearch.Run(qsim).Result.ToArray();

1. In this classical driver, wrap the call to quantum operation's Run in classical code that tracks the start and the end time. Here is C# code for it:
Stopwatch stopWatch = new Stopwatch();
stopWatch.Start();
// ... call quantum operation here ...
stopWatch.Stop();
TimeSpan ts = stopWatch.Elapsed;
string elapsedTime = String.Format("{0:00}:{1:00}:{2:00}.{3:00}",
ts.Hours, ts.Minutes, ts.Seconds,
ts.Milliseconds / 10);
Console.WriteLine("RunTime " + elapsedTime);

You can do any pre-processing before the call to quantum code and any post-processing afterwards, and they won't count towards the execution time.

You can see the full example here.

Note: This was suppose to be a comment but it got too long so I just put it here as an answer.

I don't think Microsoft has successfully built their quantum machine, and therefore it would not be be possible to do that, without having to run your program through actual quantum gate. The run time you get by seeing seeing how fast you can execute the simulation is heavily depend on your classical computer resources. The run time on the quantum circuits depend on the number of gates and depth of the circuit, and how fast you can implement those gates. For IBM's devices, the sampling time of single qubit gate is around 100-150 ns (not sure, but you can look it up and they are different for different qubits), and the CNOT gate time is around 1000 ns (again, they are different on which two qubit you operate on). Then you can work out the math of how long it takes to execute your ciruit. You probably can write small script to do this. Between the gates, you may want to add some buffer time as well, maybe 20ns or so.


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### How to recover RSA private key in a coredump of ssh-agent - Sapheads HackJam 2009 Challenge 6

$file core.6261 core.6261: ELF 32-bit LSB core file Intel 80386, version 1 (SYSV), SVR4-style, from 'ssh-agent' As you can see, this is a coredump of ssh-agent. This is from the manpage of ssh-agent: ssh-agent is a program to hold private keys used for public key authentication (RSA, DSA). The idea is that ssh-agent is started in the beginning of an X-session or a login session, and all other windows or programs are started as clients to the ssh-agent program. The manpage doesn't tell where ssh-agent stores private keys, but one can guess that it would store them in its memory. When ssh-agent crashes, all of its memory content would be written to the coredump. So I guess the idea of challenge 6 is to recover the private key of the criminals stored in the coredump, and use that private key to SSH into 67.202.60.164 which may give me the key for challenge 7. This is why Sapheads challenges are better than other CTFs. They don't try to distract us by giving false trails. They give good trails which can be deduced using logical thinking. This challenge also greatly resembles real world scenario which in turn makes it much more interesting than other non-sense challenges somehow created only to show that their authors are smarter than others. Okie enough bullshit, let's get back to challenge 6. As whats from Sapheads suggests, I take a look at static variables of ssh-agent.c which are defined like below: typedef struct identity { TAILQ_ENTRY(identity) next; Key *key; char *comment; u_int death; u_int confirm; } Identity; typedef struct { int nentries; TAILQ_HEAD(idqueue, identity) idlist; } Idtab; /* private key table, one per protocol version */ Idtab idtable[3]; int max_fd = 0; /* pid of shell == parent of agent */ pid_t parent_pid = -1; u_int parent_alive_interval = 0; /* pathname and directory for AUTH_SOCKET */ char socket_name[MAXPATHLEN]; char socket_dir[MAXPATHLEN]; So right above socket_name are 3 integers, and then comes idtable which contains pointers to Identity structures, which in turn contains pointers to Key structures. This is how a Key structure looks like (it's in openssh/key.h): struct Key { int type; int flags; RSA *rsa; DSA *dsa; }; Ah RSA structure! This is how it looks like (it's in openssl/rsa.h): struct rsa_st { /* The first parameter is used to pickup errors where * this is passed instead of aEVP_PKEY, it is set to 0 */ int pad; long version; const RSA_METHOD *meth; /* functional reference if 'meth' is ENGINE-provided */ ENGINE *engine; BIGNUM *n; BIGNUM *e; BIGNUM *d; BIGNUM *p; BIGNUM *q; BIGNUM *dmp1; BIGNUM *dmq1; BIGNUM *iqmp; /* be careful using this if the RSA structure is shared */ CRYPTO_EX_DATA ex_data; int references; int flags; /* Used to cache montgomery values */ BN_MONT_CTX *_method_mod_n; BN_MONT_CTX *_method_mod_p; BN_MONT_CTX *_method_mod_q; /* all BIGNUM values are actually in the following data, if it is not * NULL */ char *bignum_data; BN_BLINDING *blinding; BN_BLINDING *mt_blinding; }; typedef bignum_st BIGNUM; struct bignum_st { BN_ULONG *d; /* Pointer to an array of 'BN_BITS2' bit chunks. */ int top; /* Index of last used d +1. */ /* The next are internal book keeping for bn_expand. */ int dmax; /* Size of the d array. */ int neg; /* one if the number is negative */ int flags; }; As one can expect, a RSA structure contains n, e, d, p, q and some other data. We need to get these numbers, which are stored in BIGNUM structures, out of the coredump. So the next thing to do is know where idtable is in the coredump. I load the coredump into bless, my favourite hex editor (click on the image to zoom in): As you can see, at offset 0x13b0 is socket_name. Based on the analysis of the last paragraph, we can guess that max_fd, parent_pid, and parent_alive_interval are stored in 12 bytes between offset 0x13a4 and 0x13af. Then one can ask, what are those 16 null bytes from offset 0x1394 to 0x13a3? I don't know. If you know, please drop me a line. But I know for sure, idtable is stored in 36 bytes from offset 0x1370 to 0x1393, just right before those weird 16 null bytes. The idtable array has 3 entries, one for each SSH protocol version. Each entry is an structure whose length is 12 bytes (hence 36 bytes for the whole array). When ssh-agent starts, each Idtab is initiated like this: TAILQ_INIT(&idtable.idlist); idtable.nentries = 0; where TAILQ_INIT is a macro defined in sys/queue.h as following: #define TAILQ_INIT(head) do { (head)->tqh_first = NULL; (head)->tqh_last = &(head)->tqh_first; } while (/*CONSTCOND*/0) So after initiation, an Idtab structure would contain 1 zero integer for nentries, 1 null pointer for idlist.tqh_first, and 1 pointer for idlist.tqh_last which points back to idlist.tqh_first. Looking at the coredump, one can see that the first two Idtab entries of idtable don't contain any key information because their nentries is 0. This is as expected, since SSH protocol 1 and 1.1 are long deprecated. The last entry of idtable is the Idtab of SSH protocol version 2. As one can see, its nenetries is 1, and we can guess that the pointer 0x0806E3B8 is pointing to an identity structure which contains 4 pointers and 2 integers. Let's see if this is the case:$ gdb /usr/bin/ssh-agent core.6261 Core was generated by `/usr/bin/ssh-agent'. [New process 6261] #0 0xb7f55424 in __kernel_vsyscall () (gdb) x/6x 0x0806E3B8 0x806e3b8: 0x00000000 0x0806297c 0x0806e158 0x0806e3a8 0x806e3c8: 0x00000000 0x00000000 (gdb) x/s 0x0806e3a8 0x806e3a8: "id_rsa" As you can see, the 4th pointer is comment. So the 3rd pointer should be key, i.e. it should point to a Key structure which contains 2 integers and 2 pointerss: (gdb) x/4x 0x0806e158 0x806e158: 0x00000001 0x00000000 0x0806e170 0x00000000 If everything is correct, the pointer 0x0806e170 should point to a RSA structure: (gdb) x/22x 0x0806e170 0x806e170: 0x00000000 0x00000000 0xb7ed5840 0x00000000 0x806e180: 0x0806e210 0x0806e228 0x0806e240 0x0806e288 0x806e190: 0x0806e270 0x0806e2b8 0x0806e2a0 0x0806e258 0x806e1a0: 0x00000000 0x00000000 0x00000001 0x0000000e 0x806e1b0: 0x00000000 0x00000000 0x00000000 0x00000000 0x806e1c0: 0x0806e950 0x00000000 How to be sure this is a RSA structure? Is there any known value to test? Fortunately, the answer is yes. If this is a RSA structure, the 6th pointer 0x0806e228 should point to a BIGNUM structure containing the value of the e parameter which should be 0x23, the default value that ssh-keygen uses for e. Let's see: (gdb) x/5x 0x0806e228 0x806e228: 0x0806e2e0 0x00000001 0x00000001 0x00000000 0x806e238: 0x00000001 (gdb) x/2x 0x0806e2e0 0x806e2e0: 0x00000023 0xb7d39150 Yay! We got it! The next step is to extract all the parameters. It was not as easy as it sounds though. I spent quite a lot of time to read out the value of these parameters due to my ignorant of big-edian and little-edian storage. But I managed to get them out eventually. I generated the RSA private key from p and q, and used it to SSH into 67.202.60.164 which indeed gave me the key for challenge 7. I got my first breakthrough he he he. That's it. Thanks for reading. Err...but how to generate RSA private key from n, d, e, p and q? I'm glad that you ask. Tools like openssl can not help in this case. You must write your own tool. I suggest you taking a look at ASN.1. There's a very good tutorial here. If you understand ASN.1, I'm pretty sure you'd know how to generate RSA private key from its parameters. You can use pyasn1 which is a very good ASN.1 library for Python. I can't release my tool because it's part of the upcoming CTF that I'm organizing. After that CTF, I'll update this post with the tool.


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Next: , Previous: , Up: Definitions   [Contents][Index]

### 12.9 \newtheorem

Synopses:

\newtheorem{name}{title}
\newtheorem{name}{title}[numbered_within]
\newtheorem{name}[numbered_like]{title}

Define a new theorem-like environment. You can specify one of numbered_within and numbered_like, or neither, but not both.

The first form, \newtheorem{name}{title}, creates an environment that will be labelled with title; see the first example below.

The second form, \newtheorem{name}{title}[numbered_within], creates an environment whose counter is subordinate to the existing counter numbered_within, so this counter will be reset when numbered_within is reset. See the second example below.

The third form \newtheorem{name}[numbered_like]{title}, with optional argument between the two required arguments, creates an environment whose counter will share the previously defined counter numbered_like. See the third example.

This command creates a counter named name. In addition, unless the optional argument numbered_like is used, inside of the theorem-like environment the current \ref value will be that of \thenumbered_within (see \ref).

This declaration is global. It is fragile (see \protect).

Arguments:

name

The name of the environment. It is a string of letters. It must not begin with a backslash, \. It must not be the name of an existing environment, and the command name \name must not already be defined.

title

The text to be printed at the beginning of the environment, before the number. For example, ‘Theorem’.

numbered_within

Optional; the name of an already defined counter, usually a sectional unit such as chapter or section. When the numbered_within counter is reset then the name environment’s counter will also be reset.

If this optional argument is not used then the command \thename is set to \arabic{name}.

numbered_like

Optional; the name of an already defined theorem-like environment. The new environment will be numbered in sequence with numbered_like.

Without any optional arguments the environments are numbered sequentially. The example below has a declaration in the preamble that results in ‘Definition 1’ and ‘Definition 2’ in the output.

\newtheorem{defn}{Definition}
\begin{document}
\section{...}
\begin{defn}
First def
\end{defn}

\section{...}
\begin{defn}
Second def
\end{defn}

This example has the same document body as the prior one. But here \newtheorem’s optional argument numbered_within is given as section, so the output is like ‘Definition 1.1’ and ‘Definition 2.1’.

\newtheorem{defn}{Definition}[section]
\begin{document}
\section{...}
\begin{defn}
First def
\end{defn}

\section{...}
\begin{defn}
Second def
\end{defn}

In the next example there are two declarations in the preamble, the second of which calls for the new thm environment to use the same counter as defn. It gives ‘Definition 1.1’, followed by ‘Theorem 2.1’ and ‘Definition 2.2’.

\newtheorem{defn}{Definition}[section]
\newtheorem{thm}[defn]{Theorem}
\begin{document}
\section{...}
\begin{defn}
First def
\end{defn}

\section{...}
\begin{thm}
First thm
\end{thm}

\begin{defn}
Second def
\end{defn}


text:
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# check symmetric matrix python

If we sum all elements that need to be saved from all rows, we get the following result: $$1 + 2 + \cdots + N = (1 + N) \cdot \frac{N}{2}$$. C program to check if a matrix is symmetric or not: we find the transpose of the matrix and then compare it with the original matrix. There is no need to pass the number of columns since symmetric matrices are square. A Skew Symmetric Matrix or Anti-Symmetric Matrix is a square matrix whose transpose is negative to that of the original matrix. Roy likes Symmetric Logos. This method is __len__() and its only responsibility is to return the _size attribute: Until now, we were able to create a symmetric matrix with all elements initialized to zero and get its size. Input elements in matrix A.; Find transpose of matrix A, store it in some variable say B.; Check if matrix A is equal to its transpose A T then it is symmetric matrix otherwise not. A Square Matrix is said to be symmetric if it is equal to its transpose. Therefore, the elements above the diagonal do not have to be stored. The complete source code of the implemented SymmetricMatrix class, alongside with unit tests and the benchmark script, is available on GitHub. C Program To Check whether Matrix is Skew Symmetric or not. Similarly as in the previous case, to get the desired element from the matrix, the position has to be converted to a proper index to the underlying storage. Join our Blogging forum. » Embedded C Now in the given question we have make a program that takes a matrix as input and checks if it is symmetric or skew symmetric or none of them. In terms of elements of matrices: M (i, j) = M (j, i) Following is a python code for demonstrating how to check for Symmetric Matrix. __init__() firstly checks if the provided size is valid. In the following part of this post, a Python implementation of a symmetric matrix is explained step by step along with its usage. Definition and Usage. More: » C#.Net » JavaScript Logic: To find whether the matrix is symmetric or not we need to compare the original matrix with its transpose. Mathematical Python Eigenvalues and Eigenvectors Type to start searching ... Symmetric Matrices. ... # Python 3 program to check # whether given matrix Thus, this symmetric matrix implementation is suitable in circumstances where memory usage is a bigger problem than processor power. However, this is not very useful in real life. » C++ » CSS The symmetric_difference() method returns a set that contains all items from both set, but not the items that are present in both sets.. Here two matrices are given. For the third row, the situation is a little bit complicated because the elements from all the previous rows have to be summed. A = (a ij) then the symmetric condition becomes a ij = −a ji. If the left subtree and the right subtree of the root of a Binary tree are mirror images of each other, then the Binary Tree is called a Symmetric Tree. » DBMS Next, access times for writing to the entire matrix are computed for both matrix types. » About us Later, this matrix needs to be shared between several processes. Transpose of a matrix is achieved by exchanging indices of rows and columns. Algorithm: If your input is not symmetric/Hermitian, eigh will return incorrect eigenstuff. First, we will create a square matrix of order 3X3 using numpy library. This is a demo video to get program to check whether a given square matrix is symmetric or not. The following equation shows how a dot product of Matrix A and A T can result in a symmetric matrix. Which defaults to 1e-10. In linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space. A tree will be said to be symmetric if it is the same when we take the mirror image of it. We can see that the symmetric matrix can save approximately 50% of memory space. » Java » News/Updates, ABOUT SECTION ... We have to check whether the tree is a symmetric tree or not. » Machine learning It is not a big chapter but it is important to understand the next ones. Before running the script with the cProfile module, only the relevant parts were present. If it is not, the ValueError exception is raised. If you are familiar with the Python implementation of list, you may know that list does not contain elements that you insert into it. » Java Run-length encoding (find/print frequency of letters in a string), Sort an array of 0's, 1's and 2's in linear time complexity, Checking Anagrams (check whether two string is anagrams or not), Find the level in a binary tree with given sum K, Check whether a Binary Tree is BST (Binary Search Tree) or not, Capitalize first and last letter of each word in a line, Greedy Strategy to solve major algorithm problems. » CS Basics Therefore, the elements above the diagonal do not have to be stored. Therefore, a better solution when choosing the underlying data structure is leaving space for users to choose the type of the storage according to their requirements. The cumtime column informs us about the cumulative time spent in this function and all sub-functions during all calls. » DBMS » HR Numpy is a Python library which provides various routines for operations on arrays such as mathematical, logical, shape manipulation and many more. Transpose is only defined for a square matrix. Therefore, to obtain the matrix size, we wish that the following code could be used: To actuate the previous code, another magic method has to be implemented. » Android For identically two matrix should be equal, number of rows and columns in both the matrix should be equal and the corresponding elements should also be equal. » Embedded Systems In terms of elements of matrices: M(i, j) = M(j, i). for all indices and .. Every square diagonal matrix is symmetric, since all off-diagonal elements are zero. The task is to check whether the matrix is horizontal symmetric, vertical symmetric or both. Every element above this mirror is reflected to an element under this mirror. The elements in both matrices are 64b integers. » Feedback Accordingly, for the $$N$$-th row, $$N$$ elements need to be saved. The experiments ran on an Intel Quad-Core i7-4700HQ (6M Cache, 2.40 GHz) processor. Therefore, for the first row only one element has to be stored, for the second row two elements are saved and so on. In linear algebra, if the matrix and its transpose are equal, then the matrix is symmetric (MT = M). In the case of the second row, the calculated index is $$1 + column$$, since the previous row contains only one element. For a symmetric matrix A, A T = A. A square matrix as sum of symmetric and skew-symmetric matrix ? tf = issymmetric (A) returns logical 1 (true) if square matrix A is symmetric; otherwise, it returns logical 0 (false). Defining Matrix using Numpy; Transpose Matrix; Here, we will learn how to create a symmetric matrix using a non-symmetric matrix? » O.S. » Privacy policy, STUDENT'S SECTION » C++ STL © https://www.includehelp.com some rights reserved. A matrix is called symmetric if $$a_{ij}$$ is equal to $$a_{ji}$$. Solved programs: Symmetric matrix can be obtain by changing row to column and column to row. Therefore, another magic method, particularly the __getitem__() method, has to be implemented. calculate the correct index into the underlying storage. An example of such a matrix is shown below. In this post, a Python implementation of such a matrix is described. If an argument for this parameter is not passed, then list will be used as the storage type. S = A.A T. Python code creating symmetric matrix » Internship » C Following is a python code for demonstrating how to check for Symmetric Matrix. Logic to check symmetric matrix. i want to check if the matrix is symmetric or not by using nested loops and display a certain message if it is or not. It is smaller than $$size^2$$. However, we can treat list of a list as a matrix. Linear Algebra using Python, Linear Algebra using Python | Symmetric Matrices: Here, we are going to learn about the symmetric matrices and its implementation in Python. Singular values are important properties of a matrix. In Python, when an assignment to sm[1, 1] is executed, the interpreter calls the __setitem__() magic method. Submitted by Anuj Singh, on May 26, 2020. As mentioned previously, symmetric matrices can be used to represent distance or adjacency matrices. C program to check if a matrix is symmetric or not. For the second row, the number of elements in the previous row and column part of the (row, column) pair is enough. Python Program to check if two given matrices are identical. In mathematics, particularly in linear algebra, a skew-symmetric matrix (also known as an antisymmetric or antimetric) is a square matrix A which is transposed and negative. » Cloud Computing In linear algebra, if the matrix and its transpose are equal, then the matrix is symmetric (MT = M). Certainly, it will not work since array.array is not supposed to be shared by different processes. For now, only one special method has to be written, particularly the __init__() method, which takes a single parameter called size. Note: The symmetry of a matrix can only be determined when it is a square matrix. » Content Writers of the Month, SUBSCRIBE This service is done by the _get_index() method for which the last part of this section is devoted. & ans. » C Write a C+ Program to Check Matrix is a Symmetric Matrix with an example. In this tutorial, we are going to learn how to check if a given Binary Tree is a Symmetric Tree in C++ using an Iterative approach. The singular values are the lengths of the semi-axes. All code was written, tested and profiled in Python 3.4. This computation is performed five times and then the average result is calculated. Below is a program to check a square matrix is symmetric or not. This leads us to think if the used list is the best data structure for the storage. » CS Organizations » C So, the index for the (2, column) position is $$1 + 2 + column$$. » Subscribe through email. Python Programming Server Side Programming. » Ajax » C++ ... Symmetric Tree in Python. Program to find whether the given Square Matrix is symmetric or not For understanding the above output, only three columns are important for us, namely ncalls, cumtime and filename:lineno(function). Now, when we have the index, we can use the __setitem__() method provided by the underlying storage that can be called simply as self._data[index] = value: For obtaining an element from the matrix, we will proceed in a similar way. To implement this improvement only small changes are necessary in the __init__()method. Later on, the implementation of this method will be shown. If A is a symmetric matrix, then A = A T and if A is a skew-symmetric matrix then A T = – A.. Also, read: As mentioned pre… Languages: Firstly, memory usage is compared. And this tolerance is too small for that type. We also need to write to and read from the matrix. » Data Structure Interview que. scipy.linalg.ldl¶ scipy.linalg.ldl (A, lower = True, hermitian = True, overwrite_a = False, check_finite = True) [source] ¶ Computes the LDLt or Bunch-Kaufman factorization of a symmetric/ hermitian matrix. Symmetric matrix can be obtain by changing row to column and column to row. Hence, the memory requirements are higher for list than, for example, for array.array that stores the elements directly. Otherwise, the user passes his storage type during the matrix creation like in the following example: The above create_storage() returns an array holding 64b integers that can be shared by different processes. The main advantage of using a symmetric matrix in comparison with a classic matrix lies in smaller memory requirements. 0 -b -c b 0 -d c d 0 is the general form of a skew-symmetric matrix. It would be nice to have a possibility to use a standard Python way for gaining the matrix size, which is the len() function. The matrix is said to be horizontal symmetric if the first row is same as the last row, the second row is … The result of this experiment can be seen in the table below. Since only elements under and on the diagonal are stored and the whole matrix is saved in a one-dimensional data storage, a correct index to this storage needs to be calculated. If the entry in the ith row and . The matrix diagonal can be seen as a mirror. » Certificates » Networks To save space, only elements under and on the diagonal need to be saved. To clarify the second part, particularly the calculation of the index into the used storage, the above picture and the following table will be used: Note that for the first row, the column part of the (row, column) pair is sufficient to use as index to the underlying storage. Firstly, one parameter, namely create_storage, is added with default value set to None. » Web programming/HTML » C Home » Given a 2D binary matrix of N rows and M columns. We’ll write a program in C to find the matrix is symmetric or not. Here, We’ll check whether the given matrix is symmetrical or not. $$0 + 1 + 2 + 3 + \cdots + row + column$$, convert a position above the diagonal into a proper position below the diagonal and. To create the numpy matrix, numpy.zeros() is called. When creating a symmetric matrix, array.array() is used as the underlying storage. Both matrices have the same order. In this and subsequent sections, I show a particular usage first and then I show the implementation. U, S, V = svd (dat, full_matrices = 0) #rank checking S = S #this is different from matlab, as the diagonal component is automatically obtained in np svd, for matlab, we need diag(S) For now, assume that the _get_index() method returns this index. Therefore, for the (row, column) position the correct index is $$1 + 2 + 3 + \cdots + row + column$$. While it is less efficient to use eig to calculate all of the eigenvalues and check their values, this method is more flexible since you can also use it to check whether a matrix is symmetric positive semi-definite. The overhead is due to internal workings of Python and computing indexes to the underlying storage. 線型代数学における対称行列（たいしょうぎょうれつ、英: symmetric matrix）は、自身の転置行列と一致するような正方行列を言う。 対称行列 - Wikipedia. The second printed matrix below it is v, whose columns are the eigenvectors corresponding to the eigenvalues in w. Meaning, to the w[i] eigenvalue, the corresponding eigenvector is the v[:,i] column in matrix v. In NumPy, the i th column vector of a matrix v is extracted as v[:,i] So, the eigenvalue w[0] goes with v[:,0] w[1] goes with v[:,1] » C++ This parameter specifies the number of rows. Ad: » Java Geometrically, a matrix $$A$$ maps the unit sphere in $$\mathbb{R}^n$$ to an ellipse. From these two trees, the first one is symmetric, but the second one is not. Align the center of logo with the origin of Cartesian plane. : The matrix diagonal can be seen as a mirror. You are given a binary matrix … Python doesn't have a built-in type for matrices. For example: A = [[1, 4, 5], [-5, 8, 9]] We can treat this list of a list as a matrix having 2 rows and 3 columns. ... Let's check that the eigenvectors are orthogonal to each other: v1 = evecs[:,0] # First column is the first eigenvector print(v1) Indeed, it contains pointers to these elements. But the difference between them is, the symmetric matrix is equal to its transpose whereas skew-symmetric matrix is a matrix whose transpose is equal to its negative.. Description. Python Matrix. Of course, there are other data structures that are more memory efficient than list. Web Technologies: manifold.MDS by default would compute the distance matrix with the same data type as the input. » Contact us » Linux » DOS The following code shows the implementation: It is worth noting the size of the _data storage used to store the matrix. Thanks to this rule, an $$N \times N$$ symmetric matrix needs to store only $$(N + 1) \cdot \frac{N}{2}$$ elements instead of $$N^2$$ elements needed to be stored in case of a classic matrix. : Any matrix can be the symmetric matrix if the original matrix is equal to the transpose of that matrix. a b c b e d c d f is the general form of a symmetric matrix. To explain the computation of the number of elements, suppose that we have a $$N \times N$$ symmetric matrix. In computer science, symmetric matrices can be utilized to store distances between objects or represent as adjacency matrices for undirected graphs. These are well-defined as $$A^TA$$ is always symmetric, positive-definite, so its eigenvalues are real and positive. From the following table, we can see that the average access time for the implemented symmetric matrix is much worse than the average access time for the numpy matrix: The reasons behind the slow access time for the symmetric matrix can be revealed by the cProfile module. So, the question is which one should be used. Suppose that we chose array.array instead of list during the symmetric matrix implementation. Therefore, the first part comparing memory requirements and all parts using the numpy code are not included in the profiling. The following source code shows how to create a $$4 \times 4$$ symmetric matrix: To make this code runnable, the SymmetricMatrix class has to be implemented. Now if the colored pixels of the logo are symmetric about both X-axis and Y-axis, then the logo is symmetric. Method 2: Check Eigenvalues. To check whether a matrix A is symmetric or not we need to check whether A = A T or not. I'd suggest having eigh check that the input is symmetric/Hermitian. When we have the correct index, the element on this position in the underlying storage is returned: Now, it is time to show how _get_index() is implemented. » SEO If no special demands are present then list can be used as the default storage type. Thanks to the finite arithmetic progression, this expression can be simplified as follows: $$0 + 1 + 2 + \cdots + row + column = (0 + row) \cdot \frac{row + 1}{2} + column$$. The source code of this method can be broken down into two steps that have to be executed in the provided order: If the given position, (row, column), is above the diagonal, then row is swapped with column, since every element above the diagonal has its counterpart exactly at the (column, row) position. Then the function _smacof_single calls check_symmetric without a tol argument. # Approaching the problem An example of such a matrix is shown below. We will see other type of vectors and matrices in this chapter. Creation of a Square Matrix in Python. » Node.js Aptitude que. Are you a blogger? Finally, the implementation of calculating the index into the underlying storage is shown in the following source code: Now, we have a working implementation of a symmetric matrix. » Facebook » Articles the problem is that it's displaying the message after comparing each element of the original matrix with the ones in the transposed or inversed matrix.. i want the message to be displayed after both matrices are compared!. The first one, named ncalls, represents how many times the function from filename:lineno(function) was called. » Python » C# » Java Seems easy to accomplish, here's a suggestion for the beginning of eigh: if not numpy.allclose(a, numpy.asmatrix(a).H): raise ValueError('expected symmetric or Hermitian matrix,\ Program to check if a matrix is symmetric Last Updated: 04-06-2018 A square matrix is said to be symmetric matrix if the transpose of the matrix is same as the given matrix. » DS If the input matrix contains float32, then the distance matrix does too. & ans. » SQL How to check whether a logo is symmetric? This means that it satisfies the equation A = −A T. If the entry in the i-th row and j-th column is a ij, i.e. Otherwise, a function that takes one parameter, particularly the size of the storage, and returns the created storage is expected: To provide a comparison between the introduced symmetric matrix and a matrix created via the numpy module, I have written a benchmark script that uses a $$4000 \times 4000$$ matrix to show memory requirements and average access times for the implemented symmetric matrix and the numpy matrix. As can be seen from the output, the time is spent mostly in __setitem__() and _get_index(). » Puzzles We have seen in 2.3some interesting kind of matrices. Since the main motivation for using this type of matrix is memory efficiency, the question that may emerged is if a more memory efficient implementation can be made. 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