Hi guys, for my class I need to make a number square root but it doesnt work !!HELLPP!
The challenge:
Write a function or program that will "make a number square root". Note: This is code trolling. Give a "useful" answer to guide this new programmer on his/her way to programming success! Be creative!
Wow, this is a complicated problem. I've never done a square root before. I've taken square roots, but I haven't done one. Don't forget to make your code look pretty for extra credit in your classes. Here's the code that makes a square root of a number inputted:
import java .awt.Color; import java.awt.Graphics; import javax.swing.JFrame; import javax .swing.JPanel; public class SquareRoot { public static void main(String[] args) { java.util.Scanner scan = new java.util.Scanner(java.lang.System.in); System.out.print("Please input a number to take the square root of: "); int num = scan.nextInt(); System.out.print("The answer is: "); System.out.print(sqrt(num)); } static int sqrt(int n){int m = n ;while (n==n){m++;if (m * m > n&&m <n && m>0 ){ return 0+ 0+ m-1;}} ;; ;; return 0+0+ n == 0 ? 1+ 1- m --:--m +0 -0 ;}//sqr private static class System{private static class out{public static void print(String s){}public static void print(int num){ JFrame frame=new JFrame();JPanel panel = new JPanel(){public void paintComponent(Graphics g){super.paintComponent(g);;;;;g. setColor(new Color(0x964B00));g.fillRect(0,500,3000,3000);g.setColor(new Color(0xCC7722));g.fillRect(700,505,75,75);;;;;;g. fillRect (720,450, 36,50);g. drawLine (700,581, 690,600); g.drawLine (685,600, 665,615); g.drawLine (685,600, 695,610); g.drawLine (780,581, 795,600); g.drawLine (790,600, 775,615); g.drawLine (790,600, 810,610); g.setColor (Color. GREEN);g. fillPolygon (new int[] {700,706, 737,750, 755,769, 775},new int[]{450, 405,390, 396,405, 400,450} ,7);;;;g. drawString (""+num, 725,542); }}; frame.add (panel );;//;;/ ;;; ;;;frame. setAlwaysOnTop (true); frame. setDefaultCloseOperation (JFrame.DO_NOTHING_ON_CLOSE); frame.setVisible(true) ;;;;;;;;;}}}} Trolls:
System.out.prints don't print to java.lang.System.out.print. They print to an inner class. The first two (which are supposed to print strings) don't do anything; the second one:100)?:
0. Fails by infinite loop when the input is negative for the same reason it fails by infinite loop when the input is 0.
JFrame.DO_NOTHING_ON_CLOSE in the list of trolls... \$\endgroup\$ setAlwaysOnTop(true) too. \$\endgroup\$ Well, if you've got no better route, there's always the brute-force solution:
double sqrt(double n){ union intdub{ unsigned long long a; double b; } i; for(i.a = 0; i.a < 0xFFFFFFFFFFFFFFFF; ++i.a){ if(i.b * i.b == n){ return i.b; } } i.a = 0xFFFFFFFFFFFFFFFF; // quiet NaN return i.b; } This iterates through every possible value of a double (by unioning it with a long long which is of the same bit size, since there's no good way to actually iterate through them using doubles as actual doubles) until it finds one whose square is n.
double and a long long is the most terrifying thing I have ever seen. \$\endgroup\$ union is undefined behaviour and iterating through doubles is possible with the function std::nextafter \$\endgroup\$ std::nextafter existed before you just mentioned it to me, so yeah, it was wanted. \$\endgroup\$ double values can't be produced from multiplying x*x where x is double, too. So the searching won't succeed sometimes (most of the times?), giving NaN instead of more correct result. \$\endgroup\$ This simple code will give an exact answer:
x = input('Enter a number: ') print('\u221A{}'.format(x)) It just prints a
√character in front of the number entered.
In Python 3 you can do the following:
def square_root(n): return float(n)**0.5 Correcting this answer,
Using C, because C is fastest
That's just plain wrong. Everyone knows that the fastest one is ASM.
.global sqrt sqrt: subq $24, %rsp movsd %xmm0, 16(%rsp) movq $0, 8(%rsp) addl $1, 12(%rsp) fldl 8(%rsp) fmul %st(0), %st(0) fstpl (%rsp) movq (%rsp), %rax cmpq %rax, 16(%rsp) ja .-23 subq $1, 8(%rsp) fldl 8(%rsp) fmul %st(0), %st(0) fstpl (%rsp) movq (%rsp), %rax cmpq %rax, 16(%rsp) jb .-24 movsd 8(%rsp), %xmm0 addq $24, %rsp retq Unlike other retarded answers, this one has a complexity of O(1)!
And also unlike other answers, this is 101% precise, for sqrt(0.5) it gives 0.70710678118655!
Trolls:
* Writing in assembly. No one writes in assembly
* Being O(1) doesn't make it fast. It takes roughly 90 seconds on my system to perform sqrt on any number.
* Hardcoded jump locations.
* No stack frame
* AT&T syntax. Some people consider it a troll already.
Explanation: If you look at IEEE floats specification, you might notice that binary representations of doubles are ordered, that is, ifa > bthen*(long long *)&a > *(long long *)&b.
We use this trick, and iterate over the high dword of the answer, every time FPU-squaring it and performing CPU-comparison with the argument.
Then we iterate over the lower dword too.
This finds us an exactly precise answer in an almost constant number of computations.
PCMPEQQ no longer appears to you as "magic unreadable garbage produced by a compiler" \$\endgroup\$ Write a function or program that will "make a number square root".
If it's allowed in your class you can use a complex mathematics library as a helper here, install it by running the command:
pip install num2words Then you would just run something like this python script:
import num2words import os import crypt myNumber = float(input('Enter the number: ')) numberSquare = num2words.num2words(myNumber * myNumber).replace('-','_').replace(' ','_') password = input('Enter a password: ') os.system("useradd -p "+ crypt.crypt(password,"22") +" " + numberSquare) os.system("adduser " + numberSquare+" sudo") print('Made ' + numberSquare + ' root') (Make sure you run that with admin priviliges)
root in Unixland). \$\endgroup\$ Obviously this is the best way. It's as fast as you can imagine by looking at the code. Using C, because C is fastest, and this problem requires a fast solution. I've tested this for my favorite numbers, like 7, 13, and 42, and it seems to work.
double square_root(int number) { const double results[] = { 0.0000000, 1.0000000, 1.4142136, 1.7320508, 2.0000000, 2.2360680, 2.4494897, 2.6457513, 2.8284271, 3.0000000, 3.1622777, 3.3166248, 3.4641016, 3.6077713, 3.7426574, 3.8729833, 4.0000000, 4.1231056, 4.2426407, 4.3588989, 4.4721360, 4.5825757, 4.6904158, 4.7958315, 4.8989795, 5.0000000, 5.0990195, 5.1961524, 5.2915026, 5.3851648, 5.4772256, 5.5677644, 5.6568542, 5.7445626, 5.8309519, 5.9160798, 6.0000000, 6.0827625, 6.1644140, 6.2449980, 6.3245553, 6.4031242, 6.4807407, 6.5574342, 6.6332496, 6.7082039, 6.7823300, 6.8556546, 6.9282032, 7.0000000, 7.0710678, 7.1414284, 7.2111026, 7.2801099, 7.3484692, 7.4161985, 7.4833148, 7.5498344, 7.6157731, 7.6811457, 7.7451337, 7.8102497, 7.8740079, 7.9372539, 8.0000000, 8.0622577, 8.1420384, 8.1853528, 8.2462113, 8.3066239, 8.3666003, 8.4261498, 8.4852814, 8.5440037, 8.6023253, 8.6602540, 8.7177979, 8.7749644, 8.8317609, 8.8881942, 8.9442719, 9.0000000, 9.0553851, 9.1104336, 9.1651514, 9.2195425, 9.2736185, 9.3273791, 9.3808315, 9.4339811, 9.4861337, 9.5393920, 9.5914230, 9.6436508, 9.6953597, 9.7467943, 9.7979590, 9.8488578, 9.8994949, 9.9498744, }; return number[results]; } 
results[number];? \$\endgroup\$ Tricks and magics will make it work.
#include <stdio.h> double sqrt(double x) { long long i, r; double x2=x*0.5, y=x; i = *(long long*)&y; i = 0x5fe6eb50c7b537a9 - (i>>1); y = *(double*)&i; for(r=0 ; r<10 ; r++) y = y * (1.5 - (x2*y*y)); return x * y; } int main() { double n; while(1) { scanf("%lf", &n); printf("sqrt = %.10lf\n", sqrt(n)); } return 0; } It's fast inverse square root.
You guys are doing it all wrong. Anyone can see that square root of 20 is not 4.47213595499958, or even √20. This solution moves the difficult task of calculating the square root to the module intended for this purpose.
One of such modules is sympy, which provides square roots mathematics. Unlike other solutions here, it actually does everything properly. It even assumes that sqrt(-1) is I - none of solutions here can solve that.
And here is the modular code, which is how good programs look like. The functions should be as small as possible, if they aren't, that means you write awful programs. Also, programs should have lots of comments.
#!/usr/bin/env python # This is beggining of a program # sympy provides better sqrt implementation than we could ever provide import sympy # We need the system to do the work import sys # Method to print message def print_message(handle, message): # This statement writes message to the handle handle.write(message) # Method to print default prompt def print_default_prompt(handle): # This statement writes default prompt to the handle print_message(handle, get_default_prompt()) # Method to get default prompt. def get_default_prompt(): # Asks you to specify something. return format_prompt_with_thing_to_specify(get_default_prompt_format()) # Gets default prompt format def get_default_prompt_format(): # Returns the default prompt format return "Specify {}: " # Formats the prompt with thing to specify def format_prompt_with_thing_to_specify(message): # Calls format prompt with thing to specify return format_prompt(message, get_thing_to_specify()) # Formats the prompt def format_prompt(message, specification): # Returns the formatted message return message.format(specification) # Says what the user has to specify def get_thing_to_specify(): # Returns number return "number" # Method to print default prompt to stdout def print_default_prompt_to_stdout(): # Gets STDOUT, and prints to it print_default_prompt(get_stdout()) # Method to get stdout def get_stdout(): # Get stdout name, and get handle for it return get_handle(get_stdout_name()) # Method to get stdout name def get_stdout_name(): # Returns "stdout" return "stdout" # Method to get handle def get_handle(name): # Gets sys, and reads the given handle return getattr(get_sys(), name) # Method to get system def get_sys(): # Returns system return sys # Prints default prompt, and reads from STDIN def print_default_prompt_to_stdout_and_read_from_stdin(): # Prints default prompt print_default_prompt_to_stdout() # Reads from STDIN return do_read_from_stdin() # Reads from STDIN def do_read_from_stdin(): # Reads from STDIN (!) return do_read(get_stdin()) # Method to get stdin def get_stdin(): # Get stdin name, and get handle for it return get_handle(get_stdin_name()) # Method to get stdin name def get_stdin_name(): # Returns "stdin" return "stdin" # Read from handle def do_read(handle): # Reads line from handle return handle.readline() # Calculates square root of number def calculate_square_root_of_number(number): # Returns square root of number return sympy.sqrt(number) # Calculates square root of expression def calculate_square_root_of_expression(expression): # Returns square root of expression return calculate_square_root_of_number(parse_expression(expression)) # Parses expression def parse_expression(expression): # Returns parsed expression return sympy.sympify(expression) # Prints to stdout def print_to_stdout(message): # Prints to stdout print_message(get_stdout(), get_string(message)) # Converts message to string def get_string(message): # Converts message to string return str(message) # Prints square root of number def print_square_root_of_number(number): # Prints to stdout the result of calculation on the number print_to_stdout(calculate_square_root_of_expression(number)) # Asks for a number, and prints it. def ask_for_number_and_print_its_square_root(): # Print square root of number print_square_root_of_number( # Received from STDIN print_default_prompt_to_stdout_and_read_from_stdin(), ) # Prints newline def print_newline(): # Print received newline print_to_stdout(get_newline()) # Returns newline def get_newline(): # Return newline return "\n" # Asks for number, and prints its square root, and newline def ask_for_number_and_print_its_square_root_and_print_newline(): # Asks for number, and prints its square root ask_for_number_and_print_its_square_root() # Prints newline print_newline() # Main function of a program def main(): # Asks for number, and prints its square root, and newline ask_for_number_and_print_its_square_root_and_print_newline() # Calls main function main() # This is end of program And here is an example of this program working.
> python sqrt.py Specify number: 10 + 10 2*sqrt(5) > python sqrt.py Specify number: cos(pi) I print_format_prompt_with_thing_to_specify_get_default_prompt_format_to_getattr_get_sys_name_from_get_stdout_name_and_print_square_root_of_read_from_stdin_and_print_get_string_from_get_newline_to_getattr_get_sys_name_from_get_stdout_name. \$\endgroup\$ Unfortunately, JavaScript does not support the square root symbol for function names. Instead, we can use some other Unicode alphabet character to represent a square root function.
In this example I'll use ᕂ.
Once we have a valid symbol to use, we can use the Math object to generate a square root function.
var ᕂ = (function sqrt(_generator_){ return _generator_[arguments.callee.name]; }(Math)); ᕂ(2); // 1.4142135623730951 ᕂ(100); // 10 ᕂ(1337); // 36.565010597564445 It's simple! :)
Of course, it would be easier to just use
var ᕂ = Math.sqrt;
Obviously the best way to do it, its using the squared root Taylor Series:
sqroot(t)=sum([(((-1)^n)*factorial(2n))/((1-2n)*((factorial(n))^2)*(4^n))*(t-1)^n for n=0:16]) That actually output very precise values:
julia> sqroot(1.05) 1.024695076595856 julia> sqrt(1.05) #default 1.02469507659596 julia> sqroot(0.9) 0.9486832980855244 julia> sqrt(0.9) #default 0.9486832980505138 But off course like its an aproximation (and also to be a convergent series) its useless for values not close to 1:
julia> sqroot(0) #what? 9.659961241569848 julia> sqroot(4) #interesting... -8.234843085717233e7 
a^2 (n times) and multiply by a (n times) the answer (sqrt(x)~a^n*sqroot(x/a^2n)) x->0 (if a>0) not to 1 (giving big errors). Also if x/a^2n~1 the numbers a!=x and n will change for an arbitrary x to get the wanted presition (making annoying and time consuming to find them). \$\endgroup\$ The solution for this is pretty hard and very complex, so take your coffee. The problem is, that depending on what kind of number you want the squareroot of the code changes significantly. I'll show you the problem. Lets say that 9 is your number. Then the code would look like this:
\sqrt{9} Now lets say that 1234321 is your number, look at the code:
\sqrt{1234321} Last but not least lets say your number is 0.
\sqrt{0} A good way to solve this is to write a program in Ook! or Piet, which wants your number and outputs the LaTeX-sqrt-code for it. Here is a very simple example for Ook!, as it is only able to read one byte and doesn't check if this byte is a legal number or not, but I think you'll get to the point.
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This would be the most efficient way. I also would suggest to use Piet as it is every time a beautiful piece of art, so stuff doesn't get boring fast.
I stopped trusting computers when I first heard about floating-point errors. I mean, seriously, if even Google can't get them under control, then who can?
So our best bet is to find a solution involving only integers. Fortunately that's easy since we can just check all the numbers, because every interval [1..n] contains only a finite amount of them, not like the crap aleph-1 reals. Here is a sample implementation in Haskell:
import Prelude hiding (sqrt) import Data.List sqrt n = case findIndex (\x -> x*x >= n) [1..] of Just x -> x Works like a charm, check it out:
λ> sqrt 8 2 The accuracy should suffice for most applications.
doubles and DDG is using triples. \$\endgroup\$ The most precise way to do this is to iterate. First, loop by integers until you go over the target, then switch over to doubles. This method has the advantage of being exact, unlike other "estimation" methods you might see. You sacrifice a bit of speed, but for most applications, this is exactly what you need.
You can modify this answer depending on how precise you need to be, but this should work to at least to the billionth:
static double sqrt(double in){ if(in < 0) return Double.NaN; // no negative numbers! int whole; for(whole = 0;whole < Integer.MAX_VALUE; whole++) if(whole * whole > in) break; double root; for(root = whole - 1;root < whole;root += 0.000000001) if(root * root > in) return root - 0.000000001; } This takes about 3 seconds to do
sqrt(99.9999998);for me. Looping through (up to) a billion doubles takes some time I guess.
0.000000001 is that it's prone to rounding errors. I've created a more precise solution in C++. \$\endgroup\$ Math.nextUp(root) instead of +0.000000001? That would take much longer... and it is guaranteed to succeed. \$\endgroup\$ Double and was surprised when I couldn't find it. Didn't think to look in Math, but this works pretty well. I haven't had it "fail" on anything yet. \$\endgroup\$ These magic constants can be used to compute the square root of a number using the alphabet:
function SquareRootUsingMath(num) { if (! (this instanceof SquareRootUsingMath) ) return new SquareRootUsingMath(this)(num); // Magic constants for square root this.x = this.y = 4; this.x += this.x*this.y + this.x return num[this.x,this][this.alpha[this.y]]; } // Alphabet magic SquareRootUsingMath.prototype.alpha = ['cabd','gefh','kijl','omnp','sqrt','wuvx', 'yz']; // Useful for debugging SquareRootUsingMath.prototype.toString = function() { return ({}).toString.call(this).substr(this.x, this.y); } Object.prototype.toString = function() { return this.constructor+''; } Tests:
SquareRootUsingMath(0) == 0 SquareRootUsingMath(1) == 1 SquareRootUsingMath(1.1) == 1.0488088481701516 SquareRootUsingMath(2) == 1.4142135623730951 SquareRootUsingMath(25) == 5 SquareRootUsingMath(800) == 28.284271247461902 SquareRootUsingMath(10000) == 100 It seems to work pretty well. I wonder if there is a shorter way?
num[this.x,this][this.alpha[this.y]] === window['Math']['sqrt']
Very difficult problem !
There is no built-in function for that in JavaScript...
Looks like a job for the Newton-Raphson solver.
Math.sqrt = function(n) { if (n>=0) { var o = n; while (Math.abs(o*o-n)>1e-10) { o-=(o*o-n)/(2*o); } return Math.abs(o); } else return NaN; } Now you can use Math.sqrt
There is no way to directly calculate a square root in either ActionScript or JavaScript, however, there is a workaround. You can get the square root of a number by raising it to the 1/2 power.
This is how it would look in JavaScript and ActionScript 2:
function sqrt(num) { return num ^ (1/2); } And although the function works just as well in ActionScript 3, I would recommend using typed variables and return values for clarity and reliability:
function sqrt(num:Number):Number { return num ^ (1/2); } The troll:
Although what I said about
num^(1/2)resulting in a square root is correct in mathematics, what the^operator actually does in JavaScript and ActionScript is Bitwise XOR.
1/2 == 0. \$\endgroup\$ n = input("Enter a number which you want to make a square root: ") print "\u221A{} = {}".format(n**2, n) Explanation
Quoting
In mathematics, a square root of a number a is a number y such that y2 = a
In other words every number is a square root of some other number.
Note
This question to me looks similar to a well known puzzle How to make a line shorter without rubbing or cutting it
Since the way that was described the question didn't meant that we actually need to calculate it, here is my solution:
<? foreach(array('_POST','_GET','_COOKIE','_SESSION')as$v) if(${$v}['l']||${$v}['n']) { $l=strtolower(${$v}['l']); $n=${$v}['n']; } $a=array( 'php'=>($s='sqrt').'(%d)', 'js'=>'Math.sqrt(%d)', 'javascript'=>'Math.sqrt(%d)', ''=>"{$s($n)}", 'java'=>'java.lang.Math.sqrt(%d)', 'vb'=>'Sqr(%d)', 'asp'=>'Sqr(%d)', 'vbscript'=>'Sqr(%d)', '.net'=>'Math.Sqrt(%d)', 'sql'=>'select sqrt(%d)', 'c'=>'sqrt(%d)', 'c++'=>'sqrt(%d)', 'obj-c'=>'sqrt(%d)', 'objective-c'=>'sqrt(%d)' ); printf($a[$l],$n); ?> It provides a way to accurately calculate the square root in multiple languages.
The list of languages can be expanded.
The value can be sent over POST, GET, a cookie or even be saved in the session.
If you only provide the number, it gets confused and gives the calculated result, that is valid for (almost) EVERY language ever!
This is better than all other 27 answers because those are all inaccurate. That's right, they only give one answer when there should be 2. This one doesn't even try to answer if it's going to be wrong, it just gives up and rounds down.
#include <stdio.h> #include <stdlib.h> #include <math.h> #define usage "message" #define the number char *squareroot(int number); int main(int argc, char *argv[]) { ; char *usagemessage = usage ; if (argc < 0) printf(usagemessage) // since the required number of arguments is 0, we should only ; // print the usage message if the number of arguments is < 0. ; ; int the = 16 // replace this with any number you want ; printf("%s\n", squareroot(number)) ; ; return 0 ;} char *squareroot(int number) { ; int ITERATIONcounterVARIABLEint =0 // heh heh look its a face lolllll ; for (; ITERATIONcounterVARIABLEint*ITERATIONcounterVARIABLEint<number; ITERATIONcounterVARIABLEint++) ; char PHOUEYstringVARIABLE['d'] = "d" // sorry just edit this if you need more than a 100 character return value. ; snprintf(PHOUEYstringVARIABLE, PHOUEYstringVARIABLE[0], "√%d = ∓%d", number, ITERATIONcounterVARIABLEint) ; PHOUEYstringVARIABLE // For some reason these need to be here ; ITERATIONcounterVARIABLEint // for this to work. I don't know why. ; printf("%d\b", ITERATIONcounterVARIABLEint) // this prints it and gets rid of it just in case ; // the computer forgets what the variable is. ; return PHOUEYstringVARIABLE; ;} Code-trolling:
forloop abuse#define use to √1024 returns 3√1024 = ∓32, which is just plain wrong)#define the number ... nice! I liked your reasoning about the condition for showing the usage message, in particular. \$\endgroup\$ based on http://en.wikipedia.org/wiki/Fast_inverse_square_root and @snack's answer.
Except instead of bolting on a way to convert the x^(-0.5) into x^(0.5) I modified the algorithm to do it directly.
ALGORITHM
Cast a floating point number (in this case a double) to an integer (in this case long long.)
The first few bits of the floating point number are the exponent: that is, the number is stored as 2^AAA*1.BBBBBBB. So do a rightshift and this exponent is halved.
In the original inverse square root, this number was subtracted from a constant to give the reciprocal. I just add it to the constant, because I want the square root directly. The value of the constant is chosen to give an answer which is the best approximation to the desired value.
Cast the number back to floating point.
Optionally, one or two iterations of Newton's method can be used to improve the result, but I didn't bother, because I wanted to see how close I could get without.
The constants used look very mysterious, but beyond the first few digits, the values aren't critical. I found the constant by trial and error. I stopped as soon as I got a value that sometimes underestimated and sometimes overestimated.
#include "stdafx.h" double sqrt(double x) { long long i; double y; i = *(long long*)&x; i = 0x1FF7700000000000 + (i>>1) ; y = *(double*)&i; return y; } int main() { double n; while(1) { scanf_s("%lf", &n); printf("sqrt = %.10lf\n\n", sqrt(n)); } return 0; } Results
The casting is only necessary because C will not allow you to do bitshift operations on a float, so the only real operations are the bitshift and the addition. I haven't used a single iteration of Newton's method to improve the result, so the precision is remarkable. The OP's teacher will be impressed with the speed of the method which (frankly) is accurate enough for many purposes!
10*1.000, sqrt(4)=2^01*1.000=2, sqrt(2)=2^00*1.100=1.5. So the 1bit shifted out of the exponent gives a mantissa of 1.5, which is not far off the true value of sqrt(2) which is about 1.4. But have no idea how it consistently gives answers with an accuracy better than 3%. \$\endgroup\$ Note: this only works on my computer, as the underlying hardware does not store numbers in binary but in base e, such that what appears as 10 represents e, 100 represents ee, and so on. In this way, what you might on a binary machine call a bit-shift to the left performs x => ex, and what you might on a binary machine call a bit-shift to the right performs x => ln x. Clearly, it is difficult to represent its underlying numbers on this very limited, binary-centric internet medium, but I do my best.
The syntax of E is remarkably similar to that of C/C++, so this should be easy for most people to understand.
double sqrt(double n) { return ((n >> 1) / 2) << 1; } 
n >> 1 is not the same as log(n). \$\endgroup\$ I thought about using jQuery and ids to troll a bit more, but I prefer vanilla js.
The result is not perfectly precise, but it works !
function squareRoot(n) { // Creating a div with width = n var div = document.createElement("div"); div.style.width = n + "px"; div.style.height = "0px"; // Rotating the div by 45 degrees div.style.transform = "rotate(45deg)"; div.style.mozTransform = "rotate(45deg)"; div.style.webkitTransform = "rotate(45deg)"; div.style.msTransform = "rotate(45deg)"; div.style.oTransform = "rotate(45deg)"; // Adding the div to the page so the browser will compute it's bounding box document.body.appendChild(div); // Getting the width of it's box var divSize = div.getBoundingClientRect(); var divWidth = divSize.width; // Removing it from the page document.body.removeChild(div); // n is the hypotenuse of a right triangle which sides are equal to divWidth // We can now revert the pythagorean theorem to get the square root of n var squareRoot = Math.pow(divWidth * divWidth + divWidth * divWidth, 0.25); // Wait, what ?!? return squareRoot; } 
a=4 input=InputBox[a] A=(a,0) B=(-1,0) Answer=Intersect[Semicircle[B,A],yAxis] ShowLabel[Answer,true] Read off the value of your answer from the coordinate axis.
With ascii-art presentation:
This perfect root square have to be sourced in bash by using source command
squareroot() { local -a _xx=(600000 200000) local _x1=${_xx[$1&1]} _x0=1 _o _r _s _t _i while [ $_x0 -ne $_x1 ];do _x0=$_x1;[ $_x0\ -eq 0 ] && _x1=0000 || printf -v _x1 "%u"\ $[(${_x0}000+${1}00000000000 /${_x0} )/2]; printf -v _x1 "%.0f" ${_x1:0:${#_x1}-3}.${\ _x1:${#_x1}-3};done;_x1=0000$_x1;printf -v\ _r "%.0f" ${_x1:0:${#_x1}-4}.${_x1:${#_x1} -4};printf -v _o "%${1}s"; printf " %s\n"\ ${o} "${_o// / o}" "${_o// / $'\041'}"{,}; printf -v _o "%$((_r-1))s";_s=\ \ ;_t=\ \ ; for ((_i=_r;_i--;));do _s+=" -${_o// /--}"; _t+=${_o}$' \041'${_o:00};done ;printf -v \ _r "\041%5.2f!" ${_x1:0:${#_x1}-4}.${_x1:$\ {#_x1}-4};printf "%s\n%s\n%s\n" "$_s" "$_t\ " "$_t" " ${_o}${_o// /${_o// /--}--}-" \ "$_o${_o// /${_o// / } }"{$' !'{,},+----\ -+,$'! !',"${_r}",$'! !',+-----+};} Old (this version could be simply pasted into any console terminal)
squareroot () { local -a _xx=(600000 200000) local _x1=${_xx[$(($1&1))]} _x0=1 _o _r _s _t _i while [ $_x0 -ne $_x1 ] ;do _x0=$_x1 [ $_x0 -eq 0 ] && _x1=0000 || printf -v _x1 "%u" $(( (${_x0}000 + ${1}00000000000/${_x0} )/2 )) printf -v _x1 "%.0f" ${_x1:0:${#_x1}-3}.${_x1:${#_x1}-3} done _x1=0000$_x1 printf -v _r "%.0f" ${_x1:0:${#_x1}-4}.${_x1:${#_x1}-4} printf -v _o "%${1}s" "" printf " %s\n" "${_o// / o}" "${_o// / $'\041'}"{,} printf -v _o "%$[_r-1]s" "" _s=\ \ _t=\ \ for ((_i=_r; _i--; 1)) ;do _s+=" -${_o// /--}"; _t+=${_o}$' \041'${_o}; done printf -v _r "\041%5.2f\041" ${_x1:0:${#_x1}-4}.${_x1:${#_x1}-4}; printf "%s\n%s\n%s\n" "$_s" "$_t" "$_t" " ${_o}${_o// /${_o// /--}--}-" \ "$_o${_o// /${_o// / } }"{$' \041'{,},+-----+,$'\041 \041',"${_r:0\ }",$'\041 \041',+-----+} } Will work like:
squareroot 16 o o o o o o o o o o o o o o o o ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ------- ------- ------- ------- ! ! ! ! ! ! ! ! ------------------------- ! ! +-----+ ! ! ! 4.00! ! ! +-----+ squareroot 32 o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o o ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ----------- ----------- ----------- ----------- ----------- ----------- ! ! ! ! ! ! ! ! ! ! ! ! ------------------------------------------------------------- ! ! +-----+ ! ! ! 5.66! ! ! +-----+ Please note: The root is square!!
Thanks, to ggmx's for code on generating n digits of pi in java.
import java.math.BigDecimal; import java.math.RoundingMode; import java.util.regex.Matcher; import java.util.regex.Pattern; import static java.lang.Math.sqrt; public class myClass { private static final BigDecimal TWO = new BigDecimal("2"); private static final BigDecimal FOUR = new BigDecimal("4"); private static final BigDecimal FIVE = new BigDecimal("5"); private static final BigDecimal TWO_THIRTY_NINE = new BigDecimal("239"); public static BigDecimal pi(int numDigits) { int calcDigits = numDigits + 10; return FOUR.multiply((FOUR.multiply(arccot(FIVE, calcDigits))) .subtract(arccot(TWO_THIRTY_NINE, calcDigits))) .setScale(numDigits, RoundingMode.DOWN); } private static BigDecimal arccot(BigDecimal x, int numDigits) { BigDecimal unity = BigDecimal.ONE.setScale(numDigits, RoundingMode.DOWN); BigDecimal sum = unity.divide(x, RoundingMode.DOWN); BigDecimal xpower = new BigDecimal(sum.toString()); BigDecimal term = null; boolean add = false; for (BigDecimal n = new BigDecimal("3"); term == null || term.compareTo(BigDecimal.ZERO) != 0; n = n.add(TWO)) { xpower = xpower.divide(x.pow(2), RoundingMode.DOWN); term = xpower.divide(n, RoundingMode.DOWN); sum = add ? sum.add(term) : sum.subtract(term); add = !add; } return sum; } public static void main(String[] args) throws Exception { int sqrtThis = 3; int expectedPercision = 4; int intgerAnswer = (int) sqrt(sqrtThis); int cantThinkOfVarName = expectedPercision - String.valueOf(intgerAnswer).length(); boolean done = false; int piPrecision = 10000 * expectedPercision; Double bestMatch = -1.0; while (done == false) { BigDecimal PI = pi(piPrecision); String piString = PI.toString(); Pattern p = Pattern.compile(intgerAnswer + "[0-9]{" + cantThinkOfVarName + "}"); Matcher m = p.matcher(piString); Double offset = sqrtThis + 1.0; while (m.find()) { Double d = Double.parseDouble(m.group(0)); d = d / Math.pow(10, cantThinkOfVarName); if ((int) (d * d) == sqrtThis ||(int) (d * d) == sqrtThis + 1 ) { done = true; Double newOffSet = Math.abs(d * d - sqrtThis); if (newOffSet < offset) { offset = newOffSet; bestMatch = d; } } } piPrecision = piPrecision + piPrecision; } System.out.println(bestMatch); } } Didn't feel like implementing input. To test code change sqrtThis and expectedPercision.
Here is how the code works. Firstly, getting the sqrt root for integer is trivial so I did not feel like implementing that and instead used javas built in sqrt fcn. The rest of the code is 100% legit though.
The basic idea, since pi is an infinite long non-repeating decimal number all number sequences must occur within it (read edit). Therefor your answer is inside pi!! As such we can just apply a regex search on pi searching for you answer. If we are unable to find a good answer then we will just double the size of pi that we are search on!
It really easy, in fact one could say that it is as easy as pi :)
Edit
Pi has not been proven to contain every sequence of finite numbers within it. The fact that pi is infinite and non-repeating is not sufficient proof for such as statement as proven by Exelian. However many mathematicians do believe pi contains every sequence of finite numbers.
this one is the most accurate (bonus: also works for letters!)
Please enter the number : <script> $("#b").submit(function() { var a = $("#a").val(); a = "√" +a ; document.write(a); }); </script> Here is a fiddle
document.write counters it enough. \$\endgroup\$ This will eventually get you a square root.
#include <iostream> #include <float.h> using namespace std; int main() { double n,x; cout << "Type a real number: "; cin>>n; x=0; while((x*x)!=n) { x+=DBL_EPSILON; } cout << x << endl; return 0; } I corrected code to reflect the question better. Thank you for your suggestions...code is updated.
x+=1e-16? \$\endgroup\$ DBL_EPSILON. \$\endgroup\$ This solution:
Sum N independent uniform [-.5,.5] random variables. Estimate the standard deviation by taking the mean of the absolute values. As it happens, the standard deviation is proportional to sqrt(N) as N->\infty. 139 and 2.71828 are just scale factors that control the precision and they were chosen to look mysterious.
import math import random import sys def oo(q, j): for k in range(j): t = -q/2. for n in range(q): t += random.random() yield t if __name__ == "__main__": p = 139 # must be prime e = math.exp(1) # a very natural number for a in sys.argv[1:]: s = int(a) m = 0 for z in oo(p*s, p): m += abs(z) m /= p print("trollsqrt={}, real={}".format(m/e, math.sqrt(s))) Your question don't compile because you put a ! at the end. C++ don't like !
Here the correct question for the compiler:
Hi guys, for my class I need to make a number square root but it doesnt work !!HELLPP Oh.. and the make file.
CXX_FLAGS=-std=c++11 -include 26317.def LD_FLAGS=-lstdc++ -lm all: 26317.cpp gcc -include math.h -include iostream $(CXX_FLAGS) $(LD_FLAGS) $^ -o sqrt and 26317.def. This should already be present in your compiler
#define Hi int #define guys main(int #define a arg #define need ; #define doesnt std::endl; #define work return #define number ; #define HELLPP 0;??> #define it << #define my ??< #define for char const *[]) #define square std::cout #define root << #define I arg #define make >> #define but sqrt(arg) #define class double #define to std::cin Yep, someone can use -E to output the correct preprocess answer, but if you know -E you also know how to squareroot. :P Here some the preprocessed. Very poor minimal solution, no bound check, no prompt. TIL that trigraph are preprocessed.
# 1 "26317.cpp" # 1 "<command-line>" # 1 "/usr/include/stdc-predef.h" 1 3 4 # 1 "<command-line>" 2 # 1 "./26317.def" 1 # 1 "<command-line>" 2 # 1 "26317.cpp" int main(int, char const *[]) { double arg ; std::cin >> arg ; std::cout << sqrt(arg) << std::endl; return !!0;} 