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  • how to calculate a bill from several tables on mysql?

    - by Audel
    I'm using mysql to create an hotel booking system, but i am struggling a little bit to calculate the final bill. I need a SELECT command to get data from several tables and make some calculations. Basically I just need to get the 'night cost' from a table called 'room_types'. Then, use DATEDIFF function to get the difference of days between the 'checkin' and 'checkout' columns in the table 'room_booking' and multiply the difference with the night cost and display the total. These are the tables I would be using: are room_booking, room_types, booking, and room. One booking may have several room bookings, so Im looking for a table that looks something like this: +------------+------------+---------------+------------------+ | bookingid | Room price | nights stayed | total room price | +------------+------------+---------------+------------------+ | B001 | 30.00 | 4 | 120.00 | +------------+------------+---------------+------------------+ | B001 | 40.00 | 3 | 120.00 | +------------+------------+---------------+------------------+ booking id comes from table 'booking' room price from 'room_types', 'nights stayed' is calculated from the table room_booking, using the datediff command between checkin and checkout . I hope i was clear

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  • comparing two angles

    - by Elazar Leibovich
    Given four points in the plane, A,B,X,Y, I wish to determine which of the following two angles is smaller ?ABX or ?ABY. I'd rather not use cos or sqrt, in order to preserve accuracy. In the case where A=(-1,0),B=(0,0), I can compare the two angles ?ABX and ?ABY, by calculating the dot product of the vectors X,Y, and watch it's sign. What I can do in this case is: Determine whether or not ABX turns right or left If ABX turns left check whether or not Y and A are on the same side of the line on segment BX. If they are - ?ABX is a smaller than ABY. If ABX turns right, then Y and A on the same side of BX means that ?ABX is larger than ?ABY. But this seems too complicated to me. Any simpler approach?

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  • Trouble with javascript subtraction

    - by rockinthesixstring
    I'm working on a simple subtraction problem, but unfortunately it keeps returning NaN Here is the function function subtraction(a, b) { var regexp = /[$][,]/g; a = a.replace(regexp, ""); b - b.replace(regexp, ""); var _a = parseFloat(a); var _b = parseFloat(b); return _a - _b; } And here is how I'm calling it. txtGoodWill.value = subtraction(txtSellingPrice.value, txtBalanceSheet.value); The numbers that get submitted to the function are ONLY Currency (IE: $2,000 or $20, etc) Now I know that I cannot subtract numbers with a $ or a ,, but I can't for the life of me figure out why they are getting evaluated in the equasion.

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  • Multiple outliers for two variable linear regression

    - by Dave Jarvis
    Problem Building on my previous question, the "extreme" outliers in the following graph are somewhat obvious: Question Given: T - Set of all temperatures Y - Set of all years ST - Sum of temperatures. SY - Sum of years. N - Number of elements T(n) - Temperature of the nth element in the temperature set How would you implement an efficient MySQL stored procedure or user-defined function (UDF) to determine if T(n) is an outlier? (If such an implementation already exists, that would be good to know as well.) Related Sites I am slowly working through these sites to get a better understanding of the problem: Multiple Outliers Detection Procedures in Linear Regression M-estimator Measure of Surprise for Outlier Detection Ordinary Least Squares Linear Regression Many thanks!

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  • JavaScript Image zoom with CSS3 Transforms, How to calculate Origin? (with example)

    - by Sunday Ironfoot
    I'm trying to implement an image zoom effect, a bit like how the zoom works with Google Maps, but with a grid of fix position images. I've uploaded an example of what I have so far here: http://www.dominicpettifer.co.uk/Files/MosaicZoom.html (uses CSS3 transforms so only works with Firefox, Opera, Chrome or Safari) Use your mouse wheel to zoom in/out. The HTML source is basically an outer div with an inner-div, and that inner-div contains 16 images arranged using absolute position. It's going to be a Photo Mosaic basically. I've got the zoom bit working using CSS3 transforms: $(this).find('div').css('-moz-transform', 'scale(' + scale + ')'); ...however, I'm relying on the mouse X/Y position on the outer div to zoom in on where the mouse cursor is, similar to how Google Maps functions. The problem is that if you zoom right in on an image, move the cursor to the bottom/left corner and zoom again, instead of zooming to the bottom/left corner of the image, it zooms to the bottom/left of the entire mosaic. This has the effect of appearing to jump about the mosaic as you zoom in closer while moving the mouse around, even slightly. That's basically the problem, I want the zoom to work exactly like Google Maps where it zooms exactly to where your mouse cursor position is, but I can't get my head around the Maths to calculate the transform-origin: X/Y values correctly. Please help, been stuck on this for 3 days now. Here is the full code listing for the mouse wheel event: var scale = 1; $("#mosaicContainer").mousewheel(function(e, delta) { if (delta > 0) { scale += 1; } else { scale -= 1; } scale = scale < 1 ? 1 : (scale > 40 ? 40 : scale); var x = e.pageX - $(this).offset().left; var y = e.pageY - $(this).offset().top; $(this).find('div').css('-moz-transform', 'scale(' + scale + ')') .css('-moz-transform-origin', x + 'px ' + y + 'px'); return false; });

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  • Zip Code Radius Search question...

    - by KnockKnockWhosThere
    I'm wondering if it's possible to find all points by longitude and latitude within X radius of one point? So, if I provide a latitude/longitude of -76.0000, 38.0000, is it possible to simply find all the possible coordinates within (for example) a 10 mile radius of that? I know that there's a way to calculate the distance between two points, which is why I'm not clear as to whether this is possible... Because, it seems like you need to know the center coordinates (-76 and 38 in this case) as well as the coordinates of every other point in order to determine whether it falls within the specified radius... Is that right?

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  • Small-o(n^2) implementation of Polynomial Multiplication

    - by AlanTuring
    I'm having a little trouble with this problem that is listed at the back of my book, i'm currently in the middle of test prep but i can't seem to locate anything regarding this in the book. Anyone got an idea? A real polynomial of degree n is a function of the form f(x)=a(n)x^n+?+a1x+a0, where an,…,a1,a0 are real numbers. In computational situations, such a polynomial is represented by a sequence of its coefficients (a0,a1,…,an). Assuming that any two real numbers can be added/multiplied in O(1) time, design an o(n^2)-time algorithm to compute, given two real polynomials f(x) and g(x) both of degree n, the product h(x)=f(x)g(x). Your algorithm should **not** be based on the Fast Fourier Transform (FFT) technique. Please note it needs to be small-o(n^2), which means it complexity must be sub-quadratic. The obvious solution that i have been finding is indeed the FFT, but of course i can't use that. There is another method that i have found called convolution, where if you take polynomial A to be a signal and polynomial B to be a filter. A passed through B yields a shifted signal that has been "smoothed" by A and the resultant is A*B. This is supposed to work in O(n log n) time. Of course i am completely unsure of implementation. If anyone has any ideas of how to achieve a small-o(n^2) implementation please do share, thanks.

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  • Practical Uses of Fractals in Programming

    - by Sami
    Fractals have always been a bit of a mystery for me. What practical uses (beyond rendering to beautiful images) are there for fractals in the various programming problem domains? And please, don't just list areas that use them. I'm interested in specific algorithms and how fractals are used with those algorithms to solve something in practice. Please at least give a short description of the algorithm.

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  • Fastest way to list all primes below N in python

    - by jbochi
    This is the best algorithm I could come up with after struggling with a couple of Project Euler's questions. def get_primes(n): numbers = set(range(n, 1, -1)) primes = [] while numbers: p = numbers.pop() primes.append(p) numbers.difference_update(set(range(p*2, n+1, p))) return primes >>> timeit.Timer(stmt='get_primes.get_primes(1000000)', setup='import get_primes').timeit(1) 1.1499958793645562 Can it be made even faster? EDIT: This code has a flaw: Since numbers is an unordered set, there is no guarantee that numbers.pop() will remove the lowest number from the set. Nevertheless, it works (at least for me) for some input numbers: >>> sum(get_primes(2000000)) 142913828922L #That's the correct sum of all numbers below 2 million >>> 529 in get_primes(1000) False >>> 529 in get_primes(530) True EDIT: The rank so far (pure python, no external sources, all primes below 1 million): Sundaram's Sieve implementation by myself: 327ms Daniel's Sieve: 435ms Alex's recipe from Cookbok: 710ms EDIT: ~unutbu is leading the race.

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  • Projecting a targetting ring using direct3d

    - by JohnB
    I'm trying to draw a "targetting ring" on the ground below a "unit" in a hobby 3d game I'm working on. Basically I want to project a bright red patterned ring onto the ground terrain below the unit. The only approach I can think of is this - Draw the world once as normal Draw the world a second time but in my vertex shader I have the world x,y,z coordinates of the vertex and I can pass in the coordinates of the highlighted unit - so I can calculate what the u,v coordinates in my project texture should be at that point in the world for that vertex. I'd then use the pixel shader to pick pixels from the target ring texture and blend them into the previously drawn world. I believe that should be easy, and should work but it involves me drawing the whole visible world twice as it's hard to determine exactly which polygons the targetting ring might fall onto. It seems a big overhead to draw the whole world twice, once for the normal lit textured ground, and then again just to draw the targetting ring. Is there a better approach that I'm missing?

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  • Why are there 3 conflicting OpenCV camera calibration formulas?

    - by John
    I'm having a problem with OpenCV's various parameterization of coordinates used for camera calibration purposes. The problem is that three different sources of information on image distortion formulae apparently give three non-equivalent description of the parameters and equations involved: (1) In their book "Learning OpenCV…" Bradski and Kaehler write regarding lens distortion (page 376): xcorrected = x * ( 1 + k1 * r^2 + k2 * r^4 + k3 * r^6 ) + [ 2 * p1 * x * y + p2 * ( r^2 + 2 * x^2 ) ], ycorrected = y * ( 1 + k1 * r^2 + k2 * r^4 + k3 * r^6 ) + [ p1 * ( r^2 + 2 * y^2 ) + 2 * p2 * x * y ], where r = sqrt( x^2 + y^2 ). Assumably, (x, y) are the coordinates of pixels in the uncorrected captured image corresponding to world-point objects with coordinates (X, Y, Z), camera-frame referenced, for which xcorrected = fx * ( X / Z ) + cx and ycorrected = fy * ( Y / Z ) + cy, where fx, fy, cx, and cy, are the camera's intrinsic parameters. So, having (x, y) from a captured image, we can obtain the desired coordinates ( xcorrected, ycorrected ) to produced an undistorted image of the captured world scene by applying the above first two correction expressions. However... (2) The complication arises as we look at OpenCV 2.0 C Reference entry under the Camera Calibration and 3D Reconstruction section. For ease of comparison we start with all world-point (X, Y, Z) coordinates being expressed with respect to the camera's reference frame, just as in #1. Consequently, the transformation matrix [ R | t ] is of no concern. In the C reference, it is expressed that: x' = X / Z, y' = Y / Z, x'' = x' * ( 1 + k1 * r'^2 + k2 * r'^4 + k3 * r'^6 ) + [ 2 * p1 * x' * y' + p2 * ( r'^2 + 2 * x'^2 ) ], y'' = y' * ( 1 + k1 * r'^2 + k2 * r'^4 + k3 * r'^6 ) + [ p1 * ( r'^2 + 2 * y'^2 ) + 2 * p2 * x' * y' ], where r' = sqrt( x'^2 + y'^2 ), and finally that u = fx * x'' + cx, v = fy * y'' + cy. As one can see these expressions are not equivalent to those presented in #1, with the result that the two sets of corrected coordinates ( xcorrected, ycorrected ) and ( u, v ) are not the same. Why the contradiction? It seems to me the first set makes more sense as I can attach physical meaning to each and every x and y in there, while I find no physical meaning in x' = X / Z and y' = Y / Z when the camera focal length is not exactly 1. Furthermore, one cannot compute x' and y' for we don't know (X, Y, Z). (3) Unfortunately, things get even murkier when we refer to the writings in Intel's Open Source Computer Vision Library Reference Manual's section Lens Distortion (page 6-4), which states in part: "Let ( u, v ) be true pixel image coordinates, that is, coordinates with ideal projection, and ( u ~, v ~ ) be corresponding real observed (distorted) image coordinates. Similarly, ( x, y ) are ideal (distortion-free) and ( x ~, y ~ ) are real (distorted) image physical coordinates. Taking into account two expansion terms gives the following: x ~ = x * ( 1 + k1 * r^2 + k2 * r^4 ) + [ 2 p1 * x * y + p2 * ( r^2 + 2 * x^2 ) ] y ~ = y * ( 1 + k1 * r^2 + k2 * r^4 ] + [ 2 p2 * x * y + p2 * ( r^2 + 2 * y^2 ) ], where r = sqrt( x^2 + y^2 ). ... "Because u ~ = cx + fx * u and v ~ = cy + fy * v , … the resultant system can be rewritten as follows: u ~ = u + ( u – cx ) * [ k1 * r^2 + k2 * r^4 + 2 * p1 * y + p2 * ( r^2 / x + 2 * x ) ] v ~ = v + ( v – cy ) * [ k1 * r^2 + k2 * r^4 + 2 * p2 * x + p1 * ( r^2 / y + 2 * y ) ] The latter relations are used to undistort images from the camera." Well, it would appear that the expressions involving x ~ and y ~ coincided with the two expressions given at the top of this writing involving xcorrected and ycorrected. However, x ~ and y ~ do not refer to corrected coordinates, according to the given description. I don't understand the distinction between the meaning of the coordinates ( x ~, y ~ ) and ( u ~, v ~ ), or for that matter, between the pairs ( x, y ) and ( u, v ). From their descriptions it appears their only distinction is that ( x ~, y ~ ) and ( x, y ) refer to 'physical' coordinates while ( u ~, v ~ ) and ( u, v ) do not. What is this distinction all about? Aren't they all physical coordinates? I'm lost! Thanks for any input!

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  • Efficient algorithm to generate all solutions of a linear diophantine equation with ai=1

    - by Ben
    I am trying to generate all the solutions for the following equations for a given H. With H=4 : 1) ALL solutions for x_1 + x_2 + x_3 + x_4 =4 2) ALL solutions for x_1 + x_2 + x_3 = 4 3) ALL solutions for x_1 + x_2 = 4 4) ALL solutions for x_1 =4 For my problem, there are always 4 equations to solve (independently from the others). There are a total of 2^(H-1) solutions. For the previous one, here are the solutions : 1) 1 1 1 1 2) 1 1 2 and 1 2 1 and 2 1 1 3) 1 3 and 3 1 and 2 2 4) 4 Here is an R algorithm which solve the problem. library(gtools) H<-4 solutions<-NULL for(i in seq(H)) { res<-permutations(H-i+1,i,repeats.allowed=T) resum<-apply(res,1,sum) id<-which(resum==H) print(paste("solutions with ",i," variables",sep="")) print(res[id,]) } However, this algorithm makes more calculations than needed. I am sure it is possible to go faster. By that, I mean not generating the permutations for which the sums is H Any idea of a better algorithm for a given H ?

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  • Rot13 for numbers.

    - by dreeves
    EDIT: Now a Major Motion Blog Post at http://messymatters.com/sealedbids The idea of rot13 is to obscure text, for example to prevent spoilers. It's not meant to be cryptographically secure but to simply make sure that only people who are sure they want to read it will read it. I'd like to do something similar for numbers, for an application involving sealed bids. Roughly I want to send someone my number and trust them to pick their own number, uninfluenced by mine, but then they should be able to reveal mine (purely client-side) when they're ready. They should not require further input from me or any third party. (Added: Note the assumption that the recipient is being trusted not to cheat.) It's not as simple as rot13 because certain numbers, like 1 and 2, will recur often enough that you might remember that, say, 34.2 is really 1. Here's what I'm looking for specifically: A function seal() that maps a real number to a real number (or a string). It should not be deterministic -- seal(7) should not map to the same thing every time. But the corresponding function unseal() should be deterministic -- unseal(seal(x)) should equal x for all x. I don't want seal or unseal to call any webservices or even get the system time (because I don't want to assume synchronized clocks). (Added: It's fine to assume that all bids will be less than some maximum, known to everyone, say a million.) Sanity check: > seal(7) 482.2382 # some random-seeming number or string. > seal(7) 71.9217 # a completely different random-seeming number or string. > unseal(seal(7)) 7 # we always recover the original number by unsealing.

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  • GCD function in matlab

    - by SalemFayad
    hi, i am looking for a way to implement the "gcd" function used in matlab in another language but i really cant understand the way it functions. it says in http://www.mathworks.com/access/helpdesk/help/techdoc/ref/gcd.html that: "[G,C,D] = gcd(A,B) returns both the greatest common divisor array G, and the arrays C and D, which satisfy the equation: A(i).*C(i) + B(i).*D(i) = G(i)." but it says nothing about how it calculates C and D. i would be grateful if someone has a clearer idea about this subject! thanks:)

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  • What's a good library for parsing mathematical expressions in java?

    - by CSharperWithJava
    I'm an Android Developer and as part of my next app I will need to evaluate a large variety of user created mathematical expressions and equations. I am looking for a good java library that is lightweight and can evaluate mathematical expressions using user defined variables and constants, trig and exponential functions, etc. I've looked around and Jep seems to be popular, but I would like to hear more suggestions, especially from people who have used these libraries before.

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  • which euler rotations can i use ?

    - by melis
    i have two cartesian coordinates. There are xyz and BIG XYZ. I want to make these are paralel each other.forexample , x paralel to X ,y paralel to Y and z paralel to Z. I use rotation matris but I have a lot of different rotation matris . for example I have 3D point in xyz cartesien coordinates and its called A. and I want to change cartesien coordinate to BIG XYZ and find the same 3D point in this coordinates its called B.Until now it is okay. But when I used different rotational matris , points were changed.what can I do? Which Euler rotations can i use?

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  • How to calculate both positive and negative angle between two lines?

    - by Jaanus
    There is a very handy set of 2d geometry utilities here. The angleBetweenLines has a problem, though. The result is always positive. I need to detect both positive and negative angles, so if one line is 15 degrees "above" or "below" the other line, the shape obviously looks different. The configuration I have is that one line remains stationary, while the other line rotates, and I need to understand what direction it is rotating in, by comparing it with the stationary line. EDIT: in response to swestrup's comment below, the situation is actually that I have a single line, and I record its starting position. The line then rotates from its starting position, and I need to calculate the angle from its starting position to current position. E.g if it has rotated clockwise, it is positive rotation; if counterclockwise, then negative. (Or vice versa.) How to improve the algorithm so it returns the angle as both positive or negative depending on how the lines are positioned?

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  • Simple 3x3 matrix inverse code (C++)

    - by batty
    What's the easiest way to compute a 3x3 matrix inverse? I'm just looking for a short code snippet that'll do the trick for non-singular matrices, possibly using Cramer's rule. It doesn't need to be highly optimized. I'd prefer simplicity over speed. I'd rather not link in additional libraries. Primarily I was hoping to have this on Stack Overflow so that I wouldn't have to hunt around for it or rewrite from scratch again next time.

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  • Number Algorithm

    - by James
    I've been struggling to wrap my head around this for some reason. I have 15 bits that represent a number. The bits must match a pattern. The pattern is defined in the way the bits start out: they are in the most flush-right representation of that pattern. So say the pattern is 1 4 1. The bits will be: 000000010111101 So the general rule is, take each number in the pattern, create that many bits (1, 4 or 1 in this case) and then have at least one space separating them. So if it's 1 2 6 1 (it will be random): 001011011111101 Starting with the flush-right version, I want to generate every single possible number that meets that pattern. The # of bits will be stored in a variable. So for a simple case, assume it's 5 bits and the initial bit pattern is: 00101. I want to generate: 00101 01001 01010 10001 10010 10100 I'm trying to do this in Objective-C, but anything resembling C would be fine. I just can't seem to come up with a good recursive algorithm for this. It makes sense in the above example, but when I start getting into 12431 and having to keep track of everything it breaks down.

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  • Mathematics for Computer Science Students

    - by Ender
    To cut a long story short, I am a CS student that has received no formal Post-16 Maths education for years. Right now even my Algebra is extremely rusty and I have a couple of months to shape up my skills. I've got a couple of video lectures in my bookmarks, consisting of: Pre-Calculus Algebra Calculus Probability Introduction to Statistics Differential Equations Linear Algebra My aim as of today is to be able to read the CLRS book Introduction to Algorithms and be able to follow the Mathematical notation in that, as well as being able to confidently read and back-up any arguments written in Mathematical notation. Aside from these video lectures, can anyone recommend any good books to help teach someone wishing to go from a low-foundation level to a more advanced level of Mathematics? Just as a note, I've taken a first-year module in Analytical Modelling, so I understand some of the basic concepts of Discrete Mathematics. EDIT: Just a note to those that are looking to learn Linear Algebra using the Video Lectures I have posted up. Peteris Krumins' Blog contains a run-through of these lecture notes as well as his own commentary and lecture notes, an invaluable resource for those looking to follow the lectures too.

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  • Howto project a planar polygon on a plane in 3d-space

    - by sum1stolemyname
    I want to project my Polygon along a vector to a plane in 3d Space. I would preferably use a single transformation matrix to do this, but I don't know how to build a matrix of this kind. Given the plane's parameters (ax+by+cz+d), the world coordinates of my Polygon. As stated in the the headline, all vertices of my polygon lie in another plane. the direction vector along which to project my Polygon (currently the polygon's plane's normal vector) goal -a 4x4 transformation matrix which performs the required projection, or some insight on how to construct one myself

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  • Calculating negative fractions in Objective C

    - by Mark Reid
    I've been coding my way through Steve Kochan's Programming in Objective-C 2.0 book. I'm up to an exercise in chapter 7, ex 4, in case anyone has the book. The question posed by the exercise it will the Fraction class written work with negative fractions such as -1/2 + -2/3? Here's the implementation code in question - @implementation Fraction @synthesize numerator, denominator; -(void) print { NSLog(@"%i/%i", numerator, denominator); } -(void) setTo: (int) n over: (int) d { numerator = n; denominator = d; } -(double) convertToNum { if (denominator != 0) return (double) numerator / denominator; else return 1.0; } -(Fraction *) add: (Fraction *) f { // To add two fractions: // a/b + c/d = ((a * d) + (b * c)) / (b * d) // result will store the result of the addition Fraction *result = [[Fraction alloc] init]; int resultNum, resultDenom; resultNum = (numerator * f.denominator) + (denominator * f.numerator); resultDenom = denominator * f.denominator; [result setTo: resultNum over: resultDenom]; [result reduce]; return result; } -(Fraction *) subtract: (Fraction *) f { // To subtract two fractions: // a/b - c/d = ((a * d) - (b * c)) / (b * d) // result will store the result of the addition Fraction *result = [[Fraction alloc] init]; int resultNum, resultDenom; resultNum = numerator * f.denominator - denominator * f.numerator; resultDenom = denominator * f.denominator; [result setTo: resultNum over: resultDenom]; [result reduce]; return result; } -(Fraction *) multiply: (Fraction *) f { // To multiply two fractions // a/b * c/d = (a*c) / (b*d) // result will store the result of the addition Fraction *result = [[Fraction alloc] init]; int resultNum, resultDenom; resultNum = numerator * f.numerator; resultDenom = denominator * f.denominator; [result setTo: resultNum over: resultDenom]; [result reduce]; return result; } -(Fraction *) divide: (Fraction *) f { // To divide two fractions // a/b / c/d = (a*d) / (b*c) // result will store the result of the addition Fraction *result = [[Fraction alloc] init]; int resultNum, resultDenom; resultNum = numerator * f.denominator; resultDenom = denominator * f.numerator; [result setTo: resultNum over: resultDenom]; [result reduce]; return result; } -(void) reduce { int u = numerator; int v = denominator; int temp; while (v != 0) { temp = u % v; u = v; v = temp; } numerator /= u; denominator /= u; } @end My question to you is will it work with negative fractions and can you explain how you know? Part of the issue is I don't know how to calculate negative fractions myself so I'm not too sure how to know. Many thanks.

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