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• Deploying Excel 2003 VSTO Workbook using Sharepoint for versioning control

  - by KClough

  I have an excel 2003 vsto workbook that I would like to make available via sharepoint for version control. Ideally it could be checked in/out by non-developers for tweaking excel equations, and I would be able to deploy the compiled dlls somewhere else when I need to update the managed VSTO code. I understand I may need to use some clickonce functionality as well so when a user first views the sheet they get all the necessary full-trust permissioning. Also, it is my understanding that for a user to use the vsto functionality in an excel 2003 vsto workbook, they must have the compiled dll in thier GAC, is this true? When testing I get trust exceptions otherwise.

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• Graphing a line and scatter points using Matplotlib?

  - by Patrick O'Doherty

  Hi guys I'm using matplotlib at the moment to try and visualise some data I am working on. I'm trying to plot around 6500 points and the line y = x on the same graph but am having some trouble in doing so. I can only seem to get the points to render and not the line itself. I know matplotlib doesn't plot equations as such rather just a set of points so I'm trying to use and identical set of points for x and y co-ordinates to produce the line. The following is my code from matplotlib import pyplot import numpy from pymongo import * class Store(object): """docstring for Store""" def __init__(self): super(Store, self).__init__() c = Connection() ucd = c.ucd self.tweets = ucd.tweets def fetch(self): x = [] y = [] for t in self.tweets.find(): x.append(t['positive']) y.append(t['negative']) return [x,y] if __name__ == '__main__': c = Store() array = c.fetch() t = numpy.arange(0., 0.03, 1) pyplot.plot(array[0], array[1], 'ro', t, t, 'b--') pyplot.show() Any suggestions would be appreciated, Patrick

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• Make UILabel show "No results" instead of "nan"

  - by Mike Rychev

  I have a small app, where user can make some calculations and solve equations. For example, if in a square equation discriminant is less than zero, the x1 and x2 values are "nan", so when I assign x1 and x2 values to UILabels they show "nan" as well. Writing a lot of if's like if(D<0) [label setText:[NSString stringWithFormat: @"No solutions"]]; Doesn't help-there are too many cases. I want to check if after [label setText:[NSString stringWithFormat: @"%f", x]]; label's value is "nan", the label's value will be set to @"No solutions". Doing simple if(label==@"nan") { //code } doesn't help. Thanks in advance!

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• how to apply Discrete wavelet transform on image

  - by abuasis

  I am implementing an android application that will verify signature images , decided to go with the Discrete wavelet transform method (symmlet-8) the method requires to apply the discrete wavelet transform and separate the image using low-pass and high-pass filter and retrieve the wavelet transform coefficients. the equations show notations that I cant understand thus can't do the math easily , also didn't know how to apply low-pass and high-pass filters to my x and y points. is there any tutorial that shows you how to apply the discrete wavelet transform to my image easily that breaks it out in numbers? thanks alot in advance.

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• How to overcome an apparent REST vs. DRY dilemma in rails?

  - by Chris

  A rails app I'm working on features examples of quadratic equations. Obviously, these are all of a common structure: ax^2 + bx + c = 0. I don't want to store every single example of these. I'd rather generate them from a template. Storing hundreds of possible versions of this structure seems highly wasteful and un-DRY. On the other hand, if I generate them, I can't access them again reliably as I could if they were simply multiple database objects. I'm sure there must be a way to overcome this, but I'm still learning rails and I'm obviously not grasping something here. Thanks.

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• How can I write a Template.pm filter to generate PNG output from LaTeX source code?

  - by Sinan Ünür

  I am looking for a way of generating PNG images of equations from LATEX source code embedded in templates. For example, given: [% FILTER latex_display ] \begin{eqnarray*} \max && U(x,y) \\ \mathrm{s.t.} && p_x x + p_y y \leq I \\ && x \geq 0, y \geq 0 \end{eqnarray*} [% END %] I would like to get the output: <div class="latex display"><img src="equation.png" width="x" height="y"></div> which should ultimately display as: I am using ttree to generate documents offline. I know about Template::Plugin::Latex but that is geared towards producing actual documents out of LATEX templates. Any suggestions?

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• What is the best Java numerical method package?

  - by Bob Cross

  I am looking for a Java-based numerical method package that provides functionality including: Solving systems of equations using different numerical analysis algorithms. Matrix methods (e.g., inversion). Spline approximations. Probability distributions and statistical methods. In this case, "best" is defined as a package with a mature and usable API, solid performance and numerical accuracy. Edit: derick van brought up a good point in that cost is a factor. I am heavily biased in favor of free packages but others may have a different emphasis.

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• multidimensional vector rotation and angle computation -- how?

  - by macias

  Input: two multidimensional (for example dim=8) vectors a and b. I need to find out the "directed" angle (0-2*Pi, not 0-Pi) between those vectors a and b. And if they are not parallel I need to rotate vector b in plane a,b by "directed" angle L. If they are parallel, plane does not matter, but angle of rotation is still the same L. For 2d and 3d this is quite easy, but for more dimensions I am lost, I didn't find anything on google, and I prefer using some already proved&tested equations (avoiding errors introduced by my calculations :-D). Thank you in advance for tips, links, etc. Edit: dimension of the space is the same as dimension of the vectors.

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• Explain a block of crazy JS code inside Sizzle(the CSS selector engine)

  - by Andy Li

  So, here is the function for pre-filtering "CHILD": function(match){ if ( match[1] === "nth" ) { // parse equations like 'even', 'odd', '5', '2n', '3n+2', '4n-1', '-n+6' var test = /(-?)(\d*)n((?:\+|-)?\d*)/.exec( match[2] === "even" && "2n" || match[2] === "odd" && "2n+1" || !/\D/.test( match[2] ) && "0n+" + match[2] || match[2]); // calculate the numbers (first)n+(last) including if they are negative match[2] = (test[1] + (test[2] || 1)) - 0; match[3] = test[3] - 0; } // TODO: Move to normal caching system match[0] = done++; return match; } The code is extracted from line 442-458 of sizzle.js. So, why is the line var test = ..., have the exec inputing a boolean? Or is that really a string? Can someone explain it by splitting it into a few more lines of code?

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• programming question about numbers

  - by Syom

  it's a task, on which i think during whole day. i have four integer numbers a, b, c, d, and ineger x[1,40]. i must write a program, which can find the values of {a,b,c,d}, for which one of following equations is true for any (x<=40, x=1) x=a or x=b or x=a+b or x=a+b+c+d or x+a=c+d or x+a+b=c+d or ... x+a+b+c=d or ... it is very difficult to explain what i want to say. maybe the example will be helpful. example: if x=17, by {a=1,b=2,c=5,d=15} i can write x+a+b=c+d whole question is to present any x[1,40] by {a,b,c,d}. hope that you understand what i want to say, and thanks in advance

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• Preallocate memory for a program in Linux before it gets started

  - by Fyg

  Hi, folks, I have a program that repeatedly solves large systems of linear equations using cholesky decomposition. Characterising is that I sometimes need to store the complete factorisation which can exceed about 20 GB of memory. The factorisation happens inside a library that I call. Furthermore, this matrix and the resulting factorisation changes quite frequently and as such the memory requirements as well. I am not the only person to use this compute-node. Therefore, is there a way to start the program under Linux and preallocate free memory for the process? Something like: $: prealloc -m 25G ./program

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• Expert system for writing programs?

  - by aaa

  I am brainstorming an idea of developing a high level software to manipulate matrix algebra equations, tensor manipulations to be exact, to produce optimized C++ code using several criteria such as sizes of dimensions, available memory on the system, etc. Something which is similar in spirit to tensor contraction engine, TCE, but specifically oriented towards producing optimized rather than general code. The end result desired is software which is expert in producing parallel program in my domain. Does this sort of development fall on the category of expert systems? What other projects out there work in the same area of producing code given the constraints?

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• Interpolating height for a point inside a grid based on a discrete height function.

  - by fastrack20

  Hi, I have been wracking my brain to come up with a solution to this problem. I have a lookup table that returns height values for various points (x,z) on the grid. For instance I can calculate the height at A, B, C and D in Figure 1. However, I am looking for a way to interpolate the height at P (which has a known (x,z)). The lookup table only has values at the grid intervals, and P lies between these intervals. I am trying to calculate values s and t such that: A'(s) = A + s(C-A) B'(t) = B + t(P-B) I would then use the these two equations to find the intersection point of B'(t) with A'(s) to find a point X on the line A-C. With this I can calculate the height at this point X and with that the height at point P. My issue lies in calculating the values for s and t. Any help would be greatly appreciated.

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• Solving linear system over integers with numpy

  - by A. R. S.

  I'm trying to solve an overdetermined linear system of equations with numpy. Currently, I'm doing something like this (as a simple example): a = np.array([[1,0], [0,1], [-1,1]]) b = np.array([1,1,0]) print np.linalg.lstsq(a,b)[0] [ 1. 1.] This works, but uses floats. Is there any way to solve the system over integers only? I've tried something along the lines of print map(int, np.linalg.lstsq(a,b)[0]) [0, 1] in order to convert the solution to an array of ints, expecting [1, 1], but clearly I'm missing something. Could anyone point me in the right direction?

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• Calculating the null space of a matrix

  - by Ainsworth

  I'm attempting to solve a set of equations of the form Ax = 0. A is known 6x6 matrix and I've written the below code using SVD to get the vector x which works to a certain extent. The answer is approximately correct but not good enough to be useful to me, how can I improve the precision of the calculation? Lowering eps below 1.e-4 causes the function to fail. from numpy.linalg import * from numpy import * A = matrix([[0.624010149127497 ,0.020915658603923 ,0.838082638087629 ,62.0778180312547 ,-0.336 ,0], [0.669649399820597 ,0.344105317421833 ,0.0543868015800246 ,49.0194290212841 ,-0.267 ,0], [0.473153758252885 ,0.366893577716959 ,0.924972565581684 ,186.071352614705 ,-1 ,0], [0.0759305208803158 ,0.356365401030535 ,0.126682113674883 ,175.292109352674 ,0 ,-5.201], [0.91160934274653 ,0.32447818779582 ,0.741382053883291 ,0.11536775372698 ,0 ,-0.034], [0.480860406786873 ,0.903499596111067 ,0.542581424762866 ,32.782593418975 ,0 ,-1]]) def null(A, eps=1e-3): u,s,vh = svd(A,full_matrices=1,compute_uv=1) null_space = compress(s <= eps, vh, axis=0) return null_space.T NS = null(A) print "Null space equals ",NS,"\n" print dot(A,NS)

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• 2D pixel array rotation and scaling

  - by Betamoo

  Hi I am working on C# program to process images ( given as int[,] ).. So I have 2D array of pixels, and I need to rotate them around a point, then scale them down to fit the original array.. I already found articles about using matrix to transform to a point and rotate then transform back.. What remains is to scale the resultant image to fit an array of original size.. How that can be done? (preferably with 2 equations one for x and one for y ) Thanks in advance

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• Issue using Python to solve the Coin problem [closed]

  - by challarao

  I'm attempting to solve a problem commonly known as the Coin problem, but using McNuggets. McNuggets come in boxes containing either 6, 9, or 20 nuggets. I want to write a python script that uses Diophantine equations to determine if a given number of McNuggets n can be exactly purchased in these groupings. For example: 26 McNuggets -- Possible: 1 6-pack, 0 9-packs, 1 20-pack 27 McNuggets -- Possible: 0 6-packs, 3 9-packs, 0 20-packs 28 McNuggets -- Not possible This is my current attempt at writing the solution in Python, but the output is incorrect and I'm not sure what's wrong. n=input("Enter the no.of McNuggets:") a,b,c=0,0,0 count=0 for a in range(n): if 6*a+9*b+20*c==n: count=count+1 break else: for b in range(n): if 6*a+9*b+20*c==n: count=count+1 break else: for c in range(n): if 6*a+9*b+20*c==n: count=count+1 break if count>0: print "It is possible to buy exactly",n,"packs of McNuggetss",a,b,c else: print "It is not possible to buy"

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• How to convert string to integer?

  - by user1260584

  So I'm having a hard time with my situation and need some advice. I'm trying to convert my two Strings that I have into integers, so that I can use them in math equations. Here is what I tried, however it brings me an error in the app. ' equals.setOnClickListener(new View.OnClickListener() { public void onClick(View arg0) { // TODO Auto-generated method stub num1 = edit.getText().toString(); num2 = edit.getText().toString(); int first = Integer.parseInt(num1); int second = Integer.parseInt(num2); edit.setText(first + second); } }); Is there something that I am doing wrong? Thank you for any help. EDIT: Yes this is Java. num1 and num2 are strings that I have previously named. What do you mean by trim?

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• Difference between the terms Material & Effect

  - by codey

  I'm making an effect system right now (I think, because it may be a material system... or both!). The effects system follows the common (e.g. COLLADA, DirectX) effect framework abstraction of Effects have Techniques, Techniques have Passes, Passes have States & Shader Programs. An effect, according to COLLADA, defines the equations necessary for the visual appearance of geometry and screen-space image processing. Keeping with the abstraction, effects contain techniques. Each effect can contain one or many techniques (i.e. ways to generate the effect), each of which describes a different method for rendering that effect. The technique could be relate to quality (e.g. high precision, high LOD, etc.), or in-game-situation (e.g. night/day, power-up-mode, etc.). Techniques hold a description of the textures, samplers, shaders, parameters, & passes necessary for rendering this effect using one method. Some algorithms require several passes to render the effect. Pipeline descriptions are broken into an ordered collection of Pass objects. A pass provides a static declaration of all the render states, shaders, & settings for "one rendering pipeline" (i.e. one pass). Meshes usually contain a series of materials that define the model. According to the COLLADA spec (again), a material instantiates an effect, fills its parameters with values, & selects a technique. But I see material defined differently in other places, such as just the Lambert, Blinn, Phong "material types/shaded surfaces", or as Metal, Plastic, Wood, etc. In game dev forums, people often talk about implementing a "material/effect system". Is the material not an instance of an effect? Ergo, if I had effect objects, stored in a collection, & each effect instance object with there own parameter setting, then there is no need for the concept of a material... Or am I interpreting it wrong? Please help by contributing your interpretations as I want to be clear on a distinction (if any), & don't want to miss out on the concept of a material if it should be implemented to follow the abstraction of the DirectX FX framework & COLLADA definitions closely.

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• Which is the fastest way to move 1Petabyte from one storage to a new one?

  - by marc.riera

  First of all, thanks for reading, and sorry for asking something related to my job. I understand that this is something that I should solve by myself but as you will see its something a bit difficult. A small description: Now Storage = 1PB using DDN S2A9900 storage for the OSTs, 4 OSS , 10 GigE network. (lustre 1.6) 100 compute nodes with 2x Infiniband 1 infiniband switch with 36 ports After Storage = Previous storage + another 1PB using DDN S2A 990 or LSI E5400 (still to decide) (lustre 2.0) 8 OSS , 10GigE network 100 compute nodes with 2x Infiniband Previous experience: transfered 120 TB in less than 3 days using following command: tar -C /old --record-size 2048 -b 2048 -cf - dir | tar -C /new --record-size 2048 -b 2048 -xvf - 2>&1 | tee /tmp/dir.log So , big problem here, using big mathematical equations I conclude that we are going to need 1 month to transfer the data from one side to the new one. During this time the researchers will need to step back, and I'm personally not happy with this. I'm telling you that we have infiniband connections because I think that may be there is a chance to use it to transfer the data using 18 compute nodes (18 * 2 IB = 36 ports) to transfer the data from one storage to the other. I'm trying to figure out if the IB switch will handle all the traffic but in case it just burn up will go faster than using 10GigE. Also, having lustre 1.6 and 2.0 agents on same server works quite well, with this there is no need to go by 1.8 to upgrade the metadata servers with two steps. Any ideas? Many thanks Note 1: Zoredache, we can divide it in two blocks (A)600Tb and (B)400Tb. The idea is to move (A) to new storage which is lustre2.0 formated, then format where (A) was with lustre2.0 and move (B) to this lustre2.0 block and extend with the space where (B) was. This way we will end with (A) and (B) on separate filesystems, with 1PB each.

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• Converting PDF eBooks into a Kindle format

  - by Ender

  Over the past couple of years I've amassed quite a collection of guides, tutorials and ebooks in PDF format. A lot of these are quite useful for work, especially PDF documentation, and rather than have to be at a computer every time I want to read how to do something in Sitecore or to read through a software testing ebook I'd like to do it on my brand-spanking-new Kindle. However, even though there is now a native PDF reader on the Kindle due to the nature of PDF's they are practically unreadable. The text doesn't wrap due to how PDF's are sized and so far after a bunch of Google searches I've yet to find a viable solution to get my PDF's converted into a readable Kindle format. Sometimes these books have code or pictures/tables in them, but most of the time they're text-heavy and to be honest I'd be surprised if there wasn't a free tool to handle the converting of PDF to one of the (seemingly many) Kindle formats. So, can anyone help me out with this? EDIT: I've tried Calibre, and have checked their forums to play with some of the advanced settings, yet the solutions available seem to be extremely poor, especially if the book you're attempting to read contains equations, code, or anything outside of plain text. I've also tried Amazon's conversion service, which wasn't much help with such documents. The best way I have found so far is to build the entire thing over again in ePub or RTF format and convert to MOBI from there. This works for text-heavy books with tables, but anything technical still isn't covered. Can anyone help with this?

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• Problems implementing a screen space shadow ray tracing shader

  - by Grieverheart

  Here I previously asked for the possibility of ray tracing shadows in screen space in a deferred shader. Several problems were pointed out. One of the most important problem is that only visible objects can cast shadows and objects between the camera and the shadow caster can interfere. Still I thought it'd be a fun experiment. The idea is to calculate the view coordinates of pixels and cast a ray to the light. The ray is then traced pixel by pixel to the light and its depth is compared with the depth at the pixel. If a pixel is in front of the ray, a shadow is casted at the original pixel. At first I thought that I could use the DDA algorithm in 2D to calculate the distance 't' (in p = o + t d, where o origin, d direction) to the next pixel and use it in the 3D ray equation to find the ray's z coordinate at that pixel's position. For the 2D ray, I would use the projected and biased 3D ray direction and origin. The idea was that 't' would be the same in both 2D and 3D equations. Unfortunately, this is not the case since the projection matrix is 4D. Thus, some tweak needs to be done to make this work this way. I would like to ask if someone knows of a way to do what I described above, i.e. from a 2D ray in texture coordinate space to get the 3D ray in screen space. I did implement a simple version of the idea which you can see in the following video: video here Shadows may seem a bit pixelated, but that's mostly because of the size of the step in 't' I chose. And here is the shader: #version 330 core uniform sampler2D DepthMap; uniform vec2 projAB; uniform mat4 projectionMatrix; const vec3 light_p = vec3(-30.0, 30.0, -10.0); noperspective in vec2 pass_TexCoord; smooth in vec3 viewRay; layout(location = 0) out float out_AO; vec3 CalcPosition(void){ float depth = texture(DepthMap, pass_TexCoord).r; float linearDepth = projAB.y / (depth - projAB.x); vec3 ray = normalize(viewRay); ray = ray / ray.z; return linearDepth * ray; } void main(void){ vec3 origin = CalcPosition(); if(origin.z < -60) discard; vec2 pixOrigin = pass_TexCoord; //tex coords vec3 dir = normalize(light_p - origin); vec2 texel_size = vec2(1.0 / 600.0); float t = 0.1; ivec2 pixIndex = ivec2(pixOrigin / texel_size); out_AO = 1.0; while(true){ vec3 ray = origin + t * dir; vec4 temp = projectionMatrix * vec4(ray, 1.0); vec2 texCoord = (temp.xy / temp.w) * 0.5 + 0.5; ivec2 newIndex = ivec2(texCoord / texel_size); if(newIndex != pixIndex){ float depth = texture(DepthMap, texCoord).r; float linearDepth = projAB.y / (depth - projAB.x); if(linearDepth > ray.z + 0.1){ out_AO = 0.2; break; } pixIndex = newIndex; } t += 0.5; if(texCoord.x < 0 || texCoord.x > 1.0 || texCoord.y < 0 || texCoord.y > 1.0) break; } } As you can see, here I just increment 't' by some arbitrary factor, calculate the 3D ray and project it to get the pixel coordinates, which is not really optimal. Hopefully, I would like to optimize the code as much as possible and compare it with shadow mapping and how it scales with the number of lights. PS: Keep in mind that I reconstruct position from depth by interpolating rays through a full screen quad.

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• How much is a subscriber worth?

  - by Tom Lewin

  This year at Red Gate, we’ve started providing a way to back up SQL Azure databases and Azure storage. We decided to sell this as a service, instead of a product, which means customers only pay for what they use. Unfortunately for us, it makes figuring out revenue much trickier. With a product like SQL Compare, a customer pays for it, and it’s theirs for good. Sure, we offer support and upgrades, but, fundamentally, the sale is a simple, upfront transaction: we’ve made this product, you need this product, we swap product for money and everyone is happy. With software as a service, it isn’t that easy. The money and product don’t change hands up front. Instead, we provide a service in exchange for a recurring fee. We know someone buying SQL Compare will pay us $X, but we don’t know how long service customers will stay with us, or how much they will spend. How do we find this out? We use lifetime value analysis. What is lifetime value? Lifetime value, or LTV, is how much a customer is worth to the business. For Entrepreneurs has a brilliant write up that we followed to conduct our analysis. Basically, it all boils down to this equation: LTV = ARPU x ALC To make it a bit less of an alphabet-soup and a bit more understandable, we can write it out in full: The lifetime value of a customer equals the average revenue per customer per month, times the average time a customer spends with the service Simple, right? A customer is worth the average spend times the average stay. If customers pay on average $50/month, and stay on average for ten months, then a new customer will, on average, bring in $500 over the time they are a customer! Average spend is easy to work out; it’s revenue divided by customers. The problem comes when we realise that we don’t know exactly how long a customer will stay with us. How can we figure out the average lifetime of a customer, if we only have six months’ worth of data? The answer lies in the fact that: Average Lifetime of a Customer = 1 / Churn Rate The churn rate is the percentage of customers that cancel in a month. If half of your customers cancel each month, then your average customer lifetime is two months. The problem we faced was that we didn’t have enough data to make an estimate of one month’s cancellations reliable (because barely anybody cancels)! To deal with this data problem, we can take data from the last three months instead. This means we have more data to play with. We can still use the equation above, we just need to multiply the final result by three (as we worked out how many three month periods customers stay for, and we want our answer to be in months). Now these estimates are likely to be fairly unreliable; when there’s not a lot of data it pays to be cautious with inference. That said, the numbers we have look fairly consistent, and it’s super easy to revise our estimates when new data comes in. At the very least, these numbers give us a vague idea of whether a subscription business is viable. As far as Cloud Services goes, the business looks very viable indeed, and the low cancellation rates are much more than just data points in LTV equations; they show that the product is working out great for our customers, which is exactly what we’re looking for!

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• Is there a problem with scrollTop in Chrome?

  - by Shaun

  I am setting scrollTop and scrollLeft for a div that I am working with. The code looks like this: div.scrollLeft = content.cx*scalar - parseInt(div.style.width)/2; div.scrollTop = content.cy*scalar - parseInt(div.style.height)/2; This works just fine in FF, but only scrollLeft works in chrome. As you can see, the two use almost identical equations and as it works in FF I am just wondering if this is a problem with Chrome? Update: If I switch the order of the assignments then scrollTop will work and scrollLeft won't. <div id="container" style = "height:600px; width:600px; overflow:auto;" onscroll = "updateCenter()"> <script> var div = document.getElementById('container'); function updateCenter() { svfdim.cx = (div.scrollLeft + parseFloat(div.style.width)/2)/scalar; svfdim.cy = (div.scrollTop + parseFloat(div.style.height)/2)/scalar; } function updateScroll(svfdim, scalar, div) { div.scrollTop = svgdim.cy*scalar - parseFloat(div.style.height)/2; div.scrollLeft = svgdim.cx*scalar - parseFloat(div.style.width)/2; } function resizeSVG(Root) { Root.setAttribute("height", svfdim.height*scalar); Root.setAttribute("width", svfdim.width*scalar); updateScroll(svgdim, scalar, div); } </script>

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• Mathematically Find Max Value without Conditional Comparison

  - by Cnich

  ----------Updated ------------ codymanix and moonshadow have been a big help thus far. I was able to solve my problem using the equations and instead of using right shift I divided by 29. Because with 32bits signed 2^31 = overflows to 29. Which works! Prototype in PHP $r = $x - (($x - $y) & (($x - $y) / (29))); Actual code for LEADS (you can only do one math function PER LINE!!! AHHHH!!!) DERIVDE1 = IMAGE1 - IMAGE2; DERIVED2 = DERIVED1 / 29; DERIVED3 = DERIVED1 AND DERIVED2; MAX = IMAGE1 - DERIVED3; ----------Original Question----------- I don't think this is quite possible with my application's limitations but I figured it's worth a shot to ask. I'll try to make this simple. I need to find the max values between two numbers without being able to use a IF or any conditional statement. In order to find the the MAX values I can only perform the following functions Divide, Multiply, Subtract, Add, NOT, AND ,OR Let's say I have two numbers A = 60; B = 50; Now if A is always greater than B it would be simple to find the max value MAX = (A - B) + B; ex. 10 = (60 - 50) 10 + 50 = 60 = MAX Problem is A is not always greater than B. I cannot perform ABS, MAX, MIN or conditional checks with the scripting applicaiton I am using. Is there any way possible using the limited operation above to find a value VERY close to the max?

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