We all know, if we divide (1.0/0.0) we will get +Infinity and if we divide (-1.0/0.0) we will get -Infinity! I want to know How computers calculate this value, internally ??
PS: I wondering why I didn't find any question regarding this in any of StackExchange community!
Short answer: It will detect that you divide by 0 and not do the actual division.
Long answer here: https://softwareengineering.stackexchange.com/questions/311155/will-a-computer-attempt-to-divide-by-zero
Related
I'm sure someone else has asked this but my Google foo is failing me and I cannot find it.
When I divide more than once in an equation like this:
this.active[i].pos(last.pos()+(last.width()/2)+10+(this.active[i].width()/2));
"/2)+10+(this.active[i].width()/" will come up with regular expression formatting(all orange) in the editor which is driving me insane. :(
Is there a way I can change my settings to prevent this? I do not use regular expression at all, so disabling it's formatting entirely in the editor would be acceptable.
Can anyone provide, or point me towards, an answer?
If you found it on Google, I would appreciate learning your search terms.
Thank you. :)
I've been searching the web for about 45 minutes trying to find a solution to this very question when I came across this question here on stack overflow. I almost started a bounty on it but decided I'd see if I could figure it out myself.
I came up with two possible solutions, both of which are much simpler than I though they would be.
Solution 1: Separate the formula into two sections that can be stored in variables and added together when needed. For example, I happened to be writing a formula for a surface area calculation, which formed a regular expression and returned the incorrect answer:
return [(this.base * this.height)/2] + [(this.perimeter * this.slant)/2];
I split the formula at the + and stored them in variables:
var a = (this.base * this.height)/2;
var b = (this.perimeter * this.slant)/2;
return a + b
This solution worked just fine. But then I started thinking that there had to be a simpler solution I was over-looking which led me to:
Solution 2: Dividing by 2 is the same as multiplying by 0.5 (duh!) . In my case - and in almost any case - dividing by 2 and multiplying by 0.5 will get you the same answer. My code then looked like this:
return [(this.base * this.height) * 0.5] + [(this.perimeter * this.slant) * 0.5];
I tested both, and both work, though obviously solution 2 is more efficient (less code).
The only time I could imagine needing to use solution 1 is if you're dividing by a very long number or a prime number (dividing by 3 gives you a more accurate result than multiplying by 0.33).
Anyway, I know you posted this question months ago and probably either came up with a solution or moved on, but I figured I'd post this answer anyway as a reference for any future issues with the same idea.
(Also, this is in JavaScript but I can't imagine something this simple is any different in a similar language).
Im making a calculator and its already finished and released but now i want to update it and give it an percentage button and i want it to work.. But i have damn no idea how i will count the percentage of 2 variables..
i did try the % operator but that seemed to be something totaly else.. Didn't realy workthe way i wanted..
Anyone know how i would do this?
And let say how would the user work on a calculator? When you type on a calculator i want to count 20 out of 100% percent.. How would this work in an application.?
The language you are using would be helpful, but 99% of the time the % operator is the modulus operator which finds the remainder of a division of the 2 values.
http://en.wikipedia.org/wiki/Modulo_operation
I think you might be making this harder than it has to be. How would you figure out a percentage on paper? Now write an algorithm that follows the steps you just completed. I'll leave that as an exercise for the poster.
I'd think you want to divide the two values and then multiply by 100 to get the percent. For example, let's take 5 out of 10 as an example:
100*5/10 = 500/10 = 50 which is 50% or half of the value.
% is for modulo, which is the remainder after division. For example
5 % 3 == 2
You want / unless your language is really esoteric.
If you want x% of y, just multiple y by (x/100).
Also, the other answers are spot on about % being the modulo operator, which returns the remainder of a division operation.
The % operator is actually mod (i.e. remainder after division). You could always try:
int result = var * 0.2 (20% of your variable)
Let me know if this answers your question.
I have question that comes from a algorithms book I'm reading and I am stumped on how to solve it (it's been a long time since I've done log or exponent math). The problem is as follows:
Suppose we are comparing implementations of insertion sort and merge sort on the same
machine. For inputs of size n, insertion sort runs in 8n^2 steps, while merge sort runs in 64n log n steps. For which values of n does insertion sort beat merge sort?
Log is base 2. I've started out trying to solve for equality, but get stuck around n = 8 log n.
I would like the answer to discuss how to solve this mathematically (brute force with excel not admissible sorry ;) ). Any links to the description of log math would be very helpful in my understanding your answer as well.
Thank you in advance!
http://www.wolframalpha.com/input/?i=solve%288+log%282%2Cn%29%3Dn%2Cn%29
(edited since old link stopped working)
Your best bet is to use Newton;s method.
http://en.wikipedia.org/wiki/Newton%27s_method
One technique to solving this would be to simply grab a graphing calculator and graph both functions (see the Wolfram link in another answer). Find the intersection that interests you (in case there are multiple intersections, as there are in your example).
In any case, there isn't a simple expression to solve n = 8 log₂ n (as far as I know). It may be simpler to rephrase the question as: "Find a zero of f(n) = n - 8 log₂ n". First, find a region containing the intersection you're interested in, and keep shrinking that region. For instance, suppose you know your target n is greater than 42, but less than 44. f(42) is less than 0, and f(44) is greater than 0. Try f(43). It's less than 0, so try 43.5. It's still less than 0, so try 43.75. It's greater than 0, so try 43.625. It's greater than 0, so keep going down, and so on. This technique is called binary search.
Sorry, that's just a variation of "brute force with excel" :-)
Edit:
For the fun of it, I made a spreadsheet that solves this problem with binary search: binary‑search.xls . The binary search logic is in the second data column, and I just auto-extended that.
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In trying to solve a particular Project Euler question, I ran into difficulties with a particular mathematical formula. According to this web page (http://www.mathpages.com/home/kmath093.htm), the formula for determining the probability for rolling a sum, T, on a number of dice, n, each with number of sides, s, each numbered 1 to s, can be given as follows:
alt text http://www.freeimagehosting.net/uploads/8294d47194.gif
After I started getting nonsensical answers in my program, I started stepping through, and tried this for some specific values. In particular, I decided to try the formula for a sum T=20, for n=9 dice, each with s=4 sides. As the sum of 9 4-sided dice should give a bell-like curve of results, ranging from 4 to 36, a sum of 20 seems like it should be fairly (relatively speaking) likely. Dropping the values into the formula, I got:
alt text http://www.freeimagehosting.net/uploads/8e7b339e32.gif
Since j runs from 0 to 7, we must add over all j...but for most of these values, the result is 0, because at least one the choose formulae results are 0. The only values for j that seem to return non-0 results are 3 and 4. Dropping 3 and 4 into this formula, I got
alt text http://www.freeimagehosting.net/uploads/490f943fa5.gif
Which, when simplified, seemed to go to:
alt text http://www.freeimagehosting.net/uploads/603ca84541.gif
which eventually simplifies down to ~30.75. Now, as a probability, of course, 30.75 is way off...the probability must be between 0 and 1, so something has gone terribly wrong. But I'm not clear what it is.
Could I misunderstanding the formula? Very possible, though I'm not clear at all where the breakdown would be occuring. Could it be transcribed wrong on the web page? Also possible, but I've found it difficult to find another version of it online to check it against. Could I be just making a silly math error? Also possible...though my program comes up with a similar value, so I think it's more likely that I'm misunderstanding something.
Any hints?
(I would post this on MathOverflow.com, but I don't think it even comes close to being the kind of "postgraduate-level" mathematics that is required to survive there.)
Also: I definitely do not want the answer to the Project Euler question, and I suspect that other people that my stumble across this would feel the same way. I'm just trying to figure out where my math skills are breaking down.
According to mathworld (formula 9 is the relevant one), the formula from your source is wrong.
The correct formula is supposed to be n choose j, not n choose T. That'll really reduce the size of the values within the summation.
The mathworld formula uses k instead of j and p instead of T:
Take a look at article in wikipedia - Dice.
The formula here looks almost similar, but have one difference. I think it will solve your problem.
I'm going to have to show my ignorance here.... Isn't 9 choose 20 = 0? More generally, isn't n choose T going to always be 0 since T>=n? Perhaps I'm reading this formula incorrectly (I'm not a math expert), but looking at de Moive's work, I'm not sure how this formula was derived; it seems slightly off. You might try working up from Moive's original math, page 39, in the lemma.
I'm trying to write a program that will help someone study for the GRE math. As many of you may know, fractions are a big part of the test, and calculators aren't allowed. Basically what I want to do is generate four random numbers (say, 1-50) and either +-/* them and then accept an answer in fraction format. The random number thing is easy. The problem is, how can I 1) accept a fractional answer and 2) ensure that the answer is reduced all the way?
I am writing in ASP.NET (or jQuery, if that will suffice). I was pretty much wondering if there's some library or something that handles this kind of thing...
Thanks!
have a look at
http://www.geekpedia.com/code73_Get-the-greatest-common-divisor.html
http://javascript.internet.com/math-related/gcd-lcm-calculator.html
Since fractions are essentially divisions you can check to see if the answer is partially correct by performing the division on the fraction entries that you're given.
[pseudocode]
if (answer.contains("/"))
int a = answer.substring(1,answer.instanceof("/"))
int b = answer.substring(answer.instanceof("/"))
if (a/b == expectedAnswer)
if (gcd(a,b) == 1)
GOOD!
else
Not sufficiently reduced
else
WRONG!
To find out whether it's reduced all the way, create a GCD function which should evaluate to the value of the denominator that the user supplied as an answer.
Learn Python and try fractions module.