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  • Recursively determine average value

    - by theva
    I have to calculate an average value of my simulation. The simulation is ongoing and I want (for each iteration) to print the current average value. How do I do that? I tried the code below (in the loop), but I do not think that the right value is calculated... int average = 0; int newValue; // Continuously updated value. if(average == 0) { average = newValue; } average = (average + newValue)/2; I also taught about store each newValue in an array and for each iteration summarize the whole array and do the calculation. However, I don't think that's a good solution, because the loop is an infinity loop so I can't really determine the size of the array. There is also a possibility that I am thinking too much and that the code above is actually correct, but I don't think so...

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  • Write number N in base M

    - by VaioIsBorn
    I know how to do it mathematically, but i want it now to do it in c++ using some easy algorithm. Is is possible? The question is that i need some methods/ideas for writing a number N in base M, for example 1410 in base 3: (14)10 = 2*(3^0) + 1*(3^1) + 1*(3^2) = (112)3 etc.

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  • Solve Physics exercise by brute force approach..

    - by Nils
    Being unable to reproduce a given result. (either because it's wrong or because I was doing something wrong) I was asking myself if it would be easy to just write a small program which takes all the constants and given number and permutes it with a possible operators (* / - + exp(..)) etc) until the result is found. Permutations of n distinct objects with repetition allowed is n^r. At least as long as r is small I think you should be able to do this. I wonder if anybody did something similar here..

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  • Special simple random number generator

    - by psihodelia
    How to create a function, which on every call generates a random integer number? This number must be most random as possible (according to uniform distribution). It is only allowed to use one static variable and at most 3 elementary steps, where each step consists of only one basic arithmetic operation of arity 1 or 2. Example: int myrandom(void){ static int x; x = some_step1; x = some_step2; x = some_step3; return x; } Basic arithmetic operations are +,-,%,and, not, xor, or, left shift, right shift, multiplication and division. Of course, no rand(), random() or similar staff is allowed.

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  • Number of 0/1 colorings of a m X n rectangle which have no monochromatic subrectangles with both dimension greater than 1.

    - by acbruptenda
    A m x n rectangular matrix is give, and each cell is to be filled with 0/1 colour. I have to find number of colorings possible so that there is no monochromatic coloured (same colour) sub-rectangle whose both dimension is greater than 1 (eg - 2x2, 2x3,4x3) I have found a slightly different version of it here But they have said nothing about the algorithm. So, I am looking for an algorithm here !!

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  • log (1-var) operation in C

    - by heike
    I am trying to code this algorithm. I am stuck in the part of log((1.0-u)/u))/beta; As I understand, I can not get the result of this in C, as it will always return me with negative value log (returning imaginary value). Tried to print the result of log(1-5) for instance, it gives me with Nan. How can I get the result of double x = (alpha - log((1.0-u)/u))/beta then? Would appreciate for any pointers to solve this problem. Thank you

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  • Permutations with extra restrictions

    - by Full Decent
    I have a set of items, for example: {1,1,1,2,2,3,3,3}, and a restricting set of sets, for example {{3},{1,2},{1,2,3},{1,2,3},{1,2,3},{1,2,3},{2,3},{2,3}. I am looking for permutations of items, but the first element must be 3, and the second must be 1 or 2, etc. One such permutation that fits is: {3,1,1,1,2,2,3} Is there an algorithm to count all permutations for this problem in general? Is there a name for this type of problem? For illustration, I know how to solve this problem for certain types of "restricting sets". Set of items: {1,1,2,2,3}, Restrictions {{1,2},{1,2,3},{1,2,3},{1,2},{1,2}}. This is equal to 2!/(2-1)!/1! * 4!/2!/2!. Effectively permuting the 3 first, since it is the most restrictive and then permuting the remaining items where there is room.

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  • Minimizing distance to a weighted grid

    - by Andrew Tomazos - Fathomling
    Lets suppose you have a 1000x1000 grid of positive integer weights W. We want to find the cell that minimizes the average weighted distance.to each cell. The brute force way to do this would be to loop over each candidate cell and calculate the distance: int best_x, best_y, best_dist; for x0 = 1:1000, for y0 = 1:1000, int total_dist = 0; for x1 = 1:1000, for y1 = 1:1000, total_dist += W[x1,y1] * sqrt((x0-x1)^2 + (y0-y1)^2); if (total_dist < best_dist) best_x = x0; best_y = y0; best_dist = total_dist; This takes ~10^12 operations, which is too long. Is there a way to do this in or near ~10^8 or so operations?

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  • Given a vector of maximum 10 000 natural and distinct numbers, find 4 numbers(a, b, c, d) such that

    - by king_kong
    Hi, I solved this problem by following a straightforward but not optimal algorithm. I sorted the vector in descending order and after that substracted numbers from max to min to see if I get a + b + c = d. Notice that I haven't used anywhere the fact that elements are natural, distinct and 10 000 at most. I suppose these details are the key. Does anyone here have a hint over an optimal way of solving this? Thank you in advance! Later Edit: My idea goes like this: '<<quicksort in descending order>>' for i:=0 to count { // after sorting, loop through the array int d := v[i]; for j:=i+1 to count { int dif1 := d - v[j]; int a := v[j]; for k:=j+1 to count { if (v[k] > dif1) continue; int dif2 := dif1 - v[k]; b := v[k]; for l:=k+1 to count { if (dif2 = v[l]) { c := dif2; return {a, b, c, d} } } } } } What do you think?(sorry for the bad indentation)

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  • Calculating a parabola: What am I doing wrong? [closed]

    - by Nils
    I was following this thread and copied the code in my project. Playing around with it turns out that it seems not to be very precise. Recall the formula: y = ax^2 + bx +c Since the first given point I have is at x1 = 0, we already have c=y1 . We just need to find a and b. Using: y2 = ax2^2 + bx2 +c y3 = ax3^2 + bx3 +c Solving the equations for b yields: b = y/x - ax - cx Now setting both equations equal to each other so b falls out y2/x2 - ax2 - cx2 = y3/x3 - ax3 - cx3 Now solving for a gives me: a = ( x3*(y2 - c) + x2*(y3 - c) ) / ( x2*x3*(x2 - x3) ) (is that correct?!) And then using again b = y2/x2 - ax2 - cx2 to find b. However so far I haven't found the correct a and b coeffs. What am I doing wrong?

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  • PHP: Trying to come up with a "prev" and "next" link

    - by fwaokda
    I'm displaying 10 records per page. The variables I have currently that I'm working with are.. $total = total number of records $page = whats the current page I'm displaying I placed this at the top of my page... if ( $_GET['page'] == '' ) { $page = 1; } //if no page is specified set it to `1` else { $page = ($_GET['page']); } // if page is specified set it Here are my two links... if ( $page != 1 ) { echo '<div style="float:left" ><a href="index.php?page='. ( $page - 1 ) .'" rev="prev" >Prev</a></div>'; } if ( !( ( $total / ( 10 * $page ) ) < $page ) ) { echo '<div style="float:right" ><a href="index.php?page='. ( $page + 1 ) .'" rev="next" >Next</a></div>'; } Now I guess (unless I'm not thinking of something) that I can display the "Prev" link every time except when the page is '1'. How can make it where the "Next" link doesn't show on the last page though?

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  • Print all ways to sum n integers so that they total a given sum.

    - by noghead
    Im trying to come up with an algorithm that will print out all possible ways to sum N integers so that they total a given value. Example. Print all ways to sum 4 integers so that they sum up to be 5. Result should be something like: 5 0 0 0 4 1 0 0 3 2 0 0 3 1 1 0 2 3 0 0 2 2 1 0 2 1 2 0 2 1 1 1 1 4 0 0 1 3 1 0 1 2 2 0 1 2 1 1 1 1 3 0 1 1 2 1 1 1 1 2

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  • How long would this have to go on for...

    - by Pieman
    I have the Pi formulae -Well one of them... 1 - 1/3 + 1/5 - 1/7 etc. How long would it take to get to like 1000 S.F correct? -Well, not how long, how big would the denominator be? -I have it updating 4 times in one refresh: http://zombiewrath.com/pi.php So the section above would be done in one refresh, then 7 to 13 in another etc. Answer this maths question please :) Also how can I get the 10,002 length variable onto 'seperate lines'? -I want it to fill 100% screen width -no scrolling needed (well downwards only)

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  • Calculating percent "x/y * 100" always results in 0?

    - by Patrick Beninga
    In my assignment i have to make a simple version of Craps, for some reason the percentage assignments always produce 0 even when both variables are non 0, here is the code. import java.util.Random; Header, note the variables public class Craps { private int die1, die2,myRoll ,myBet,point,myWins,myLosses; private double winPercent,lossPercent; private Random r = new Random(); Just rolls two dies and produces their some. public int roll(){ die1 = r.nextInt(6)+1; die2 = r.nextInt(6)+1; return(die1 + die2); } The Play method, this just loops through the game. public void play(){ myRoll = roll(); point = 0; if(myRoll == 2 ||myRoll == 3 || myRoll == 12){ System.out.println("You lose!"); myLosses++; }else if(myRoll == 7 || myRoll == 11){ System.out.println("You win!"); myWins++; }else{ point = myRoll; do { myRoll = roll(); }while(myRoll != 7 && myRoll != point); if(myRoll == point){ System.out.println("You win!"); myWins++; }else{ System.out.println("You lose!"); myLosses++; } } } This is where the bug is, this is the tester method. public void tester(int howMany){ int i = 0; while(i < howMany){ play(); i++; } bug is right here in these assignments statements winPercent = myWins/i * 100; lossPercent = myLosses/i* 100; System.out.println("program ran "+i+" times "+winPercent+"% wins "+ lossPercent+"% losses with "+myWins+" wins and "+myLosses+" losses"); } }

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  • Why do we need to estimate the true position in Kalman filters?

    - by Kalla
    I am following a probably well-known tutorial about Kalman filter here From these lines of code: figure; plot(t,pos, t,posmeas, t,poshat); grid; xlabel('Time (sec)'); ylabel('Position (feet)'); title('Figure 1 - Vehicle Position (True, Measured, and Estimated)') I understand that x is the true position, y is measured position, xhat is estimated position. Then, if we can compute x (this code: x = a * x + b * u + ProcessNoise;), why do we need to estimated x anymore?

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  • Scale 2D coordinates and keep their relative euclidean distances intact?

    - by eiaxlid
    I have a set of points like: pointA(3302.34,9392.32), pointB(34322.32,11102.03), etc. I need to scale these so each x- and y-coordinate is in the range (0.0 - 1.0). I tried doing this by first finding the largest x value in the data set (maximum_x_value), and the largest y value in the set (minimum_y_value). I then did the following: pointA.x = (pointA.x - minimum_x_value) / (maximum_x_value - minimum_x_value) pointA.y = (pointA.y - minimum_y_value) / (maximum_y_value - minimum_y_value) This changes the relative distances(?), and therefore makes the data useless for my purposes. Is there a way to scale these coordinates while keeping their relative distances the intact?

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  • How to map a long integer number to a N-dimensional vector of smaller integers (and fast inverse)?

    - by psihodelia
    Given a N-dimensional vector of small integers is there any simple way to map it with one-to-one correspondence to a large integer number? Say, we have N=3 vector space. Can we represent a vector X=[(int32)x1,(int32)x2,(int32)x3] using an integer (int48)y? The obvious answer is "Yes, we can". But the question is: "What is the fastest way to do this and its inverse operation?"

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