Hi so I am trying to create an if statement where...if the sprite has been touched...then something will happen
However, for some reason, I am getting an expected identified error and can't for the life of me figure out why?
Thank you for any help and knowledge you can provide meKKInput* input = [KKInput sharedInput];
KKInput* input = [KKInput sharedInput];
if([input isAnyTouchOnNode:kan touchPhase:[KKTouchPhaseAny]])
{
for ( int x = 105.0f; x < 100000 ; ++x)
{
for ( int y = 50.0f; y < 100000 ; ++y)
{
kan.position = CGPointMake((x),(y));
}
}
}
The square brackets around KKTouchPhaseAny (probably an enum) must be removed.
Related
So I tried to write a code that finds the largest palindromic number from two (3 spaces long) multiplied numbers. Does my code work fine or are there no palindromes for this?
function checkPalindrom(str) {
return str === str.split('').reverse().join('');
}; //Declares the funciton to check if a string is a palindrome
var x = 999;
var y = 999;
var z = 0;
var n = z.toString(); //Declares that n is the string of z
for (i=0; i<899; i++) { //For loop: counts from 0 to 899
x*y===z; //Is this correct? z is set equal to x*y
if(checkPalindrom(n) === true) { //If n is a palindrome,
console.log(n); //Write out the palindrome
} else {
x-=1; //subtract 1 from x and run again
}
};
Also, what is the best way to check for all combinations of 3 digit numbers? Because right now I am just checking for any number from 100 to 999, but I actually need to check for all combinations...
Your post has a few problems, as well as multiple questions in it. I'll try to hone in on the major stuff but, as this is a fairly standard type of Programming 101 homework question, I'm not going to give you an exact answer right out.
First off, there are three different 'equals' in javascript, =, ==, and ===. A single = is an assignment operator and it always works from right to left. Thus,
var x = 2;
assigns the value of 2 to the variable x. In your code,
x*y === z;
has a couple of problems. First off, it is backwards. Secondly, it uses === but should be using =.
z = x*y;
That is what you were trying to put here.
In javascript, == and === are both comparitives. The triple === adds type comparison and is stronger but generally unnecessary. In almost all cases, == is sufficient. But, what it does is compare the values like inside an if statement:
if(x == 2)
This just checks if the value of x is equal to the value of 2, but the values themselves do not change.
Ok, for your other question: "number from 100 to 999, but I actually need to check for all combinations..."
The best way to handle this is a double loop:
var z;
for(var x = 100; x < 1000; x++)
for(var y = x; y < 1000; y++)
z = x*y;
This will first let x = 100, then check 100 * every number from 100 to 999. Then you let x = 101 and check 101* every number from 101 to 999.
function checkPalindrom(str) {
return str === str.split('').reverse().join('');
}; //Declares the funciton to check if a string is a palindrome
var x;
var y;
var z;
var n;
var max = 0;
for (x=999; x >= 100; x--) {
for (y=999; y >= 100; y--) {
z = x*y;
n = z.toString();
if(checkPalindrom(n) === true && max < z) {
console.log(n);
max = z;
}
}
}
I did make algorithm for creating Bézier Curve with Objective-C and Cocos2D. Here is my code
-(int)factorial:(int)x{
int sum=1;
int i;
if(x == 0){
return 1;
}else{
for(i=1;i<x;i++){
sum = sum*i;
}
return sum;
}
}
-(int)binomialCoefficient:(int)n:(int)i{
int sum;
//NSLog([NSString stringWithFormat:#"fac n-i=%f\n", fach] );
sum = [self factorial:n]/([self factorial:i]*[self factorial:(n-i)]);
return sum;
}
-(float)convertT:(int)t{
return t*(0.001);
}
-(float)power:(float)a:(int)b{
int i;
float hasil=1;
for(i=0;i<b;i++){
hasil = hasil*a;
}
return hasil;
}
-(float)bernstein:(float)t:(int)n:(int)i{
float sum = 0;
sum = [self binomialCoefficient:n:i]*[self power:t :i]*[self power:(1-t) :(n-i)];
//NSLog([NSString stringWithFormat:#"yeah"]);
return sum;
}
and for implementation you just put an array of x and y and access it. For example to draw a single dot in control curve I did it like this
float myPx = px[i];
float myPy = py[i];
posx = posx+([self bernstein:theT :banyak-1 :i]*myPx);
posy = posy+([self bernstein:theT :banyak-1 :i]*myPy);
Yes, this code doesn't give the perfect nice line, but I try to draw it dot by dot.
It works well, but the problem arise when I try to use 3 dots. The middle dot for curving the lines didn't behave like what I expected. For example if I put 3 dots in these coordinates:
a(100,200)
b(250,250)
c(500,200)
It didn't curving up but curving down. If I want to put it straight I have to put it all the way higher.
Am I do it wrong in syntax or data types? Or is it just my algorithm?
Thanks in advance
Best Regards
(sorry for my bad english)
The factorial loop should be
for ( i = 1 ; i <= x ; i++ )
instead of
for ( i = 1 ; i < x ; i++ )
I'm new to the OpenNI and I'm trying to create a simple ImageGenerator that just display a pure color image, say white, I modified the “SampleModule” and in the UpdateData() method I assign the *pPixel value with 255. The UpdateData() method is as following
XnStatus SampleImage::UpdateData()
{
XnStatus nRetVal = XN_STATUS_OK;
XnUInt8* pPixel = m_pImageMap;
for (XnUInt y = 0; y < 300; ++y)
{
for (XnUInt x = 0; x < 400; ++x, ++pPixel)
{
*pPixel = (XnUInt8)255;
}
}
m_nFrameID++;
m_nTimestamp += 1000000 / SUPPORTED_FPS;
// mark that data is old
m_bDataAvailable = FALSE;
return (XN_STATUS_OK);
}
The code compile fine, and I could register it with nireg, but when I try to read the image pixel value from the data generated by the module I got some strange value (not 255 as I expected), I use the following code to read the pixel value.
const XnUInt8* pImageMap = mImageGenerator.GetImageMap();
for (XnUInt y = 0; y < 300; ++y)
{
for (XnUInt x = 0; x < 400; ++x, ++pImageMap)
{
cout << (int)*pImageMap << endl;
}
}
and also when I run the “NiViewer” the program still say it can't find the image node, but the “SampleModule” can be find as a depth.
Any advice would be appreciate.
Thanks a million,
Haolin Wei
Check if you did following things:
1. set color format, i.e, rgb (or YUV)
2. set correct value for each pixel in the updataData(), i.e. r=255,g=255,b=255
i'm making a simple iPhone game using cocos2d-iphone. I have an array of fiends, the "fiendSet" which has to navigate around a field full of obstacles. I spent the last three nights trying to get my A* pathfinding to work. I found the actual A* implementation here on stackoverflow and it works brilliantly. However, once i try to move my fiends around i run into trouble.
Each of my fiends has a CGPoint called motionTarget which contains the x and y values for where the fiend has to go. If only set the positions x and y to absolute values once a second, it works, like so:
-(void) updateFiendPositions:(ccTime)dt {
for (MSWFiend *currFiend in fiendSet) {
currFiend.position = ccp(currFiend.motionTarget.x,currFiend.motionTarget.y);
}
}
However, this doesn't look very nice, the fiends just "jump" 20px each second instead of animating nicely. I only implemented this as a placeholder method to verify the pathfinding. Now i want smooth animation. This is what i did:
-(void) updatePositions:(ccTime) dt {
for (MSWFiend *currFiend in fiendSet) {
if (currFiend.motionTarget.x != -1 && currFiend.motionTarget.y != -1) {
float x,y;
if ((int)floor(currFiend.position.x) < (int)floor(currFiend.motionTarget.x)) {
x = currFiend.position.x+(currFiend.speed*dt);
}
if ((int)floor(currFiend.position.x) > (int)floor(currFiend.motionTarget.x)) {
x = currFiend.position.x-(currFiend.speed*dt);
}
if (abs((int)floor(currFiend.position.x)-(int)floor(currFiend.motionTarget.x)) < 2) {
x = currFiend.motionTarget.x;
}
if ((int)floor(currFiend.position.y) < (int)floor(currFiend.motionTarget.y)) {
y = currFiend.position.y+(currFiend.speed*dt);
}
if ((int)floor(currFiend.position.y) > (int)floor(currFiend.motionTarget.y)) {
y = currFiend.position.y-(currFiend.speed*dt);
}
if (abs((int)floor(currFiend.position.y)-(int)floor(currFiend.motionTarget.y)) < 2) {
y = currFiend.motionTarget.y;
}
currFiend.position = ccp(x,y);
}
}
}
This works great for fiends moving in one direction. As soon as a fiend is supposed to go around a bend, trouble starts. Instead of for example first going up, then right, then down; my fiends will combine the up/right motion into one, they are "cutting corners". I only want my fiends to move either north/south OR east/west for each position update, not both. In other words, i don't want to animate changes to x and y simultaneously. I hope this explanation is clear enough..
I'm pretty sure i have a logic error somewhere.. i just havn't been able to figure it out for the last three sleepless nights after work.. Help!
You have to keep track of each node in the path to the target. That way you only animate the motion to the next node. Also you can use a CCMoveTo action instead on doing the animation yourself.
#Aleph thanks for your suggestion. I found that it was the code which determines when to assign a new motionTarget, that was faulty, not the code i posted to begin with. When you mentioned keeping track of each nodes position, i thought of my motionTarget determination code and found the error after 2-3 hours. I ended up doing it like this:
-(void) updatePositions:(ccTime) dt {
for (MSWFiend *currFiend in fiendSet) {
int fiendGX,fiendGY,targetGX,targetGY,dGX,dGY;
float x,y,snappedX,snappedY;
BOOL snappedIntoPosition = FALSE;
fiendGX = (int)round(100.0f*(currFiend.position.x/20));
fiendGY = (int)round(100.0f*(currFiend.position.y/20));
targetGX = (int)round(100.0f*(currFiend.motionTarget.x/20));
targetGY = (int)round(100.0f*(currFiend.motionTarget.y/20));
snappedX = currFiend.position.x;
snappedY = currFiend.position.y;
dGX = abs(fiendGX-targetGX);
dGY = abs(fiendGY-targetGY);
float snappingThreshold; //1 = snap when 0.1 from motionTarget.
snappingThreshold = currFiend.speed/10;
if (dGX < snappingThreshold && dGY < snappingThreshold) {
snappingThreshold = currFiend.motionTarget.x;
snappingThreshold = currFiend.motionTarget.y;
int newPathStep;
newPathStep = currFiend.pathStep + 1;
currFiend.pathStep = newPathStep;
}
int gX,gY;
gX = [[currFiend.path objectAtIndex:currFiend.pathStep] nodeX];
gY = (tileMap.mapSize.height-[[currFiend.path objectAtIndex:currFiend.pathStep] nodeY])-1;
currFiend.motionTarget = ccp(gX*20,gY*20); //Assign motion target to the next A* node. This is later used by the position updater.
if (currFiend.motionTarget.x != -1 && currFiend.motionTarget.y != -1) {
x = currFiend.motionTarget.x;
y = currFiend.motionTarget.y;
if ((int)floor(currFiend.position.x) < (int)floor(currFiend.motionTarget.x)) {
//Move right
x = snappedX+(currFiend.speed*dt);
if (x > currFiend.motionTarget.x) {
x = currFiend.motionTarget.x;
}
y = snappedY;
}
if ((int)floor(currFiend.position.x) > (int)floor(currFiend.motionTarget.x)) {
//Move left
x = snappedX-(currFiend.speed*dt);
if (x < currFiend.motionTarget.x) {
x = currFiend.motionTarget.x;
}
y = snappedY;
}
if ((int)floor(currFiend.position.y) < (int)floor(currFiend.motionTarget.y)) {
//Move up
y = snappedY+(currFiend.speed*dt);
if (y > currFiend.motionTarget.y) {
y = currFiend.motionTarget.y;
}
x = snappedX;
}
if ((int)floor(currFiend.position.y) > (int)floor(currFiend.motionTarget.y)) {
//Move down
y = snappedY-(currFiend.speed*dt);
if (y < currFiend.motionTarget.y) {
y = currFiend.motionTarget.y;
}
x = snappedX;
}
}
currFiend.position = ccp(x,y);
}
}
How many possible combinations of the variables a,b,c,d,e are possible if I know that:
a+b+c+d+e = 500
and that they are all integers and >= 0, so I know they are finite.
#Torlack, #Jason Cohen: Recursion is a bad idea here, because there are "overlapping subproblems." I.e., If you choose a as 1 and b as 2, then you have 3 variables left that should add up to 497; you arrive at the same subproblem by choosing a as 2 and b as 1. (The number of such coincidences explodes as the numbers grow.)
The traditional way to attack such a problem is dynamic programming: build a table bottom-up of the solutions to the sub-problems (starting with "how many combinations of 1 variable add up to 0?") then building up through iteration (the solution to "how many combinations of n variables add up to k?" is the sum of the solutions to "how many combinations of n-1 variables add up to j?" with 0 <= j <= k).
public static long getCombos( int n, int sum ) {
// tab[i][j] is how many combinations of (i+1) vars add up to j
long[][] tab = new long[n][sum+1];
// # of combos of 1 var for any sum is 1
for( int j=0; j < tab[0].length; ++j ) {
tab[0][j] = 1;
}
for( int i=1; i < tab.length; ++i ) {
for( int j=0; j < tab[i].length; ++j ) {
// # combos of (i+1) vars adding up to j is the sum of the #
// of combos of i vars adding up to k, for all 0 <= k <= j
// (choosing i vars forces the choice of the (i+1)st).
tab[i][j] = 0;
for( int k=0; k <= j; ++k ) {
tab[i][j] += tab[i-1][k];
}
}
}
return tab[n-1][sum];
}
$ time java Combos
2656615626
real 0m0.151s
user 0m0.120s
sys 0m0.012s
The answer to your question is 2656615626.
Here's the code that generates the answer:
public static long getNumCombinations( int summands, int sum )
{
if ( summands <= 1 )
return 1;
long combos = 0;
for ( int a = 0 ; a <= sum ; a++ )
combos += getNumCombinations( summands-1, sum-a );
return combos;
}
In your case, summands is 5 and sum is 500.
Note that this code is slow. If you need speed, cache the results from summand,sum pairs.
I'm assuming you want numbers >=0. If you want >0, replace the loop initialization with a = 1 and the loop condition with a < sum. I'm also assuming you want permutations (e.g. 1+2+3+4+5 plus 2+1+3+4+5 etc). You could change the for-loop if you wanted a >= b >= c >= d >= e.
I solved this problem for my dad a couple months ago...extend for your use. These tend to be one time problems so I didn't go for the most reusable...
a+b+c+d = sum
i = number of combinations
for (a=0;a<=sum;a++)
{
for (b = 0; b <= (sum - a); b++)
{
for (c = 0; c <= (sum - a - b); c++)
{
//d = sum - a - b - c;
i++
}
}
}
This would actually be a good question to ask on an interview as it is simple enough that you could write up on a white board, but complex enough that it might trip someone up if they don't think carefully enough about it. Also, you can also for two different answers which cause the implementation to be quite different.
Order Matters
If the order matters then any solution needs to allow for zero to appear for any of the variables; thus, the most straight forward solution would be as follows:
public class Combos {
public static void main() {
long counter = 0;
for (int a = 0; a <= 500; a++) {
for (int b = 0; b <= (500 - a); b++) {
for (int c = 0; c <= (500 - a - b); c++) {
for (int d = 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2656615626.
Order Does Not Matter
If the order does not matter then the solution is not that much harder as you just need to make sure that zero isn't possible unless sum has already been found.
public class Combos {
public static void main() {
long counter = 0;
for (int a = 1; a <= 500; a++) {
for (int b = (a != 500) ? 1 : 0; b <= (500 - a); b++) {
for (int c = (a + b != 500) ? 1 : 0; c <= (500 - a - b); c++) {
for (int d = (a + b + c != 500) ? 1 : 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2573155876.
One way of looking at the problem is as follows:
First, a can be any value from 0 to 500. Then if follows that b+c+d+e = 500-a. This reduces the problem by one variable. Recurse until done.
For example, if a is 500, then b+c+d+e=0 which means that for the case of a = 500, there is only one combination of values for b,c,d and e.
If a is 300, then b+c+d+e=200, which is in fact the same problem as the original problem, just reduced by one variable.
Note: As Chris points out, this is a horrible way of actually trying to solve the problem.
link text
If they are a real numbers then infinite ... otherwise it is a bit trickier.
(OK, for any computer representation of a real number there would be a finite count ... but it would be big!)
It has general formulae, if
a + b + c + d = N
Then number of non-negative integral solution will be C(N + number_of_variable - 1, N)
#Chris Conway answer is correct. I have tested with a simple code that is suitable for smaller sums.
long counter = 0;
int sum=25;
for (int a = 0; a <= sum; a++) {
for (int b = 0; b <= sum ; b++) {
for (int c = 0; c <= sum; c++) {
for (int d = 0; d <= sum; d++) {
for (int e = 0; e <= sum; e++) {
if ((a+b+c+d+e)==sum) counter=counter+1L;
}
}
}
}
}
System.out.println("counter e "+counter);
The answer in math is 504!/(500! * 4!).
Formally, for x1+x2+...xk=n, the number of combination of nonnegative number x1,...xk is the binomial coefficient: (k-1)-combination out of a set containing (n+k-1) elements.
The intuition is to choose (k-1) points from (n+k-1) points and use the number of points between two chosen points to represent a number in x1,..xk.
Sorry about the poor math edition for my fist time answering Stack Overflow.
Just a test for code block
Just a test for code block
Just a test for code block
Including negatives? Infinite.
Including only positives? In this case they wouldn't be called "integers", but "naturals", instead. In this case... I can't really solve this, I wish I could, but my math is too rusty. There is probably some crazy integral way to solve this. I can give some pointers for the math skilled around.
being x the end result,
the range of a would be from 0 to x,
the range of b would be from 0 to (x - a),
the range of c would be from 0 to (x - a - b),
and so forth until the e.
The answer is the sum of all those possibilities.
I am trying to find some more direct formula on Google, but I am really low on my Google-Fu today...