i have this code in if statement , and i think it is very big , how can i change this by short code and for infinity i.e that i want the if statement to work on two and after two by sixteen i.e 2 += 16 , forever , i know that my question might be not understandable
but please help me
Take the current count and subtract 2. If the number is divisible by 16 (the remainder when you do modulo division is zero), then your statement is true.
if ((count - 2) % 16 == 0)
{
_secondBall.hidden = YES;
}
This can be reduced to a mathematical problem. What you want is to divide by 16 and see if there is no 'remainder'. The 'modulo' operator gives you this remainder. You can Google it if you want to understand. So, as others have quickly posted before me, modulo in Objective-C is done via %. So x % 16 means: divide x by 16 and return the remainder (or: subtract 16 from x until x is smaller than 16). x % 16 will be a number between 0 and 15. Always. When x % 16 is 0, it means x is dividable by 16.
Since you want to take action when count is 2 + 16 * n, you want to subtract 2 from count first. Like so if ( ( count - 2 ) % 16 == 0). Or you can do this, which is shorter but perhaps less easy to understand: if ( count % 16 == 2 ).
Try this:
if ((count - 2) % 16 == 0)
Use this syntax
if ((count-2) % 16 == 0)
Related
I would like to encode three keyboard modifiers (CTRL, ALT, SHIFT) + the ASCII code of the pressed key into a single value. This falls naturally into the category of bitmasks.
One way I could do this is that the sender encodes each key as the following:
CTRL: 1000
ALT: 10000
SHIFT: 100000
KeyCode: 1-255
For example, if I were to click all modifiers + the last key in the ascii table, I would get:
100000 + 10000 + 1000 + 255 = 111255. The receiver side it would then be possible to do substraction and check if the number goes below 0 as such:
has_shift = X - 100000 < 0
has_alt = X - 10000 < 0
has_ctrl = X - 1000 < 0
if has_shift
X -= 100000
if has_alt
X -= 10000
if has_ctrl
X -= 1000
keyCode = X (the remainder)
Surely enough, I find this horrible and would assume that this could be done in a far better using bit-shift or something in that ballpark. How could this possibly be done better?
Instead add 256, 512, and 1024 respectively for ctrl, alt, shift. Then use the and operator in whatever language you're using (missing from question tags) to extract the modifiers and code. In C and many languages, that operator is &. So X & 1024 is not zero if shift was pressed. X & 255 is the character code.
I'm trying to solve the FizzBuzz game.
I need to check if a number is divisible by 3. So, when we take a number and divide it by 3, we need that operation to have NO REST, or 0 REST.
The solution given to me is this:
def fizzbuzz(number)
if number % 3 == 0
return "Fizz"
end
Why does they propose to use the "%"symbol? Why not the "/"symbol? Or both work as the same?
How should I check if the division has NO REST?
division operator / - gives the quotient of the division whatever the remainder of the division is. So you cannot determine if a number is perfectly divisible (remainder = 0) or not perfectly divisible (with non-zero remainder) using a division operator (/).
10 / 3
#=> 3
modulo operator % - gives the remainder of the division. If perfectly divisible, the output is 0, if not-perfectly divisible the output is non-zero value.
10 % 3
#=> 1
In your case number % 3 == 0 is true only if number is divisible by 3 with 0 remainder (i.e if number passed into the method frizzbuzz is a multiple of 3 like -12, -3, 3, 6, 9, etc )
Ruby has 4 division operators.
divmod returns the division and the remainder
number = 15
number.divmod(7) # => 2, 1
modulo alias % returns the remainder only
number = 15
number % 7 # => 1
number.modulo(7) # => 1
div alias / returns the integer division if both operands are integers, and floating-point division if either operand is a float.
number = 15
number / 7 # => 2
number.div(7) # => 2
1.0 * number / 7 # => 2.142857142857143
fdiv always returns a full precision floating-point division
number = 15
number.fdiv(7) # => 2.142857142857143
% is the Modulus - Divides left hand operand by right hand operand and returns remainder. \ is just the Divider. No Rest means that x % y == 0.
lets take an example to understand better:
number1 = 12
number2 = 13
lets see if number1 and number2 is devisible by 4 ok?
number1 / 4 = 3 and the rest = 0
number2 / 4 = 3 and the rest = 1
so the "/" operation let us know the result of devision operation
and the "%" operation let us know the rest of devision operation
so if we take our examples the number1 is devisible by 3 because
number1 % 3 = 0 ( the rest )
i want to generate a series of number through looping.
my series will contain numbers like 0,3,5,8,10,13,15,18 and so on.
i try to take reminder and try to add 2 and 3 but it wont work out.
can any one please help me in generating this series.
You can just use an increment which toggles between 3 and 2, e.g.
for (i = 0, inc = 3; i < 1000; i += inc, inc = 5 - inc)
{
printf("%d\n", i);
}
It looks like the the sequence starts at zero, and uses increments of 3 and 2. There are several ways of implementing this, but perhaps the simplest one would be iterating in increments of 5 (i.e. 3+2) and printing two numbers - position and position plus three.
Here is some pseudocode:
i = 0
REPEAT N times :
PRINT i
PRINT i + 3
i += 5
The iteration i=0 will print 0 and 3
The iteration i=5 will print 5 and 8
The iteration i=10 will print 10 and 13
The iteration i=15 will print 15 and 18
... and so on
I was pulled in with the tag generate-series, which is a powerful PostgreSQL function. This may have been tagged by mistake (?) but it just so happens that there would be an elegant solution:
SELECT ceil(generate_series(0, 1000, 25) / 10.0)::int;
generate_series() returns 0, 25, 50, 75 , ... (can only produces integer numbers)
division by 10.0 produces numeric data: 0, 2.5, 5, 7.5, ...
ceil() rounds up to your desired result.
The final cast to integer (::int) is optional.
SQL Fiddle.
I need to round a number, let's say 543 to either the hundreds or the tens place. It could be either one, as it's part of a game and this stage can ask you to do one or the other.
So for example, it could ask, "Round number to nearest tens", and if the number was 543, they would have to enter in 540.
However, I don't see a function that you can specify target place value to round at. I know there's an easy solution, I just can't think of one right now.
From what I see, the round function rounds the last decimal place?
Thanks
To rounding to 100's place
NSInteger num=543;
NSInteger deci=num%100;//43
if(deci>49){
num=num-deci+100;//543-43+100 =600
}
else{
num=num-deci;//543-43=500
}
To round to 10's place
NSInteger num=543;
NSInteger deci=num%10;//3
if(deci>4){
num=num-deci+100;//543-3+10 =550
}
else{
num=num-deci;//543-3=540
}
EDIT:
Tried to merge the above in one:
NSInteger num=543;
NSInteger place=100; //rounding factor, 10 or 100 or even more.
NSInteger condition=place/2;
NSInteger deci=num%place;//43
if(deci>=condition){
num=num-deci+place;//543-43+100 =600.
}
else{
num=num-deci;//543-43=500
}
You may just use an algorithm in your code:
For example, lets say that you need to round up a number to hundred's place.
int c = 543
int k = c % 100
if k > 50
c = (c - k) + 100
else
c = c - k
To round numbers, you can use the modulus operator, %.
The modulus operator gives you the remainder after division.
So 543 % 10 = 3, and 543 % 100 = 43.
Example:
int place = 10;
int numToRound=543;
// Remainder is 3
int remainder = numToRound%place;
if(remainder>(place/2)) {
// Called if remainder is greater than 5. In this case, it is 3, so this line won't be called.
// Subtract the remainder, and round up by 10.
numToRound=(numToRound-remainder)+place;
}
else {
// Called if remainder is less than 5. In this case, 3 < 5, so it will be called.
// Subtract the remainder, leaving 540
numToRound=(numToRound-remainder);
}
// numToRound will output as 540
NSLog(#"%i", numToRound);
Edit: My original answer was submitted before it was ready, because I accidentally hit a key to submit it. Oops.
If I am given a floating point number but do not know beforehand what range the number will be in, is it possible to scale that number in some meaningful way to be in another range? I am thinking of checking to see if the number is in the range 0<=x<=1 and if not scale it to that range and then scale it to my final range. This previous post provides some good information, but it assumes the range of the original number is known beforehand.
You can't scale a number in a range if you don't know the range.
Maybe what you're looking for is the modulo operator. Modulo is basically the remainder of division, the operator in most languages is is %.
0 % 5 == 0
1 % 5 == 1
2 % 5 == 2
3 % 5 == 3
4 % 5 == 4
5 % 5 == 0
6 % 5 == 1
7 % 5 == 2
...
Sure it is not possible. You can define range and ignore all extrinsic values. Or, you can collect statistics to find range in run time (i.e. via histogram analysis).
Is it really about image processing? There are lots of related problems in image segmentation field.
You want to scale a single random floating point number to be between 0 and 1, but you don't know the range of the number?
What should 99.001 be scaled to? If the range of the random number was [99, 100], then our scaled-number should be pretty close to 0. If the range of the random number was [0, 100], then our scaled-number should be pretty close to 1.
In the real world, you always have some sort of information about the range (either the range itself, or how wide it is). Without further info, the answer is "No, it can't be done."
I think the best you can do is something like this:
int scale(x) {
if (x < -1) return 1 / x - 2;
if (x > 1) return 2 - 1 / x;
return x;
}
This function is monotonic, and has a range of -2 to 2, but it's not strictly a scaling.
I am assuming that you have the result of some 2-dimensional measurements and want to display them in color or grayscale. For that, I would first want to find the maximum and minimum and then scale between these two values.
static double[][] scale(double[][] in, double outMin, double outMax) {
double inMin = Double.POSITIVE_INFINITY;
double inMax = Double.NEGATIVE_INFINITY;
for (double[] inRow : in) {
for (double d : inRow) {
if (d < inMin)
inMin = d;
if (d > inMax)
inMax = d;
}
}
double inRange = inMax - inMin;
double outRange = outMax - outMin;
double[][] out = new double[in.length][in[0].length];
for (double[] inRow : in) {
double[] outRow = new double[inRow.length];
for (int j = 0; j < inRow.length; j++) {
double normalized = (inRow[j] - inMin) / inRange; // 0 .. 1
outRow[j] = outMin + normalized * outRange;
}
}
return out;
}
This code is untested and just shows the general idea. It further assumes that all your input data is in a "reasonable" range, away from infinity and NaN.