I need to find the first multiple for a number starting from a base number. For example: The first multiple of 3 from 7 is 9. My first attempt was to do this:
multiple = baseNumber
while(multiple%number !=0 )
multiple++
At the end, "multiple" will have the first multiple of number after baseNumber. The problem is that when number becomes too large, the number of iterations becomes too many. So my question is: is there a faster way to do this?
If everything is guaranteed to be positive, try
multiple = baseNumber + number - 1;
multiple -= (multiple % number);
That does it in constant time.
First, we add number - 1 to make sure that we have a number at least as large as the next multiple but smaller than the one after that. Then we subtract the remainder of the division by number to make sure we have the desired multiple.
If baseNumber can be negative (but number still positive), we face the problem that multiple % number may be negative if multiple < 0, so the above could skip a multiple of number. To avoid that, we can use e.g.
remainder = multiple % number;
if (remainder < 0) remainder += number;
multiple -= remainder;
If branching is too expensive, we can avoid the if at the cost of two divisions instead of one,
multiple -= (number + (multiple % number)) % number;
Generally, the if seems preferable, though.
If number can be negative, replace it with its absolute value first.
Note: The above returns, as the original code does, baseNumber if that is already a multiple of number. If that isn't desired, remove the - 1 in the first line.
try this (Requires INTEGER division):
multiple = ((base/number) + 1) * number;
7/3 = 2. 3*(2+1) = 9.
You have an edge case where the baseNumber already is a multiple of number, which you will have to test using the modulus operation.
Why do you need a loop?
multiple = (floor(number/baseNumber)+1)*baseNumber
while(multiple * number < baseNumber)
multiple++;
so for baseNumber = 3, number = 7, your multiple is 3;
though, something tells me bignums are about to show up in here.
Related
I have a simple problem, but because this "programming language" I am using is 32-bit and only supports basic functions such as addition, subtraction, multiplication, division, and concatenation (literally that's it), I am having some trouble.
For the input, I have a 16 digit number like so: 3334,5678,9523,4567
I want to then subtract 2 other random 16 digit numbers from this number and check if the first and last digits are 1.
For example, if the two other numbers are 1111,1111,1111,1111 and 1234,5678,9123,4565.
My final number would be: 0988,8888,9288,8891.
Here, the last number is 1, but the first number is 0, so the test would fail.
The issue is with 32-bit systems, there are massive errors due to not enough precision provided by the bits. What are some ways to bypass this issue?
If you're using a language like C or Java you should be able to use a long to create a 64 bit integer. If that's not possible you could divide the numbers into two 32 bit numbers, one to hold the upper half and one to hold the lower half.
Something like this:
//Each half is 8 digits to represent 8 of the 16
//Because of this each half should be less than 100000000
int upperHalf = 33345678;
int lowerHalf = 95234567;
//randomInt represents a function to generate a random
//integer equal to or greater than 0 and less than the
//argument passed to it
int randUpperHalf = randomInt(100000000);
int randLowerHalf = randomInt(100000000);
int lowerHalf = lowerHalf - randLowerHalf;
//If lowerHalf was a negative number you need to borrow from the upperHalf
if (lowerHalf < 0) {
upperHalf = upperHalf - 1;
lowerHalf = lowerHalf + 100000000;
}
upperHalf = upperHalf - randUpperHalf;
//Check that the first and last digits are 1
if ((upperHalf / 100000000) == 1 && (lowerHalf % 10) == 1) {
//The first and last digits are 1
}
Edit: Comments have been added to explain the code better. (lowerHalf % 2) == 1 has been changed to (lowerHalf % 10) == 1 and should now be able to tell if the number ends in a 1.
I am new to objective C and trying to understand arc4random().
There are so many conflicting explanations on the web. Please clear my confusion, which of the following is correct:
// 1.
arc4random() % (toNumber - fromNumber) + fromNumber;
OR
//2.
arc4random() % ((toNumber - fromNumber) + 1) + fromNumber;
//toNumber-fromNumbers are any range of numbers like random # between 7-90.
This code will get you a random number between 7 and 90.
NSUInteger random = 7 + arc4random_uniform(90 - 7);
Use arc4random_uniform to avoid modulo bias.
Adam's answer is correct. However, just to clarify the difference between the two, the second one raises the possible range by one to make the range inclusive. The important thing to remember is that modulo is remainder division, so while there are toNumber possible outcomes, one of them is zero (if the result of arc4random() is a multiple of toNumber) and toNumber itself can not be the remainder.
// 1.
arc4random() % (10 - 5) + 5;
This results in a range of 0 + 5 to 4 + 5, which is 5 to 9.
//2.
arc4random() % ((10 - 5) + 1) + 5;
This results in a range of 0 + 5 to (4 + 1) + 5, which is 5 to 10.
Neither is correct or incorrect if you wish to use modulo. One is exclusive of the upper range while the other is inclusive of the upper range. However, if you think about how remainder division works and think of the pool of numbers returned by any PRNG in terms of cycles the length of your total range, then you'll realize that if the range does not divide evenly into the maximum range of the pool you'll get biased results. For instance, if arc4random() returned a result from 1 to 5 (it doesn't, obviously) and you wanted a number from 0 to 2, and you used arc4random() % 3, these are the possible results.
1 % 3 = 1
2 % 3 = 2
3 % 3 = 0
4 % 3 = 1
5 % 3 = 2
Note that there are two ones and two twos, but only one zero. This is because our range of 3 does not evenly divide into the PRNG's range of 5. The result is that (humorously enough) PRNG range % desired range numbers at the end of the cycle need to be culled because they are "biased"–the numbers themselves aren't really biased, but it's easier to cull from the end. Failing to do this results in the lower numbers of the range becoming more likely to appear.
We can cull the numbers by calculating the upper range of the numbers we can generate, modulo it with the desired range and then pull those numbers off of the end. By "pull those numbers off of the end" I really mean "loop infinitely until we get a number that isn't one of the end numbers".
Some would say that's bad practice; you could theoretically loop forever. In practice, however, the expected number of retries is always less than one since the modulo bias is never more than half the pool (usually much less than that) of the PRNG's numbers. I once wrote a wrapper for rand using this technique.
You can see an example of this in the source for OpenBSD, where arc4random_uniform calls arc4random in a loop until a number is determined to be clean.
I am using arc4random to generate a random number. I would like to generate a number between 1-9. How can I exclude 0?
int r = arc4random() % (9);
NSLog(#"Random Number: %d",r);
int r = (arc4random() % 8) + 1
You can use arc4random_uniform(), as in
arc4random_uniform(9) + 1
Generally, to generate a number between lo and hi (inclusive), you use:
arc4random_uniform(hi - lo + 1) + lo
If you don't want to use arc4random_uniform() and want to stick with arc4random(), noting that the resulting value from using modulus formula is not uniformly distributed, use
(arc4random() % (hi - lo + 1)) + lo
int r = arc4random() % 8 + 1;
See other answers (e.g., one from me) for why you probably don't want to use % for this task, and what you should use instead.
You could simply try repeatedly until you get a number in the range you want, throwing out numbers you don't want. This has the fancy name "acceptance-rejection method" in math. It's simple, and it works.
In case you're worried that this could take a long time, in theory it could. But this approach is used all the time in statistical applications. The probability of going through a while-loop N times decreases rapidly as N increases, so the average number of times through the loop is small.
I know the modulus (%) operator calculates the remainder of a division. How can I identify a situation where I would need to use the modulus operator?
I know I can use the modulus operator to see whether a number is even or odd and prime or composite, but that's about it. I don't often think in terms of remainders. I'm sure the modulus operator is useful, and I would like to learn to take advantage of it.
I just have problems identifying where the modulus operator is applicable. In various programming situations, it is difficult for me to see a problem and realize "Hey! The remainder of division would work here!".
Imagine that you have an elapsed time in seconds and you want to convert this to hours, minutes, and seconds:
h = s / 3600;
m = (s / 60) % 60;
s = s % 60;
0 % 3 = 0;
1 % 3 = 1;
2 % 3 = 2;
3 % 3 = 0;
Did you see what it did? At the last step it went back to zero. This could be used in situations like:
To check if N is divisible by M (for example, odd or even)
or
N is a multiple of M.
To put a cap of a particular value. In this case 3.
To get the last M digits of a number -> N % (10^M).
I use it for progress bars and the like that mark progress through a big loop. The progress is only reported every nth time through the loop, or when count%n == 0.
I've used it when restricting a number to a certain multiple:
temp = x - (x % 10); //Restrict x to being a multiple of 10
Wrapping values (like a clock).
Provide finite fields to symmetric key algorithms.
Bitwise operations.
And so on.
One use case I saw recently was when you need to reverse a number. So that 123456 becomes 654321 for example.
int number = 123456;
int reversed = 0;
while ( number > 0 ) {
# The modulus here retrieves the last digit in the specified number
# In the first iteration of this loop it's going to be 6, then 5, ...
# We are multiplying reversed by 10 first, to move the number one decimal place to the left.
# For example, if we are at the second iteration of this loop,
# reversed gonna be 6, so 6 * 10 + 12345 % 10 => 60 + 5
reversed = reversed * 10 + number % 10;
number = number / 10;
}
Example. You have message of X bytes, but in your protocol maximum size is Y and Y < X. Try to write small app that splits message into packets and you will run into mod :)
There are many instances where it is useful.
If you need to restrict a number to be within a certain range you can use mod. For example, to generate a random number between 0 and 99 you might say:
num = MyRandFunction() % 100;
Any time you have division and want to express the remainder other than in decimal, the mod operator is appropriate. Things that come to mind are generally when you want to do something human-readable with the remainder. Listing how many items you could put into buckets and saying "5 left over" is good.
Also, if you're ever in a situation where you may be accruing rounding errors, modulo division is good. If you're dividing by 3 quite often, for example, you don't want to be passing .33333 around as the remainder. Passing the remainder and divisor (i.e. the fraction) is appropriate.
As #jweyrich says, wrapping values. I've found mod very handy when I have a finite list and I want to iterate over it in a loop - like a fixed list of colors for some UI elements, like chart series, where I want all the series to be different, to the extent possible, but when I've run out of colors, just to start over at the beginning. This can also be used with, say, patterns, so that the second time red comes around, it's dashed; the third time, dotted, etc. - but mod is just used to get red, green, blue, red, green, blue, forever.
Calculation of prime numbers
The modulo can be useful to convert and split total minutes to "hours and minutes":
hours = minutes / 60
minutes_left = minutes % 60
In the hours bit we need to strip the decimal portion and that will depend on the language you are using.
We can then rearrange the output accordingly.
Converting linear data structure to matrix structure:
where a is index of linear data, and b is number of items per row:
row = a/b
column = a mod b
Note above is simplified logic: a must be offset -1 before dividing & the result must be normalized +1.
Example: (3 rows of 4)
1 2 3 4
5 6 7 8
9 10 11 12
(7 - 1)/4 + 1 = 2
7 is in row 2
(7 - 1) mod 4 + 1 = 3
7 is in column 3
Another common use of modulus: hashing a number by place. Suppose you wanted to store year & month in a six digit number 195810. month = 195810 mod 100 all digits 3rd from right are divisible by 100 so the remainder is the 2 rightmost digits in this case the month is 10. To extract the year 195810 / 100 yields 1958.
Modulus is also very useful if for some crazy reason you need to do integer division and get a decimal out, and you can't convert the integer into a number that supports decimal division, or if you need to return a fraction instead of a decimal.
I'll be using % as the modulus operator
For example
2/4 = 0
where doing this
2/4 = 0 and 2 % 4 = 2
So you can be really crazy and let's say that you want to allow the user to input a numerator and a divisor, and then show them the result as a whole number, and then a fractional number.
whole Number = numerator/divisor
fractionNumerator = numerator % divisor
fractionDenominator = divisor
Another case where modulus division is useful is if you are increasing or decreasing a number and you want to contain the number to a certain range of number, but when you get to the top or bottom you don't want to just stop. You want to loop up to the bottom or top of the list respectively.
Imagine a function where you are looping through an array.
Function increase Or Decrease(variable As Integer) As Void
n = (n + variable) % (listString.maxIndex + 1)
Print listString[n]
End Function
The reason that it is n = (n + variable) % (listString.maxIndex + 1) is to allow for the max index to be accounted.
Those are just a few of the things that I have had to use modulus for in my programming of not just desktop applications, but in robotics and simulation environments.
Computing the greatest common divisor
Determining if a number is a palindrome
Determining if a number consists of only ...
Determining how many ... a number consists of...
My favorite use is for iteration.
Say you have a counter you are incrementing and want to then grab from a known list a corresponding items, but you only have n items to choose from and you want to repeat a cycle.
var indexFromB = (counter-1)%n+1;
Results (counter=indexFromB) given n=3:
`1=1`
`2=2`
`3=3`
`4=1`
`5=2`
`6=3`
...
Best use of modulus operator I have seen so for is to check if the Array we have is a rotated version of original array.
A = [1,2,3,4,5,6]
B = [5,6,1,2,3,4]
Now how to check if B is rotated version of A ?
Step 1: If A's length is not same as B's length then for sure its not a rotated version.
Step 2: Check the index of first element of A in B. Here first element of A is 1. And its index in B is 2(assuming your programming language has zero based index).
lets store that index in variable "Key"
Step 3: Now how to check that if B is rotated version of A how ??
This is where modulus function rocks :
for (int i = 0; i< A.length; i++)
{
// here modulus function would check the proper order. Key here is 2 which we recieved from Step 2
int j = [Key+i]%A.length;
if (A[i] != B[j])
{
return false;
}
}
return true;
It's an easy way to tell if a number is even or odd. Just do # mod 2, if it is 0 it is even, 1 it is odd.
Often, in a loop, you want to do something every k'th iteration, where k is 0 < k < n, assuming 0 is the start index and n is the length of the loop.
So, you'd do something like:
int k = 5;
int n = 50;
for(int i = 0;i < n;++i)
{
if(i % k == 0) // true at 0, 5, 10, 15..
{
// do something
}
}
Or, you want to keep something whitin a certain bound. Remember, when you take an arbitrary number mod something, it must produce a value between 0 and that number - 1.
So I thought that negative numbers, when mod'ed should be put into positive space... I cant get this to happen in objective-c
I expect this:
-1 % 3 = 2
0 % 3 = 0
1 % 3 = 1
2 % 3 = 2
But get this
-1 % 3 = -1
0 % 3 = 0
1 % 3 = 1
2 % 3 = 2
Why is this and is there a workaround?
result = n % 3;
if( result < 0 ) result += 3;
Don't perform extra mod operations as suggested in the other answers. They are very expensive and unnecessary.
In C and Objective-C, the division and modulus operators perform truncation towards zero. a / b is floor(a / b) if a / b > 0, otherwise it is ceiling(a / b) if a / b < 0. It is always the case that a == (a / b) * b + (a % b), unless of course b is 0. As a consequence, positive % positive == positive, positive % negative == positive, negative % positive == negative, and negative % negative == negative (you can work out the logic for all 4 cases, although it's a little tricky).
If n has a limited range, then you can get the result you want simply by adding a known constant multiple of 3 that is greater that the absolute value of the minimum.
For example, if n is limited to -1000..2000, then you can use the expression:
result = (n+1002) % 3;
Make sure the maximum plus your constant will not overflow when summed.
We have a problem of language:
math-er-says: i take this number plus that number mod other-number
code-er-hears: I add two numbers and then devide the result by other-number
code-er-says: what about negative numbers?
math-er-says: WHAT? fields mod other-number don't have a concept of negative numbers?
code-er-says: field what? ...
the math person in this conversations is talking about doing math in a circular number line. If you subtract off the bottom you wrap around to the top.
the code person is talking about an operator that calculates remainder.
In this case you want the mathematician's mod operator and have the remainder function at your disposal. you can convert the remainder operator into the mathematician's mod operator by checking to see if you fell of the bottom each time you do subtraction.
If this will be the behavior, and you know that it will be, then for m % n = r, just use r = n + r. If you're unsure of what will happen here, use then r = r % n.
Edit: To sum up, use r = ( n + ( m % n ) ) % n
I would have expected a positive number, as well, but I found this, from ISO/IEC 14882:2003 : Programming languages -- C++, 5.6.4 (found in the Wikipedia article on the modulus operation):
The binary % operator yields the remainder from the division of the first expression by the second. .... If both operands are nonnegative then the remainder is nonnegative; if not, the sign of the remainder is implementation-defined
JavaScript does this, too. I've been caught by it a couple times. Think of it as a reflection around zero rather than a continuation.
Why: because that is the way the mod operator is specified in the C-standard (Remember that Objective-C is an extension of C). It confuses most people I know (like me) because it is surprising and you have to remember it.
As to a workaround: I would use uncleo's.
UncleO's answer is probably more robust, but if you want to do it on a single line, and you're certain the negative value will not be more negative than a single iteration of the mod (for example if you're only ever subtracting at most the mod value at any time) you can simplify it to a single expression:
int result = (n + 3) % 3;
Since you're doing the mod anyway, adding 3 to the initial value has no effect unless n is negative (but not less than -3) in which case it causes result to be the expected positive modulus.
There are two choices for the remainder, and the sign depends on the language. ANSI C chooses the sign of the dividend. I would suspect this is why you see Objective-C doing so also. See the wikipedia entry as well.
Not only java script, almost all the languages shows the wrong answer'
what coneybeare said is correct, when we have mode'd we have to get remainder
Remainder is nothing but which remains after division and it should be a positive integer....
If you check the number line you can understand that
I also face the same issue in VB and and it made me to forcefully add extra check like
if the result is a negative we have to add the divisor to the result
Instead of a%b
Use: a-b*floor((float)a/(float)b)
You're expecting remainder and are using modulo. In math they are the same thing, in C they are different. GNU-C has Rem() and Mod(), objective-c only has mod() so you will have to use the code above to simulate rem function (which is the same as mod in the math world, but not in the programming world [for most languages at least])
Also note you could define an easy to use macro for this.
#define rem(a,b) ((int)(a-b*floor((float)a/(float)b)))
Then you could just use rem(-1,3) in your code and it should work fine.