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.
First of all I see the number of strings as the following:
1 (epsilon 0 length string) + 3 (pick one letter) + 9 (3 options for first letter, 3 options for second)
For a total of 13 strings. Now as far as I know a language can pick any combination of this for example l1 = {ab,a,ac} l2 = {c}
I'm not sure how to calculate the total number of languages there could be here. Any Help?
So you have a set with 13 elements. A particular language could be any subset of this set. How many subsets does this set have?
This is called the power set of that set, and it has 213 elements.
Cardinality of character set, say d = 3.
Total words possible of length (<= k), say w = (d^(k+1) - 1)/(d-1) = 13.
Total languages possible = Power Set {Each word can be included or not} = 2^w = 8192.
I'm testing the speed of some functions so I made a test to run the functions over and over again and I stored the results in an array. I needed them to be sorted by the size of the array I randomly generated. I generate 100 elements. Merge sort to the rescue! I used this link to get me started.
The section of code I'm focusing on:
private void mergesort(int low, int high) {
// check if low is smaller then high, if not then the array is sorted
if (low < high) {
// Get the index of the element which is in the middle
int middle = low + (high - low) / 2;
// Sort the left side of the array
mergesort(low, middle);
// Sort the right side of the array
mergesort(middle + 1, high);
// Combine them both
merge(low, middle, high);
}
}
which translated to VB.NET is
private sub mergesort(low as integer, high as integer)
' check if low is smaller then high, if not then the array is sorted
if (low < high)
' Get the index of the element which is in the middle
dim middle as integer = low + (high - low) / 2
' Sort the left side of the array
mergesort(low, middle)
' Sort the right side of the array
mergesort(middle + 1, high)
' Combine them both
merge(low, middle, high)
end if
end sub
Of more importance the LOC that only matters to this question is
dim middle as integer = low + (high - low) / 2
In case you wanna see how merge sort is gonna run this baby
high low high low
100 0 10 0
50 0 6 4
25 0 5 4
12 0 12 7
6 0 10 7
3 0 8 7
2 0 :stackoverflow error:
The error comes from the fact 7 + (8 - 7) / 2 = 8. You'll see 7 and 8 get passed in to mergesort(low, middle) and then we infinite loop. Now earlier in the sort you see a comparison like this again. At 5 and 4. 4 + (5 - 4) / 2 = 4. So essentially for 5 and 4 it becomes 4 + (1) / 2 = 4.5 = 4. For 8 and 7 though it's 7 + (1) / 2 = 7.5 = 8. Remember the numbers are typecasted to an int.
Maybe I'm just using a bad implementation of it or my typecasting is wrong, but my question is: Shouldn't this be a red flag signaling something isn't right with the rounding that's occuring?
Without understanding the whole algorithm, note that VB.NET / is different than C# /. The latter has integer division by default, if you want to truncate decimal places also in VB.NET you have to use \.
Read: \ Operator
So i think that this is what you want:
Dim middle as Int32 = low + (high - low) \ 2
You are correct in your diagnosis: there's something inconsistent with the rounding that's occurring, but this is entirely expected if you know where to look.
From the VB.NET documentation on the / operator:
Divides two numbers and returns a floating-point result.
This documentation explicitly states that , if x and y are integral types, x / y returns a Double. So, 5 / 2 in VB.NET would be expected to be 2.5.
From the C# documentation on the / operator:
All numeric types have predefined division operators.
And further down the page:
When you divide two integers, the result is always an integer.
In the case of C#, if x and y are integers, x / y returns an integer (rounded down). 5 / 2 in C# is expected to return 2.
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 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.