Just doing some basic modulo operations and trying to wrap my head around the below operations with questions marks.
0%5 // 0 - Totally understand
1%5 // 1 ?
2%5 // 2 ?
3%5 // 3 ?
4%5 // 4 ?
5%5 // 0 - Totally understand
Perhaps I'm thinking in the wrong way. For example 1/5 would return a Double of 0.2 and not a single integer so how does it return a remainder of 1?
I understand these. It makes sense but the above I can't wrap my head around.
9%4 // 1
10%2 // 0
10%6 // 4
Be great if someone could explain this. Seems I'm having a brain fart. Source of learning.
From the same Basic Operators page that you link to:
The remainder operator (a % b) works out how many multiples of b will fit inside a and returns the value that is left over (known as the remainder).
Specifically for 1 % 5:
5 doesn't fit in 1, so it fits 0 times.
This means that 1 can be described as
1 = (5 * multiplier) + remainder
Since the multiplier is 0, the remainder is 1
1 = (5 * 0) + remainder
1 = remainder
If we instead look at 6 % 5 the remainder is also 1. This is because 5 fit in 6 one time:
6 = (5 * multiplier) + remainder
6 = (5 * 1) + remainder
6-5 = remainder
1 = remainder
This / the division operator when you say 1/5 if division is in integer it'll give 0 , but this 1.0/0.5 when you make it in Double , it'll give 0.2
but % the modulo operator when you say 1%5 = 1 because you have 1 = 0*5 + 1 which means that 1 has zero number of 5 and the reminder is 1
I would like to know where I can read about algorithms for solving this problem efficiently:
Four directions allowed: up, down, left, right
Cells containing zero can't be visited.
Visiting the same cell twice is illegal.
Moves wraps around the edges:
(first row is connected with last row)
(first col is connected with last col)
Example, 5x5 and 5 steps:
9 1 3 1 9
6 3 2 4 1
0 7 * 7 7
5 4 9 4 9
7 9 1 5 5
Starting point: *
Solution: down,left,down,left,down. That is 9 + 4 + 9 + 7 + 9 = 38
[9] 1 3 1 9
6 3 2 4 1
0 7 * 7 7
5 [4][9] 4 9
[7][9] 1 5 5
This problem is probably not related to:
Finding the maximum sub matrix
Dynamic programming
You specified in comments that you wanted a sub-second way of finding the best value 20-step path out of a 5x5 matrix. I've implemented a basic recursive search tree that does this. Ultimately, the difficulty of this is still O(3^k), but highly saturated cases like yours (21 out of 24 allowed nodes visited) will solve much faster because the problem simplifies to "skip the n*n-z-k-1 lowest value cells" (in this case, n=5, z=1 and k+1 = 21; the winning path skips three 1's).
The problem instance in your question solves in 0.231seconds on a 3 year old i5 laptop and about half a second on ideone.com. I've provided code here http://ideone.com/5kOyxq (note that 'up' and 'down' are reversed because of the way I input the data).
For less saturated problems you may need to add a Bound/Cut method. You can generate a Bound as follows:
First, run over the NxN matrix and collect the K highest value elements (can be done in N² log K) and sort them by max-first. Then, cumulatively calculate the value UB[t] = SUM[i::0->t] SortedElements[i]. So, any t-length path has a UB of UB[t] (max t elements).
At step T, the current Branch's UB is UB[t]. If ValueSoFar[T] + UB[K-T] <= BestPathValue, you can stop that branch.
There may be better ways, but this should be sufficient for reasonably sized matrices and path lengths.
Game or puzzle. Given a matrix, number of steps and a sum, find the path.
Would be nice if there is a real world application for this, but i haven't found it.
Games tend to "burn in" knowledge in young brains, so why not burn in something useful, like addition?
#include<iostream>
#include<climits>
#define R 3
#define C 3
int MAX(int x, int y, int z);
int Max_Cost(int cost[R][C], int m, int n)
{
if (n < 0 || m < 0)
return INT_MIN;
else if (m == 0 && n == 0)
return cost[m][n];
else
return cost[m][n] + MIN( Max_Cost(cost, m-1, n-1),
Max_Cost(cost, m-1, n),
Max_Cost(cost, m, n-1)
);
}
int MAX(int x, int y, int z)
{
return max(max(x, y), z);
}
int main()
{
int cost[R][C] = { {3, 2, 5},
{5, 8, 2},
{9, 7, 1}
};
cout<<Max_Cost(cost, 2, 1);
return 0;
}
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 need to divide one int into 2 other int's. the first int is not constant so one problem would be, what to do with odd numbers because I only want whole numbers. For example, if int = 5, then int(2) will = 2 and int(3) will = 3. Any help will greatly be appreciated.
Supposing you want to express x = a + b, where a and b are as close to x/2 as possible:
a = ceiling(x / 2.0);
b = floor(x / 2.0);
That's pseudo code, you have to find out the actual functions for floor and ceiling from your library. Make sure the division is performed as floating point numbers.
As pure integers:
a = x / 2 + (x % 2 == 0 ? 0 : 1);
b = x / 2
(This may be a bit fishy for negative numbers, because it'll depend on the behaviour of division and modulo for negative numbers.)
You can try ceil and floor functions from math to produce results like 2 and 3 for odd inputs;
int(2)=ceil(int/2); //will produce 3 for input 5
int(3)=floor(int/2); //will produce 2 for input 5
Well my answer is not in Objective-C but i guess you could translate this easily.
My idea is:
part1 = source_number div 2
part2 = source_number div 2 + (source_number mod 2)
This way the second number will be bigger if the starting number is an odd number.
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.