int myRand(int min, int max) {
...
}
Are there better parameter names that makes it clear that max is exclusive, not inclusive. I.e. the function generates a random number in the range [min, max)? How about lower and upper?
I would use [min, supremum) and (infimum, max].
Related
Hi this is my first ever program I'm tryin to write in android studio/Kotlin and I'm not sure how to proceed.
so in my program i have a few math tasks to do and it does it fine but what I need to do now is separate part of the answer then covert it then print out both parts
for example if my answer was 1.5232 i would like to convert the decimal part of the answer to a string that matches a range if its in it. the ranges I have are in the .0000 area so I would like to limit the decimal range too.
so final result would look like this
1 (whatever my string in range is)
I hope I included enough info thank you in advance.
The first part of the task is to split the number into the integer and fractional components:
val input = 1.5232
val integer = input.toInt() // rounds DOWN to nearest smaller Int
val fractional = input % 1.0 // The remainder when dividing by 1.0 is the fraction
The strategy I would use to round to the nearest fractional value given a certain precision is to multiply by that precision, and round to the nearest integer. That would give you the numerator, and the precision would be the denominator:
val denominator = 8 // fractional precision
val numerator = (fractional * denominator).roundToInt() // rounds up or down to nearest Int
Then to put it together, you can use a string template:
val result = "$integer $numerator/$denominator"
println(result)
Simplifying the fraction would be another task if you need that. You can find various algorithms for finding greatest common divisor of two numbers. Use one of those and divide the numerator and denominator by that value.
I have a list of elements to be searched in a dataset of variable lengths. I have tried binary search and I found it is not always efficient when the objective is to search a list of elements.
I did the following study and conclude that if the number of elements to be searched is less than 5% of the data, binary search is efficient, other wise the Linear search is better.
Below are the details
Number of elements : 100000
Number of elements to be searched: 5000
Number of Iterations (Binary Search) =
log2 (N) x SearchCount=log2 (100000) x 5000=83048
Further increase in the number of search elements lead to more iterations than the linear search.
Any thoughts on this?
I am calling the below function only if the number elements to be searched is less than 5%.
private int SearchIndex(ref List<long> entitylist, ref long[] DataList, int i, int len, ref int listcount)
{
int Start = i;
int End = len-1;
int mid;
while (Start <= End)
{
mid = (Start + End) / 2;
long target = DataList[mid];
if (target == entitylist[listcount])
{
i = mid;
listcount++;
return i;
}
else
{
if (target < entitylist[listcount])
{
Start = mid + 1;
}
if (target > entitylist[listcount])
{
End = mid - 1;
}
}
}
listcount++;
return -1; //if the element in the list is not in the dataset
}
In the code I retun the index rather than the value because, I need to work with Index in the calling function. If i=-1, the calling function resets the value to the previous i and calls the function again with a new element to search.
In your problem you are looking for M values in an N long array, N > M, but M can be quite large.
Usually this can be approached as M independent binary searches (or even with the slight optimization of using the previous result as a starting point): you are going to O(M*log(N)).
However, using the fact that also the M values are sorted, you can find all of them in one pass, with linear search. In this case you are going to have your problem O(N). In fact this is better than O(M*log(N)) for M large.
But you have a third option: since M values are sorted, binary split M too, and every time you find it, you can limit the subsequent searches in the ranges on the left and on the right of the found index.
The first look-up is on all the N values, the second two on (average) N/2, than 4 on N/4 data,.... I think that this scale as O(log(M)*log(N)). Not sure of it, comments welcome!
However here is a test code - I have slightly modified your code, but without altering its functionality.
In case you have M=100000 and N=1000000, the "M binary search approach" takes about 1.8M iterations, that's more that the 1M needed to scan linearly the N values. But with what I suggest it takes just 272K iterations.
Even in case the M values are very "collapsed" (eg, they are consecutive), and the linear search is in the best condition (100K iterations would be enough to get all of them, see the comments in the code), the algorithm performs very well.
DISCLAIMER: Rather theoretical question here, not looking for a correct answere, just asking for some inspiration!
Consider this:
A function is called repetitively and returns integers based on seeds (the same seed returns the same integer). Your task is to find out which integer is returned most often. Easy enough, right?
But: You are not allowed to use arrays or fields to store return values of said function!
Example:
int mostFrequentNumber = 0;
int occurencesOfMostFrequentNumber = 0;
int iterations = 10000000;
for(int i = 0; i < iterations; i++)
{
int result = getNumberFromSeed(i);
int occurencesOfResult = magic();
if(occurencesOfResult > occurencesOfMostFrequentNumber)
{
mostFrequentNumber = result;
occurencesOfMostFrequentNumber = occurencesOfResult;
}
}
If getNumberFromSeed() returns 2,1,5,18,5,6 and 5 then mostFrequentNumber should be 5 and occurencesOfMostFrequentNumber should be 3 because 5 is returned 3 times.
I know this could easily be solved using a two-dimentional list to store results and occurences. But imagine for a minute that you can not use any kind of arrays, lists, dictionaries etc. (Maybe because the system that is running the code has such a limited memory, that you cannot store enough integers at once or because your prehistoric programming language has no concept of collections).
How would you find mostFrequentNumber and occurencesOfMostFrequentNumber? What does magic() do?? (Of cause you do not have to stick to the example code. Any ideas are welcome!)
EDIT: I should add that the integers returned by getNumber() should be calculated using a seed, so the same seed returns the same integer (i.e. int result = getNumber(5); this would always assign the same value to result)
Make an hypothesis: Assume that the distribution of integers is, e.g., Normal.
Start simple. Have two variables
. N the number of elements read so far
. M1 the average of said elements.
Initialize both variables to 0.
Every time you read a new value x update N to be N + 1 and M1 to be M1 + (x - M1)/N.
At the end M1 will equal the average of all values. If the distribution was Normal this value will have a high frequency.
Now improve the above. Add a third variable:
M2 the average of all (x - M1)^2 for all values of xread so far.
Initialize M2 to 0. Now get a small memory of say 10 elements or so. For every new value x that you read update N and M1 as above and M2 as:
M2 := M2 + (x - M1)^2 * (N - 1) / N
At every step M2 is the variance of the distribution and sqrt(M2) its standard deviation.
As you proceed remember the frequencies of only the values read so far whose distances to M1 are less than sqrt(M2). This requires the use of some additional array, however, the array will be very short compared to the high number of iterations you will run. This modification will allow you to guess better the most frequent value instead of simply answering the mean (or average) as above.
UPDATE
Given that this is about insights for inspiration there is plenty of room for considering and adapting the approach I've proposed to any particular situation. Here are some thoughts
When I say assume that the distribution is Normal you should think of it as: Given that the problem has no solution, let's see if there is some qualitative information I can use to decide what kind of distribution would the data have. Given that the algorithm is intended to find the most frequent number, it should be fine to assume that the distribution is not uniform. Let's try with Normal, LogNormal, etc. to see what can be found out (more on this below.)
If the game completely disallows the use of any array, then fine, keep track of only, say 10 numbers. This would allow you to count the occurrences of the 10 best candidates, which will give more confidence to your answer. In doing this choose your candidates around the theoretical most likely value according to the distribution of your hypothesis.
You cannot use arrays but perhaps you can read the sequence of numbers two or three times, not just once. In that case you can read it once to check whether you hypothesis about its distribution is good nor bad. For instance, if you compute not just the variance but the skewness and the kurtosis you will have more elements to check your hypothesis. For instance, if the first reading indicates that there is some bias, you could use a LogNormal distribution instead, etc.
Finally, in addition to providing the approximate answer you would be able to use the information collected during the reading to estimate an interval of confidence around your answer.
Alright, I found a decent solution myself:
int mostFrequentNumber = 0;
int occurencesOfMostFrequentNumber = 0;
int iterations = 10000000;
int maxNumber = -2147483647;
int minNumber = 2147483647;
//Step 1: Find the largest and smallest number that _can_ occur
for(int i = 0; i < iterations; i++)
{
int result = getNumberFromSeed(i);
if(result > maxNumber)
{
maxNumber = result;
}
if(result < minNumber)
{
minNumber = result;
}
}
//Step 2: for each possible number between minNumber and maxNumber, count occurences
for(int thisNumber = minNumber; thisNumber <= maxNumber; thisNumber++)
{
int occurenceOfThisNumber = 0;
for(int i = 0; i < iterations; i++)
{
int result = getNumberFromSeed(i);
if(result == thisNumber)
{
occurenceOfThisNumber++;
}
}
if(occurenceOfThisNumber > occurencesOfMostFrequentNumber)
{
occurencesOfMostFrequentNumber = occurenceOfThisNumber;
mostFrequentNumber = thisNumber;
}
}
I must admit, this may take a long time, depending on the smallest and largest possible. But it will work without using arrays.
For example I have a float 55.2f and want to round it down such that the result can be divided by two without rest.
So 55.2 would become 54 as that is the nearest smaller "step" that can be divided by two. Is there a function for this or must I write my own algorithm?
If your result must remain a float, you can do:
float f=55.2f;
f=floorf(f/2.f)*2.f;
First convert to an integral type, such as int or long, and then clear the lowest bit.
float f = 55.2f;
int i = (int)f & ~1;
Explanation
~ means the bitwise inverse, i.e. all the 0 bits become 1, and vice versa.
So, if 1 has the bit pattern
0...0001
then ~1 is
1...1110
(Here I'm using ... to represent all the in-between bits depending on how big an integer is on your platform.)
When you & (bitwise AND) your integer with 1...1110, you are preserving the value of each bit apart from the lowest bit, which is forced to 0. See this description of the bitwise AND operator if you still don't get it.
By forcing the lowest bit to be 0, you are rounding the number down to the nearest even number.
You can write your own algorithm, for example with bitwise operators.
The following code works with clearing the last bit of your number. An even number has indeed the last bit not set.
int
f(float x)
{
return (int)x & ~1;
}
How about long int f = lrintf(x / 2);, where x is your float?
You could also just say int f = x / 2;, but some people have argued that that's more expensive, because the C standard mandates a specific rounding mode which may or may not be native to the CPU. The lrintf function on the other hand uses the CPU's native rounding mode. You need to #include <math.h>.
This is a part of a book I'm reading to learn Objective-C.
The following defines a macro called MAX that gives the maximum of two
values: #define MAX(a,b) ( ((a) > (b)) ? (a) : (b) )
And then there are some exercises in the book that asks the reader to define a macro (MIN) to find the minimum of two values and another that asks to define a macro called MAX3 that gives the maximum of 3 values. I think these two definitions will look similar to MAX, but I don't understand how the MAXformula finds the maximum value. I mean if I just did this
int limits = MAX (4,8)
It'll just assign limits the value of 8. What does that have to do with finding a variable's maximum value?
I think you are confusing value and variable. The macro example you listed expands to a comparison between two values and returns the greater of the two values (i.e. which is greater, a or b). So you are right, int limits = MAX(4,8) just assigns 8 to limits and has nothing to do with finding the maximum value you can store in limits.
The header limits.h defines many values like INT_MAX that will tell you information about the min/max values of variable types on your system.
To break it apart:
The declaration:
#define MAX(a,b)
If a is greater than b, use a else use b:
( ((a) > (b)) ? (a) : (b) )
Then to create a MIN expression, use a similar form:
#define MIN(a,b) ( ((a) < (b)) ? (a) : (b) )
^
Then to create a MAX3 expression, you can combine them:
#define MAX3(a,b,c) ( MAX(a, MAX(b,c)) )
Specifically, this macro's intended to be used with scalars (C builtins) which can be compared using < or >. If you passed an objc variable, it would result in comparison of addresses and MAX would return the one with the higher address (it would be very rare if you actually wanted to compare addresses of objc instances).
Also note that this is the classic example of how macros can bite you. With macros, the preprocessor simply expands (textual copy/paste) the parameters in place, so: int limits = MAX (4,8) literally expands to int limits = (4 > 8 ? 4 : 8). If you write MAX(x,++y), then y will be incremented twice if y is greater than or equal to x because it expands to: int limits = (x > ++y ? x : ++y).
generally, you will use a MAX() or MIN() macro to get whichever is the higher/lower of a pair of variables, or of a variable and a constant, or even a pair of macro constants or other non-literal constant expressions. you generally won't supply 2 literal constants as you have done in your question.
Algorithm for max (Objective-C)
// get max value
- (float)maxValue:(NSArray *)arrValue
{
float maxValue = 0.0;
for (NSString *value in arrValue) {
float compareValue = [value floatValue];
if (compareValue > maxValue) {
maxValue = compareValue;
}
}
return maxValue;
}
NSArray *number=[NSArray arrayWithObjects:[NSNumber numberWithFloat:57.02], [NSNumber numberWithFloat:55.02], [NSNumber numberWithFloat:45.02], nil];
NSLog(#"%f", [self maxValue:number]);
result 57.020000