I currently have code in objective C that can pull out an integer's most significant digit value. My only question is if there is a better way to do it than with how I have provided below. It gets the job done, but it just feels like a cheap hack.
What the code does is that it takes a number passed in and loops through until that number has been successfully divided to a certain value. The reason I am doing this is for an educational app that splits a number up by it's value and shows the values added all together to produce the final output (1234 = 1000 + 200 + 30 + 4).
int test = 1;
int result = 0;
int value = 0;
do {
value = input / test;
result = test;
test = [[NSString stringWithFormat:#"%d0",test] intValue];
} while (value >= 10);
Any advice is always greatly appreciated.
Will this do the trick?
int sigDigit(int input)
{
int digits = (int) log10(input);
return input / pow(10, digits);
}
Basically it does the following:
Finds out the number of digits in input (log10(input)) and storing it in 'digits'.
divides input by 10 ^ digits.
You should now have the most significant number in digits.
EDIT: in case you need a function that get the integer value at a specific index, check this function out:
int digitAtIndex(int input, int index)
{
int trimmedLower = input / (pow(10, index)); // trim the lower half of the input
int trimmedUpper = trimmedLower % 10; // trim the upper half of the input
return trimmedUpper;
}
Related
Example, I have 2 ints and 1 output.
int IntNumberOne = 100;
int IntNumberTwo = 30;
NSLog(#"");
I would like the output to display just ONE of the two ints, and randomize every time this program is started.
How would that be done?
Use arc4random() function to pick number randomly:
NSLog(#"%d", arc4random()%2 ? IntNumberOne : IntNumberTwo);
I have two text boxes and user can input 2 positive integers (Using Objective-C). The goal is to return a random value between the two numbers.
I've used "man arc4random" and still can't quite wrap my head around it. I've came up with some code but it's buggy.
float lowerBound = lowerBoundNumber.text.floatValue;
float upperBound = upperBoundNumber.text.floatValue;
float rndValue;
//if lower bound is lowerbound < higherbound else switch the two around before randomizing.
if(lowerBound < upperBound)
{
rndValue = (((float)arc4random()/0x100000000)*((upperBound-lowerBound)+lowerBound));
}
else
{
rndValue = (((float)arc4random()/0x100000000)*((lowerBound-upperBound)+upperBound));
}
Right now if I put in the values 0 and 3 it seems to work just fine. However if I use the numbers 10 and 15 I can still get values as low as 1.0000000 or 2.000000 for "rndValue".
Do I need to elaborate my algorithm or do I need to change the way I use arc4random?
You could simply use integer values like this:
int lowerBound = ...
int upperBound = ...
int rndValue = lowerBound + arc4random() % (upperBound - lowerBound);
Or if you mean you want to include float number between lowerBound and upperBound? If so please refer to this question: https://stackoverflow.com/a/4579457/1265516
The following code include the minimum AND MAXIMUM value as the random output number:
- (NSInteger)randomNumberBetween:(NSInteger)min maxNumber:(NSInteger)max
{
return min + arc4random_uniform((uint32_t)(max - min + 1));
}
Update:
I edited the answer by replacing arc4random() % upper_bound with arc4random_uniform(upper_bound) as #rmaddy suggested.
And here is the reference of arc4random_uniform for the details.
Update2:
I updated the answer by inserting a cast to uint32_t in arc4random_uniform() as #bicycle indicated.
-(int) generateRandomNumberWithlowerBound:(int)lowerBound
upperBound:(int)upperBound
{
int rndValue = lowerBound + arc4random() % (upperBound - lowerBound);
return rndValue;
}
You should avoid clamping values with mod (%) if you can, because even if the pseudo-random number generator you're using (like arc4random) is good at providing uniformly distributed numbers in its full range, it may not provide uniformly distributed numbers within the restricted modulo range.
You also don't need to use a literal like 0x100000000 because there is a convenient constant available in stdint.h:
(float)arc4random() / UINT32_MAX
That will give you a random float in the interval [0,1]. Note that arc4random returns an integer in the interval [0, 2**32 - 1].
To move this into the interval you want, you just add your minimum value and multiply the random float by the size of your range:
lowerBound + ((float)arc4random() / UINT32_MAX) * (upperBound - lowerBound);
In the code you posted you're multiplying the random float by the whole mess (lowerBound + (upperBound - lowerBound)), which is actually just equal to upperBound. And that's why you're still getting results less than your intended lower bound.
Objective-C Function:
-(int)getRandomNumberBetween:(int)from to:(int)to
{
return (int)from + arc4random() % (to-from+1);
}
Swift:
func getRandomNumberBetween(_ from: Int, to: Int) -> Int
{
return Int(from) + arc4random() % (to - from + 1)
}
Call it anywhere by:
int OTP = [self getRandomNumberBetween:10 to:99];
NSLog(#"OTP IS %ld",(long)OTP);
NSLog(#"OTP IS %#",[NSString stringWithFormat #"%ld",(long)OTP]);
For Swift:
var OTP: Int = getRandomNumberBetween(10, to: 99)
In Swift:
let otp = Int(arc4random_uniform(6))
Try this.
i am not sure about my english, but i need to get the unit digit of an integer.
WITHOUT complex algorithm but with some API or another trick.
for example :
int a= 53;
int b=76;
this i add because i almost always dont "meet the quality standards" to post! its drive me crazy! please , fix it ! it took me 10 shoots to post this,and other issue also.
i need to get a=3 and b=6 in a simple smart way.
same about the other digit.
thanks a lot .
here is how to split the number into parts
int unitDigit = a % 10; //is 3
int tens= (a - unitDigit)/10; //is 53-3=50 /10 =5
You're looking for % operator.
a=a%10;//divides 'a' by 10, assigns remainder to 'a'
WARNING
here is how to divine the number into parts
int unitDigit = a % 10; //is 3
int tens= (a - unitDigit)/10; //is 53-3=50 /10 =5
this answer is totally incorrect. It may work only in a number of cases. For example try to get the first digit of 503 via this way
It seems the simplest answer (but not very good in performance):
int a = ...;
int digit = [[[NSString stringWithFormat:#"%d", a] substringToIndex:1] intValue]; //or use substringWithRange to get ANY digit
Modulo operator will help you (as units digit is a reminder when number is divided by 10):
int unitDigit = a % 10;
The following code "gets" the digits of a given number and counts how many of them divide the number exactly.
int findDigits(long long N){
int count = 0;
long long newN = N;
while(newN) // kinda like a right shift
{
int div = newN % 10;
if (div != 0)
if (N % div == 0) count++;
newN = newN / 10;
}
return count;
}
I'm using
for (int i = 1, i<100, i++)
int i = arc4random() % array count;
but I'm getting repeats every time. How can I fill out the chosen int value from the range, so that when the program loops I will not get any dupe?
It sounds like you want shuffling of a set rather than "true" randomness. Simply create an array where all the positions match the numbers and initialize a counter:
num[ 0] = 0
num[ 1] = 1
: :
num[99] = 99
numNums = 100
Then, whenever you want a random number, use the following method:
idx = rnd (numNums); // return value 0 through numNums-1
val = num[idx]; // get then number at that position.
num[idx] = val[numNums-1]; // remove it from pool by overwriting with highest
numNums--; // and removing the highest position from pool.
return val; // give it back to caller.
This will return a random value from an ever-decreasing pool, guaranteeing no repeats. You will have to beware of the pool running down to zero size of course, and intelligently re-initialize the pool.
This is a more deterministic solution than keeping a list of used numbers and continuing to loop until you find one not in that list. The performance of that sort of algorithm will degrade as the pool gets smaller.
A C function using static values something like this should do the trick. Call it with
int i = myRandom (200);
to set the pool up (with any number zero or greater specifying the size) or
int i = myRandom (-1);
to get the next number from the pool (any negative number will suffice). If the function can't allocate enough memory, it will return -2. If there's no numbers left in the pool, it will return -1 (at which point you could re-initialize the pool if you wish). Here's the function with a unit testing main for you to try out:
#include <stdio.h>
#include <stdlib.h>
#define ERR_NO_NUM -1
#define ERR_NO_MEM -2
int myRandom (int size) {
int i, n;
static int numNums = 0;
static int *numArr = NULL;
// Initialize with a specific size.
if (size >= 0) {
if (numArr != NULL)
free (numArr);
if ((numArr = malloc (sizeof(int) * size)) == NULL)
return ERR_NO_MEM;
for (i = 0; i < size; i++)
numArr[i] = i;
numNums = size;
}
// Error if no numbers left in pool.
if (numNums == 0)
return ERR_NO_NUM;
// Get random number from pool and remove it (rnd in this
// case returns a number between 0 and numNums-1 inclusive).
n = rand() % numNums;
i = numArr[n];
numArr[n] = numArr[numNums-1];
numNums--;
if (numNums == 0) {
free (numArr);
numArr = 0;
}
return i;
}
int main (void) {
int i;
srand (time (NULL));
i = myRandom (20);
while (i >= 0) {
printf ("Number = %3d\n", i);
i = myRandom (-1);
}
printf ("Final = %3d\n", i);
return 0;
}
And here's the output from one run:
Number = 19
Number = 10
Number = 2
Number = 15
Number = 0
Number = 6
Number = 1
Number = 3
Number = 17
Number = 14
Number = 12
Number = 18
Number = 4
Number = 9
Number = 7
Number = 8
Number = 16
Number = 5
Number = 11
Number = 13
Final = -1
Keep in mind that, because it uses statics, it's not safe for calling from two different places if they want to maintain their own separate pools. If that were the case, the statics would be replaced with a buffer (holding count and pool) that would "belong" to the caller (a double-pointer could be passed in for this purpose).
And, if you're looking for the "multiple pool" version, I include it here for completeness.
#include <stdio.h>
#include <stdlib.h>
#define ERR_NO_NUM -1
#define ERR_NO_MEM -2
int myRandom (int size, int *ppPool[]) {
int i, n;
// Initialize with a specific size.
if (size >= 0) {
if (*ppPool != NULL)
free (*ppPool);
if ((*ppPool = malloc (sizeof(int) * (size + 1))) == NULL)
return ERR_NO_MEM;
(*ppPool)[0] = size;
for (i = 0; i < size; i++) {
(*ppPool)[i+1] = i;
}
}
// Error if no numbers left in pool.
if (*ppPool == NULL)
return ERR_NO_NUM;
// Get random number from pool and remove it (rnd in this
// case returns a number between 0 and numNums-1 inclusive).
n = rand() % (*ppPool)[0];
i = (*ppPool)[n+1];
(*ppPool)[n+1] = (*ppPool)[(*ppPool)[0]];
(*ppPool)[0]--;
if ((*ppPool)[0] == 0) {
free (*ppPool);
*ppPool = NULL;
}
return i;
}
int main (void) {
int i;
int *pPool;
srand (time (NULL));
pPool = NULL;
i = myRandom (20, &pPool);
while (i >= 0) {
printf ("Number = %3d\n", i);
i = myRandom (-1, &pPool);
}
printf ("Final = %3d\n", i);
return 0;
}
As you can see from the modified main(), you need to first initialise an int pointer to NULL then pass its address to the myRandom() function. This allows each client (location in the code) to have their own pool which is automatically allocated and freed, although you could still share pools if you wish.
You could use Format-Preserving Encryption to encrypt a counter. Your counter just goes from 0 upwards, and the encryption uses a key of your choice to turn it into a seemingly random value of whatever radix and width you want.
Block ciphers normally have a fixed block size of e.g. 64 or 128 bits. But Format-Preserving Encryption allows you to take a standard cipher like AES and make a smaller-width cipher, of whatever radix and width you want (e.g. radix 2, width 16), with an algorithm which is still cryptographically robust.
It is guaranteed to never have collisions (because cryptographic algorithms create a 1:1 mapping). It is also reversible (a 2-way mapping), so you can take the resulting number and get back to the counter value you started with.
AES-FFX is one proposed standard method to achieve this. I've experimented with some basic Python code which is based on the AES-FFX idea, although not fully conformant--see Python code here. It can e.g. encrypt a counter to a random-looking 7-digit decimal number, or a 16-bit number.
You need to keep track of the numbers you have already used (for instance, in an array). Get a random number, and discard it if it has already been used.
Without relying on external stochastic processes, like radioactive decay or user input, computers will always generate pseudorandom numbers - that is numbers which have many of the statistical properties of random numbers, but repeat in sequences.
This explains the suggestions to randomise the computer's output by shuffling.
Discarding previously used numbers may lengthen the sequence artificially, but at a cost to the statistics which give the impression of randomness.
The best way to do this is create an array for numbers already used. After a random number has been created then add it to the array. Then when you go to create another random number, ensure that it is not in the array of used numbers.
In addition to using secondary array to store already generated random numbers, invoking random no. seeding function before every call of random no. generation function might help to generate different seq. of random numbers in every run.
I'm looking for a way to generate combinations of objects ordered by a single attribute. I don't think lexicographical order is what I'm looking for... I'll try to give an example. Let's say I have a list of objects A,B,C,D with the attribute values I want to order by being 3,3,2,1. This gives A3, B3, C2, D1 objects. Now I want to generate combinations of 2 objects, but they need to be ordered in a descending way:
A3 B3
A3 C2
B3 C2
A3 D1
B3 D1
C2 D1
Generating all combinations and sorting them is not acceptable because the real world scenario involves large sets and millions of combinations. (set of 40, order of 8), and I need only combinations above the certain threshold.
Actually I need count of combinations above a threshold grouped by a sum of a given attribute, but I think it is far more difficult to do - so I'd settle for developing all combinations above a threshold and counting them. If that's possible at all.
EDIT - My original question wasn't very precise... I don't actually need these combinations ordered, just thought it would help to isolate combinations above a threshold. To be more precise, in the above example, giving a threshold of 5, I'm looking for an information that the given set produces 1 combination with a sum of 6 ( A3 B3 ) and 2 with a sum of 5 ( A3 C2, B3 C2). I don't actually need the combinations themselves.
I was looking into subset-sum problem, but if I understood correctly given dynamic solution it will only give you information is there a given sum or no, not count of the sums.
Thanks
Actually, I think you do want lexicographic order, but descending rather than ascending. In addition:
It's not clear to me from your description that A, B, ... D play any role in your answer (except possibly as the container for the values).
I think your question example is simply "For each integer at least 5, up to the maximum possible total of two values, how many distinct pairs from the set {3, 3, 2, 1} have sums of that integer?"
The interesting part is the early bailout, once no possible solution can be reached (remaining achievable sums are too small).
I'll post sample code later.
Here's the sample code I promised, with a few remarks following:
public class Combos {
/* permanent state for instance */
private int values[];
private int length;
/* transient state during single "count" computation */
private int n;
private int limit;
private Tally<Integer> tally;
private int best[][]; // used for early-bail-out
private void initializeForCount(int n, int limit) {
this.n = n;
this.limit = limit;
best = new int[n+1][length+1];
for (int i = 1; i <= n; ++i) {
for (int j = 0; j <= length - i; ++j) {
best[i][j] = values[j] + best[i-1][j+1];
}
}
}
private void countAt(int left, int start, int sum) {
if (left == 0) {
tally.inc(sum);
} else {
for (
int i = start;
i <= length - left
&& limit <= sum + best[left][i]; // bail-out-check
++i
) {
countAt(left - 1, i + 1, sum + values[i]);
}
}
}
public Tally<Integer> count(int n, int limit) {
tally = new Tally<Integer>();
if (n <= length) {
initializeForCount(n, limit);
countAt(n, 0, 0);
}
return tally;
}
public Combos(int[] values) {
this.values = values;
this.length = values.length;
}
}
Preface remarks:
This uses a little helper class called Tally, that just isolates the tabulation (including initialization for never-before-seen keys). I'll put it at the end.
To keep this concise, I've taken some shortcuts that aren't good practice for "real" code:
This doesn't check for a null value array, etc.
I assume that the value array is already sorted into descending order, required for the early-bail-out technique. (Good production code would include the sorting.)
I put transient data into instance variables instead of passing them as arguments among the private methods that support count. That makes this class non-thread-safe.
Explanation:
An instance of Combos is created with the (descending ordered) array of integers to combine. The value array is set up once per instance, but multiple calls to count can be made with varying population sizes and limits.
The count method triggers a (mostly) standard recursive traversal of unique combinations of n integers from values. The limit argument gives the lower bound on sums of interest.
The countAt method examines combinations of integers from values. The left argument is how many integers remain to make up n integers in a sum, start is the position in values from which to search, and sum is the partial sum.
The early-bail-out mechanism is based on computing best, a two-dimensional array that specifies the "best" sum reachable from a given state. The value in best[n][p] is the largest sum of n values beginning in position p of the original values.
The recursion of countAt bottoms out when the correct population has been accumulated; this adds the current sum (of n values) to the tally. If countAt has not bottomed out, it sweeps the values from the start-ing position to increase the current partial sum, as long as:
enough positions remain in values to achieve the specified population, and
the best (largest) subtotal remaining is big enough to make the limit.
A sample run with your question's data:
int[] values = {3, 3, 2, 1};
Combos mine = new Combos(values);
Tally<Integer> tally = mine.count(2, 5);
for (int i = 5; i < 9; ++i) {
int n = tally.get(i);
if (0 < n) {
System.out.println("found " + tally.get(i) + " sums of " + i);
}
}
produces the results you specified:
found 2 sums of 5
found 1 sums of 6
Here's the Tally code:
public static class Tally<T> {
private Map<T,Integer> tally = new HashMap<T,Integer>();
public Tally() {/* nothing */}
public void inc(T key) {
Integer value = tally.get(key);
if (value == null) {
value = Integer.valueOf(0);
}
tally.put(key, (value + 1));
}
public int get(T key) {
Integer result = tally.get(key);
return result == null ? 0 : result;
}
public Collection<T> keys() {
return tally.keySet();
}
}
I have written a class to handle common functions for working with the binomial coefficient, which is the type of problem that your problem falls under. It performs the following tasks:
Outputs all the K-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters. This method makes solving this type of problem quite trivial.
Converts the K-indexes to the proper index of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle. My paper talks about this. I believe I am the first to discover and publish this technique, but I could be wrong.
Converts the index in a sorted binomial coefficient table to the corresponding K-indexes.
Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers.
The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to perform the 4 above methods. Accessor methods are provided to access the table.
There is an associated test class which shows how to use the class and its methods. It has been extensively tested with 2 cases and there are no known bugs.
To read about this class and download the code, see Tablizing The Binomial Coeffieicent.
Check out this question in stackoverflow: Algorithm to return all combinations
I also just used a the java code below to generate all permutations, but it could easily be used to generate unique combination's given an index.
public static <E> E[] permutation(E[] s, int num) {//s is the input elements array and num is the number which represents the permutation
int factorial = 1;
for(int i = 2; i < s.length; i++)
factorial *= i;//calculates the factorial of (s.length - 1)
if (num/s.length >= factorial)// Optional. if the number is not in the range of [0, s.length! - 1]
return null;
for(int i = 0; i < s.length - 1; i++){//go over the array
int tempi = (num / factorial) % (s.length - i);//calculates the next cell from the cells left (the cells in the range [i, s.length - 1])
E temp = s[i + tempi];//Temporarily saves the value of the cell needed to add to the permutation this time
for(int j = i + tempi; j > i; j--)//shift all elements to "cover" the "missing" cell
s[j] = s[j-1];
s[i] = temp;//put the chosen cell in the correct spot
factorial /= (s.length - (i + 1));//updates the factorial
}
return s;
}
I am extremely sorry (after all those clarifications in the comments) to say that I could not find an efficient solution to this problem. I tried for the past hour with no results.
The reason (I think) is that this problem is very similar to problems like the traveling salesman problem. Until unless you try all the combinations, there is no way to know which attributes will add upto the threshold.
There seems to be no clever trick that can solve this class of problems.
Still there are many optimizations that you can do to the actual code.
Try sorting the data according to the attributes. You may be able to avoid processing some values from the list when you find that a higher value cannot satisfy the threshold (so all lower values can be eliminated).
If you're using C# there is a fairly good generics library here. Note though that the generation of some permutations is not in lexicographic order
Here's a recursive approach to count the number of these subsets: We define a function count(minIndex,numElements,minSum) that returns the number of subsets of size numElements whose sum is at least minSum, containing elements with indices minIndex or greater.
As in the problem statement, we sort our elements in descending order, e.g. [3,3,2,1], and call the first index zero, and the total number of elements N. We assume all elements are nonnegative. To find all 2-subsets whose sum is at least 5, we call count(0,2,5).
Sample Code (Java):
int count(int minIndex, int numElements, int minSum)
{
int total = 0;
if (numElements == 1)
{
// just count number of elements >= minSum
for (int i = minIndex; i <= N-1; i++)
if (a[i] >= minSum) total++; else break;
}
else
{
if (minSum <= 0)
{
// any subset will do (n-choose-k of them)
if (numElements <= (N-minIndex))
total = nchoosek(N-minIndex, numElements);
}
else
{
// add element a[i] to the set, and then consider the count
// for all elements to its right
for (int i = minIndex; i <= (N-numElements); i++)
total += count(i+1, numElements-1, minSum-a[i]);
}
}
return total;
}
Btw, I've run the above with an array of 40 elements, and size-8 subsets and consistently got back results in less than a second.