Related
I have some array of integers that I generate randomly for example between 0 and 9.
I have a set of constraints that I need to apply to this generation.
For example :
1) Constraint 1 -> the numbers at even position in the array should be 0 or 1.
2) Constraint 2 -> the array should have at least one 0 and at least one 1 on even values
etc ...
What I do for now :
I generate randomly the numbers array.
Then for each even position, I randomly pick between 0 and 1.
Then I check that for each even value, I have at least one 0 and at least one 1. If not, I regerate all the values with the constraint above (Then for each even position, I randomly pick between 0 and 1.) until I have something which works (in a do for).
However, this works good because these are very simple constraints.
3) Constraint 3 -> the sum of the differences between the numbers should be superior to a certain value.
etc ...
The issue I have is that it is inter dependent constraints problems and I dont want to encapsulate do while in another do while, etc ... when I add one constraint.
What would be the proper way to achieve this in the cleanest way possible?
Edit:
I realized I was not clear at all... My apologies.
I edited the constraints to make it easier to understand.
My code looks like that (typed on notepad++ might be mistakes) :
std::vector<int> myVector;
int N = 100; // vector contains 100 values
do{
myVector.clear();
for(int i=0;i<N;i++){
if(i%2==0){
myVector.push_back(rand() % 2);
}
else{
myVector.push_back(rand() % 10);
}
}
}while(!doesContains0and1(myVector));
bool MyClass::doesContains0and1(std::vector<int> avector)
{
bool returnVal = true;
for(int i=0;i<avector.size;i++){
if(i%2==0){
if(!avector.contains(0) || !avector.contains(1){
returnVal = false;
}
}
}
return returnVal;
}
The 3) constraint means, if I have for example :
0 5 1 7 0 9 0 1 1
that there is a constraint on abs(5-0) + abs (1-5) + abs(7-1) + ... etc > a certain value
These constraints are examples, I am more looking for some methodology than pure code :)
Thanks for your help !
Let me take a stab at this...
int sum;
for (int i = 0 ; i < array_size ; i++){
if (!(i % 10)) sum = 0; // resets the sum to 0
if(i % 2){ //even or odd
//odd
sum += array[i] = rand() % 2; // 1 or 0
if ((!(i % 10)) && (sum == 0 || sum == 5))
array[i] = sum?0:1;
}else{
//even
array[i] = rand();
}
}
This follows the first 2 constraints. The sum will count the number of 1's in each subset of 10. All 0's will yield a sum of 0 and all 1's will yield a sum of 5. It will set the last element accordingly.
I'm not sure how to interpret the 3rd constraint, but hopefully this gets you started
What is the best method to find the number of digits of a positive integer?
I have found this 3 basic methods:
conversion to string
String s = new Integer(t).toString();
int len = s.length();
for loop
for(long long int temp = number; temp >= 1;)
{
temp/=10;
decimalPlaces++;
}
logaritmic calculation
digits = floor( log10( number ) ) + 1;
where you can calculate log10(x) = ln(x) / ln(10) in most languages.
First I thought the string method is the dirtiest one but the more I think about it the more I think it's the fastest way. Or is it?
There's always this method:
n = 1;
if ( i >= 100000000 ) { n += 8; i /= 100000000; }
if ( i >= 10000 ) { n += 4; i /= 10000; }
if ( i >= 100 ) { n += 2; i /= 100; }
if ( i >= 10 ) { n += 1; }
Well the correct answer would be to measure it - but you should be able to make a guess about the number of CPU steps involved in converting strings and going through them looking for an end marker
Then think how many FPU operations/s your processor can do and how easy it is to calculate a single log.
edit: wasting some more time on a monday morning :-)
String s = new Integer(t).toString();
int len = s.length();
One of the problems with high level languages is guessing how much work the system is doing behind the scenes of an apparently simple statement. Mandatory Joel link
This statement involves allocating memory for a string, and possibly a couple of temporary copies of a string. It must parse the integer and copy the digits of it into a string, possibly having to reallocate and move the existing memory if the number is large. It might have to check a bunch of locale settings to decide if your country uses "," or ".", it might have to do a bunch of unicode conversions.
Then finding the length has to scan the entire string, again considering unicode and any local specific settings such as - are you in a right->left language?.
Alternatively:
digits = floor( log10( number ) ) + 1;
Just because this would be harder for you to do on paper doesn't mean it's hard for a computer! In fact a good rule in high performance computing seems to have been - if something is hard for a human (fluid dynamics, 3d rendering) it's easy for a computer, and if it's easy for a human (face recognition, detecting a voice in a noisy room) it's hard for a computer!
You can generally assume that the builtin maths functions log/sin/cos etc - have been an important part of computer design for 50years. So even if they don't map directly into a hardware function in the FPU you can bet that the alternative implementation is pretty efficient.
I don't know, and the answer may well be different depending on how your individual language is implemented.
So, stress test it! Implement all three solutions. Run them on 1 through 1,000,000 (or some other huge set of numbers that's representative of the numbers the solution will be running against) and time how long each of them takes.
Pit your solutions against one another and let them fight it out. Like intellectual gladiators. Three algorithms enter! One algorithm leaves!
Test conditions
Decimal numeral system
Positive integers
Up to 10 digits
Language: ActionScript 3
Results
digits: [1,10],
no. of runs: 1,000,000
random sample: 8777509,40442298,477894,329950,513,91751410,313,3159,131309,2
result: 7,8,6,6,3,8,3,4,6,1
CONVERSION TO STRING: 724ms
LOGARITMIC CALCULATION: 349ms
DIV 10 ITERATION: 229ms
MANUAL CONDITIONING: 136ms
Note: Author refrains from making any conclusions for numbers with more than 10 digits.
Script
package {
import flash.display.MovieClip;
import flash.utils.getTimer;
/**
* #author Daniel
*/
public class Digits extends MovieClip {
private const NUMBERS : uint = 1000000;
private const DIGITS : uint = 10;
private var numbers : Array;
private var digits : Array;
public function Digits() {
// ************* NUMBERS *************
numbers = [];
for (var i : int = 0; i < NUMBERS; i++) {
var number : Number = Math.floor(Math.pow(10, Math.random()*DIGITS));
numbers.push(number);
}
trace('Max digits: ' + DIGITS + ', count of numbers: ' + NUMBERS);
trace('sample: ' + numbers.slice(0, 10));
// ************* CONVERSION TO STRING *************
digits = [];
var time : Number = getTimer();
for (var i : int = 0; i < numbers.length; i++) {
digits.push(String(numbers[i]).length);
}
trace('\nCONVERSION TO STRING - time: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
// ************* LOGARITMIC CALCULATION *************
digits = [];
time = getTimer();
for (var i : int = 0; i < numbers.length; i++) {
digits.push(Math.floor( Math.log( numbers[i] ) / Math.log(10) ) + 1);
}
trace('\nLOGARITMIC CALCULATION - time: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
// ************* DIV 10 ITERATION *************
digits = [];
time = getTimer();
var digit : uint = 0;
for (var i : int = 0; i < numbers.length; i++) {
digit = 0;
for(var temp : Number = numbers[i]; temp >= 1;)
{
temp/=10;
digit++;
}
digits.push(digit);
}
trace('\nDIV 10 ITERATION - time: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
// ************* MANUAL CONDITIONING *************
digits = [];
time = getTimer();
var digit : uint;
for (var i : int = 0; i < numbers.length; i++) {
var number : Number = numbers[i];
if (number < 10) digit = 1;
else if (number < 100) digit = 2;
else if (number < 1000) digit = 3;
else if (number < 10000) digit = 4;
else if (number < 100000) digit = 5;
else if (number < 1000000) digit = 6;
else if (number < 10000000) digit = 7;
else if (number < 100000000) digit = 8;
else if (number < 1000000000) digit = 9;
else if (number < 10000000000) digit = 10;
digits.push(digit);
}
trace('\nMANUAL CONDITIONING: ' + (getTimer() - time));
trace('sample: ' + digits.slice(0, 10));
}
}
}
This algorithm might be good also, assuming that:
Number is integer and binary encoded (<< operation is cheap)
We don't known number boundaries
var num = 123456789L;
var len = 0;
var tmp = 1L;
while(tmp < num)
{
len++;
tmp = (tmp << 3) + (tmp << 1);
}
This algorithm, should have speed comparable to for-loop (2) provided, but a bit faster due to (2 bit-shifts, add and subtract, instead of division).
As for Log10 algorithm, it will give you only approximate answer (that is close to real, but still), since analytic formula for computing Log function have infinite loop and can't be calculated precisely Wiki.
Use the simplest solution in whatever programming language you're using. I can't think of a case where counting digits in an integer would be the bottleneck in any (useful) program.
C, C++:
char buffer[32];
int length = sprintf(buffer, "%ld", (long)123456789);
Haskell:
len = (length . show) 123456789
JavaScript:
length = String(123456789).length;
PHP:
$length = strlen(123456789);
Visual Basic (untested):
length = Len(str(123456789)) - 1
conversion to string: This will have to iterate through each digit, find the character that maps to the current digit, add a character to a collection of characters. Then get the length of the resulting String object. Will run in O(n) for n=#digits.
for-loop: will perform 2 mathematical operation: dividing the number by 10 and incrementing a counter. Will run in O(n) for n=#digits.
logarithmic: Will call log10 and floor, and add 1. Looks like O(1) but I'm not really sure how fast the log10 or floor functions are. My knowledge of this sort of things has atrophied with lack of use so there could be hidden complexity in these functions.
So I guess it comes down to: is looking up digit mappings faster than multiple mathematical operations or whatever is happening in log10? The answer will probably vary. There could be platforms where the character mapping is faster, and others where doing the calculations is faster. Also to keep in mind is that the first method will creats a new String object that only exists for the purpose of getting the length. This will probably use more memory than the other two methods, but it may or may not matter.
You can obviously eliminate the method 1 from the competition, because the atoi/toString algorithm it uses would be similar to method 2.
Method 3's speed depends on whether the code is being compiled for a system whose instruction set includes log base 10.
For very large integers, the log method is much faster. For instance, with a 2491327 digit number (the 11920928th Fibonacci number, if you care), Python takes several minutes to execute the divide-by-10 algorithm, and milliseconds to execute 1+floor(log(n,10)).
import math
def numdigits(n):
return ( int(math.floor(math.log10(n))) + 1 )
Regarding the three methods you propose for "determining the number of digits necessary to represent a given number in a given base", I don't like any of them, actually; I prefer the method I give below instead.
Re your method #1 (strings): Anything involving converting back-and-forth between strings and numbers is usually very slow.
Re your method #2 (temp/=10): This is fatally flawed because it assumes that x/10 always means "x divided by 10". But in many programming languages (eg: C, C++), if "x" is an integer type, then "x/10" means "integer division", which isn't the same thing as floating-point division, and it introduces round-off errors at every iteration, and they accumulate in a recursive formula such as your solution #2 uses.
Re your method #3 (logs): it's buggy for large numbers (at least in C, and probably other languages as well), because floating-point data types tend not to be as precise as 64-bit integers.
Hence I dislike all 3 of those methods: #1 works but is slow, #2 is broken, and #3 is buggy for large numbers. Instead, I prefer this, which works for numbers from 0 up to about 18.44 quintillion:
unsigned NumberOfDigits (uint64_t Number, unsigned Base)
{
unsigned Digits = 1;
uint64_t Power = 1;
while ( Number / Power >= Base )
{
++Digits;
Power *= Base;
}
return Digits;
}
Keep it simple:
long long int a = 223452355415634664;
int x;
for (x = 1; a >= 10; x++)
{
a = a / 10;
}
printf("%d", x);
You can use a recursive solution instead of a loop, but somehow similar:
#tailrec
def digits (i: Long, carry: Int=1) : Int = if (i < 10) carry else digits (i/10, carry+1)
digits (8345012978643L)
With longs, the picture might change - measure small and long numbers independently against different algorithms, and pick the appropriate one, depending on your typical input. :)
Of course nothing beats a switch:
switch (x) {
case 0: case 1: case 2: case 3: case 4: case 5: case 6: case 7: case 8: case 9: return 1;
case 10: case 11: // ...
case 99: return 2;
case 100: // you get the point :)
default: return 10; // switch only over int
}
except a plain-o-array:
int [] size = {1,1,1,1,1,1,1,1,1,2,2,2,2,2,... };
int x = 234561798;
return size [x];
Some people will tell you to optimize the code-size, but yaknow, premature optimization ...
log(x,n)-mod(log(x,n),1)+1
Where x is a the base and n is the number.
Here is the measurement in Swift 4.
Algorithms code:
extension Int {
var numberOfDigits0: Int {
var currentNumber = self
var n = 1
if (currentNumber >= 100000000) {
n += 8
currentNumber /= 100000000
}
if (currentNumber >= 10000) {
n += 4
currentNumber /= 10000
}
if (currentNumber >= 100) {
n += 2
currentNumber /= 100
}
if (currentNumber >= 10) {
n += 1
}
return n
}
var numberOfDigits1: Int {
return String(self).count
}
var numberOfDigits2: Int {
var n = 1
var currentNumber = self
while currentNumber > 9 {
n += 1
currentNumber /= 10
}
return n
}
}
Measurement code:
var timeInterval0 = Date()
for i in 0...10000 {
i.numberOfDigits0
}
print("timeInterval0: \(Date().timeIntervalSince(timeInterval0))")
var timeInterval1 = Date()
for i in 0...10000 {
i.numberOfDigits1
}
print("timeInterval1: \(Date().timeIntervalSince(timeInterval1))")
var timeInterval2 = Date()
for i in 0...10000 {
i.numberOfDigits2
}
print("timeInterval2: \(Date().timeIntervalSince(timeInterval2))")
Output
timeInterval0: 1.92149806022644
timeInterval1: 0.557608008384705
timeInterval2: 2.83262193202972
On this measurement basis String conversion is the best option for the Swift language.
I was curious after seeing #daniel.sedlacek results so I did some testing using Swift for numbers having more than 10 digits. I ran the following script in the playground.
let base = [Double(100090000000), Double(100050000), Double(100050000), Double(100000200)]
var rar = [Double]()
for i in 1...10 {
for d in base {
let v = d*Double(arc4random_uniform(UInt32(1000000000)))
rar.append(v*Double(arc4random_uniform(UInt32(1000000000))))
rar.append(Double(1)*pow(1,Double(i)))
}
}
print(rar)
var timeInterval = NSDate().timeIntervalSince1970
for d in rar {
floor(log10(d))
}
var newTimeInterval = NSDate().timeIntervalSince1970
print(newTimeInterval-timeInterval)
timeInterval = NSDate().timeIntervalSince1970
for d in rar {
var c = d
while c > 10 {
c = c/10
}
}
newTimeInterval = NSDate().timeIntervalSince1970
print(newTimeInterval-timeInterval)
Results of 80 elements
0.105069875717163 for floor(log10(x))
0.867973804473877 for div 10 iterations
Adding one more approach to many of the already mentioned approaches.
The idea is to use binarySearch on an array containing the range of integers based on the digits of the int data type.
The signature of Java Arrays class binarySearch is :
binarySearch(dataType[] array, dataType key) which returns the index of the search key, if it is contained in the array; otherwise, (-(insertion point) – 1).
The insertion point is defined as the point at which the key would be inserted into the array.
Below is the implementation:
static int [] digits = {9,99,999,9999,99999,999999,9999999,99999999,999999999,Integer.MAX_VALUE};
static int digitsCounter(int N)
{
int digitCount = Arrays.binarySearch(digits , N<0 ? -N:N);
return 1 + (digitCount < 0 ? ~digitCount : digitCount);
}
Please note that the above approach only works for : Integer.MIN_VALUE <= N <= Integer.MAX_VALUE, but can be easily extended for Long data type by adding more values to the digits array.
For example,
I) for N = 555, digitCount = Arrays.binarySearch(digits , 555) returns -3 (-(2)-1) as it's not present in the array but is supposed to be inserted at point 2 between 9 & 99 like [9, 55, 99].
As the index we got is negative we need to take the bitwise compliment of the result.
At last, we need to add 1 to the result to get the actual number of digits in the number N.
In Swift 5.x, you get the number of digit in integer as below :
Convert to string and then count number of character in string
let nums = [1, 7892, 78, 92, 90]
for i in nums {
let ch = String(describing: i)
print(ch.count)
}
Calculating the number of digits in integer using loop
var digitCount = 0
for i in nums {
var tmp = i
while tmp >= 1 {
tmp /= 10
digitCount += 1
}
print(digitCount)
}
let numDigits num =
let num = abs(num)
let rec numDigitsInner num =
match num with
| num when num < 10 -> 1
| _ -> 1 + numDigitsInner (num / 10)
numDigitsInner num
F# Version, without casting to a string.
I need help solving problem N from this earlier competition:
Problem N: Digit Sums
Given 3 positive integers A, B and C,
find how many positive integers less
than or equal to A, when expressed in
base B, have digits which sum to C.
Input will consist of a series of
lines, each containing three integers,
A, B and C, 2 ≤ B ≤ 100, 1 ≤ A, C ≤
1,000,000,000. The numbers A, B and C
are given in base 10 and are separated
by one or more blanks. The input is
terminated by a line containing three
zeros.
Output will be the number of numbers,
for each input line (it must be given
in base 10).
Sample input
100 10 9
100 10 1
750000 2 2
1000000000 10 40
100000000 100 200
0 0 0
Sample output
10
3
189
45433800
666303
The relevant rules:
Read all input from the keyboard, i.e. use stdin, System.in, cin or equivalent. Input will be redirected from a file to form the input to your submission.
Write all output to the screen, i.e. use stdout, System.out, cout or equivalent. Do not write to stderr. Do NOT use, or even include, any module that allows direct manipulation of the screen, such as conio, Crt or anything similar. Output from your program is redirected to a file for later checking. Use of direct I/O means that such output is not redirected and hence cannot be checked. This could mean that a correct program is rejected!
Unless otherwise stated, all integers in the input will fit into a standard 32-bit computer word. Adjacent integers on a line will be separated by one or more spaces.
Of course, it's fair to say that I should learn more before trying to solve this, but i'd really appreciate it if someone here told me how it's done.
Thanks in advance, John.
Other people pointed out trivial solution: iterate over all numbers from 1 to A. But this problem, actually, can be solved in nearly constant time: O(length of A), which is O(log(A)).
Code provided is for base 10. Adapting it for arbitrary base is trivial.
To reach above estimate for time, you need to add memorization to recursion. Let me know if you have questions about that part.
Now, recursive function itself. Written in Java, but everything should work in C#/C++ without any changes. It's big, but mostly because of comments where I try to clarify algorithm.
// returns amount of numbers strictly less than 'num' with sum of digits 'sum'
// pay attention to word 'strictly'
int count(int num, int sum) {
// no numbers with negative sum of digits
if (sum < 0) {
return 0;
}
int result = 0;
// imagine, 'num' == 1234
// let's check numbers 1233, 1232, 1231, 1230 manually
while (num % 10 > 0) {
--num;
// check if current number is good
if (sumOfDigits(num) == sum) {
// one more result
++result;
}
}
if (num == 0) {
// zero reached, no more numbers to check
return result;
}
num /= 10;
// Using example above (1234), now we're left with numbers
// strictly less than 1230 to check (1..1229)
// It means, any number less than 123 with arbitrary digit appended to the right
// E.g., if this digit in the right (last digit) is 3,
// then sum of the other digits must be "sum - 3"
// and we need to add to result 'count(123, sum - 3)'
// let's iterate over all possible values of last digit
for (int digit = 0; digit < 10; ++digit) {
result += count(num, sum - digit);
}
return result;
}
Helper function
// returns sum of digits, plain and simple
int sumOfDigits(int x) {
int result = 0;
while (x > 0) {
result += x % 10;
x /= 10;
}
return result;
}
Now, let's write a little tester
int A = 12345;
int C = 13;
// recursive solution
System.out.println(count(A + 1, C));
// brute-force solution
int total = 0;
for (int i = 1; i <= A; ++i) {
if (sumOfDigits(i) == C) {
++total;
}
}
System.out.println(total);
You can write more comprehensive tester checking all values of A, but overall solution seems to be correct. (I tried several random A's and C's.)
Don't forget, you can't test solution for A == 1000000000 without memorization: it'll run too long. But with memorization, you can test it even for A == 10^1000.
edit
Just to prove a concept, poor man's memorization. (in Java, in other languages hashtables are declared differently) But if you want to learn something, it might be better to try to do it yourself.
// hold values here
private Map<String, Integer> mem;
int count(int num, int sum) {
// no numbers with negative sum of digits
if (sum < 0) {
return 0;
}
String key = num + " " + sum;
if (mem.containsKey(key)) {
return mem.get(key);
}
// ...
// continue as above...
// ...
mem.put(key, result);
return result;
}
Here's the same memoized recursive solution that Rybak posted, but with a simpler implementation, in my humble opinion:
HashMap<String, Integer> cache = new HashMap<String, Integer>();
int count(int bound, int base, int sum) {
// No negative digit sums.
if (sum < 0)
return 0;
// Handle one digit case.
if (bound < base)
return (sum <= bound) ? 1 : 0;
String key = bound + " " + sum;
if (cache.containsKey(key))
return cache.get(key);
int count = 0;
for (int digit = 0; digit < base; digit++)
count += count((bound - digit) / base, base, sum - digit);
cache.put(key, count);
return count;
}
This is not the complete solution (no input parsing). To get the number in base B, repeatedly take the modulo B, and then divide by B until the result is 0. This effectively computes the base-B digit from the right, and then shifts the number right.
int A,B,C; // from input
for (int x=1; x<A; x++)
{
int sumDigits = 0;
int v = x;
while (v!=0) {
sumDigits += (v % B);
v /= B;
}
if (sumDigits==C)
cout << x;
}
This is a brute force approach. It may be possible to compute this quicker by determining which sets of base B digits add up to C, arranging these in all permutations that are less than A, and then working backwards from that to create the original number.
Yum.
Try this:
int number, digitSum, resultCounter = 0;
for(int i=1; i<=A, i++)
{
number = i; //to avoid screwing up our counter
digitSum = 0;
while(number > 1)
{
//this is the next "digit" of the number as it would be in base B;
//works with any base including 10.
digitSum += (number % B);
//remove this digit from the number, square the base, rinse, repeat
number /= B;
}
digitSum += number;
//Does the sum match?
if(digitSum == C)
resultCounter++;
}
That's your basic algorithm for one line. Now you wrap this in another For loop for each input line you received, preceded by the input collection phase itself. This process can be simplified, but I don't feel like coding your entire answer to see if my algorithm works, and this looks right whereas the simpler tricks are harder to pass by inspection.
The way this works is by modulo dividing by powers of the base. Simple example, 1234 in base 10:
1234 % 10 = 4
1234 / 10 = 123 //integer division truncates any fraction
123 % 10 = 3 //sum is 7
123 / 10 = 12
12 % 10 = 2 //sum is 9
12 / 10 = 1 //end condition, add this and the sum is 10
A harder example to figure out by inspection would be the same number in base 12:
1234 % 12 = 10 //you can call it "A" like in hex, but we need a sum anyway
1234 / 12 = 102
102 % 12 = 6 // sum 16
102/12 = 8
8 % 12 = 8 //sum 24
8 / 12 = 0 //end condition, sum still 24.
So 1234 in base 12 would be written 86A. Check the math:
8*12^2 + 6*12 + 10 = 1152 + 72 + 10 = 1234
Have fun wrapping the rest of the code around this.
Is there any nice algorithm to find the nearest prime number to a given real number? I only need to search within the first 100 primes or so.
At present, I've a bunch of prime numbers stored in an array and I'm checking the difference one number at a time (O(n)?).
Rather than a sorted list of primes, given the relatively small range targetted, have an array indexed by all the odd numbers in the range (you know there are no even primes except the special case of 2) and containing the closest prime. Finding the solution becomes O(1) time-wise.
I think the 100th prime is circa 541. an array of 270 [small] ints is all that is needed.
This approach is particularly valid, given the relative high density of primes (in particular relative to odd numbers), in the range below 1,000. (As this affects the size of a binary tree)
If you only need to search in the first 100 primes or so, just create a sorted table of those primes, and do a binary search. This will either get you to one prime number, or a spot between two, and you check which of those is closer.
Edit: Given the distribution of primes in that range, you could probably speed things up (a tiny bit) by using an interpolation search -- instead of always starting at the middle of the table, use linear interpolation to guess at a more accurate starting point. The 100th prime number should be somewhere around 250 or so (at a guess -- I haven't checked), so if (for example) you wanted the one closest to 50, you'd start about 1/5th of the way into the array instead of halfway. You can pretty much treat the primes as starting at 1, so just divide the number you want by the largest in your range to get a guess at the starting point.
Answers so far are rather complicated, given the task in hand. The first hundred primes are all less then 600. I would create an array of size 600 and place in each the value of the nearest prime to that number. Then, given a number to test, I would round it both up and down using the floor and ceil functions to get one or two candidate answers. A simple comparison with the distances to these numbers will give you a very fast answer.
The simplest approach would be to store the primes in a sorted list and modify your algorithm to do a binary search.
The standard binary search algorithm would return null for a miss, but it should be straight-forward to modify it for your purposes.
The fastest algorithm? Create a lookup table with p[100]=541 elements and return the result for floor(x), with special logic for x on [2,3]. That would be O(1).
You should sort your number in array then you can use binary search. This algorithm is O(log n) performance in worst case.
public static boolean p(int n){
for(int i=3;i*i<=n;i+=2) {
if(n%i==0)
return false;
}
return n%2==0? false: true; }
public static void main(String args[]){
String n="0";
int x = Integer.parseInt(n);
int z=x;
int a=0;
int i=1;
while(!p(x)){
a = i*(int)Math.pow(-1, i);
i++;
x+=a;
}
System.out.println( (int) Math.abs(x-z));}
this is for n>=2.
In python:
>>> def nearest_prime(n):
incr = -1
multiplier = -1
count = 1
while True:
if prime(n):
return n
else:
n = n + incr
multiplier = multiplier * -1
count = count + 1
incr = multiplier * count
>>> nearest_prime(3)
3
>>> nearest_prime(4)
3
>>> nearest_prime(5)
5
>>> nearest_prime(6)
5
>>> nearest_prime(7)
7
>>> nearest_prime(8)
7
>>> nearest_prime(9)
7
>>> nearest_prime(10)
11
<?php
$N1Diff = null;
$N2Diff = null;
$n1 = null;
$n2 = null;
$number = 16;
function isPrime($x) {
for ($i = 2; $i < $x; $i++) {
if ($x % $i == 0) {
return false;
}
}
return true;
}
for ($j = $number; ; $j--) {
if( isPrime($j) ){
$N1Diff = abs($number - $j);
$n1 = $j;
break;
}
}
for ($j = $number; ; $j++) {
if( isPrime($j) ){
$N2Diff = abs($number - $j);
$n2 = $j;
break;
}
}
if($N1Diff < $N2Diff) {
echo $n1;
} else if ($N1Diff2 < $N1Diff ){
echo $n2;
}
If you want to write an algorithm, A Wikipedia search for prime number led me to another article on the Sieve of Eratosthenes. The algorithm looks a bit simple and I'm thinking a recursive function would suit it well imo. (I could be wrong about that.)
If the array solution isn't a valid solution for you (it is the best one for your scenario), you can try the code below. After the "2 or 3" case, it will check every odd number away from the starting value until it finds a prime.
static int NearestPrime(double original)
{
int above = (int)Math.Ceiling(original);
int below = (int)Math.Floor(original);
if (above <= 2)
{
return 2;
}
if (below == 2)
{
return (original - 2 < 0.5) ? 2 : 3;
}
if (below % 2 == 0) below -= 1;
if (above % 2 == 0) above += 1;
double diffBelow = double.MaxValue, diffAbove = double.MaxValue;
for (; ; above += 2, below -= 2)
{
if (IsPrime(below))
{
diffBelow = original - below;
}
if (IsPrime(above))
{
diffAbove = above - original;
}
if (diffAbove != double.MaxValue || diffBelow != double.MaxValue)
{
break;
}
}
//edit to your liking for midpoint cases (4.0, 6.0, 9.0, etc)
return (int) (diffAbove < diffBelow ? above : below);
}
static bool IsPrime(int p) //intentionally incomplete due to checks in NearestPrime
{
for (int i = 3; i < Math.Sqrt(p); i += 2)
{
if (p % i == 0)
return false;
}
return true;
}
Lookup table whit size of 100 bytes; (unsigned chars)
Round real number and use lookup table.
Maybe we can find the left and right nearest prime numbers, and then compare to get the nearest one. (I've assumed that the next prime number shows up within next 10 occurrences)
def leftnearestprimeno(n):
n1 = n-1
while(n1 >= 0):
if isprime(n1):
return n1
else:
n1 -= 1
return -1
def rightnearestprimeno(n):
n1 = n+1
while(n1 < (n+10)):
if isprime(n1):
return n1
else:
n1 += 1
return -1
n = int(input())
a = leftnearestprimeno(n)
b = rightnearestprimeno(n)
if (n - a) < (b - n):
print("nearest: ", a)
elif (n - a) > (b - n):
print("nearest: ", b)
else:
print("nearest: ", a) #in case the difference is equal, choose min
#value
Simplest answer-
Every prime number can be represented in the form (6*x-1 and 6*X +1) (except 2 and 3).
let number is N.divide it with 6.
t=N/6;
now
a=(t-1)*6
b=(t+1)*6
and check which one is closer to N.
I have been trying to work my way through Project Euler, and have noticed a handful of problems ask for you to determine a prime number as part of it.
I know I can just divide x by 2, 3, 4, 5, ..., square root of X and if I get to the square root, I can (safely) assume that the number is prime. Unfortunately this solution seems quite klunky.
I've looked into better algorithms on how to determine if a number is prime, but get confused fast.
Is there a simple algorithm that can determine if X is prime, and not confuse a mere mortal programmer?
Thanks much!
The first algorithm is quite good and used a lot on Project Euler. If you know the maximum number that you want you can also research Eratosthenes's sieve.
If you maintain the list of primes you can also refine the first algo to divide only with primes until the square root of the number.
With these two algoritms (dividing and the sieve) you should be able to solve the problems.
Edit: fixed name as noted in comments
To generate all prime numbers less than a limit Sieve of Eratosthenes (the page contains variants in 20 programming languages) is the oldest and the simplest solution.
In Python:
def iprimes_upto(limit):
is_prime = [True] * limit
for n in range(2, limit):
if is_prime[n]:
yield n
for i in range(n*n, limit, n): # start at ``n`` squared
is_prime[i] = False
Example:
>>> list(iprimes_upto(15))
[2, 3, 5, 7, 11, 13]
I see that Fermat's primality test has already been suggested, but I've been working through Structure and Interpretation of Computer Programs, and they also give the Miller-Rabin test (see Section 1.2.6, problem 1.28) as another alternative. I've been using it with success for the Euler problems.
Here's a simple optimization of your method that isn't quite the Sieve of Eratosthenes but is very easy to implement: first try dividing X by 2 and 3, then loop over j=1..sqrt(X)/6, trying to divide by 6*j-1 and 6*j+1. This automatically skips over all numbers divisible by 2 or 3, gaining you a pretty nice constant factor acceleration.
Keeping in mind the following facts (from MathsChallenge.net):
All primes except 2 are odd.
All primes greater than 3 can be written in the form 6k - 1 or 6k + 1.
You don't need to check past the square root of n
Here's the C++ function I use for relatively small n:
bool isPrime(unsigned long n)
{
if (n == 1) return false; // 1 is not prime
if (n < 4) return true; // 2 and 3 are both prime
if ((n % 2) == 0) return false; // exclude even numbers
if (n < 9) return true; //we have already excluded 4, 6, and 8.
if ((n % 3) == 0) return false; // exclude remaining multiples of 3
unsigned long r = floor( sqrt(n) );
unsigned long f = 5;
while (f <= r)
{
if ((n % f) == 0) return false;
if ((n % (f + 2)) == 0) return false;
f = f + 6;
}
return true; // (in all other cases)
}
You could probably think of more optimizations of your own.
I'd recommend Fermat's primality test. It is a probabilistic test, but it is correct surprisingly often. And it is incredibly fast when compared with the sieve.
For reasonably small numbers, x%n for up to sqrt(x) is awfully fast and easy to code.
Simple improvements:
test 2 and odd numbers only.
test 2, 3, and multiples of 6 + or - 1 (all primes other than 2 or 3 are multiples of 6 +/- 1, so you're essentially just skipping all even numbers and all multiples of 3
test only prime numbers (requires calculating or storing all primes up to sqrt(x))
You can use the sieve method to quickly generate a list of all primes up to some arbitrary limit, but it tends to be memory intensive. You can use the multiples of 6 trick to reduce memory usage down to 1/3 of a bit per number.
I wrote a simple prime class (C#) that uses two bitfields for multiples of 6+1 and multiples of 6-1, then does a simple lookup... and if the number i'm testing is outside the bounds of the sieve, then it falls back on testing by 2, 3, and multiples of 6 +/- 1. I found that generating a large sieve actually takes more time than calculating primes on the fly for most of the euler problems i've solved so far. KISS principle strikes again!
I wrote a prime class that uses a sieve to pre-calculate smaller primes, then relies on testing by 2, 3, and multiples of six +/- 1 for ones outside the range of the sieve.
For Project Euler, having a list of primes is really essential. I would suggest maintaining a list that you use for each problem.
I think what you're looking for is the Sieve of Eratosthenes.
Your right the simples is the slowest. You can optimize it somewhat.
Look into using modulus instead of square roots.
Keep track of your primes. you only need to divide 7 by 2, 3, and 5 since 6 is a multiple of 2 and 3, and 4 is a multiple of 2.
Rslite mentioned the eranthenos sieve. It is fairly straight forward. I have it in several languages it home. Add a comment if you want me to post that code later.
Here is my C++ one. It has plenty of room to improve, but it is fast compared to the dynamic languages versions.
// Author: James J. Carman
// Project: Sieve of Eratosthenes
// Description: I take an array of 2 ... max values. Instead of removeing the non prime numbers,
// I mark them as 0, and ignoring them.
// More info: http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
#include <iostream>
int main(void) {
// using unsigned short.
// maximum value is around 65000
const unsigned short max = 50000;
unsigned short x[max];
for(unsigned short i = 0; i < max; i++)
x[i] = i + 2;
for(unsigned short outer = 0; outer < max; outer++) {
if( x[outer] == 0)
continue;
unsigned short item = x[outer];
for(unsigned short multiplier = 2; (multiplier * item) < x[max - 1]; multiplier++) {
unsigned int searchvalue = item * multiplier;
unsigned int maxValue = max + 1;
for( unsigned short maxIndex = max - 1; maxIndex > 0; maxIndex--) {
if(x[maxIndex] != 0) {
maxValue = x[maxIndex];
break;
}
}
for(unsigned short searchindex = multiplier; searchindex < max; searchindex++) {
if( searchvalue > maxValue )
break;
if( x[searchindex] == searchvalue ) {
x[searchindex] = 0;
break;
}
}
}
}
for(unsigned short printindex = 0; printindex < max; printindex++) {
if(x[printindex] != 0)
std::cout << x[printindex] << "\t";
}
return 0;
}
I will throw up the Perl and python code I have as well as soon as I find it. They are similar in style, just less lines.
Here is a simple primality test in D (Digital Mars):
/**
* to compile:
* $ dmd -run prime_trial.d
* to optimize:
* $ dmd -O -inline -release prime_trial.d
*/
module prime_trial;
import std.conv : to;
import std.stdio : w = writeln;
/// Adapted from: http://www.devx.com/vb2themax/Tip/19051
bool
isprime(Integer)(in Integer number)
{
/* manually test 1, 2, 3 and multiples of 2 and 3 */
if (number == 2 || number == 3)
return true;
else if (number < 2 || number % 2 == 0 || number % 3 == 0)
return false;
/* we can now avoid to consider multiples
* of 2 and 3. This can be done really simply
* by starting at 5 and incrementing by 2 and 4
* alternatively, that is:
* 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, ...
* we don't need to go higher than the square root of the number */
for (Integer divisor = 5, increment = 2; divisor*divisor <= number;
divisor += increment, increment = 6 - increment)
if (number % divisor == 0)
return false;
return true; // if we get here, the number is prime
}
/// print all prime numbers less then a given limit
void main(char[][] args)
{
const limit = (args.length == 2) ? to!(uint)(args[1]) : 100;
for (uint i = 0; i < limit; ++i)
if (isprime(i))
w(i);
}
I am working thru the Project Euler problems as well and in fact just finished #3 (by id) which is the search for the highest prime factor of a composite number (the number in the ? is 600851475143).
I looked at all of the info on primes (the sieve techniques already mentioned here), and on integer factorization on wikipedia and came up with a brute force trial division algorithm that I decided would do.
So as I am doing the euler problems to learn ruby I was looking into coding my algorithm and stumbled across the mathn library which has a Prime class and an Integer class with a prime_division method. how cool is that. i was able to get the correct answer to the problem with this ruby snippet:
require "mathn.rb"
puts 600851475143.prime_division.last.first
this snippet outputs the correct answer to the console. of course i ended up doing a ton of reading and learning before i stumbled upon this little beauty, i just thought i would share it with everyone...
I like this python code.
def primes(limit) :
limit += 1
x = range(limit)
for i in xrange(2,limit) :
if x[i] == i:
x[i] = 1
for j in xrange(i*i, limit, i) :
x[j] = i
return [j for j in xrange(2, limit) if x[j] == 1]
A variant of this can be used to generate the factors of a number.
def factors(limit) :
limit += 1
x = range(limit)
for i in xrange(2,limit) :
if x[i] == i:
x[i] = 1
for j in xrange(i*i, limit, i) :
x[j] = i
result = []
y = limit-1
while x[y] != 1 :
divisor = x[y]
result.append(divisor)
y /= divisor
result.append(y)
return result
Of course, if I were factoring a batch of numbers, I would not recalculate the cache; I'd do it once and do lookups in it.
Is not optimized but it's a very simple function.
function isprime(number){
if (number == 1)
return false;
var times = 0;
for (var i = 1; i <= number; i++){
if(number % i == 0){
times ++;
}
}
if (times > 2){
return false;
}
return true;
}
Maybe this implementation in Java can be helpful:
public class SieveOfEratosthenes {
/**
* Calling this method with argument 7 will return: true true false false true false true false
* which must be interpreted as : 0 is NOT prime, 1 is NOT prime, 2 IS prime, 3 IS prime, 4 is NOT prime
* 5 is prime, 6 is NOT prime, 7 is prime.
* Caller may either revert the array for easier reading, count the number of primes or extract the prime values
* by looping.
* #param upTo Find prime numbers up to this value. Must be a positive integer.
* #return a boolean array where index represents the integer value and value at index returns
* if the number is NOT prime or not.
*/
public static boolean[] isIndexNotPrime(int upTo) {
if (upTo < 2) {
return new boolean[0];
}
// 0-index array, upper limit must be upTo + 1
final boolean[] isIndexNotPrime = new boolean[upTo + 1];
isIndexNotPrime[0] = true; // 0 is not a prime number.
isIndexNotPrime[1] = true; // 1 is not a prime number.
// Find all non primes starting from 2 by finding 2 * 2, 2 * 3, 2 * 4 until 2 * multiplier > isIndexNotPrime.len
// Find next by 3 * 3 (since 2 * 3 was found before), 3 * 4, 3 * 5 until 3 * multiplier > isIndexNotPrime.len
// Move to 4, since isIndexNotPrime[4] is already True (not prime) no need to loop..
// Move to 5, 5 * 5, (2 * 5 and 3 * 5 was already set to True..) until 5 * multiplier > isIndexNotPrime.len
// Repeat process until i * i > isIndexNotPrime.len.
// Assume we are looking up to 100. Break once you reach 11 since 11 * 11 == 121 and we are not interested in
// primes above 121..
for (int i = 2; i < isIndexNotPrime.length; i++) {
if (i * i >= isIndexNotPrime.length) {
break;
}
if (isIndexNotPrime[i]) {
continue;
}
int multiplier = i;
while (i * multiplier < isIndexNotPrime.length) {
isIndexNotPrime[i * multiplier] = true;
multiplier++;
}
}
return isIndexNotPrime;
}
public static void main(String[] args) {
final boolean[] indexNotPrime = SieveOfEratosthenes.isIndexNotPrime(7);
assert !indexNotPrime[2]; // Not (not prime)
assert !indexNotPrime[3]; // Not (not prime)
assert indexNotPrime[4]; // (not prime)
assert !indexNotPrime[5]; // Not (not prime)
assert indexNotPrime[6]; // (not prime)
assert !indexNotPrime[7]; // Not (not prime)
}
}
The AKS prime testing algorithm:
Input: Integer n > 1
if (n is has the form ab with b > 1) then output COMPOSITE
r := 2
while (r < n) {
if (gcd(n,r) is not 1) then output COMPOSITE
if (r is prime greater than 2) then {
let q be the largest factor of r-1
if (q > 4sqrt(r)log n) and (n(r-1)/q is not 1 (mod r)) then break
}
r := r+1
}
for a = 1 to 2sqrt(r)log n {
if ( (x-a)n is not (xn-a) (mod xr-1,n) ) then output COMPOSITE
}
output PRIME;
another way in python is:
import math
def main():
count = 1
while True:
isprime = True
for x in range(2, int(math.sqrt(count) + 1)):
if count % x == 0:
isprime = False
break
if isprime:
print count
count += 2
if __name__ == '__main__':
main()