Loop not faster despite looping through half as many iterations? - c++

I wrote a program that searches for primes:
#include <iostream>
#include <fstream>
#include <chrono>
typedef std::chrono::high_resolution_clock Clock;
using namespace std;
int main() {
int p;
int x = 1;
int b;
int a[1000000];
bool n = false;
a[0] = 2;
a[1] = 3;
auto t1 = Clock::now();
ofstream outfile;
outfile.open("p.txt");
for (p = 3; p < 7500000; p = p + 2)
{
for (b = 0; b <= x && n == 0; b++)
{
if (p % a[b / 2] == 0)
{
n = true;
}
}
if (n == false)
{
cout << p << endl;
outfile << p << endl;
x++;
a[x] = p;
}
else
{
n = false;
}
}
auto t2 = Clock::now();
std::cout
<< std::chrono::duration_cast<std::chrono::nanoseconds>(t2 - t1).count()
<< " nanoseconds" << std::endl;
outfile.close();
}
Initially for the loop increment I had p++, but I changed that to p=p+2 because all primes essentially are odd and even numbers don't need to be checked. Problem is when I benchmarked this, there was no difference in speed between the old and new code. So what is the bottleneck in the process, if checking all the numbers is no different than checking half? And is there a better way to approach this?

Your outer loop skips half the numbers. But your inner loop tests every number twice. So you give up all your gains.
If you don't see that your inner loop does everything twice, consider that a[b/2] is the same when b is 1 as it is when b is 0.

It's this line:
for(b=0; b<=x && n==0; b++)
Once n=true; executes, the b loop exits because of the && n==0 condition. This happens with the very first test: every even number is divisible by two, which is a[0]. So for even numbers (that you include if you use p++ instead of p=p+2) the inner loop is very quick, much quicker than for a typical odd number. This explains why including them makes so little difference.

Related

How can I find prime reversed numbers?

I have to write a program to check if the entered number has these qualifications:
A number that is prime it self, the reverse of that number is also prime, and the number's digits are prime numbers too (Like this number: 7523).
If the needs meet, it has to show "yes" when you enter and run the program otherwise "no".
I know both codes for prime and reverse numbers but I don't know how to merge them.
This is the code:
#include <iostream>
#include <conio.h>
using namespace std;
void prime_check(int x) {
int a, i, flag = 1;
cin >> a;
for (i = 2; i <= a / 2 && flag == 1; i++) {
if (a % i == 0)
flag = 0;
}
if (flag == 1)
cout << "prime";
else
break;
}
int main() {
int a, r, sum = 0;
cin >> a;
while (a != 0) {
r = a % 10;
sum = (sum * 10) + r;
a = a / 10;
}
}
The program has to check each digit of the number entered to see if it is prime or not in every step, then show "yes", but it doesn't work.
Welcome to the site.
I don't know how to merge them.
void prime_check(int n) { /*code*/ }
I'd understand that you don't know how to use this.
It's very easy!
int main()
{
int i = 0;
prime_check(i);
}
If you are confused about how the program executes, you could use a debugger to see where it goes. But since using a debugger can be a bit hard at first, I would suggest to add debug prints to see how the program executes.
This line of code prints the file and line number automatically.
std::cout << __FILE__ << ":" << __LINE__ << "\n";
I'd suggest to add it at the start of every function you wish to understand.
One step further is to make it into a macro, just so that it's easy to use.
#define DEBUGPRINT std::cout << __FILE__ << ":" << __LINE__ << "\n";
Check a working example here:
http://www.cpp.sh/2hpam
Note that it says <stdin>::14 instead of the filename because it's running on a webpage.
I have done some changes to your code, and added comments everywhere I've made changes. Check it out:
#include <iostream>
#include <conio.h>
using namespace std;
bool prime_check(int x) { // I have changed the datatype of this function to bool, because I want to store if all the digits are prime or not
int i, flag = 1; // Removed the variable a, because the function is already taking x as input
for (i = 2; i <= x / 2 && flag == 1; i++) {
if (x % i == 0)
flag = 0;
}
return flag == 1;
}
int main() {
int a, r, sum = 0, original; // added original variable, to store the number added
bool eachDigit = true; // added to keep track of each digit
cin >> a;
original = a;
while (a != 0) {
r = a % 10;
eachDigit = prime_check(r); // Here Each digit of entered number is checked for prime
sum = (sum * 10) + r;
a = a / 10;
}
if (eachDigit && prime_check(original) && prime_check(sum)) // At the end checking if all the digits, entered number and the revered number are prime
cout << "yes";
else
cout<< "no";
}
For optimization, you can check if the entered number is prime or not before starting that loop, and also you can break the loop right away if one of the digits of the entered number is not prime, Like this:
#include <iostream>
#include <conio.h>
using namespace std;
bool prime_check(int x) { // I have changed the datatype of this function to bool, because I want to store if all the digits are prime or not
int i, flag = 1; // Removed the variable a, because the function is already taking x as input
for (i = 2; i <= x / 2 && flag == 1; i++) {
if (x % i == 0)
flag = 0;
}
return flag == 1;
}
int main() {
int a, r, sum = 0;
bool eachDigit = true, entered; // added to keep track of each digit
cin >> a;
entered = prime_check(a);
while (a != 0 && entered && eachDigit) {
r = a % 10;
eachDigit = prime_check(r); // Here Each digit of entered number is checked for prime
sum = (sum * 10) + r;
a = a / 10;
}
if (eachDigit && entered && prime_check(sum)) // At the end checking if all the digits, entered number and the revered number are prime
cout << "yes";
else
cout<< "no";
}
Suppose you have an int variable num which you want to check for your conditions, you can achieve your target by the following:
int rev_num = 0;
bool flag = true; // Assuming 'num' satisfies your conditions, until proven otherwise
if (prime_check(num) == false) {
flag = false;
}
else while (num != 0) {
int digit = num % 10;
rev_num = rev_num * 10 + digit;
// Assuming your prime_check function returns 'true' and 'false'
if (prime_check(digit) == false) {
flag = false;
break;
}
num /= 10;
}
if (prime_check(rev_num) == false) {
flag = false;
}
if (flag) {
cout << "Number satisfies all conditions\n";
}
else {
cout << "Number does not satisfy all conditions\n";
}
The problem is that each of your functions is doing three things, 1) inputting the number, 2) testing the number and 3) outputting the result. To combine these functions you need to have two functions that are only testing the number. Then you can use both functions on the same number, instead of inputting two different numbers and printing two different results. You will need to use function parameters, to pass the input number to the two functions, and function return values to return the result of the test. The inputting of the number and the outputting of the result go in main. Here's an outline
// returns true if the number is a prime, false otherwise
bool prime_check(int a)
{
...
}
// returns true if the number is a reverse prime, false otherwise
bool reverse_prime_check(int a)
{
...
}
int main()
{
int a;
cin >> a;
if (prime_check(a) && reverse_prime_check(a))
cout << "prime\n";
else
cout << "not prime\n";
}
I'll leave you to write the functions themselves, and there's nothing here to do the digit checks either. I'll leave you do to that.

Possible infinite loop

I think my code has an infinite loop. Can someone tell me where I went wrong?
The code is supposed to find the number of valid numbers, with a valid number being a number without a digit repeating. For example, 1212 would be a non-valid number because 1 and 2 repeated.
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main()
{
int a; int b; int count_validNums = 1; int digit; int last_digit; bool is_valid = true;
vector <int> num_list;
cout << "Enter numbers 0 < a <= b < = 10000: ";
cin >> a >> b;
// Checks for invalid input
if (a < 0 || b < 0 || a > 10000 || b > 10000) {
cout << "Invalid input";
return 1;
}
// Checks every number from the range [a,b]
for (int i = a; i <= b; i++){
last_digit = i % 10;
num_list.push_back(last_digit);
i = i / 10;
while (i != 0){
digit = i % 10;
if (find(num_list.begin(), num_list.end(), digit) != num_list.end()){
is_valid = false;
}
num_list.push_back(digit);
i = i / 10;
}
if (is_valid) count_validNums++;
}
cout << "They are " << count_validNums << " valid numbers between" << a << " and " << b << endl;
}
The inner while loop terminates when i == 0. Then the outer for loop increments it (so i == 1), then the inner loop reduces it to zero again. Then the other loop increments it, then ...
What is happening to cause the infinite loop is that you are constantly reducing the int i back down to 0. Consider these highlights:
`for(int i = a; i <= b; i++){
//stuff
while(i != 0){ //<--this forces i down to 0
//more stuff
i = i / 10;
}
//final stuff
}`
i here is all one variable, so any changes you make to it anywhere will affect it everywhere else it exists! Instead, you can try saying something like int temp = i; and then perform your operations on temp so that i remains independent, but because your for-loop terminates when i <= b and you are constantly resetting i to 0, it will never reach b.
Also, I noticed that in your check for valid numbers you verify that 0 < a,b < 10000, but later in your for-loop you seem to make the assumption that a <= b will be true. Unfortunately, your test does not ensure this, so the for-loop will immediately terminate for inputs where b < a is true (which your program currently allows) and your program will report answers that are likely incorrect. The same is true when I enter letters as input instead of numbers. You might want to revisit that portion of code.

Value assignment into array c++

I'm trying to create a array of prime numbers done by calculation. As a project to learn coding. Ultimately to build my own math library so this is something I can add onto at a variety of levels as I learn to code c++.
The following is code that works great for printing prime numbers to the screen based on the search range, but my totalPrimes iterator is stuck at 1. So each time it places the last prime found in the PrimeNumbers[1] position.
Any advice would be awesome.
#include <iostream>
#include <array>
std::array<long, 10000000> PrimeNumbers={0};
void isPrime(long x);
int main() {
for (long i = 1; i < 10; i++) {
isPrime(i);
}
for(int h = 0; h < 10; h++) {
std::cout << "\nSecond Prime is : " << PrimeNumbers[h];
}
}
void isPrime(long x) {
int count(0), totalPrimes(0);
for (long a = 1; a < x; a++) {
if ((x % a) == 0) {
count += 1;
}
}
if (count == 1) {
++totalPrimes;
std::cout << '\n' << x << " is a Prime number";
PrimeNumbers[totalPrimes] = x;
}
}
You're initializing totalPrimes to 0 every time the function runs. You would need to have totalPrimes as a global variable, or better yet (because global variables can become problematic), set it equal to the first available member of PrimeNumbers before you do anything else in that function.
Keep track of a position along with your PrimeNumbers array.
size_t nLastPos=0;
...
for(size_t x = 0; 1000 > x; ++x)
{
if(isPrime(x))
{
PrimeNumbers[nLastPos++] = x;
}
}
for(size_t i = 0; nLastPos > n; ++n)
{/* print out number PrimeNumbers[n] */ }
It looks like you're having some trouble with variable scoping. The reason for your problem (as I noted in the comment) is that totalPrimes is local, so you're creating a new integer variable and setting it to 0 every time the function is called.
However, you've made PrimeNumbers global and are having the isPrime function modify it, which doesn't look like good practice.
All of this can be fixed with a little restructuring to make the code nicer:
#include <iostream>
#include <array>
bool isPrime(long x);
int main() {
std::array<long, 10000000> PrimeNumbers={0};
int totalPrimes = 0;
for (long i = 1; i < 10; i++) {
if (isPrime(i)) {
std::cout << '\n' << i << " is a Prime number";
PrimeNumbers[totalPrimes++] = i;
}
}
for(int h = 0; h < 10; h++) {
std::cout << h << " Prime is : " << PrimeNumbers[h] << std::endl;
}
}
bool isPrime(long x) {
int count(0);
for (long a = 1; a < x; a++) {
if ((x % a) == 0) {
count += 1;
}
}
return count == 1;
}
Your program can be re-structured little bit to make it easier to follow and debug.
Don't put things in isPrime other than the logic to decide whether a number is prime. Make sure it returns a bool. This will make the function a bit simpler and easier to debug.
Use the return value of isPrime in the calling function to perform other bookkeeping tasks.
The logic you have used to check whether a number is prime is incorrect. That needs to be fixed.
Here's an updated version of your posted code.
#include <iostream>
#include <array>
#include <cmath>
std::array<long, 10000000> PrimeNumbers={0};
bool isPrime(long x);
int main()
{
int totalPrimes = 0;
for (long i = 1; i < 10; i++)
{
if ( isPrime(i) )
{
std::cout << i << " is a Prime number" << std::endl;
PrimeNumbers[totalPrimes] = i;
++totalPrimes;
}
}
}
bool isPrime(long x) {
// 1, 2, and 3 are primes.
if ( x <= 3 )
{
return true;
}
// Even numbers are not primes.
if ( x % 2 == 0 )
{
return false;
}
// Check the rest.
long end = (long)std::sqrt(x);
for (long a = 3; a < end; a += 2) {
if ((x % a) == 0)
{
return false;
}
}
return true;
}
and its output:
1 is a Prime number
2 is a Prime number
3 is a Prime number
5 is a Prime number
7 is a Prime number
9 is a Prime number
Everybody is talking about how your totalPrimes variable is reset each time the function is called, and this is obviously true. You could return the value from the function and increment it from main, you could use global variables having the variable being defined outside of the function so that it's not reset each time inside the function or you could use
A static variable!
Take a look at this simple case. I have a function called up_two which increases the value of by two each time the function is called. The static variable int value has a memory of each time the function up_two() is called which increments value by two each time. If I were to use a just an integer it would always reset the value and have it be zero, which is what I initially defined it to be.
The advantage of using a static variable is that I can count how many times a function has been called, and I can keep my counter specific to a particular function.
#include <iostream>
using namespace std;
void up_two();
int main()
{
for(int i = 0; i < 10; i++)
{
up_two();
}
return 0;
}
void up_two()
{
static int value = 0;
cout << value << endl;
value += 2;
}
This program doesn't solve the particular problem that you want to solve, but if you figure out how the static variable is working, it should make your workflow easier.
The magic line here is this:
static int value = 0;
With it like this my program prints the following:
0
2
4
6
8
10
12
14
16
18
Without the static declaration, you just get 10 lines of zeroes
which is troublesome.
Hope that helps you optimize your program the way you want it to be.

What is the purpose of the increment statement?

Why are increment statements a thing in for-loops in C++? To me it seems redundant, because you could simply put the increments inside the conditional code. Am I misunderstanding something important here?
To illustrate my question better, I'm including some pseudocode:
What is the difference between ->
for( int a = 10; a < 20; a = a + 1 )
{
cout << a << endl;
}
and
for( int a = 10; a < 20;)
{
a = a + 1
cout << a << endl;
}
It's more than mere convenience sometimes.
These are equivalent:
for (int a = 10; a < 20; a = a + 1) {
cout << a << endl;
}
for (int a = 10; a < 20; ) {
cout << a << endl;
a = a + 1;
}
But, these are not:
// this works ...
for (int a = 10; a < 20; a = a + 1) {
if (blah ...)
continue;
cout << a << endl;
}
// this doesn't
for (int a = 10; a < 20; ) {
if (blah ...)
continue;
cout << a << endl;
a = a + 1;
}
Since you're coming from python, an idiomatic for loop is like a python range, but much more powerful. Your C for loop, expressed in python would be:
for a in range(10,20,1)
It's more idiomatic to express this as:
for (a = 10; a < 20; a += 1)
Because the loop increment is 1, it's even more idiomatic to do this:
for (a = 10; a < 20; ++a)
But, for loops are:
for ([init_stmt]; [test_stmt]; [incr_stmt])
Where any of the *_stmt can be compound:
for (x = 0, y = 0; x < 10; ++x, y += 2)
Convenience.
However, your equivalent code should be:
for (int a = 10; a < 20;)
{
cout << a << endl;
a = a + 1;
}
It runs at the end of the loop body.
[ snips grumbling about quality of now deleted/ edited answers ;-) ]
This:
for (unsigned counter = 1; counter <= 10; ++counter) {
doStuff();
}
is largely equivalent to this:
unsigned counter = 1;
while (counter <= 10) {
doStuff();
++counter;
}
with the notable exception that, in the 1st case, you have the considerable benefit that counter is scoped only to within the for block and automatically goes out-of-scope as soon as it finishes - whereas with the latter, counter must remain in-scope after the loop, where it's potentially useless or even an obstacle.
(tangential: Note that C did not support within-for declaration, or any non-top-of-block declarations, until C99 - but barring extenuating circumstances, anyone not using at least C99 by now is making a questionable choice imho.)
edit: Craig also makes a very good point regarding continue - an oft-forgotten but certainly useful statement. I'm sure there are probably other differences we could conjure up.
for this example:
using namespace std;
int main(int argc, char** argv) {
for( int a = 10; a < 20;)
{
a = a + 1;
cout << a << endl;
}
return 0;
}
the output will be from 11-->20
the first example will be from 10-->19
your are putting the increment part outside the loop and this possible, but notice that the value 10 will not appear, because you are increment before printing the value of a
so in the 2nd example your printing the value and then increment, and at the end of the loop, you are quiting the loop without reaching 20, because the condition get you out the loop
executing code block before increment is the key for you, the for loop increment just after the code block is executed
Well it is not required, it is just for convenience.
In your second code, you made a little mistake which would make the code nonequivalent to the the first one.
Your increment should be at the end of loop in order to be equivalent to the first code
so it should rather be:
for( int a = 10; a < 20;)
{
cout << a << endl;
a = a + 1; //increment at the end of instructions
}
These little errors and also errors like forgetting to include your increment is why it is convenient to include the increment in the for loop.
Or you can use the while loop instead:
while (condition)
{//instructions here;}

Can't print Fibonacci series

I was writing a small snippet to get a Fibonacci number sequence depending on the user input. If the user supplies 4 as an input, it should return him the first N members of the Fibonacci sequence.
#include <iostream>
using namespace std;
int main (){
int a = 0;
int b = 1;
int c;
int n = 3;
n -= 2;
if (n == 1){
cout << a << endl;
} else {
cout << a << b << endl;
for (int i=0;i<n;i++){
c = b + a;
cout << c << endl;
a = b;
b = c;
}
}
}
However, I end up getting a 0 as an output for whatever number I supply. I have this working in PHP and I kinda miss where I've blundered. I guess I don't actually render input and output properly.
int a =0;
int n = 3;
n -= 2;
if (n == 1){
cout << a << endl;
}
You have n equal to 3, you subtract 2, thus n equal to 1, so, you enter the if body and output a, which is zero.
[EDIT]
You don't seem to get any input -as stated in a comment- in your program (you could use std::cin or std::getline() for this), but you probably mean that you have the input hard-coded, by changing the value of n by hand.
You may want to check how the Fibonacci series program is expected to work:
Fib. at Rosseta page.
Fib. with recursion
Non-recursive Fib.
After reading the links I provided above, you should be able to see that your code should be changed to this:
#include <iostream>
using namespace std;
int main (){
int a = 1;
int b = 0;
int c;
int n = 10; // "input" is 10
if (n == 0 || n == 1) { // 0 and 1 case
cout << n << endl;
} else {
for (int i = 2; i <= n; ++i) { // here you want to reach n
c = a + b;
b = a;
a = c;
}
cout << c << endl;
}
return 0;
}
However, the code above outputs only the result. You should slightly modify it to get the terms of the sequence, but I'll leave you have some fun too.
In order to really let the user input the number, change:
int n = 10;
to
int n;
std::cout << "Please, input.\n";
std::cin >> n;
However, letting user inputting must be followed by validation of the input. You see users can, by accident or not, provide input in your program, that can cause undefined behaviour.
The sequence you want is 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, ...
As I pointed out in a comment to another answer, your code does not produce a correct Fibonacci sequence. F(3) isn't the problem with your code; the problem is that you get confused between all the variables, a, b, c and use them to mean different things at once.
You also incorrectly decrement n: your code does it in the wrong place, and even if you move it to the right place, it wouldn't help as the operation would make n go negative.
Your existing Code
Let's walk through your code a bit:
int a = 0;
int b = 1;
int c;
int n = 3;
n -= 2;
Well, this is weird. We set n to 3 then immediately subtract 2, making it 1. This means that if you try to set n to 0, 1, or 2 you end up with n being a negative number. If you set it to 3, you end up with n being 1.
if (n == 1){
cout << a << endl;
}
We're in trouble right here. Remember that you subtract 2 from n which means that for n==3 you will return whatever is in a which is wrong. But even if you meant this to special-case F(1) that code is still wrong because F(1)=1.
else {
cout << a << b << endl;
for (int i=0;i<n;i++){
Remember, that we can get here with n zero or negative. Obviously in the case of n <= 0 this loop will never execute, so c will never be printed.
c = b + a;
cout << c << endl;
Here, we seem to calculate and output the next Fibonacci number by adding the two previous numbers. This should be fine.
a = b;
b = c;
And here, we keep the new Fibonacci number and its predecessor for the next loop iteration, if any.
The problems with this code are, of course, fixable. But the problem is that the existing code is confusing. It outputs all sorts of different values, and it's unclear what variable is supposed to represent.
Looking at this problem, your first instinct would be to make a function which accepts as input a number n and returns F(n) - you could call it fib or somesuch.
Reworking this
So, how to go about writing such a function? Here's a simple recursive implementation that you can use:
int fib(int n)
{
if ((n == 0) || (n == 1))
return n;
return fib(n-1) + fib(n-2);
}
Notice how this function is short, sweet and to the point. There's no need for a ton of variables, no need for complicated control structures or storing state. It almost reads like a text-based description of the Fibonacci algorithm.
Of course, it's not super-efficient and ends up redoing a lot of work. That's a legitimate criticism, but it's unlikely that there performance considerations here.
Still, perhaps you just don't like recursion. Many people think of recursion as a dirty word, and avoid it with a passion. So how about a non-recursive implementation instead? It's possible, but it's a bit more difficult to understand.
int fib (int n)
{
/* F(0) = 0 */
if (n == 0)
return 0;
int a = 0;
int b = 1;
for (int i = 2; i < n; i++)
{
int c = a + b;
a = b;
b = c;
}
/* F(n) = F(n-2) + F(n-1) */
return a + b;
}
This is a little bit more efficient and not that much more difficult to understand.
I hope that this helped.
Try this which would give you the list you needed.
#include <iostream>
using namespace std;
int fib(int num){
int ans;
if (num >2) {
ans = fib(num-1) + fib(num-2);
}
else
ans = 1;
return ans;
}
int main()
{
int num, x=1;
cin >> num;
while (num >= x) {
cout << fib(x) <<" ";
x++;
}
return 0;
}