OK, so the goal of this was to write some code for the Fibonacci numbers itself then take those numbers figure out which ones were even then add those specific numbers together. Everything works except I tried and tried to figure out a way to add the numbers up, but I always get errors and am stumped as of how to add them together. I looked elsewhere but they were all asking for all the elements in the vector. Not specific ones drawn out of an if statement.
P.S. I know system("pause") is bad but i tried a few other options but sometimes they work and sometimes they don't and I am not sure why. Such as cin.get().
P.S.S I am also new to programming my own stuff so I have limited resources as far as what I know already and will appreciate any ways of how I might "improve" my program to make it work more fluently. I also take criticism well so please do.
#include "../../std_lib_facilities.h"
int main(){
vector<int>Fibonacci;
int one = 0;
int two = 1;
int three = 0;
int i = 0;
while (i < 4000000){
i += three;
three = two + one; one = two; two = three;
cout << three << ", ";
Fibonacci.push_back(three);
//all of the above is to produce the Fibonacci number sequence which starts with 1, 2 and adds the previous one to the next so on and so forth.
//bellow is my attempt and taking those numbers and testing for evenness or oddness and then adding the even ones together for one single number.
}
cout << endl;
//go through all points in the vector Fibonacci and execute code for each point
for (i = 0; i <= 31; ++i)
if (Fibonacci.at(i) % 2 == 0)//is Fibonacci.at(i) even?
cout << Fibonacci.at(i) << endl;//how to get these numbers to add up to one single sum
system("pause");
}
Just do it by hand. That is loop over the whole array and and keep track of the cumulative sum.
int accumulator = 0; // Careful, this might Overflow if `int` is not big enough.
for (i = 0; i <= 31; i ++) {
int fib = Fibonacci.at(i);
if(fib % 2)
continue;
cout << fib << endl;//how to get these numbers to add up to one single sum
accumulator += fib;
}
// now do what you want with "accumulator".
Be careful about this big methematical series, they can explode really fast. In your case I think the calulation will just about work with 32-bit integers. Best to use 64-bit or even better, a propery BigNum class.
In addition to the answer by Adrian Ratnapala, I want to encourage you to use algorithms where possible. This expresses your intent clearly and avoids subtle bugs introduced by mis-using iterators, indexing variables and what have you.
const auto addIfEven = [](int a, int b){ return (b % 2) ? a : a + b; };
const auto result = accumulate(begin(Fibonacci), end(Fibonacci), 0, addIfEven);
Note that I used a lambda which is a C++11 feature. Not all compilers support this yet, but most modern ones do. You can always define a function instead of a lambda and you don't have to create a temporary function pointer like addIfEven, you can also pass the lambda directly to the algorithm.
If you have trouble understanding any of this, don't worry, I just want to point you into the "right" direction. The other answers are fine as well, it's just the kind of code which gets hard to maintain once you work in a team or have a large codebase.
Not sure what you're after...
but
int sum=0; // or long or double...
for (i = 0; i <= 31; ++i)
if (Fibonacci.at(i) % 2 == 0) {//is Fibonacci.at(i) even?
cout << Fibonacci.at(i) << endl;//how to get these numbers to add up to one single sum
sum+=Fibonacci.at(i);
}
// whatever
}
Related
I'm working on a GA and seem to be having problems with the tournament selection. I think this is due to the fact that I'm not comparing what I want to compare (in terms of fitness values)
srand(static_cast <unsigned> (time(0)));
population Pop;
vector<population> popvector;
vector<population> survivors;
population *ptrP;
for (int i = 0; i <= 102; i++)
{
ptrP = new population;
ptrP->generatefit;
ptrP->findfit;
popvector.push_back(*ptrP);
//include finding the persons "overall". WIP
}
cout << "The fit values of the population are listed here: " << endl;
vector<population> ::iterator it; //iterator to print everything in the vector
for (it = popvector.begin(); it != popvector.end(); ++it)
{
it->printinfo();
}
unsigned seed = std::chrono::system_clock::now().time_since_epoch().count(); // generate a seed for the shuffle process of the vector.
cout << "Beggining selection process" << endl;
shuffle(popvector.begin(), popvector.end(), std::default_random_engine(seed));
//Shuffling done to randomize the parents I will be taking.
// I also want want to pick consecutive parents
for (int i = 0; i <= 102; i = i + 3)
{
if (popvector[i] >= popvector[i++]);
}
}
Now what I think my problem is, is that when im trying to compare the Overall values (Not found yet, working on how to properly model them to give me accurate Overall fitness values) I'm not comparing what I should be.
I'm thinking that once I find the persons "Overall" I should store it in a Float vector and proceed from there, but I'm unsure if this is the right way to proceed if I wish to create a new "parent" pool, since (I think) the "parent pool" has to be part of my population class.
Any feedback is appreciated.
srand(static_cast <unsigned> (time(0)));
This is useless: you're calling std::shuffle in a form not based on std::rand:
shuffle(popvector.begin(), popvector.end(), std::default_random_engine(seed));
If somewhere else in the program you need to generate random numbers, do it via functions / distributions / engines in random pseudo-random number generation library (do not use std::rand).
Also consider that, for debugging purpose, you should have a way to initialize the random engine with a fixed seed (debug needs repeatable results).
for (int i = 0; i <= 102; i++)
Do not use magic numbers.
Why 102? If it's the population size, store it in a constant / variable (populationSize?), document the variable use and "enjoy" the fact that when you need to change the value you haven't to remember the locations where it's used (just in this simple snippet there are two distinct use points).
Also consider that the population size is one of those parameters you need to change quite often in GA.
ptrP = new population;
ptrP->generatefit;
ptrP->findfit;
popvector.push_back(*ptrP);
Absolutely consider Sam Varshavchik's and paddy's remarks.
for (int i = 0; i <= 102; i = i + 3)
{
if (popvector[i] >= popvector[i++]);
// ...
Generally it's not a good practice to change the index variable inside the body of a for loop (in some languages, not C / C++, the loop variable is immutable within the scope of the loop body).
Here you also have an undefined behaviour:
popvector[i] >= popvector[i++]
is equivalent to
operator>=(popvector[i], popvector[i++])
The order that function parameters are evaluated is unspecified. So you may have:
auto a = popvector[i];
auto b = popvector[i++];
operator>=(a, b); // i.e. popvector[i] >= popvector[i]
or
auto b = popvector[i++];
auto a = popvector[i];
operator>=(a, b); // i.e. popvector[i + 1] >= popvector[i]
Both cases are wrong.
In the first case you're comparing the same elements and the expression is always true.
In the second case the comparison probably is the opposite of what you were thinking.
Take a look at:
Undefined behavior and sequence points
What are all the common undefined behaviours that a C++ programmer should know about?
and always compile source code with -Wall -Wextra (or their equivalent).
I'm not sure to correctly understand the role of the class population. It may be that the name is misleading.
Other questions / answers you could find interesting:
C++: "std::endl" vs "\n"
http://herbsutter.com/2013/05/13/gotw-2-solution-temporary-objects/ (the section about premature pessimization)
My professor assigned homework to write a function that takes in an array of integers and sorts all zeros to the end of the array while maintaining the current order of non-zero ints. The constraints are:
Cannot use the STL or other templated containers.
Must have two solutions: one that emphasizes speed and another that emphasizes clarity.
I wrote up this function attempting for speed:
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
void sortArray(int array[], int size)
{
int i = 0;
int j = 1;
int n = 0;
for (i = j; i < size;)
{
if (array[i] == 0)
{
n++;
i++;
}
else if (array[i] != 0 && j != i)
{
array[j++] = array[i++];
}
else
{
i++;
n++;
}
}
while (j < size)
{
array[j++] = 0;
}
}
int main()
{
//Example 1
int array[]{20, 0, 0, 3, 14, 0, 5, 11, 0, 0};
int size = sizeof(array) / sizeof(array[0]);
sortArray(array, size);
cout << "Result :\n";
for (int i = 0; i < size; i++)
{
cout << array[i] << " ";
}
cout << endl << "Press any key to exit...";
cin.get();
return 0;
}
It outputs correctly, but;
I don't know what the speed of it actually is, can anyone help me figure out how to calculate that?
I have no idea how to go about writing a function for "clarity"; any ideas?
I my experience, unless you have very complicated algorithm, speed and clarity come together:
void sortArray(int array[], int size)
{
int item;
int dst = 0;
int src = 0;
// collect all non-zero elements
while (src < size) {
if (item = array[src++]) {
array[dst++] = item;
}
}
// fill the rest with zeroes
while (dst < size) {
array[dst++] = 0;
}
}
Speed comes from a good algorithm. Clarity comes from formatting, naming variables and commenting.
Speed as in complexity?
Since you are, and need, to look at all the elements in the array — and as such have a single loop going through the indexes in the range [0, N)—where N denotes the size of the input—your solution is O(N).
Further reading:
Plain English explanation of big O
Determining big O Notation
Regarding clearity
In my honest opinion there shouldn't need to be two alternatives when implementing such functionality as you are presenting. If you rename your variables to more suitable (descriptive) names your current solution should be clear enough to count as both performant and clear.
Your current approach can be written in plain english in a very clear fashion:
pseudo-explanation
set write_index to 0
set number_of_zeroes to 0
For each element in array
If element is 0
increase number_of_zeros by one
otherwise
write element value to position denoted by write_index
increase write_index by one
write number_of_zeroes 0s at the end of array
Having stated the explanation above we can quickly see that sortArray is not a descriptive name for your function, a more suitable name would probably be partition_zeroes or similar.
Adding comments could improve readability, but you current focus should lie in renaming your variables to better express the intent of the code.
(I feel your question is almost off-topic; I am answering it from a Linux perspective; I recommend using Linux to learn C++ programming; you'll adapt my advices to your operating system if you are using something else....)
speed
Regarding speed, you should have two complementary approaches.
The first (somehow "theoretical") is to analyze (i.e. think on) your algorithm and give (with some proof) its asymptotic time complexity.
The second approach (only "practical", and often pragmatical) is to benchmark and profile your program. Don't forget to compile with optimizations enabled (e.g. using g++ -Wall -O2 with GCC). Have a benchmark which runs for more than half of a second (so processes a large amount of data, e.g. several million numbers) and repeat it several times (e.g. using time(1) command on Linux). You could also measure some time inside your program using e.g. <chrono> in C++11, or just clock(3) (if you read a large array from some file, or build a large array of pseudo-random numbers with <random> or with random(3) you certainly want to measure separately the time to read or fill the array with the time to move zeros out of it). See also time(7).
(You need to process a large amount of data - more than a million items, perhaps many millions of them - because computer are very fast; a typical "elementary" operation -a machine instruction- takes less than a nanosecond, and you have lot of uncertainty on a single run, see this)
clarity
Regarding clarity, it is a bit subjective, but you might try to make your code readable and concise. Adding a few good comments could also help.
Be careful about naming: sorting is not exactly what your program is doing (it is more moving zeros than sorting the array)...
I think this is the best - Of course you may wish to use doxygen or some other
// Shift the non-zeros to the front and put zero in the rest of the array
void moveNonZerosTofront(int *list, unsigned int length)
{
unsigned int from = 0, to = 0;
// This will move the non-zeros
for (; from < length; ++from) {
if (list[from] != 0) {
list[to] = list[from];
to++;
}
}
// So the rest of the array needs to be assigned zero (as we found those on the way)
for (; to < length; +=to) {
list[to] = 0;
}
}
Basic idea: Given an array, find all the permutations of that array. Then, take each of those arrays and put it all together. Eg the array {6,5,3,4,1,2} gives you 653412. The permutations work, but I cannot get the integers.
int main ()
{
int myints[] = {2,3,4,5,6,7,8,9};
int k;
int dmartin=0;
int powof10=1;
std::cout << "The 8! possible permutations with 8 elements:\n";
do {
for(k=0; k<8; k++){
std::cout << myints[k] << ' ';
dmartin=myints[8-k-1]*powof10+dmartin;
powof10=powof10*10;
}
cout << "\n" << dmartin << "\n";
} while ( std::next_permutation(myints,myints+8) );
dmartin=0;
return 0;
}
I also have some code that works when you just have one array, but in this case there are thousands. I though I needed to reset dmartin=0 at the end of each while loop so that it didn't keep adding to the previous answer, however when I tried that I got "0" for each of my answers. Without trying to reset, I get answers that seem random (and are negative).
The problem is that you're not resetting your two variables inside your loop, so they'll continue from the values they had during the previous iteration, which will just be wrong, and will quickly overflow, giving seemingly rubbish output. Try putting this at the beginning or the end of the do-while loop:
dmartin = 0;
powof10 = 1;
But you're really overcomplicating it a lot. It would be way simpler to just build the number from the most significant digit instead of the least significant one instead. This would eliminate the need for a powof10 variable. This new for-loop would look like this:
for(k = 0; k < 8; k++){
std::cout << myints[k] << ' ';
dmartin = 10*dmartin + myints[k];
}
That won't work for long, since your integer will soon overflow.
That's probably what you are experiencing when you get negative numbers.
Using an integer to store the result does not seem the most appropriate choice to me. Why not use a string, for instance? That would save you the hassle of reinventing base10 conversion in 2014, and you could easily derive a number from the string when needed.
That won't solve the overflow problem, though.
First point: the code to take a vector of digits and turn them into a single number should almost certainly be written as a function, not just code inside the loop.
Second point: you can use std::string like a container of char, and apply normal algorithms to it.
Seem to me, the lazy way would look like this:
std::string input="23456789";
do {
std::cout<<std::stoi(input)<<"\n";
} while (std::next_permutation(input.begin(), input.end()));
The goal here was to create a program that found and output all the prime numbers between 1 and 100. I've noticed I have a tendency to complicate things and create inefficient code, and I'm pretty sure I did that here as well. The initial code is mine, and everything that I've put between the comment tags is the code given in the book as a solution.
// Find all prime numbers between 1 and 100
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int counter; // loop counter
int count_two; // counter for second loop
int val; // equals the number of count, used in division to check for primes
bool check;
check = true;
for(counter = 1; counter <= 100; counter++){
val = counter;
for(count_two = 2; count_two <= 9; count_two++){
if((val % count_two) == !(check)){
cout << val << " is a prime number.\n";
}
}
}
return 0;
}
// program didn't work properly because of needless complication; all that needs to be checked for is whether a number is divisible by two
/*
*********correct code***********
#include <iostream>
using namespace std;
int main()
{
int i, j;
bool isprime;
for(i=1; i < 100; i++) {
isprime = true;
// see if the number is evenly divisible
for(j=2; j <= i/2; j++)
// if it is, then it is not prime
if((i%j) == 0) isprime = false;
if(isprime) cout << i << " is prime.\n";
}
return 0;
}
********************************
*/
From what I can gather, I was on a reasonably correct path here. I think I complicated things with the double loop and overuse of variables, which probably led to the program working incorrectly -- I can post the output if need be, but it's certainly wrong.
My question is basically this: where exactly did I go wrong? I don't need somebody to redo this because I'd like to correct the code myself, but I've looked at this for a while and can't quite figure out why mine isn't working. Also, since I'm brand new to this, any input on syntax/readability would be helpful as well. Thanks in advance.
As it is, your code says a number is prime if it is divisible by any of the numbers from 2 to 9. You'll want a bool variable somewhere to require that it's all and not any, and you'll also need to change this line:
if((val % count_two) == !(check)){
Since check = true, this resolves as follows:
if ((val % count_two) == !true){
and
if ((val % count_two) == false){
and
if ((val % count_two) == 0){
(Notice how the value false is converted to 0. Some languages would give a compile error here. C++ converts it into an integer).
This in fact does the opposite of what you want. Instead, write this, which is correct and clearer:
if (val % count_two != 0) {
Finally, one thing you can do for readability (and convenience!) is to write i, j, and k instead of counter, count_two, and count_three. Those three letters are universally recognized by programmers as loop counters.
In addition to the points made above:
You seemed to think you didn't need to have 2 loops. You do need them both.
Currently, in your code, the upper range of the inner loop is in-dependent on the value of your outer loop. But this is not correct; you need to test divisibility up the the sqrt(outer_loop_value). You'll note in your "correct" code they use half of the outer_loop_value - this could be a performance trade off but strictly speaking you need to test up to sqrt(). But consider that your outer loop was up to 7, your inner loop is testing division all the way up to 9 and 7 is in that range. Which means 7 would be reported as not prime.
In your "correct" code the indenting makes the code harder to interpret. The inner for loop only has a single instruction. That loop loops through all possible divisors. This is unnecessary it could break out at the first point that the mod is zero. But the point is that the if(isprime) cout << i << " is prime.\n"; is happening in the outer loop, not the inner loop. In your (un-commented) code you have put that in the inner loop and this results in multiple responses per outer loop value.
Stylistically there is no need to copy the counter into a new val variable.
In my algorithm I have two values that I need to choose at random but each one has to be chosen a predetermined number of times.
So far my solution is to put the choices into a vector the correct number of times and then shuffle it. In C++:
// Example choices (can be any positive int)
int choice1 = 3;
int choice2 = 4;
int number_of_choice1s = 5;
int number_of_choice2s = 1;
std::vector<int> choices;
for(int i = 0; i < number_of_choice1s; ++i) choices.push_back(choice1);
for(int i = 0; i < number_of_choice2s; ++i) choices.push_back(choice2);
std::random_shuffle(choices.begin(), choices.end());
Then I keep an iterator to choices and whenever I need a new one I increase the iterator and grab that value.
This works but it seems like there might be a more efficient way. Since I always know how many of each value I'll use I'm wondering if there is a more algorithmic way to go about doing this, rather than just storing the values.
You are unnecessarily using so much memory. You have two variables:
int number_of_choice1s = 5;
int number_of_choice2s = 1;
Now simply randomize:
int result = rand() % (number_of_choice1s + number_of_choice2s);
if(result < number_of_choice1s) {
--number_of_choice1s;
return choice1;
} else {
--number_of_choice2s;
return choice2;
}
This scales very well two millions of random invocations.
You could write this a bit more simply:
std::vector<int> choices(number_of_choice1s, choice1);
choices.resize(number_of_choice1s + number_of_choice2s, choice2);
std::random_shuffle(choices.begin(), choices.end());
A biased random distribution will keep some kind of order over the resulting set ( the choice that was picked the most have lesser and lesser chance to be picked next ), which give a biased result (specially if the number of time you have to pick the first value is large compared to the second value, you'll endup with something like this {1,1,1,2,1,1,1,1,2}.
Here's the code, which looks a lot like the one written by #Tomasz Nurkiewicz but using a simple even/odd which should give about 50/50 chance to pick either values.
int result = rand();
if ( result & 1 && number_of_choice1s > 0)
{
number_of_choice1s--;
return choice1;
}else if (number_of_choice2s>0)
{
number_of_choice2s--;
return choice2;
}
else
{
return -1;
}