int fib(int numb){
vector<int> temp;
int str;
if(numb==0 || numb==1){
return numb;
}
else{
str=(fib(numb-1)+fib(numb-2));
temp.push_back(str);
return str;
}
for(int i=0;i<temp.size();i++){
if(temp[i]==numb){
return temp[i];
}}
Fibonacci function and it work but how do I check if the for loop part of the function really works? Its for a traversal method of finding a existing number and returning it instead of processing another recursion.
Your loop can't possibly work. It will never work. Because there's no way to get to the loop. Every code path before the loop ends with a return statement.
Walk through your code, statement by statement, and see for yourself that your code will never reach the loop.
You must process the stored elements before returning any value. More, as you store elements in vector during recursive calls, the vector temp must be static.
And the research should not be that: you should store in the vector the values, said differently, what you want it temp[i] is fib(i).
A simple way to do that is to make use that C++ allows initializing of static values through functions. You could then initialize temp to { 0, 1}, and when asked for a value, just look if the number is higher than temp.size:
if it is, compute it and store it into temp - as you compute them with values of fib(n-1) and fib(n-2), when you compute it you know that the temp vector already contains fib(n-1), and does not still contains fib(n) => you have just to push it back into temp
it not just extract it from temp
Code could be:
// return a temporary vector containing 0 and 1
std::vector<int> inifib() {
std::vector<int> t;
t.push_back(0);
t.push_back(1);
return t;
}
int fib(unsigned int numb) {
static std::vector<int> temp = inifib(); // initialize once the static temp with size 2 and values 0,1
if (numb >= temp.size()) {
int cr = fib(numb-1) + fib(numb - 2);
temp.push_back(cr); // when we are here, temp contains everything up to fib(numb - 1) - just push
}
return temp[numb];
}
Related
I need to find the Index of the minimum number in an array using a recursive function in c++ the function can only get 2 parameters: the pointer to the array and the size of it.
int smallest(int arr[], int num);
I managed to do this but with a helper variable that is static and declared outside the function here is what I got:
static int flag = 0;
int smallest(int* arr, int num) {
if (flag == num - 1)
return flag;
if (arr[num - 1] > arr[flag]) {
return smallest(arr, num - 1);
} else {
flag++;
return smallest(arr, num);
}
}
Now my question is can I do this without the static variable or any other variable other than the num? here is what I got so far:
int smallest(int arr[], int num) {
if (arr != &arr[num - 1])
if (*arr < arr[num - 1])
smallest(arr, num - 1);
else
smallest(arr + 1, num);
return num - 1;
}
It doesn't return the index of the minimum value but it does get to its adress, the function ends when the array pointer address is the same as the address of the minimum value. how can I get it to return the index?
I'm a student and I'm pretty new to C++ I appreciate the help! thanks!
===
edit:
this is originally from a homework assignment but the constraint to not use external help variables or functions is mine! and I'm curious to know if its even possible.
Because this is obviously homework, I'm not going to reveal the ACTUAL answer in entirety, but I'll give some (hopefully) good advice.
With recursion, think first of what your end condition is. That should be an array of 1 element. You return the index 0 in that case because of the array you have, it's the only element, so return the index of it.
if(num == 1)
{
return 0;
}
So then how is that useful to you? Compare it to exactly one other element. That's how you break this down. Your array turns into "one element and MANY" or "It's just one element." If it's just one, return the only value. If it's many, call yourself recursively from the second element (index 1) onward:
int restOfArraySmallest = smallest(arr+1, num-1) + 1;
You need the +1 because you've offset where it starts from. But you know the value in restOfArraySmallest is the index into YOUR arr of the smallest value of everything except the first element. Because your recursive calls don't include the first element, just the rest.
Hopefully that's enough to get you the rest of the way.
Edit: Because the OP has responded and said it wasn't essential to their homework assignment, here's the entire function:
// Recursively finds the index of the smallest element of the passed-in array
int smallest(int* arr, int num)
{
if(num <= 1)
{
return 0; // If we're in the last element, the smallest is the only element
}
// More than one element, so find the least element in all the elements
// past the first element of the array
// Note: the index returned is offset from our current view of the array,
// so add +1 to any returned result so we can index directly from it
int restOfArraySmallest = smallest(arr+1, num-1) + 1;
// Compare the first element in the array to the smallest of the entire rest
// of the array:
if(arr[0] < arr[restOfArraySmallest])
{
// The first element is smaller, it's the smallest, so return that index
return 0;
}
else
{
// The element already found is smaller, so return that index. It's already
// correctly offset
return restOfArraySmallest;
}
}
And that's it. If there's duplicate entries for smallest, it will favor the LAST one.
The trick with recursion is to NOT keep external variables. What you pass as the arguments and RETURN BACK are all the information you have. For some algorithms, it's enough.
Note, if you use recursive functions with datasets big enough, you will eventually get a Stack Overflow. Not the website, the crash type. This algorithm is pretty light in that only one extra variable and the two arguments themselves get allocated each pass, but it adds up eventually.
I have an array of values e.g. 1, 4, 7, 2.
I also have another array of values and I want to add its values to this first array, but only when they all are different from all values that are already in this array. How can I check it? I've tried many types of loops, but I always ended with an iteration problem.
Could you please tell me how to solve this problem? I code in c++.
int array1[7] = {2,3,7,1,0};
int val1 = rand() % 10;
int val2 = rand() % 10;
int array2[2] = {val1, val2};
and I am trying to put every value from array2 into array1. I tried loop
for (int x:array2)
{
while((val1 && val2) == x)
{
val1 = rand() % 10;
val2 = rand() % 10;
}
}
and many more, but still cannot figure it out. I have this problem because I may have various number of elements for array2. So it makes this "&&" solution infinite.
It is just a sample to show it more clearly, my code has much more lines.
Okay, you have a few problems here. If I understand the problem, here's what you want:
A. You have array1 already populated with several values but with space at the end.
1. How do you identify the number of entries in the array already versus the extras?
B. You have a second array you made from two random values. No problem.
You want to append the values from B to A.
2. If initial length of A plus initial length of B is greater than total space allocated for A, you have a new problem.
Now, other people will tell you to use the standard template library, but if you're having problems at this level, you should know how to do this yourself without the extra help from a confusing library. So this is one solution.
class MyArray {
public:
int * data;
int count;
int allocated;
MyArray() : data(nullptr), count(0), allocated(0) {}
~MyArray() { if (data != nullptr) free(data); }
// Appends value to the list, making more space if necessary
void add(int value) {
if (count >= allocated) {
// Not enough space, so make some.
allocated += 10;
data = (data == nullptr) malloc(allocated * sizeof(int))
: realloc)data, allocated * sizeof(int));
}
data[count++] = value;
}
// Adds value only if not already present.
void addUnique(int value) {
if (indexOf(value) < 0) {
add(value);
}
}
// Returns the index of the value, if found, else -1
int indexOf(int value) {
for (int index = 0; index < count; ++index) {
if (data[index] == value) {
return index;
}
}
return -1;
}
}
This class provides you a dynamic array of integers. It's REALLY basic, but it teaches you the basics. It helps you understand about allocation / reallocating space using old-style C-style malloc/realloc/free. It's the sort of code I was writing back in the 80s.
Now, your main code:
MyArray array;
array.add(2);
array.add(3);
array.add(7);
// etc. Yes, you could write a better initializer, but this is easy to understand
MyArray newValues;
newValues.add(rand() % 10);
newValues.add(rand() % 10);
for (int index = 0; index < newValues.count; ++index) {
array.addUnique(newValues.data[index]);
}
Done.
The key part of this is the addUnique function, which simply checks first whether the value you're adding already is in the array. If not, it appends the value to the array and keeps track of the new count.
Ultimately, when using integer arrays like this instead of the fancier classes available in C++, you HAVE TO keep track of the size of the array yourself. There is no magic .length method on int[]. You can use some magic value that indicates the end of the list, if you want. Or you can do what I did and keep two values, one that holds the current length and one that holds the amount of space you've allocated.
With programming, there are always multiple ways to do this.
Now, this is a lot of code. Using standard libraries, you can reduce all of this to about 4 or 5 lines of code. But you're not ready for that, and you need to understand what's going on under the hood. Don't use the fancy libraries until you can do it manually. That's my belief.
The two structures used in my code, one is nested
struct Class
{
std::string name;
int units;
char grade;
};
struct Student
{
std::string name;
int id;
int num;
double gpa;
Class classes[20];
};
I am trying to figure out a way to sort the structures within the all_students[100] array in order of their ID's in ascending order. My thought was, to start counting at position 1 and then compare that to the previous element. If it was smaller than the previous element then I would have a temporary array of type Student to equate it to, then it would be a simple matter of switching them places within the all_students array. However, when I print the results, one of the elements ends up being garbage numbers, and not in order. This is for an intermediate C++ class in University and we are not allowed to use pointers or vectors since he has not taught us this yet. Anything not clear feel free to ask me.
The function to sort the structures based on ID
void sort_id(Student all_students[100], const int SIZE)
{
Student temporary[1];
int counter = 1;
while (counter < SIZE + 1)
{
if (all_students[counter].id < all_students[counter - 1].id)
{
temporary[0] = all_students[counter];
all_students[counter] = all_students[counter - 1];
all_students[counter - 1] = temporary[0];
counter = 1;
}
counter++;
}
display(all_students, SIZE);
}
There are a few things wrong with your code:
You don't need to create an array of size 1 to use as a temporary variable.
Your counter will range from 1 to 100, you will go out of bounds: the indices of an array of size 100 range from 0 to 99.
The following solution uses insertion sort to sort the array of students, it provides a faster alternative to your sorting algorithm. Note that insertion sort is only good for sufficiently small or nearly sorted arrays.
void sort_id(Student* all_students, int size)
{
Student temporary;
int i = 1;
while(i < size) // Read my note below.
{
temporary = all_students[i];
int j = i - 1;
while(j >= 0 && temporary.id < all_students[j].id)
{
all_students[j+1] = all_students[j]
j--;
}
all_students[j+1] = temporary;
i++;
}
display(all_students, size);
}
Note: the outer while-loop can also be done with a for-loop like this:
for(int i = 1; i < size; i++)
{
// rest of the code ...
}
Usually, a for-loop is used when you know beforehand how many iterations will be done. In this case, we know the outer loop will iterate from 0 to size - 1. The inner loop is a while-loop because we don't know when it will stop.
Your array of Students ranges from 0, 99. Counter is allowed to go from 1 to 100.
I'm assuming SIZE is 100 (in which case, you probably should have the array count also be SIZE instead of hard-coding in 100, if that wasn't just an artifact of typing the example for us).
You can do the while loop either way, either
while(counter < SIZE)
and start counter on 0, or
while (counter < SIZE+1)
and start counter on 1, but if you do the latter, you need to subtract 1 from your array subscripts. I believe that's why the norm (based on my observations) is to start at 0.
EDIT: I wasn't the downvoter! Also, just another quick comment, there's really no reason to have your temporary be an array. Just have
Student temporary;
I overlooked the fact that I was allowing the loop to access one more element than the array actually held. That's why I was getting garbage because the loop was accessing data that didn't exist.
I fixed this by changing while (counter < SIZE + 1)
to: while (counter < SIZE )
Then to fix the second problem which was about sorting, I needed to make sure that the loop started again from the beginning after a switch, in case it needed to switch again with a lower element. So I wrote continue; after counter = 1
I wrote the following function, as an implementation of this algorithm/approach, to generate the power-set (set of all subsets) of a given string:
vector<string> getAllSubsets(string a, vector<string> allSubsets)
{
if(a.length() == 1)
{
// Base case,
allSubsets.push_back("");
allSubsets.push_back(a);
}
else {
vector<string> temp = getAllSubsets(a.substr(0,a.length()-1),allSubsets);
vector<string> with_n = temp;
vector<string> without_n = temp;
for(int i = 0;i < temp.size()-1;i++)
{
allSubsets.push_back(with_n[i] + a[a.length()-1]);
allSubsets.push_back(without_n[i]);
}
}
return allSubsets;
}
however, someone appears to be going wrong: the size of temp and allSubsets remains static from recursive call to recursive call, when they should be increasing due to the push_back() calls. is there any reason why this would take place?
It's because you have an off-by-one error. Because this occurs in your next-to-base case, you are never inserting any entries.
Since the first invalid index is temp.size(), i < temp.size() means that you will always have a valid index. Subtracting 1 means that you are missing the last element of the vector.
It's worth noting that passing allSubsets in as a parameter is kinda silly because it's always empty. This kind of algorithm simply doesn't require a second parameter. And secondly, you could be more efficient using hash sets that can perform deduplication for you simply and quickly.
I am trying to make a function that can return the prime factors of a given number in an array (or multi-set, but I'm trying to use an array).
For example, if I put in 12, I want to get 2, 2, and 3, not 2, and 3 like with a set. This is so that I can use these to see if it is a Smith number or not, so I need the numbers seperately.
Also, I am taking a recursive approach.
I have tried (to no avail) to return the array many ways, including passing an initial pointer into the code which points to a space to store the array.
I've tried just initializing the array in the function and then returning it.
From what I can tell, I can get the array back from the base case iteration and then when trying to construct a new array with size oldArray+1 to copy values to, things get messy. This is where I get lost.
From what I've read, although this isn't the most efficient implementation, I should be able to make it work.
I have a function, nextPrime(int n), which given n will give back the next prime up from that number.
See source below:
int* find(int n, int p) {
int root = (int) floor(sqrt(n));
if (p > root) {
// Base case, array gets initialized and returned
// depending on value of n and p.
if (n > 1) {
factors = new int[1];
factors[0] = n;
return factors;
}
else {
factors = new int[0];
return factors;
}
}
else
if (n%p == 0){
// Inductive step if p is a factor
int newFloor = (int) floor(n/p);
factors = find(newFloor, p);
// Initialize new array.
int* newFactors;
newFactors = new int[(sizeof(factors) / sizeof(int)) + 1];
// Add p to first slot, fill rest with contents of factors.
factors[0] = p;
for (int i = 0; i < (sizeof(factors) / sizeof(int)); i++) {
newFactors[i+1] = factors[i];
}
return newFactors;
}
else {
// Inductive step p isn't a factor of n
factors = find(n, factors, nextPrime(p));
return factors;
}
}
As I say, the error is with returning the array and using its value, but why does it seem to return OK from the first iteration?
Something like this could work. Not terribly efficient !!
void FindFactors( int number , std::vector<int>& factors )
{
for ( int i = 2; i <= number; ++i )
{
if ( number % i == 0 )
{
factors.push_back( i );
FindFactors( number / i , factors);
break;
}
}
}
int main()
{
std::vector<int> factors;
FindFactors( 121 , factors );
return 0;
}
After you call the function factors will contain only the prime factors.
You should be using std::vector for this. The main problem you have is that a pointer to an array has no way of knowing the number of items the array contains. Concretely, the part where you say sizeof(factors) is wrong. As I understand, you're expecting that to give you the number of items in the array pointed to by factors, but it really gives you the number of bytes needed to store a pointer to int.
You should be either returning a vector<int> or passing it in as a reference and updating it each time you find a factor.