Im trying to program a recursive function which will allow me to print each digit of a number on a separate line like this :
int decompose(int n){
if(n%10){
cout << n%10 << endl;
return decompose(n/10);
}
}
int main(int argc, char *argv[]){
decompose(2345);
}
OUTPUT
5
4
3
2
PROBLEM : I wanted to print on the other way
2
3
4
5
How to do that ?
void decompose(int n){
if (n>=10) //Call decompose if there are "more" digits in the number to print
decompose(n/10);
cout << n%10 << endl; //print the last digit of n
}
int main(int argc, char *argv[]){
decompose(2345);
}
Since you decompose first and print later in a number like 2345 the stack will look like this:
main()
decompose(2345)
decompose(234)
decompose(23)
decompose(2)
then as we start returning each call will print the last digit: 2 -> 3 ->4 ->5
NOTE: In your version you use the condition if(n%10){} this will be false whenever n is a multiple of 10. It means the versions with it will stop the moment a 0 digit is reached.
Try them with inputs such as 23450 or 23045
In your particular case you can just put the cout after the recursive calls and it will print in reverse order.
More generally, though, you can't always print directly in reverse order or it is not evident how to do it: in such cases you could append the output at each step to a vector of string and, once out of the function, print the vector in reverse order.
It's slightly less efficient and requires to keep in memory all the output at each step but, when these factors are negligible, it's easy to implement and understand.
Related
Hello I have a homework question I am stuck in..any hint or tips would be appreciated. the questions is:
Write a single recursive function in C++ that takes as argument a positive integer n then print n, n-1, n-2,...3,2,1,2,3,...n. How many recursive call does your algorithm makes? What is the worst case running time of your algorithm?
I am stuck in the first part. writing a recursive function that prints n, n-1, n-2,...3,2,1,2,3,...n
so far I have:
print(int n)
{
if (n==0)
return;
cout<<n<<" ";
print(n-1);
return;
}
but this only prints from n to 1
I am lost about how I would print from 2 to n using just one parameter and recursively single function.
I tried this: which gives the correct output but has a loop and has two parameters:
p and z has the same value.
void print(int p,int z)
{
if (p==0)
{
for(int x=2;x<=z; x++)
cout<<x<<" ";
return;
}
else
cout<<p<<" ";
print(p-1,z);
return;
}
any hint or tips is much appreciated thank you.
so it is working now, but I am having trouble understanding how (question in comment):
void print(int n)
{
if (n==1){
cout<<n;
return;
}
else
cout<< n;
print(n-1); // how does it get passed this? to the line below?
cout<<n; // print(n-1) takes it back to the top?
return;
}
The output you want is mirrored, so you can have this series of steps:
print num
recursive step on num-1
print num again
That's the recursive case. Now you need an appropriate base case upon which to stop the recursion, which shouldn't be difficult.
Given the pseudocode:
recursive_print(n):
if n == 1:
print 1
return
print n
recursive_print(n-1)
print n
(If you prefer, just look at your solution instead).
Let's trace it. A dot will mark where we're up to in terms of printing.
. recursive_print(3) // Haven't printed anything
3 . recursive_print(2) 3 // Print the 3
3 2 . recursive_print(1) 2 3 //Print 2
3 2 1 . 2 3 // Print 1
3 2 1 2 . 3 // Print 2
3 2 1 2 3 . // Print 3
Each unrolling of the function gives us 2 numbers on opposite sides and we build down to the "1", then go back and print the rest of the numbers.
The "unrolling" is shown in this picture:
If you strip away the functions and leave yourself with a sequence of commands, you'll get:
print 3
print 2
print 1
print 2
print 3
where each indentation signifies a different function.
Simple solution for this:
Def fxn(n):
if n <= n:
if n > 0:
print(n)
fxn(n - 1)
print(n)
Def main():
Number = 6
fxn(Number)
Main()
If you struggle understanding how this works:
Basically, each time you call a function in a recursive problem, it isn't a loop. It's as if you were on the woods leaving a trail behind. Each time you call a function inside a function, it does its thing, then it goes right back to were you called it.
In other words, whenever you call a recursive function, once the newer attempt is done, it will go right back to were it used to be.
In loops, once each step is done, it is done, but in recursive functions you can do a lot with a lot less.
Printing before the recurse call in my code is the descension step, and once its descension finished, it will progresively unfold step by step, printing the ascension value and going back to the former recurse.
It seems way harder than it is, but it is really easy once you grasp it.
The answer is simpler than you think.
A recursive call is no different from a regular call. The only difference is that the function called is also the caller, so you need to make sure you don't call it forever (you already did that). Let's think about a regular call. If you have the following code snippet:
statement1
f();
statement2
The statement1 is executed, then f is called and does it's thing and then, after f finishes, statement2 is executed.
Now, let's think about your problem. I can see that your hard work on this question from the second program you've written, but forget about it, the first one is very close to the answer.
Let's think about what your function does. It prints all numbers from n to 0 and then from 0 to n. At the first step, you want to print n, then all the numbers from n-1 to 0 and from 0 to n-1, and print another n. See where it's going?
So, you have to do something like this:
print(n)
call f(n-1)
print(n)
I hope my explanation is clear enough.
This is more of hack -- using the std::stream rather than recursion...
void rec(int n) {
if (n==1) { cout << 1; return; }
cout << n;
rec(n-1);
cout << n;
}
int main() {
rec(3);
}
prints
32123
Ok so I am having trouble understanding exactly how this program works:
#include <iostream>
using namespace std;
int getFibNumber(int fibIndex)
{
if(fibIndex < 2)
return fibIndex;
else
return getFibNumber(fibIndex - 1) + getFibNumber(fibIndex - 2);
}
int main(int argc, char** argv)
{
cout << "Enter 0-based index of desired Fibonacci number: ";
int index = 0;
cin >> index;
cout << "Fibonacci number is: " << getFibNumber(index) << endl;
return 0;
}
Specifically, what does "getFibNumber(...)" do when it reiterates(if that is the correct word)? I can't figure out what it does if the integer "fibIndex" that is passed in is greater than or equal to 2. Sorry to ask such a basic question, but I am really stumped by this and I feel like I'm missing something.
As everyone mentioned here, this is basically recursion.
Just to get a feel of how this program works, I have made the recursion tree with initial fibIndex as 5.
5 5 calls 4 and 3.
/ \
4 3 4 calls 3 and 2. 3 calls 2 and 1.
/ \ / \
3 2 2 1 1 is base case, returns 1.
/ \ / \ / \
2 1 1 0 1 0 2 is not base case. So calls 1 and 0.
/ \
1 0
This is called recursion. Instead of doing this with a loop it calls the function again, but with a different parameter. Eventually, the base condition will be true, and the function will return, causing the rest of the calls to return also. This can be a very powerful tool in the right situation.
Hi Im trying to translate this code to TI-BASIC. Im having problems with how to change for loop into while loop and also with incrementing a number in TI-BASIC.
#include <stdio.h>
int main()
{
int n, i, flag=0;
printf("Enter a positive integer: ");
scanf("%d",&n);
for(i=2;i<=n/2;++i)
{
if(n%i==0)
{
flag=1;
break;
}
}
if (flag==0)
printf("%d is a prime number.",n);
else
printf("%d is not a prime number.",n);
return 0;
}
You can efficiently use a While loop in this situation:
Input "NUMBER: ",A
1->B
3->I
√(A->D
If not(fPart(A/2
DelVar BWhile I<=D and B
fPart(A/I->B
I+2->I
End
If not(B
Disp "NOT
Disp "PRIME
In TI-Basic a While loop works as you would expect and you can have conditions for it.
Incrementing a number is as simple as saying
X+i->X
Where 'i' is the incrementer.
To change a For loop into a While loop, you'll have to set up the While loop to constantly check to see if the number and increment have passed the upper bound while increasing the increment each run through.
If you wanted to mimic i++ or ++i in TI-Basic (Using a While loop), all you would have to change would be the arrangement of the code. Please note that TI-Basic For statements always operates under ++i.
Example (i++):
0->X
While X<10
Disp X
X+1->X
End
This will display (With each number on a new line)
0 1 2 3 4 5 6 7 8 9
Example (++i):
0->X
While X<10
X+1->X
Disp X
End
This will display (With each number on a new line)
1 2 3 4 5 6 7 8 9 10
Let it be noted that TI-Basic For statements are much much faster than While loops when it comes to incrementing and should almost always be considered superior for the task.
Integrating Timtech's idea to skip even numbers effectively halves the required time to check the primality of the number with the addition of only a few extra lines.
I expanded the idea to skip multiples of two and multiples of three.
Input "Number:",X:abs(X->X
0
If not(fPart(X/2)) or not(fPart(X/3:Return
For(B,5,sqrt(X),6)
If not(fPart(X/B)) or not(fPart((X+2)/B:Return
End
1
Test Number: 1003001
Time Required: ~4 Seconds (So much better than 15 :D)
Size: 65 Bytes
I dont see why you would want to use a while loop as ti-basic has for loops:
0->F
Input "ENTER NUMBER:",N
For(I,2,int(N/2
If N/I=int(N/I
Then
int(N/2->I
1->F
End
End
If F
Then
Disp "NUMBER IS PRIME
Else
Disp "NUMBER IS NOT PRIME
End
N/I=int(N/I is a way to check for a number's remainder (another way of saying N%I==0 but ti basic does not have modulus). Another trick here is setting I to its maximum bound (int(N/2) as a sort of "break" like other languages would have
hello i have this piece of code that i coded based on some other recursion and factorial programs
but my problem is that i am really confused as to how it stored the value and kept it and then returned it at the end
int factorialfinder(int x)
{
if (x == 1)
{
return 1;
}else
{
return x*factorialfinder(x-1);
}
}
int main()
{
cout << factorialfinder(5) << endl;
}
so 5 goes in, and gets multiplied by 4 by calling its function again and again and again, then it gets to one and it returns the factorial answer
why? i have no idea how it got stored, why is return 1 returning the actual answer, what is it really doing?
Source: Image is taken from: IBM Developers website
Just take a look at the picture above, you will understand it better. The number never gets stored, but gets called recursively to calculate the output.
So when you call the fact(4) the current stack is used to store every parameter as the recursive calls occur down to factorialfinder(1). So the calculation goes like this: 5*4*3*2*1.
int factorialfinder(int x)
{
if (x == 1) // HERE 5 is not equal to 1 so goes to else
{
return 1;
}else
{
return x*factorialfinder(x-1); // returns 5*4*3*2*1 when x==1 it returns 1
}
}
Hope this helps.
Return 1 is not returning the actual answer. It's just returning the answer to calling
factorialfinder(1);
which happens in your code.
In any program, a call stack is a space in memory that is used to keep track of function calls. Space from this memory is used to store the arguments to a function, as well as the return value of that function. Whenever some function A calls another function B, A gets the return value of B from that space.
A recursive function is nothing special, it's just an ordinary function calling another function (that happens to be itself). So really, when a recursive function F calls itself, it's calling another function: F calls F', which calls F'', which calls F''', etc. It's just that F, F'', F''' etc. execute the same code, just with different inputs.
The expression if (x == 1) is there to check when this process should be stopped.
The return value of F''' is used by F''. The return value of F'' is used by F'. The return value of F' is used by F.
In Factorial of some number, the operation is (n) * (n-1) * (n-2) * .... * (1).
I've highlighted the 1; this is the condition that's being checked.
A recursive function breaks a big problem down into smaller cases.
Going over your program:
call factorialfinder with 5, result is stored as 5 * factorialfinder(4)
call factorialfinder with 4, result is stored as 5 * 4 * factorialfinder(3)
call factorialfinder with 3, result is stored as 5 * 4 * 3 * factorialfinder(2)
call factorialfinder with 2, result is stored as 5 * 4 * 3 * 2 * factorialfinder(1)
call factorialfinder with 1, result is stored as 5 * 4 * 3 * 2 * 1
in essence it combines the result of a stack of calls to factorialfinder until you hit your base case, in this case x = 1.
Well, the factorial function can be written using recursion or not, but the main consideration in the recursion is that this one uses the system stack, so, each call to the function is a item in the system stack, like this (read from the bottom to the top):
Other consideration in the recursion function is that this one has two main code piece:
The base case
The recursion case
In the base case, the recursive function returns the element that bounds the algorithm, and that stop the recursion. In the factorial this element is 1, because mathematically the factorial of number one is 1 by definition. For other numbers you don't know the factorial, because of that, you have to compute by using the formula, and one implementation of it is using recursion, so the recursive case.
Example:
The factorial of 5, the procedure is: 5*4*3*2*1 = 120, note you have to multiply each number from the top value until number 1, in other words, until the base case takes place which is the case that you already knew.
#include<iostream>
using namespace std;
int factorial(int n);
int main()
{
int n;
cout << "Enter a positive integer: ";
cin >> n;
cout << "Factorial of " << n << " = " << factorial(n);
return 0;
}
int factorial(int n)
{
if(n > 1)
return n * factorial(n - 1);
else
return 1;
}
Just for fun I created an algorithm that computes every possible combination from a given rugby score (3, 5 or 7 points). I found two methods : The first one is brute force, 3 imbricated for loops. The other one is recursion.
Problem is some combinations appear multiple times. How can I avoid that ?
My code :
#include <iostream>
using namespace std;
void computeScore( int score, int nbTryC, int nbTryNC, int nbPenalties );
int main()
{
int score = 0;
while (true)
{
cout << "Enter score : ";
cin >> score;
cout << "---------------" << endl << "SCORE = " << score << endl
<< "---------------" << endl;
// Recursive call
computeScore(score, 0, 0, 0);
}
return 0;
}
void computeScore( int score, int nbTryC, int nbTryNC, int nbPenalties )
{
const int tryC = 7;
const int tryNC = 5;
const int penalty = 3;
if (score == 0)
{
cout << "* Tries: " << nbTryC << " | Tries NT: " << nbTryNC
<< " | Penal/Drops: " << nbPenalties << endl;
cout << "---------------" << endl;
}
else if (score < penalty)
{
// Invalid combination
}
else
{
computeScore(score - tryC, nbTryC+1, nbTryNC, nbPenalties);
computeScore(score - tryNC, nbTryC, nbTryNC+1, nbPenalties);
computeScore(score - penalty, nbTryC, nbTryNC, nbPenalties+1);
}
}
One way to think about this is to realize that any time you have a sum, you can put it into some "canonical" form by sorting all the values. For example, given
20 = 5 + 7 + 3 + 5
You could also write this as
20 = 7 + 5 + 5 + 3
This gives a few different options for how to solve your problem. First, you could always sort and record all of the sums that you make, never outputting the same sum twice. This has the problem that you're going to end up repeatedly generating the same sums multiple different times, which is extremely inefficient.
The other (and much better) way to do this is to update the recursion to work in a slightly different way. Right now, your recursion works by always adding 3, 5, and 7 at each step. This is what gets everything out of order in the first place. An alternative approach would be to think about adding in all the 7s you're going to add, then all the 5's, then all the 3's. In other words, your recursion would work something like this:
Let kValues = {7, 5, 3}
function RecursivelyMakeTarget(target, values, index) {
// Here, target is the target to make, values are the number of 7's,
// 5's, and 3's you've used, and index is the index of the number you're
// allowed to add.
// Base case: If we overshot the target, we're done.
if (target < 0) return;
// Base case: If we've used each number but didn't make it, we're done.
if (index == length(kValues)) return;
// Base case: If we made the target, we're done.
if (target == 0) print values; return;
// Otherwise, we have two options:
// 1. Add the current number into the target.
// 2. Say that we're done using the current number.
// Case one
values[index]++;
RecursivelyMakeTarget(target - kValues[index], values, index);
values[index]--;
// Case two
RecursivelyMakeTarget(target, values, index + 1);
}
function MakeTarget(target) {
RecursivelyMakeTarget(target, [0, 0, 0], 0);
}
The idea here is to add in all of the 7's you're going to use before you add in any 5's, and to add in any 5's before you add in any 3's. If you look at the shape of the recursion tree that's made this way, you will find that no two paths end up trying out the same sum, because when the path branches either a different number was added in or the recursion chose to start using the next number in the series. Consequently, each sum is generated exactly once, and no duplicates will be used.
Moreover, this above approach scales to work with any number of possible values to add, so if rugby introduces a new SUPER GOAL that's worth 15 points, you could just update the kValues array and everything would work out just fine.
Hope this helps!
Each time you find a solution you could store it in a dictionary ( a set of strings for example, with strings looking like "TC-TNT-P" )
Before printing a solution you verify it was not in the dictionary.
A nested for-loop is the natural way to do this. Using recursion is just silly (as you seem to have discovered).