3n+1 Problem Solution using C++.














































3n+1 Problem Solution using C++.



The Problem Link.

Below is the description of the problem.

The 3n + 1 problem

Problems in Computer Science are often classified as belonging to a certain class of problems (e.g.,
NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose
classification is not known for all possible inputs.
Consider the following algorithm:

    1. input n
    2. print n
    3. if n = 1 then STOP
    4.     if n is odd then n <-- 3n + 1
    5.     else n <-- n/2
    6. GOTO 2

Given the input 22, the following sequence of numbers will be printed

22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1

It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input
value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has
been verified, however, for all integers n such that 0 < n < 1, 000, 000 (and, in fact, for many more
numbers than this.)
Given an input n, it is possible to determine the number of numbers printed before and including
the 1 is printed. For a given n this is called the cycle-length of n. In the example above, the cycle
length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers
between and including both i and j.

    Input

The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers
will be less than 10,000 and greater than 0.
You should process all pairs of integers and for each pair determine the maximum cycle length over
all integers between and including i and j.
You can assume that no operation overflows a 32-bit integer.

    Output

For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers
between and including i and j. These three numbers should be separated by at least one space with all
three numbers on one line and with one line of output for each line of input. The integers i and j must
appear in the output in the same order in which they appeared in the input and should be followed by
the maximum cycle length (on the same line).

    Sample Input

    1 10
    100 200
    201 210
    900 1000

    Sample Output

    1 10 20
    100 200 125
    201 210 89
    900 1000 174


Approach

To solve this Problem, We need to observe two of the following things -->

    1. The input numbers i.e. 'i' and 'j', is not necessarily in order, i.e. the first number can be greater

    than second.

    2. If we get a input, such that i > j, then we need to find in range [j, i], and print result, in the

    form, (i + " " + j + " " + maximum_cycle_length).


    Here is the simple C++ implementation of the same


#include <iostream>

using namespace std;
#define ll long long

// utility to swap to number
void swap(ll &a, ll &b) {
ll temp = a;
a = b;
b = temp;
}

// utility to find max of two numbers
ll max(ll a, ll b) {
if (a > b)
return a;
return b;
}

int main() {
ll lower_bound, upper_bound;
while (cin >> lower_bound >> upper_bound) {

/*
* Storing the values of our input integers so, that at printing
* Stage we can print the values in the same order as the input values.
*/
ll result_lower_bound = lower_bound, result_upper_bound = upper_bound;

// Checking the condition of our 2nd observation.
if (lower_bound > upper_bound) {
swap(lower_bound, upper_bound);
}

// Finding the maximum path length
ll max_path_length = 0;
for (ll i = lower_bound; i <= upper_bound; i++) {

ll n = i; // Storing i in a temporary variable n
ll current_path_length = 0; // Store the value of current path length

// Loop until the n is not equal to 1
while (n != 1) {
// If n is divisible by 2
if (n % 2 == 0)
n /= 2; // Divide n by 2

else
n = 3 * n + 1; // Change n to 3*n + 1

current_path_length++; // Increase the current Path length by 1
}

max_path_length = max(current_path_length,
max_path_length); // Storing maximum path length in max_path_length
}
// Final stage --> printing result
cout << result_lower_bound << " " << result_upper_bound << " " << max_path_length + 1 << endl;

// max_path_length + 1, because we also include the step when n has become 1.
}
}




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