PP1. Week 8. Functions
The 3n + 1 problem
Consider the following algorithm: 1. input n 2. print n 3. if n = 1 then STOP 4. if n is odd then n = 3 * n + 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 (including the 1). 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 i and j inclusive.
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 1,000,000 and greater than 0. You can assume that no operation overflows a 32-bit integer. Output
For each pair of integers i and j you should output i and j in the same order in which they appeared in the input. Then print the maximum cycle length for integers between and including i and j. For each test case print three numbers on one line, separating them by one space.
1 10 100 200 201 210 900 1000
1 10 20 100 200 125 201 210 89 900 1000 174