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.
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.
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.
http://blog.adilakhter.com/2013/02/21/1729/
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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.
The Problem
Consider the following algorithm:1. input n 2. print nGiven 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
3. if n = 1 then STOP
4. if n is odd then6. GOTO 2
5. else
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.
The 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 1,000,000 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.
The 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
public
static
void
main(String[] args)
throws
Exception {
Scanner in =
new
Scanner(System.in);
PrintWriter out =
new
PrintWriter(System.out,
true
);
// while there is some input to read
while
(in.hasNextInt()) {
int
i = in.nextInt(),
j = in.nextInt(),
from = Math.min(i, j),
to = Math.max(i, j),
max =
0
;
// loop through all the numbers
// and find the biggest cycle
for
(
int
n = from; n <= to; n++) {
max = Math.max(max, cycleLength(n));
}
out.printf(
"%d %d %d\n"
, i, j, max);
}
}
// cache for already computed cycle lengths
public
static
int
[] cache =
new
int
[
1000000
];
public
static
int
cycleLength(
long
n) {
// our base case
// 1 has a cycle length of 1
if
(n ==
1
)
return
1
;
// check if we've already cached the
// cycle length of the current number
if
(n <
1000000
&& cache[(
int
)n] !=
0
)
return
cache[(
int
)n];
// the cycle length of the current number is 1 greater
// than the cycle length of the next number
int
length =
1
+ cycleLength(next(n));
// we cache the result if the
// current number is not too big
if
(n <
1000000
)
cache[(
int
)n] = length;
return
length;
}
// a function that returns the
// next number in the sequence
public
static
long
next(
long
n) {
if
(n %
2
==
0
)
return
n /
2
;
// if n is even
else
return
3
* n +
1
;
// if n is odd
}
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