Solving C++ 3n+1 Problem - Max Cycle Length w/ Code

  • Context: Comp Sci 
  • Thread starter Thread starter chaoseverlasting
  • Start date Start date
  • Tags Tags
    C++
Click For Summary

Discussion Overview

The discussion revolves around solving the C++ implementation of the 3n+1 problem, focusing on finding the maximum cycle length. Participants explore issues related to data types, program performance, and potential flaws in the code, with a particular emphasis on handling large integer values.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that the program hangs when the input exceeds 100,000 due to integer overflow, proposing the use of larger data types like unsigned long or uint32_t.
  • Others argue that unsigned ints can handle values up to 4294967295, and that the original program fails because it does not consistently use unsigned ints in critical calculations.
  • A participant mentions that the maximum value for a signed int is 2147483647, which can lead to issues with the calculation x = 3*x + 1 when x exceeds a certain threshold.
  • Some participants discuss the size of data types, clarifying that an int is typically 32 bits on modern architectures, contrary to a misconception that it might be 2 bits.
  • There are suggestions to test simple cases manually to verify the correctness of the program's output.
  • A later reply indicates that changing the data type to unsigned long int resolved the hanging issue for one participant, allowing them to obtain the correct maximum cycle length.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate data types to use and the implications of integer overflow. While some agree on the need for larger data types, others debate the specifics of integer sizes and their behavior in the program.

Contextual Notes

There are unresolved questions regarding the handling of integer sizes across different compilers and architectures, as well as the specific conditions under which the program hangs.

Who May Find This Useful

Readers interested in C++ programming, algorithm optimization, and numerical methods in computer science may find this discussion relevant.

chaoseverlasting
Messages
1,050
Reaction score
3
Im trying to find the maximum cycle length, but the program hangs when i>100000 or so. Here's the code:

#include<iostream>

using namespace std;

int main()
{
unsigned int i, j;
int count, cmax;

int process(int x);
count<<"Enter range:\nFrom: ";
cin>>i;
count<<"\nTo: ";
cin>>j;
count<<i<<" "<<j;

for(;i<=j;i++)
{
count=process(i);
if(count>cmax)
cmax=count;

}
count<<" "<<cmax;

return 0;
}



int process(int x)
{
int count;
for(count=1;x!=1;++count)
{

if(x%2==0)
x/=2;
else
x=3*x+1;

}
return count;
}

I think the values become too large for integers to handle. I was told to assume that the values would not exceed 32 bits, aren't integers 2 bits though?
 
Physics news on Phys.org
chaoseverlasting said:
I think the values become too large for integers to handle. I was told to assume that the values would not exceed 32 bits, aren't integers 2 bits though?

Unsigned int can go from 0 to 65536. Try using unsigned long.
BTW, i don't know if there's a 32 bit fundamental datatype ; maybe you're supposed to use user defined data types like arrays
 
chaoseverlasting said:
Im trying to find the maximum cycle length, but the program hangs when i>100000 or so. Here's the code:

...

I think the values become too large for integers to handle. I was told to assume that the values would not exceed 32 bits, aren't integers 2 bits though?

Using unsigned ints is fine, using ints is not. You aren't using unsigned ints in a very critical calculation. Hint: your problem occurs with i=113383, count=120.

f(x) said:
Unsigned int can go from 0 to 65536. Try using unsigned long.
BTW, i don't know if there's a 32 bit fundamental datatype ; maybe you're supposed to use user defined data types like arrays

This is not true with most modern compilers/modern machines. Unsigned ints usually go from 0 to 4294967295 nowadays.

If you want guaranteed size, use uint32_t (#include <stdint.h>).
 
x=3*x+1;

by pen&paper...calculate when your forloop should terminate given the condition ou have stated

count's or printfs are your friend.
 
D H said:
Using unsigned ints is fine, using ints is not. You aren't using unsigned ints in a very critical calculation. Hint: your problem occurs with i=113383, count=120.

If you want guaranteed size, use uint32_t (#include <stdint.h>).

How did you find that count? And how do you use uint32_t ?
 
chaoseverlasting said:
How did you find that count?
The biggest problem is that you are not using unsigned ints in some very critical places. The maximum value for a signed int (assuming 32 bit integers) is 231-1=2147483647, which means the calculation x = 3*x + 1 will fail when x exceeds than (2147483647-1)/3=715827882. So how did I find when this occurs? I added a test for this event.

BTW, your program has another serious flaw. You should be able to do some simple cases by hand (e.g., 1, 2, 3). Do you get the right answers?

And how do you use uint32_t ?

You don't need this, but simply #include <stdint.h>.
 
u should be able to declare it as
uint32_t...it'd be like using u_int64_t or int64_t.
 
D H said:
The biggest problem is that you are not using unsigned ints in some very critical places. The maximum value for a signed int (assuming 32 bit integers) is 231-1=2147483647, which means the calculation x = 3*x + 1 will fail when x exceeds than (2147483647-1)/3=715827882. So how did I find when this occurs? I added a test for this event.

BTW, your program has another serious flaw. You should be able to do some simple cases by hand (e.g., 1, 2, 3). Do you get the right answers?



You don't need this, but simply #include <stdint.h>.

By default, arent int's 2 bits and floats 4 bits, double 8 bits and long double 16 bits?
 
I take it you mean bytes, not bits. 2 bits doesn't represent much at all: 0,1,2,3, done. 2 bytes represents 0 to 65535.

The answer is still no. An unsigned short int and an unsigned int must be able to hold all integer values from 0 to 65535 (16 bits) while an unsigned long int must be able to hold all integer values from 0 to 4294967295 (32 bits). There is nothing to stop a compiler vendor from doing more than the minimum.

In particular, an "int" is supposed to be the "natural" size for the computer in question. In the days of 186 PCs, the natural word size was 16 bits, so ints were 16 bits on those machines. The natural word size on a 32-bit architecture machine is 32 bits; ints are much more likely to be 32 bits than 16 bits on such machines. Unless you work with a dinosaur, your computer has a 32-bit architecture, minimum, and ints will be 32 bits wide.
 
  • #10
simple test to help you go along way...
sizeof(datatype)...
 
  • #11
Im sorry to bring this up again, but I've been busy with college.

Im getting the right results when I run the program up until it hangs (I checked the sample input/output of the problem). I changed the int's to unsigned long int, but I still get the same problem... the program hangs somewhere around n=110000. I can't spot the major flaw. Could you give me a hint DH?
 
  • #12
Are you using unsigned long ints everywhere you might have a problem with signed integers? (Hint: the original program is not.)
 
  • #13
Yes... that's exactly what I did, and it worked... I got 525 as the longest cycle length. Thank you so very much!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 23 ·
Replies
23
Views
9K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K