This is an old one. One of the first programming problems I ever solved. Not very difficult to get it right, not very difficult to make it fast.

The 3n + 1 Problem is problem 100 in the UVa Online Judge. Even though I include the problem description in this post, I encourage you to visit the UVa Online Judge because there you will be able to submit your solution to get it judged.

Problem

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 for any integral input value. This conjecture is the Collatz Conjecture. 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.

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.

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

Solution

Consider the following solution. Function collatz_len computes the cycle length of a given integer. Each recursive call corresponds to an assignment in the algorithm of the problem statement. The numbers in the left margin indicate execution count for a random input.

/*        -:    1: */ #include <stdio.h>
/*        -:    2: */ 
/*        -:    3: */ #define MAX(x,y) x > y ? x : y
/*        -:    4: */ #define MIN(x,y) x < y ? x : y
/*        -:    5: */ #define Si(n) scanf("%d", &n)
/*        -:    6: */ 
/* 11448555:    7: */ int collatz_len(const int n) {
/*        -:    8: */   int len;
/*        -:    9: */   if(n == 1)
/*    82704:   10: */     return 1;
/*        -:   11: */   if(n & 1)
/*  3779464:   12: */     len = collatz_len(3*n + 1) + 1;
/*        -:   13: */   else
/*  7586387:   14: */     len = collatz_len(n >> 1) + 1;
/* 11365851:   15: */   return len;
/*        -:   16: */ }
/*        -:   17: */ 
/*        1:   18: */ int main() {
/*        -:   19: */   int i, j, ii, jj, max;
/*        -:   20: */   while(Si(i) != EOF) {
/*      100:   21: */     Si(j);
/*      300:   22: */     ii = MIN(i, j);
/*      300:   23: */     jj = MAX(i, j);
/*   242570:   24: */     for(max = 0; ii<=jj; ii++)
/*   242370:   25: */       max = MAX(max, collatz_len(ii));
/*      100:   26: */     printf("%d %d %d\n", i, j, max);
/*        -:   27: */   }
/*        1:   28: */   return 0;
/*        -:   29: */ }

We reduce the number of steps taken to solve the random input by memoizing the cycle lengths. The following solution memoizes calls for n less than 1,000,000. Memoization reduces the number of calls one order of magnitude because the calls to collatz_len decreased from ~11,400,000 steps to ~900,000 and we added ~3,000,000 steps for initialization in line 26.

/*        -:    1: */ #include <stdio.h>
/*        -:    2: */ 
/*        -:    3: */ #define MAX_INPUT 1000000
/*        -:    4: */ #define MAX(x,y) x > y ? x : y
/*        -:    5: */ #define MIN(x,y) x < y ? x : y
/*        -:    6: */ #define Si(n) scanf("%d", &n)
/*        -:    7: */ 
/*        -:    8: */ int lengths[MAX_INPUT];
/*        -:    9: */ 
/*   889627:   10: */ int collatz_len(const int n) {
/*        -:   11: */   int len;
/*        -:   12: */   if(n < MAX_INPUT && lengths[n] > 0)
/*    82704:   13: */     return lengths[n];
/*        -:   14: */   if(n & 1)
/*   285459:   15: */     len = collatz_len(3*n + 1) + 1;
/*        -:   16: */   else
/*   521464:   17: */     len = collatz_len(n >> 1) + 1;
/*        -:   18: */   if(n < MAX_INPUT)
/*   445674:   19: */     lengths[n] = len;
/*   806923:   20: */   return len;
/*        -:   21: */ }
/*        -:   22: */ 
/*        1:   23: */ int main() {
/*        -:   24: */   int i, j, ii, jj, max;
/*        1:   25: */   lengths[1] = 1;
/*  2999996:   26: */   for(i = 2; i<MAX_INPUT; i++) lengths[i] = 0;
/*        -:   27: */   while(Si(i) != EOF) {
/*      100:   28: */     Si(j);
/*      300:   29: */     ii = MIN(i, j);
/*      300:   30: */     jj = MAX(i, j);
/*   242570:   31: */     for(max = 0; ii<=jj; ii++)
/*   242370:   32: */       max = MAX(max, collatz_len(ii));
/*      100:   33: */     printf("%d %d %d\n", i, j, max);
/*        -:   34: */   }
/*        1:   35: */   return 0;
/*        -:   36: */ }

Discussion

Our solution ranks 183 in the UVa Online Judge. There are submissions with reported runtimes of 0.000 seconds. I have yet to figure out a faster solution.

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