Section 8.5 The 3n + 1 Sequence
As another example of indefinite iteration, let’s look at a sequence that has fascinated mathematicians for many years. The rule for creating the sequence is to start from some positive integer, call it
n, and to generate the next term of the sequence from
n, either by halving
n is even, or else by multiplying it by three and adding 1 when it is odd. The sequence terminates when
n reaches 1.
This Python function captures that algorithm. Try running this program several times supplying different values for n.
The condition for this loop is
n != 1. The loop will continue running until
n == 1 (which will make the condition false).
Each time through the loop, the program prints the value of
n and then checks whether it is even or odd using the remainder operator. If it is even, the value of
n is divided by 2 using integer division. If it is odd, the value is replaced by
n * 3 + 1. Try some other examples.
n sometimes increases and sometimes decreases, there is no obvious proof that
n will ever reach 1, or that the program terminates. For some particular values of
n, we can prove termination. For example, if the starting value is a power of two, then the value of
n will be even each time through the loop until it reaches 1.
You might like to have some fun and see if you can find a small starting number that needs more than a hundred steps before it terminates.
Particular values aside, the interesting question is whether we can prove that this sequence terminates for all positive values of
n. So far, no one has been able to prove it or disprove it!
Think carefully about what would be needed for a proof or disproof of the hypothesis “All positive integers will eventually converge to 1”. With fast computers we have been able to test every integer up to very large values, and so far, they all eventually end up at 1. But this doesn’t mean that there might not be some as-yet untested number which does not reduce to 1.
You’ll notice that if you don’t stop when you reach one, the sequence gets into its own loop: 1, 4, 2, 1, 4, 2, 1, 4, and so on. One possibility is that there might be other cycles that we just haven’t found.
Note 8.5.2. Choosing between
for loop if you know the maximum number of times that you’ll need to execute the body. For example, if you’re traversing a list of elements, or can formulate a suitable call to
range, then choose the
So any problem like “iterate this weather model run for 1000 cycles”, or “search this list of words”, “check all integers up to 10000 to see which are prime” suggest that a
for loop is best.
By contrast, if you are required to repeat some computation until some condition is met, as we did in this 3n + 1 problem, you’ll need a
As we noted before, the first case is called definite iteration — we have some definite bounds for what is needed. The latter case is called indefinite iteration — we are not sure how many iterations we’ll need — we cannot even establish an upper bound!
Check your understanding
- The 3n+1 sequence has not been proven to terminate for all values of n.
- It has not been disproven that the 3n+1 sequence will terminate for all values of n. In other words, there might be some value for n such that this sequence does not terminate. We just have not found it yet.
- No one knows.
- That this sequence terminates for all values of n has not been proven or disproven so no one knows whether the while loop will always terminate or not.
Consider the code that prints the 3n+1 sequence in ActiveCode box 6. Will the while loop in this code always terminate for any positive integer value of n?
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