打印出python中变量之间的关系列表？

Question

I'm an absolute beginner with python and my professor is sort of doing a trial of fire approach to teaching us the language. The goal is to list the triples in the relation {(a,b,c) : a, b, and c are integers with 0 < a < b < c < 5}. Use range(5) as the source for values of a, b, and c.

I think i understand the general concept of what it is, i set the values of a,b,c fine, but I don't quite get how I can print out variables like that? I was thinking of using some sort of while loop of like while a < b < c print out (a,b,c) or something but I can't get it to work with the syntax.

Thanks so much for the help!

Answer 1

So you need to vary each of a , b , and c in the range 0-4 (which is what range(5) produces). For this you can use three nested for loops.

for a in range(5):
    for b in range(5):
        for c in range(5):

This will ensure that for each possible value of a , all possible values of b are tried, and similarly, for each possible value of b , all possible values of c are tried. In other words, the innermost (bottom-most, as they're written) loop c will go through 0-4 for each value of b , which is also going through 0-4 for each value of a , which is also going through 0-4.

Now inside these loops, you just need to check to see if the criterion your prof has given is true, and if so, print the numbers. Fortunately, the syntax your prof has given works fine in Python:

if 0 < a < b < c < 5:
    print(a, b, c)

Putting it all together, then, we have:

for a in range(5):
    for b in range(5):
        for c in range(5):
            if 0 < a < b < c < 5:
                print(a, b, c)

Now, this works fine, and will undoubtedly satisfy your professor, but there are some simple improvements you can make to this program. First, the variables each go from 0-4, but the value 0 is never going to pass the if test, because it will never be less than 0! So why even try it? You could save a bit of work by writing each range as range(1, 5) .

But... there's also the fact that if a is 1, b can never be 1 (because a < b must be true, and if a is 2, b can likewise never be 2, and so on. And also, from the other end of the range, if c is 4, b can never be 4 and satisfy the test. We can write the ranges accordingly, so that the range for b starts 1 past a and leaves off one before c , so that those combinations of values need never be tested.

Finally, your if statement can also be simplified: because of the ranges being used, you don't need to make sure 0 is less than a ; that will always be true because a starts at 1. At the same time, you know that c will always be less than 5 because the range() makes it that way, so you can remove that test too.

So you will get identical results, while doing less work, by writing it like this:

for a in range(1, 3):
    for b in range(a+1, 4):
        for c in range(b+1, 5):
            if a < b < c:
                print(a, b, c)

In this case, the extra work is not that important... both versions are so fast you will never notice the extra time the first approach took. But it's always worth thinking about how a problem can be rethought to do less work; sometimes it's the difference between a usable program and an unusable one.

In this case, the initial version goes through 125 tests. The second, more efficient version? Four. Which happens to be exactly the same as the number of correct results!

So we can take out the if statement entirely, and just print every combination, because the loops we've written make sure that only numbers that pass the test are used to begin with.

And we've done less than 1/30th of the work in the process!

Answer 2

How about this?

for a in range(5):
    for b in range(5):
        for c in range(5):
            if 0 < a < b < c < 5:
                print(a, b, c)

This will iterate over the values 0 till 4 for a , b and c and it'll print them when the condition 0 < a < b < c < 5 holds.