Help with arrays

Mart. mdekauwe at
Wed Aug 26 15:34:59 CEST 2009

On Aug 26, 3:02 am, Dave Angel <da... at> wrote:
> Stephen Fairchild wrote:
> > Philip Semanchuk wrote:
> >> On Aug 25, 2009, at 6:14 PM, Gleb Belov wrote:
> >>> Hello! I'm working on an exercise wherein I have to write a Guess The
> >>> Number game, but it's the computer who's guessing MY number. I can get
> >>> it to work, but there's one obvious problem: the computer generates
> >>> random numbers until one of them corresponds to my number, but it will
> >>> often generate one number (eg. 4) numerous times, meaning it doesn't
> >>> know that this number is invalid. What I mean is, it will sometimes
> >>> use 37 tries to guess a number out of 1 - 9, which makes no sense,
> >>> since it should only take 9 tries, at most. I was trying to find a way
> >>> to make a dynamic list of all the numbers the computer generates in
> >>> the loop and then make it re-generate the number if the previous
> >>> number is present in the list, so it doesn't keep on generating 4 (as
> >>> an example). I don't know if that makes sense... Basically, we humans
> >>> know that once something is incorrect, there's no point in trying to
> >>> use it as the answer next time, because we already know it's
> >>> incorrect. How do I go about coding this in Python? I'm still quite
> >>> new to the language so any help will be appreciated...
> >> One cheap way to do it (not necessarily efficient) is to make a list
> >> of your possible guesses (e.g. range(1,10)), use random.shuffle() to
> >> put them in random order and then run through the guesses one at a time.
> > import random
> > import time
> > l = range(1, 10)
> > while l:
> >     print l.pop(random.randint(0, len(l) - 1))
> >     time.sleep(2)
> While both of these will work well, I'd point out that a direct
> translation of your question is to use a set.  Define an empty set, and
> each time you try a number unsuccessfully, add it to the set.  Then just use
>        while x in myset:
>                x = newguess()
> to find the next guess.  This approach isn't as efficient, but it's a
> useful paradigm to understand.
> A separate question is whether the human is supposed to tell the
> computer whether the guess is high or low.  If so, you can eliminate
> many numbers on each guess.   For example, suppose the solution is 8.  A
> guess of 5 would say "too low."  Then you'd cross off now only 5 but 1-4
> as well.
> With this change the best solution changes from a random shuffle to a
> binary search.
> DaveA

That's a good point if you can define whether the computer has guessed
too low/high, then you could use the previous guess to set a bounds
and rescale the next random guess.

min + (random_number * max)

although I think random.randomint does this for you!

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