<div dir="ltr"><font size="4">Hey thanks for the great feedback guys. It has been a really crazy busy day and I haven't had a chance to carefully go over all the replies. Len, your comparison to the BigInt algorithm is interesting. I wrote that program last semester in Data Structures. You know, you've got two arrays that represent giant int values greater than 2**31 - 1, and then you use a carry variable whose initial value is 0 and then can take on values of 0 or 1 depending on the two integers being added. I can see the resemblance to what I'm trying to do here, although the problem is a tad bit different. I'll think about it!</font><div><font size="4"><br></font></div><div><font size="4">On a totally different subject, I've been messing around with Ocaml lately. Found some great YouTube tutorial videos about Ocaml and bought a couple books on it. Very interesting. Is there any connection between Python and Ocaml? Just wondering.</font></div><div><font size="4"><br></font></div><div><font size="4">Thanks guys! Your feedback is a huge help to me in my studies. Not sure if I'll ever be a big league developer/programmer, but I've learned a lot from your posts and replies. Thank you so much.</font></div><div><font size="4"><br></font></div><div><font size="4">Have a great weekend even if it's like monsoon season here in Chicago. Yuck, I hate all this rain!</font></div><div><font size="4"><br></font></div><div><font size="4">Doug.</font></div><div><font size="4"><br></font></div></div><div class="gmail_extra"><br><div class="gmail_quote">On Thu, Jul 16, 2015 at 11:17 AM, Len Wanger <span dir="ltr"><<a href="mailto:len_wanger@hotmail.com" target="_blank">len_wanger@hotmail.com</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<div><div dir="ltr">Douglas,<br><br>There are a lot of ways to do this, but since it's your homework I'll give you hints but not the code. One easy way to look at this is like implementing addition (with carrying digits) or an odometer. Create a vector/list of N digits, then loop over that list adding 1 each iteration by adding 1 to the least significant digit. If the digit set exceeds the high range for the digit (exceeds 4 in your example), then setting it to the low digit (in your example 1) and carry it the next least significant digit (i.e. add 1 to the next digit to the left). When you get to the maximum value (i.e. in your example 4 in each digit) stop.<br><br>A couple of notes:<br><br>1) I would avoid recursion here. This is an example of simple tail recursion, and can easily be unrolled into a loop. In a language with a good optimizer you could right this recursively and let the compiler take care of it, but in Python it is very inefficient. <br><br>2) This is a great place to use a generator. The list can get very large (e.g. 4 choices per digit and 10 digits is a list with 4**10 entries, which is > 1 million). A yield statement in the right place in the loop will make it a generator which would be much more efficient.<br><br>3) The previous posting that this is like a conversion to a different number base is not quite right. If you had 0 for a digit value that would be a nice way to look at it (then you could loop to base**num_digits and do a conversion to that base for each value). Since your example used digits "1" to "4" that's not quite right. <br><br>Good luck,<br><br>Len<br><br><div>> Date: Wed, 15 Jul 2015 17:15:23 -0500<br>> From: "Lewit, Douglas" <<a href="mailto:d-lewit@neiu.edu" target="_blank">d-lewit@neiu.edu</a>><br>> To: The Chicago Python Users Group <<a href="mailto:chicago@python.org" target="_blank">chicago@python.org</a>><br>> Subject: [Chicago] Printing out the Cartesian product.<span class=""><br>> <br>> Hi everyone,<br>> <br>> I need some advice on how to do something. Let's say that I want to print<br>> out the Cartesian product of the integers 1 to 4. It would look something<br>> like this:<br>> <br>> 1, 1, 1, 1<br>> 1, 1, 1, 2<br>> 1, 1, 1, 3<br>> 1, 1, 1, 4<br>> 1, 1, 2, 1<br>> 1, 1, 2, 2<br>> 1, 1, 2, 3<br>> 1, 1, 2, 4<br>> 1, 1, 3, 1<br>> 1, 1, 3, 2<br>> 1, 1, 3, 3, etc, etc until finally we have<br>> ..............<br>> 4, 4, 4, 3<br>> 4, 4, 4, 4<br>> <br>> This is NOT a permutation because repetition is allowed. Nor is it a<br>> combination because for example 1, 1, 1, 2 is distinct from 2, 1, 1, 1. I<br>> believe this is technically called a Cartesian product.<br>> <br>> I know of two ways to do this:<br>> <br>> Method #1:<br>> <br></span>> *for a in range(1, 5):*<br>> * for b in range(1, 5):*<br>> * for c in range(1, 5):*<br>> * for d in range(1, 5):*<br>> * print( a, end = ", " )*<br>> * print( b, end = ", " )*<br>> * print( c, end = ", " )*<br>> * print( d, end = "\n" )*<span class=""><br>> <br>> <br>> The above works just fine! But there's a problem. What if let's say I<br>> want the Cartesian product of all the integers from 1 to 10? That means<br>> working with 10 nested for-loops! That's insane! It's tedious to write<br>> and probably even worse for the person who is reading it. So this method<br>> is limited to smaller examples.<br>> <br>> Method #2:<br>> <br>> Cheat by using Python's itertools package! (Or is it a library? What's<br>> the difference between a package and a library?)<br>> <br>> import itertools<br>> <br>> carProduct = itertools.product([1, 2, 3, 4], repeat = 4)<br>> <br></span>> *try:*<br>> * while True:*<br>> * print( next(carProduct) )*<br>> *except StopIteration:*<br>> * pass*<span class=""><br>> <br>> This method is short and sweet! It works great! But it's really cheating<br>> in the sense that the programmer is taking advantage of code that has<br>> already been written. (I have nothing against doing that! But in a CS<br>> class if I write down that answer on an exam do you think the professor is<br>> going to give me any credit? Probably not!!! LOL! )<br>> <br>> Is there some other way? I'm guessing that I would need a for-loop that<br>> contains a recursive function. Or.... maybe a recursive function that<br>> contains a for-loop? I really don't know. I've been struggling with this<br>> for a couple days now and I can't think of a solution. Can someone please<br>> enlighten me?<br>> <br>> Gratefully,<br>> <br>> Douglas.<br><br></span></div> </div></div>
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