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I am quite new to programming, and was wondering if someone might be able to help me get the following working in python:

Define $\Phi_mx(mn+r)=mx(n)+\frac{r}{m}(x(n+1)-x(n))$, where we consider x(0)=0, r is always positive and less than m, and all values of n are positive integers. $\Phi_m$ takes in a string of numbers and outputs another string. How might I program this?

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Programmers.SE doesn't accept that type of markup. Could you rewrite it... and consider taking a stab at writing some pseudocode for it? –  MichaelT Jul 4 '13 at 23:10
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Looks like a homework problem to me. –  vy32 Jul 4 '13 at 23:37
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@vy32: homework-related problems are fine as long as the person asking the question is trying to master programming, and not get the homework done for him. –  9000 Jul 4 '13 at 23:39
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It actually isn't homework related, I'm a math student trying to automate this to make some other work using it faster. –  user95751 Jul 4 '13 at 23:49
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Could you take another crack at trying to explain it and walk us through what it is doing? THere are quite a few values there that appear to be interrelated. Also please note that none of the stack exchanges are a code writing service, if you are looking for the code you might want to try writing some first and then ask about the problems you are having in the code. –  MichaelT Jul 5 '13 at 0:12
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closed as off-topic by Glenn Nelson, MichaelT, GlenH7, BЈовић, Dynamic Jul 5 '13 at 13:44

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1 Answer

up vote 2 down vote accepted

OK, some quick intro for you.

To store ready-made finite sequences, use lists:

x = [0, 1, 2, 2, 3, 3, 3, 3]

This lets you easily address each sequence element by index in brackets, 0 being the first index, so x[0] == 0 and x[4] == 3.

Your function Ф could be defined as

def F(n, m, r, x):
  return m * x[n] + (r / m) * (x[n + 1] - x[n])

So F(3, 2.0, 1.5, x) == 4.75.

To see the function applied to the entire list x, you can do something like this:

# we use len() - 1 because len(x) is the number of elements in x,
# and the last valid k for x[k] is len(x) - 1;
# F uses x[n+1], so n must not exceed len(x) - 2.
# But range(n) gives us numbers 0..n-1, so the last n we generate
# is (len(x) - 1) - 1, exactly what we need. 
for n in range(len(x) - 1):
  print n, F(n, 2.0, 1.5, x)

Now go read Python in 5 minutes and then Dive Into Python to understand things better. Though a math student might find Haskell more compelling, if not as easy :)

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I see what you mean. Is there a way though for Python to read the whole list at once and give separate output for each integer? Or even better (though I should read up on it for this one I would guess) a variety of m's? –  user95751 Jul 5 '13 at 0:30
 
I actually have several math friends that like Prolog –  jozefg Jul 5 '13 at 0:48
 
See the updated answer. –  9000 Jul 5 '13 at 0:48
 
@jozefg: Prolog is nice, but usually it's a key from a different door than this. Einstein's houses puzzle is a good fit for Prolog. –  9000 Jul 5 '13 at 0:50
 
@9000 Thanks, that makes good sense. The one last hurdle I'll have to overcome is varying r, since r should in some sense depend on a value a. That is, we seek x(a)=x(mn+r), so for example x(5) = x(2x2+1). This gives us n and r, but we only actually knew a and any previously calculated terms. –  user95751 Jul 5 '13 at 4:21
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