current community

your communities

more stack exchange communities

Stack Exchange
tour help
careers 2.0
Take the 2-minute tour ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 1 down vote favorite

I am a bit stuck.

I am trying to calculate the distance to move gears of different sizes so they align next to each other. I think its some sort of recursive call, but I can't figure out how to write it. I have a list of radii in the order of the gears

radiusList=[6,16,14,20,24,28]

so the 1st gear = movement distance = 0
2nd gear movement distance = 0 + 6 + 16
3rd gear=0 +6 + 16 +16 + 14
4th gear = 0 +6 + 16 + 16 + 14 + 14 + 20
5th gear = 0 +6 + 16 + 16 + 14 + 14 + 20, +20 ,+ 24
6th gear = 0 +6 + 16 + 16 + 14 + 14 + 20, +20 ,+ 24 + 24 +28 etc...

The other thing is I need it to update - now matter on the radius size, and the number of gears. But can't get my head around how to approach it. Any help would be greatly appreciated. Thank you.

UPDATE : Thanks everyone, I wrote something like this in the end. Sees a bit long winded though.

def createmovesequence():
if numberofcogs == 0 :
    void
if numberofcogs == 1:
    newmovelist.append(0)
if numberofcogs == 2:
    newmovelist.append(0)
    newmovelist.append(radiusList[0])
    newmovelist.append(radiusList[1])
if numberofcogs >= 3:
    newmovelist.append(0)
    newmovelist.append(radiusList[0])
    #newmovelist.append(radiusList[1])
    for i in range(2, len(radiusList)):
        newmovelist.append(radiusList[i-1])
        newmovelist.append(radiusList[i-1])
    newmovelist.append(radiusList[-1])

# elif numberofcogs != len(radiusList): # print 'error'
print newmovelist

createmovesequence()

My only other idea was something along the lines of a for loop with lot of if statements...

share|improve this question
add comment

3 Answers

up vote 0 down vote

This is an interesting problem. Here's a solution without recursion:

>>> gears = [6,16,14,20,24,28] 

>>> def distance(n, radius):
...    return sum(gears[:n] + gears[1:n-1])

>>> distance(1, gears) == 6
True

>>> distance(2, gears) == 6 + 16
True

>>> distance(3, gears) == 6 + 16 + 16 + 14
True

>>> distance(4, gears) == 6 + 16 + 16 + 14 + 14 + 20
True
share|improve this answer
add comment
up vote 0 down vote 
def dist(n, radiusList):
  if n <= 1:
    return 0
  return dist(n-1, radiusList) + radiusList[n-1] + radiusList[n-2]
share|improve this answer
    
Second call to dist should take 2 parameters. It's giving the wrong answer anyway, though. –  hcalves Apr 2 '13 at 20:12
    
Fixed. Silly me. –  Andreas Haferburg Apr 2 '13 at 22:14
add comment
up vote 0 down vote

The loop you probably wanted to see is:

gear = 4    # which gear
total = 0   # distance. we start from zero

# for each idx such that 0 <= idx < gear…
for idx in xrange(gear):
    total += radiusList[idx]
    total += radiusList[idx+1]

Note how the loop does nothing in case gear = 0 (because there's no such number that would be smaller than zero and yet at least equal to zero). This is probably the most explicit notation, but hcalves' one is shorter and learning it will be helpful for you too.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.