1(a)  m = len(s)/2
      s[:m]==s[m:]

1(b) One character in s[1:] is different from s[0], but the rest are the same as s[0]

1(c) 'axbxxc'

2 (a) s1 = 2+4 = 6
      s2 = 1+3 = 4
      so s2-s1 is -2
   
      s1 = 2+4+...+ 98
      s2 = 1+3+   + 97 + 99
      so s2 - s1 is 99-49 = 50

(b) for k in rand(21):
       print 5*k

(c) t is a string made up of those characters in s that appear only pnce

3(a) (s.count('N')==s.count('S')) and (s.count('E')== s.count('W'))

 (b) count=0
     L = len(s)
     for k in range(L-4):
       if isCycle(s[k:k+4]):
          count +=1
     return count

4(b) (2*3)/2 evaluates to 3
      2*(3/2) evaluates to 2

 (c)   1  10   0
      10   1  12
       2  10  22
       1  12   0

  (d) u is a local variable. A variable that a function may use to "do its thing"
      x and y are parameters. How the "input" is provided to the function.

5.    # Draw a circle with radius r and center at (x,0)
      x = x + r + .75*r # The new x
      r = r*.75   # The new r

6. if n<=9;
      return '00' + str(s)
   elif n<=99:
      return '0' + str(n)
   else
      return str(n)