def where(s,c):
    '''Given string s and single-character string c, return the
    position of the first occurrence of c in s.  If c is not in s,
    return -1.'''  

    # Do not use any string methods, but you can use functions in __builtins__

    if not c in s:
        return -1
    else:
        for i in range(len(s)):
            if s[i] == c:
                return i

def every_ith(l,i):
    '''Given list l and int i > 0, return a new list that contains every ith element
       of l, starting with index 0.'''

    # You can assume that i is > 0.  You don't need to perform error checking on that.
    
    new_l = []
    pos = 0
    while pos < len(l):
        new_l.append(l[pos])
        pos += i
    return new_l

def duplicates(l):
    '''Return True if list l contains adjacent elements with the same
    value, and False otherwise'''

    # Use a while loop rather than a for-loop 

    i = 0
    while i+1 < len(l):
        if l[i] == l[i+1]:
            return True
        i += 1
    return False    

print where("abcdeddw x",'d'), "sb", 3
print where("a",'a'), "sb", 0
print where("abcdeddw x",'x'), "sb", 9
print where("abcdeddw x",'z'), "sb", -1
print where("",'z'), "sb", -1 

print every_ith([10,11,12,13,14,15,16,17],3), "sb", [10,13,16]
print every_ith([10,11,12,13,14,15,16,17],10), "sb", [10]
print every_ith([],2), "sb", []
print every_ith([1,2,3],1), "sb", [1,2,3]

# You decide on a good set of test cases here for duplicates
print duplicates([0,1,2,3,"hello","hello",34.5]),"sb", True
print duplicates([0,1,2,3,"hello","hello"]),"sb", True
print duplicates([0,0,1,2,3,"hello","hello"]),"sb", True
print duplicates([0,0]),"sb", True
print duplicates([0]),"sb", False
print duplicates([]),"sb", False
print duplicates([0,1,2,3]),"sb", False