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def gap_binsearch(l, o=0):
    """binary search for a gap in a sequence

    takes a list l and an offset for start in the original list o"""
    #protect against zero element list
    if(len(l) == 0):
        return 1
    #if we recurse to a one element list, we have searched to the right or left edge of the list, and must return 1 or l[0]+1
    if(len(l) == 1):
        if(o != 0):
            return l[0] +1
        else:
            return 1
    mid = len(l)/2
    left = mid-1
    #if the left side contains a gap, we must recurse until the right side contains it or no items are left
    if(l[left] != o + (left + 1)):
        return gap_binsearch(l[:mid],o)
    #if both the left and right side have no gap, recurse on the right
    elif(l[mid] == o + (mid + 1)):
        return gap_binsearch(l[mid:],o+mid)
    #if the gap exists to the right of the midpoint, and not to the left, we have found our lowest gap
    else:
        return l[left]+1