#Medium boolean logic puzzles -- if else and or not

#We want to make a row of bricks that is goal inches long. We have a number of small bricks (1 inch each) and big
# bricks (5 inches each). Return True if it is possible to make the goal by choosing from the given bricks.
# This is a little harder than it looks and can be done without any loops. See also: Introduction to MakeBricks
def make_bricks(small, big, goal):
  return (goal%5)<=small and (goal-(big*5))<=small

#Given 3 int values, a b c, return their sum. However, if one of the values is the same as another of the values,
# it does not count towards the sum.
def lone_sum(a, b, c):
  sum = a+b+c
  if a==b==c:
    return 0
  elif a==b:
    return c
  elif a==c:
    return b
  elif b==c:
    return a
  else:
    return sum

#Given 3 int values, a b c, return their sum. However, if one of the values is 13 then it does not count towards
# the sum and values to its right do not count. So for example, if b is 13, then both b and c do not count.
def lucky_sum(a, b, c):
  if a==13:
    return 0
  elif b==13:
    return a
  elif c==13:
    return a+b
  else:
    return a+b+c

#Given 3 int values, a b c, return their sum. However, if any of the values is a teen -- in the range 13..19 inclusive
# -- then that value counts as 0, except 15 and 16 do not count as a teens. Write a separate helper
# "def fix_teen(n):"that takes in an int value and returns that value fixed for the teen rule. In this way,
# you avoid repeating the teen code 3 times (i.e. "decomposition"). Define the helper below and at the same indent
# level as the main no_teen_sum().
def no_teen_sum(a, b, c):
    def fix_teen(n):
        if n in [13, 14, 17, 18, 19]:
            return 0
        else:
            return n

    return fix_teen(a) + fix_teen(b) + fix_teen(c)

#For this problem, we'll round an int value up to the next multiple of 10 if its rightmost digit is 5 or more,
# so 15 rounds up to 20. Alternately, round down to the previous multiple of 10 if its rightmost digit is less than 5,
# so 12 rounds down to 10. Given 3 ints, a b c, return the sum of their rounded values. To avoid code repetition,
# write a separate helper "def round10(num):" and call it 3 times. Write the helper entirely below and at the same
# indent level as round_sum().
def round_sum(a, b, c):
    return round10(a) + round10(b) + round10(c)


def round10(num):
    if num % 10 < 5:
        return num - (num % 10)
    else:
        return num + (10 - num % 10)

#Given three ints, a b c, return True if one of b or c is "close" (differing from a by at most 1), while the other is
# "far", differing from both other values by 2 or more. Note: abs(num) computes the absolute value of a number.
def close_far(a, b, c):
  if abs(a-b)<=1:
    return(abs(a-c)>=2 and abs(c-b)>=2)
  elif abs(a-c)<=1:
    return(abs(a-b)>=2 and abs(c-b)>=2)

#We want make a package of goal kilos of chocolate. We have small bars (1 kilo each) and big bars (5 kilos each).
# Return the number of small bars to use, assuming we always use big bars before small bars. Return -1
# if it can't be done.
def make_chocolate(small, big, goal):
    if goal >= 5 * big:
        remainder = goal - 5 * big
    else:
        remainder = goal % 5

    if remainder <= small:
        return remainder

    return -1

#In6days






