If a donut shaped house has two doors to the outside and three doors to the inner courtyard then it's possible to end up back at your starting place by walking through all five doors of the house?
No, There are two ways to solve this problem the shorter and the longer one; The short uses the rule of even and odd from Discrete Math. Add the numbers of exits and enterings and if the total sum is even you gona end up on the same side that you started otherwise if it's odd then you end up on the other side!! Proven Mathematically!!! You can use this rule when you work with lines and dots… the longer explanation is to waste your time and go over it.....which of course can lead you to mistakes sometimes..... Start outside Go through door 1 into the house, go through door 2 into the courtyard. Leave the courtyard by door 3. You are now in the house and can either ..... ... go through the door 4 into the courtyard, but if you do there is no unused door to exit it by ... go through the door 4 to the outside, but if you do you have no unused door to re-enter the house to go through the last door into the courtyard. Start in the courtyard Go through door 1 into the house, leave the house by door 2 to the outside, Re-enter the house by door 3, leave by door 4 into the courtyard and... you are back in the courtyard and haven't used door 5 If you leave the house by door 2 into the courtyard, you must use door 3 to get back into the house and this leaves you with no unused door to re-enter the courtyard. Start in the house Go into the courtyard by door 1 and return by door 2. If you go outside and re-enter you can only use door 5 once and this will leave you in the courtyard.