In review, my brothers, David and Michael, finally came out to visit me in California before they both started their respective school years - Hah, suckers! They hijacked my car during the day to do fun touristy things while I worked, such as visiting San Francisco (to take a romantic tour of Alcatraz and watch the Cubs beat the SF Giants), Berkeley (to visit Jeff Noh and Top Dog), and Santa Cruz (to check out the beach and local offerings of "green trees" and "wacky-bo-jacky").
Over the weekend, we all headed up to South Lake Tahoe to do some climbing and camping. I'm now convinced that when any group is biased with males, conversation will tend toward topics related to farting, pooping, and beer. When it didn't (and oh how I savored these brief moments), we shared bad jokes and silly riddles to pass the time. And as a reward for reading this far, I'll share two simple riddles with you. First to answer will probably be rewarded with a life of fame, glory, and riches beyond your wildest expectations. Probably.
1. How bright are you?
You are positioned in front of a closed door and three light switches, each of which is wired to a light bulb in a room behind the closed door. Your task is to determine which light switch controls which light bulb, but there is a catch! You can position the switches in any arrangement you want prior to opening the door, but once you open the door to peer into the room and look at the light bulbs, you can no longer adjust the light switches and must determine the wirings. How do you do it?
2. A darker riddle.
A man dressed all in black is driving down a black paved road in his black car. There are no street lights on and he has neither his headlights on or any lights turned on in his car. Out of nowhere, a black cat jolts out in front of his car and he slams on the breaks just in time. How did the man see the cat?
5 comments:
Spoiler warning.
#1
Assuming you know which direction for the switches is on, and we're dealing with incandescent bulbs:
Turn two of the switches on. Leave the switches on for several minutes. Turn off one of the two switches that you've turned on. Open the door. The light that is still on is linked to the on switch. Feel the two bulbs that are off. The one that is warm is tied to the switch that you turned on then off. The cold bulb is linked to the switch you never flipped.
#2
I laughed when I saw this, I remember this one from third grade. It's day time and the sun in out.
Ok, now I'll have to come up with some riddles to post on my blog.
Ryan wins!! I'll have to think of harder ones now.
Nice pictures. Here's a somewhat harder riddle.
A psychotic prison warden decides to play a game with his prisoners. There's a room in the prison with nothing but a binary switch (either on or off). The switch itself has no functionality other than being in the on or off state.
The game goes as follows. In each step of the game, the warden will lead a prisoner into the room. The prisoner is then allowed to modify the switch if desired and then led out.
While in the room, the current prisoner can also declare that all the prisoners have visited the room. If all the prisoners have indeed visited the room, then they are all set free. Otherwise, they all die.
The prisoners are allowed to discuss a strategy beforehand. But once the game starts, there is no discussion. Devise a strategy such that the prisoners are guaranteed to be set free.
Some other conditions:
- The warden can lead the prisoners in any order. So for example, the same prisoner could be led in the room multiple times before the another prisoner is taken.
- As long as the game keeps going, for any finite K, each prisoner is guaranteed to be led into the room K times in a finite number of game steps.
- The prisoners cannot use the passage of time to calculate how many game steps have elapsed.
- The switch is initially in the off state at the beginning of the game.
- The switch can only be in an on or off state, it has no other functionality.
- The prisoners are not allowed to modify the room in any other way.
First stab at it, which I'm not sure is the most efficient strategy:
Assume there are n prisoners (which the prisoners can determine the value of during their strategy session). They nominate one prisoner to be the captain (Captain Crazy?), who will have the responsibility of telling the warden when each prisoner has entered the room.
For each of the n-1 other prisoners, if they enter the room and the switch is on, they leave the room and switch untouched. If they enter the room and the switch is off AND they have never turned the switch on, they turn the switch on. Once a prisoner has turned the switch on, he never does it again and always leaves the room with the switch untouched.
The captain's task is to keep count of how many prisoners have entered the room. Whenever he enters and the light is on, he increases his count by one and switches the light off. If the switch is off, the captain leaves the room untouched. Once the captain counts to n-1, he can safely declare all prisoners have entered the room (since he is the nth prisoner and he necessarily has entered the room).
How's that?
Correct!
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