Seven Jack O'Lanterns
Logic puzzles require you to think. You will have to be logical in your reasoning.
It appears that you have angered the spirit of Halloween by failing to revere the Great Pumpkin, and now a curse has befallen you. On the walkway to your house is a Ward of Seven Jack O'Lanterns arranged in a circle. If midnight comes and any of the seven are still lit, a dark reaper and seven dark horses with seven dark riders shall visit thy abode. They shall surround thy domicile and, while circling it, they will proceed to pelt thy dwelling with eggs and cream of shaving. And come morn there will be a great mess to be reckoned with. Verily. So you better get those lanterns out.
You quickly discover something odd about these lanterns. When you blow out the first one, the lanterns on either side extinguish as well! But there is more. If you blow out a lantern adjacent to one that is extinguished, the extinguished one(s) will relight. It seems that blowing on any lantern will change the state of three - the one you blew on and its two neighbors. Finally, you can blow on an extinguished lantern and it will relight, and its neighbors will light/extinguish as applicable. After trying once and finding all seven lit again, you decide, being the excellent puzzler, you sit down and examine this closer. But hurry, I hear the beating of many hooves...
Answer
If you examine the setup carefully, you'll note a number of facts which make the puzzle easier to solve by deduction. First, blowing on a lantern is a commutative property; blowing on lanterns 1, 5, then 3 is the same as blowing on 3, then 1, and then 5. No matter what order the lanterns are blown on, if the same lanterns are blown on the same number of times, the result won't change. For that reason, blowing on a lantern twice is as good as not blowing on it at all. And three times is as good as one time. So, it seems that it should be able to be done in seven steps or less.
What else can we tell about the solution? Since each operation changes the state of three lanterns, and there are 7 lanterns, and each lantern must change its state an odd number of times, it's a safe bet that there will need to be an odd number of steps. We can easily see it can't be done in 1 or 3 steps, so it must be 5 or 7. Trying 5 steps comes up with 3 different patterns that are not symmetrical and fail to leave all lanterns extinguished. So that leaves 7 steps and to your surprise, based on the commutative property, the easiest solution is to blow on each one in order! So doing this, the Great Pumpkin has decided to give you a treat for figuring this out and you find all seven lanterns full of candy the next morning! Congratulations!
Hide
Comments
rugbys_girl
Jan 26, 2005
| Very nice teaser. I have seen something similar done with a wall of bricks where the four bricks surrounding the first brick become recessed.....FUN!  |
kri_kri  
Jan 26, 2005
| This was hard to me because i have a short attention span and ummm what was i talking about?     |
alex_jane
Jan 27, 2005
| nice one but is a bit long  |
suzygirl
Jan 28, 2005
| This is one of my favorites. I didn't really look at it the way you did though, I just figured out what pattern they had to be in to extinguish them all, and then I got to that point. It is interesting though because it took me exactly seven steps. Nice one! |
waker
Jan 29, 2005
| Actually it can be done in seven steps
Imagine these are the seven (B=burning, O=Out)
Start: BBBBBBB
#1: OOOBBBB
#2: OOBOBBB
#3: OOBBOBB
#4: OOBBBOO
#5: OOOOOOO |
waker
Jan 29, 2005
| whups, I meant 5 steps |
waker
Jan 29, 2005
| again whups i did it wrong I thought i had checked it, but i guess i didn't can somebody delete these three posts? |
fishmed  
Jan 29, 2005
| I just requested the last 3 were removed for you. I am glad you guys liked it. |
2sexy4wrds
Jan 31, 2005
| u add way to much information... just tell me wat i need to know and get it over with... i get lost when there is too much info |
kellgo
Feb 01, 2005
| Very well written and entertaining.
Fabulous Job!!! |
fishmed
Feb 01, 2005
| Thanks. Glad you liked it. |
cnmne
Feb 02, 2005
| Nothing wrong with a good storyline. |
theketchupwins
Feb 09, 2005
| loved this one! i guess i'm one of few who likes the funny story line?
took me a little while to figure out. nice work.
|
theketchupwins
Feb 09, 2005
| loved this one! i guess i'm one of few who likes the funny story line?
took me a little while to figure out. nice work.
|
rose_rox
Feb 11, 2005
| Like Peanuts? |
fishmed
Feb 11, 2005
| Of course I do! |
short_stuff780
Feb 13, 2005
| great! long! but one of the bests  |
fishmed
Feb 13, 2005
| Thanks. |
juggleboy502
Feb 28, 2005
| too long... i got lost on the first paragraph  |
rotwyla98
May 10, 2005
| heh. I blew out the two corner ones and then the middle one  |
sueintexas
May 15, 2005
| I loved the story and the challenge |
fishmed
May 16, 2005
| rotwyla, they are in a circle, so there are no 'corner' ones. Nice thought, though.  |
JCDuncan
May 31, 2005
| I like this teaser, but I loved the answer. Explaining it in with the commutative power is wonderful. I did answer it, but with trial and error and I did not come close to understanding the mathmatical proof for solving. (actually the answer was the first attempt I made) |
Vigo95
Mar 25, 2006
| WHAT ???
yah ... thanks alot ! now i'm scared of jack-o-lanterns ! |
paul726
May 21, 2006
| Nicely done, Fish! Kudos! |
nu_rob_roy
Jul 29, 2006
| I'm really lost, who or what is Jack O'Lanterns? But seems like a well liked teasers. |
fishmed
Aug 11, 2006
| Jack O' Lanterns are hollowed-out pumpkins that have faces carved into them and then are lit from inside with a candle. Usually used for decorations during Halloween. |
geekgirljess
Feb 07, 2007
| Nice ripoff, this was originally posted at www.greylabyrinth.com back in 1998. |
fishmed
Feb 08, 2007
| I actually got it from my father. I am not sure where he got it from. |
LeafFan4life
Jun 05, 2007
| this one was cleverly nice |
HarryPutter
Jun 10, 2007
| nice one to the good teaser |
4demo
Jun 18, 2007
| Fun teaser but quite difficult! |
scallio
Jan 18, 2008
| This was funny, entertaining and unique... I loved it!
I cut 7 little squares from paper, drew a little lit candle on the one side and left the other blank then arranged them in a circle, lit side up. When I altered one, I flipped the two adjacent squares. Wasn't too difficult. Trick was to "blow out" all but one then blow out the two middles of the lit ones.
Anyway, a lot of fun! |
javaguru
Feb 28, 2009
| Excellent teaser! I don't generally like too much extraneous story, but I thought it was amusing enough and not too long, so it was fine.
I missed the commutive property, but came up with another line of reasoning to determine that seven steps were the minimum.
Each step toggles three candles. You need an odd multiple of seven toggles in order to toggle the seven candles off. The least common multiple between three and seven is 3 x 7 = 21, so seven steps is the minimum. Then I just guessed that you could blow them out in order and was mildly surprised when is worked!
|
(user deleted)
Feb 20, 2010
| It is easier than you describe. All lanterns need to be switched an odd number of times, so the total number of switches must fall in the sequence 7, 21, 35, 49 etc. but since 3 are switched with each blow the total must also be divisible by 3. The first such number in the sequence is 21, meaning 7 blows are required (7x3=21). By blowing a lantern once and the lanterns either side once as well the lantern in question ends up being blown out. It is clear from this that each must be blown once because all are blown once, and for each lantern both either side are blown once - meaning they all end up being out. |
princess2007
Feb 22, 2013
| I didn't really analyze very much. Just tried it out and got it on the first try with 7 blows. Reading the answer and finding out that the order of the blows didn't really matter made me laugh... Great teaser! |
Back to Top
| |
|
