A simple fuse problem

Fuses: A, B1, B2

A = 10 minutes
B1, B2 = 30+x minutes (x may be negative)

T = 0
Light one end of the 10-minute fuse (A).
Light both ends of B1.

T = 10
Fuse A is burned up.
Fuse B1 has 10+x minutes left.
Fuse B2 has 30+x minutes left.
Extinguish one end of B1.
Light both ends of B2.

T=20+x
Fuse B1 has burned up.
Fuse B2 has (30+x)-2*(10+x)=10-x minutes left.
Extinguish one end of B2.

T=30
Fuse B2 has burned up.

This one was actually more difficult for me.

Hope you liked it

Note the riddle doesn't work if |x|>10 which is why I specified +- 5 min.

I enjoyed both of them. Thanks for sharing. Do you have others like them?

Yes, I do have another one - let me find it. I did an complete study on the 3 fuse burning problem, my notes are somewhere....This particular case stood out from the rest because it was the only one (and believe me there are a lot of possibilities) where if two of the fuses burnt for the same time their times cancelled each other. I just thought it was so neat I had to share it with someone and I'm glad I found someone who appreciated it - but to be honest with you I never expected you to find the answer so quickly.

To be honest with you, I had convinced myself that there was no solution. So, I wrote a program to compute all possible time intervals that could be measured under the given conditions.

I made this up. Hopefully it isn't too easy.

You are in a locked room with a switch, ample matches, and three fuses (non-uniform burn rates) that burn for exactly 8, 24, and 40 minutes.

To unlock the door, you must throw the switch for exactly 19 minutes.

You may start your process at 12:00. How soon can you unlock the door?

fuse A = 8min; B= 24min; C=40min

light both ends of A, one end of B, one end of C.

T= 4min
A is gone
B has 20min left
C has 36min left
light other end of C
turn switch ON

T= 22min
B has 2min left
C is gone
light other end of B

T=23min
B is gone
turn switch OFF

switch has been on 23-4= 19 minutes

u can ge out after23 minutes :-)

it took me a while :-/ ....dont have a program haha so i did it by trial and error

Thoh did you hide the answer?

Mark - 23 minutes. Good one!! Should I post the answer? Seems like thoh has hidden his somehow - don't want to spoil the fun for everyone else. How do you do it with a computer program - excel? I have scraps of paper eveywhere - I use the computer a lot but still like to use the old fashion pen and paper.

Here is another one.

3 fuses - X burns for twice as long as Y.
Y burns for more than twice as long as Z
Z burns for 20 minutes.

Measure a time of exactly 60 minutes.

How to edit the title? - All fuse riddles here.

kurma, i just made my text color white so i dont spoil the answer for others...picked it up from mark

still cant figure that one cuz i cant get my times to cancel


you have two hour-glasses. one measures 3 min and the other 7 min. you need to cook an egg for exactly 5 min. how?

[size=8]HOUR GLASSES
My notation is top/bottom.

7/0, 3/0
4/3, 0/3 flip 3 to get 4/3, 3/0
1/6, 0/3 flip 3 to get 1/6, 3/0
0/7, 2/1 flip 7 to get 7/0, 2/1
5/2, 0/3 put egg in pan
0/7, 0/3 remove egg from pan[/size]

so i see u can cook my 5min egg in 12 minutes.
im too hungry for that. i want my egg in 8 minutes.

does it have to be 60 minutes in one chunk? b/c i have:

[size=7]
20min = Z; (40 + y)min= Y; (80+2y)min = X

T=0
light one end of Z
light both ends of Y
start timing

T=20
Z is gone
Y has y min left
stop timing
extinguish one end of Y
light both ends of X

T=20+y
Y is gone
X has 80min left
start timing

T=60+y
X is gone
stop timing

timing for a total of 20 + 40 min = 60min
[/size]

Forgot to mention that you both got my problem right.

[size=8]HOUR GLASSES
T=0 7/0, 3/0
T=3 4/3, 0/3 flip 3 to get 4/3, 3/0 put egg in pan
T=6 1/6, 0/3 flip 3 to get 1/6, 3/0
T=7 0/7, 2/1 flip 3 to get 0/7, 1/2
T=8 0/7, 0/3 remove egg from pan

It's funny how much easier some problems are when you know what the answer needs to be.[/size]

[size=8]20, 40+x, 80+2x

T=0
20, 40+x, 80+2x
Light one end of 20, both ends of 40+x, and one end of 80+2x

T=20
0, x, 60+2x
Extinguish one end of 40+x, and light the other end of 80+2x

T=20+x
0, 0, 60
Extinguish one end of 80+2x
Start 60 minute timer

T=80+x
0, 0, 0
End 60 minute timer

I did this one by hand - no computer.[/size]

To answer the previous question about how I'm using a computer, I'm not using a spreadsheet. I wrote a program to calculate all possible elapsed times by recursively trying every combination of lighting 0, 1, or 2 ends of each fuse. At least one fuse is eliminated after each round; so the search isn't very deep.

Also, it was nice to see an hourglass problem. They offer a bit of a different twist than the fuse problems. Water pouring problems are also similar in nature to these.

oh so thats how that fuse one is done....

and i have my egg

Fuses & Hourglasses
After doing these problems, I thought it would be cool to combine fuses and hourglasses. I've modified my program to deal with both items. The number of states rises dramatically, and there are typically many ways to measure a certain period of time.

Here are a couple of problems. I hope they're not too easy. I'll continue to search for a worthy one.

In both cases, the goal is to measure 11.75 minutes from the time you start.

#1
Two fuses: 8 and 4 minutes
Two hourglasses: 5 and 3 minutes

#2
Two fuses: 5 and 3 minutes
Two hourglasses: 8 and 4 minutes

CASE 1:

8 min fuse = A; 4 min fuse = B; 5 min hr-glass = C; 3 min hr-glass= D

T=0
light one end of A
flip over C
flip over D

T=3min
A has 5 min left
C has 2 min left
D is empty
flip over D

T=5min
A has 3 min left
C is empty
D has 1min left (and 2min on bottom)
light other end of A
light one end of B
lay D on its side

T=6.5min
A is gone
B has 2.5min left
light other end of B

T=7.75min
B is gone
flip over C (has 5min)
flip D upright so it has 2min left

T=9.75min
C has 3 min left
D is empty
flip C over so it has 2min left

T=11.75
C is empty


CASE 2:

5 min fuse = A; 3 min fuse = B; 8 min hr-glass = C; 4 min hr-glass= D

T=0
light both ends of A
light one end of B
flip over C
flip over D

T=2.5min
A is gone
B has .5min left
C has 5.5min left
D has 1.5min left
light other end of B

T=2.75min
B is gone
C has 5.25min left
D has 1.25min left
lay D on its side

T=8min
C is gone
flip over C (has 8min)
flip D upright so has 1.25min

T=9.25min
C has 6.75min
D is gone
flip over C (has 1.25min)
flip over D (has 4min)

T=10.5min
C is gone
D has 2.75min left and 1.25 in bottom
flip D over

T=11.75min
D is gone

Very good! However, what if you were required to do both problems in four steps?

T=0 do something
T=x do something
T=y do something
T=z do something
T=11.75 done