ZENO'S COFFEEHOUSE Results #7
The Seventh Coffeehouse Challenge
Here's the original problem, followed by the results. Enjoy!
CHALLENGE: CHARLES' TIME PROBLEM
Charles has been frustrated at every turn, and a waiter suggested the following, succinct, problem for Maggie (and for himself!):
Using
standard hourglass-type measuring devices, what is the quickest way to measure nine minutes with a four-minute glass and a seven-minute glass?
RESULTS
Here's one by a former student, Roy Kilgard, who's listed below among the entries:
From: Roy Kilgard
Come now, Barnette, this one's easy!!!
To time nine minutes with a four minute and seven minute timer:
Start both timers simultaneously.
When the four minute runs out, start it again.
When the seven minute runs out, begin timing nine minute interval.
Let the remaining minute fall through the four minute and run it twice
more.
7-4=3
4-3=1
1+4+4=9
Simple enough.
-Roy
THANKS TO ALL THE ENTRANTS! KEEP IT UP!
From: Danny Sharpe
Hi, Ron.
You probably don't remember me but I took a couple of courses from you
back around 1980 +/-. Found your
departmental page -- it looks nice -- and thought I'd try the puzzle.
If you start both hourglasses simultaneously at time 0 and flip each one
promptly when it finishes, you can
measure off any number of minutes between 0 and 10 (and beyond; I
didn't enumerate them past 10). The time
line below shows when each hourglass gets flipped. Using time 0 as your
initial start time, compute the
elapsed time between the current start time and each flip of either hour
glass, until either you find an
elapsed time of 9 minutes (the solution) or you find an elapsed time
greater than 9 minutes, in which case
pick the next higher available start time and try again.
Counting time 0 as a flip then it is 9 minutes from the second flip of the 7-
minute hourglass at time 7
to the fifth flip of the 4-minute hourglass at time 16. The total time it took
to do this was 16 minutes,
which should be minimal unless there's some trick to it I didn't see.
Time Flips Start time End time Elapsed time
0 4,7 0 0 0
1 | 0 4 4
2 | 0 7 7
3 | 0 8 8
4 4 0 12 12 (Bigger than 9 -- start
over)
5 | 4 4 0
6 | 4 7 3
7 7 4 8 4
8 4 4 12 8
9 | 4 14 10 (Bigger than 9 -- start
over)
10 | 7 7 0
11 | 7 8 1
12 4 7 12 5
13 | 7 14 7
14 7 7 16 9 (Bingo!)
15 |
16 4
17 |
18 |
19 |
20 4
21 7
22 |
23 |
24 4
25 |
26 |
27 |
28 4,7
From: Karl Hahn
First, there is no "quickest way" to measure nine minute. By
definition, if you measure nine minutes, it will take you exactly
nine minutes. There is no quicker or slower way than that.
Now to the solution. I had hoped there would be a trick, like
having to lay one of the glasses on its side for some period, or
something like that, but no -- the answer is very straightforward.
Assume both glasses have start with all the sand in the bottom.
1) At time of 0 minutes, invert both glasses.
2) At time of 4 minutes the 4 minute glass has run out. The
7 minute glass still has 3 minutes left to run. Invert the 4 minute
glass.
3) At time of 7 minutes the 7 minute glass runs out. The 4 minute
glass still has 1 minute to run. Invert the 7 minute glass.
4) At time of 8 minutes, the 4 minute glass runs out again. The
7 minute glass has 6 minutes to run, but has exactly 1 minute of
sand accumulated in the bottom. Invert the 7 minute glass.
5) At time of 9 minutes, the 7 minute glass runs out.
Now, if what you meant by the "quickest way" is, in fact, the
method with the fewest inversions, then logic tells me that my
solution is that one. Reasoning is as follows. At each step you
have a choice of inverting either of the glasses, both of the glasses,
or doing nothing, for a total of 4 choices.
Here is the decision tree.
If at step 1 I invert only the 7 minute glass, then at 7 minutes
I am left with a 4 minute timer and a 7 minute timer, giving me
totals of 11 or 14 minutes. So that is out.
If at step 1 I only invert the 4 minute glass, then at 4 minutes
I can invert either or both of the glasses. If I invert only
the 7 minute glass, nothing happens until 11 minutes, so that
is out. If I invert only the 4 minute glass, then at 8 minutes
I have a 4 minute timer and a 7 minute timer, for totals of 12 and
15 minutes, so that's out.
That leaves only inverting both at step 1. If I do that, then
when the 4 minute glass is done, I can:
Do nothing. No measurement is them made after 4 minutes, so that's
no good.
Invert the 7 minute glass only. It will take 4 minutes to run out,
so at 8 minutes I will be left with a 4 minute timer and a 7 minute
timer for totals again of 12 and 15 minutes. No good.
Or at step 2 I can invert the both glasses. Same result.
Or at step 2 I can invert the 4 minute glass only. By process of
elimination, that is what I must do.
The next glass to run out will be the 7 minute glass at 7 minutes.
At that point (step 3) I can:
Do nothing. The 4 minute glass runs out at 8 minutes, leaving me
again with a 4 minute timer and a 7 minute timer for totals of
12 and 15 minutes. No good.
Invert the only the 4 minute glass. It will then run out at 10
minutes, which is no good.
Invert both glasses. Again, nothing will happen until 10 minutes,
which is no good.
By elimination again we have that we must invert only the 7 minute
glass. The next thing to happen will be that the 4 minute glass
runs out at 8 minutes. At that point, there is 1 minute of sand
in the bottom of the 7 minute glass, and the path to the solution
is obvious.
Since at every point I eliminated all possibilbities except the
one I ended up taking, the path I took must therefore be the
path of minimal inversions.
Take care,
Karl
From: "William R. Wagenseller ;-{>"
Organization: Heald Colleges
Fastest time: 9 minutes
Number of operations: 4
(1) Turn over both timers.
(2) When the 4 min timer runs out, turn it over
- TIME SINCE LAST ACTION = 4 MINUTES
(3) When the 7 min timer runs out, turn it over
- TIME SINCE LAST ACTION = 3 MINUTES
- TOTAL ELAPSE TIME = 7 MINUTES
(4) When the 4 min timer runs out, turn the 7 MINUTE TIMER over
- TIME SINCE LAST ACTION = 1 MINUTE
- TOTAL ELAPSE TIME = 8 MINUTES
... When the 7 min timer runs out, PROBLEM SOLVED...
- TIME SINCE LAST ACTION = 1 MINUTE
- TOTAL ELAPSE TIME = 9 MINUTES
From: eosm@aqua.civag.unimelb.EDU.AU (Emma Osman)
Nine minutes
Start both off at the same time. At four minutes, turn the four-minute one
over. At seven minutes, turn the seven-minute one over. At eight minutes,
the seven minute one has been going for one minute. Stop it and turn it
over,
to re-measure the one minute. At the end of that, you are at nine minutes.
Voila!
I acheived this result by being morally certain that the answer was going
to
turn out to be "nine", then adding and subbing fours and sevens till I got it
Emma Osman
From: John Orthwein
Let T be the set of whole numbers, representing time. At t=0, both
hourglasses
are turned over. A t=4, the 4-minute hourglass is empty and there is three
minute's worth of sand in the 7-minute hourglass. At this instant, the
4-minute hourglass is turned over and started again. At t=7, the 4-minute
hourglass now has one minute's worth of sand left. At t=8, one minute's
worth
of sand is in the bottom of the 7-minute hourglass and the 4-minute
hourglass
is empty. At t=8, turn the 7-minute hourglass over. When the sand runs
out of
the 7-minute's hourglass, t=9.
One could now even measure time to any discreet minute by exploiting the
fact
that one hourglass is empty and the other has one minute's worth of sand
in the
bottom.
From: THHACKETT@vassar.edu
The way I would solve this problem is as follows:
Start both hour-glasses at the same time. As each one completes,
immediately re-start it. When the longer (seven-minute) hour glass
completes for the first time, start the nine-minute interval. At
that point, the four-minute hour-glass has one minute left on its
second iteration. When the four-minute glass completes two more
complete iterations, the nine-minute interval will have completed
(sixteen minutes from the start of the whole thing).
In order to arrive at this solution, I simply laid-out the seven and
four minute intervals on a number line and found the first instance
of a difference of nine. Intuitively I feel this is the shortest
nine-mintue measurement, but I'm not yet sure how to prove it.
Tom
From: Michael Feld
Organization: Philosophy Dept., University of Manitoba
1) The easy, obvious answer is:
Invert the 4-minute sandclock; when it is half-run, invert the 7-
minute
sandclock; or reverse the order.
Of course, this assumes keen eyes and a symmetrical sandclock.
2) The classical wise-guy answer is:
"Oh mister time-keeping person, would you like to trade your cheap
electronic
time-keeping machine for this nice pair of sand-clocks?"
3) My stab at a real answer:
Run both sand-clocks simultaneously,thus:
7 4
- -
0 0
at the end of four minutes, the clocks will look like this:
3 0
- -
4 4
Invert the smaller, i.e., the 4-minute glass, while letting the larger, i.e., the
7-minute glass, continue to run: thus,
3 4
- -
4 0
At the end of a further three minutes, the glasses will look like this:
0 1
- -
7 3
Next, invert the larger, 7-minute glass, while permitting the smaller, 4-
minute glass
to continue to flow. Thus,
7 1
- -
0 3
At the end of one minute, the glasses will appear:
1 0
- -
6 4
Invert only the larger, 7-minute glass, *AND START YOUR TIMING*.
At the end of one
minute, the glasses will appear:
0 0
- -
7 4
Next, invert only the smaller, 4-minute glass, *AND CONTINUE THE
TIMING*; then, fo
course, invert the 4-minute glass again.
The last nine minutes of the process give you what you want. To reach
those last nine
minutes, you have used an additional eight minutes.
So: the answer I reach is: 17 minutes in all
Best,
Michael Feld
--
---------------------
From: Mike Knight
Subject: Hourglass puzzle
Excellent page and puzzle. Here is my answer.
Solution: Label the 4-minute hourglass 'A', and
the seven-minute glass 'B'. Start with both glasses,
A and B, sand in the bottom. Flip both. Allow A to
drain completely. At this time, 4 minutes have gone by.
3 minutes remain, however, in the B hourglass. Flip the
A glass only. Allow the B hourglass to finish draining.
7 minutes have elapsed. One minute remains in the A
glass. Flip the B hourglass. When A has drained, the 8
minute mark, flip B. Allow B to drain, measuring the
last minute.
Unfortunately, I can't provide a competent explaination
of why this works. I simply know that we must find some
way of splitting the total time into increments that we
can measure with our hourglasses. The 4+3+1+1=9 was the
idea that worked for me, other than trying to tell when
half the sand had run out in the 4-minute hourglass,
then starting the 7-minute glass. Also, in the real
world, reaction time will play a part in our going
slightly over the nine-minute mark. But, in puzzledom,
human error need not apply.
From: George Drapeau
Hello,
I've never visited this page before (Zeno's Coffeehouse) but came to it
via a research page at Cornell on Project Zeno. Anyway, if the puzzle
is serious, then here is my answer:
0:00 Start both hourglasses, the 4-minute and 7-minute glass.
4:00 When 4 minutes have elapsed, notice that the larger glass will have
3 minutes' worth of sand in it. Immediately turn over the
now-empty 4-minute hourglass and let the larger one continue to
drain.
7:00 The large hourglass has now drained and the small glass has drained
3 minutes' worth, leaving one minute left. Immediately turn over
the larger hourglass, let the small one drain.
8:00 The small hourglass has drained a second time. Toss it out; you no
longer need it. Notice that the larger hourglass has drained
exactly one minute. Immediately turn it over and drain that
minute.
9:00 Problem solved! Time for coffee.
From: scott haney
Well, the *quickest* way would take exactly 9 minutes. :)
I don't have time to check if this is the shortest, but it should work:
start both hourglasses
restart the 4-minute glass when it runs out
restart the 7-minute glass when it runs out
when the 4-minute glass runs out, restart the
7-minute glass immediately! It should run for
1 minute. When it runs out, 9 minutes have passed.
--
Scott Haney rhaney@cacd.rockwell.com
(319)395-8281
Received: from ballard.staff.umkc.edu by axp2.umkc.edu (MX V4.1 AXP)
with SMTP;
Wed, 12 Jun 1996 14:56:57 CST
I'm not sure that the challenge is to outfox a paradox here,
but it seems to me that the fastest way to time nine minutes with
a four minute hourglass and a seven minute hourglass, asuming perfect
attention and instantaneous flipping, starting from zeroed hourglasses,
is to
0 minutes 1: flip 'em both
4 minutes 2: flip the empty four minute hourglass, to time to 8 minutes.
7 minutes 3: flip the now empty seven minute hourglass
8 minutes 4: flip the seven minute hourglass that has just accumulated
one minute of sand, which takes us to
9 minutes.
<~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~\_/~>
David_Nicol dnicol@cctr.umkc.edu pob45163_kcmo64171
(816)_235_1187
"beware of the leopard"
From: mkeef@ix.netcom.com (MARV KEEFER)
Subject: Timimg Puzzle
Start both hourglasses.
Turn over the four minute glass (four minutes.)
Turn over the seven minute glass (seven minutes.)
Turn over the seven minute glass when the four minute glass finishes
(eight minutes.)
Nine minutes have passed when the sand of the seven minute glass
finishes draining.
From: Ken Duisenberg
Nine minutes can be measured in nine minutes using a four-minute and
seven-minute hourglass:
Start both hourglasses 0 minutes
Restart the 4-minute glass 4 minutes
Restart the 7-minute glass 7 minutes
When the 4-minute glass runs
out, turn over the 7-minute
glass (it ran one minute) 8 minutes
When the 7-minute glass runs
out, the timing is done. 9 minutes
Ken Duisenberg
Roseville, CA
From: shack@esinet.net (Shack Toms)
Assuming that both timers start with all of the sand on one side,
it can be done in 13 minutes. Otherwise I may have to wait up
to an additional 3 and a half minutes to get the timers ready to
start.
The deductive principle is this, an hourglass can measure any
interval of time and then duplicate that interval by letting the
sand flow in reverse. So if I can get one timer in an initial
state (all sand on one side) and a known time interval on one
side of the other timer, I can duplicate that known time
interval again and again, with one hourglass measuring the known
interval while the other hourglass returns to its initial state.
To be more explicit, at t=0 I start both timers. When the 4
minute timer runs out (at t=4) 3 minutes of sand remains in the
seven minute timer. Now I can apply the above principle to
measure three multiples of 3 minutes. I invert the 4 minute
timer, to measure the 3 minutes remaining in the 7 minute timer.
At t=7, the seven minute timer has emptied. I invert both timers
so that with the 7 minute timer I can measure the 3 minute
interval "stored" in the 4 minute timer. At t=10, the 4 minute
timer is back in its initial state and 3 minutes of sand has
drained from the 7 minute timer. I invert the 7 minute timer.
At t=13 (at the end of my 9 minute interval that began at t=4)
the 3 minutes of sand has now drained back out of the 7 minute
timer.
Shack Toms
shack@esinet.net
From: "Antonio Giosue'"
First, start both hourglasses.
When the 4m. hourglass ends, start it again.(4 minutes)
When the 7m. hourglass ends, start it again.(7 minutes)
When the 4m. hourglass ends, turn the 7 m. hourglass.(8 minutes)
When the tm. hourglass ends we have measured nine minutes.(9 minutes)
From: Mark Young
Organization: Acadia University
9+7 = 16 = 4*4, so...
T-7 minutes: Start both timers
T-3 minutes: Flip the 4 (it just ran out)
T: Begin measured nine minutes (the 7 just ran out)
T+1: Flip the 4 (it just ran out)
T+5: Flip the 4 (it just ran out)
T+9: End measured 9 minutes (the 4 just ran out)
...mark young
From: Michael Faron
If the 9 minutes to be measured needs to be contiguous, then the
"quickest"
way i have found needs 7 minutes to get to the 9 minute interval....
at the same time, start the 4 and 7 minute hourglass
when the 4 has emptied, immediately invert it. the 7 continues with
3
minutes left.
when the 7 finishes its last 3 minutes, there is one minute remaining
in
the 4. immediately flip the 4. this one minute begins the
9 minute interval.
follow the one minute with two 4 minute hourglass intervals. thats 9
minutes.
if the 9 minute interval can be broken up, the "quickest" way i have found
needs 5 minutes in addition to the 9 minutes desired....
at the same time, start the 4 and 7 minute hourglass.
when the 4 has emptied, immediately invert it. the 3 minutes left in
the 7
begin the 9 minute interval.
when the 3 minutes has finished, immediately flip both hourglasses.
1
minute is left in the 4 and begins to empty. the full 7 is emptying as well.
when the 1 minute has emptied, there are 6 minutes left left in the 7
minute hour glass
add these 6 onto the previous three and you get the desired 9.
thats all i can come up with
mfaron@sinet.com
From: "Peter W. Woodruff"
Dear Ron:
Lovely site! Let t be the time; it must either be nine minutes since
some time we turned the four-minute glass or since some time we turned
the
seven minute glass, and it must be marked as the expiration of the other
glass. Put another way, the beginning of the interval must be marked by
one glass, the end by the other. Thus we need either that the end is both
4f + 9 and 7s or both 7s+9 and 4f for some f and s. Calculation (or
experimentation) yields that the minimum values for f and s in the first
case are 3 and 3, in the second case 1 and 4; the first yields a time of
21, the second of 16, so 16 minutes is the shortest (of course, this
assumes that we can start the process we want to measure at any time; if
the start time is fixed, there is no solution).
--Peter
From: Francis Varela
Using standard hourglass-type measuring devices, what is the quickest
way
to measure nine minutes with a four-minute glass and a seven-minute
glass?
Solution:
Time 0 Step 1. Set both the 4-min. and the 7-min. glasses.
After 4 min. Step 2. Invert the 4-min. glass.
After 3 min. Step 3. The 7-min. glass is used up; start counting.
After 1 min. Step 4. Invert the 4-min. glass.
After 4 min. Step 5. Invert the 4-min. glass.
After 4 min. Step 6. The 4-min. glass is used up; 9 min. have been
counted.
The 9-minute period is counted starting with Step 3. The whole procedure
requires 16 minutes to perform.
Francis Varela
From: Roy Kilgard
Come now, Barnette, this one's easy!!!
To time nine minutes with a four minute and seven minute timer:
Start both timers simultaneously.
When the four minute runs out, start it again.
When the seven minute runs out, begin timing nine minute interval.
Let the remaining minute fall through the four minute and run it twice
more.
7-4=3
4-3=1
1+4+4=9
Simple enough.
-Roy
From: JohnD_Halter
The best way to time 9 minutes takes 16. YOu will have to flip the 4-
minute
glass so that it may consume the sand four times. The 7-minute glass will
be flipped only to begin the process, and the timing of the 9 minute span
should begin when this glass runs out.
Thanks for the challenge. I'll be looking forward to the next one!
JOhn
From: Terrence Nolley
I don't think it really makes a difference what combination of
glasses
you use. The process will still take nine minutes regardless of the
combination.
From: "Robert N. Kelly"
Empty the four minute hourglass. Fill the four minute hourglass with the
seven minute
hourglass giving yourself three minutes left in the Seven minute hourglass.
Turn the
seven minute hourglass over three times for nine minutes.
If this is illegal then
Turn both hourglasses over at the same time.
At the end of four minutes turn the four minute hourglass back over.
At the end of the seven minute hourglass turn the four and the Seven
minute hourglass
back over.
{one minute is left in the four minute hourglass.}
At the end of that minute turn the seven minute hourglass over again.
(one minute is left in the seven minute hourglass}
Since the seven minute hourglass expired fully once, and you used the
minute left in the
four minute hourglass on the third turn over and you measured a minute
of sand in the
seven minute hour glass and turned that over you have measured NINE
minutes. Seven plus
one plus one.
From: monlyn
Subject: 9 MINUTES
TI think the best way would be to do 1, 7-minute, then = of the 4-minute
hourglasses.There is no "quicker" way, for 9 minutes is nine minutes! Just
MY opinion,only........Monica
From: Ed Russell
Deduction:
1. The result cannot be obtained without measuring a time
interval of 1, 2, or 5 by subtraction.
2. The subtraction process must be a continuous use of
both the 4 and 7 glasses.
3. This leads to the differences in the series 7, 14, ...
and 4, 8, 12, ... => (7-4=3), (7-8=-1).... Therefore of
the time intervals specified in (1), 1 is the first to be
measured.
4. The total time for measuring 9 minutes: 8 minutes to
measure 1 minute, plus 2 times 4 minutes = 16.
-- Ed Russell
ehr@synchronicity.com
From: Chad Miller
Start the four- and seven-minute glasses at the same time. When the four-
minute glass runs out, flip it. When the seven-minute glass ends, start
timing; there is one minute left in the four-minute glass. When it finishes,
flip the four-minute glass twice, adding eight more minutes. Viola!
Chad, ex-phicyberite
http://www.surfsouth.com/~cmiller/
0400
Date: Mon, 1 Jul 1996 10:46:47 -0400 (EDT)
From: Rex DeVane
There is no quick way to measure nine minutes - it will take nine minutes
any way you attempt to measure it.
The following is one method using the seven and four minute hourglass
type measuring devices:
1. Turn both the seven and the four minute glasses over to start them
running
2. As soon as the four minute glass is empty turn it over to start a new
four minutes. Note that the seven minute glass is still running and has
only 3 minutes left.
3. When the seven minute glass runs out turn it over to start a new seven
minutes. Note that when we you turn the seven minute glass over you
have
only one minute left in the four minute glass.
4. When the four minute glass runs out turn the seven minute glass back
over. Note that the seven minute glass only had time to drop 1 minute of
sand.
5. When the seven minute glass is empty you have measured nine minutes!
From: Rex DeVane
Subject: Charles' time problem
To: rbarnett@grits.valdosta.peachnet.edu
Message-ID:
MIME-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII
Content-Length: 883
Status: O
X-Status:
There is no quick way to measure nine minutes - it will take nine minutes
any way you attempt to measure it.
The following is one method using the seven and four minute hourglass
type measuring devices:
1. Turn both the seven and the four minute glasses over to start them
running
2. As soon as the four minute glass is empty turn it over to start a new
four minutes. Note that the seven minute glass is still running and has
only 3 minutes left.
3. When the seven minute glass runs out turn it over to start a new seven
minutes. Note that when we you turn the seven minute glass over you
have
only one minute left in the four minute glass.
4. When the four minute glass runs out turn the seven minute glass back
over. Note that the seven minute glass only had time to drop 1 minute of
sand.
5. When the seven minute glass is empty you have measured nine minutes!
From: Math/CS Lab
Turn both hour glasses over and let the 4 min. one run out.
Stop the 7 min one as soon as the 4 min runs out.
You have 3 min left on the 7 min one and 0 min on th e4 min one.
Restart both hour glasses and let the 7 min one run out.
You have 0 min on the 7 min one and 1 min on the 4 min one.
Restart both hourglasses again.
You'll have 6 min on the 7 min glass and 3 min on the 4 min glass.
Start the 7 min glass and when it runs out start the 4 min glass.
You get 9 minutes!
Laura
Received: from reid.phil.indiana.edu (reid.phil.indiana.edu From: Karen
Brown
Turn both glasses over. When the four minute one runs down, turn it over
again. When the seven minute one runs out, the nine minutes begins.
Allow the four minute glass to run out three times [1 minute + 4 minutes
+ 4 minutes].
From: "Christoffer J. Anderson"
This is actually a pretty easy one. Here goes:
(1) Start both timers going simultaneously.
(2) When the Four-Minute-Timer (4MT) is done, flip it over and leave the
Seven-Minute-Timer (7MT) to keep going its last 3 minutes. At
this point, we have elapsed 4 minutes, and there is 3 minutes
left on the 7MT.
(3) When the 7MT is done, flip it over. At this point, we have elapsed
7 minutes, and there is 1 minute left on the 4MT.
(4) When the 4MT runs out, there will have accrued 1 minute's worth of
sand in the bottom of the 7MT. Flip the 7MT over when the 4MT
is done. At this point, we have elapsed 8 minutes, and there is
1 minute left in the 7MT.
(5) When the 7MT has run its 1 minute's worth of sand out, we have
elapsed exactly 9 minutes.
Gimme another!
Christoffer J. Anderson
Colorado State University
Department of Philosophy
Received: from Iris (193.219.56.3) by mbiserv.fermentas.lt
0. both glasses begin.
4. revert 4-glass. in 7-glass 3 min remaining.
7. 7-glass end, revert 4-glass with 1 min. remaining. start count.
8. revert 4-glass.
12. revert 4-glass.
16. end count.
is 16 min the limit?
--
v. tiknius
vt@fermentas.lt
http://iris.fermentas.lt
From: SBIJ101@aol.com
1)Turn over both timers
2)When the four minute timer runs out turn it back over while the
remaining
three minutes runs out on the other timer.
TOTAL 4min
3)When the seven minute timer is empty turn it back over while the
remaining minute runs out on the four minute timer
TOTAL 7min
4)When the four minute timer is empty turn the seven minute timer
back over
TOTAL
8min
5)When the seven minute timer is empty you have measured -----------
TOTAL
9min
From: Gregory Reihman
> Using standard hourglass-type measuring devices, what is the
> quickest way to measure nine minutes with a four-minute glass and a
> seven-minute glass?
One can measure eight minutes by using the four-minute glass twice in
succession. Conveniently, this is one minute longer than the time
measured by the seven-minute timer, and one minute shorter than the total
desired time. So, the desired nine minutes can be measured by first
beginning the clocks at the same time, then flipping the four-minute
clock when it runs out, and then flipping the seven-minute clock when it
(the seven-minute clock) is finished. Finally, when the four-minute
clock runs out a second time (at this point one minute's worth of sand
will have fallen in the seven-minute clock), flip the seven minute clock.
When this minute's worth of sand falls, a total of nine minutes will have
elapsed.
This is assuming that the sand falls at a constant rate throughout the
seven-minutes of time (or, at least, that the first and last minute's
worth of sand are the same). If, for example, the sand were to fall
faster near the beginning (when most of the sand is still on top) than
near the end (when most of the sand is at the bottom), then it will take
more than a minute for the "minute's worth of sand" to fall in the final
step, and the time elapsed in the above procedure will be more than nine
minutes. If the sand does fall at a constant rate, the procedure will
work.
From: Christopher Hitchcock
Re: New Zeno challenge
You can time nine minutes in nine minutes. Start both hourglasses at
minute zero. When the four minute glass is finished, turn it over.
When the seven minute glass is finished, turn it over again. When the
four minute glass finishes a second time (and eight minutes have
elapsed) there will be one minute's worth of sand in the bottom of the
seven minute glass. Turn it over, and it will take one more minute for
the sand to fall through. Voila--nine minutes.
From: MattSwarm@aol.com
A fast way to measure nine minutes with a 4 minute and a 7 minute
hour-glass type device takes 9 minutes. There are very few ways to do it
faster.
Minute 00: Start both 4 and 7 minute timer.
Minute 04: Turn over empty 4 minute timer.
Minute 07: Turn over empty 7 minute timer.
Minute 08: 4 Minute timer has finished, 7 minute timer has 1 minute of
sand in bottom. Turn over 7 minute timer.
Minute 09: Aforementioned 1 minute of sand has drained to bottom.
Question: What integer periods, if any, CANNOT be measured using
this
and similar methods?
Matt Swarm
mattswarm@aol.com
From: Michael Britton
simple enough, although it's eaven simpler to be trapped in the 3-minute
pit. (i.e., trying to use the 7-4=3 and 3x3=9 properties... I'm sure
some respondent will provide this solution in full. it requres 4
minutes preparation.)
I'll use a wierd notation here: t=__ is the time from start;
| x|
><
| y| is an hourglass with x minutes of sand in the top and y minutes
of sand in the bottom.
the first hourglass will be the 4-minute one, and the second will be
the 7-minute one.
so initially, we start both hourglasses running:
t=0 | 4| | 7|
>< ><
| 0| | 0|
4 minutes later, the first hourglass runs out, so flip it over:
t=4 | 0| | 3| ---\ | 4| | 3|
>< >< > >< ><
| 4| | 4| ---/ | 0| | 4|
3 minutes later, the other hourglass runs out, so flip it over:
t=7 | 1| | 0| ---\ | 1| | 7|
>< >< > >< ><
| 3| | 7| ---/ | 3| | 0|
1 minute after that, the first hourglass runs out again. this time,
flip the second hourglasses over again.
t=8 | 0| | 6| ---\ | 0| | 1|
>< >< > >< ><
| 4| | 1| ---/ | 4| | 6|
1 minute later, the second hourglass runs out of sand again. when this
happens, 9 minutes have expired.
t=9 | 0| | 0|
>< ><
| 4| | 7|
and there you go!
Mike Britton
From: ryanearehart@juno.com
To: RBARNETT@GRITS.VALDOSTA.PEACHNET.EDU
IINTERESTING QUESTION MY SOLUTION TAKES 12 MINUTES TO
SET UP BEFORE YOU
CAN MEASURE YOUR NINE MINUTES. fIRST YOU START THEM
BOTH GOING AT THE
SAME TIME AND WHEN THE 4 MINUTES FINISHES KNOCK OVER
THE SEVEN SO THAT IT HAS 3 MINUTES IN ONE HALF AND
FOUR IN THE OTHER. USE THE 3 MINUTES TO
REDUCE THE 4 MINUTE ONE TO ONE MINUTE AND 3 MINUTES -
- THEN TAKE THE 1
MINUTE AND REDUCE THE SEVEN TO SIX AND ONE. FROM THE
SIX DEDUCT FOUR
LEAVING YOU WITH 2 MINUTES AND 5 MINUTES LEFT IN THE
BIG ONE AND 4
MINUTES IN THE SMALLL ONE USE THE 4 AND THE FIVE
TOGETHER AND I THINK
THAT ADDS UP TO NINE!
{4} - {7}
{4}-{3/4} 4MIN
{1/3}-{7} 3MIN
{4}-{6/1} 1MIN
{4}-{5/2} 4MIN
ELAPSED TIME = 12MINUTES
4+5=9
Received: from xyp29p1.ltec.net (xyp29p1.ltec.net [204.96.104.11]) by
iac1.ltec.net (8.6.10/8.6.10) with SMTP id UAA39373 for Ok, start the 7
minute glass and the four minute glass at the same time.
When the four minute glass runs out, start it again. When the seven
minute glass runs out, begin timing. When the four minute glass runs
out, one minute will have elapsed. Start the one minute glass again.
When it runs out, five minutes will have elapsed. When it runs out
again, you will have measured nine minutes.
Matt Seaman
From: "Michael O'Leary"
1.Flip both 4 and 7 hourglasses over.
2.When the 4 glass is done, flip both 4 and 7 over (with 3 minutes left
in 3.
3. When the 7 (with 3 min lifet in it is done, flip over the 4 (with one
minute left) and the 7 glass.
4. When the 4 (with one minute left is done, flip the 7 over. It will
have 1 minute left in it.
That will take 9 minutes.
--
_\|/_
O O
+----------------oOO-(_)-OOo-----------------------+
Michael O'Leary
From: Lauri Kultti
Organization: University of Helsinki
First turn both glasses. When the four-minutes glass is done, the
seven-minutes glass has three to go. Turn the four-minutes glass again.
When the seven minutes-glass (with its three minutes left) is done, the
four-minutes glass has one to go. START THE MEASURING NOW. Let
the
remaining one minute (of the four-minutes glass) go. Then simply use the
four-minutes glass twice. So, you have your nine minutes.
From: Tony Reardon
16 mins elapsed. To make 9 minutes you either need a 2 minute interval to
add to 7 or a 1 minute interval to add to 4 plus 4. 2 minutes cannot be
derived until 14 minutes have elapsed but 1 minute can be derived after 7
minutes. There are a couple of ways of doing this but the simplest is to
start both timers. After 4 minutes, restart the 4 minute timer. After 7
minutes restart the 4 minute timer which will run for 1 minute. Then use
the 4 minute timer twice more to get 9 minutes.
From: Wayne BARNES
The process will take a total of 25 minutes. Begin by
flipping both glasses. When one glass is finished note the
amount of time remaining in the other. After 16 minutes there
will be 5 minutes remaining in the 7 minute hour-glass. You
can now begin measuring the required 9 minutes by turning the
4 minute hour-glass after this five minutes has expired.
From: Dave Evans
Hi, The quickest way I can figure is to have one 4 minute hourglass
and three 7 minute hourglasses. Turn them all over at once, when the
4 minute glass is done, turn 2 of the seven minute glasses on their
sides, so they stop flowing. The nine minutes starts now. When the 1st
seven minute glass is done, restart the 2nd, when the second is done,
restart the 3rd. Since each had 3 minutes left (7-4=3), when the
third is done, 9 minutes has elapsed. Grand total of 13 minutes to
measure 9.