6 minute measurement with string puzzler
How do you measure 6 minutes? I’m baffled.
Ask GeorgeSanJose:
Bad puzzlercsolution. A pendulum? With a fuse? An hour long fuse would be too long and too irregular to be used as a pendulum.
My solution was to have 10 flame fronts working on the fuse. The endpoints and 4 points in the middle. Whenever the flame fronts met and extinguished themselves then light another burning segment in the middle. 1 flame front burns the fuse in one hour. 10 would do it in six minutes.
But remember, the fuse can burn unevenly. It could burn up all but one inch in one second, and take 59 min, 59 sec to burn that last inch.
Yes that wouldn’t matter. As long as ten flame fronts are going constantly it would work. In reality that would be hard to do but it’s more realistic than a pendulum.
Ah, now I understand your method (I think). So, you could solve the “15 minute” version by using 4 flame fronts like this: On one fuse, light both ends and one point in the middle, thus producing 4 flame fronts. When one of the two pair of flames meet and extinguish each other, light the middle of the remaining burning segment. When another pair of flame fronts meet, again light the middle of the remaining burning segment. Repeat the last step until the two pair of flame fronts meet simultaneously and there’s your 15 minutes. Right?
I do believe the pendulum solution is easily doable. I took a 3 oz table knife and hung it on a 3 ft long string and measured the period with a stopwatch over 10 periods to get about 2 sec per period. It appears that it wouldn’t keep swinging for 30 min, so you’d have to give it a nudge every 10 min or so to keep it going.
@kenberthiaume … that’s a very clever idea. +1 for you. I think you are right, your method would time out 6 minutes as long as you could keep all 10 points burning simultaneously. Keeping all 10 points burning at the same time until the 6 minutes are up might be a practical problem, depending on how unevenly the string burns. But so would keeping the pendulum moving.
@kenberthiaume …The puzzle specifically says that the string can burn unevenly. If one of the short peices has most of the hour in it (as Insighful said), that piece would burn for longer than 6 minutes. The other pieces would have already finished burning. You can’t keep all ten ends burning the whole time because some would burn up early. This is another time when averaging doesn’t work.
I think the OP already addressed your point @David L . In their first May 3 posting. It was perhaps a little unclear maybe. Think of it this way. Cut the string into 5 equal length pieces first. Then light both ends of each segment. So you have 10 flame fronts burning at once. If all segments burn the same rate, they’ll all extinguish at the same time and you are done. 6 minutes. If one segment burns faster than another, it will burn itself out while another segment continues to burn. So OP would watch for this, then at that time light the middle of an existing segment, effectively creating another segment to replace the extinguished one. The idea is to always have 5 segments burning at all times.
Don’t see how that gets to 6 minutes. Five could burn in 1 minute, then what?
4 could burn in 1 minute. I think that is what you mean. If that happened, the 5th segment would still be burning at both ends. It would be re-cut into 5 segments, and both ends of each segment lit again. There’s some practical problems, but I think that is the idea.
yes, GeorgeSanJose, that’s my thinking exactly. It also has the benefit of measuring 6 minutes from the start, whereas the pendumlum has to go one full hour and then you divide your number of pendulum swings by 10. You also don’t have to light the exact middle of a segment, just anywhere will do. Although keeping 10 flame fronts going at once would be tricky, and you’d lose time lighting a new front, etc. But restarting a pendulum after it’s petered out would cause error too.
You don’t restart the pendulum, you nudge it every once and a while as it slows down (similar to a pendulum clock that nudges it every cycle). With any skill, very little error would result.
Convinced there would be a solution to this puzzler, I delayed learning the answer for two weeks while I worked on the problem. I have come up with a solution which I believe also works, although somewhat involved.
Divide one length of fuse into five segments, preferably with one or two longer than the others, and divide the other into six segments, also with one or two preferably longer than the others. If all 11 segments are ignited at the same time, The total burn time for the five segment group will be 12 minutes, and the total burntime for the six segment group will be 10 minutes. Once the 10 minute burn time is reached, the total time remaining of the fuse lengths in the five segment group, If burned separately, will be a total of 12 minutes.
The trick is to have all of the fuse segments burn simultaneously, even though they are of different lengths. To do this, within each group, whenever a segment is exhausted, the second end of the longest remaining fuse in that group must be ignited, and so on so that there are always five, or six burning ends in the respective group. Then, when all of the fuse in the six segment group is exhausted, all of the fuses in the five segment group must be extinguished and all of the fuse segments connected together. Then, this remaining fuse segment, having a 12 minute duration, can be ignited from both ends, and when it is extinguished, six minutes will have elapsed.k
This one is clever and just came up again.
Let’s assume you cut the string in 10, 100, or even 1000 pieces. You light 1 piece then when it’s burned your immediately light another and so on. It will be done in an hour since it’s the same as burning the entire string sequentially.
Now for illustration, let’s arrange the 1000 pieces in the exact order they were on the string, we’re going to light both “ends” but instead of one continuous string, it will be our neatly arranged line of tiny string pieces.
So we will start by lighting two pieces at opposite ends, when one is extinguished we light another, always maintaining two.
If this example was successful, it should be clear that this is the same as lighting one string at both ends. Three only difference is the string is now in a bunch of pieces and you’re given more work to do.
That’d be measuring 30 minutes as we know.
So with 1000 pieces and you maintain 1 always burning, that’s 60 minutes. You maintain 2 always burning, that’s 30 minutes…
What if you make sure 3 are always lit? That’s 20 minutes by the same logic. Our solution is in site!
When there is only two pieces remaining, light the other end of one. When it’s only one piece, cut it in half, immediately lighting the other end and so on. You need to have 3 at all times as much as possible.
Not convinced? Try to target 15 minutes. Imagine again the same line and think about the original 15 minute solution. You’ll find it’s the same thing again, just broken up into little string pieces.
So now we can hit our 6 minute mark with a lot of work!
So now you make sure 10 are always lit and busily work around using the above strategy cutting and lighting. When it’s at 9 pieces, you light the other end of one making it 10 flames. When it’s at 8 pieces, you have 2 burning at both ends maintaining the 10 flames.
When it gets down to 4 pieces, you cut one in half to get back to 5 and light the ends to get to 10.
It’s hard to execute but it works and doesn’t introduce anything new, such as a pendulum or anything else.
If you assume cutting is possible, cutting time is zero and lighting time is zero, it’s an exact solution. I think those are all fine assumptions for a math puzzler.
Also you get to keep the second string as a souvenir, it wasn’t needed
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