Have you ever measured time without using sto$αεpwatches? When you move in the sun and sky, can →λyou measure the time ca→↑refully by watching the wobble of the s♣↓↕∑un? I know this is not the most unforgettable stΩΩatement in the world, but ♣λit is absolutely true. The‍  puzzle is a dream ex↕→ercise for your brain. Therefore, you will be wis✘​e to try to solve at least seve &♣δral different types of puzzles at least onc÷§♠e a week. And the best part is thatαα₹♦ this mental exercise is very interestαπing.

To help you start to ≤Ω get more confused, today we ha>λπδve to see a great intellectual c♥♦hallenge I met recently. The problem is all a÷ ∏€bout time, and how to ε±measure time intervals in a ve$→¥ry unusual way: inste÷ ad of using stopwatch, it is to measure the tim≈©≈±e by burning a special string. H£§δow does it work? What is a braiβ₹§n Trailer?

How do you measure 45 minutes without s σ♥topwatch?

Imagine that you have given several string★ε↑s of different length and thicknγ∑♣ess. Not only is the length and width of a sing ¥le piece different, but also the width of eaπ¶★βch piece is uneven alon>φg its length. In other words, when you go fro÷↑₽βm one end to the other, the♣↕δy become thicker and thicker, in differ↓λ♦ent places and different places.Although all wo ±∏rks are different, they have one thing iπ★n common: if you light one end, it usual'☆​♥ly takes an hour to burn to the other end'☆β. However, as each piece becomes thicker and thi$Ωβ≈nner, a given string d←§£oes not necessarily burn★λγ at an average speed. I mean, if you want st •βopwatch to record a rope in 30 minutes,δ>•≈ it doesn't necessarily burn half the&​≥γ length - we can say t™'hat it burns the whole length '∏↕ in an entire hour.

This is the setting. The follα&÷owing question is: can you πσδ♦use these strings (as many as∑π possible) to measure the interval o₹σ∏✘f 45 minutes? And, if possible, what would yo‍✔↓u do? Just like every puzzle, if you tr✘¶∏y before finding the ♣±≥answer, it will become more inter ×®esting. So I encourage you to sus♥©β>pend for a few minutes and let it g¶© o forever. Then, when you are ready, keep loo✔≠©king for answers.

This is a simple problem←±★↔. Before we solve today&φ™™‍#39;s problems, let¶₹≠9;s imagine a slightly simpler j× igsaw where all the stri™♠βngs we give are burnin•φ'g at a uniform rate. In this case, you can simpl>¥₽y fold the single string into one half to sol∞>σve the problem, and then☆₹​' fold it in half to make the chord cr≠™eases the entire length of 1/4, 1↓ ♥/2 and 3/4. Because the st≈ ←ring is burning at a consistent speed, the &>¶φ45 minute time to meaδ≥sure is only to ignite one end of the flame ©↑"‍string, and then wait until the mark on ↔↓≠£the other end of the 3/4; is it simple e→→λπnough, if it can do so, do not £¶need stopwatch?

But, sadly, the performance of st₹✔ring pieces we gave in this puzzle is not so g₽<ood, because they do not burn at♣$ uniform speed. This means that we ca<•n not simply break a chord into 1/4 to cβφ∞alculate the time. But "‌±that does not mean that we are not l λucky... Because we can be smarter.

Here's what we need ≤↓∏to do: we are not just lighting one end π≥ of the line, but lighting both ends. W'γhat does this have to do with u₹δs? Then, the uneven thickness of a str↔↓€πing means that we can't sa€∏×y how fast the fire can burn from one enλ↓βd to the other at any given time. Neverthel"∏±ess, we do know that after two or tα hree minutes, the two flames that we >¥$★have to burn together must get ‌←together and meet each ≤€β€other.

This means that we now have a £Ω✔φway to use stopwatch to measure ♣γ♦the interval of 30 minutes every 1 hours. O™₹f course, what we really want is to measure the γ✔‍time interval of 45 minutes.↕" So we haven't finished it yet.•∞ What are we doing now? Before answering this€€ question, if you haven'¶÷¥;t figuring out this question©↑♦£, I encourage you to stop₹♥≠ and wait for a while, then go on.

OK, so what can we do to measure the time inter←φval between 45 minute€™$£s instead of 30 or 60 minutes? The trick is Ω≈↕that when we light the first str≥Ω♣ing from both ends, we need to light another st™<λring from one end. Then, when the two f↓≤lames of the first string are gathered ©αtogether, we need to ignite the second ends  β≈ of the half burnt side according to time. A€<♣★t this point, the second flame of≈↔εδ the string will start to burn to the other, ≤πand after the other 15 minut∏"₩₽es, they will meet in the middle, and yo↓ βu can also use stopwatch to calculate.

So let's put all ≠↔§this together, and we can see§∏ that after we start the whole proces®€s of setting up the string, the second string♥δ✘‍s need a total of 30 + 15 or 45 minutes for the e&↑nd of the burning. Therefore±φ, by using some clever ideas, β₹we managed to measure the interval of 45 ≠γminutes. It's cool, right? So we can measure©→ time without the use of stopwatch, ₹ but these techniques are not good, and we÷β≥↓ have to use stopwatch to measure time, aε♠γεnd the stopwatch is i ≥β÷rreplaceable.