Following on from the recent blindfolded near the great Wall of China puzzle.

This time, two wise guys are blindfolded and disoriented and standing together exactly 1 mile from the Great Wall of China. they need to reach to Great Wall as soon as possible.

They move exactly 1 miles/hour (no less or no more) and they can communicate with a transmitter.

How long will it take for them to reach the Great Wall in the optimally worst case? (both needs to be there at the end.)

Since no one could solve it yet, I want to comment about it: @f" is the closest one to the answer still not the best optimal worst case!

If noone could solve in 2 days, I will share the real solution.

  • $\begingroup$ @sjaakbakker did u even read the question? i already said that it is a following question, it is not a duplicatr ffs. $\endgroup$
    – Oray
    Commented Jan 27, 2016 at 10:31
  • 3
    $\begingroup$ At this rythm we'll soon have a barman, a french and a dinosaur blindfolded and disoriented near the great wall of china ;) $\endgroup$ Commented Jan 27, 2016 at 11:05
  • $\begingroup$ @RiddlerNewComer all the riddles are not the same? always the same methodology but the words are different? this question's answer is not like the following questions' answer over 2... partially u need to think differently at least. $\endgroup$
    – Oray
    Commented Jan 27, 2016 at 11:09
  • 1
    $\begingroup$ Worst case? One gets eaten by murder squirrels after stumbling around near the original location for several hours. The other one falls off of a cliff six months later, still shouting over the transmitter: "Hello, are you there, hello, are you there, helloooooooooooo...." -No honestly, wait, it could probably be a lot worse than that. $\endgroup$
    – Culme
    Commented Jan 27, 2016 at 14:42
  • 1
    $\begingroup$ @culme, that is why it says optimally worst case. $\endgroup$ Commented Jan 27, 2016 at 16:35

7 Answers 7


If at any point during this strategy one person encounters the wall, the other person stops whatever they were doing and instead walks perpendicularly directly towards the wall.

Let the starting point have coordinates of $(0,0)$. In Phase 1 of the strategy, one person walks in a straight line to $(\frac{1}{\sqrt{3}},-1)$, and the other walks to $(-\frac{1}{\sqrt{3}},-1)$.

The people then proceed straight to $(\frac{\sqrt{3}}{2},-\frac{1}{2})$ and $(-\frac{\sqrt{3}}{2},-\frac{1}{2})$ (they will not find the wall while doing this).

In Phase 2 each person walks in an arc of $\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}$ radians (approximately 12.886 degrees)* around the circle, and then continues on a tangent until reaching the line $x+y=\sqrt{2}$ or $-x+y=\sqrt{2}$.

Then, the people walk along the lines $x+y=\sqrt{2}$ and $-x+y=\sqrt{2}$ until they reach $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ and $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ (again, they will not find the wall until they reach these points). In Phase 3, they walk along the circle towards $(0,1)$, which they will reach at a time of $\sqrt3+\frac{8\sqrt2-2\sqrt7}{5}+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{12}\approx3.94679$ hours if the wall is not encountered earlier.

The worst case if the wall is found during Phase 1 is $1+\frac{2}{\sqrt3}\approx2.15470$ hours.

The worst case if the wall is found during Phase 2 is when the wall is at an angle of $\frac{1}{2}\sin^{-1}\frac{3}{4}$ radians (approximately 24.295 degrees)* from the vertical. The farther person reaches it at a time of $1+\sqrt3+\frac{\sqrt7}{2}+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}\approx4.27892$ hours.

The worst case if the wall is found during Phase 3 is $1+\sqrt3+\frac{8\sqrt2-2\sqrt7}{5}+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}\approx4.16139$ hours (if the wall is found immediately after they reach the circle, the farther person takes $1$ hour to reach it).

Overall, the worst possible case takes $\boxed{4.27892}$ hours.

Diagrams: green circle shows where the wall is found, green diamond is where the other person reaches it.

Worst case for phase 2 (and worst case overall):

Worst case for phase 3:

Latest possible wall encounter:

*Where does this strange value $\sin^{-1}\frac{3}{4}$ come from?

Name the starting point $A$ and the unit circle $O$. Suppose that the people are following paths that are mirror images across the x-axis. Call the horizontal tangent ($y=-1$) $l$. Suppose the wall (call it $w$ is at an angle of $\theta$ to the vertical. Consider which choice of path minimizes the time so that:

  • Both people touch $l$.
  • Then, one person touches $w$ without going inside $O$ (if they go inside the circle, they are missing some tangents).
  • Then, the other person stops following the mirror-image path and heads to $w$.

Note that if we let $v$ be the reflection of $w$ across the x-axis, the time for the last step (second person moves to $w$) is the same as if the first person moved to $v$ instead.

Applying the reflective technique I used in this answer, we now reflect $A$ over $l$ to get $A^\prime$, $v$ over $w$ to get $v^\prime$, and then $A^\prime$ and $O$ over $v^\prime$ to get $A^{\prime\prime}$ and $O^\prime$:


Now what is half the length of the shortest path that passes from $A^\prime$ to $A^{\prime\prime}$ while staying outside $O$ and $O^\prime$?

If $\frac{\pi}{9}<\theta<\frac{\pi}{6}$, the path looks like this:


$ABA^\prime$ is a right triangle with $AB=1$ and $AA^\prime=2$, so $\angle A^\prime AB=\frac{\pi}{3}$ and $A^\prime B=\sqrt3$. $AC$ is parallel to $v^\prime$, so $\angle A^\prime AC=3\theta$ and $\angle BAC=3\theta-\frac{\pi}{3}$. But what is $CD$?


Reflect $A$ over $w$ to get $E$, and construct rectangle $CDGF$ with $E$ on $FG$. $\angle AEG=\pi-2\theta$ because $EG$ is perpendicular to $v^\prime$ and $AE$ is perpendicular to $w$. $AE=2$, and $EG=1$. So $CD=FG=EG+EF=1+2\cos{2\theta}$.

In total, the path length from $A^\prime$ to $D$ is $\sqrt3+3\theta-\frac{\pi}{3}+1+2\cos{2\theta}$. When we take the derivative and set it to zero, we get $3-4\sin{2\theta}=0$, and so the required path length is maximized at $\theta=\frac{1}{2}\sin^{-1}\frac{3}{4}$ with a value of $\sqrt3+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}+1+\frac{\sqrt7}{2}\approx4.27982$, as desired.

  • $\begingroup$ What software do you use to draw these kind of images? $\endgroup$
    – Ivo
    Commented Jan 27, 2016 at 13:23
  • 2
    $\begingroup$ @IvoBeckers I used GeoGebra (app.geogebra.org). $\endgroup$
    – f''
    Commented Jan 27, 2016 at 13:23
  • $\begingroup$ There is still a better answer. $\endgroup$
    – Oray
    Commented Jan 27, 2016 at 14:54
  • $\begingroup$ This one is very close to the answer though (the answer that I found) I think you just did a small mistake with the logic. I found less than 4.3835. $\endgroup$
    – Oray
    Commented Jan 27, 2016 at 18:39
  • 1
    $\begingroup$ I was quite surprised to find the closed form for that number: $1+\sqrt3+\frac{\sqrt7}{2}+\frac{3}{2}\sin^{-1}\frac{3}{4}-\frac{\pi}{3}$. Never would have guessed that :P $\endgroup$
    – f''
    Commented Jan 30, 2016 at 5:54

A slight improvement on Electric_monk's answer:

each one starts off 90 degrees from eachtother for $\sqrt2$ miles straight. Then they walk for 1 mile straight in such a direction so they are exactly 2 miles apart and on opposite sides of the starting point, then they follow the circle of radius 1 mile until they meet each other. Like this:

enter image description here

I believe the worse case is here when they are the 2 miles apart and one of them finds it, which means that the answer is $1+\sqrt2 +2 \approx 4.41$

  • $\begingroup$ sorry, there is for sure better answer. $\endgroup$
    – Oray
    Commented Jan 27, 2016 at 18:40
  • $\begingroup$ What software do you use to draw these kind of images? $\endgroup$ Commented Jan 27, 2016 at 22:54
  • $\begingroup$ @user1717828 check the comments on f'''s answer ;) got it from him $\endgroup$
    – Ivo
    Commented Jan 27, 2016 at 22:56

4.57 hours.

They walk 1 mile in the same direction, then one walks clockwise and the other walks counter-clockwise along an arc of 1 mile radius. Worst case: They started parallel to the wall. In which case, one of them would reach the wall when he completes a quarter circle and then the other would have to walk 2 miles to the first guy.

  • $\begingroup$ unfortunately no, there is a better answer. but good thinking though. $\endgroup$
    – Oray
    Commented Jan 27, 2016 at 11:50

The other answer is shorter, but relies upon both being able to hold a very precise arc while blind. If they can do that, great, however walking in a straight line may be easier.

In which case the answer is approximately 6 hours. Assuming they're able to maintain orientation on a straight course, and can accurately turn to a desired heading, and can accurately track time.

They can walk the complete bounding box of a 1 mile radius circle in that time:

  1. For the first hour, they should both walk together in the same direction.
  2. Then one turns 90 degrees to the right and the other 90 degrees to the left.
  3. Both walk straight for the next hour.
  4. Each makes another 90 degree turn in the same direction (left turns left again, right turns right again).
  5. Both walk straight for the next two hours.
  6. Repeat step #4.
  7. Both walk straight for the next hour, and which point they are reunited.
  8. If at any point after step #1 one of them encounters the wall, they can transmit to the other that they've found it; the one receiving the message will 1) execute a 180 degree turn if messaged during hour 2, or 2) execute a 90 degree turn if messaged during hours 3 or 4, or 3) continue on their current heading otherwise.

In the worst-case scenario, the wall will be encountered at one of the corners reached at the end of the 4th hour. The other participant will need to walk 2 hours to reach that spot.

However, I say "approximately" 6 hours because the bounding box is larger than the 1 mile radius within which some portion of the wall should coincide (at a tangent, I assume, because otherwise some portion of the wall is less than 1 mile away). So if the wall is encountered at a corner of the bounding box, there's a reasonable chance that the second participant will encounter a different portion of the wall somewhat before they reach the far corner (i.e. in slightly less than 6 hours). Both will have reached the wall, but they will be at different points along it.


It will take them 6 hours, 8 minutes, and 30 seconds (Pi+3 hours). I'm assuming that they are capable of traveling in a perfect arc.

Both travelers walk in opposite directions away from each other for one hour. In the best case scenario, one would reach the wall at this point and radio the other. Otherwise, both would head clockwise around the circle. In the worst case scenario, the wall was immediately to the left of one of the wise guys and the other one has to walk half the radius of the circle (taking Pi hours) to reach the wall. They then radio the other, who is on the other side of the circle, to traverse the diameter back.

1 hour to reach the circle, Pi hours to traverse half the circle, 2 hours for the one who didn't reach it to cut back across; Pi+3 hours.

  • $\begingroup$ Nice solution, but there are faster ones! Welcome to Puzzling.SE, hope to see you around more! c: $\endgroup$
    – Deusovi
    Commented Jan 31, 2016 at 4:18

I think both of them can reach the wall in less than 4 hours.

For starters, they begin to walk in the opposite directions. Let us name them as $c$ for closer one and $f$ for further.

Case 1

After 1 hour, if $c$ finds the wall, then $f$ turns back 180 degrees and walks 2 hour. 1 + 2 = 3 hours.

enter image description here

Case 2

If neither finds the wall after 1 hour, then both of them again turns 90 degrees in the opposite directions and walk 1 more hour.

Case 2.1
If the $c$ heads in the right direction, then he will find the wall less than 1 hour. $f$ will compute the angle of his direction, correct his direction and walk through the wall, which will take less than 2 hours.

enter image description here

Case 2.2
If $c$ again heads in the wrong direction, he will not find wall after 1 hour. However, now $f$ will be closer to the wall. $f$ turns another 90 degrees and keeps on walking. $c$ waits $f$ to reach the wall, and then himself walks through the wall.

enter image description here

  • $\begingroup$ What if the wall is on the bottom of your last diagram instead? Then neither of them will find it. $\endgroup$
    – f''
    Commented Jan 31, 2016 at 20:25
  • $\begingroup$ then you just swap c with f $\endgroup$
    – padawan
    Commented Jan 31, 2016 at 20:51
  • $\begingroup$ But how do they know which one is c (who should stay still) and who is f (who should turn and keep walking)? $\endgroup$
    – f''
    Commented Jan 31, 2016 at 22:06
  • $\begingroup$ it does not matter. all the cases can be mirrored. one of them will be c and the other will be f with respect to the wall's place $\endgroup$
    – padawan
    Commented Jan 31, 2016 at 22:39
  • $\begingroup$ Yes, but if they don't know who is c and who is f, how do they determine who should walk and who should wait? $\endgroup$
    – f''
    Commented Jan 31, 2016 at 22:44

The answer of two blindfolded men question

NOTE: Before checking out the solution of this one, please check the original question first to figure out this one.

First of all, this question has totally different type of solution than recent blindfolded near the great Wall of China puzzle. I will explain the solution as simple as possible and step by step since the solution is longer.

The critical difference of this problem is that if one man encounters the wall on the way, the other man stops and walk perpendicularly to the wall since both needs to be at the wall when the time is up. To do that, you have to consider the perpendicular distance between one man and the other man’s wall all the time and since the solution is symmetric, it does not matter who encounters the wall first.

I believe the idea is clear at this point. Let’s come to the solution:

Figure 1 Figure-1

The first part is the easiest part since the distance between both guys would not affect the solution much, so they need to move on the arc to cover the area as much as possible. As the original question, this one is pretty the same. We need to move out of the arc until where it covers as much as possible depending on the distance you get over:

Eq-1: $maximize\ \left [2a-(\frac{1}{\cos(a)}-1)-\tan(a)\right ] \ where\ a<\pi/2$

In this question, I tried to explain where cosine and tangent comes from on the Figure 1. Since those are basic maths, I will not get into much detail here and in the further. But this equation explains itself well and please ask if you still do not understand any equation/figure later. As a result;

You may see the result here: Wolfram Alpha Solution for minimum a

As a result $a=\pi/6$

I do not consider the distance between these guys here, I will explain it later but you may easily guess that the distance between these men are not much at the beginning.

The total distance they moved from the beginning to the end can be found easily from the figure:


So after taking the fastest way to the arc again, how much you need to move on the arc? That part is the hardest part since you need to start to consider the distance between these guys now!

These guys is supposed to move on the arc and move on the tangent after a while to complete all possible points on it, but now they need to consider the distance between them and the perpendicular distance of the wall of the other man as well:

Eq-2: $\frac{1}{\cos(a)}+\tan(a)+b+\tan(c_{1})+\left [\frac{cos(3c_{1}+4a+2b-\pi)}{cos(c_{1})}+1 \right ]$

Let me explain the equation briefly, I think everything is clear except the bracket equation. The bracket equation is actually the distance between one guy and the other guy’s wall on the way at any point.

enter image description here Figure-2

In F2 (please zoom into the figure), c represents the distance you are planning to take on the tangent and after that point you will move back to the arc again.

Why do we need to move on the tangent again? The reason is actually secure the distance between these guys, even it seems the distance is getting bigger, it is not since you cover faster than moving on the arc.

And $c_1$ represents while moving on the tangent to $c$, what would be the distance to the wall if the other one finds the wall: We have to consider every single point on the way to $c$ point. To do that, we need to minimize $b$ where $c<c_1$ and $c1<\pi/2$;

$\frac{1}{\cos(a)}+\tan(a)+b+\tan(c_{1})+\left [\frac{cos(3c_{1}+4a+2b-\pi)}{cos(c_{1})}+1 \right ]\ where\ 0<d<\pi/2$

As result: Wolfram Alpha Solution for b

We got: $b=0.224896$.

Before explain what c should be, we need to consider point d in that part to minimize the distance between these men as below:

enter image description here Figure-3

In Figure 3- you will notice a dark angle ($\mu$), if we can find that angle, we can find the perpendicular distance to the wall when a man encounter the wall just after finishing the arc move after $c$, so;

$\mu = \pi/2 - \left [2\pi-4c-4a-2b-\pi/2 \right ]=4a+2b+4c-\pi$

as a result the total distance to the wall included everything;

minimum distance to the wall: $\frac{1}{cos(a)}+\tan(a)+b+2tan(c)+\cos(4a+2b+4c-\pi)+1$

we know $a$ and $b$ and if we minimize it with wolfram alpha: Wolfram Alpha Result for $c$ as a minimum

so $c=0.687035$

Since we have found the best $a$,$b$ and $c$, let's find out the worst case from this route. Actually there is two critical points where you can find the worst case, one of them is in between $b$ point and $c$, the other one is point $d$:

$b$ -> $c$: Wolfram Alpha Result for $c_1 - 4.27982

with maximization of; $maximize\ \frac{1}{cos(\pi/6)}+\tan(\pi/6)+(b=0.224896)+tan(c_1)+cos(3c_1+2\times0.224896-\pi/3)/cos(c)+1\ where\ 0<c_1<0.687035=c$

$d$: Wolfram Alpha Result for d - 4.04970

As result: the optimal worst case for 2 blindfolded men is 4.27982

  • $\begingroup$ You still haven't shown a case that makes my solution worse than 4.27982. There is enough room between 4.04970 and 4.27982 that my choice of $c$, even if doesn't always produce the best route, still matches the optimum worst case. $\endgroup$
    – f''
    Commented Feb 1, 2016 at 19:51
  • $\begingroup$ In particular, the path only needs to be the same up to the point where the worst case finds the wall. After that, there is a significant amount of leeway where the path can be changed without making the worst-case time greater than 4.27982. $\endgroup$
    – f''
    Commented Feb 1, 2016 at 19:57
  • $\begingroup$ @f'' that's exactly what I meant in your solution's comment... You found out the optimum worst case but your path was wrong because of $d$ point that I pointed out in my solution. But you are right about you can choose $d$ value between 4.0497 to 4.27982, it would not change the result, but your $d$ value was not in this range. That's why I did not accept your answer as a solution but still gave you +50 rep! :) $\endgroup$
    – Oray
    Commented Feb 1, 2016 at 20:00
  • $\begingroup$ Your calculations seem to have my $c$ as 24.295 degrees, but that wasn't what I chose for $c$. The method I described makes $c$ such that $2a+b+2c=\frac{3\pi}{4}$. The actual value of $c$ in my method is approximately 0.54205 radians or 31.057 degrees. You can check that with this value of $c$, the time for $d$ is 4.16139, which is still within the range. $\endgroup$
    – f''
    Commented Feb 1, 2016 at 20:10
  • $\begingroup$ 24.295 degrees is a completely different number, which describes what the angle of the wall is in the worst case. It has nothing to do with the path. $\endgroup$
    – f''
    Commented Feb 1, 2016 at 20:12

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