6
$\begingroup$

In the puzzling community, the French mathematician Édouard Lucas is well-known as the inventor of the Towers of Hanoi. Beyond the Towers of Hanoi, Lucas' recreational math book "Récréations mathématiques" contains a wealth of other puzzles, and in particular the following question:

Daily at noon a ship departs from Le Havre bound for New York and conversely, another ship leaves New York bound for Le Havre. The crossing lasts $7$ days and $7$ nights. During the passage to New York, how many Le Havre-bound ships will the New York-bound vessel meet, with today as its date of departure?

$\endgroup$
  • 4
    $\begingroup$ they leave simultaneously? not according to time zones? $\endgroup$ – JMP Mar 31 '16 at 9:59
  • 2
    $\begingroup$ Do they arrive before or after noon ? $\endgroup$ – Fabich Mar 31 '16 at 10:00
  • 2
    $\begingroup$ @GordonK, not the case, as the use of the words 'the' and 'its' imply that you are talking about only the New York bound vessel $\endgroup$ – SGR Mar 31 '16 at 10:02
  • 4
    $\begingroup$ The puzzle is poorly formulated. Questions left open: do the ships leave at noon local time or noon Le Havre time? What does "7 days and 7 nights" mean? Is it a fixed number of hours or does it mean that every journey always ends at noon local time? How long does the ship stay in port? Does it stay for a day and a night or does it essentially bounce against the dock, arriving and immediately departing at noon? And what does "during the passage" mean? Does the passage end on arrival so we have the ship being both "during passage" and "idle", or does it always count as "during passage"? $\endgroup$ – MichaelK Apr 1 '16 at 9:27
  • 3
    $\begingroup$ SPOILER ALERT: I might also add that the main point of the riddle — the mind trap that you are supposed to dodge — is that you do not meet only 7 ships but you actually meet 2 ships per day & night. Once you figure that out you have essentially "solved" it because that thing is what you are supposed to realize. It seems Lucas did not bother the nitty gritty stuff about what happens in the ports, time zones, and all those other ambiguities; they are just minor details. The point of the puzzle is to arrive at the conclusion "We meet 2 ships per day". $\endgroup$ – MichaelK Apr 1 '16 at 11:47

13 Answers 13

6
$\begingroup$

How many Le Havre-bound ships will the New York-bound vessel meet, with today as its date of departure?

15

Reason:

7 : each passing day a ship will depart, which will be crossed along the way, this part is easy.
+ 6 : when you depart, there are already ships on their way. From these, one of them will be docking at Le Havre; one casting at New York; and six will be currently at sea (which you will cross along your way).
+ 2 : when you depart, there will be a ship that is docked when you cast off. Also, when you dock a ship will be cast out.

So: 7 + 6 + 2 = 15

$\endgroup$
12
$\begingroup$

Well, I'm a bit confused but I'd say

$15$, $13$ or $1$

With the following reasoning

A ship leaves each port every 24 hours. Since they are travelling in opposite directions with exactly the same average speed then every ship making the passage can expect to meet a ship travelling in the opposite direction on average every 12 hours. If we assume that the timing is exactly 7 24-hour days and we can count as "meeting" the zero time instant when one ship is arriving and the other departing, then the answer must be 15. I believe this is the intended answer.

If we disregard the two endpoints, the answer is 13.

The last phrase as it is rendered here could be interpreted to mean "how many ships with today's departure date will the New York-bound ship meet". Only one ship from New York has today as its date of departure. The ship from Le Havre will meet it. A google search suggests that this is not the intended solution. I assume the original was written in French and that this is merely a slightly awkward translation. Nevertheless, I give this as a possible solution to the problem as stated for the sake of completeness.
Note that it is never stated that ships leave New York at noon local time nor is the arrival time actually mentioned. I am assuming that "conversely" actually means "simultaneously". Since Édouard Lucas lived before the institution of daylight saving time in New York and Le Havre, it is barely possible to interpret "conversely" as meaning that ships depart at noon local time at both ends. If, then, the journey is 7 24-hour days then ships will arrive at New York at noon Le Havre time and depart New York at Noon local time making 14 meetings another possible (but unreasonable in my opinion) solution. If you go way, way out on a limb you could argue about the meaning of "7 days and 7 nights" and claim that the passage from New York to Le Havre (aided, perhaps, by the Gulf Stream) actually takes fewer hours than the passage in the opposite direction exactly matching the difference in time zones so that all ships arrive and depart at noon local time. This, of course, brings us back to 15 meetings in a somewhat interesting way. That, I sincerely hope, is my last word on the subject.

$\endgroup$
  • 1
    $\begingroup$ I agree that the the 'today as its date of departure' requirement leas to answer 'only one'. Perhaps it could mean something else? $\endgroup$ – florisla Mar 31 '16 at 12:30
  • $\begingroup$ Cannot be 1 because that would violate the "daily" requirement. $\endgroup$ – MichaelK Mar 31 '16 at 15:49
  • 2
    $\begingroup$ @MichaelKarnerfors - the "1" comes from interpreting the question as meaning "how many ships that departed today will it meet?" Obviously, since only 1 of those daily ships left today, it is the only one that counts. But it seems more likely to me that the phrase refers to the NY-bound ship, and is just to say that the puzzle is about how many LH-bound ships this ship meets, not the NY-bound ships leaving on other days. $\endgroup$ – Paul Sinclair Mar 31 '16 at 17:02
  • $\begingroup$ Ah, I see what you mean... "with today as its date of departure" being applies to the Le Havre bound ship instead of the New York bound. $\endgroup$ – MichaelK Mar 31 '16 at 18:02
  • $\begingroup$ I interpreted that as somewhat awkward wording for "how many Le Havre-bound ships will the New York-bound vessel with today as its date of departure meet"... $\endgroup$ – psmears Mar 31 '16 at 21:24
6
$\begingroup$

Answer:

14

Assumptions:

  • A one-way journey takes exactly 7 x 24 hours.
  • The ships leave at noon, local time.
  • The ships leave at the first noon that takes place after arriving.

Reasoning;

Noon in New York happens 6 hours after noon in Le Havre. This means that 6 hours after a ship leaves Le Havre, a ship sets out from New York. Also, just as a ship leaves Le Havre, a ship is arriving in New York. At 11:55 our ship is waiting to depart. At the same time, 7 ships are west-bound, in transit to New York. One is 5 minutes away from arriving in New York. The others are 1, 2, 3, 4, 5 and 6 days — plus 5 minutes — from arriving in New York. Also, in the east-bound lane, another 7 ships are heading for Le Havre. One is 6 hours from arriving. The others are 1, 2, 3, 4, 5 and 6 days — plus 6 hours — from arriving in Le Havre. Our ship must meet all of these. That makes 7 + 7 = 14 ships.

EDIT: The reason for these assumptions is that without them, we have a wide host of ambiguities that need to be resolved.

  1. When does the New York ship leave? Is it noon New York time or noon Le Havre time (i.e. 0600 in the morning in New York)?

  2. What does it mean that a passing takes "7 days and 7 nights"? Is it 7 x 24 hours or does it mean that the journey always ends at noon, local time?

  3. Do the ships idle in port or not? Do they stay for unloading and reloading, or do they depart instantly?

  4. What does "meet during the passing" mean? If the ship can be in two states — Passing and Idle — when does it move from one state to the other? Because this would mean that a ship that departs a port at noon while another arrives at the same time do not meet during the passing, because they go from Idle - Passing to Passing - Idle at the same instant... they are never Passing-Passing at the same time. Or are they?! Is it perhaps three states: Idle, berthing/unberthing (does this count as "during the passing" or not), and Passing? And if that is the case... when does berthing/unberthing start and end?!

All of these ambiguities are resolved with the assumptions I make, and it even makes sense from a realistic point of view...

A ship that arrives in New York has from 0600 to 1200 to unload/disembark and reload/embark. A ship that arrives in Le Havre at 1800 in the evening has time to let the stevedores go home for a good night's sleep, then come back at 0600 in the morning and start unloading/reloading the ship in the same time that the New York stevedores have. Also it makes sense that travelling the same distance takes the same amount of time. And finally it makes the schedule make sense: "The service departs every day at noon. No no good Sir/Madam, don't bother with time zones, it's noon local time. Easy to remember, just look at the nearest public clock and you will know").

For the record I'm a software developer and these kinds of ambiguities is what makes reading requirements specifications so much "fun"...

$\endgroup$
  • $\begingroup$ You are assuming that the ship arrives, unloads/loads and returns immediately so that you pass that ship again, correct? $\endgroup$ – Trenin Mar 31 '16 at 14:21
  • $\begingroup$ No. A ship arrives in NY at 06:00 (local time) in the morning. It unloads and reloads, leaving at 12:00. A ship that arrives in Le Havre docks at 18:00, unloading, reloading and leaving 18 hours later at 12:00 local time. $\endgroup$ – MichaelK Mar 31 '16 at 14:27
  • $\begingroup$ Interesting. I took "7 days and 7 nights" to mean that if I leave at a certain time locally, I will arrive at the same time locally 7 days later. Thus, arriving at noon. $\endgroup$ – Trenin Mar 31 '16 at 14:49
  • $\begingroup$ I disagree. 'A day and a night' I assume is a measure of time, equivalent of 24 hours. $\endgroup$ – MichaelK Mar 31 '16 at 14:52
  • 1
    $\begingroup$ Yes - travelling in those directions take a different amount of time if you travel for a day and night. Other answers take this into account and it isn't a problem at all. $\endgroup$ – Trenin Mar 31 '16 at 17:28
5
$\begingroup$

Answer:

13

Because

The crossing takes 7 days and 7 nights. If a ship leaves at noon, we can assume that it will take a few hours less than 7*24 hours, because at the end of the 7th night it's not yet noon (whatever time you consider as the end of the night).

So, at the time of the departure, we have 6 ships on their way to Le Havre and 1 ship that is departing now (or in 6 hours, if we consider the timezone difference). There's a ship that docked a few hours ago at Le Havre and I don't think we can say that this is met.

During the 7 days and 7 nights of crossing, we have 6 ships that will depart each day (because we already considered the one departing the first day). When we dock at New York it's dawn (whether we consider the timezone or not, it's before noon) so the ship departing at noon isn't met.
I wasn't quite sure yet, even after posting the answer, so I tried to make a graph to help me out: graph

$\endgroup$
4
$\begingroup$

Answer:

15


Reasoning:

For each of the $7$ days the ship travels to New York, there would be one ship returning to Le Havre, giving us a total of $7$ ships.
But that is not all. The reason being that some ships are already on their way back to Le Havre from New York, giving us a total of $7$ more ships.
There is also one extra ship that gets crossed: the one about to leave New York just as our ship reaches there, giving a total of $15$ ships.


Generalisation:

If the ship takes $n$ days to travel to its destination, then number of similar ships encountered on the way is $2n+1$.

$\endgroup$
  • 1
    $\begingroup$ Good answer. I like the fact that you generalized. It's important to do that with problems like these. $\endgroup$ – DevilApple227 Mar 31 '16 at 12:18
  • $\begingroup$ Isn't that generalization only true if the similar ships leave at once-a-day intervals? $\endgroup$ – cr0 Mar 31 '16 at 13:43
  • $\begingroup$ @cr0 I suppose it was clearly written in the question that the ships leave each day at noon. $\endgroup$ – Dragonemperor42 Mar 31 '16 at 13:46
  • $\begingroup$ Yep...I guess I was trying to take your generalization farther than this specific question. @Shkeil attempted a more generalized equation for this it looks like: 2T/H + 1 = ships encountered en route on a voyage where T is trip length in hours and H is gap between departures of 2 boats from the same dock. So if the departure gab is cut in half to 12, we'd have 2(24)/12 + 1 = 29 encounters instead of 15. $\endgroup$ – cr0 Mar 31 '16 at 13:56
  • $\begingroup$ (Which makes me recognize, halving H the gab between departures will result in doubling the encounters, minus 1. ...Math!) $\endgroup$ – cr0 Mar 31 '16 at 13:57
4
$\begingroup$

The answer is:

Zero

Because

Any east-bound ship would not be taking the same route as a west-bound ship. They would each take a route where the ocean currents are in their favor. Since the currents in any given location cannot be favorable for both east- and west-bound ships, there is no way they would ever be in the same location.

As further proof of this:

The journey takes exactly 7 days and 7 nights - both ways. If the ships were on the same route, one ship would have a shorter trip than the other.

Yay, lateral thinking. But this is what would probably happen in reality.

$\endgroup$
  • $\begingroup$ Depends on how we define "meet". If we define "meet" as "crossing a longitude at the same moment, no matter their latitude", then you don't need to worry about sea lanes. $\endgroup$ – MichaelK Mar 31 '16 at 14:16
  • $\begingroup$ I thought about this too... however, I hope the calculation-puzzle tag in the post isn't there by chance $\endgroup$ – Alessandro Mar 31 '16 at 14:20
  • $\begingroup$ There could be a port on the way there where they both temporarily stay. Imagine a figure 8 round trip route. $\endgroup$ – Superbest Mar 31 '16 at 22:34
4
$\begingroup$

Naive answer

There are $24 \times 7$ hours in "7 days and 7 nights". The ships leave every 24 hours, but since they are travelling towards each other, you will encounter one every 12 hours. This means you will see 14 ships on the trip.

Taking into account both ends

Lets say the eastbound ships arrive at 12:01pm and the westbound ships leave at 12:00pm. This avoids meeting a ship just as the westbound ship departs. Thus, the eastbound ships must depart at 12:01pm as well. The first ship will be encountered 30 seconds after departure since it is due to arrive very shortly. When the westbound ship arrives, it will get there a minute before the westbound ship leaves, so that ship is not met. In this case, you count the ship at one end, but not the other, so it also gives 14 meetings.

Taking into account time zones

If they each leave at their own noon and arrive at local noon, then eastbound ships are faster than westbound ships. Westbound ships will take $168+6=174$ hours and eastbound will take $168-6=162$ hours.

Their speed towards each other, however, is the same, so their meeting frequency remains at one every 12 hours. The time these occur will always be noon or midnight local time, but that requires time zones to be very accurate, which they aren't. Regardless, there are still 14 meetings.

$\endgroup$
  • $\begingroup$ On your third answer, it's never said the ship arrive at noon. Only they leave at noon. $\endgroup$ – Shkeil Mar 31 '16 at 14:24
  • $\begingroup$ @Shkeil And it takes exactly 7 days and 7 nights to cross, so it must arrive at noon as well. $\endgroup$ – Trenin Mar 31 '16 at 14:25
3
$\begingroup$

As every answer already told it, the answer is:

15

But I'll use an other proof, a more mathematical one.

There is a gap of H hours between 2 boats from the same dock. But since they go on opposite side, a ship from New York have to wait only H/2 hours after meet a ship from Le Havre to see another one (Half part from the ship itself, other part from the ship it'll meet).

The trip last T hours, so the ship will meet 2T/H ship +1, the one it meet at departure (or arrival).

With H=24 and T = 7x24, we have the solution: 2x7+1 = 15 ships.

$\endgroup$
  • $\begingroup$ You are counting the ship at arrival and departure to be "met". $\endgroup$ – Trenin Mar 31 '16 at 14:23
  • $\begingroup$ @Trenin Yes I do, even if they don't meet on the ocean, they still meet on the docks. The question isn't restrictive about it. $\endgroup$ – Shkeil Mar 31 '16 at 14:26
  • $\begingroup$ OK. Just clarifying. In my interpretation, I chose to count one as a meeting and not the other. $\endgroup$ – Trenin Mar 31 '16 at 14:50
2
$\begingroup$

The question fails to take into account the fact that, leaving 3/31, the ships will be heading into heavy iceberg season. (Remember 14 Apr 1912?)

As such, as a result of a Le Havre bound ship hitting an ice berg and requiring assistance three days in, all shipping ceased until they could assess the danger. As such, only nine ships were met all told (the six already out, the damaged ship, the two that left before the damaged ship was damaged). Technically two of those were no longer Le Havre bound, as they returned to New York, but I think they still count given their original destination - otherwise seven is correct.

$\endgroup$
1
$\begingroup$

The answer is

15 ships.

At the moment where the ship takes off, six New York-Le Havre ships are already at sea, one is docking in Le Havre and one is casting off in New York. At noon of every day of the journey a new ship will be launched. Thus the ship must encounter every ship that was already at sea (6 ships) and each ship that was launched during its journey (including the one that launched at the same time as itself, so 7 more ships).

So the ship will pass

  • 13 other ships at sea, and
  • also encounter 1 ship that docked as it cast off, and
  • another 1 ship that cast off as it docked.
$\endgroup$
0
$\begingroup$

My answer is

1

Because

During the passage to New York, how many Le Havre-bound ships will the New York-bound vessel meet, with today as its date of departure?
The bolded part restricts the question to applying only to ships that depart today. There is only one.

$\endgroup$
0
$\begingroup$

Answer:

2

Reasoning:

Based on the logic of Darrel Hoffman, but noting that it meets one when it leaves (the one just arriving from New York), and another when it arrives (the one just leaving for Le Havre).

$\endgroup$
0
$\begingroup$

My Answer is

16 if Time Zone is not a factor - Because starting at the beginning, there will be a Le Havre at the dock when NY leaves, then over the course of 7 days and 7 nights, the NY will run into 14 ships, because if the NY and Le Havre are traveling at the same speed, then the NY ship will see the Le Havre every 12 hours (Ships depart daily, each ship moves the same speed, will meet each other halfway) Finally as the ship arrives at the end of Night 7 beginning of Day 8, it will see the last Le Havre ship, leaving port, giving us 16.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.