6
$\begingroup$

This variation of the classical Farmer Goat puzzle was in a puzzle contest at my college two years ago.
Here it goes.

Mr. Bruce Wayne is very fond of parties. Well, he’s more fond of girls, but, let’s just say that he’s fond of parties.
So, he plans a party at an island off the coast of Gotham Bay. All his guests have already reached. And we all know, Mr. Wayne likes to make an entrance. So, he decides to go in the chopper along with two supermodels. But unfortunately for Mr. Wayne, the chopper can hold only two people.

To add insult to the injury, the Mayor arrives with his two assistants to get a ride in the chopper. And there’s Mr. Fox, but with a secret case. Never mind, let’s just go to the party. But it seems someone doesn't want Mr. Wayne to enjoy tonight because there is no pilot to fly the chopper. And only Mr. Wayne, The Mayor and Mr. Fox know how to fly the chopper.

Oh no, wait! That’s not it. Mayor’s assistants won't stay with Mr. Wayne without their Mayor (I don’t know Bruce Wayne freaks them out); the supermodels won’t stay with the Mayor without Bruce (they think the Mayor is a pervert but Bruce isn't!) and Mr. Fox would never leave the case.

Let’s revise the situation.

The passenger list is:

  1. Bruce Wayne
  2. Bruce's 2 supermodels
  3. Mayor
  4. His 2 assistants
  5. Mr. Fox
  6. The secret case


And the constraints are:

  • The capacity of the chopper is two persons.
  • The secret case counts as a person (what, it's important!).
  • Only Bruce, Mayor or Mr. Fox can fly the chopper.
  • The Supermodels won’t stay with the Mayor without Bruce; Assistants won’t stay with Bruce without their Mayor and Mr. Fox won’t leave the case alone.

So much for making an entrance, Mr. Wayne! All things said and done, we still have a party to attend to. Now that you’ve understood the situation, help Mr. Wayne and his friends (and the case, too) reach the party.

I think I was able to solve this one, can you solve this puzzle?

$\endgroup$
  • 1
    $\begingroup$ Can the helicopter carry a pilot plus 2 passengers, or 2 people including the pilot? $\endgroup$ – frodoskywalker Aug 26 '15 at 6:39
  • $\begingroup$ It can carry at most two passengers, one of which is the pilot himself. $\endgroup$ – ABcDexter Aug 26 '15 at 6:41
  • 1
    $\begingroup$ Mr. Fox would never leave the case or Mr. Fox won’t leave the case alone ? $\endgroup$ – The random guy Aug 26 '15 at 7:22
  • 1
    $\begingroup$ He can leave the case only if Bruce was with it. $\endgroup$ – ABcDexter Aug 26 '15 at 7:28
  • $\begingroup$ This looks very similar (but slightly different) from this puzzle: brainden.com/crossing-river.htm (see the River Crossing Game), which disallows Dad with the girls even if Dad goes in the next trip $\endgroup$ – justhalf Aug 26 '15 at 13:17
8
$\begingroup$

Let me give it a try: Let's call the models B1 and B2 since they belong to B and the assistants M1 and M2 and the case F1.

B and M
B back

B and B1
B and M back

B and B2
B back

M and B
M back

F and F1
B back

B and M
M back

M and M1
B and M back

M and M2
M back

finally M and B

| improve this answer | |
$\endgroup$
  • $\begingroup$ This is it. Nota : they are some other possibilities because of Mr Fox and his case (they can reach the island at the beginning, or at the end of the process for example) $\endgroup$ – The random guy Aug 26 '15 at 7:41
  • $\begingroup$ @IvoBeckers well done. $\endgroup$ – ABcDexter Aug 26 '15 at 7:51
  • $\begingroup$ @Therandomguy No, Mr. Fox can't go in the beginning as that would leave the Case "alone" at the other side. $\endgroup$ – ABcDexter Aug 26 '15 at 7:52
  • 2
    $\begingroup$ I don't get the conditions or this answer. After the very first two steps, it seems like a contradiction. "B and M, B back" But now B is on the side with M1 and M2 without M there... $\endgroup$ – user12713 Aug 26 '15 at 9:49
  • 1
    $\begingroup$ You are right but as @ABcDexter said on comment, assistants won't stay with Bruce. If he leave the next turn it is okay $\endgroup$ – The random guy Aug 26 '15 at 10:42
5
$\begingroup$

With Bruce Wayne B, Supermodels (S1 and S2), Mayor M, Assistants (A1 and A2), Mr Fox F and the Secret Case C.

Note: From the comments, Bruce can be with the Case while Mr Fox flys the chopper.

Notation: (pilot) and passenger

(F) and S1
(F) back

(F) and S2
(F) back

(M) and B (Bruce Wayne gets flown in by the Mayor to his waiting Supermodels).
(M) back

(F) and C
(F) back

(F) and M
(F) back

(F) and A1
(F) back

Finally, (F) and A2

I think that should work.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This sounds good with this rule about the case and Batm...Bruce. +1 for doing less travels. (ps : Poor Mr.Fox that do all the work D: ) $\endgroup$ – The random guy Aug 26 '15 at 14:24
  • $\begingroup$ Yes this also works :-) $\endgroup$ – ABcDexter Aug 27 '15 at 3:45
3
$\begingroup$

I think this conundrum cannot be solved. Given the constraints, only Batman (B), Mayor (M) and Mr. Fox (F) can pilot the chopper.

  • If F goes first, he will travel with his case. And because he won't leave his case on the other side, so he will also travel back with his case. This makes F a useless candidate to undertake the first journey.
  • If B pilots the first trip, he can only take one of his supermodels or M along with him (F doesn't travel without the case, so he cannot be a co-passenger). If he takes one supermodel, the other will freak out about staying with M on the same side. Taking M along seems like the only alternative. If both M and B go, then one of them has to return with the chopper, and either the supermodels or assistants will freak out, depending on who returns with the chopper.
  • The same scenario as above plays out if M pilots the first trip.

So in the light of the current constraints, I'd say that no one can go to the party.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Well, it's written assistants won't stay with Bruce, so that's the way. $\endgroup$ – ABcDexter Aug 26 '15 at 7:24
  • 8
    $\begingroup$ Batman? Who said anything about Batman? This is Bruce Wayne we're talking about here. Sheesh. $\endgroup$ – Kingrames Aug 26 '15 at 12:13
  • 2
    $\begingroup$ Oops, I guess I leaked a long guarded secret. $\endgroup$ – CodeNewbie Aug 26 '15 at 12:17
1
$\begingroup$

@IvoBeckers is technically correct, but more realistically:

B and M
B back

B and B1
B and M back

B and B2
B back

M and B
M back

F and F1
B knocks out F (because no innocent person was ever named Mr. Fox), "confiscates" the contents of F1, then makes his grand entrance with his 2 supermodels. Sure M wants to go to the party, but that is hardly B's problem.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ But the question was asking for something technically correct, not "realistic"! $\endgroup$ – Rand al'Thor Aug 26 '15 at 20:51
  • $\begingroup$ @randal'thor You are right. Mayor's assistants must be respected if not Mayor ;-) $\endgroup$ – ABcDexter Aug 27 '15 at 3:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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