Please answer its important to me

[mohammedhmandour]

You are an expert on paranormal activity and have been hired to locate a spirit haunting an old resort hotel. Strong signs indicate that the spirit lies behind one of four doors. The inscriptions on each door read as follows:

Door A: It’s behind B or C
Door B: It’s behind A or D
Door C: It’s in here
Door D: It’s not in here

Your psychic powers have told you three of the inscriptions are false, and one is true. Behind which door will you find the spirit?

[Bazzy]

Does it relate to C++?

[jmc]

ROFL a riddle, i take door D (only B is true). ^^

[jsmith]

If your psychic powers are good enough to tell you that three are false and only one is true, then
why couldn't your psychic powers just give you the right answer?

[jmc]

Maybe because most people with psychic powers have a mental retardation?

[PiMaster]

This doesn't really have much to do with programming, more game design.... Maybe this riddle is more suited for Q&A sites like Y!A?

[Zhuge]

I don't see any reason we can't just open all the doors... >_>

[Duthomhas]

I think the point was to use logic and find the contradictions

[Bazzy]

Why didn't he ask at the reception of the hotel?

[chrisname]

This is easy.

A can't be true because if A is true then D is false but D being false contradicts A being true.
B must be true because if B is true then D is false, and if D is false then it must be behind D.
C is false because C being true would conflict with D being false
D is false because D being true conflicts with B being false.

It's B.

[Athar]

It's D.

C/D can't be true because then either A or B would also be true.
A can't be true because then either C or D would be true.
B is true. It can't be behind A because D would be true, so it is D, for which there are no contradictions.

[helios]

I agree with Athar, but I prefer this reasoning:

A=b || c
B=a || d
C=c
D=!d

If D, then !A && !B && !C.
!A && !B == !(b || c) && !(a || d) == !b && !c && !a && !d.
If D isn't the correct answer, then no door is the correct answer.

[chrisname]

@Athar,
Yeah, B is true like I said:

I wrote:

B must be true because if B is true then D is false, and if D is false then it must be behind D.

[jmc]

Why didn't he ask at the reception of the hotel?

The best reply I've seen so far :)) thx Bazzy^^

[jsmith]

I prefer the reasoning that says that the spirit is behind the door which has only one reference.
Both B and D refer to A; A and D refer to B; A, C and D refer to C; only B refers to D. Therefore
the spirit is behind D.

[Bazzy]

I prefer the reasoning that says that it's not a question to be posted in general C++ forum

[chrisname]

I prefer the reasoning that says that Bazzy is probably right and this thread ought to be deleted.

[m4ster r0shi]

Let's make it relate to C++ ^_^

#include <iostream>
using namespace std;

//the spirit is in...
bool A, B, C, D;

//inscriptions
bool IA() {return B||C;}
bool IB() {return A||D;}
bool IC() {return C;}
bool ID() {return !D;}

bool check()
{
    return IA()+IB()+IC()+ID()==1;
}

int main()
{
    char door;

    for (;;)
    {
        A=1;
        if (check()) {door='A'; break;}

        A=0;B=1;
        if (check()) {door='B'; break;}

        B=0;C=1;
        if (check()) {door='C'; break;}

        C=0;D=1;
        if (check()) {door='D'; break;}
    }

    cout << "The spirit is behind door ";
    cout << door << endl;

    cin.get();
    return 0;
}

[helios]

I don't like it.

#include <iostream>

//inscriptions
inline int IA(int a){ return a==1 || a==2; }
inline int IB(int a){ return a==0 || a==3; }
inline int IC(int a){ return a==2; }
inline int ID(int a){ return a!=3; }

bool check(int a){
    return IA(a)+IB(a)+IC(a)+ID(a)==1;
}

int main(){
    int door=-1;
    char doors[]="ABCD";
    for (int a=0;a<4 && door<0;a++)
        if (check(a))
            door=a;
    if (door<0)
        std::cerr <<"Not found.\n";
    else
        std::cout <<"The spirit is behind door "<<doors[door]<<std::endl;
    return door>=0;
}

[m4ster r0shi]

helios wrote:

I don't like it.

I don't care.