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The ruler of the island of wishes to establish diplomatic relations with your country (the population of the island are all either Honestants, who speak only the truth, or Swindlecants, who always lie). He sends four envoys: an Admiral, a Bishop, a Consul, and a Duke, whose names are Edwards, Fitzroy, Gerrard, and Harris, though perhaps not in that order. The envoys introduce themselves to you:

Envoy 1: "My name is Edwards. Fitzroy is an Admiral."

Envoy 2: "My name is Fitzroy. There are at least two Swindlecants in this delegation."

Envoy 3: "My name is Gerrard. If Harris is a Bishop, then Edwards is a Swindlecant."

Envoy 4: "I am the Consul. My name is not Harris."

Well, that's enough chit-chat. You have a job to do, which is to introduce the envoys one at a time to your king, using their correct names and official titles. Any mistakes or omissions would be politically embarrassing, so naturally you would be beheaded for your incompetence. Ah! Here comes the king now...

Stating the obvious:

Multiple statements are not logical conjunctions, in other words, each statement made by a Swindlecant must be a lie, not just one of them.

Each envoy has one and only one official title (Admiral/Bishop/Consul/Duke).

Edited by octopuppy
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I find 13 solutions.

Envoy 2 is an Honestant.

Otherwise 1234=HSHH.

Envoy 4 is Fitzroy. Envoy 4 is both Admiral and Consul.

Envoy 1 is a Swindlecant.

Otherwise 1234=HHSS.

Envoy 1 is Edwards. Envoy 2 is Fitzroy. Both Envoy 3 and Envoy 4 are Harris.

Envoy 3 is a Swindlecant.

Otherwise 1234=SHHS.

Envoy 2 is Fitzroy. Envoy 3 is Gerrard. Both Envoy 1 and Envoy 4 are Harris.

Thus 1234=SHSx:

Envoy 1: My name is not Edwards. Fitzroy is not an Admiral.

Envoy 2: My name is Fitzroy. There are at least 2 Swindlecants here.

Envoy 3: My name is not Gerrard. Edwards is a Swindlecant. Harris is a Bishop.

Envoy 4 is an Honestant => I am the Consul. My name is not Harris:

1234=SHSH

1234=HFEG

1!=B. 2!=A. 4=C.

Duke could be any of Envoys 1, 2, 3: [3 soltuions.]

1234=DBAC.

1234=ADBC.

1234=ABDC.

Envoy 4 is a Swindlecant => I am not the Consul. My name is Harris.

1234=SHSS

1234=GFEH

2!=A. 4!=B. 4!=C.

Duke could be any of the 4 Envoys: [10 solutions.]

1234=DBCA. 1234=DCBA.

1234=BDCA. 1234=CDBA.

1234=BCDA. 1234=CBDA.

1234=ABCD. 1234=CBAD. 1234=ACBD. 1234=BCAD.

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I find 13 solutions.
Envoy 2 is an Honestant.

Otherwise 1234=HSHH.

Envoy 4 is Fitzroy. Envoy 4 is both Admiral and Consul.

Envoy 1 is a Swindlecant.

Otherwise 1234=HHSS.

Envoy 1 is Edwards. Envoy 2 is Fitzroy. Both Envoy 3 and Envoy 4 are Harris.

Envoy 3 is a Swindlecant.

Otherwise 1234=SHHS.

Envoy 2 is Fitzroy. Envoy 3 is Gerrard. Both Envoy 1 and Envoy 4 are Harris.

Thus 1234=SHSx:

Envoy 1: My name is not Edwards. Fitzroy is not an Admiral.

Envoy 2: My name is Fitzroy. There are at least 2 Swindlecants here.

Envoy 3: My name is not Gerrard. Edwards is a Swindlecant. Harris is a Bishop.

Envoy 4 is an Honestant => I am the Consul. My name is not Harris:

1234=SHSH

1234=HFEG

1!=B. 2!=A. 4=C.

Duke could be any of Envoys 1, 2, 3: [3 soltuions.]

1234=DBAC.

1234=ADBC.

1234=ABDC.

Envoy 4 is a Swindlecant => I am not the Consul. My name is Harris.

1234=SHSS

1234=GFEH

2!=A. 4!=B. 4!=C.

Duke could be any of the 4 Envoys: [10 solutions.]

1234=DBCA. 1234=DCBA.

1234=BDCA. 1234=CDBA.

1234=BCDA. 1234=CBDA.

1234=ABCD. 1234=CBAD. 1234=ACBD. 1234=BCAD.

So close, bonanova, but I think you got a bit lost in the middle there. Take another look at Envoy 3...

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Nice one bonanova. I must say I am quite pleased with the format for this puzzle. The possibilities are enormous, and maybe I wimped out by leaving a trail of breadcrumbs. Who knows, I might do a harder version, or someone else might ;)

Here's the complete solution for anyone who's stuck:

Is Envoy 2 lying? If so, there is at most one Swindlecant, which must be Envoy 2, so all the others must be Honestants. This would mean that Envoy 2’s name is Harris and Envoy 4’s name is Fitzroy. But Envoy 4 (Fitzroy) says he is the Consul and Envoy 1 says Fitzroy is the Admiral, so that gives us a contradiction.

Therefore Envoy 2 must be an Honestant, and at least two of the others are Swindlecants.

Next consider the status of Envoys 1 and 3. They cannot both be Honestants as this would leave only one Swindlecant. If one is an Honestant and the other a Swindlecant, the Swindlecant’s name must be Harris, as there would be no other choices of name. But that would make Envoy 4 an Honestant (“My name is not Harris”), so we would still have only one Swindlecant.

The only way to have two Swindlecants is if Envoys 1 and 3 are both Swindlecants.

From Envoy 3’s statement we can now infer that Harris is a Bishop, and Edwards is an Honestant.

The only way Edwards could be an Honestant is if Envoy 4 is Edwards.

Gerrard can’t be Envoy 3 so he must be Envoy 1. So Envoy 3 is Harris.

Edwards is the Consul. Harris is the Bishop. Fitzroy (from Envoy 1’s statement) may not be the Admiral, so he must be the Duke. So Gerrard is the Admiral.

This gives us:

Envoy 1: Admiral Gerrard (a Swindlecant)

Envoy 2: Duke Fitzroy (an Honestant)

Envoy 3: Bishop Harris (a Swindlecant)

Envoy 4: Consul Edwards (an Honestant)

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I think the following is the solution :

Envoy 1 is Gerrard, the Admiral.

Envoy 2 is Fitzroy, the Duke.

Envoy 3 is Harris, the Bisop.

Envoy 4 is Edward, the Consul.

Yes? No?

This is my first time answering a riddle on brainden, so I'm not going to check, but wait to be corrected (or told that I'm right :) )

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I think the following is the solution :

Envoy 1 is Gerrard, the Admiral.

Envoy 2 is Fitzroy, the Duke.

Envoy 3 is Harris, the Bisop.

Envoy 4 is Edward, the Consul.

Yes? No?

This is my first time answering a riddle on brainden, so I'm not going to check, but wait to be corrected (or told that I'm right :) )

You're absolutely right. Well done and welcome to Brainden!
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