A hard liar/truthteller type puzzle

[tepgn]

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Then again, with your lineup, it looks like you may be talking about the line method and not the pair method, so I may just be confused here.

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It's actually the 'line' method mixed with the 'pair' method [img]/images/graemlins/smile.gif[/img].

[BluffBlank]

wouldn't the spies have masks on and the knights have helmets and plate mail?

are there female knights/spies in this mix? I'd just spend my 150 questions asking them for a date. That'd seem to be more productive, since there's no way I'm figuring out how to do this.

[metsandfinsfan]

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wouldn't the spies have masks on and the knights have helmets and plate mail?

are there female knights/spies in this mix? I'd just spend my 150 questions asking them for a date. That'd seem to be more productive, since there's no way I'm figuring out how to do this.

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2

[BluffBlank]

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wouldn't the spies have masks on and the knights have helmets and plate mail?

are there female knights/spies in this mix? I'd just spend my 150 questions asking them for a date. That'd seem to be more productive, since there's no way I'm figuring out how to do this.

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2

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a new high for me!

[His Boy Elroy]

i missed you bluff

[img]/images/graemlins/heart.gif[/img]

[kokiri]

<font color="blue">"does that cute chick fancy me?"
<font color="red">"no"
<font color="blue">"Godammit you spy!"

"Next! Does that cute chick fancy me?"
<font color="red">"No"
<font color="blue">"Spy!"
.
.
.</font>

[Galwegian]

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i think that the stephen H/mute method works.

Summary:
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You stand everyone in line.

You ask a person about the next guy in line. If he says spy, you call them a spy pair and send them to the beach. If he says knight, you have the start of a knight chain.

Continue this until someone answers knight.

You then recursively ask the guy at the end of the knight chain about the next guy in line, eliminating spy pairs, or growing the chain in the same way as the first question.
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After 51 questions, you will have S spy pairs and a knight-chain of 51-S in length.

1) You know that, at best, the spy pair population is split 50:50 knights and spies. It could have more spies than knights, but contains, at most S knights.

2) you also know that there are at least 51 knights in the total population

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Let me use the notation xSy for "x says y is a spy", and xKy for "x says Y is a knight".

Suppose we try to follow the scheme you outline. The following could happen

1K2, 2K3, 3S4 (so (3,4) is a spy pair), 2S5 (so (2,5) is a spy pair), 1S6 (so (1,6) is a spy pair). So after 5 questions we have 3 spy pairs and a knight chain of length 0. Extending this idea, we see that after 51 questions we might end up with a bunch of spy pairs and a knight chain of length 0..... busto

[Galwegian]

Hold on. Now I think that your method might work - not sure. I need to think more about it.

[mute]

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Extending this idea, we see that after 51 questions we might end up with a bunch of spy pairs and a knight chain of length 0..... busto

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But when you have 49 spy pairs, the two remaining would have to be knights, right?

[Galwegian]

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Extending this idea, we see that after 51 questions we might end up with a bunch of spy pairs and a knight chain of length 0..... busto

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But when you have 49 spy pairs, the two remaining would have to be knights, right?

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I don't think that you must necessarily end up with enough spy pairs after 51 questions. Example, using my earlier notation

1k2,2k3,3k4..........24k25,25k26,26s27,25s28,24s29 ......
....2s51,1s52. Now you have asked 51 questions and you have only 26 spy pairs (namely (1,52),(2,51),...,(26,27) ) and you have no knight chain