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✏️ Fixed logic

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tomit4 2026-05-12 14:32:21 -07:00
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chapter_1/1_1/investigate_1_1_1.md
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@ -11,32 +11,30 @@ Troll 3: Either we are all knaves, or at least one of us is a knight.
Which troll is which?
Spend a few minutes thinking about the Investigate problem above. What could you
conclude if you knew Troll 1 really was a knave (i.e., their statement was
false)? Share your initial thoughts on this.
A:
Let's think this through step by step by assuming the first Troll is lying.
If "If I am a knave, then there are exactly two knights here." = False:
Then: Troll 1 = knave, BUT that doesn't meant that there are exactly two
knights. In other words, the other two could be either knaves or knights. But we
know that this troll is a knave.
So we know he is lying. Although he is indeed a knave, as he claims, we cannot
trust that his presumption, that there are exactly two knights here is true,
because we can assume he is lying from the "Try it" assumption.
Moving on, we have:
Moving on we have Troll 2:
Troll 2: "Troll 1 is lying"
Troll 2: Troll 1 is lying.
If Troll 2 is lying, then Troll 1 is not lying (i.e. a knight), and this
invalidates our previous assumption that Troll 1 is lying, and therefore we
still don't know if the other two trolls are knights.
This troll is a knight, since we know that Troll 1 is a knave
Moving on to troll 3:
Troll 3: Either we are all knaves, or at least one of us is a knight.
Troll 3: "Either we are all knaves, or at least one of us is a knight."
If we assume troll 3 is also lying, then all three trolls are knaves, and we
have solved their riddle and pass, because he is no knight.
These are just my initial thoughts on this based off the prompt that we could
start just thinking about this if Troll 1 were lying.
$$ \therefore $$
Troll 3 is also a knight, since we know there is one knight here, we know that
all the Trolls are not knaves, and his second presumption, that there is _at
least one_ knight holds true, as now we have 2. Interestingly, if he had said
"exactly one of us is a knight", this problem would become inconsistent and
unsolvable.