Logic

[Bokkom]

Call out to all you razor sharp critical thinkers out there.

If A, then B
If B, then C
If C, then D

If all the above statements are true, which of the following must also be true:

A. If D, then A.
B. If not B, then not C.
C. If not D, then not A.
D. If D, then E.
E. If not A, then not D.

What's the best option and, more importantly, what is the logical reasoning underpinning it?

[CrazyIslander]

I'll be the first to be proven dumb.

B n D are correct because they follow the sequence of the alphabet.

[Raggs]

C is correct at a guess. At no point are we told that you cannot have a C without a B etc. Could have J and C and D (have to have a D if there's a C).

If there's no D then there cannot be an A since any A must produce a D.

[Ymx]

None of them are 100% true.

[Ymx]

The only knowns are that an early letter positive ensures a later letter positive, but not a late letter positive effects an earlier letter. So the later letters have no effect on the earlier letters. So discard all those ones. And obviously discard the earlier letter negative ones as they have no rule on outcome of later letters. And E is never mentioned.

So that discounts all of them

[Ymx]

.

[Koalabyte]

B. If not B, then not C.

At no point are we told D to A or D to E. Those are assumptions.

[Ymx]

I’m sticking with my original assertion that none are right

[Ymx]

Changed it again.

C

If D is false, A can’t be true. If A was true D would be.

[Sandstorm]

Thursday

[Fangle]

I’ve decided to see this as a Venn diagram with A on the outside which includes B etc, to have D in the centre ring, without any overlapping of these concentric ring.

Which in this case means that B must be the correct answer.

[Kiap]

I like Fangle's method; would go (B) and (E).

Another question:

[Caley_Red]

Would need to know whether it's a close-looped system or not as the answer would vary in that case.

[Bokkom]

Changed it again.

C

If D is false, A can’t be true. If A was true D would be.

Correct.

[Kiap]

Changed it again.

C

If D is false, A can’t be true. If A was true D would be.

Correct.

Ah yes.

The Venn diagram from above should have A in the centre ... to D on the outer.

[CrazyIslander]

Why is it just an alphabetical sequence? Do you guys think those letters fell randomly in that order?

[Raggs]

Changed it again.

C

If D is false, A can’t be true. If A was true D would be.

Correct.

Where's my correct? I got it right earlier and without posting 10 different answers to get there!

[clydecloggie]

I like Fangle's method; would go (B) and (E).

Another question:

it's B.

[Bokkom]

Changed it again.

C

If D is false, A can’t be true. If A was true D would be.

Correct.

Where's my correct? I got it right earlier and without posting 10 different answers to get there!

Apologies! Missed that.
Raggs is indeed the winner here.

[Munch]

Call out to all you razor sharp critical thinkers out there.

If A, then B
If B, then C
If C, then D

If all the above statements are true, which of the following must also be true:

A. If D, then A.
B. If not B, then not C.
C. If not D, then not A.
D. If D, then E.
E. If not A, then not D.

What's the best option and, more importantly, what is the logical reasoning underpinning it?

Gutted I missed this at the time, lovely thread idea.

Just for the record I want everyone to know that I thought it was C.

[lilyw]

I’ve decided to see this as a Venn diagram with A on the outside which includes B etc, to have D in the centre ring, without any overlapping of these concentric ring.

Which in this case means that B must be the correct answer.

The OP doesn't say that B is the only condition for C; i.e. B is a sufficient but not necessary condition for C. Your Venn Diagram is wrong - C is not completely contained within B.

[Munch]

I like Fangle's method; would go (B) and (E).

Another question:

I'I think this is unanswerable because it is a recursive question; a question about itself. i.e.

• If only 1 of the answers is correct, then that means the chances are 1 in 4, or 25%.

• However, if the answer's value is 25%, then the chance of getting that value at random is 2 in 4, or 50%.

• However, if the answer is 50%, then the chances are 1 in 4.

This brings us back to the start.

[Kiap]

I like Fangle's method; would go (B) and (E).

Another question:

I think this is unanswerable because it is a recursive question; a question about itself. i.e.

• If only 1 of the answers is correct, then that means the chances are 1 in 4, or 25%.

• However, if the answer's value is 25%, then the chance of getting that value at random is 2 in 4, or 50%.

• However, if the answer is 50%, then the chances are 1 in 4.

This brings us back to the start.

So you're saying it's 0%, or B?

[CrazyIslander]

So if not D then not C,B as well?

[Raggs]

So if not D then not C,B as well?

Correct. If you had one of those, you'd definitely have a D.

[Woddy]

I don't agree that it's [answer] C, not necessarily at least.

If it has to be within A to D: not D could be C (or B). If it's C (or B), then we know it could be A, so not A would be wrong. If it's possible to be outside A to D, then all bets are off on not D meaning anything.

[Raggs]

I don't agree that it's [answer] C, not necessarily at least.

If it has to be within A to D: not D could be C (or B). If it's C (or B), then we know it could be A, so not A would be wrong. If it's possible to be outside A to D, then all bets are off on not D meaning anything.

Who says it has to be within A to D?

If there's C, there must be a D. If there's a B, there must be a C and then must be a D.

You cannot have an A without having a D.

[Ymx]

Changed it again.

C

If D is false, A can’t be true. If A was true D would be.

Correct.

Where's my correct? I got it right earlier and without posting 10 different answers to get there!

Ha ha.

I imagine it got ignored because you started with an at a guess. And the OP wanted definitive proof.

[Woddy]

ah, by "not D" you mean "D is not possible". I read that you meant D as a specific resolution point, such "not D" simply meant "[the letter you are looking at,] it's not D".

With that and if your rules above are the only rules, then I get it. Still feels a bit loose.

[Ymx]

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in the first day. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?


There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."

[Plato’sCave]

C. If not D, then not A.

Because A ultimately inherits as an equal or a subset of D.

[Munch]

I like Fangle's method; would go (B) and (E).

Another question:

I think this is unanswerable because it is a recursive question; a question about itself. i.e.

• If only 1 of the answers is correct, then that means the chances are 1 in 4, or 25%.

• However, if the answer's value is 25%, then the chance of getting that value at random is 2 in 4, or 50%.

• However, if the answer is 50%, then the chances are 1 in 4.

This brings us back to the start.

So you're saying it's 0%, or B?

Saying B would give the same result.

• If B is the answer, then the chances of randomly guessing it are 1 in 4, or 25%.

• However, if the answer is 25%, then the chance of picking that answer is 2 in 4, or 50%

• However, if the answer is 50%, then the chances are 1 in 4.

This brings us back to the start.

[Munch]

Call out to all you razor sharp critical thinkers out there.

If A, then B
If B, then C
If C, then D

If all the above statements are true, which of the following must also be true:

A. If D, then A.
B. If not B, then not C.
C. If not D, then not A.
D. If D, then E.
E. If not A, then not D.

What's the best option and, more importantly, what is the logical reasoning underpinning it?

I posted a quick reply, thinking that we had moved on from this, since some folks are still discussing it I thought I'd explain my reasoning.

The answer is C because:

A. You can start at B or C and still get D.
B. You can start with A and get to C: A=B=C.
D. We have no rule linking D to E. While it is very likely, there are other things that could occur, such as D=A. So we can't say that it must be true.
E. You can start with B or C and and still get to D.

[CrazyIslander]

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in the first day. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?


There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."

Obviously a blue eyed person leaves otherwise what's the point of the Guru saying it. Or Im guessing a certain number will leave the first night because they would've deduced that there's only two types of eye colours in the island apart from Guru. They will guess their eye colour 50/50 chance. Therefore 100 people will leave on the first night. 100 will leave the second night.

[Raggs]

It adds up.

If there was 1 blue eyed person they'd leave on the first night. No one else's eyes are blue ergo mine are.

If 2 they'd leave on the second night. First day they'd each see one another. After the first night they could conclude that as the other person didn't leave they must have also seen a blue eyed person.

If 3 then you'd see two blue eyed people, however as they did not leave on the 2nd night there must be another (you), and so you all leaveon the 3rd night.

So one hundred blue eyed people would leave together on the hundredth night.

Don't think anyone else can leave.

[Clogs]

23.

[Ymx]

It adds up.

If there was 1 blue eyed person they'd leave on the first night. No one else's eyes are blue ergo mine are.

If 2 they'd leave on the second night. First day they'd each see one another. After the first night they could conclude that as the other person didn't leave they must have also seen a blue eyed person.

If 3 then you'd see two blue eyed people, however as they did not leave on the 2nd night there must be another (you), and so you all leaveon the 3rd night.

So one hundred blue eyed people would leave together on the hundredth night.

Don't think anyone else can leave.

Absolutely right.

Raggs is on fire.

[.OverThere]

On the first night every single person heads to the ferry and says I have blue eyes, so all blue eyes are removed. The following night everyone heads up and this time indicates brown, as this is what they mostly see remaining. The poor old guru is left behind until he can guess correctly.

[Ymx]

The interesting part of the puzzle for me is
All the islanders could already see there were at least one islander with blue eyes, so what difference did it make that the guru said something they already knew? Or wouldn’t it have mattered.

Presumably on day 101 the brown eyes would go too, seeing that the blue eyes had figured out who they were.

[Raggs]

The interesting part of the puzzle for me is
All the islanders could already see there were at least one islander with blue eyes, so what difference did it make that the guru said something they already knew? Or wouldn’t it have mattered.

Presumably on day 101 the brown eyes would go too, seeing that the blue eyes had figured out who they were.

The guru, by naming only a single colour, confirms that colour exists, which is vital for the process. If he didn't exist, I couldn't know if my eyes were blue, just because someone else's were and they hadn't left.

The final 100 brown eyed people cannot use the same logic, they can be sure their eyes aren't blue, since the blues left too early for them to be, but they then can't know if they brown or green (or purple, or red) eyes. Only the confirmed colour can exist, and I believe, only confirming one colour allows it to work. Saying I see brown and blue eyes, wouldn't allow you to know if your eyes were blue or brown, and so the key day 1 logic, doesn't work.

EDIT - Hmm, need to think more about what would happen if the Guru said I see brown eyes and blue eyes. Maybe the logic could still hold up. However, for the original riddle, I'm not sure you could be sure of your own eye colour with the facts given.


EDIT EDIT - Spitballing. I see brown and blue eyes.

1 brown, 1 blue 100 red 1 green.

Brown leaves since he knows he's brown. Blue leaves since he knows he's blue.

1 brown, 2 blue, 100 red, 1 green.

Brown leaves night 1. 2 blue leave on night 2 following original logic.

2 brown, 2 blue, 100 red, 1 green.

All 4 leave on night 2.

3 brown, 2 blue, 100 red, 1 green

2 blue leave night 2, 3 brown leave night 3.

I think the Guru's a plum. If he'd said I see brown and blue eyes. Everyone else could have gotten off the Island (but for the Guru).