There are a group of 5 good logic puzzles  HatsGreen-eyed guruPerfectly logical piratesPrisoners with HatsLeprechauns. (I think they are good), and their responses need a place. Here it is.

1. July 3, 2009 pm31 8:12 pm 8:12 pm

Hats

• July 3, 2009 pm31 8:27 pm 8:27 pm

By Sara:

Sal’s hat is red.

If Sal’s hat were white, any of the possibilities for Pal’s and Ral’s hats would result in one of them knowing their own hat. If either Pal’s or Ral’s hat were also white, the other would look at two white hats and know his own was red. If both Pal’s and Ral’s hats were red, Pal would be unable to tell what color his own hat was, but Ral, looking at a white and a red hat, and knowing that Pal also must have looked at a white and red hat (since they both saw Sal’s white hat, and Pal could not answer about his own hat), would know his own hat was red. Since neither Pal nor Ral was able to tell what color their hat was, Sal’s hat cannot be white.

2. July 3, 2009 pm31 8:12 pm 8:12 pm

Green-eyed Guru

• July 3, 2009 pm31 8:39 pm 8:39 pm

By Dr. Rick:

OK, let me see if I can get this one… I’m not great at these, which makes them fun.

Let’s suppose there is only one blue-eyed islander (me). I looks around, see everyone else has brown eyes, and leave the first night.

Now let’s suppose there are two (one’s me). I can see one person with blue eyes. If he were the only one, he’d leave the first night. He’s didn’t, so he’s not – nobody else I can see is blue-eyed, so I must be. I leave the second night, and so does he (having reasoned identically).

If there’s three, when the two I can see didn’t leave the second night, we all leave the third.

So – the blue-eyed boys all leave on night 100. Each brown-eyed islander was ready to leave on night 101 if they hadn’t – thinking his eyes were blue. But they went, so his eyes were brown. So on night 101 everybody else leaves (except the sleeping guru, of course).

• July 12, 2009 am31 10:22 am 10:22 am

Almost. The Islanders don’t know how many of each eye color there is. And they don’t know that there are only 3 colors….

Which is to say, your little finishing bit is wrong. Once the blue-eyed folks leave, everyone knows that their own eyes are not blue, but nothing about brown or green or grey or purple…. So nothing else happens.

3. July 3, 2009 pm31 8:12 pm 8:12 pm

Perfectly Logical Pirates

• July 3, 2009 pm31 8:31 pm 8:31 pm

By Mr. Burke:

32, 34, 34, 0 and 0.

A gets lots of gold and gets to live.
No one gets more gold than B so he votes aye
No one gets more gold than C so he votes aye.
Ayes win 3 to 2.

Alternatively, A convinces B that it would be harder for B to get 3 out of the 4 remaining votes than for A to get 3 out of 5. B will only live if he votes aye. So, conceivably, B gets a little less to stay alive and C gets more to buy his vote.
32, 34, 34, 0 and 0. A gets lots of gold and gets to live. No one gets more gold than B so he votes aye No one gets more gold than C so he votes aye. Ayes win 3 to 2. Alternatively, A convinces B that it would be harder for B to get 3 out of the 4 remaining votes than for A to get 3 out of 5. B will only live if he votes aye. So, conceivably, B gets a little less to stay alive and C gets more to buy his vote.

(x, why?)
http://mrburkemath.blogspot.com

• July 3, 2009 pm31 8:40 pm 8:40 pm

By Dr. Rick:

Let’s see… I haven’t read the article, but if I get it right feel free to hide it (and any other solutions of mine!) so others can play.

If it gets down to just D and E, E will vote “no”, be the sole survivor and get everything. So, if it comes to C’s turn D will vote “yes” (he wants to live) and E will vote “no” (not that it matters). So C would propose 100, 0, 0.

Thus C will vote against B if it comes to his turn, regardless, and D and E will vote for him as long as B gives them more than nothing. So B would propose 98, 0, 1, 1 and win three-one.

So – B will definitely vote against A. C, D and E would each vote for him if they get more than the 0, 1, 1 they’ll get if they get him killed. He needs only two votes, so he should give 1 to C, 2 to either D or E, and none to B and the other of D, E.

So A should propose 97, 0, 1, 2, 0 or 97, 0, 1, 0, 2 and will win the vote 3-2.

4. July 3, 2009 pm31 8:13 pm 8:13 pm

Prisoners with Hats

• July 4, 2009 pm31 5:20 pm 5:20 pm

I am pretty sure I have the solution to this one – I had seen a similar puzzle before and had it explained, so I don’t feel my solution’s entirely “mine”, unlike the others. I am struggling to express it as clearly as I want to, though, so I’ll come back to that later.

A is wearing 16, B 26 (we know) and C 10.

• July 4, 2009 pm31 5:40 pm 5:40 pm

OK, here we go.

As each prisoner takes his turn, he knows the hats the other two are wearing and he knows that his own is either their sum or their difference. He also knows that certain ratios of hat numbers A:B:C are impossible, either because his own hat would then be zero or because they would have been deduced by other prisoners already. If the two hats he can see match one of those possibilities, he will then know what his is (it turns out he can only deduce his own hat number if it’s the highest). We list who plays each turn, and in which ratios the hat numbers would be if he could deduce his own.

A – 2:1:1 (if B and C had the same numbers, he knows his own isn’t zero so it must be the sum)

B – 1:2:1, 2:3:1 (as before for A, and if A is in, say 4 and C’s in, say, 2, B is in 2 or 6. If he’s in 2, A would be free by now, so he’s in 6).

C – 1:1:2, 1:2:3, 2:1:3, 2:3:5.

A – 3:2:1, 4:3:1, 3:1:2, 5:2:3, 4:1:3, 8:3:5.

B – 1:3:2, 1:4:3, 2:5:3, 2:7:5, 3:4:1, 4:5:1, 3:5:2, 5:8:3. 4:7:3, 8:13:5.

Since B did deduce he was in 26. it was one of these last line of ratios – and the only one that divides 26 in the second place is 8:13:5. So the hats were 16:26:10.

• July 4, 2009 pm31 5:45 pm 5:45 pm

Hat’s off to you!

Notice anything interesting about those ratios? (ahem)

• July 10, 2009 am31 8:30 am 8:30 am

Yes :). It’s like, y’know, some rabbits, out of a hat…

5. July 4, 2009 pm31 12:34 pm 12:34 pm

Leprechauns

• July 4, 2009 pm31 5:28 pm 5:28 pm

This one is new to me, and I like it very much – especially because it’s easy enough I could give it to a decent class and expect them to kill it :). I think this is the second easiest, after #1.

The oldest leprechaun would rather wait as long as possible, and will always vote no.

If there are two left, the younger will vote split, and it’ll be split.

If there are three, the two older will vote no and it’ll roll down to three.

If there are four, the two younger will both vote to split (or they’d get nothing), and it’ll split.

And so on. Any time the number younger than the last “split point” is less than half, they’ll all get voted out. If they constitute a half, they’ll be able to split it. So it splits on the first power of two it reaches – 512 leprechauns.