Re: Logic Puzzles

SamVS Fri 05 Jun 2015, 23:07

Guess:

SamVS Fri 05 Jun 2015, 23:13

Spoiler:

Jamie Sat 06 Jun 2015, 20:42

If the King used 2 black hats, the Prince in the white hat would know immediately he had a white on, and would win.

If the King only used 1 black hat, the black hat Prince would have no chance. One white hat Prince (call him Jamie), would see one white and one black hat; and know that the Prince he can see with the white hat, cannot see 2 black hats (because, if he could, he would immediately declare himself to have a white hat). Because the other white hat prince is not declaring himself to have a white hat; our Prince Jamie, surmises that he himself must have a white hat.

Let's just hope the other white hatted prince isn't just a little slow, and has indeed seen 2 black hats! Or doesn't like the Princess, and for some unknown reason, wants Prince Jamie dead (by tricking him in to wrongly declaring his own hat colour).

I'm not sure how 'fairness' enterers in to it!?

SamVS Sat 06 Jun 2015, 21:12

Let's just hope the other white hatted prince isn't just a little slow, and has indeed seen 2 black hats! Or doesn't like the Princess, and for some unknown reason, wants Prince Jamie dead (by tricking him in to wrongly declaring his own hat colour).

I'm not sure how 'fairness' enterers in to it!?

The puzzle says:

You also know that the king is a man of his word, and he has said that the test is a fair test of intelligence and bravery.

It wouldn't be "a fair test" if one prince could see two black hats and declare himself the winner before the other princes had any information to go on. Nor if there was one black hat and that prince had no way of working it out.

Jamie Sat 06 Jun 2015, 21:18

Ah yeah. Only way it could be fair would be to have 3 white hats! Smile

SamVS Sat 06 Jun 2015, 21:22

Our answer is better anyway Smile

If it was about fairness, why even put the hats on? The smartest prince would have immediately called the king out on his shenanigans.

I worked this one out before:

The 100 Coins

There are 10 sets of 10 coins. You know how much the coins should weigh. You know all the coins in one set of ten are exactly a hundredth of an ounce off, making the entire set of ten coins a tenth of an ounce off. You also know that all the other coins weight the correct amount. You are allowed to use an extremely accurate digital weighing machine only once.

How do you determine which set of 10 coins is faulty?

Hint:

Jamie Sat 06 Jun 2015, 21:36

1 from set 1
2 from set 2
3 from set 3
...
10 from set 10

Jamie Sat 06 Jun 2015, 21:38

Say we know each coin is X weight off.

Following on from my previous post; if the total weight off (considering the total number of balls we weigh) is 1X, it's set 1, if 2X, it's set 2,...

SamVS Mon 08 Jun 2015, 00:16

We've almost finished all the puzzles on the site? Do you all feel more cleverer?

Flipping Coins

There are twenty coins sitting on the table, ten are currently heads and tens are currently tails. You are sitting at the table with a blindfold and gloves on. You are able to feel where the coins are, but are unable to see or feel if they heads or tails. You must create two sets of coins. Each set must have the same number of heads and tails as the other group. You can only move or flip the coins, you are unable to determine their current state. How do you create two even groups of coins with the same number of heads and tails in each group?

Don't let it keep you up too late, Jamie!

Jamie Mon 08 Jun 2015, 00:29

Move any 10 coins to the side (pile a). Flip all 10 remaining coins (pile b). Both pikes have same number pf heads and tails.

SamVS Mon 08 Jun 2015, 07:47

Did you skip a step or seven?

Jamie Mon 08 Jun 2015, 09:26

Nope, unless I misunderstood the problem?

You have 20 coins, 10 heads up, 10 tails up. If you take any 10 of those coins (not being able to discern heads from tails), and flip those 10 coins. You will have 2 piles of ten coins, each containing the same number of heads, and the same number of tails (e.g. pile A, 6 heads, 4 tails; pile B, 6 heads, 4 tails).

Maybe I'm missing something?, it did seem a bit easy...

SamVS Mon 08 Jun 2015, 09:36

Whoops, yeah, I realized what you meant just after posting. You got it.

Jamie Mon 08 Jun 2015, 11:53

DAMN MONKEY wrote:Ten people land on a deserted island. There they find lots of coconuts and a monkey. During their first day they gather coconuts and put them all in a community pile. After working all day they decide to sleep and divide them into ten equal piles the next morning.

That night one castaway wakes up hungry and decides to take his share early. After dividing up the coconuts he finds he is one coconut short of ten equal piles. He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him.

Later another castaway wakes up hungry and decides to take his share early. On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's slightly bloodied coconut. The monkey conks the second man on the head and kills him.

One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself. What is the smallest number of possible coconuts in the pile, not counting the monkeys?

Jamie Mon 08 Jun 2015, 11:55

I solved this already, it's not too tricky...

Jamie Mon 08 Jun 2015, 12:08

This one seems a bit harder...

A pack of 9 cards is numbered 1 to 9. One card is placed on the forehead of each of four logicians. Each of them can see the other three numbers, but not their own. In turn, each states whether or not he knows his own number. If not, he announces the sum of two of the numbers he can see.

One game went as follows. Alf: “No, 14”. Bert: “Yes”. Chris: “No, 7. Dave: “No”- but before Dave could continue, Alf had worked out his own number.

What was Dave’s number?

Jamie Mon 08 Jun 2015, 12:24

Kicking off...

To make 14 = 9/5 or 8/6.
To make 7 = 6/1 or 5/2 or 4/3.

Responses we get:
A = "14"
B = "yes"
C = "no, 7"
D = "no,...<interrupted by 'A'>"

A ... can see one of the pair combinations making 14 on 2 of B,C or D.

B ... must know that he is the other half of the pair making 14, so he must see one card of the 9/5 or 8/6 pairs; on C or D (and know that he has the other number in that pair).

C ... can see one of the pair combinations making 7. We know it can't be on B (as he already worked out he must be part of the pair making 14), so A and D have either 6/1 5/2 or 4/3.

D ... says he doesn't know his own number, which is enough information for A to fill in the blacks and determine his own number.

I think it could be scraps of paper time, and drawing things in grids! Shocked

BeardyTom Mon 08 Jun 2015, 13:43

Jamie wrote:This one seems a bit harder...

A pack of 9 cards is numbered 1 to 9. One card is placed on the forehead of each of four logicians. Each of them can see the other three numbers, but not their own. In turn, each states whether or not he knows his own number. If not, he announces the sum of two of the numbers he can see.

One game went as follows. Alf: “No, 14”. Bert: “Yes”. Chris: “No, 7. Dave: “No”- but before Dave could continue, Alf had worked out his own number.

What was Dave’s number?

And now I've just been sat here mumbling to myself like a crazy person for the past 20 minutes. Thanks Jamie!

I'm stuck on how Bert could be sure of his number. He knows either B+C or B+D or C+D is 14. If he sees that C+D=14 then I don't think he can know what his own number is out of the 6 remaining possibilities so he must know that his number is part of a pair totaling 14 with either C or D. He then rules out one of them in some way (maybe one of them is <5 or totals 14 with A) but then I get stuck. Surely C and D know that B saying he knew his number implies that his number added to one of theirs totals 14. Therefore one of C and D will see that the other doesn't sum with B's to make 14 and will conclude that theirs does.

Except they didn't. Each said he didn't know his own number. So either I've missed something or Dave the logician wasn't paying attention and muffed his answer.

There must be some way that B could know his own number without the others being certain that he knew he was part of a pair totaling 14 with C or D. I don't see how

Jamie Mon 08 Jun 2015, 15:59

I'm stuck on how Bert could be sure of his number. He knows either B+C or B+D or C+D is 14. If he sees that C+D=14 then I don't think he can know what his own number is out of the 6 remaining possibilities so he must know that his number is part of a pair totaling 14 with either C or D.

What's interesting, is that (presumably) B works out he is part of a pair adding up to 14. However, neither C or D, know that they could be the other half of that pair totally 14 (they both say 'No'); maybe B has a number that can be part of a pair adding up to 14, or part of another pair adding up to 7 (that would means B is a 6).

Jamie Mon 08 Jun 2015, 16:28

The trick is, I think, finding a combination, where D would not know his own number. So D cannot be the only match for 7 or 14; or; could match either 7 or 14.

Jamie Mon 08 Jun 2015, 16:36

Most likely, it's D can match with either 7 or 14, if so, D would be 5 or 6.

So...

A=2 B=9 C=? D=5
A=1 B=8 C=? D=6

Was looking at...

A=1 B=8 C=2 D=6

A says 14 (8+6)
B says he is number 8. D sees D is 6, so he has to be 8.
C says 7 (1+6)

However, A would then know he is 1 (as C said 7, and D is 6, A could only be 1). It looks like A only knows his number, because D doesn't know his.

Jamie Mon 08 Jun 2015, 16:50

A=2 B=9 C=1 D=5

Oh no! That's not good! C says he can see 7, which A would not consider C's number; A would just look at D, and know that he (A) is 2.

PaulC Mon 08 Jun 2015, 17:09

Jamie wrote:A=2 B=9 C=1 D=5

Oh no! That's not good! C says he can see 7, which A would not consider C's number; A would just look at D, and know that he (A) is 2.

Also, D says "No", but if he was the other side of the 14 sum, he should know it.

I think I'm finally stumped. I will tender my resignation to the Smug Bastard Society forthwith.

Jamie Mon 08 Jun 2015, 18:10

Does this work?

a=5 b=6 c=8 d=1

No, it's won't. When c says '7' it can only be the 6 and 1.

I wrote a quick program to just blast through each possible combination a to d having unique values 1 to 9 (no duplicates), then removing impossible combinations; got it down to 12. None of them look good. Sad

May have to concede defeat, unless anyone else has any ideas?

PaulC Mon 08 Jun 2015, 18:13

Jamie wrote:Does this work?

a=5 b=6 c=8 d=1

No, because C would be able to determine their number in the same way that B did.

Jamie Mon 08 Jun 2015, 18:28

Link to where I found this problem...

http://www.folj.com/bb/viewtopic.php?t=74&sid=f36e99b5213dce1c9d3c1863c6491cc7

They don't solve it either. Note, the bottom post, someone says...

D is about to answer (he knows his number) but is interrupted by A who knows his number.

...which seems wrong, because D didn't know his number (he said "No,..")

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