Re: Logic Puzzles

Jamie on 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,..")

SamVS on Tue 09 Jun 2015, 12:52

Solution:: This puzzle is dumb

BeardyTom on Tue 09 Jun 2015, 15:53

Sam wrote:
Solution:
This puzzle is dumb

I think you've got it Sam. Well done!

Jamie on Tue 09 Jun 2015, 18:26

I wonder if the problem was stated incorrectly? It seems the original source is...

Sunday Times Brain Teaser 12th Oct 08

Wonder if it's possible to get back issues?

SamVS on Tue 09 Jun 2015, 18:52

http://forums.xkcd.com/viewtopic.php?f=3&p=967155

Seems that The Times made a mistake?

steveygee on Wed 10 Jun 2015, 10:25

I ran a D&D adventure about a year ago in which the adventurers spent an hour trying to solve puzzles that i had made purposefully unsolvable whilst I sat with a smug smile on my face smoking a cigarette like a James Bond villian. I wonder if this is how the puzzle setter at the Times is feeling now?

SamVS on Wed 10 Jun 2015, 18:32

If we've all recovered from the previous ordeal: have you guys tried this one yet?:

The Warden

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?

BeardyTom on Thu 11 Jun 2015, 10:26

I presume the 100 wise men of the earlier problem revolted against the stark raving mad king but the rebellion failed and there are now only 23 survivors. The king put them in prison in the charge of his equally stark raving mad brother.

Anyway, here's where I am with this. I can give the prisoners a simple set of instructions which will mean that at some point one of them will be sure that either all of them or all but one of them has been to the room. I'm 90% sure there's a tweak to the system that will get me that extra one (rather than having to rip up this method and try something totally different) but I haven't seen it yet.

Not-quite-there method:: One of the prisoners is designated as the counter. Each of the others will have one rule to follow: the first time they go in the room and see that switch 2 is in the down position, they will switch it to up. Otherwise they will toggle switch 1. Each time the counter goes in the room and sees switch 2 in the up position he will switch it to down. When he has done this 22 times he knows that either a) all 22 other prisoners have been in and switched 2 to up, or b) Only 21 of the others have been in but switch 2 started in the up position.

Jamie on Thu 11 Jun 2015, 22:20

I'm stumped.

Think you could be on the right track though Tom. Just that last missing bit of the jig-saw. I think it could be something to do with the fact, that at some point, they'll all have gone in to the room the same number of times. Your solution doesn't seem to make use of that information, but, can't help but feel it's significant, otherwise, why would they mention it?

Not that I have any ideas though...

BeardyTom on Fri 12 Jun 2015, 10:59

Jamie wrote:Think you could be on the right track though Tom. Just that last missing bit of the jig-saw. I think it could be something to do with the fact, that at some point, they'll all have gone in to the room the same number of times. Your solution doesn't seem to make use of that information, but, can't help but feel it's significant, otherwise, why would they mention it?

I thought that too. It seemed like such a tantalising nugget of information but it turned out to be a red herring. I think this is the final answer now.

Solution:: As above, there is one counter and the other 22 follow this procedure: The first *two* times they see switch 2 in the down position, they switch it up. Otherwise they just flip switch 1.

The counter flips switch 2 down every time he sees it up and when he's done that 44 times he knows that everyone else has been in. In fact he knows that most of them have been in at least twice. It's possible that one of them may have only been in once but that's enough.

The flipping of switch 1 is completely irrelevant to tracking the prisoners. It's just there for not flipping switch 2. It would be an equivalent problem to say there's one switch and the prisoners can choose whether to flip it or not each time.

Jamie on Fri 12 Jun 2015, 13:16

Brilliant, well worked out Tom!

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