Logic Puzzles

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Post  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,..")

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Post  SamVS on Tue 09 Jun 2015, 12:52

Solution:
This puzzle is dumb
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Post  BeardyTom on Tue 09 Jun 2015, 15:53

Sam wrote:
Solution:
This puzzle is dumb

I think you've got it Sam. Well done!
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Post  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?
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Post  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?
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Post  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?
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Post  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?
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Post  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.
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Post  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...
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Post  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.
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Post  Jamie on Fri 12 Jun 2015, 13:16

Brilliant, well worked out Tom!
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Post  BeardyTom on Fri 24 Apr 2020, 11:08

In this question it is assumed that two half-brothers equals one full brother and therefore, for example, a person with just one full brother can be said to have twice as many brothers as a person with just one half-brother.

During our father’s somewhat complicated life, he has had children with three different women (Joan, Rose and Amanda) and therefore our generation of the family, consisting of four boys (Pete, Mike, Frank and Tony) and two girls (Louise and Kate), includes several full sibling and half-sibling relationships. From the information given below, can you determine which mother had which children?

• Tony has as many sisters as brothers.
• Mike has three times as many brothers as Kate has sisters.
• Kate has more brothers than anyone else has.
• Pete is Rose's son.
• Louise is Amanda’s daughter.


Last edited by BeardyTom on Fri 24 Apr 2020, 15:12; edited 1 time in total
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Post  JamesETT on Fri 24 Apr 2020, 14:20

I beleive I've got the answer but dont k is how to post in spoiler mode
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Post  BeardyTom on Fri 24 Apr 2020, 14:57

Code:
[spoiler="SPOILER"]Here is some text.[/spoiler]

Will produce:
SPOILER:
Here is some text.
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Post  JamesETT on Fri 24 Apr 2020, 15:03

SPOILER:

Since Kate has more brothers than anyone the only possible brother configurations are 310, 220 or 211 with Kate being with the highest number and louise being with a lower number of full brothers ergo kate and louise aren't full sisters. Therefore Tony's formula is kate and louise equals Pete and Mike and Frank. Kate and louise is equal to 1 or 1.5 sisters however Pete Mike and Frank cant fall any lower than 1.5 so tony does not have any full brothers but is the full blooded brother of louise. Mike has as many brothers as Kate's sisters which is 3 times .5 so 1.5 this means mike also has no full brothers. The only possible brother configuration left is 211. So in conclusion Amanda's children are louise and tony, Joan is the mother of mike, and rose gave birth to pete frank and kate



Last edited by JamesETT on Fri 24 Apr 2020, 15:13; edited 1 time in total
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Post  BeardyTom on Fri 24 Apr 2020, 15:04

JamesETT wrote:
SPOILER:
Bob is your uncle.

Well that is *an* answer...
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Post  BeardyTom on Fri 24 Apr 2020, 15:11

I'm just going to be a bit annoying and slightly change the question. It doesn't change the logic particularly, but makes it a bit more satisfying, I feel. Also doesn't allow a bit of meta logic to give part of the solution.
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Post  JamesETT on Fri 24 Apr 2020, 15:15

Just posted that to test without getting it wrong edited to inc,use my actual a swer
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Post  BeardyTom on Fri 24 Apr 2020, 15:24

JamesETT wrote:
SPOILER:

Since Kate has more brothers than anyone the only possible brother configurations are 310, 220 or 211 with Kate being with the highest number and louise being with a lower number of full brothers ergo kate and louise aren't full sisters. Therefore Tony's formula is kate and louise equals Pete and Mike and Frank. Kate and louise is equal to 1 or 1.5 sisters however Pete Mike and Frank cant fall any lower than 1.5 so tony does not have any full brothers but is the full blooded brother of louise. Mike has as many brothers as Kate's sisters which is 3 times .5  so 1.5 this means mike also has no full brothers. The only possible brother configuration left is 211. So in conclusion Amanda's children are louise and tony, Joan is the mother of mike, and rose gave birth to pete frank and kate


Looks good to me.

JamesETT wrote:Just posted that to test without getting it wrong edited to inc,use my actual a swer

I knew that. Also, given that I changed all the names of my real family for this puzzle, I suppose I could also change my Uncle Jim's name and actually have Bob as my uncle.
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Post  Samuel_C on Mon 27 Apr 2020, 20:17

Just had a go, I got the same answer as James (I checked my answer against his as I was pretty sure I'd got it right).

Good fun, thanks Tom (or Mike, Pete, Frank or Tony..)
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