
Author 
Message 





25 
Posts: 598 Joined: 29 Mar 2010 Last Visit: 30 Apr 2016
Location: In my garden, enjoying life  


Posted: Wed 13 Jul, 2011 


On Puce's riddle:
Click here to see the hidden message (It might contain spoilers)
Working from the other way: If today is friday, then tomorrow is saturday. Yesterday = 'tomorrow' so yesterday is saturday... Today must be Sunday
Am I right?



 
 
   




23

Posts: 255 Joined: 09 Jan 2010 Last Visit: 22 Aug 2018
 


Posted: Wed 13 Jul, 2011 


Marvin:
Click here to see the hidden message (It might contain spoilers)
Stupid me. But no, it isn't Wedesday either.
Blenderman:
Click here to see the hidden message (It might contain spoilers)
Yes, you are correct.



 
 
   





Posts: 1971 Joined: 31 Aug 2006 Last Visit: 28 Feb 2019
LD count: 15
 


Posted: Wed 13 Jul, 2011 


Still on Puce's riddle
Click here to see the hidden message (It might contain spoilers)
I disagree. The way I read the riddle, it cannot be Sunday. Let's assume today is Sunday. Then "tomorrow" would be Monday. And if yesterday := Monday then today would be Tuesday. Not Friday. Or where is my error?
The way Puce and Blenderman intepreted it, "tomorrow" is viewed in the hypothetical time that today would be Friday. For me, "tomorrow" is viewed as if it was in the current real timeline.
Now if today is really Wednesday, then "tomorrow" would be Thursday. If we assign yesterday Thursday, then "today would be Friday", exactly as it is in the riddle...



 
 
   




23

Posts: 255 Joined: 09 Jan 2010 Last Visit: 22 Aug 2018
 


Posted: Wed 13 Jul, 2011 


Click here to see the hidden message (It might contain spoilers)
Well, the way I translated it, I tried to mean "tomorrow" in the hypothetical sense while "yesterday" exists as normal. To the extend of my knowledge, the ifclause conveys that meaning.
Now the way I understand your argument, you solve it this way: Today is Friday. Yesterday is then Thursday. Since we said "yesterday was tomorrow", then the day we're looking for is yesterday for Thursday, which is Wednesday.
Looking at it the other way around, I see this: The man claims that if yesterday was tomorrow, then today would be Friday. So let's assume that it in fact is Wednesday. Then yesterday (which is Tuesday) should be tomorrow. Then that day would be Monday. Which is not Friday.
The solution I and Blenderman have goes like this: Today is Friday. We had made yesterday tomorrow, and tomorrow is then Saturday. Since it is also yesterday from the actual day, today is Sunday.
To verify this: Let's assume today is Sunday. We had said "if yesterday was tomorrow", so it means "if Saturday was tomorrow", which would point today as Friday.
I think the part we disagree on is this: Yesterday is not always tomorrow and tomorrow is not always yesterday.



 
 
   





Posts: 1971 Joined: 31 Aug 2006 Last Visit: 28 Feb 2019
LD count: 15
 


Posted: Wed 13 Jul, 2011 


Click here to see the hidden message (It might contain spoilers)
I think the main problem is that we both thought that the solution would be unique. But due to the ambiguity of the riddle itself: Is tomorrow or yesterday in the hypothetical time (and the other one in the current time). Because as you pointed it out, depending on HOW you interpret the actual riddle the solution would either be Sunday or Wednesday. I wouldn't say either is wrong.
And speculating which one (between yesterday and tomorrow) was meant in a hypothetical way and which in the real one, I guess this is more a matter of taste and personal interpretation that based on a linguistic rule. At least as far as my English goes...



 
 
   




25 
Posts: 598 Joined: 29 Mar 2010 Last Visit: 30 Apr 2016
Location: In my garden, enjoying life  


Posted: Wed 13 Jul, 2011 


Click here to see the hidden message (It might contain spoilers)
Well, I'm sorry to say Marvin, I really do think you are wrong. The key word being 'if'. It's not yesterday = tomorrow and therefore you can turn it around, even though I did (quite wrongly) use the = sign in my explanation.
The riddle says: 'If yesterday was tomorrow'. That isn't the same as 'If tomorrow was yesterday'. There is truth and there is imagining. The true values are 'today' and therefore 'yesterday', the other words are made up... that's why the word 'if' is used. If you work from the other side, from the 'if' side... so taking 'friday' and than trying to get the answer like I did, the statement would be 'If tomorrow was yesterday'.
Let's reason backwards from our answers. It is sunday. If yesterday (saturday) was tomorrow, then tomorrow would be saturday. Then it would be friday. That is correct.
It's wednesday. If yesterday (tuesday) was tomorrow, then tomorrow would be tuesday. Then it would be monday.
You simply have to keep track of what is before the 'if' and what comes after it.



 
 
   




23

Posts: 255 Joined: 09 Jan 2010 Last Visit: 22 Aug 2018
 


Posted: Wed 13 Jul, 2011 


^ Blenderman put what I actually tried to mean in better words.



 
 
   




20 
Posts: 332 Joined: 03 May 2011 Last Visit: 26 Jan 2017
LD count: 38ish
 


Posted: Wed 13 Jul, 2011 


Paul is 20 years old in 1980 but only 15 years old in 1985. How is this possible?
and i have one more
he who makes it, sells it
he who buys it, doesnt use it,
he who uses it, doesnt know it
define "it"
Current LD goal(s): shared dream, spend a week surviving in the wilderness, FLY!@!@



 
 
   




33 
Posts: 69 Joined: 23 Jul 2011 Last Visit: 01 Mar 2019
LD count: 0...
 


Posted: Sun 24 Jul, 2011 


Click here to see the hidden message (It might contain spoilers)
For your first, the only solution I can come up with is that Paul is a character in a movie. In 1980, the character is 20, and in 1985 they make a prequel where he is 15.
Your second, however, is a coffin. He who makes the coffin, sells it. He who buys it is purchasing it for a recently passed loved one. And if you are using it... well, you are dead, so you couldn't possibly know thet you are.
Uh, this is the only riddle I can think of:
What walks on four legs in the morning, walks on two legs tn the afternoon, and three legs in the evening?
Current LD goal(s): Improve dream recall to at least 3 in one week



 
 
   




25 
Posts: 598 Joined: 29 Mar 2010 Last Visit: 30 Apr 2016
Location: In my garden, enjoying life  


Posted: Sun 24 Jul, 2011 


Click here to see the hidden message (It might contain spoilers)
I think Paul might be living BC... so as Paul get's older the years get smaller... towards 0



 
 
   





Posts: 669 Joined: 28 Jun 2012 Last Visit: 24 Mar 2019
LD count: 11.8 estimate
 

This was from [part VIII]... 
Posted: Sat 30 Jun, 2012 


Alvin wrote: 
You are given 12 balls. All of them are of equal weight except one. One ball is either lighter or heavier than the others, you don't know which. You are given the task of finding this ball. You are given a scale to help you in your task but you may only use it 3 times.
How are you supposed to find the ball that has a different weight than the others? 
It took me a few days to get it, and an hour or so to write it out, but here's my solution.
Click here to see the hidden message (It might contain spoilers)
1st WEIGH: Separate the balls into 3 groups with 4 balls each. Weigh two of the groups, group A against group B. Group C is not weighed.
Case 1: They weigh the same. In this case, you now know the eight balls from groups A and B are normal, and the odd ball is one of the 4 balls in group C.
2nd WEIGH: Separate group C into 3 subgroups: 211. Weigh the group of two balls (group a) against a third ball and one of the normal balls (group b). Make sure you remember which is the normal ball. The remaining ball from the group C would not be weighed.
Case 1.1: They weigh the same. All balls on the scale are normal. The remaining ball from group C is thus the odd ball. No need to use the scale again, unless you want to find out if it's heavier or lighter than the normal balls.
Case 1.2: They weigh differently. In this case, note which group was heavier. The odd ball is either one of the three balls.
3rd WEIGH: Weigh the two balls from group a against each other.
Case 1.2.1: They weigh the same. The odd ball is the one from group b.
Case 1.2.2: They weigh differently. You now know the odd ball was from group a. If group a was heavier, then the odd ball is the heavier one. Otherwise, the odd ball is the lighter one.
Case 2: They weigh differently. You now know that the 4 balls in group C are all normal. The odd ball is either in group A or B. You also know which group is heavier.
2nd WEIGH: Split group A into two groups of two. Add a normal ball to the first group, and a ball from group B to the second group; thus weigh two balls from group A + one normal ball (group a) against the remaining two balls from group A + one ball from group B (group b). Three balls from group B will be left over.
Case 2.1: They weigh the same. The odd ball is one of the three remaining balls from group B. Since you know whether group B was heavier or lighter, you now know if the odd ball is heavier or lighter.
3rd WEIGH: Weigh two of the three balls in the leftover group.
Case 2.1.1: They weigh the same. The odd ball is the third ball.
Case 2.1.2: They weigh differently. If group B was heavier, then the odd ball is the heavier one. Otherwise the odd ball is the lighter one.
Case 2.2: They weigh differently.
Case 2.2.1: Group A was heavier and group a is now lighter (or viceversa). You now know that the odd ball was one of the two balls from group A which were in group b, and that the odd ball is heavier (or lighter).
3rd WEIGH: Weigh those two balls against each other. The heavier (or lighter) ball is the odd ball.
Case 2.2.2: Group A was heavier and group a is still heavier (or lighter). There are two possibilities, either one of the two balls in group A that were also in group a is the odd ball and is heavier than normal, or the ball from group B which was added to group b is the odd ball and it is lighter.
3rd WEIGH: Weigh the two balls from group A that were also in group a against each other.
Case 2.2.2.1: They weigh the same. The odd ball is the ball from group B which was added to group b.
Case 2.2.2.2: They weigh differently. The heavier (or lighter) ball is the odd ball.
Last edited by Alot on Sun 17 Feb, 2013; edited 1 time in total



 
 
   





Posts: 812 Joined: 19 May 2005 Last Visit: 18 Mar 2019
 


Posted: Sat 30 Jun, 2012 


Haha. I had to search my own posts with 'balls' as a keyword to find my old post again. Wow. That was from five years ago. Time sure does fly. I did a quick check in that part to see if my riddle was solved. That doesn't seem to be the case.
I had forgotten the solution to the riddle myself so I decided to work it out again. Happy to say I managed to solve it so I can compare my solution with yours. Reading up on it, all I can say is... Your solution is complicated :rolleyes: At least, it is definitely different from mine. I'll post my solution later and you can check that too.
After spending a longer than anticipated time understanding your solution with a pen and paper, I can confidently say that you are correct Thanks for solving a 5 year old post.



 
 
   





Posts: 812 Joined: 19 May 2005 Last Visit: 18 Mar 2019
 


Posted: Sun 01 Jul, 2012 


Here's my solution to my riddle.
Click here to see the hidden message (It might contain spoilers)
Code: 
Group the balls into three groups of four. We'll call these groups A, B, C.
Additonal notes: A[1] would stand for the first ball in group A. A[1,2] would stand for balls 1 and 2 of group A. '=' would stand for an equal reading while '!=' stands for not equal.
1st Weigh: A and B
IF =
We know that the odd ball is in C
2nd Weigh: C[1] and C[2]
IF =
We know that the odd ball is in C[3,4]
3rd Weigh: C[1] and C[3]
IF =
We know that C[4] is the odd ball
IF !=
We know that C[3] is the odd ball
IF !=
We know that the odd ball is in C[1,2]
3rd Weigh: C[1] and C[3]
IF =
We know that C[2] is the odd ball
IF !=
We know that C[1] is the odd ball
IF !=
We know that the odd ball is in A or B
For simplicity's sake, assume that the lighter group is A
2nd Weigh: A[1]+B[1,2] and A[2]+B[3,4]
IF =
We know that the odd ball is in A[3,4]
We know that the odd ball is lighter than the others
(Remember that A was the lighter group)
3rd Weigh: A[3] and A[4]
We know that the lighter ball is the odd ball
IF !=
Take note of which group is lighter and which group is heavier
We know that the odd ball is either one of the B balls in the heavier side or the A ball in the lighter side
(Remember that the A group was lighter group in the 1st Weigh)
IF A[1]+B[1,2] is lighter
We know that the odd ball is in B[3,4] or is A[1]
3rd Weigh: B[3] and B[4]
IF =
We know that A[1] is the odd ball
IF !=
We know that the heavier ball is the odd ball
(Remember that A[2]+B[3,4] weighed heavier in the 2nd Weigh)
IF A[2]+B[3,4] is lighter
We know that the odd ball is in B[1,2] or is A[2]
3rd Weigh: B[1] and B[2]
IF =
We know that A[2] is the odd ball
IF !=
We know that the heavier ball is the odd ball
(Remember that A[1]+B[1,2] weighed heavier in the 2nd Weigh)




 
 
   





Posts: 669 Joined: 28 Jun 2012 Last Visit: 24 Mar 2019
LD count: 11.8 estimate
 


Posted: Sun 01 Jul, 2012 


^^ very nice solution Alvin!
Very elegant



 
 
   





Posts: 669 Joined: 28 Jun 2012 Last Visit: 24 Mar 2019
LD count: 11.8 estimate
 


Posted: Tue 09 Jan, 2018 


There are 25 race horses that each run at different speeds. You can only race 5 horses at a time, but you have no stopwatch or any way to time them. However, you know that all the horses always run at constant speeds. (i.e. The fastest horse will always be the fastest in every race.)
What is the minimum number of races you need to determine the fastest 3 horses?
Sample way of determining the fastest 3 horses (but not necessarily the optimal way):
Click here to see the hidden message (It might contain spoilers)
Race 1: race 5 horses
Race 2: race the fastest three horses from Race 1 with 2 new horses
Race 3: race the fastest three horses from Race 2 with 2 new horses
Continue until all 25 horses have been raced. The fastest 3 in the last race are the fastest 3 of all.
Total races needed: 11 < correct answer is less than or equal to 11



 
 
   
 
 