The riddle of the sfinx

[Xinto 18]

I was going to say, go back and try the gold/silver handled door (the one you didn't first time) as that would have different options. =P

[EgoReaper 9]

I was going to say, go back and try the gold/silver handled door (the one you didn't first time) as that would have different options. =P

But in the riddle it didn't matter what door you chose, implying the end result would always be the same. Similar to some riddles I've heard before.

[Demyndra 6]

Mr. Dodgers volunteered to be the chief organizer for the world’s teatherball championships. There were 657 contestants from around the world. The tournament was set up whereby the winner would advance and the loser would be eliminated. Since there was an odd number of participants, the initial pairing leaves one player out. That player gets a bye and automatically advances to the next round. How many matches will Mr. Dodgers have to schedule to determine the teatherball champion?

[Shine 144]

656 players in the first set, 2 players per match makes 656/2 = 328 matches

I can do the math further but these should be the only matches he has to plan, the rest will sort out itself.

so my answer is 328.

[Demyndra 6]

656 players in the first set, 2 players per match makes 656/2 = 328 matches

I can do the math further but these should be the only matches he has to plan, the rest will sort out itself.

so my answer is 328.

328 is incorrect, however, you are on the right track.

[Caveman 0]

The answer is equal to the number of players minus 1, 656 in this case.

 

The solution is to show first that 1 player results in 0 matches, then show that in all cases, if we add a player, we add a single match.

 

 

It's sorta evident that for 1 player, there will be 0 matches.  Auto winner!

 

If we add a player to an odd pool of players, they'll just play the bye in that first round.  After that it's the same counts as before. So it only adds one match.

Ex: First round of 5 would be 2 1v1 matches w/ a bye resulting in 3 players in next round.

First round of 6 would be 3 1v1 matches resulting in 3 players in next round.

 

The next part is a bit more complicated and depends on the fact that taking a number > 1 and repeatedly dividing by 2 will eventually result in an odd number at some point.

 

If we add a player to an even pool of players, then they will get a bye to the next round.  This doesn't add a match in this round, but adds a player to the next round.

    If the next round would have been odd before, then we are adding a player to an odd pool and the first rule above applies, adding 1 match.

    If the next round would have been even, then *this* rule applies and we go again.

As stated above, you must eventually reach an odd case.

So the even case eventually reduces to the odd case, which adds one match.

Ex: If we have 8 players, it's:

  Round 1: 8 players = 4 matches

  Round 2: 4 players = 2 matches

  Round 3: 2 players = 1 match

If we add a player, then

  Round 1 is now 9 players, adds a bye, no extra match.

  Round 2 is now 5 players, adds a bye, no extra match

  Round 3 is now 3 players, adds a bye, no extra match

  Round 4 is now 2 players, this is the extra match.

 

 

 

 

 

[Demyndra 6]

The answer is equal to the number of players minus 1, 656 in this case.

 

The solution is to show first that 1 player results in 0 matches, then show that in all cases, if we add a player, we add a single match.

 

 

It's sorta evident that for 1 player, there will be 0 matches.  Auto winner!

 

If we add a player to an odd pool of players, they'll just play the bye in that first round.  After that it's the same counts as before. So it only adds one match.

Ex: First round of 5 would be 2 1v1 matches w/ a bye resulting in 3 players in next round.

First round of 6 would be 3 1v1 matches resulting in 3 players in next round.

 

The next part is a bit more complicated and depends on the fact that taking a number > 1 and repeatedly dividing by 2 will eventually result in an odd number at some point.

 

If we add a player to an even pool of players, then they will get a bye to the next round.  This doesn't add a match in this round, but adds a player to the next round.

    If the next round would have been odd before, then we are adding a player to an odd pool and the first rule above applies, adding 1 match.

    If the next round would have been even, then *this* rule applies and we go again.

As stated above, you must eventually reach an odd case.

So the even case eventually reduces to the odd case, which adds one match.

Ex: If we have 8 players, it's:

  Round 1: 8 players = 4 matches

  Round 2: 4 players = 2 matches

  Round 3: 2 players = 1 match

If we add a player, then

  Round 1 is now 9 players, adds a bye, no extra match.

  Round 2 is now 5 players, adds a bye, no extra match

  Round 3 is now 3 players, adds a bye, no extra match

  Round 4 is now 2 players, this is the extra match.

656 is correct... it was my riddle and I feel insanely stupid after reading your answer lol

[airacaz 1]

so easy a caveman could do it ;)

[Caveman 0]

Induction can be fun.    So, I suppose it's my question?  Well, I'm not very creative, so how about I just extend Demyndra's question to double elimination!  

 

We still have 657 players, but now, when they lose, they get added to the next round of the loser's bracket.  If they lose a match there, they are done.  The last match of the tournament is the undefeated champion vs. the challenger, i.e. the winner of the losers bracket.  

 

We'll use the same 'bye' mechanics in the loser's bracket for an odd number of players.  

It's possible that the champion vs. challenger match can involve multiple matches since the champion has to lose twice to lose, but we'll just call that a single "match".  

 

How many matches will the poor overworked Mr. Dodger's have to set up now?  This will teach him to volunteer!    

[Demyndra 6]

1,312...I think I did this right.

Winner's bracket is unchanged... there are still 656 matches to get the winner out of that bracket.

Everyone who loses, plays again to find a winner from the losers bracket, so that is another 655 matches to be played.

Then one last match for the winner versus challenger.

[Caveman 0]

Yup, that's right.  A double elimination tournament is effectively 2 merged single elimination tournaments.

 

What I find really interesting is that it doesn't matter at what stage in the tournament you add someone, it still results in exactly 1 match (in single elimination) or 2 (in double elimination).  I don't think that's exactly intuitive.

 

For example, normally you would do both brackets somewhat simultaneously.  For example for a 10 entrant tourney:

Rnd WB               LB

   1   5v5             

   2   2v2 + bye    2v2 + bye

   3   1v1 + bye    2v2 + bye

   4   1v1              2v2

   5   bye              1v1 + bye

   6   bye              1v1

   7             1v1

 

Total matches = 18

 

But the math says it should be the same separately where we get all the losers together before playing a single game.

 

Rnd WB               LB

   1   5v5             

   2   2v2 + bye    5

   3   1v1 + bye    7

   4   1v1              8

   5   bye              9

 

Now  the 9 in the losers bracket play it out.

 

Rnd LB  

   1   4v4 + bye             

   2   2v2 + bye

   3   1v1 + bye

   4   1v1   

 

Plus the challenge match 1v1.

 

Of course, the first method is superior for fairness because it preseeds the players in the losers bracket according to when they lost, but the way total matches equal out is still pretty neat.

 

 

 

 

 

[Shine 144]

I still think only the first 328 matches should be planned. after that its a simple tree upwards and you do not have to plan those since all the matches will sort themselves out.

first round there are 328 matches and 1 player left

second round there are 164 matches and 1 player left

third round there are 82 matches and 1 player left

fourth round there are 41 matches and one player left

fifth round there are 21 matches and no players left

sixth round there are 10 matches and one player left

seventh round there are 5 matches and one player left

eighth round there are 3 matches no players left

ninth round there is 1 match and 1 player left

10th round is the the grand finale between the winner and spare of round 9.

328 + 164 + 82 + 41 + 21 + 10 + 5 + 3 + 1 + 1 = 656 so there will indeed be 656 matches. but everything from round 2 and up does not have to be planned.

I'm not trying to get the turn or something, I am just under the impression that when you ask how many matches have to be set up, its only 328.

Cheers.

P.S. this is only speaking of the winners tree obviously, it does however account for the losing side aswell.

Edited September 14, 2013 by Shine

[EgoReaper 9]

656 contestants leaves 328 winners with 1 bye (328 matches there)
328 contestants leaves 164 winners with 1 bye (164 matches there)
164 contestants leaves 82 winners with 1 bye (82 matches there)
82 contestants leaves 41 winners with 1 bye (41 matches)
42 contestants (bye added in) leaves 21 winners (21 matches)
20 contestants leaves 10 winners with 1 bye (10 matches)
10 contestants leaves 5 winners with 1 bye (5 matches)
6 contestants leaves 3 winners (3 matches)
1 match without the bye, 1 match with the bye

328+164+82+41+21+10+5+3+2 = 656 matches.

Interesting answer.
Edit: sorry, didn't see the new page lol, was still answering demy's question
 

Edited September 14, 2013 by EgoReaper

[Caveman 0]

I still think only the first 328 matches should be planned. after that its a simple tree upwards and you do not have to plan those since all the matches will sort themselves out.

first round there are 328 matches and 1 player left

second round there are 164 matches and 1 player left

third round there are 82 matches and 1 player left

fourth round there are 41 matches and one player left

fifth round there are 21 matches and no players left

sixth round there are 10 matches and one player left

seventh round there are 5 matches and one player left

eighth round there are 3 matches no players left

ninth round there is 1 match and 1 player left

10th round is the the grand finale between the winner and spare of round 9.

328 + 164 + 82 + 41 + 21 + 10 + 5 + 3 + 1 + 1 = 656 so there will indeed be 656 matches. but everything from round 2 and up does not have to be planned.

I'm not trying to get the turn or something, I am just under the impression that when you ask how many matches have to be set up, its only 328.

Cheers.

P.S. this is only speaking of the winners tree obviously, it does however account for the losing side aswell.

 

Either 1) you are setting up the full bracket sheets, in which case you are planning all those matches for the next rounds even though you don't yet know who the players are by name, but rather by result, i.e. this match is played by winner of match 1 and winner of match 2.

Or 2) You are setting them up for each round.

 

Having run various types of tournaments, you plan *everything*.   For example, for a tetherball tourney, you would have to assign the tetherball pole to be used for each match.  There's usually a limited number.  Optimizing equipment usage is probably another interesting question, but I think all these math questions which I personally like are not technically "riddles".

 

Besides it's Demyndra's turn to ask.

Edited September 14, 2013 by Caveman

[Nemo 27]

Math explosion... I'm just a poor humanities student, have some mercy!

[Shine 144]

I guess you are right caveman. I checked back to the original post and it said scheduled, not planned. I guess you should schedule each match apart. so the answer would then indeed be more. I only tried to think in terms of riddles. and often answers are not what you expect. so I hoped my answer was right.

as I said in that original answer was that I could do the math further, but I chose not to since I would kinda give two answers.

too bad, I tried ^^

[Caveman 0]

There's a bit of a pause, so I'll go ahead and throw another one out there.  It's only a bit mathy, but I think the result is surprising.

 

Crazy Jim decides that it's silly that a globe has a line on the equator but there isn't one in real life.  Being crazy, his solution is to wrap the earth with a big rope.  You'd think he'd have trouble with that, but it's surprising how many people will ignore a crazy guy with a rope!  So he gets it done, but then decides that it would be nice to be able to limbo under it world-wide.  So he wants to raise it up 3 feet at each point.  This will make the rope longer, but the question is how much?  You can assume that the earth is a perfect circle with a diameter of 7918 miles.

Edited September 14, 2013 by Caveman

[Xinto 18]

2*Pi*radius

2*Pi*(radius + 3 feet) = 2*Pi*radius + 2*Pi*3feet

 

so how much longer is 2*Pi*3 feet = 6Pi feet or 18.84955592153875943077586029968 feet to 31 significant figures

[EgoReaper 9]

2*Pi*radius

2*Pi*(radius + 3 feet) = 2*Pi*radius + 2*Pi*3feet

 

so how much longer is 2*Pi*3 feet = 6Pi feet or 18.84955592153875943077586029968 feet to 31 significant figures

I want whatever calculator you're using, mine doesn't get nearly as man sig figs.

[Xinto 18]

I want whatever calculator you're using, mine doesn't get nearly as man sig figs.

Hahaha it was just the windows calc in scientific mode =)

[Shine 144]

I have no idea how much feet go in a mile, being somewhat of a scientist I hate (and refuse to use) the imperial system. I can give you the calculation using the metric system, which I will do. this does make the answer wrong but the calculation right.

diameter = 7918 kilometers

rope raisal all around = 1 meter (I beleive 3 feet should be about 1 meter)

which makes the circumference of the earth 7918000 meters * pi (for which I will use only 3.14)

7,918,000 * 3.14 = 24,862,520 meters

now we want to raise the rope 1 meter all around, meaning the diameter goes up by 2 meters.

7,918,002 * 3.14 = 24,862,526

the difference between the two circumferences is 6 meters, which is your answer.

in order to make the rope raise 1 meter all around you would need to make the rope 6 meters longer.

I used to do this riddle a bit different:

if you have a rope tight around the earth, and you make it 1 meter longer, how high can you raise it. which is surprisingly high.

anywasy, 6 meters is the answer (using the metric system with the same values rather than calculating from imperial to metric)

[Xinto 18]

Since the calculation for circumferance (2*Pi*R) is commutative and the conversion factors would be put in and then cancelled out, You can use the imperial form.

 

But I fully agree, metric is the way it should be.

 

Note 6 meters is approximately equal to 18.8 foot using your conversion approximation.

 

If we do agree that that is right, and it would be my turn, Shine can take it for me =)

Edited September 15, 2013 by Xinto

[Shine 144]

well, I guess it is right (and I noted that my answer was pretty much the same as yours).

I'm just going to go ahead and post the next riddle without confirmation because I cannot think of anything being wrong with either of our answers.

 

I did not actually get to read all of them but I think this one has not been around yet.

 

you are in a hallway,

there are three lightswitches,

one of them turns on a lightbulb,

the others do nothing,

you cannot see the lightbulb since it is in a room upstairs,

you can try all of the switches as many times as you like,

when you walk into the room you must know which switch belongs to the lightbulb,

you may not walk out of the room to try other switches.

once you are in the room, you must have your answer.

 

how do you find out which switch belongs to the lightbulb?

[airacaz 1]

since there are 3 switches you only have to turn two on to find out which one controls the light

the switches (1,2,3) control one light

first you take switch one and turn it on wait for a while for the light bulb to heat up, hopefully the light bulb isn't florescent or LED so that you can notice the change otherwise you'll have to be quick getting upstairs to check it

so after waiting for it to heat up turn the first switch off and then turn the second switch on and then hurry upstairs to look at the light bulb if it is on then the second switch controls the light if the bulb is off then you have to check the temperature of the light bulb if it is warm then switch 1 is the controlling switch if it is cold then switch 3 is the controlling switch

[Shine 144]

very good. and yes, in this scenario that lightbulb heats up so you can deduce it from that.

 

This is a very good lesson to teach people to use more of their senses. most people only see and hear to draw conclusions, whilst tasting feeling and smelling are just as valuable.

 

Airacraz, you're up