Challenge Can You Solve These Challenging Riddles and Win a Prize?

[Orodruin]

If you are in position 1, then the first victim must be as position:

$2^{m+1} +2-N$

E.g. $N = 1000$, first victim at $26$

This equates to first victim at $1$ you need to be at $976$.

You need to be at position $2(N-2^m)$ if the first victim is at position 1.

This depends on the direction you count in. I specifically stated I count in the direction of lower numbers. Your solution counts the other way.

 

[PeroK]

This depends on the direction you count in. I specifically stated I count in the direction of lower numbers. Your solution counts the other way.

Sorry, I missed that. Yours was a very neat solution in any case.

 

[CynicusRex]

So #4 hasn't been solved by Votingmachine?

 

[micromass]

Yep, he did solve it.

So the winners are:
mfb
RUber
Orodruin
PeroK
votingmachine
fresh_42
Zarqon
martinbn

I will ask @Greg Bernhardt to choose one of these who wins the prize. Thank you all for playing!

 

[Orodruin]

You can remove me from contention. Apart from looking rigged if a mentor would win, the thought process itself is enough reward.

 

[Greg Bernhardt]

I assigned each member a number and then ran a random number generator. @RUber wins!

 

[Orodruin]

I assigned each member a number and then ran a random number generator. @RUber wins!

Congratulations @RUber , well deserved.

 

[RUber]

Thanks!

 

[PeroK]

Just an afterthought for #3 to avoid counting backwards. I'll do it with $N = 11$ members, so that $2^m = 8$. And we want position 1 to survive.

First note that for $8$ members, you will survive by starting at $2$

$1, 2, 3, 4, 5, 6, 7,8$

The evens go first, then the odds (3,7,5). As long as N is a power of 2, position 1 survives.
You need to place the remaining three members so that they go first, followed by $2$. So you do:

$1, 2,3,4,5,6,X1,7,X2,8,X3$

And start with $X1$.

In general, $X1$ will be in position:

$N - 2(N-2^m) +2 = 2^{m+1}- N +2$

 

[jedishrfu]

If it wasnt mentioned earlier the 1000 people circle is a variation of the Josephus problem. Josephus proposed this among his leaders rather than surrender to the Romans as it was forbidden to take your own life you were to instead kill the one next to you.

In the end, only Josephus was left standing, subsequently captured by the Romans and he then aided them in putting down the Jewish revolt.

As a reward, he was freed and became the one to chronical the history of the time. The two Roman generals, father and son both ascended to Roman emperors Vaspasian and Titus.

It was always suspected that he rigged the game knowing the best place to stand ie he selected the first player to start the game.

The moral of the story is knowing the right math can save your life.

https://en.wikipedia.org/wiki/Josephus

 

[micromass]

The game with the coins is called nim. It has a very deep and interesting theory: https://en.wikipedia.org/wiki/Nim A very neat introduction to this and other games can be found here: http://www.math.ucla.edu/~tom/Game_Theory/comb.pdf

 

[jack action]

I'm late in the game as I don't come in this section often, but I don't feel a good answer has been given for #6:

Dividing a cake fairly between two people is easy: let one person cut the cake in two pieces, let the other choose one of the pieces. How would you divide a cake fairly between three people?

• Person 1 cuts the cake into 3 pieces (each of the 3 pieces should satisfy him/her equally);
• Person 2 chooses 2 pieces (each of the 2 pieces should satisfy him/her equally);
• Case 1: Person 3 chooses 1 piece from person 2;
• Case 2: Person 3 chooses the piece from person 1, then person 1 chooses 1 piece from person 2.

This will ensure that everyone can choose what he/she consider a satisfying piece of the cake between the 3 available pieces, size-wise and/or quality-wise.

The «winning» solutions aren't satisfying, especially if we consider that 2 persons could conspire against the other one:

Let one person cut the cake in three pieces, the other two pick first, and the one who cut gets the remaining piece. The only way for the cutter to maximize his piece to 1/3 of the cake, is to cut it fairly, since any other cut than three 1/3 pieces will always leave him with a smaller piece.

This is not necessarily fair because the third person can only choose between 2 choices (whatever the second person did not choose). So person 1 could decide to make a very large piece and 2 small ones, agree with person 2 to pick the large one and person 3 is stuck with a choice between 2 small pieces such that the others share the large piece and one small piece together.

One cut what he thinks is a 1/3, if the other two are happy, it's his part and the problem is reduced to two people. If one of them is not happy he cuts a piece of the "1/3" of the first person to make it what he, the second person, thinks is a 1/3, if the third person is happy the second takes the piece and the problem is reduced to two people. If the third person is not happy cuts a what he thinks is a third and the problem is reduced to two people.

Although this may work size-wise, it is kind of messy if the second and third persons both cut a piece of the first "1/3". How do we decide who gets to keep the two small pieces to «complete» their share? To fairly do it, I guess one of the remaining person would have to cut each of the 3 remaining pieces in 2 parts and the other one would choose 1 part from each piece. We now have 7 pieces in total. Really messy.

 

[mfb]

especially if we consider that 2 persons could conspire against the other one:

That is still possible in your approach: if 1 makes one piece very good, 3 can pick it, and 2 cannot get it.

 

[jack action]

That is still possible in your approach: if 1 makes one piece very good, 3 can pick it, and 2 cannot get it.

You're right. Then I guess someone will have to eat a multi-piece share!