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Riddle Answering; Someone asks a riddle, the one who answeres correctly has to ask the next.
Topic Started: Aug 20 2010, 11:31 AM (12,843 Views)
amginE
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ky's solution is correct.

here's my version of the diagram
win/loss ratio on first line
current amount of money +- amount you should bet that round on second line

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V
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Man, so tedious. Too much written work for me I guess.
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kycse
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Alright, Since I'm more of a riddler, I'll use a riddle I don't have the answer to either. I'll probably know if the answer is right or wrong though, and this one is, every now and then, on my mind for quite some time now.

Suppose you have 100 coins. 96 of them are heavy and 4 of them are light. Nothing is known regarding the proportion of their weights. You want to find at least one genuine (heavy) coin. You are allowed to use a weight balance twice. How do you find it?

Assumptions:

Heavy coins all have the same weight; same for the light coins.

The weight balance compares the weight of two sides on the balance instead of giving numerical measurement of weights.
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Tetsuki
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I know a similar riddle. The difference is that theirs only 1 with a different weight and you have to find out witch it is and if it is lighter or heavier. And you have a total of 12 and can weight 3 times.

Okay, perhaps the answer is similar.
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Tetsuki
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Now that I thought about it abit I think it is impossible with just wighting them 2 times. because if when you wight them the first time and 2 or 1 light coins are in each you dont have enough to find out.
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V
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Closest I get is:
3 on each side, if one side is higher ignore it, other side has either 1 light or 0 light coins. Take one out and compare the remaining, if the same then both are heavy, other wise the lower one is heavy.

But this doesn't account for 1 or 2 light coins on both sides.

EDIT: Well it does account for 2 light coins on both sides
Edited by V, May 29 2011, 05:51 PM.
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kycse
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See, that one isn't that easy, but I recall that it can be solved. Therefore, do your best~ (I'll too of course)
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Tetsuki
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I think I got it.
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Tetsuki
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But it's abit complicated. Don't know if I can explain it in english correctly- We will see.

For the first time you weight 2-2:

In case it is uneven:

You use the heavier site 1-1:

If even both must be heavier ones, because the 4 lighter ones could only be distributed: 2-0; 1-0; 2-1

If uneven it is the heavier one.

In case it is even:

Now it get's a little tricky.

For the second weighting you use, one the left site, one of the coins from the left and one from the right from the previous weighting. On the right site the second coin of the previous left site and a coin you havend used jyet.

So you get:

(1+1) - (1+1)

In case they are even:

Because before they were even their can only be light coins in the combinations: (0-0), (1-1) or (2-2)

In the first case every coin would be heavy.

In the third case they can't be weight the some, because all 4 light coins were already used.

In the second case:

Either both coins are on the left, but because we have the second of the first weight they cant be even so it is not possible that both are on the left.
So that leaves only 1 light one on the left and on the right and in that case the second and the other added coin would be heavy.

So those 2 are certainly HEAVY

In case it is uneven:

(1+1) - (1+1)

We still have the options of (0-0), (1-1) or (2-2)

In the first case the added coin would be lighter and the right side would be lighter.

In the third case the left side would be lighter and the added coin would be heavy.

In the second case (this stupid 1-1 is the problem in all the weights...):

If the left is lighter:

only possible if both are lighter or if only the left is lighter. In both cases the secon coin of thefirst weigt would be heavy.

If the right is lighter:

the left left coin must be heavy because only if both right ones are light ones it is possible or at least the left coin on the right must be.


Hahaha... I just noticed the flaw in this plan...
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Tetsuki
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I know that I am only one step away from the answer...
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amginE
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kycse, are you certain that there exists a correct answer to this?
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kycse
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I'm 100% certain that there is a correct answer. I could even give you the link to it if you want. If I recall correctly this was a task in a russian math contest or something.
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Tetsuki
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Anyone got an idea how we caqn change the last weighting so my way functions. If anyone understood what I did.
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V
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I don't see how it is possible due to the possibility of having 1 or 2 light coins on both sides.
There isn't any 2 step way to distinguish between having 1 light coin on both sides or 2 light coins on both sides.
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Tetsuki
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That's why I am trying to find a way where in both cases the weight would be on the same site and not the other. Now I have to find a way to do it.
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amginE
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i got the answer. let's see how long it takes for you guys to figure it out =)
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Tetsuki
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omg

no tell us.
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amginE
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thirty three coins on the left, another thirty three on the right.

if one side is heavier than the other, then you know for certain that the heavier side has at most ONE light coin. you can then just take two random coins from the heavier side and compare their weights. if they are the same weight, you know both are heavy. if they are different, then obviously the heavier coin is heavier.

if both sides are equal, there's three possibilities:

1) 33 heavy coins on each side. (the third group has 30 heavy coins and 4 light coins)
2) 32 heavy coins and 1 light coin on each side. (the third group has 32 heavy coins and 2 light coins)
3) 31 heavy coins and 2 light coins on each side. (the third group has 34 heavy coins)

now, take one coin from one of the piles of 33 and add it to the other pile of 33 coins, making a pile of 34 coins.
for the second weighing, put this new group of 34 on the left, and the 34 coins from the third group that didn't get weighed the first time on the right.

the possibilities are
1)
left side: 33 heavy coins plus 1 added heavy coin.
right side: 30 heavy coins and 4 light coins
result: LEFT side is heavier

2a)
left side: 32 heavy coins and 1 light coin plus one added coin of unknown weight, which happened to be a heavy coin.
right side: 32 heavy coins and 2 light coins
result: LEFT side is heavier

2b)
left side: 32 heavy coins and 1 light coin plus one added coin of unknown weight, which happened to be a light coin.
right side: 32 heavy coins and 2 light coins
result: balance is EQUAL

3)
left side: 31 heavy coins and 2 light coins plus one added coin of unknown weight (doesn't matter which).
right side: 34 heavy coins
result: RIGHT side is heavier

only possibility #3 would result in the right side being heavier, so if that's the result, you know that all the coins on the right side are heavy.

only possibility #2b could result in a tie. so if that's the result, you know that the 32 coins not on the balance during the second weighting are heavy. (since it was originally a pile of 33 with one light coin, but that light coin was taken away)

if the left side is heavier, then it could be either possibility #1 or possibility #2a. in either case, the coin that was added had to be a heavy coin regardless, and that's the only coin you can be certain is heavy.

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apoptoxin4869
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kycse
May 29 2011, 05:10 AM
I just got it, amgine you are great! Equivalence classes all the way do the trick! I'll write another diagram, cause visualization works really well with this :D

Okay, let's do this:

First our approach with win or loss was more or less totally wrong (well, at least in my approach it was.) We need to think in # of wins and # of losses. Therefore 0/0 will mean, 0 wins and 0 losses.
Okay, let us visualize what we know then:

Posted Image
Original Size

Okay. This is the Graph, left means win, right means loss. The circled states are our final states, when we get there we've either won or lost, and therefore we know how much gold we have at those states. for 4/X states that's 4000 and for X/4 states that's 0. And we know that we have 2000 at the beginning of this, the 0/0 state. Now we can calculate all the other states by adding the values of their two childs and divide them by two.
Example:
3/3's childs are 4/3 and 3/4 -> 4000 + 0 / 2 = 2000.

Posted Image
Original Size
edit: Just noticed that 0/1 is 1375 and not 1325...

This is what the graph looks like when we write the "current amount of money" under the respective state. To calculate the bet per round you have to subtract from the value of the current state the value of it's right child:
0/0 - 0/1 = 2000 - 1375 = 675, so 675 gold is the bet in the first round. For all states that makes:

250: 3/0, 0/3
500: 2/0, 0/2, 3/1, 1/3
625: 0/0, 1/0, 0/1
750: 1/1, 2/1, 1/2
1000: 2/2, 3/2, 2/3
2000: 3/3

And then we will end up at either 0 or 4000.
from handwriting youre a guy?????????????????????????????????????????????????? my handwriting is bubbly xDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD sry iz bit off topic
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kycse
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amgine, yep works, your turn.

@apo: I am... But that's still with my pen, if I take a biro it tends to get scratchy... If you want you can start some kind of "Handwriting Comparing" Thread, though please refrain from commenting on that in this thread again :D (And I always wondered how people, who write bubbly / kind of cursively, write fast... even if I write slowly it ends up nearly illegible sometimes)
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amginE
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This might be even more difficult than kycse's...

You have 9 pieces of legendary equipment:

Arcane Armlet of Agility
Brilliant Brooch of the Battlemage
Clairvoyant's Circlet of Cognition
Demonic Dragonhide of the Druid
Enchanted Earrings of Evasion
Fortuitous Footgear of the Falcon
Glorious Gauntlets of the Gorgon
Holy Helm of the Hermit
Iron Leggings of the Iguana

Each gives an amazing stat boost and are coveted by all the magic-users of Second Life. However, one of them was rumored to be crafted by the Dictator of Life himself, and is considerably better than the others (roughly 1.5x better). You really want to know which item is it, but you have no clue. The only way is to test them out in battle...

However, only the highest leveled mages can wear such legendary items, and nobody on your own team qualifies. You put out a request on the notice board that you want to hire some high level mages, and got a response from Enigma team, a trio of elven sorcerers: Xenophon, Yggdrasil, and Zerocity.

Each member of the Enigma team specializes in a particular school of magic. One is a pro at Psionic Pressure, another is a master of the Mind Meld, and the third is the most skillful Spiritual Striker around. Their levels are all more or less equal, so in an even fight, it will just depend on who has the better gear. If, for example, one sorcerer wears four legendary items and the other wears three, then the one with four will be sure to win. If both sides wear four items, but one side has the godly item crafted by the Dictator, then the one with the godly item will clearly win. If both sides wear three legendary items each, and neither has the godly item, then it will result in a tie.

The only exception is that when the Mind Melder and the Spiritual Striker face off, their attacks will actually collide and create a wave of Chaos Chakra, which deals immense damage to both sides, and the outcome (win, loss or tie) is really just a matter of pure luck, regardless of what equipment both sides are using.

Upon hearing this, you ask them to reveal which one specializes in which, so that you can avoid such an outcome. Unfortunately, they refuse, since that is their closely guarded secret.

Given these conditions, what is the minimum number of duels that you need to set up between the members of Enigma team in order for you to figure out which item is the one created by the Dictator?

In your answer, you should specify something like:
duel 1: Xenophon wearing the Armlet and the Earring duels Zerocity wearing the Footgear and Gauntlet
duel 2: ...
duel 3: ...
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Pikastache
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I have a question about the question listed above. xD (That sounds a bit weird.) Do you know which of the hired mages has which ability? If so, then you would only need 2 of them to test out the equipment. Also, it would lower the amount of times needed to ! the combinations with a certain equation. ;)
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amginE
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No, they won't tell you that info, it's their "closely guarded secret".
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Pikastache
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Okay, I think I might have got it. Not completely sure though.
Duel 1: X vs Y (Without weapons)
<In the worst case scenario there will be a chaos wave. Therefore, you can chose one of them to battle Z so when testing the items there is no underlying variables.>
<If there is no Chaos wave then just use X and Y>
Duel 2: The two duelers above who did not create the chaos wave. (Let's say in this case X and Y)
X: Wears a,b,c,d
Y: Wears e,f,g,h
Leave "i" untouched.
<Best case scenario: There will be a tie and "i" Is the godly item>
<Worst case scenario: One of them wins.>
Duel 3: Assuming worst case scenario and Let's just say X wins.
X: Wears a,b
Y: Wears c,d
Duel 4: Let's keep assuming X wins.
X: Wears a
Y: Wears b
<Between the two items one of them will be the godly item.>

Answer: Best case Scenario is 2 Duels. Worst case Scenario is 4 Duels. *onion hero pose*

**I apologize for my previous post. I must have skimmed over the part where they decide not to disclose their powers.
Edited by Pikastache, Jun 2 2011, 11:16 PM.
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amginE
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chaos wave is a completely psychic/mental phenomenon between the minds of the two duelers, you can't tell whether or not it has occurred.
sorry for not mentioning it earlier
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daystar
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hmmm.... check they're HP?
Edited by daystar, Jun 4 2011, 10:18 PM.
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kycse
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Okay, I'm not confident that this solution is the one with the minimal number of duels, but it works:

Duel 1: X fights Y with X having accessories a to d and Y having e to h. If X or Y is stronger proceed with Duel 3, And we know that those two are a save combination. If it's even proceed with Duel 2.

Duel 2: Y fights Z, Y has a to d, Z has e to h. If it's even, then the accessory i is the accessory you search for (Since either Duel 1 or Duel 2 would have to be unevenly matched). If Y or Z is stronger, you know that this combination is save.

Before Duel 3: By now you know which at least one set of mages, who are save with each other, therefore I'll refer to them as S1 and S2. You also know that there is one set of accessories in which the one you search for is in, either a-d or e-h, therefore I'll just call them a1 to a4 now. (Since it's a hassle to write down all the possibilities although the same method is used...)

Duel 3: S1 vs. S2, with S1 wearing a1 and S2 wearing a2. If S1 wins, then a1 is the accessory, if S2 is stronger than a2 is the accessory. If both are even, proceed with Duel 4.

Duel 4: S1 vs. S2, with S1 wearing a3 and S2 wearing a4. Either S1 or S2 is stronger now. Take the respective accessory then.

best case: 2, worst case: 4... However, I don't think the worst case 4 is the best solution to this, there's probably some way to get it done with 3.

Nvm, I should read the whole question first... I thought chaos wave is always a tie.
Edited by kycse, Jun 5 2011, 08:46 AM.
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BabyMonkey28
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I have an answer that will use 10 battles at most, but 9 in most cases. But something tells me that that's too much..... Oh well! Here it is!

1. X using A against Y using B
2. Y using C against Z using D
3. Z using E against X using F
4. X using G against Y using H
5. Y using A against Z using I
6. Z using B against X using C
7. X using D against Y using E
8. Y using F against Z using G
9. Z using H against X using I
10. I made a really big chart of possible outcomes (with the chaos results written as random). So once you got the outcome of these battles, you would record it and compare it to the chart. Any of the possible outcomes could point to either 1 or 2 weapons as possibly being the legendary ones. If it’s one possibility, you’re all set. But in the rare case that there are two possible, you would need to continue with 1 more battle. So, using the chart again, it’ll show at least pair of fighters that would definitely not cause the chaos thingy. So use that pair to test those two possible weapons, and the winner holds the legendary weapon.

Is that right?
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amginE
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BabyMonkey: The duelers can wear more than one equipment at a time. Would that help you reduce the number of battles?
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BabyMonkey28
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Probably, but... meh... I don't have time right now to do it out. :(
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daystar
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i'm thinking but nothing yet
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BabyMonkey28
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Ok, so I got it with 6 battles this time. You need all 6 no matter what, but no more.

1. X using BEHI vs Y using CDFG
2. Y using AEHI vs Z using CDFG
3. Z using AEFH vs X using BDGI
4. X using ABCH vs Y using EFGI
5. Y using ABDG vs Z using CFHI
6. Z using ACDG vs X using BEHI

I'm still wondering if 6 is too much though....
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amginE
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yes, still too many
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Tetsuki
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How about lifting it?
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apoptoxin4869
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Y NO ONE POST WHEN I GONE??????????? IS DEDDDDDDDD DDDDDDDDD: WE NEED NEW RIDDLEEEEEEEEEEEEEEEEEE NOOOOOOOOOOOOOOOOOO ONE ANSWERSSSSSSSSSSSSSSSSSSSS
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