# Tag Info

## New answers tagged strategy

2

I have found a better answer than this answer by just playing with it: with not sure this is optimal though, but most likely. gonna write a program if noone does that until I wrote :D

8

Edit: my answer was inadequate. Only downvotes now please. @Oray asked to follow up his answer and I found a fault in my previous work, resulting in My original and obsolete answer was

8

Found a solution requiring First, label the coins A,B,C,D,E,F,G,H,I,J. Then for the first 4 weighings, weigh: For: For (Update: fixed mistake, thanks to comments): For: For: In any case, And as has been pointed out, there are 360 initial configurations, and $360 > 243 = 3^5$, so at least 6 weighings are required.

0

I think 6 will workout.Let us divide the total coins in set of 3,3,3,1. Now when you compare 3(1),3(2)/ 3(1),3(3)/3(2)3(3) ypu will get various weights increasing order. I will use mathematical forms to represent the possible weights. Possibilities are +,0,- +,+,0 ++,0,- +,+,- 0,0,+ By internal comparison you will be able to get in 6 minimum chances.

2

2

Not sure if this is optimal, but with my strategy I need at most balancing actions. We use the following notation: Here is the tactic: There is no way that this step takes fewer turns using this tactic because: Now that we figured out which case we have, we can do this: We are done.

4

I have a method that will identify the fakes in weighings: Start by There are three possible outcomes: Let's handle these separately, starting with the easiest case that has Next, let's handle the case with And finally, there could be Now, if only we were able to shave off one weighing from the "only 1 unbalanced" case..

6

Not a solution but the theoretical minimum weighings required is 6. There are 360 combinations of coins. 10 locations for the smaller fake and then 2/9 locations for the larger fakes: 10*nchoosek(9,2) = 360. Each weighing provides a ternary bit of information. At a minimum, 6 weighings are sufficient to learn all fake coin locations: 3^6 = 729 > 360. ...

7

Lets try it in natural language or abbreviated

0

2

Lets give it a try. We name the coins $HHLGGGGGGG$ (Heavier Lighter, Genuine) Weighing 1 Weighing 2-5 After making Weighings 2-5 you have a pattern with 4 results H (Higher), L (Lower), E (Equal), ordering is not important Now, if your Pattern is ... So go on with what is left Solution

8

@Oray congrats this question is glorious!

7

Upper Bound Given 10 people, there are 45 possible pairings. This would mean that there is an upper bound of 45 items. However, the first pair to enter the dungeon will be the same level (level 1), so there will be no item awarded. Similarly, the last pair to enter the dungeon will have gone with all 8 others first and both will be at level 9. Their ...

11

Between 10 friends, there are only 45 unique pairs. Each friend will participate exactly 9 times, which means after 45 rounds everyone will be level 10. Since it's impossible to get an item on the first & last run, there is a theoretical maximum of 43 items you can get. However, due to some restrictions, it seems to me like it's impossible to get this ...

7

Let's start low: 40. Let's name the players A,B,C,D,E,F,G,H,I,J and define "n-next" of a player to be the player who is positioned n places afterwards, with round wrapping. Eg, the 2-next of A is C, and the 3-next of J is C. First, AB go in to become level 2. Then, for eight rounds, the player who was 1-next of his partner in the previous round goes in ...

2

A completely sideways answer - but possibly only one try... Here you need to cheat a little, and cause yourself some sort of injury to get some blood - smear some of that blood on one card, and then another - keep doing this until you find two that have different shades. The red "filter" cannot restore invisible (to you) colour, but it can hide some of the ...

0

I can do it in

13

I can prove a lower bound of The first step is to notice that the question is equivalent to The reason is: and Then we observe and So there are a few cases. If If If

13

Here is one way to win almost all the candies: I think this solution is optimal, in the sense that you cannot end with more candies.

24

It can be done in Solution

16

Also not sure whether this is optimal but I can do this in First, In the worst case, Finally,

1

Not sure if this is optimal, but: Then worst case, after So you're guaranteed to find two green on your

-2

If I can get the cards back after an attempt Otherwise,

11

This is like a reverse puzzle to the well-known Here in reverse: A remark as a thinking outside the box solution:

13

I believe I can open it in The reasoning behind it is:

5

A note: @Reinier, besides having an awesome palindromic username, also solved this puzzle first (my solve was still independent). I like my explanation better (but of course I'm biased), but if you upvote my answer, don't forget to upvote Reinier's as well. So, it's somewhat apparent that we need to find Here's my basic strategy: When we present each bag ...

12

I can ensure that I receive at least To get this amount of coins, I could apply the following strategy: My friend can also ensure that I do not get more coins than this in the following way: So it appears that I should have chosen a fairer game to distribute the coins...

1

@noedne's answer from the beginning struck me as a good one, but I have recently come to the conclusion that it is also the optimal answer. I'll be first to admit that I don't understand everything going on, but I feel the need to at least try and explain it. Everything below is what I said the first time-around. It is of no importance. Sorry I'm late to ...

1

I think @Trenin has nailed it. Here's a slightly different formulation where we prove it by induction. We can get 1 point in a row with each of us playing the same number of moves ($m_1=1$) and one end free of other points. Suppose we can get $k$ points in a row after we've each played $m_k$ moves with one end free of other points. Required to show: that we ...

7

First of all, lets present a strategy for $N=4$. $N=4$ $O$ can win in 7 moves (13 total). To start, the first two moves are arbitrary due to rotation and stretching, so let us assume we have the following; 1. $O_1=[0,0]$ 2. $X_1=[-1,-1]$ Now, $O$ will pick a line not on the $O_1,X_1$ line and place another point. $X$, meanwhile, should place a ...

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