Question

HELP WITH STRATEGY PROBLEM!!!!!

Answer 1

Idk how to even try :O

Answer 2

I know the answer (my friend gave it to me) but i dont know how to do it

Answer 3

The answer is 1-6-2-10-3-7-4-9-5-8

Answer 4

Sorry I have no clue how to even explain it.

Answer 5

@kmeis002 ?

Answer 6

If i understand the question correctly, you are saying that with each card face up, you place n on the bottom, where n is the value of the card?

Answer 7

yeah i think.

Answer 8

you have to know the order in which the cards will be placed.

Answer 9

The answer is 1-6-2-10-3-7-4-9-5-8 but I have to explain it

Answer 10

Explain why that order works or how the solution can be formed?

Answer 11

Both

Answer 12

they are the same thing

Answer 13

Well, explaining the latter will give us the former, but explaining the former doesnt necessarily give us the logic behind the solution, just a verification that it works, I'm working on it, but I am a bit uncertain if i understand the game completely. Like when is a rule broken? Can you put cards you cycle for through the cards after they've gone on the bottom before removing one through the deck (I assume you can since this solution has to), but I just dont see what you are trying to avoid in this game.

Answer 14

oh, you want the cards placed down in order, 1, 2, 3, 4, 5.. I see

Then the question is, do you put n cards on the bottom where n is the card you revealed or when n is how many cards you already removed

Answer 15

Oh, okay, this was simpler than I was thinking... once second

Answer 16

:D

Answer 17

Quick explanation, not so rigourous, but:

We will cycle through the cards, as each one is removed. Lets define a cycle as having passed through all the cards completely before starting over. So we can show that for each cycle, we will can count how many cards are left over recursively:

\[ a_{n+1} = a_n -\lfloor\frac{a_n}{2}\rfloor \] \[ a_0 = 10 \]
where an is the number of cards

So, after Cycle A, we've removed 5, only 5 cards remain, but now 2 will be skipped over, so we will remove 3, leave 2. Cycle C will have two cards left, 1 will be skipped over and the final Cycle D, will only retain 1 card.
We know that cycle A will have to follow this pattern in the stack
\[ \left\{A_1, *, A_2, * , A_3, *, A_4, *, A_5,* \right\} \]
Then B must start the new cycle, there should only be 3 B's acording to our sequence
\[ \left\{A_1, B_1, A_2, * , A_3, B_2, A_4, *, A_5,B_3 \right\} \]
C1 starts the new cycle, and there should only be 1
\[ \left\{A_1, B_1, A_2, C_1 , A_3, B_2, A_4, *, A_5,B_3 \right\} \]

And the last space left is D1

\[ \left\{A_1, B_1, A_2, C_1 , A_3, B_2, A_4, D_1, A_5,B_3 \right\} \]

Since a is the first cycle, A1-5 must be 1-5
B1-B3 must be 6-8,
C1 must be 9
D1 must be 10, place those in your set and done.

There is probably a much more elegant solution, though. If I think of it I will repost.

Answer 18

Once again, typos all day long, the recursive formula should be the ceiling function

\[ a_{n+1} = a_n - \lceil \frac{a_n}{2}\rceil \]

Note that
\[\lceil \frac{a_n}{2}\rceil = \] the amount of cards removed from cycle n+1

Answer 19

So how would you use this formula. I dont really understand?

Answer 20

Well I accidentally deleted my explanation, but that formula can show you how many cards are left after a cycle has passed: so for 10

\[a_0 = 10 \] \[ a_1 = 10 - \lceil \frac{10}{2}\rceil = 10 - 5 = 5\]
\[ a_2 = 5 - \lceil \frac{5}{2}\rceil = 10 - 3 = 2\]
\[ a_3 = 2- \lceil \frac{2}{2}\rceil = 2 - 1 = 1\]
\[ a_4 = 1- \lceil \frac{1}{2}\rceil = 1 - 1 = 0\]

So this will have 4 cycles, (A, B, C ,D)
A will have 5, B will have 3 cards, C will have 1, and D will have 1.
A goes from 1-5
B from 6-8
C is 9
D is 10

Then arrange your cycles in the order of the rules of the game.

Answer 21

This can solve any size deck of cards with the same rules.

Answer 22

I feeel stupid. Is there another way to explain this other than writing an equation?

Answer 23

Like can an organized list be used?

Answer 24

Ah, alright, the equation is basically saying, each cycle removes half of the cards left over, rounded up. So you start with 10, the first cycle will remove half of 10 and leave us with 5, we can call this cycle A.

Cycle B removes 5/2 rounded up (3) and leave us with 2
Cycle C removes 2/2 (1) and leaves us with 1
D completes the last card.

Now just order the cycles according to the rules and replace with ascending order cards.
{A, B, A, C, A, B, A, D, A, B}

Answer 25

Yes, except the an/2 most be the ceiling function instead of the floor

Answer 26

is that a joke?

Answer 27

No, the ceiling function is defined as rounding up a number and the floor rounds down.

\[ \lceil ceiling\rceil \]
\[ \lfloor floor\rfloor \]

Answer 28

Ohh okay xD I never knew that lol xP

Answer 29

Thanks a lot :) Have a great day!

Answer 30

Not a problem, thats for the nice puzzle.

Answer 31

:)

Answer 32

Gonna tell my teacher you helped me. I dont wanna take all the credit. What is your name?

Answer 33

Karl Meister

Answer 34

Thanks again Karl :)

Answer 35

You're welcome, glad I could be of some assistance.

Answer 36

look Karl I wrote ur name on it!!! @kmeis002