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This question actually relates to a video game, I came across the scenario and I realized I had no idea how to go about solving something like this or even what branch of mathematics it falls under.

Anyway, the simplified version is as follows:

There are two types of currencies, I will call them currency A and currency B.

You start out with an income of 512 of currency A per second and an income of 36 of currency B per second.

The goal is to get to 1 million or more of each type of currency at the same time as fast as possible.

Although, in order to speed up your progress you can invest in things that will give you more currency. The things you can invest in and their bonus income are as follows:

Format: Option name, cost, bonus income

  • A1, 33 B, +1 A/s (s is seconds)
  • A2, 257 B, +8 A/s
  • A3, 1025 B, +32 A/s
  • A4, 4097 B, +128 A/s
  • A5, 16385 B, +512 A/s
  • A6, 65537 B, +2048 A/s
  • A7, 262145 B, +8192 A/s
  • A8, 1000001 B, +32768 A/s
  • B1, 4096 A, +1 B/s
  • B2, 20480 A, +6 B/s
  • B3, 81920 A, +36 B/s
  • B4, 262144 A, +216 B/s
  • B5, 1000000 A, +1296 B/s

Using these investment options, what is the fastest way to obtain 1 million or more of both currency A and currency B at the same time?

I am most concerned with what TYPE of question this is and HOW to solve it rather than the actual answer. Any help on this would be appreciated.

Note: Since I am not sure what branch of mathematics this is I put optimization down as the tag, if this tag is incorrect I would appreciate it if someone could add the correct tag.

Aletheies
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  • What's the cost time / currency of the options mentioned? – Nathaniel Bubis Jun 17 '13 at 03:03
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    Can you get an option more than once? If so, getting A1 eight times seems a better deal than getting A2. – vadim123 Jun 17 '13 at 03:04
  • "Game theory" is an easy answer to what type of problem this is; but that doesn't say much about how to solve it. If I were to approach this problem, my first attempt wouldn't involve reaching for any specialized mathematical tool. I would start by just trying to apply logic and cleverness. Speculating a bit further, I think a solution might involve a combination of computer search of strategies, and the use of logic to narrow the search space. – Nick Thomas Jun 17 '13 at 03:14
  • @vadim123 yes, you can. Although some things that I didn't mention in order to keep the problem simple is that each option A takes up an amount of space on the map and there's a finite amount of space, so it reaches a point where it's better to get 1 A7 rather than 8192 A1's. Although since the ratio between all of the A1-7 options is roughly the same I don't think it would have much impact on the final answer. – Aletheies Jun 17 '13 at 03:28
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    This sounds like a typical dynamic programming problem as well. – gt6989b Jun 17 '13 at 03:30
  • @NickThomas Thanks, I added the game theory tag. And I was hoping a sort of trial and error wasn't the only method. Any tips you might have on narrowing the search space? – Aletheies Jun 17 '13 at 03:31
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    One thing that usually holds about this type of problem is that whatever buildings you build can only help you more if you build them earlier, so there is never a reason to delay building something that you've decided you will need. The question is, which combination of final buildings will be achieved most quickly? I would recommend first solving the simpler problem, where $C=A=B$ is the only resource, and using just one of the two charts. – vadim123 Jun 17 '13 at 03:33
  • This isn't game theory - there is only one player - game theory is a stategic game between 2 or more players. – Dale M Jul 02 '13 at 06:24

2 Answers2

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One possible approach is as follows. Write a computer program which evaluates all the possible strategies on the first turn (buy nothing, buy A1, buy A2, etc.). Order the strategies according to the following relation: say $S_1 \leq S_2$ if the amount of $A$ you have after $S_1$ is no more than the amount of $A$ you have after $S_2$; and the amount of $B$ you have after $S_1$ is no more than the amount you have after $S_2$; and your income of $A$ after $S_1$ is no more than your income of $B$ after $S_2$; and your income of $B$ after $S_1$ is no more than your income of $B$ after $S_2$. Intuitively, if $S_1 \leq S_2$, then $S_1$ can't possibly be a better strategy than $S_2$.

Now you have a preordered set of strategies. Essentially you want to take all of the "maximal elements" of this set. Actually the issue is subtler, because you can have a set of strategies which are all just as good each other in every dimension; if you regard "being just as good as each other in every dimension" as an equivalence relation on strategies, then the set of equivalence classes has a naturally induced partial order. Take one representative from each equivalence class which is maximal in this order; these are your "maximal elements."

Those maximal elements are the strategies which move on to the next round. In the next round, evaluate all of the possible ways of making the second move from the given strategies. Order these two-turn strategies as before, and again take a set of "maximal elements" of the resulting set. Move on to the third turn with those strategies. Continue like this. When you first find a strategy which reaches the goal, you have your answer.

My hope is that this process takes a reasonable amount of computation time, because taking only the best strategies from each round keeps the search space from blowing up exponentially. That is, my hope is that the set of maximal elements is always reasonably small, and in particular doesn't increase exponentially in size. However, I am not certain this would happen. Does anybody have any idea?

Nick Thomas
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    You are right. One wants to search a tree and the preorder you cite will prune it, probably very effectively. If it takes too long (unlikely) one could notice that B's are harder to come by than A's,so one could create a heuristic total order by making a quantity $A+nB$ with $n$ pretty large. This will prune the tree much more rapidly, but with some risk, so try various $n$'s Depending on the industriousness and nefariousness of the puzzle setter, I would expect the coefficients to be chosen so this will fail. – Ross Millikan Jun 17 '13 at 03:53
  • I really have no experience with this sort of programming, do you know of any resources that can help me with this? – Aletheies Jun 17 '13 at 04:32
  • What kind of programming experience do you have at the moment? – Nick Thomas Jun 17 '13 at 04:33
  • @NickThomas I took AP Computer Science two years ago, so I know a minimal amount of java – Aletheies Jun 17 '13 at 04:36
  • Unfortunately I'm not the person to ask about programming learning resources. :-/ I could write the program, but I wouldn't know where to learn how. Maybe try googling around, or cs.stackexchange.com? – Nick Thomas Jun 17 '13 at 05:03
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Let $S$ be your salary, $C$ be the cost of making investments, $R$ be the return, $V(t)$ be your cash (a 1x2 matrix), $P(t)$ be your portfolio (an 8x2 matrix) and $I(t)$ your investment strategy (an 8x2 matrix) all at time $t$. Then

$$S=\begin{bmatrix} 512&36\\ \end{bmatrix}$$

$$C=\begin{bmatrix} 4096&20480&81920&262144&10000000&0&0&0\\ 33&257&1025&4097&16385&65537&262145&1000001\\ \end{bmatrix}$$

$$R=\begin{bmatrix} 1&8&32&128&512&2048&8192&32768\\ 1&6&36&216&1296&0&0&0\\ \end{bmatrix}$$

$$V(0)=0_{1,2}$$

$$P(0)=0_{8,2}$$

And

$$P(t+1)=P(t)+I(t)$$

$$V(t+1)=V(t)+S-CI(t)+R(P(t)+I(t))$$

subject to the proviso that all elements of $V(t)\ge CI(t)$.

Now all you have to do is solve this partial differential equation and fine the minimum $T$ such that all the elements of $V(T)\ge1,000,000$ while at least one of the elements of $V(T-1)\lt 1,000,000$.

Dale M
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