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Neopia's Fill in the Blank News Source | 2nd day of Relaxing, Yr 22
The Neopian Times Week 43 > Articles > Advanced Codebreakers Strategy

Advanced Codebreakers Strategy

by conundrumsolver

SPACE STATION - I read the article by sassysarah656 about Codebreakers with great interest. It is a good starting point for playing Codebreaker, but it might not the fastest strategy.

So, as an amateur mathematician, I have decided to add my tips and hints to sassysarah's for the enjoyment and benefit of everyone. With my strategy, you are guaranteed to be able to continue to play Codebreakers as long as you want, getting as high a score as you wish (of course, keep in mind that the maximum amount of Neopoints you can earn is 500).

To pique your interest, using my sequence of initial moves, getting the code in five moves is a very common occurrence. Getting the code in four moves also happens with some frequency. There is no guess work involved, only logic.

I will use the following notation for Codebreaker moves:

1) rwgy = 1:2

The move number is followed by the sequence of stones. In this sequence, r = red, w = white, y = yellow, o = orange, p = purple, g = green.

After the sequence, I write down the result as a pair of numbers after the equal sign:

1:2 means 1 red and 2 white slashes (1 stone correct and in the right position and 2 stones correct but *not* in the right position).

Now, sassysarah's strategy is to basically do the following:

1) rrrr
2) wwww
3) yyyy
4) oooo

and so on. While this gives us valuable information, the downside of this strategy is that it can take as many as six moves to simply discover the set of stones *without* gaining any information about their possible locations.

Instead, my initial set of moves are the following:

1) wrrr
2) oyyy
3) gppp

In three moves, we can discover a lot of information. For example, suppose after the first move, we get:

1) wrrr = 1:2

If we get this result, it can only mean that the final code includes two red stones and a white stone, and that the one of the red stones is in the first position!

Writing this in the notation:

1) wrrr = 1:2 implies rXXX where the XXX includes another r and w.

Why is that? Well, since we know that the code contains three of the stones from wrrr, it can only be one of two possibilities: "rrr" or "wrr". But if it is "rrr" then we must have either: 2:1 or 3:0 as the result, and thus it *must* be "wrr". Continuing on with our logic, since we have *two* white slashes, then one of the red stones *must* be in the first position. Otherwise, we would not have 2 white slashes, but at most 1.

So, not only do we glean information about the stones in the set, we also find out their only logical locations.

Let's play through a typical game to illustrate how to apply this system. Suppose the code that we have to break is "gygw".

1) wrrr = 0:1
2) oyyy = 1:0
3) gppp = 1:0

Now, at this point, we can start constructing reasonable codes that match all three results:

(a) Since the first result was 0:1, we surmise that we have a "w" instead of an "r" (in fact, the chances are three times as likely for the "w" to be the one producing the 0:1 instead of the "r").

(b) Since we only got three hits, we also know that we must have two "o" or two "g" stones (the reason for that is simply this: if there were two "y" or "p" stones, we would have four hits in the three moves, not three).

So, we guess "oowp", which matches all the results so far, but gives us:

4) oowp = 0:1

Do not be discouraged by these misses! This is *great* information. Let us reason together:

(c) We now know that only 1 of "o", "w" and "p" are correct.

(d) If we assume that "o" is the one, then we must assume that "g" is not in the set (by the results of move 2 and 3). But this means, by (b) above, that we have "oo" in the set, which would produce at least 2 hits! Therefore, we must have "gg" in the set.

So, at this point, we *know* that the code is: gXXX where XXX contains g.

(e) Since we know that "g" is in the set, then "p" is NOT (by the result of move 3), so the only remaining possibility is "w". So now we know, the code must be of the form gXYZ, where XYZ contains "g" and "w", but Y != "w".

(f) Now, since we know the first place is "g", we also know (from the result of move 2) that our code *must* contain "y".

We now know that the code must be of the form gXYZ, where XYZ contains "gyw", but Y != "w". This means there are only 4 possible codes:

gwgy, gwyg, ggyw and gygw

So, we guess the first:

5) gwgy = 2:2

This result now knocks out "ggyw" and leaves us with "gwyg" and "gygw" as the only possibilities.

6) gwyg = 1:3

Now, we know that the only possibility is:

7) gygw = 4:0

Success!!! The complete game was:

1) wrrr = 0:1
2) oyyy = 1:0
3) gppp = 1:0
4) oowp = 0:1
5) gwgy = 2:2
6) gwyg = 1:3
7) gygw = 4:0

Compare this to the seemingly simpler initial moves we talked about earlier:

1) rrrr = 0:0
2) wwww = 1:0
3) yyyy = 1:0
4) oooo = 0:0
5) gggg = 2:0

At this point we know the code is a combination of "wygg", but we have 12 equally possible combinations:

wygg, wgyg, wggy, ywgg, ygwg, yggw, gwyg, gwgy, ggwy, ggyw, gygw, gywg

The game can easily continue as follows:

6) wygg = 2:2
7) wggy = 1:3
8) ywgg = 1:3
9) gygw = 4:0

As can be seen, the simpler method leads us to a maximum of 9 tries, while the advanced strategy gives a maximum of 7 tries (but many times you can deduce the code in five or six tries).

So, to summarise the strategy for Codebreakers:

* Use the following set of initial moves:

1) wrrr
2) oyyy
3) gppp

* If at any point (even at move 2), the total number of results is 4, then you can start guessing in earnest. There is no need at that point to continue with the normal moves.

* Use logic to eliminate stone colour combinations and stone-placement options (see the example above).

Applying these steps, you can be sure to win at Codebreakers for as long as you want!

Have fun and I wish you good skill in your application of logic.

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