Update notes to code 6

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RunasSudo 2016-05-29 21:32:32 +09:30
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@ -42,7 +42,7 @@ At the foothills, the first area of the RPG:
Required opcodes: All except `halt`
## Code 5 (Twisty passages)
After falling down the bridge, there is an empty lantern to the `east`. To the west, there is a passage, where one can take the `ladder` down or venture into the `darkness`. Attempting to venture into the `darkness` at this point will result in being eaten by Grues.
After falling down the bridge, there is an `empty lantern` to the `east`, which should be `take`n. To the `west`, there is a passage, where one can take the `ladder` down or venture into the `darkness`. Attempting to venture into the `darkness` at this point will result in being eaten by Grues.
Taking the `ladder` down, then traversing `west`, `south`, `north`, a code is obtained:
@ -53,3 +53,40 @@ Taking the `ladder` down, then traversing `west`, `south`, `north`, a code is ob
You take note of this and keep walking.
There is also a can, which can be `take`n and `use`d to fill the lantern, which can then be `use`d to become lit, and which will keep away Grues.
## Code 6 (Dark passage, ruins and teleporter)
Returning to the fork at the passage and venturing to the `darkness`, we now `continue` `west`ward to the ruins. The `north` door is locked, but dotted elsewhere around the ruins are a `red coin` (2 dots), `corroded coin` (triangle = 3 sides), `shiny coin` (pentagon = 5 sides), `concave coin` (7 dots) and `blue coin` (9 dots) which should be `take`n and `look`ed at, and the equation:
_ + _ * _^2 + _^3 - _ = 399
### Mathematical!
In other words, we seek a solution to the equation *a* + *bc*<sup>2</sup> + *d*<sup>3</sup> - e = 399, where {*a*, *b*, *c*, *d*, *e*} = {2, 3, 5, 7, 9}.
Synacor being, of course, a programming-orientated exercise, the usual response to this problem is to code up a quick program to loop through all 5! = 120 permutations of the coins to find which one satisfies the equation. Being a mathematician, however – could you tell from the italics? – I will solve this the [*proper* way](https://xkcd.com/435/): with a scientific calculator and some thinking (\*insert insufferably smug expression here\*).
Firstly, note that -7 ≤ *a**e* ≤ 7, so 392 ≤ *bc*<sup>2</sup> + *d*<sup>3</sup> ≤ 406. Furthermore, since *bc*<sup>2</sup> is always positive, *d*<sup>3</sup> ≤ 406. We can thus rule out *d* = 9, so *d* ≤ 7, *d*<sup>3</sup> ≤ 343 and *bc*<sup>2</sup> ≤ 63, also ruling out *c* = 7 and *c* = 9. Furthermore, since 392 ≤ *bc*<sup>2</sup> + *d*<sup>3</sup>, *d*<sup>3</sup> ≥ 329. Since we already knew *d* ≤ 7, *d* = 7.
Consequently, 49 ≤ *bc*<sup>2</sup> ≤ 63. It is trivial to check that no solutions satisfy this for *c* = 2 and *c* = 3, so *c* = 5 and *b* = 2. The values of *a* and *e* are now easily found.
The solution is:
9 + 2 * 5^2 + 7^3 - 3 = 399
**BOOM!**
(Well, actually all I did the first time was get up to *d* ≤ 7, then use guess and check.)
The coins should therefore be `use`d in the order: `blue`, `red`, `shiny`, `concave`, `corroded`.
Proceed to the `north` door and `use` the `teleporter` to obtain the code:
== Ruins ==
Things of interest here:
- teleporter
> use teleporter
You activate the teleporter! As you spiral through time and space, you think you see a pattern in the stars...
............
After a few moments, you find yourself back on solid ground and a little disoriented.