Added Solution for 2020 day 14

2020/day14_docking_data/challenge.txt

@@ -51,6 +51,81 @@ To initialize your ferry's docking program, you need the sum of all values left
 
 Execute the initialization program. *What is the sum of all values left in memory after it completes?* (Do not truncate the sum to 36 bits.)
 
-To begin, [get your puzzle input](14/input).
+Your puzzle answer was `6559449933360`.
 
-Answer:
+\--- Part Two ---
+----------
+
+For some reason, the sea port's computer system still can't communicate with your ferry's docking program. It must be using *version 2* of the decoder chip!
+
+A version 2 decoder chip doesn't modify the values being written at all. Instead, it acts as a [memory address decoder](https://www.youtube.com/watch?v=PvfhANgLrm4). Immediately before a value is written to memory, each bit in the bitmask modifies the corresponding bit of the destination *memory address* in the following way:
+
+* If the bitmask bit is `0`, the corresponding memory address bit is *unchanged*.
+* If the bitmask bit is `1`, the corresponding memory address bit is *overwritten with `1`*.
+* If the bitmask bit is `X`, the corresponding memory address bit is *floating*.
+
+A *floating* bit is not connected to anything and instead fluctuates unpredictably. In practice, this means the floating bits will take on *all possible values*, potentially causing many memory addresses to be written all at once!
+
+For example, consider the following program:
+
+```
+mask = 000000000000000000000000000000X1001X
+mem[42] = 100
+mask = 00000000000000000000000000000000X0XX
+mem[26] = 1
+
+```
+
+When this program goes to write to memory address `42`, it first applies the bitmask:
+
+```
+address: 000000000000000000000000000000101010  (decimal 42)
+mask:    000000000000000000000000000000X1001X
+result:  000000000000000000000000000000X1101X
+
+```
+
+After applying the mask, four bits are overwritten, three of which are different, and two of which are *floating*. Floating bits take on every possible combination of values; with two floating bits, four actual memory addresses are written:
+
+```
+000000000000000000000000000000011010  (decimal 26)
+000000000000000000000000000000011011  (decimal 27)
+000000000000000000000000000000111010  (decimal 58)
+000000000000000000000000000000111011  (decimal 59)
+
+```
+
+Next, the program is about to write to memory address `26` with a different bitmask:
+
+```
+address: 000000000000000000000000000000011010  (decimal 26)
+mask:    00000000000000000000000000000000X0XX
+result:  00000000000000000000000000000001X0XX
+
+```
+
+This results in an address with three floating bits, causing writes to *eight* memory addresses:
+
+```
+000000000000000000000000000000010000  (decimal 16)
+000000000000000000000000000000010001  (decimal 17)
+000000000000000000000000000000010010  (decimal 18)
+000000000000000000000000000000010011  (decimal 19)
+000000000000000000000000000000011000  (decimal 24)
+000000000000000000000000000000011001  (decimal 25)
+000000000000000000000000000000011010  (decimal 26)
+000000000000000000000000000000011011  (decimal 27)
+
+```
+
+The entire 36-bit address space still begins initialized to the value 0 at every address, and you still need the sum of all values left in memory at the end of the program. In this example, the sum is *`208`*.
+
+Execute the initialization program using an emulator for a version 2 decoder chip. *What is the sum of all values left in memory after it completes?*
+
+Your puzzle answer was `3369767240513`.
+
+Both parts of this puzzle are complete! They provide two gold stars: \*\*
+
+At this point, you should [return to your Advent calendar](/2020) and try another puzzle.
+
+If you still want to see it, you can [get your puzzle input](14/input).