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+--- 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. 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?
+
+