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+--- Part Two ---
+
+After some careful analysis, you believe that exactly one instruction is corrupted.
+
+Somewhere in the program, either a jmp is supposed to be a nop, or a nop
+is supposed to be a jmp. (No acc instructions were harmed in the corruption of this boot code.)
+
+The program is supposed to terminate by attempting to execute an instruction immediately
+after the last instruction in the file. By changing exactly one jmp or nop, you can repair the
+boot code and make it terminate correctly.
+
+For example, consider the same program from above:
+
+nop +0 acc +1 jmp +4 acc +3 jmp -3 acc -99 acc +1 jmp -4 acc +6
+
+If you change the first instruction from nop +0 to jmp +0, it would create a single-instruction
+infinite loop, never leaving that instruction. If you change almost any of the jmp instructions, the
+program will still eventually find another jmp instruction and loop forever.
+
+However, if you change the second-to-last instruction (from jmp -4 to nop -4), the program
+terminates! The instructions are visited in this order:
+
+nop +0 | 1 acc +1 | 2 jmp +4 | 3 acc +3 | jmp -3 | acc -99 | acc +1 | 4 nop -4 | 5 acc
++6 | 6
+
+After the last instruction (acc +6), the program terminates by attempting to run the instruction
+below the last instruction in the file. With this change, after the program terminates, the
+accumulator contains the value 8 (acc +1, acc +1, acc +6).
+
+Fix the program so that it terminates normally by changing exactly one jmp (to nop) or nop (to jmp).
+What is the value of the accumulator after the program terminates?
+
+