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Peter Norvig
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@ -784,7 +784,7 @@
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"source": [
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"# [Day 5](https://adventofcode.com/2017/day/5): A Maze of Twisty Trampolines, All Alike\n",
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"\n",
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"Let's first make sure we can read the data okay:"
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"Let's first make sure we can read the data/program okay:"
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]
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},
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{
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@ -795,7 +795,7 @@
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{
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"data": {
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"text/plain": [
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"[0, 2, 0, 0, -2, -2, -1, -4, -5, -6]"
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"(0, 2, 0, 0, -2, -2, -1, -4, -5, -6)"
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]
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},
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"execution_count": 18,
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@ -804,17 +804,17 @@
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}
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],
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"source": [
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"def jumps(): return [int(x) for x in Input(5)]\n",
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"program = mapt(int, Input(5))\n",
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"\n",
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"jumps()[:10]"
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"program[:10]"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Now I'll make a little interpreter, `run`, which takes a list, `M` for memory, as input,\n",
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"and maintains a program counter `pc`, and does the incrementing as described in the puzzle,\n",
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"Now I'll make a little interpreter, `run`, which takes a program, loads it into memory,\n",
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" and executes the instruction, maintaining a program counter, `pc`, and doing the incrementing/branching as described in the puzzle,\n",
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"until the program counter is out of range:"
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]
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},
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@ -835,16 +835,17 @@
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}
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],
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"source": [
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"def run(M):\n",
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" pc = 0\n",
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" for steps in count_from(1):\n",
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"def run(program):\n",
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" memory = list(program)\n",
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" pc = steps = 0\n",
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" while pc in range(len(memory)):\n",
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" steps += 1\n",
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" oldpc = pc\n",
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" pc += M[pc]\n",
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" M[oldpc] += 1\n",
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" if pc not in range(len(M)):\n",
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" pc += memory[pc]\n",
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" memory[oldpc] += 1\n",
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" return steps\n",
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" \n",
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"run(jumps())"
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"run(program)"
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]
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},
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{
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@ -853,7 +854,7 @@
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"source": [
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"**Part Two:**\n",
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"\n",
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"Part Two seems tricky, so I'll include an optional argument, `verbose`, to print out info as we go, to make sure I've got it right on a small example:"
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"Part Two seems tricky, so I'll include an optional argument, `verbose`, and check if the printout it produces matches the example in the puzzle description:"
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]
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},
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{
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@ -889,14 +890,15 @@
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}
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],
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"source": [
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"def run2(M, verbose=False):\n",
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" pc = 0\n",
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" for steps in count_from(1):\n",
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"def run2(program, verbose=False):\n",
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" memory = list(program)\n",
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" pc = steps = 0\n",
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" while pc in range(len(memory)):\n",
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" steps += 1\n",
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" oldpc = pc\n",
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" pc += M[pc]\n",
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" M[oldpc] += (-1 if M[oldpc] >= 3 else 1)\n",
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" if verbose: print(steps, pc, M)\n",
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" if pc not in range(len(M)):\n",
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" pc += memory[pc]\n",
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" memory[oldpc] += (-1 if memory[oldpc] >= 3 else 1)\n",
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" if verbose: print(steps, pc, memory)\n",
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" return steps\n",
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" \n",
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"run2([0, 3, 0, 1, -3], True)"
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@ -926,7 +928,179 @@
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}
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],
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"source": [
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"run2(jumps())"
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"run2(program)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Thanks to [Clement Sreeves](https://github.com/ClementSreeves) for the suggestion of making a distinction between the `program` and the `memory`. In my first version, `run` would mutate the argument, which was OK for a short exercise, but not best practice for a reliable API."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# [Day 6](https://adventofcode.com/2017/day/6): Memory Reallocation "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"I had to read the puzzle description carefully, but then it is pretty clear what to do. I'll keep a set of previously seen configurations, which will all be tuples. But in the function `spread`, I want to mutate the configuration of banks, so I will convert to a list at the start, then convert back to a tuple at the end."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 22,
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"metadata": {},
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"outputs": [],
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"source": [
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"banks = vector('4\t10\t4\t1\t8\t4\t9\t14\t5\t1\t14\t15\t0\t15\t3\t5')\n",
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"\n",
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"def realloc(banks):\n",
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" \"How many cycles until we reach a configuration we've seen before?\"\n",
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" seen = {banks}\n",
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" for cycles in count_from(1):\n",
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" banks = spread(banks)\n",
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" if banks in seen:\n",
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" return cycles\n",
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" seen.add(banks)\n",
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" \n",
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"def spread(banks):\n",
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" \"Find the area with the most blocks, and spread them evenly to following areas.\"\n",
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" banks = list(banks)\n",
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" maxi = max(range(len(banks)), key=lambda i: banks[i])\n",
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" blocks = banks[maxi]\n",
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" banks[maxi] = 0\n",
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" for i in range(maxi + 1, maxi + 1 + blocks):\n",
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" banks[i % len(banks)] += 1\n",
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" return tuple(banks)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 23,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"(2, 4, 1, 2)"
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]
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},
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"execution_count": 23,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"spread((0, 2, 7, 0))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 24,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"5"
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]
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},
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"execution_count": 24,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"realloc((0, 2, 7, 0))"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"These tests look good; let's solve the problem:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 25,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"12841"
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]
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},
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"execution_count": 25,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"realloc(banks)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**Part Two:** Here I will just replace the `set` of `seen` banks with a `dict` of `{bank: cycle_number}`; everything else is the same, and the final result is the current cycle number minus the cycle number of the previously-seen tuple of banks."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 26,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"4"
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]
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},
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"execution_count": 26,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"def realloc2(banks):\n",
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" \"When we hit a cycle, what is the length of the cycle?\"\n",
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" seen = {banks: 0}\n",
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" for cycles in count_from(1):\n",
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" banks = spread(banks)\n",
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" if banks in seen:\n",
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" return cycles - seen[banks]\n",
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" seen[banks] = cycles\n",
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"\n",
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"realloc2((0, 2, 7, 0))"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 27,
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"8038"
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]
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},
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"execution_count": 27,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"realloc2(banks)"
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]
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}
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],
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