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Peter Norvig
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@ -411,7 +411,7 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"This was an easy warmup puzzle. I forgot that Advent of Code was starting at 9:00PM my time, so I started late, but even if I had started on time, I doubt I would have been fast enough to score points. (I created a recurring calendar reminder so I'll be more likely to start on time in the future.)"
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"This was an easy warmup puzzle. "
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]
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},
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{
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@ -501,9 +501,9 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"This day was also very easy. In Part One, I was slowed down by a typo: I had `\"=\"` instead of `\"-\"` in `\"max(row) - min(row)\"`. I was confused by Python's misleading error message, which said `\"SyntaxError: keyword can't be an expression\"`. Later on, Alex Martelli explained to me that the message meant that in `abs(max(row)=...)` it thought that `max(row)` was a keyword argument to `abs`. \n",
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"This day was also very easy. In Part One, I was slowed down by a typo: I had `\"=\"` instead of `\"-\"` in `\"max(row) - min(row)\"`. I was confused by Python's misleading error message, which said `\"SyntaxError: keyword can't be an expression\"`. Later on, Alex Martelli explained to me that the message meant that in `abs(max(row)=...)` it thought that `max(row)` was a keyword argument to `abs`, as in `abs(x=-1)`.\n",
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"\n",
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"In Part Two, note that to check that `a/b` is an exact integer, I used `a // b == a / b`, which I think is more clear and less error-prone than the expression one would typically use here, `a % b == 0`."
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"In Part Two, note that to check that `a/b` is an exact integer, I used `a // b == a / b`, which I think is more clear and less error-prone than the expression one would typically use here, `a % b == 0`, which requires you to think about two things: division and the modulus operator (is it `a % b` or `b % a`?)."
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]
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},
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{
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@ -653,7 +653,7 @@
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"outputs": [],
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"source": [
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"def spiralsums():\n",
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" \"Yield the values of a spiral where each point has the sum of the 8 neighbors.\"\n",
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" \"Yield the values of a spiral where each square has the sum of the 8 neighbors.\"\n",
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" value = defaultdict(int)\n",
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" for p in spiral():\n",
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" value[p] = sum(value[q] for q in neighbors8(p)) or 1\n",
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@ -775,7 +775,158 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"It took me less than five minutes, and my preparation with `Input` and `canon` helped, but I was still too slow to score any points."
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"It took me less than five minutes, and my preparation with `Input` and `canon` helped, but I was still too slow to score any points--the top 20 took less than two minutes each."
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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 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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]
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},
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{
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"cell_type": "code",
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"execution_count": 18,
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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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"[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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"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 jumps(): return [int(x) for x in Input(5)]\n",
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"\n",
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"jumps()[: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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"until the program counter is out of range:"
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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": 19,
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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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"364539"
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]
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},
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"execution_count": 19,
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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 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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" 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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" return steps\n",
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" \n",
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"run(jumps())"
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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:**\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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]
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},
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{
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"cell_type": "code",
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"execution_count": 20,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"1 0 [1, 3, 0, 1, -3]\n",
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"2 1 [2, 3, 0, 1, -3]\n",
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"3 4 [2, 2, 0, 1, -3]\n",
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"4 1 [2, 2, 0, 1, -2]\n",
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"5 3 [2, 3, 0, 1, -2]\n",
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"6 4 [2, 3, 0, 2, -2]\n",
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"7 2 [2, 3, 0, 2, -1]\n",
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"8 2 [2, 3, 1, 2, -1]\n",
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"9 3 [2, 3, 2, 2, -1]\n",
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"10 5 [2, 3, 2, 3, -1]\n"
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]
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},
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{
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"data": {
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"text/plain": [
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"10"
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]
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},
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"execution_count": 20,
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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 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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" 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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" return steps\n",
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" \n",
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"run2([0, 3, 0, 1, -3], True)"
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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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"That looks right, so I can solve the puzzle:"
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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": 21,
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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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"27477714"
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]
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},
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"execution_count": 21,
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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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"run2(jumps())"
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]
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}
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],
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