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@@ -4,7 +4,7 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"<div align=\"right\" style=\"text-align: right\"><i>Peter Norvig<br>April 2015<br>Python 3: Feb 2019<br>Steve's bus: Apr 2020</i></div>\n",
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"<div align=\"right\" style=\"text-align: right\"><i>Peter Norvig<br>April 2015<br>Python 3: Feb 2019<br>Steve's bus: Apr 2020<br>Mad Cheryl: May 2020</i></div>\n",
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"\n",
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"# When is Cheryl's Birthday?\n",
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"\n",
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@@ -482,7 +482,7 @@
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],
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"source": [
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"times = '06:32 06:43 06:50 07:17 07:46 08:19 08:32 09:17 09:19 09:50'.split()\n",
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"dates = [t.replace(':', ' ') for t in times]\n",
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"dates = [time.replace(':', ' ') for time in times]\n",
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"\n",
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"cheryls_birthday()"
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]
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@@ -493,6 +493,47 @@
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"source": [
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"Steve took the 8:32 bus."
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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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"# Another New Puzzle: Evil Mad Scientist Cheryl\n",
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"\n",
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"\n",
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"\n",
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"Again, we can solve this problem just by changing the format of `dates`:"
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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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"{'C 3'}"
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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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"pads = 'A2 A3 A6 B4 B5 C1 C3 D1 D2 D4'.split()\n",
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"dates = [L+' '+N for L,N in pads]\n",
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"\n",
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"cheryls_birthday()"
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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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"Mad scientist Cheryl was refering to pad C3. (But may I point out that this Cheryl is not actually a mad scientist, just a [mad engineer](https://www.evilmadscientist.com/2015/evil-mad-engineers/). A true mad scientist would kill 25 people and use the other 25 as a control group.)"
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]
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}
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],
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"metadata": {
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