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Peter Norvig authored and 
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@@ -25,17 +25,17 @@
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" \n",
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"Now the first person chooses one of the four houses (which are all acceptable). We'll indicate an occupied house by a `1`, so the four equiprobable choices are:\n",
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"\n",
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" 1...\n",
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" .1..\n",
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" ..1.\n",
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" ...1\n",
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" Choice 0: 1...\n",
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" Choice 1: .1..\n",
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" Choice 2: ..1.\n",
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" Choice 3: ...1\n",
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" \n",
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"When a house is occupied, any adjacent houses become unacceptable. We'll indicate that with a `0`:\n",
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"\n",
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" 10..\n",
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" 010.\n",
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" .010\n",
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" ..01\n",
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" Choice 0: 10..\n",
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" Choice 1: 010.\n",
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" Choice 2: .010\n",
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" Choice 3: ..01\n",
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" \n",
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"In all four cases, a second occupant has a place to move in, but then there is no place for a third occupant. \n",
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"The occupancy is 2 in all cases, and thus the *expected occpancy* is 2. The occupancy fraction, or *density*, is 2/*N* = 2/4 = 1/2.\n",
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@@ -45,35 +45,35 @@
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"\n",
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"With *N*=7 houses, there are 7 equiprobable choices for the first occupant:\n",
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"\n",
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" 10.....\n",
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" 010....\n",
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" .010...\n",
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" ..010..\n",
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" ...010.\n",
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" ....010\n",
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" .....01\n",
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" Choice 0: 10.....\n",
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" Choice 1: 010....\n",
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" Choice 2: .010...\n",
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" Choice 3: ..010..\n",
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" Choice 4: ...010.\n",
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" Choice 5: ....010\n",
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" Choice 6: .....01\n",
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" \n",
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"Now we'll add something new: the lengths of the **runs** of consecutive acceptable houses to the left and right of the chosen house:\n",
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"\n",
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" 10..... runs = (0, 5)\n",
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" 010.... runs = (0, 4)\n",
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" .010... runs = (1, 3)\n",
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" ..010.. runs = (2, 2)\n",
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" ...010. runs = (3, 1)\n",
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" ....010 runs = (4, 0)\n",
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" .....01 runs = (5, 0)\n",
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" Choice 0: 10..... runs = (0, 5)\n",
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" Choice 1: 010.... runs = (0, 4)\n",
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" Choice 2: .010... runs = (1, 3) \n",
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" Choice 3: ..010.. runs = (2, 2) \n",
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" Choice 4: ...010. runs = (3, 1) \n",
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" Choice 5: ....010 runs = (4, 0) \n",
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" Choice 6: .....01 runs = (5, 0) \n",
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" \n",
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"# Defining `occ(n)`\n",
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"\n",
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"This gives me a key hint as to how to analyze the problem. I'll define `occ(n)` to be the expected number of occupied houses in a row of `n` houses. So:\n",
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"\n",
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" 10..... runs = (0, 5); occupancy = occ(0) + 1 + occ(5)\n",
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" 010.... runs = (0, 4); occupancy = occ(0) + 1 + occ(4)\n",
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" .010... runs = (1, 3); occupancy = occ(1) + 1 + occ(3)\n",
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" ..010.. runs = (2, 2); occupancy = occ(2) + 1 + occ(2)\n",
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" ...010. runs = (3, 1); occupancy = occ(3) + 1 + occ(1)\n",
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" ....010 runs = (4, 0); occupancy = occ(4) + 1 + occ(0)\n",
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" .....01 runs = (5, 0); occupancy = occ(5) + 1 + occ(0) \n",
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" Choice 0: 10..... runs = (0, 5); occupancy = occ(0) + 1 + occ(5)\n",
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" Choice 1: 010.... runs = (0, 4); occupancy = occ(0) + 1 + occ(4)\n",
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" Choice 2: .010... runs = (1, 3); occupancy = occ(1) + 1 + occ(3)\n",
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" Choice 3: ..010.. runs = (2, 2); occupancy = occ(2) + 1 + occ(2)\n",
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" Choice 4: ...010. runs = (3, 1); occupancy = occ(3) + 1 + occ(1)\n",
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" Choice 5: ....010 runs = (4, 0); occupancy = occ(4) + 1 + occ(0)\n",
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" Choice 6: .....01 runs = (5, 0); occupancy = occ(5) + 1 + occ(0) \n",
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" \n",
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"So we can say that `occ(n)` is:\n",
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"\n",
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@@ -88,9 +88,7 @@
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {
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"collapsed": false
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},
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"metadata": {},
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"outputs": [],
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"source": [
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"from statistics import mean\n",
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@@ -115,44 +113,47 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Let's check that `occ(4)` is 2, as we computed it should be:"
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"Let's check that `occ(4)` is 2, as we computed it should be, and for other small values, up to 7:"
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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": 30,
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"metadata": {
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"collapsed": false
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},
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"execution_count": 2,
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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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"2.0"
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"{0: 0,\n",
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" 1: 1,\n",
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" 2: 1,\n",
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" 3: 1.6666666666666667,\n",
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" 4: 2,\n",
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" 5: 2.466666666666667,\n",
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" 6: 2.888888888888889,\n",
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" 7: 3.323809523809524}"
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]
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},
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"execution_count": 30,
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"execution_count": 2,
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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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"occ(4)"
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"{n: occ(n) for n in range(8)}"
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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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"And that `runs(7)` is what we described above:"
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"That looks good, although I can't prove the value above 4 are ccorrect. Now check that `runs(7)` is what we described above:"
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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": 3,
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"metadata": {
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"collapsed": false
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},
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"metadata": {},
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"outputs": [
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{
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"data": {
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@@ -173,75 +174,15 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Let's check on `n = 7`:"
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"# Dynamic Programming Version of `occ`\n",
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"\n",
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"The computation of `occ(n)` makes multiple calls to `occ(n-1), occ(n-2),` etc. To avoid re-computing the same calls over and over, we will modify `occ` to save previous results using [dynamic programming](https://en.wikipedia.org/wiki/Dynamic_programming). I could implement that in one line with the decorator [`@functools.lru_cache`](https://docs.python.org/3/library/functools.html#functools.lru_cache), but then I would have to worry about the recursion limit. Instead I will explicitly manage a list, `cache`, such that `cache[n]` holds `occ(n)`:"
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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": 4,
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"metadata": {
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"collapsed": false
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"3.3238095238095235"
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]
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},
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"execution_count": 4,
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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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"occ(7)"
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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": 5,
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"metadata": {
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"collapsed": false
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},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"0.47482993197278905"
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]
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},
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"execution_count": 5,
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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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"density(7)"
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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 seems reasonable, but I don't know for sure that it is correct. "
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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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"# Dynamic Programming Version of `occ`\n",
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"\n",
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"The computation of `occ(n)` makes multiple calls to `occ(n-1), occ(n-2),` etc. To avoid re-computing the same calls over and over, we will modify `occ` to save previous results using [dynamic programming](https://en.wikipedia.org/wiki/Dynamic_programming). I could implement that in one line with the decorator [`@functools.lru_cache`](https://docs.python.org/3/library/functools.html#functools.lru_cache), but instead I will explicitly manage a list, `cache`, such that `cache[n]` holds `occ(n)`:"
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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": 6,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"def occ(n, cache=[0, 1]):\n",
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@@ -262,10 +203,8 @@
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},
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{
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"cell_type": "code",
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"execution_count": 32,
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"metadata": {
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"collapsed": false
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},
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"execution_count": 5,
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"metadata": {},
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"outputs": [
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{
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"data": {
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@@ -273,7 +212,7 @@
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"True"
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]
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},
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"execution_count": 32,
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"execution_count": 5,
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"metadata": {},
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"output_type": "execute_result"
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}
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@@ -284,46 +223,42 @@
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},
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{
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"cell_type": "code",
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"execution_count": 33,
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"metadata": {
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"collapsed": false
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},
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"execution_count": 6,
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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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"0.47482993197278905"
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"3.323809523809524"
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]
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},
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"execution_count": 33,
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"execution_count": 6,
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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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"density(7)"
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"occ(7)"
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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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"Let's make sure the caching makes computation faster the second time:"
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"Let's make sure the caching makes computation pretty fast the first time, and *very* fast the second time:"
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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": 9,
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"metadata": {
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"collapsed": false
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},
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"execution_count": 7,
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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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"CPU times: user 6.23 s, sys: 115 ms, total: 6.34 s\n",
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"Wall time: 6.97 s\n"
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"CPU times: user 2.68 s, sys: 13.5 ms, total: 2.7 s\n",
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"Wall time: 2.71 s\n"
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]
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},
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{
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@@ -332,7 +267,7 @@
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"864.9617138385324"
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]
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},
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"execution_count": 9,
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"execution_count": 7,
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"metadata": {},
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"output_type": "execute_result"
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}
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@@ -343,17 +278,15 @@
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},
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{
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"cell_type": "code",
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"execution_count": 10,
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"metadata": {
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"collapsed": false
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},
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"execution_count": 8,
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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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"CPU times: user 5 µs, sys: 0 ns, total: 5 µs\n",
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"Wall time: 9.06 µs\n"
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"CPU times: user 4 µs, sys: 0 ns, total: 4 µs\n",
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"Wall time: 5.01 µs\n"
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]
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},
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{
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@@ -362,7 +295,7 @@
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"864.9617138385324"
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]
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},
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"execution_count": 10,
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"execution_count": 8,
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"metadata": {},
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"output_type": "execute_result"
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}
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@@ -382,14 +315,13 @@
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},
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{
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"cell_type": "code",
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"execution_count": 11,
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"metadata": {
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"collapsed": false
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},
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"execution_count": 9,
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"metadata": {},
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"outputs": [],
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"source": [
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"%matplotlib inline\n",
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"import matplotlib.pyplot as plt\n",
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"plt.style.use('fivethirtyeight')\n",
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"\n",
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"def plot_density(ns):\n",
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" \"Plot density(n) for each n in the list of numbers ns.\"\n",
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@@ -400,10 +332,8 @@
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},
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{
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"cell_type": "code",
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"execution_count": 12,
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"metadata": {
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"collapsed": false
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},
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"execution_count": 10,
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"metadata": {},
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"outputs": [
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{
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"data": {
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@@ -411,15 +341,15 @@
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"0.4353323288377046"
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]
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},
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"execution_count": 12,
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"execution_count": 10,
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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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"data": {
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||||
"image/png": 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truncated
|
||||
"image/png": 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truncated
|
||||
"text/plain": [
|
||||
"<matplotlib.figure.Figure at 0x1043b8eb8>"
|
||||
"<matplotlib.figure.Figure at 0x11202c0b8>"
|
||||
]
|
||||
},
|
||||
"metadata": {},
|
||||
@@ -439,10 +369,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 13,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 11,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -450,15 +378,15 @@
|
||||
"0.46203174603174607"
|
||||
]
|
||||
},
|
||||
"execution_count": 13,
|
||||
"execution_count": 11,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
},
|
||||
{
|
||||
"data": {
|
||||
"image/png": 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truncated
|
||||
"image/png": 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truncated
|
||||
"text/plain": [
|
||||
"<matplotlib.figure.Figure at 0x10c5ac940>"
|
||||
"<matplotlib.figure.Figure at 0x1120519e8>"
|
||||
]
|
||||
},
|
||||
"metadata": {},
|
||||
@@ -478,10 +406,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 14,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 12,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -489,15 +415,15 @@
|
||||
"0.43240754751464183"
|
||||
]
|
||||
},
|
||||
"execution_count": 14,
|
||||
"execution_count": 12,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
},
|
||||
{
|
||||
"data": {
|
||||
"image/png": 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truncated
|
||||
"image/png": "iVBORw0KGgoAAAANSUhEUgAAAeMAAAEtCAYAAAA7lQj8AAAABHNCSVQICAgIfAhkiAAAAAlwSFlz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 truncated
|
||||
"text/plain": [
|
||||
"<matplotlib.figure.Figure at 0x10ca875c0>"
|
||||
"<matplotlib.figure.Figure at 0x1149702e8>"
|
||||
]
|
||||
},
|
||||
"metadata": {},
|
||||
@@ -523,7 +449,7 @@
|
||||
"\n",
|
||||
"# lim <sub>n → ∞</sub> density(n)\n",
|
||||
"\n",
|
||||
"[Thes puzzle](http://fivethirtyeight.com/features/can-you-solve-the-puzzle-of-your-misanthropic-neighbors/)\n",
|
||||
"[The puzzle](http://fivethirtyeight.com/features/can-you-solve-the-puzzle-of-your-misanthropic-neighbors/)\n",
|
||||
"is to figure out the limit of `density(n)` as `n` goes to infinity. The plot above makes it look like 0.432+something, but we can't answer the question just by plotting; we'll need to switch modes from *computational* thinking to *mathematical* thinking.\n",
|
||||
"\n",
|
||||
"At this point I started playing around with `density(n)`, looking at various values, differences of values, ratios of values, and ratios of differences of values, hoping to achieve some mathematical insight. Mostly I got dead ends.\n",
|
||||
@@ -532,10 +458,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 15,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 13,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -543,7 +467,7 @@
|
||||
"0.0014849853757253895"
|
||||
]
|
||||
},
|
||||
"execution_count": 15,
|
||||
"execution_count": 13,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -563,10 +487,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 16,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 14,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -574,7 +496,7 @@
|
||||
"0.0007424926878626947"
|
||||
]
|
||||
},
|
||||
"execution_count": 16,
|
||||
"execution_count": 14,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -592,10 +514,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 17,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 15,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -603,7 +523,7 @@
|
||||
"2.0"
|
||||
]
|
||||
},
|
||||
"execution_count": 17,
|
||||
"execution_count": 15,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -621,10 +541,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 18,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 16,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -632,7 +550,27 @@
|
||||
"2.0"
|
||||
]
|
||||
},
|
||||
"execution_count": 18,
|
||||
"execution_count": 16,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"n = 250; diff(n, 2*n) / diff(2*n, 4*n)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 17,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"2.0"
|
||||
]
|
||||
},
|
||||
"execution_count": 17,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -641,28 +579,6 @@
|
||||
"n = 500; diff(n, 2*n) / diff(2*n, 4*n)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 19,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"1.9999999999992524"
|
||||
]
|
||||
},
|
||||
"execution_count": 19,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"n = 1000; diff(n, 2*n) / diff(2*n, 4*n)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
@@ -682,24 +598,22 @@
|
||||
"\n",
|
||||
" density(n) = A + B / n \n",
|
||||
" \n",
|
||||
"would fit the patterm. I can try a `curve_fit` to estimate the parameters *A* and *B*:"
|
||||
"would fit the patterm. I can try a `scipy.optimize.curve_fit` to estimate the parameters *A* and *B*:"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 20,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 18,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"array([[ 7.64810902e-34, -2.42654280e-31],\n",
|
||||
" [ -2.42654280e-31, 4.67778766e-28]])"
|
||||
"array([[ 4.70916321e-33, -1.30373854e-30],\n",
|
||||
" [ -1.30373854e-30, 1.44885627e-27]])"
|
||||
]
|
||||
},
|
||||
"execution_count": 20,
|
||||
"execution_count": 18,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -707,7 +621,7 @@
|
||||
"source": [
|
||||
"from scipy.optimize import curve_fit\n",
|
||||
"\n",
|
||||
"Ns = list(range(100, 10001, 100))\n",
|
||||
"Ns = list(range(100, 5001, 100))\n",
|
||||
"\n",
|
||||
"def f(n, A, B): return A + B / n\n",
|
||||
"\n",
|
||||
@@ -725,18 +639,16 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 21,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 19,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"(0.43233235838169343, 0.29699707514520357)"
|
||||
"(0.43233235838169332, 0.29699707514503415)"
|
||||
]
|
||||
},
|
||||
"execution_count": 21,
|
||||
"execution_count": 19,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -754,10 +666,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 22,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 20,
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": [
|
||||
"def estimated_density(n): return A + B / n"
|
||||
@@ -772,18 +682,16 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 23,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 21,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"3.8857805861880479e-16"
|
||||
"6.106226635438361e-16"
|
||||
]
|
||||
},
|
||||
"execution_count": 23,
|
||||
"execution_count": 21,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -801,7 +709,7 @@
|
||||
"\n",
|
||||
"We now have a plausible answer to the puzzle:\n",
|
||||
"\n",
|
||||
"> lim<sub style=\"font-size:large\"> <tt>n</tt>→∞</sub> <tt>density(n) ≅ A = 0.43233235838169343</tt>"
|
||||
"> lim<sub style=\"font-size:large\"> <tt>n</tt>→∞</sub> <tt>density(n) ≅ A = 0.43233235838169</tt>"
|
||||
]
|
||||
},
|
||||
{
|
||||
@@ -812,7 +720,7 @@
|
||||
"\n",
|
||||
"Theis answer is empirically strong (15 decimal places of accuracy) but theoretically weak: we don't have a **proof**, and we don't have an explanation for **why** the density function has this form. We need some more mathematical thinking. \n",
|
||||
"\n",
|
||||
"I didn't have any ideas, so I looked to see if anyone else had written something about the number 0.43233235838169343. I tried several searches and found two interesting formulas:\n",
|
||||
"I didn't have any ideas, so I looked to see if anyone else had written something about the number 0.432332358169 I tried several searches and found two interesting formulas:\n",
|
||||
"\n",
|
||||
"- Search: [`[0.4323323]`](https://www.google.com/search?q=0.4323323) Formula: `sinh(1) / exp(1)` [(Page)](http://arxiv.org/pdf/1310.4360.pdf)\n",
|
||||
"- Search: [`[0.432332358]`](https://www.google.com/search?q=0.432332358) Formula: `0.5(1-e^(-2))` [(Page)](http://www.actuarialoutpost.com/actuarial_discussion_forum/archive/index.php/t-52095.html)\n",
|
||||
@@ -822,18 +730,16 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 24,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 22,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"(0.43233235838169365, 0.43233235838169365, 0.43233235838169343)"
|
||||
"(0.43233235838169365, 0.43233235838169365, 0.43233235838169332)"
|
||||
]
|
||||
},
|
||||
"execution_count": 24,
|
||||
"execution_count": 22,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -844,8 +750,6 @@
|
||||
"S = sinh(1) / exp(1)\n",
|
||||
"E = 0.5 * (1 - e ** (-2))\n",
|
||||
"\n",
|
||||
"assert S == E\n",
|
||||
"\n",
|
||||
"S, E, A"
|
||||
]
|
||||
},
|
||||
@@ -933,9 +837,9 @@
|
||||
"\n",
|
||||
"Is our implementation of `occ(n)` correct? I think it is, but I can anticipate some objections and answer them:\n",
|
||||
"\n",
|
||||
"- In `occ(n)`, is it ok to start from all empty houses, rather than considering layouts of partially-occupied houses? Yes, I think it is ok, because the problem states that initially all houses are empty, and each choice of a house breaks the street up into runs of acceptable houses, flanked by unacceptable houses. If we get the computation right for a run of `n` acceptable houses, then we can get the whole answer right. A key point is that the chosen first house breaks the row of houses into 2 runs of *acceptable* houses, not 2 runs of *unoccupied* houses. If it were unoccupied houses, then we would have to also keep track of whether there were occupied houses to the right and/or left of the runs. By considering runs of acceptable houses, eveything is clean and simple.\n",
|
||||
"- *In `occ(n)`, is it ok to start from all empty houses, rather than considering layouts of partially-occupied houses?* Yes, because the problem states that initially all houses are empty, and each choice of a house breaks the street up into runs of acceptable houses, flanked by unacceptable houses. If we get the computation right for a run of `n` acceptable houses, then we can get the whole answer right. A key point is that the chosen first house breaks the row of houses into 2 runs of *acceptable* houses, not 2 runs of *unoccupied* houses. If it were unoccupied houses, then we would have to also keep track of whether there were occupied houses to the right and/or left of the runs. By considering runs of acceptable houses, eveything is clean and simple.\n",
|
||||
"\n",
|
||||
"- In `occ(7)`, if the first house chosen is 2, that breaks the street up into runs of 1 and 3 acceptable houses. There is only one way to occupy the 1 house, but there are several ways to occupy the 3 houses. Shouldn't the average give more weight to the 3 houses, since there are more possibilities there? No, I don't think so. We are caclulating occupancy, and there is a specific number (5/3) which is the expected occupancy of 3 houses; it doesn't matter if there is one combination or a million combinations that contribute to that expected value, all that matters is what the expected value is.\n",
|
||||
"- *In `occ(7)`, if the first house chosen is 2, that breaks the street up into runs of 1 and 3 acceptable houses. There is only one way to occupy the 1 house, but there are several ways to occupy the 3 houses. Shouldn't the average give more weight to the 3 houses, since there are more possibilities there?* No. We are caclulating occupancy, and there is a specific number (5/3) which is the expected occupancy of 3 houses; it doesn't matter if there is one combination or a million combinations that contribute to that expected value, all that matters is what the expected value is.\n",
|
||||
"\n",
|
||||
"\n"
|
||||
]
|
||||
@@ -948,19 +852,17 @@
|
||||
"\n",
|
||||
"\n",
|
||||
"A simulation can add credence to our implementation of `density(n)`, for two reasons:\n",
|
||||
"- The code for the simulation can have a more direct corresponance to the problem statement.\n",
|
||||
"- The code for the simulation can have a more direct correspondance to the problem statement.\n",
|
||||
"- When two very different implementations get the same result, that is evidence supporting both of them.\n",
|
||||
"\n",
|
||||
"\n",
|
||||
"The simulation will start with an empty set of occupied houses. Following Lamping's suggestion, we sample <i>n</i> houses (to get a random ordering) and go through them, occupying just the ones that have no neighbor\" "
|
||||
"The simulation will start with an empty set of occupied houses. We then go through the houses in random order, occupying just the ones that have no neighbor. (Note that `random.sample(range(n), n)` gives a random ordering of `range(n)`.)"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 25,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 23,
|
||||
"metadata": {},
|
||||
"outputs": [],
|
||||
"source": [
|
||||
"import random\n",
|
||||
@@ -988,28 +890,27 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 26,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 24,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
" n simul density estimated\n",
|
||||
" 25 0.444 0.444 0.444\n",
|
||||
" 50 0.438 0.438 0.438\n",
|
||||
"100 0.435 0.435 0.435\n",
|
||||
"200 0.434 0.434 0.434\n",
|
||||
"400 0.433 0.433 0.433\n"
|
||||
" n simulate density estimated\n",
|
||||
" 10 0.4621 0.4620 0.4620\n",
|
||||
" 25 0.4442 0.4442 0.4442\n",
|
||||
" 50 0.4383 0.4383 0.4383\n",
|
||||
" 100 0.4354 0.4353 0.4353\n",
|
||||
" 200 0.4339 0.4338 0.4338\n",
|
||||
" 400 0.4331 0.4331 0.4331\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
"source": [
|
||||
"print(' n simul density estimated')\n",
|
||||
"for n in (25, 50, 100, 200, 400):\n",
|
||||
" print('{:3} {:.3} {:.3} {:.3}'\n",
|
||||
"print(' n simulate density estimated')\n",
|
||||
"for n in (10, 25, 50, 100, 200, 400):\n",
|
||||
" print('{:4d} {:.4f} {:.4f} {:.4f}'\n",
|
||||
" .format(n, simulated_density(n), density(n), estimated_density(n)))"
|
||||
]
|
||||
},
|
||||
@@ -1017,17 +918,15 @@
|
||||
"cell_type": "markdown",
|
||||
"metadata": {},
|
||||
"source": [
|
||||
"We got perfect agreement (at least to 3 decimal places), suggesting that either our three implementations are all correct, or we've made mistakes in all three. \n",
|
||||
"We get good agreement (at least to 3 decimal places), suggesting that either our three implementations are all correct, or we've made mistakes in all three. \n",
|
||||
"\n",
|
||||
"The `simulate` function can also give us insights when we look at the results it produces:"
|
||||
]
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 27,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 25,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -1035,7 +934,7 @@
|
||||
"[0, 2, 4, 6]"
|
||||
]
|
||||
},
|
||||
"execution_count": 27,
|
||||
"execution_count": 25,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -1053,24 +952,22 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 28,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 26,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
"text/plain": [
|
||||
"Counter({(0, 2, 4, 6): 3303,\n",
|
||||
" (0, 2, 5): 1255,\n",
|
||||
" (0, 3, 5): 959,\n",
|
||||
" (0, 3, 6): 783,\n",
|
||||
" (1, 3, 5): 1426,\n",
|
||||
" (1, 3, 6): 974,\n",
|
||||
" (1, 4, 6): 1300})"
|
||||
"Counter({(0, 2, 4, 6): 3220,\n",
|
||||
" (0, 2, 5): 1214,\n",
|
||||
" (0, 3, 5): 1029,\n",
|
||||
" (0, 3, 6): 827,\n",
|
||||
" (1, 3, 5): 1417,\n",
|
||||
" (1, 3, 6): 976,\n",
|
||||
" (1, 4, 6): 1317})"
|
||||
]
|
||||
},
|
||||
"execution_count": 28,
|
||||
"execution_count": 26,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -1099,10 +996,8 @@
|
||||
},
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 29,
|
||||
"metadata": {
|
||||
"collapsed": false
|
||||
},
|
||||
"execution_count": 27,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"data": {
|
||||
@@ -1110,7 +1005,7 @@
|
||||
"'ok'"
|
||||
]
|
||||
},
|
||||
"execution_count": 29,
|
||||
"execution_count": 27,
|
||||
"metadata": {},
|
||||
"output_type": "execute_result"
|
||||
}
|
||||
@@ -1135,9 +1030,9 @@
|
||||
" occupied = set(occupied) # coerce to set\n",
|
||||
" for house in range(n):\n",
|
||||
" if house in occupied:\n",
|
||||
" assert (house - 1) not in occupied and (house + 1) not in occupied\n",
|
||||
" assert not ((house - 1) in occupied or (house + 1) in occupied)\n",
|
||||
" else:\n",
|
||||
" assert (house - 1) in occupied or (house + 1) in occupied\n",
|
||||
" assert (house - 1) in occupied or (house + 1) in occupied\n",
|
||||
"\n",
|
||||
"test()"
|
||||
]
|
||||
@@ -1172,9 +1067,9 @@
|
||||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.5.1"
|
||||
"version": "3.5.3"
|
||||
}
|
||||
},
|
||||
"nbformat": 4,
|
||||
"nbformat_minor": 0
|
||||
"nbformat_minor": 1
|
||||
}
|
||||
Reference in new issue
Block a user