Move FizzBuzz exercise from chapter 01 to 03
This commit is contained in:
parent
dda246284b
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2 changed files with 61 additions and 99 deletions
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@ -275,78 +275,6 @@
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"for number in [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]:\n",
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" print(...)"
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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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"### Fizz Buzz"
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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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"The kids game [Fizz Buzz](https://en.wikipedia.org/wiki/Fizz_buzz) is said to be often used in job interviews for entry level positions. However, opinions vary as to how good of a test it actually is ([source](https://news.ycombinator.com/item?id=16446774)).\n",
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"\n",
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"In its simplest form, a group of people start counting upwards in an alternating fashion. Whenever a number is divisible by $3$, the person must say \"Fizz\" instead of the number. The same holds for numbers divisible by $5$ when the person must say \"Buzz\". If a number is divisible by both numbers, one must say \"FizzBuzz\". Probably, this game would also make a good drinking game with the \"right\" beverages.\n",
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"\n",
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"With just Chapter 1, we actually do not yet know all of Python's language constructs we need to write an implementation of Fizz Buzz in a Pythonic way. Yet, we will tweak what we know a bit and make it work."
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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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"**Q11.1**: First, create a list `numbers` with the numbers from 1 through 100. You could type all numbers manually but there is of course a smarter way. The built-in [range()](https://docs.python.org/3/library/functions.html#func-range) may be useful here. Read how it works in the documentation. To make the output of [range()](https://docs.python.org/3/library/functions.html#func-range) a `list` object, you have to \"wrap\" it with the [list()](https://docs.python.org/3/library/functions.html#func-list) built-in (i.e., `list(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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"numbers = ..."
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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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"**Q11.2**: Loop over the `numbers` list and replace numbers for which one of the two (or both) conditions apply with text strings `\"Fizz\"`, `\"Buzz\"`, or `\"FizzBuzz\"` using the indexing operator `[...]` and the assignment statement `=`. In the chapter we saw that Python starts indexing with `0` as the first element. So in each iteration of the `for`-loop you have to determine the *index* as well as checking the actual `number`.\n",
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"\n",
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"Also note that for numbers divisible by both $3$ and $5$ we need some sort of a \"third\" condition check: As we only know about the `if` statement so far (and have not heard about `elif` and `else` from Chapter 3), there will be three `if` statements in total within the loop. And the order of them matters!\n",
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"\n",
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"Hint: Is there a single condition that checks for both $3$ and $5$?"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"for number in numbers:\n",
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" ..."
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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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"**Q11.3**: Create a loop that prints out either the number or any of the Fizz Buzz substitutes. Do it in such a way that we do not end up with 100 lines of output here."
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"for number in numbers:\n",
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" ..."
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]
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}
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],
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"metadata": {
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@ -211,11 +211,6 @@
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n"
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]
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},
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@ -235,28 +230,36 @@
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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"source": [
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"discounted_price(...)"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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"source": [
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"discounted_price(...)"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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"source": [
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"discounted_price(...)"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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"source": [
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"discounted_price(...)"
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]
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},
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{
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"cell_type": "markdown",
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@ -270,14 +273,18 @@
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": []
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"source": [
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"discounted_price(...)"
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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": null,
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"metadata": {},
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"outputs": [],
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"source": []
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"source": [
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"discounted_price(...)"
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]
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},
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{
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"cell_type": "markdown",
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@ -304,14 +311,23 @@
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Fizz Buzz revisited"
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"### Fizz Buzz"
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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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"When you worked on the Fizz Buzz exercise in Chapter 1, you actually did not know about the `elif` and `else` keywords yet. Well, now you do."
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"The kids game [Fizz Buzz](https://en.wikipedia.org/wiki/Fizz_buzz) is said to be often used in job interviews for entry level positions. However, opinions vary as to how good of a test it actually is ([source](https://news.ycombinator.com/item?id=16446774)).\n",
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"\n",
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"In its simplest form, a group of people start counting upwards in an alternating fashion. Whenever a number is divisible by $3$, the person must say \"Fizz\" instead of the number. The same holds for numbers divisible by $5$ when the person must say \"Buzz\". If a number is divisible by both numbers, one must say \"FizzBuzz\". Probably, this game would also make a good drinking game with the \"right\" beverages."
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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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"**Q9.1**: First, create a list `numbers` with the numbers from 1 through 100. You could type all numbers manually but there is of course a smarter way. The built-in [range()](https://docs.python.org/3/library/functions.html#func-range) may be useful here. Read how it works in the documentation. To make the output of [range()](https://docs.python.org/3/library/functions.html#func-range) a `list` object, you have to wrap it with the [list()](https://docs.python.org/3/library/functions.html#func-list) built-in (i.e., `list(range(...))`)."
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]
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},
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{
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@ -320,16 +336,20 @@
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"metadata": {},
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"outputs": [],
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"source": [
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"numbers = list(range(1, 101))"
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"numbers = ..."
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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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"**Q9**: Copy and paste your answer to **Q11.2** in Chapter 1 here and instead of three consecutive `if` statements re-write it with *one* compound `if` statement.\n",
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"**Q9.2**: Loop over the `numbers` list and replace numbers for which one of the two (or both) conditions apply with text strings `\"Fizz\"`, `\"Buzz\"`, or `\"FizzBuzz\"` using the indexing operator `[]` and the assignment statement `=`.\n",
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"\n",
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"This code will then be a lot more robust as the order of the three `if` statements cannot be screwed up."
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"In Chapter 1 we saw that Python starts indexing with `0` as the first element. Keep that in mind.\n",
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"\n",
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"So in each iteration of the `for`-loop you have to determine an `index` variable as well as checking the actual `number` for its divisors.\n",
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"\n",
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"Hint: the order of the conditions is important!"
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]
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},
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{
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@ -338,17 +358,31 @@
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"metadata": {},
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"outputs": [],
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"source": [
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n",
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"\n"
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"for number in numbers:\n",
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" ...\n",
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" ...\n",
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" ...\n",
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" ...\n",
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" ...\n",
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" ...\n",
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" ..."
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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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"**Q9.3**: Create a loop that prints out either the number or any of the Fizz Buzz substitutes. Do it in such a way that we do not end up with 100 lines of output here."
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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": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"for number in numbers:\n",
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" print(...)"
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]
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}
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],
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