{
"cells": [
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"cell_type": "code",
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"source": [
"# By submitting this lab assignment, I agree to the following:\n",
"# â€œAggies do not lie, cheat, or steal, or tolerate those who doâ€\n",
"# â€œI have not given or received any unauthorized aid on this assignmentâ€\n",
"#\n",
"# Name: YOUR NAME\n",
"# Section: YOUR SECTION NUMBER\n",
"# Assignment: Lab 4\n",
"# Date: DAY MONTH YEAR"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"#### Computation Lab\n",
"\n",
"## Lab4 Conditionals and Loops"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Part 1 Conditionals and for Loops"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"***\n",
"__Q1:__ Using a for-loop and if statements, generate the list __(odd5)__ of odd numbers in the range -20 to 20 that divide either by 5 or by 3.\n",
"\n",
"__Hint:__ Expected output is `odd5 = [-15, -9, -5, -3, 3, 5, 9, 15]`"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"output_type": "stream",
"name": "stdout",
"text": [
   "[-15, -9, -5, -3, 3, 5, 9, 15]\n"
]
}
],
"source": [
"### Your code here\n",
"\n",
"odd5 = []\n",
"\n",
"item = -19\n",
"while item < 20:\n",
" abs_ = abs(item) \n",
" if abs_ % 5 == 0 or abs_ % 3 == 0:\n",
"    odd5.append(item)\n",
"    pass\n",
"\n",
" item = item + 2\n",
" pass\n",
"\n",
"\n",
"\n",
"\n",
"print(odd5)\n",
"    \n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"***\n",
"__Q2:__ Write an __if-elif-else__ statement that evaluates and prints the outcomes of applying the following comparison tests on two variables $a$ and $b$. \n",
"\n",
" if (a == b) print a is equal to b\n",
" if (a > b) print a is larger than b\n",
" if (a >= b) print a is larger or equal to b\n",
" if (a < b) print a is smaller than b\n",
" if (a <= b) print a is smaller or equal to b\n",
"\n",
" \n",
"Try your code for `a=3 and b=10`"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [
{

Time to sort! You will implement 4 sorting algorithms: mergesort, q...