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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "id": "75778ca0",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "slide"
+ }
+ },
+ "source": [
+ "# UE5 Fundamentals of Algorithms\n",
+ "## Lecture 4-5-6: Programming strategies\n",
+ "### Ecole Centrale de Lyon, Bachelor of Science in Data Science for Responsible Business\n",
+ "#### Romain Vuillemot\n",
+ "<center><img src=\"figures/Logo_ECL.png\" style=\"width:300px\"></center>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "f0c7488c",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Outline\n",
+ "- Definitions of programming strategies\n",
+ "- Divide and conquer\n",
+ "- Greedy algorithms\n",
+ "- Dynamic programming"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b3ce5394",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "# Programming strategies\n",
+ "\n",
+ "> A programming strategy are algorithms aimed at solving a specific problem in a precise manner.\n",
+ "\n",
+ "Examples of Strategies:\n",
+ "\n",
+ "- **Divide and Conquer:** Divide a problem into simpler sub-problems, solve the sub-problems, and then combine the solutions to solve the original problem.\n",
+ "\n",
+ "- **Dynamic Programming:** Solve a problem by breaking it down into sub-problems, calculating and memorizing the results of sub-problems to avoid unnecessary recomputation.\n",
+ "\n",
+ "- **Greedy Algorithm:** Make a series of choices that seem locally optimal at each step to find a solution, with the hope that the result will be globally optimal as well.\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "bd22ed47",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "slide"
+ }
+ },
+ "source": [
+ "# Divide and conquer\n",
+ "\n",
+ "> The **Divide and Conquer** strategy involves breaking a complex problem into smaller, similar subproblems, solving them recursively, and then combining their solutions to address the original problem efficiently.\n",
+ "\n",
+ "1. **Divide:** Divide the original problem into subproblems of the same type.\n",
+ "\n",
+ "2. **Conquer:** Solve each of these subproblems recursively.\n",
+ "\n",
+ "3. **Combine:** Combine the answers appropriately.\n",
+ "\n",
+ "_It is very close to the recursive approach_\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "d12319a6",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "\n",
+ "### Examples of divide and conquer algorithms:\n",
+ "\n",
+ "- Binary search\n",
+ "- Quick sort and merge sort\n",
+ "- Map Reduce\n",
+ "- Others: Fast multiplication (Karatsuba)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "fbb1b64c",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Binary search\n",
+ "\n",
+ "_Given a sorted list, find or insert a specific value while keeping the order._\n",
+ "\n",
+ "<img src=\"figures/recherche-dichotomique.png\" style=\"width:500px\">\n",
+ "\n",
+ "See [the notebook](03-lists-search-sort.ipynb)."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "bba7c4c6",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Quick sort\n",
+ "\n",
+ "Recursive sorting algorithm which works in two steps:\n",
+ "\n",
+ "1. select a pivot element \n",
+ "2. partitioning the array into smaller sub-arrays, then sorting those sub-arrays.\n",
+ "\n",
+ "<img src=\"figures/quicksort.png\" style=\"height:400px\">"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "400d3619",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Merge sort\n",
+ "\n",
+ "Divide an array recursively into two halves (based on a _pivot_ value), sorting each half, and then merging the sorted halves back together. This process continues until the entire array is sorted.<br> Complexity: $O(n log(n))$.\n",
+ "\n",
+ "<img src=\"figures/tri-fusion.png\" style=\"width:500px\">"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "7d33da6c",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Map reduce\n",
+ "\n",
+ "Divide a large dataset into smaller chunks and processes them independantly. Two main steps: \n",
+ "- the Map stage, where data is filtered and transformed into key-value pairs\n",
+ "- the Reduce stage, where data is aggregated and the final result is produced.\n",
+ "<img src=\"figures/Mapreduce.png\" style=\"width:700px\">"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "a590fbe9",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Map reduce (without map reduce..)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "9f163e56",
+ "metadata": {},
+ "source": [
+ "_Calculate the sum of squares values from a list of numerical values._"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 49,
+ "id": "ba28ddd1",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 60,
+ "id": "f34730fa",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "[(1, 1), (4, 2), (9, 3), (16, 4), (25, 5), (36, 6), (49, 7), (64, 8), (81, 9), (100, 10)]\n",
+ "385\n"
+ ]
+ }
+ ],
+ "source": [
+ "result = {}\n",
+ "for num in data:\n",
+ " square = num * num\n",
+ " result[square] = num\n",
+ "\n",
+ "final_result = list(result.items())\n",
+ "\n",
+ "print(final_result)\n",
+ "print(sum([x[0] for x in final_result]))"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "261c7a2c",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Map reduce (Python)\n",
+ "\n",
+ "1. Divide the problem in sub-problems\n",
+ "2. Apply the mapping function\n",
+ "3. Reduce the results"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 69,
+ "id": "9ef9a5cc",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "385\n"
+ ]
+ }
+ ],
+ "source": [
+ "data = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n",
+ "\n",
+ "def mapper(numbers):\n",
+ " result = []\n",
+ " for num in numbers: # calculate the squares\n",
+ " result.append((num, num * num))\n",
+ " return result\n",
+ "\n",
+ "def reducer(pairs):\n",
+ " result = {}\n",
+ " for key, value in pairs: # sums the squares\n",
+ " if key in result:\n",
+ " result[key] += value \n",
+ " else:\n",
+ " result[key] = value\n",
+ " return result.items()\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "975a4b17",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Map reduce (Python)\n",
+ "\n",
+ "1. Divide the problem in sub-problems\n",
+ "2. Apply the mapping function\n",
+ "3. Reduce the results"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "id": "77215ac8",
+ "metadata": {},
+ "outputs": [
+ {
+ "ename": "NameError",
+ "evalue": "name 'data' is not defined",
+ "output_type": "error",
+ "traceback": [
+ "\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
+ "\u001b[0;31mNameError\u001b[0m Traceback (most recent call last)",
+ "Cell \u001b[0;32mIn[5], line 2\u001b[0m\n\u001b[1;32m 1\u001b[0m chunk_size \u001b[38;5;241m=\u001b[39m \u001b[38;5;241m2\u001b[39m\n\u001b[0;32m----> 2\u001b[0m chunks \u001b[38;5;241m=\u001b[39m [data[i:i\u001b[38;5;241m+\u001b[39mchunk_size] \u001b[38;5;28;01mfor\u001b[39;00m i \u001b[38;5;129;01min\u001b[39;00m \u001b[38;5;28mrange\u001b[39m(\u001b[38;5;241m0\u001b[39m, \u001b[38;5;28mlen\u001b[39m(\u001b[43mdata\u001b[49m), chunk_size)]\n\u001b[1;32m 4\u001b[0m mapped_data \u001b[38;5;241m=\u001b[39m [mapper(chunk) \u001b[38;5;28;01mfor\u001b[39;00m chunk \u001b[38;5;129;01min\u001b[39;00m chunks] \n\u001b[1;32m 6\u001b[0m grouped_data \u001b[38;5;241m=\u001b[39m {}\u001b[38;5;66;03m# map\u001b[39;00m\n",
+ "\u001b[0;31mNameError\u001b[0m: name 'data' is not defined"
+ ]
+ }
+ ],
+ "source": [
+ "\n",
+ "chunk_size = 2\n",
+ "chunks = [data[i:i+chunk_size] for i in range(0, len(data), chunk_size)]\n",
+ "\n",
+ "mapped_data = [mapper(chunk) for chunk in chunks] \n",
+ "\n",
+ "grouped_data = {}# map\n",
+ "for chunk in mapped_data:\n",
+ " for key, value in chunk:\n",
+ " if key in grouped_data:\n",
+ " grouped_data[key].append(value)\n",
+ " else:\n",
+ " grouped_data[key] = [value]\n",
+ "\n",
+ "reduced_data = [reducer(list(grouped_data.items()))] # reduce\n",
+ "result = sum([x[1][0] for x in final_result])\n",
+ "\n",
+ "print(result)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "be8744c6",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Discussion on Divide and Conquer\n",
+ "\n",
+ "- Similarities with recursion by dividing a problem in a sub-problem\n",
+ "\n",
+ "- But with a combination step (which may hold most of the code difficulty)\n",
+ "\n",
+ "- Can be implemented in a non-recursive way\n",
+ "\n",
+ "- $n log(n)$ complexity when split the problem and solves each split"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c1f7b96a",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "slide"
+ }
+ },
+ "source": [
+ "# Greedy algorithms\n",
+ "\n",
+ "> Algorithms that make a locally optimal choice.\n",
+ "\n",
+ "### Examples:\n",
+ "\n",
+ "- Change-making problem\n",
+ "- Knapsack problem\n",
+ "- Maze solving\n",
+ "- Graph coloring"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "17867aaf",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Example: Change-making problem\n",
+ "\n",
+ "\n",
+ "$Q_{opt}(S,M) = min \\ \\sum_{i=1}^n x_i$.\n",
+ " \n",
+ "$S$: all the available coins\n",
+ " \n",
+ "$M$: amount\n",
+ " \n",
+ "Greedy solution:\n",
+ "\n",
+ "1. Sort the coins in descending order\n",
+ "\n",
+ "2. Initialize a variable to count coins used\n",
+ "\n",
+ "3. Substrack the number of coins used (if limited)\n",
+ "\n",
+ "4. Continue this process until amount becomes zero.\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "adb8552d",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Example: Change-making problem (Python)\n",
+ "\n",
+ "_Greedy solution to return the minimal number of coins necessary._"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 82,
+ "id": "e900b357",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "coins = [1, 2, 5]\n",
+ "amount = 11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 83,
+ "id": "b4e91e95",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [],
+ "source": [
+ "def coin_change_greedy(coins, amount):\n",
+ " coins.sort(reverse=True) # important! sort in descending order\n",
+ " \n",
+ " coin_count = 0\n",
+ " remaining_amount = amount\n",
+ " \n",
+ " for coin in coins:\n",
+ " while remaining_amount >= coin:\n",
+ " remaining_amount -= coin\n",
+ " coin_count += 1\n",
+ " \n",
+ " if remaining_amount == 0:\n",
+ " return coin_count\n",
+ " else:\n",
+ " return -1"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 81,
+ "id": "131184bf",
+ "metadata": {},
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "3\n"
+ ]
+ }
+ ],
+ "source": [
+ "print(coin_change_greedy(coins, amount)) # 3 (11 = 5 + 5 + 1)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "3ab88980",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Example: Change-making problem (Python)\n",
+ "\n",
+ "_Greedy solution that returns the **list of coins** used._\n",
+ "\n",
+ "\n",
+ "Tip: use a list with the same structure as coins."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 87,
+ "id": "9401818b",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Minimum coins needed: 6\n",
+ "Coins used: [2, 1, 0, 3]\n"
+ ]
+ }
+ ],
+ "source": [
+ "def coin_change_greedy(coins, amount):\n",
+ " coins.sort(reverse=True) \n",
+ " \n",
+ " coin_count = 0\n",
+ " remaining_amount = amount\n",
+ " used_coins = [0] * len(coins)\n",
+ " \n",
+ " for i, coin in enumerate(coins):\n",
+ " while remaining_amount >= coin:\n",
+ " remaining_amount -= coin\n",
+ " coin_count += 1\n",
+ " used_coins[i] += 1 \n",
+ " \n",
+ " if remaining_amount == 0:\n",
+ " return coin_count, used_coins\n",
+ " else:\n",
+ " return -1, []\n",
+ "\n",
+ "coins = [25, 10, 5, 1]\n",
+ "amount = 63\n",
+ "min_coins, coins_used = coin_change_greedy(coins, amount)\n",
+ "\n",
+ "print(f\"Minimum coins needed: {min_coins}\")\n",
+ "print(\"Coins used:\", coins_used)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "1ae1863a",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Example: Change-making problem (Python)\n",
+ "\n",
+ "_Greedy solution that returns the **list of coins** used from **a limited availability of coins**._\n",
+ "\n",
+ "\n",
+ "Tip: use a list of coins availability of same structure as coins."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 92,
+ "id": "8a936fe3",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "coins = [25, 10, 5, 1]\n",
+ "amount = 63\n",
+ "coin_availability = [1, 2, 3, 4]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 91,
+ "id": "1700c684",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Minimum coins needed: 9\n",
+ "Coins used: [1, 2, 3, 3]\n"
+ ]
+ }
+ ],
+ "source": [
+ "def coin_change_greedy(coins, amount, coin_availability):\n",
+ " coins.sort(reverse=True)\n",
+ "\n",
+ " coin_count = 0\n",
+ " remaining_amount = amount\n",
+ " used_coins = [0] * len(coins)\n",
+ " \n",
+ " for i, coin in enumerate(coins):\n",
+ " while remaining_amount >= coin and used_coins[i] < coin_availability[i]:\n",
+ " remaining_amount -= coin\n",
+ " coin_count += 1\n",
+ " used_coins[i] += 1\n",
+ " \n",
+ " if remaining_amount == 0:\n",
+ " return coin_count, used_coins\n",
+ " else:\n",
+ " return -1, []\n",
+ "\n",
+ "min_coins, coins_used = coin_change_greedy(coins, amount, coin_availability)\n",
+ "\n",
+ "\n",
+ "print(f\"Minimum coins needed: {min_coins}\")\n",
+ "print(\"Coins used:\", coins_used)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "7aeac877",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Discussion on Greedy algorithms\n",
+ "\n",
+ "- Often considered as an _heuristic_\n",
+ "- Easy to understand, implement and communicate\n",
+ "- They often lead to non-optimal solution"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "5f36ec7b",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "slide"
+ }
+ },
+ "source": [
+ "# Dynamic programming\n",
+ "\n",
+ "> **Dynamic programming** involves breaking down a problem into subproblems, *solving* these subproblems, and *combining* their solutions to obtain the solution to the original problem. The steps are as follows:\n",
+ "\n",
+ "1. Characterize the structure of an optimal solution.\n",
+ "2. Define the value of an optimal solution recursively.\n",
+ "3. Reconstruct the optimal solution from the computations.\n",
+ "\n",
+ "Notes :\n",
+ "- Applies to problems with optimal substructure.\n",
+ "- Also applies to problems where solutions are often interrelated (distinguishing it from divide and conquer).\n",
+ "- Utilizes a memoization approach, involving storing an intermediate solution (e.g., in a table).\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b125a618",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "### Examples of dynamic programming algorithms\n",
+ "\n",
+ "- Fibonacci Sequence\n",
+ "- Rod Cutting\n",
+ "- Sequence Alignment, Longest Subsequence Finding\n",
+ "- Shortest Path Findin"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "3e3de7c4",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Fibonnacci (reminder)\n",
+ "\n",
+ "To calculate the $n$-th number in the Fibonacci sequence, which is determined as follows:\n",
+ "\n",
+ "latex\n",
+ "Copy code\n",
+ "$fib(n) = fib(n-1) + fib(n-2)$, $n \\in \\mathbb{N}$\n",
+ "Where the sequence starts with 1, 1, and then continues as 2, 3, 5, 8, 13, 21, and so on, to find the 9th number ($n = 9$).\n",
+ "\n",
+ "Let's calculate the 9th Fibonacci number step by step:\n",
+ "\n",
+ "$fib(1) = 1$\n",
+ "\n",
+ "$fib(2) = 1$\n",
+ "\n",
+ "$fib(3) = fib(2) + fib(1) = 1 + 1 = 2$\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "cced97f7",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Fibonnacci (naive)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "id": "7bc7701a",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "def fib(n):\n",
+ " if n < 2:\n",
+ " return n\n",
+ " else:\n",
+ " return fib(n - 1) + fib(n - 2)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "81eb2956",
+ "metadata": {},
+ "source": [
+ "Call tree (for $n = 6$):"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b588ca8b",
+ "metadata": {},
+ "source": [
+ "<img src=\"figures/fibonacci-tree.png\" style=\"width:400px\">"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "42b6ecd1",
+ "metadata": {},
+ "source": [
+ "Requires to calculate the same F-value multiple times."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "fc74f400",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Fibonnacci (dynamic programming)\n",
+ "\n",
+ "_Optimized using a `lookup` table, which is a data structure to memoize values that have already been computed._"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 97,
+ "id": "9eae506b",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "6-th Fibonacci number is 8\n"
+ ]
+ }
+ ],
+ "source": [
+ "def fib(n, lookup):\n",
+ " if n == 0 or n == 1:\n",
+ " lookup[n] = n\n",
+ "\n",
+ " if lookup[n] is None:\n",
+ " lookup[n] = fib(n - 1, lookup) + fib(n - 2, lookup)\n",
+ "\n",
+ " return lookup[n]\n",
+ "\n",
+ "def main():\n",
+ " n = 6\n",
+ "\n",
+ " lookup = [None] * (n + 1)\n",
+ " result = fib(n, lookup)\n",
+ " print(f\"{n}-th Fibonacci number is {result}\")\n",
+ "\n",
+ "if __name__==\"__main__\": \n",
+ " main() "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "ac5e42aa",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Rod cutting \n",
+ "\n",
+ "_Given a list of cuts and prices, identify the optimal cuts. Given the example below, what is the best cutting strategy for a rod of size 2?_\n",
+ "\n",
+ "<img src=\"figures/rod-cutting.png\" style=\"width:500px\">"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "2f933396",
+ "metadata": {},
+ "source": [
+ "| size (i) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n",
+ "|--------------|---|---|---|---|---|---|---|---|\n",
+ "| price (pi) | 1 | 5 | 8 | 9 |10 |17 |17 |20 |\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "47119bfb",
+ "metadata": {},
+ "source": [
+ "Solution: $2$ with $5 + 5 = 10$."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "61aa819a",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Rod cutting: check a solution\n",
+ "\n",
+ "Given the previous table of size and price, check the cost of a given solution by defining a function `check_rod_cutting(prices, n)`."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 27,
+ "id": "c42b61d4",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [],
+ "source": [
+ "def check_rod_cutting(prices, n):\n",
+ " table = [0] * (n + 1)\n",
+ "\n",
+ " for i in range(1, n + 1):\n",
+ " max_price = float('-inf')\n",
+ " for j in range(1, i + 1):\n",
+ " max_price = max(max_price, prices[j] + table[i - j])\n",
+ " table[i] = max_price\n",
+ "\n",
+ " return table[n]"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 30,
+ "id": "9a3bd3d3",
+ "metadata": {},
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The maximum total price for a rod of length 2 is 5\n"
+ ]
+ }
+ ],
+ "source": [
+ "prices = [0, 1, 5, 8, 9, 10, 17, 17, 20]\n",
+ "n = 2\n",
+ "\n",
+ "max_total_price = check_rod_cutting(prices, n)\n",
+ "print(f\"The maximum total price for a rod of length {n} is {max_total_price}\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "151c0a39",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Rod cutting (brute force)\n",
+ "\n",
+ "Let's solve the rod cutting problem using a brute force (naive) approach.\n",
+ "\n",
+ "1. define a value function\n",
+ "2. identify a base case\n",
+ "3. identify a recursion mechanism"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "id": "6e7a3a0d",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [],
+ "source": [
+ "def cut_brute_force(n, t):\n",
+ " if n == 0:\n",
+ " return 0\n",
+ " max_valeur = float('-inf')\n",
+ " for i in range(1, n + 1):\n",
+ " valeur_courante = t[i] + coupe_brute_force(n - i, t)\n",
+ " max_valeur = max(max_valeur, valeur_courante)\n",
+ " return max_valeur"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 34,
+ "id": "1ee43707",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "The maximum value for a rod of length 2 is 5.\n"
+ ]
+ }
+ ],
+ "source": [
+ "lengths = [0, 1, 2, 3, 4, 5, 6, 7, 8]\n",
+ "values = [0, 1, 5, 8, 9, 10, 17, 17, 20]\n",
+ "rod_length = 2\n",
+ "max_value = coupe_brute_force(rod_length, values)\n",
+ "print(f\"The maximum value for a rod of length {rod_length} is {max_value}.\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "1596f825",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Rod cutting (dynamic programming)\n",
+ "\n",
+ "\n",
+ "General case:\n",
+ "\n",
+ "- Cutting a rod of length $i$ optimally.\n",
+ "- Cutting a rod of length $(n - i)$ optimally.\n",
+ "\n",
+ "\n",
+ "<img src=\"figures/rod-cutting-tree.png\" style=\"width:500px\">\n",
+ "\n",
+ "General case: $V_{n} = max_{1 \\leq i \\leq n} (p_i + V_{n - i})$ "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "be5d9a15",
+ "metadata": {},
+ "source": [
+ "| size (i) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |\n",
+ "|--------------|---|---|---|---|---|---|---|---|\n",
+ "| price (pi) | 1 | 5 | 8 | 9 |10 |17 |17 |20 |\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b2845bd5",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "$V_{3} = \\max_{1 \\leq i \\leq 3} (p_i + V_{3 - i})$\n",
+ "\n",
+ "Let's calculate `V_3` step by step for each possible value of `i`:\n",
+ "\n",
+ "1. If `i = 1`, we cut the rod into two pieces: one of length 1 and one of length 2.\n",
+ " - $V_1 = p_1 = 2$\n",
+ " - $V_{3 - 1} = V_2$\n",
+ "\n",
+ "2. If `i = 2`, we cut the rod into two pieces: one of length 2 and one of length 1.\n",
+ " - $V_2 = p_2 = 5$\n",
+ " - $V_{3 - 2} = V_1$\n",
+ "\n",
+ "3. If `i = 3`, we cut the rod into one piece of length 3.\n",
+ " - $V_3 = p_3 = 9$\n",
+ " - $V_{3 - 3} = V_0$ (Assuming that $V_0 = 0$ as a base case.)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "eb092761",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "Now, we can calculate the values for `V_2` and `V_1` recursively using the same formula:\n",
+ "\n",
+ "For `V_2`:\n",
+ "$$V_2 = \\max(p_1 + V_1, p_2 + V_0) = \\max(2 + V_1, 5 + 0) = \\max(2 + 2, 5 + 0) = \\max(4, 5) = 5$$\n",
+ "\n",
+ "For `V_1`:\n",
+ "$$V_1 = \\max(p_1 + V_0) = \\max(2 + 0) = 2$$\n",
+ "\n",
+ "So, `V_2` is 5 and `V_1` is 2.\n",
+ "\n",
+ "Now, we can calculate `V_3` using the values of `V_2` and `V_1`:\n",
+ "\n",
+ "$$V_3 = \\max(p_1 + V_2, p_2 + V_1, p_3 + V_0) = \\max(2 + 5, 5 + 2, 9 + 0) = \\max(7, 7, 9) = 9$$"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "0e097cbf",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Rod cutting (dynamic programming)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 36,
+ "id": "5812781c",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "fragment"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Max size cut 22 8\n"
+ ]
+ }
+ ],
+ "source": [
+ "INT_MIN = 0\n",
+ "\n",
+ "def cutRod(price, n): \n",
+ "\n",
+ " # init cache tables\n",
+ " val = [0 for x in range(n+1)] \n",
+ " val[0] = 0\n",
+ " \n",
+ " for i in range(1, n+1): \n",
+ " max_val = INT_MIN \n",
+ " for j in range(i): \n",
+ " max_val = max(max_val, price[j] + val[i-j-1]) \n",
+ " val[i] = max_val \n",
+ " \n",
+ " return val[n] \n",
+ " \n",
+ "if __name__==\"__main__\": \n",
+ " arr = [1, 5, 8, 9, 10, 17, 17, 20] \n",
+ " size = len(arr) \n",
+ " print(\"Max size cut \" + str(cutRod(arr, size)), len(arr) ) "
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "fd3e3af7",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Lessons on dynamic programming\n",
+ "\n",
+ "- It is necessary to study each problem on a case-by-case basis.\n",
+ "\n",
+ "- Storing a large number of partial results, which requires significant memory usage.\n",
+ "\n",
+ "- Suitable for only certain problems (min, max, counting the number of solutions).\n"
+ ]
+ }
+ ],
+ "metadata": {
+ "celltoolbar": "Slideshow",
+ "kernelspec": {
+ "display_name": "Python 3 (ipykernel)",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.10.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}
diff --git a/07-stacks-queues.ipynb b/07-stacks-queues.ipynb
new file mode 100644
index 0000000000000000000000000000000000000000..e41de9135de4815f4639c2a95a4bb5861bfb8cf5
--- /dev/null
+++ b/07-stacks-queues.ipynb
@@ -0,0 +1,671 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "id": "a4973a08",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "slide"
+ }
+ },
+ "source": [
+ "# UE5 Fundamentals of Algorithms\n",
+ "## Lecture 7: Stacks and queues\n",
+ "### Ecole Centrale de Lyon, Bachelor of Science in Data Science for Responsible Business\n",
+ "#### Romain Vuillemot\n",
+ "<center><img src=\"figures/Logo_ECL.png\" style=\"width:300px\"></center>\n",
+ "\n",
+ "\n",
+ "---"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "ffa36fea",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Outline\n",
+ "- Definitions\n",
+ "- Stacks\n",
+ "- Queues\n",
+ "- Priority queues"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "cd2e76d7",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Defintions\n",
+ "\n",
+ "> Stacks and queues allow the manipulation of values (or objects) sequentially. They have many operations, the main ones are: addition (push) and removal (pop), but with different order strategies:\n",
+ "\n",
+ "<img src=\"figures/stacks-queues.png\" style=\"width:500px\">\n",
+ "\n",
+ "- `stacks` follow the Last-In, First-Out (LIFO) principle\n",
+ "- `queues`follows the First-In, First-Out (FIFO) principle\n",
+ "\n",
+ "Note that stacks and queues define the operations and their results, but not their implementation."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "cc2c4270",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Operations\n",
+ "\n",
+ "- `empty()`: Checks for emptiness.\n",
+ "- `full()`: Checks if it's full (if a maximum size was provided during creation).\n",
+ "- `get()`: Returns (and removes) an element.\n",
+ "- `push()`: Adds an element.\n",
+ "- `size()`: Returns the size of the list.\n",
+ "- `reverse()`: Reverses the order of elements.\n",
+ "- `peek()`: Returns an element (without removing it)."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c0885103",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "### Stacks\n",
+ "\n",
+ "> A stack is an abstract data type that follows the Last-In, First-Out (LIFO) principle\n",
+ "\n",
+ "- It supports operations on a collection of elements.\n",
+ "- The element inserted last is at the _head_.\n",
+ "- Easily achievable with a simple list! See this [Python tutorial](https://docs.python.org/3/tutorial/datastructures.html)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "363737b7",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "### Stacks (using lists)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 39,
+ "id": "6326e146",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "stack = [3, 4, 5]\n",
+ "stack.append(6) # push\n",
+ "stack.append(7)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 40,
+ "id": "6604a331",
+ "metadata": {},
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "[3, 4, 5, 6, 7]\n",
+ "[3, 4, 5, 6]\n",
+ "[3, 4]\n",
+ "4\n"
+ ]
+ }
+ ],
+ "source": [
+ "print(stack)\n",
+ "stack.pop() # get\n",
+ "print(stack)\n",
+ "stack.pop()\n",
+ "stack.pop()\n",
+ "print(stack)\n",
+ "print(stack[-1]) # peek"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "075dc32b",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "### Stacks (using modules)\n",
+ "\n",
+ "https://docs.python.org/3/library/queue.html"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 41,
+ "id": "b467ca83",
+ "metadata": {},
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "4 3 2 1 0 "
+ ]
+ }
+ ],
+ "source": [
+ "import queue\n",
+ "pile = queue.LifoQueue()\n",
+ "\n",
+ "for i in range(5): pile.put(i)\n",
+ "\n",
+ "while not pile.empty(): \n",
+ " print(pile.get(), end=\" \")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "697d2be8",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "### Stacks (using OOP)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "id": "8ae9a611",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "class Stack():\n",
+ " def __init__(self, values = []):\n",
+ " self.__values = []\n",
+ " for v in values:\n",
+ " self.push(v)\n",
+ "\n",
+ " def push(self, v):\n",
+ " self.__values.append(v)\n",
+ " return v\n",
+ "\n",
+ " def get(self):\n",
+ " v = self.__values.pop()\n",
+ " return v\n",
+ "\n",
+ " def display(self):\n",
+ " for v in self.__values:\n",
+ " print(v)\n",
+ "\n",
+ " def size(self):\n",
+ " return len(self.__values)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "a0901fcb",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "### Stacks (using OOP)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 42,
+ "id": "c61545dc",
+ "metadata": {},
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "A\n",
+ "B\n",
+ "C\n"
+ ]
+ }
+ ],
+ "source": [
+ "data = [\"A\", \"B\", \"C\"]\n",
+ "\n",
+ "s = Stack()\n",
+ "for d in data:\n",
+ " s.push(d)\n",
+ " e = s.pop()\n",
+ " print(e)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "9145ef5e",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Queues\n",
+ "\n",
+ "> A stack is an abstract data type that follows the First-In, First-Out (FIFO) principle\n",
+ "\n",
+ "- Similar to a Srtack\n",
+ "- But the returned element is the first one inserted"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "724ac3b7",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Queues (list)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 44,
+ "id": "b2cc2dd0",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "queue = [3, 4, 5]\n",
+ "queue.append(6)\n",
+ "queue.append(7) # push"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 45,
+ "id": "1157237e",
+ "metadata": {},
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "[3, 4, 5, 6, 7]\n",
+ "[4, 5, 6, 7]\n",
+ "[6, 7]\n",
+ "6\n"
+ ]
+ }
+ ],
+ "source": [
+ "print(queue) \n",
+ "queue.pop(0) # get\n",
+ "print(queue)\n",
+ "queue.pop(0)\n",
+ "queue.pop(0)\n",
+ "print(queue)\n",
+ "print(queue[0]) # peek"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "78d3504a",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Queues (module)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 33,
+ "id": "a6392890",
+ "metadata": {},
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "0 1 2 3 4 "
+ ]
+ }
+ ],
+ "source": [
+ "import queue\n",
+ "\n",
+ "q = queue.Queue()\n",
+ "\n",
+ "for i in range (5): q.put(i)\n",
+ "\n",
+ "while not q.empty(): \n",
+ " print(q.get(), end=\" \")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "c2269f3b",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Priority queues\n",
+ "\n",
+ "> A **priority queue** is a queue (or stack or list) that returns an element based on the characteristics of a variable (priority).\n",
+ "\n",
+ "- For a quantitative variable, it's the minimum or maximum of the queue. For other types of variables (e.g., categories), any order relation is valid.\n",
+ "\n",
+ "- Queues can exhibit the same behavior but have a different internal state: either constantly updated or updated after reads/writes.\n",
+ "\n",
+ "- The internal state can be preserved with a sorting function, thus optimizing the complexity of the data structure.\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "a46609ab",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Priority queues (module)\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 46,
+ "id": "6a31eba0",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "from heapq import heapify, heappush, heappop\n",
+ "heap = [10, 8, 1, 2, 4, 9, 3, 4, 7]\n",
+ "heapify(heap)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 47,
+ "id": "0fbe9868",
+ "metadata": {},
+ "outputs": [
+ {
+ "data": {
+ "text/plain": [
+ "[1, 2, 3, 4, 4, 9, 10, 8, 7]"
+ ]
+ },
+ "execution_count": 47,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "heap"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 48,
+ "id": "c72a1070",
+ "metadata": {},
+ "outputs": [
+ {
+ "data": {
+ "text/plain": [
+ "1"
+ ]
+ },
+ "execution_count": 48,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "heappop(heap)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 49,
+ "id": "be80320d",
+ "metadata": {},
+ "outputs": [
+ {
+ "data": {
+ "text/plain": [
+ "[2, 4, 3, 4, 7, 9, 10, 8]"
+ ]
+ },
+ "execution_count": 49,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "heap"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 50,
+ "id": "2aa86e2e",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "heappush(heap, 5)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 51,
+ "id": "5d5f290a",
+ "metadata": {},
+ "outputs": [
+ {
+ "data": {
+ "text/plain": [
+ "[2, 4, 3, 4, 7, 9, 10, 8, 5]"
+ ]
+ },
+ "execution_count": 51,
+ "metadata": {},
+ "output_type": "execute_result"
+ }
+ ],
+ "source": [
+ "heap"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "b841cc1f",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ "## Priority queues (using OOP)"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 56,
+ "id": "e34b0c73",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "class PriorityQueue(object): \n",
+ " def __init__(self): \n",
+ " self.__queue = [] \n",
+ " \n",
+ " def __str__(self): \n",
+ " return ' '.join([str(i) for i in self.__queue]) \n",
+ " \n",
+ " def isEmpty(self): \n",
+ " return len(self.__queue) == 0\n",
+ " \n",
+ " def insert(self, data): \n",
+ " self.__queue.append(data) \n",
+ " \n",
+ " def size(self): \n",
+ " return len(self.__queue)\n",
+ "\n",
+ " def delete(self): \n",
+ " min = 0\n",
+ " for i in range(0,len(self.__queue)): \n",
+ " if self.__queue[i][2] < self.__queue[min][2]: \n",
+ " min = i \n",
+ " item = self.__queue[min] \n",
+ " del self.__queue[min] \n",
+ " return item \n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 60,
+ "id": "0e326166",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "1\n",
+ "7\n",
+ "12\n",
+ "14\n"
+ ]
+ }
+ ],
+ "source": [
+ "import queue\n",
+ "\n",
+ "myQueue = queue.PriorityQueue()\n",
+ "\n",
+ "# Insert elements into the priority queue\n",
+ "myQueue.put(12)\n",
+ "myQueue.put(1)\n",
+ "myQueue.put(14)\n",
+ "myQueue.put(7)\n",
+ "\n",
+ "# Print the contents of the priority queue\n",
+ "while not myQueue.empty():\n",
+ " print(myQueue.get())"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "id": "01f5f437",
+ "metadata": {
+ "slideshow": {
+ "slide_type": "subslide"
+ }
+ },
+ "source": [
+ " ## Improvements\n",
+ "\n",
+ "- Handle empty lists"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "id": "5aa5a7ad",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "\n",
+ "def dequeue(self):\n",
+ " if not self.is_empty():\n",
+ " return self.items.pop(0)\n",
+ " else:\n",
+ " raise IndexError(\"Queue is empty\")\n",
+ " "
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "id": "89407366",
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "import queue\n",
+ "q = queue.Queue(5000)\n",
+ "# q.put([1,2,3,4])\n",
+ "q.put([2,3,4,5])\n",
+ "try:\n",
+ " [x1, x2, x3, x4] = q.get(block=True)\n",
+ "except queue.Empty:\n",
+ " print(\"Problème : aucun élément dans la file\")\n",
+ "else:\n",
+ " print([x1, x2, x3, x4])\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "id": "1bf32fdb",
+ "metadata": {},
+ "outputs": [],
+ "source": []
+ }
+ ],
+ "metadata": {
+ "celltoolbar": "Slideshow",
+ "kernelspec": {
+ "display_name": "Python 3 (ipykernel)",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.10.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}