\\[\nA = \\frac12 \\left| \\sum_{i = 1}^{n} x_iy_{i + 1} - y_i x_{i + 1} \\right|\\]<\/p>\n
In the formula above, \\(P_{n + 1}\\)<\/span> is \\(P_1\\)<\/span>.<\/p>\n The formula is super practical and easy to compute, which I find amusing given that it works for any (simple) polygon!<\/p>","summary":"Today I learned about the shoelace formula to compute the area of arbitrary simple polygons.","date_modified":"2025-10-20T22:34:56+02:00","tags":["mathematics","geometry"],"image":"\/user\/pages\/02.blog\/04.til\/133.shoelace-formula\/thumbnail.webp"},{"title":"TIL #132 \u2013 Double factorial","date_published":"2025-09-25T11:47:00+02:00","id":"https:\/\/mathspp.com\/blog\/til\/double-factorial","url":"https:\/\/mathspp.com\/blog\/til\/double-factorial","content_html":" Today I learned about the double factorial.<\/p>\n\n The double factorial, \\(n!!\\)<\/span>, is defined as the product of all positive integers less than or equal to \\(n\\)<\/span> that have the same parity as \\(n\\)<\/span>.<\/p>\n Some examples:<\/p>\n When reading about it in a book authored by a friend of mine, the identity \\((2n)!! = 2^n n!\\)<\/span> was also presented, and I'll prove it now by induction.\nFor \\(n = 1\\)<\/span>, we have \\(2!! = 2 = 2^1 \\times 1!\\)<\/span>, which is true.\nNow, assuming the identity holds up to \\(n\\)<\/span>, we show it holds for \\(n + 1\\)<\/span>:<\/p>\n \\[\n\\begin{align}\n(2(n + 1))!! &= (2(n + 1)) \\times (2n)!! \\\\\n&= (2(n + 1)) \\times 2^n n! \\\\\n&= (n + 1) \\times 2^{n + 1} n! \\\\\n&= 2^{n + 1} (n + 1)!\n\\end{align}\\]<\/p>\n Done!<\/p>","summary":"Today I learned about the double factorial.","date_modified":"2025-10-20T22:34:56+02:00","tags":["combinatorics","arithmetic","induction","mathematics"],"image":"\/user\/pages\/02.blog\/04.til\/132.double-factorial\/thumbnail.webp"},{"title":"TIL #126 \u2013 Hash of infinity","date_published":"2025-07-20T22:18:00+02:00","id":"https:\/\/mathspp.com\/blog\/til\/hash-of-infinity","url":"https:\/\/mathspp.com\/blog\/til\/hash-of-infinity","content_html":" Today I learned about a Python Easter Egg hidden in the hash of two special float values.<\/p>\n\n Today I learned that the values In case you can't tell, those are the first few digits of the mathematical constant \\(\\pi\\)<\/span>:<\/p>\n I learned about this during the CPython panel<\/a> held at EuroPython 2025, after the hosts \u0141ukasz & Pablo asked the panel if they know what the hash of This article walks through the code used to animate a tree fractal.<\/p>\n\n I really enjoy fractals and for the longest time I've been meaning to draw the classical tree fractal that starts with the main trunk which then successively splits into two branches, that themselves can be seen as the full tree.<\/p>\n Today I quickly put together the code to do this and then I animated the rotation of the tree to make sure it's easy to see the self-similarity of the tree.<\/p>\n This is the animated GIF that I produced:<\/p>\n To draw the tree I wrote a recursive function (obviously!) that draws a trunk and then recursively draws the remainder of the tree.\nTo “recursively draw the remainder of the tree” I needed to compute the two trunks that stem from the trunk I just drew, so to do that I:<\/p>\n The recursive function looks like this:<\/p>\n One important thing that I ended up doing, and that explains the weird calculations inside the call to The full script to draw the tree looks like this:<\/p>\n To make it easier to animate the tree I first started by splitting the function Reverse-engineering the program from “Chronospatial Computer”, day 17 of Advent of Code 2024.<\/p>\n\n The “Chronospatial Computer” is from Advent of Code 2024, day 17<\/a>, a problem that entertained me for a couple of hours.<\/p>\n My input file looked like this:<\/p>\n To read the input and parse it I used a context manager and a couple of calls to Part 1 required me to simulate a series of simple instructions that operate on three registers.\nWhen I read the problem statement, I decided I wanted to use some ideas from functional programming.\nSo, what I did was to separate each operator (there are 8) into three parts:<\/p>\n By using First, the computation part of each operation:<\/p>\n In the lambda functions above, when we use The operation Then, I wrote a list with all the functions that update the state of the program:<\/p>\n This uses the dunder method Finally, all operators move the program pointer by two positions except the operator Solving “Bridge Repair”, from day 7 of Advent of Code 2024, in 4ms with Python with a simple deductive algorithm.<\/p>\n\n Today I solved the problem from Advent of Code 2024, day 7<\/a> in 4ms using Python.\nMy first solution used brute-force and ran in 15 seconds.\nIt was very easy to implement and I got the job done quite quickly.\nBut then I thought about using a smarter algorithm, and I got a solution that's over 4,000 times faster and that runs in under 4ms.\nI will describe both solutions in this article.<\/p>\n (It should be obvious by now, but this article contains spoilers!)<\/p>\n In short, given a target integer and a list of integers, can we use addition, multiplication, and concatenation, to manipulate the list of integers to equal the target integer?\nThere is the added restriction that the operations are always evaluated from left to right and the integers cannot be reordered.<\/p>\n Here are some examples that are possible:<\/p>\n To make \\(5\\)<\/span> with \\(2\\)<\/span> and \\(3\\)<\/span> we can do \\(5 = 2 + 3\\)<\/span>.\nTo make \\(6\\)<\/span> with \\(2\\)<\/span> and \\(3\\)<\/span> we can multiply the two numbers and to make \\(23\\)<\/span> with \\(2\\)<\/span> and \\(3\\)<\/span> we can concatenate the two digits.\nFinally, to make \\(94\\)<\/span> with the integers \\(2\\)<\/span>, \\(4\\)<\/span>, \\(9\\)<\/span>, and \\(5\\)<\/span>, we start by multiplying, then we concatenate, and then we add.\nIf we use \\(||\\)<\/span> to represent concatenation, then \\(94 = ((2 \\times 4) || 9) + 5\\)<\/span>.<\/p>\n To determine if a certain list of numbers can be manipulated to match the target value, we can write a function that generates all distinct sequences of operators and then tries to use each sequence of operators in turn.<\/p>\n Using Advent of Code provides an input file with 850 tests<\/a> and this code takes slightly over 15 seconds to classify all 850 tests on my machine.\nNow, I will show you how you can speed this up by a factor of 2,000.<\/p>\n The faster algorithm is conceptually very simple.\nDisappointingly simple.\nInstead of working from the front to the back, trying random operations and seeing if they work out in the end, we work from the back to the front.<\/p>\n For example, consider the target The...<\/p>","summary":"Solving \u201cBridge Repair\u201d, from day 7 of Advent of Code 2024, in 4ms with Python with a simple deductive algorithm.","date_modified":"2025-07-23T16:49:02+02:00","tags":["algorithms","mathematics","programming","python","recursion"],"image":"\/user\/pages\/02.blog\/solving-bridge-repair-in-4ms-with-python\/thumbnail.webp"},{"title":"Problem #066 \u2013 regex crossword","date_published":"2024-08-27T12:00:00+02:00","id":"https:\/\/mathspp.com\/blog\/problems\/regex-crossword","url":"https:\/\/mathspp.com\/blog\/problems\/regex-crossword","content_html":" Can you solve this crossword where all hints are regular expressions?<\/p>\n\n At EuroPython 2024<\/a> I watched a lightning talk about the art of puzzle solving.\nIn it, the speaker showed the regex crossword puzzle that you can see above, that I took from this URL<\/a>.\nYou are supposed to fill the hexagonal grid with letters so that all of the regular expressions shown match.<\/p>\n If you don't know regular expressions, the puzzle only uses a simple subset of the syntax:<\/p>\n You can look this up and then you will be able to solve the challenge.\nYou can also use the site regex101<\/a> to help you check what each regular expression means.<\/p>\n Give it some thought!<\/p>\n<\/div>\n In the spirit of the puzzle hunting community, I will not share my solved grid here.\nIf you need help, feel free to email me<\/a> and we can talk it over.<\/p>\n P.S. if I understood correctly, in the puzzle hunting community you're not supposed to just fill in the grid.\nIn some way, somehow, you are supposed to be able to extract an English word or phrase from that filled puzzle without any extra information, and then you check that you got it correctly by \u201cchecking your answer spoiler-free\u201d in the website where I took the regex crossword from<\/a>\nI am still stuck in that step!<\/p>","summary":"Can you solve this crossword where all hints are regular expressions?","date_modified":"2025-07-23T16:49:02+02:00","tags":["mathematics","regex"],"image":"\/user\/pages\/02.blog\/03.problems\/p066-regex-crossword\/thumbnail.webp"},{"title":"Animating a rotating spiral","date_published":"2024-07-10T00:00:00+02:00","id":"https:\/\/mathspp.com\/blog\/animating-a-rotating-spiral","url":"https:\/\/mathspp.com\/blog\/animating-a-rotating-spiral","content_html":" With a couple of loops and a bit of maths you can create a rotating spiral.<\/p>\n\n\n
Hash of infinity<\/a><\/h2>\n
float(\"inf\")<\/code> and float(\"-inf\")<\/code> have two very special hashes:<\/p>\n>>> hash(float(\"inf\"))\n314159\n>>> hash(float(\"-inf\"))\n-314159<\/code><\/pre>\n>>> import math\n>>> math.pi\n3.141592653589793\n## ^^^^^<\/code><\/pre>\n-1<\/code> was<\/a>.<\/p>","summary":"Today I learned about a Python Easter Egg hidden in the hash of two special float values.","date_modified":"2025-10-20T22:34:56+02:00","tags":["algorithms","mathematics","python","programming"],"image":"\/user\/pages\/02.blog\/04.til\/126.hash-of-infinity\/thumbnail.webp"},{"title":"Animating a tree fractal","date_published":"2025-01-12T19:10:00+01:00","id":"https:\/\/mathspp.com\/blog\/animating-a-tree-fractal","url":"https:\/\/mathspp.com\/blog\/animating-a-tree-fractal","content_html":"![\"Animated](\"\/user\/pages\/02.blog\/animating-a-tree-fractal\/_tree.gif?decoding=auto&fetchpriority=auto\")
<\/p>\nDrawing the tree<\/a><\/h2>\n
def tree(screen, start, to, depth=0):\n if depth >= 15:\n return\n sx, sy = start\n tx, ty = to\n pygame.draw.line(\n screen,\n BLACK,\n (WIDTH \/\/ 2 + sx, HEIGHT \/\/ 2 - sy),\n (WIDTH \/\/ 2 + tx, HEIGHT \/\/ 2 - ty),\n 1,\n )\n direction = np.array([[tx - sx], [ty - sy]])\n for angle in [pi \/ 3, -pi \/ 3]:\n rot = np.array(\n [\n [cos(angle), -sin(angle)],\n [sin(angle), cos(angle)],\n ]\n )\n new_direction = ((rot @ ((direction) \/ 1.7))).reshape((2,))\n tree(screen, to, to + new_direction, depth + 1)<\/code><\/pre>\npygame.draw.line<\/code>, is that I decided to do all calculations as if the coordinate system were the standard one from maths class: the origin is at the centre of the pygame window and the values of the coordinate y increase by going up, not by going down.\nThen, I need to convert from this sytem to the pygame system right before drawing, which is why I add (WIDTH \/\/ 2, HEIGHT \/\/ 2)<\/code> to the points we're drawing and then for the y coordinate I need to use the symmetric value (-y<\/code> instead of y<\/code>) to compensate for the fact that the y axis is flipped.<\/p>\nfrom math import sin, cos, pi\nimport sys\n\nimport numpy as np\nimport pygame\nimport pygame.locals\n\nWIDTH, HEIGHT = 600, 400\n\nWHITE = (255, 255, 255)\nBLACK = (0, 0, 0)\n\nscreen = pygame.display.set_mode((WIDTH, HEIGHT))\n\ndef tree(screen, start, to, depth=0):\n ...\n\ndef main() -> None:\n screen.fill(WHITE)\n tree(screen, np.array((0, -200)), np.array((0, 0)))\n pygame.display.flip()\n pygame.image.save(screen, \"tree.png\")\n\n while True:\n for event in pygame.event.get():\n if event.type == pygame.locals.QUIT:\n pygame.quit()\n sys.exit(0)\n\nif __name__ == \"__main__\":\n main()<\/code><\/pre>\nAnimating the tree<\/a><\/h2>\n
tree<\/code> in two:<\/p>\nParsing the input<\/a><\/h2>\n
Register A: 64012472\nRegister B: 0\nRegister C: 0\n\nProgram: 2,4,1,7,7,5,0,3,1,7,4,1,5,5,3,0<\/code><\/pre>\nreadline<\/code>:<\/p>\nwith open(\"input.txt\", \"r\") as f:\n register_a = int(f.readline().split()[-1])\n register_b = int(f.readline().split()[-1])\n register_c = int(f.readline().split()[-1])\n _ = f.readline()\n program = [int(num) for num in f.readline().split()[-1].split(\",\")]\n\nprint(program, register_a, register_b, register_c)<\/code><\/pre>\nSolving part 1 with functional programming<\/a><\/h2>\n
lambda<\/code> functions, dunder methods<\/a>, and currying with functools.partial<\/code><\/a>, each list below represents one of the three parts of each opcode.<\/p>\nregisters = [0, 1, 2, 3, register_A, register_B, register_C]\nA, B, C = 4, 5, 6 # Indices of the named registers in the list `registers`.\n\ncomputations = [\n lambda o: registers[A] \/\/ pow(2, registers[o]), # ADV\n lambda o: registers[B] ^ o, # BXL\n lambda o: registers[o] % 8, # BST\n lambda o: ..., # JNZ\n lambda o: registers[B] ^ registers[C], # BXC\n lambda o: registers[o] % 8, # OUT\n lambda o: registers[A] \/\/ pow(2, registers[o]), # BDV\n lambda o: registers[A] \/\/ pow(2, registers[o]), # CDV\n]<\/code><\/pre>\no<\/code> in isolation, we're using the operand as a literal operand, whereas the list registers<\/code> maps an operand into its combo operand.\nBy using this list, we can map the numbers 0 through 3 to themselves and the indices 4, 5, and 6, to the registers A, B, and C, respectively, without having to use a conditional statement.<\/p>\nJNZ<\/code> has a lambda function that does nothing because there is no proper computation for this operator.<\/p>\nfrom functools import partial\n\noutput = []\nstate_updates = [\n partial(registers.__setitem__, A),\n partial(registers.__setitem__, B),\n partial(registers.__setitem__, B),\n lambda v: ...,\n partial(registers.__setitem__, B),\n output.append,\n partial(registers.__setitem__, B),\n partial(registers.__setitem__, C),\n]<\/code><\/pre>\n__setitem__<\/code><\/a> and the function functools.partial<\/code><\/a> to create a function that accepts a single value and that writes that value to the correct register in the list registers<\/code>.<\/p>\nJNZ...<\/code><\/p>","summary":"Reverse-engineering the program from \u201cChronospatial Computer\u201d, day 17 of Advent of Code 2024.","date_modified":"2025-07-23T16:49:02+02:00","tags":["algorithms","mathematics","modular arithmetic","programming","python","recursion"],"image":"\/user\/pages\/02.blog\/reverse-engineering-the-chronospatial-computer\/thumbnail.webp"},{"title":"Solving \u201cBridge Repair\u201d in 4ms with Python","date_published":"2024-12-07T19:00:00+01:00","id":"https:\/\/mathspp.com\/blog\/solving-bridge-repair-in-4ms-with-python","url":"https:\/\/mathspp.com\/blog\/solving-bridge-repair-in-4ms-with-python","content_html":"Problem statement<\/a><\/h2>\n
5: 2 3\n6: 2 3\n23: 2 3\n94: 2 4 9 5<\/code><\/pre>\nBrute-force solution, 15 seconds<\/a><\/h2>\n
itertools.product<\/code>, we can generate all sequences of distinct operators quite easily:<\/p>\nfrom itertools import product\nfrom operator import add, mul\n\nvalid_operators = [\n add,\n mul,\n lambda x, y: int(f\"{x}{y}\")\n]\n\ndef is_valid(target, operands):\n operator_sequences = product(valid_operators, repeat=len(operands) - 1)\n start, *head = operands\n for operators in operator_sequences:\n total = start\n for operator, operand in zip(operators, head):\n total = operator(total, operand)\n if total == target:\n return True\n return False\n\nprint(is_valid(94, [2, 4, 9, 5]))<\/code><\/pre>\nDeducing valid operations, 4 milliseconds<\/a><\/h2>\n
94<\/code> and the list of integers [2, 4, 9, 5]<\/code>.\nIf this is a possible case, then the final operation must be an addition.\nWhy?<\/p>\nProblem statement<\/a><\/h2>\n
![\"Puzzle](\"\/user\/pages\/02.blog\/03.problems\/p066-regex-crossword\/_puzzle.webp\" "\\"The")
\n
.<\/code>;<\/li>\n|<\/code>;<\/li>\n?<\/code>, +<\/code>, and *<\/code>;<\/li>\n[...]<\/code> and negated character sets with [^...]<\/code>; and<\/li>\n()<\/code> and group references with \\1<\/code>, \\2<\/code>, etc.<\/li>\n<\/ul>\nSolution<\/a><\/h2>\n