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# left inverse surjective

intros a'. If every "A" goes to a unique "B", and every "B" has a matching "A" then we can go back and forwards without being led astray. Surjective Function. g f = 1A is equivalent to g(f(a)) = a for all a ∈ A. Question: Prove That: T Has A Right Inverse If And Only If T Is Surjective. _\square A right inverse of f is a function: g : B ---> A. such that (f o g)(x) = x for all x. Inverse / Surjective / Injective. given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Next story A One-Line Proof that there are Infinitely Many Prime Numbers; Previous story Group Homomorphism Sends the Inverse Element to the Inverse … Qed. Hence, it could very well be that $$AB = I_n$$ but $$BA$$ is something else. Then we may apply g to both sides of this last equation and use that g f = 1A to conclude that a = a′. then f is injective iff it has a left inverse, surjective iff it has a right inverse (assuming AxCh), and bijective iff it has a 2 sided inverse. We are interested in nding out the conditions for a function to have a left inverse, or right inverse, or both. Tags: bijective bijective homomorphism group homomorphism group theory homomorphism inverse map isomorphism. iii) Function f has a inverse iff f is bijective. (a) Apply 4 (c) and (e) using the fact that the identity function is bijective. We want to show, given any y in B, there exists an x in A such that f(x) = y. (e) Show that if has both a left inverse and a right inverse , then is bijective and . Suppose g exists. Thus, to have an inverse, the function must be surjective. Thus setting x = g(y) works; f is surjective. The image on the left has one member in set Y that isn’t being used (point C), so it isn’t injective. The rst property we require is the notion of an injective function. distinct entities. record Surjective {f ₁ f₂ t₁ t₂} {From: Setoid f₁ f₂} {To: Setoid t₁ t₂} (to: From To): Set (f₁ ⊔ f₂ ⊔ t₁ ⊔ t₂) where field from: To From right-inverse-of: from RightInverseOf to-- The set of all surjections from one setoid to another. LECTURE 18: INJECTIVE AND SURJECTIVE FUNCTIONS AND TRANSFORMATIONS MA1111: LINEAR ALGEBRA I, MICHAELMAS 2016 1. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. Surjection vs. Injection. - destruct s. auto. i) ⇒. On A Graph . Math Topics. Equivalently, f(x) = f(y) implies x = y for all x;y 2A. Suppose $f\colon A \to B$ is a function with range $R$. PropositionalEquality as P-- Surjective functions. for bijective functions. Suppose f is surjective. Sep 2006 782 100 The raggedy edge. destruct (dec (f a')). There won't be a "B" left out. Nov 19, 2008 #1 Define $$\displaystyle f:\Re^2 \rightarrow \Re^2$$ by $$\displaystyle f(x,y)=(3x+2y,-x+5y)$$. The composition of two surjective maps is also surjective. Suppose f has a right inverse g, then f g = 1 B. Let f : A !B. We say that f is injective if whenever f(a 1) = f(a 2) for some a 1;a 2 2A, then a 1 = a 2. We say that f is surjective if for all b 2B, there exists an a 2A such that f(a) = b. intros A B a f dec H. exists (fun b => match dec b with inl (exist _ a _) => a | inr _ => a end). A: A → A. is defined as the. The same argument shows that any other left inverse b ′ b' b ′ must equal c, c, c, and hence b. b. b. Figure 2. Pre-University Math Help. (See also Inverse function.). "if a function is injective but not surjective, then it will necessarily have more than one left-inverse ... "Can anyone demonstrate why this is true? If h is a right inverse for f, f h = id B, so f is surjective by problem 4(e). In this case, the converse relation $${f^{-1}}$$ is also not a function. map a 7→ a. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. Read Inverse Functions for more. Similarly the composition of two injective maps is also injective. The inverse function g : B → A is defined by if f(a)=b, then g(b)=a. Let $f \colon X \longrightarrow Y$ be a function. It follows therefore that a map is invertible if and only if it is injective and surjective at the same time. Theorem right_inverse_surjective : forall {A B} (f : A -> B), (exists g, right_inverse f g) -> surjective … to denote the inverse function, which w e will define later, but they are very. If a function $$f$$ is not surjective, not all elements in the codomain have a preimage in the domain. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. (b) has at least two left inverses and, for example, but no right inverses (it is not surjective). Let f : A !B. The identity map. See the answer. This example shows that a left or a right inverse does not have to be unique Many examples of inverse maps are studied in calculus. 1.The map f is injective (also called one-to-one/monic/into) if x 6= y implies f(x) 6= f(y) for all x;y 2A. Proof. Let b ∈ B, we need to find an element a … An invertible map is also called bijective. Discrete Math: Jan 19, 2016: injective ZxZ->Z and surjective [-2,2]∩Q->Q: Discrete Math: Nov 2, 2015 Can someone please indicate to me why this also is the case? Show transcribed image text. The function is surjective because every point in the codomain is the value of f(x) for at least one point x in the domain. If y is in B, then g(y) is in A. and: f(g(y)) = (f o g)(y) = y. Thus, π A is a left inverse of ι b and ι b is a right inverse of π A. Secondly, Aluffi goes on to say the following: "Similarly, a surjective function in general will have many right inverses; they are often called sections." A function $g\colon B\to A$ is a pseudo-inverse of $f$ if for all $b\in R$, $g(b)$ is a preimage of $b$. Definition (Iden tit y map). Behavior under composition. T o define the inv erse function, w e will first need some preliminary definitions. Showing f is injective: Suppose a,a ′ ∈ A and f(a) = f(a′) ∈ B. Thus f is injective. Thread starter Showcase_22; Start date Nov 19, 2008; Tags function injective inverse; Home. id. Peter . Let A and B be non-empty sets and f: A → B a function. In other words, the function F maps X onto Y (Kubrusly, 2001). When A and B are subsets of the Real Numbers we can graph the relationship. This problem has been solved! Forums. unfold injective, left_inverse. De nition 2. A function is called to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. It means that each and every element “b” in the codomain B, there is exactly one element “a” in the domain A so that f(a) = b. here is another point of view: given a map f:X-->Y, another map g:Y-->X is a left inverse of f iff gf = id(Y), a right inverse iff fg = id(X), and a 2 sided inverse if both hold. Prove That: T Has A Right Inverse If And Only If T Is Surjective. a left inverse must be injective and a function with a right inverse must be surjective. Showcase_22. We will show f is surjective. Expert Answer . Implicit: v; t; e; A surjective function from domain X to codomain Y. What factors could lead to bishops establishing monastic armies? A surjective function, also called a surjection or an onto function, is a function where every point in the range is mapped to from a point in the domain. De nition. Proof. So let us see a few examples to understand what is going on. Introduction to the inverse of a function Proof: Invertibility implies a unique solution to f(x)=y Surjective (onto) and injective (one-to-one) functions Relating invertibility to being onto and one-to-one Determining whether a transformation is onto Simplifying conditions for invertibility Showing that inverses are linear. F or example, we will see that the inv erse function exists only. Let f: A !B be a function. (Note that these proofs are superfluous,-- given that Bijection is equivalent to Function.Inverse.Inverse.) Any function that is injective but not surjective su ces: e.g., f: f1g!f1;2g de ned by f(1) = 1. Function has left inverse iff is injective. De nition 1.1. ... Bijective functions have an inverse! Prove that: T has a right inverse if and only if T is surjective. For instance, if A is the set of non-negative real numbers, the inverse map of f: A → A, x → x 2 is called the square root map. 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A. such that (f o g)(x) = x for all x. Inverse / Surjective / Injective. given $$n\times n$$ matrix $$A$$ and $$B$$, we do not necessarily have $$AB = BA$$. Next story A One-Line Proof that there are Infinitely Many Prime Numbers; Previous story Group Homomorphism Sends the Inverse Element to the Inverse … Qed. Hence, it could very well be that $$AB = I_n$$ but $$BA$$ is something else. Then we may apply g to both sides of this last equation and use that g f = 1A to conclude that a = a′. then f is injective iff it has a left inverse, surjective iff it has a right inverse (assuming AxCh), and bijective iff it has a 2 sided inverse. We are interested in nding out the conditions for a function to have a left inverse, or right inverse, or both. Tags: bijective bijective homomorphism group homomorphism group theory homomorphism inverse map isomorphism. iii) Function f has a inverse iff f is bijective. (a) Apply 4 (c) and (e) using the fact that the identity function is bijective. We want to show, given any y in B, there exists an x in A such that f(x) = y. (e) Show that if has both a left inverse and a right inverse , then is bijective and . Suppose g exists. Thus, to have an inverse, the function must be surjective. Thus setting x = g(y) works; f is surjective. The image on the left has one member in set Y that isn’t being used (point C), so it isn’t injective. The rst property we require is the notion of an injective function. distinct entities. record Surjective {f ₁ f₂ t₁ t₂} {From: Setoid f₁ f₂} {To: Setoid t₁ t₂} (to: From To): Set (f₁ ⊔ f₂ ⊔ t₁ ⊔ t₂) where field from: To From right-inverse-of: from RightInverseOf to-- The set of all surjections from one setoid to another. LECTURE 18: INJECTIVE AND SURJECTIVE FUNCTIONS AND TRANSFORMATIONS MA1111: LINEAR ALGEBRA I, MICHAELMAS 2016 1. A function is bijective if and only if has an inverse November 30, 2015 De nition 1. Surjection vs. Injection. - destruct s. auto. i) ⇒. On A Graph . Math Topics. Equivalently, f(x) = f(y) implies x = y for all x;y 2A. Suppose $f\colon A \to B$ is a function with range $R$. PropositionalEquality as P-- Surjective functions. for bijective functions. Suppose f is surjective. Sep 2006 782 100 The raggedy edge. destruct (dec (f a')). There won't be a "B" left out. Nov 19, 2008 #1 Define $$\displaystyle f:\Re^2 \rightarrow \Re^2$$ by $$\displaystyle f(x,y)=(3x+2y,-x+5y)$$. The composition of two surjective maps is also surjective. Suppose f has a right inverse g, then f g = 1 B. Let f : A !B. We say that f is injective if whenever f(a 1) = f(a 2) for some a 1;a 2 2A, then a 1 = a 2. We say that f is surjective if for all b 2B, there exists an a 2A such that f(a) = b. intros A B a f dec H. exists (fun b => match dec b with inl (exist _ a _) => a | inr _ => a end). A: A → A. is defined as the. The same argument shows that any other left inverse b ′ b' b ′ must equal c, c, c, and hence b. b. b. Figure 2. Pre-University Math Help. (See also Inverse function.). "if a function is injective but not surjective, then it will necessarily have more than one left-inverse ... "Can anyone demonstrate why this is true? If h is a right inverse for f, f h = id B, so f is surjective by problem 4(e). In this case, the converse relation $${f^{-1}}$$ is also not a function. map a 7→ a. Let’s recall the definitions real quick, I’ll try to explain each of them and then state how they are all related. Read Inverse Functions for more. Similarly the composition of two injective maps is also injective. The inverse function g : B → A is defined by if f(a)=b, then g(b)=a. Let $f \colon X \longrightarrow Y$ be a function. It follows therefore that a map is invertible if and only if it is injective and surjective at the same time. Theorem right_inverse_surjective : forall {A B} (f : A -> B), (exists g, right_inverse f g) -> surjective … to denote the inverse function, which w e will define later, but they are very. If a function $$f$$ is not surjective, not all elements in the codomain have a preimage in the domain. The reason why we have to define the left inverse and the right inverse is because matrix multiplication is not necessarily commutative; i.e. (b) has at least two left inverses and, for example, but no right inverses (it is not surjective). Let f : A !B. The identity map. See the answer. This example shows that a left or a right inverse does not have to be unique Many examples of inverse maps are studied in calculus. 1.The map f is injective (also called one-to-one/monic/into) if x 6= y implies f(x) 6= f(y) for all x;y 2A. Proof. Let b ∈ B, we need to find an element a … An invertible map is also called bijective. Discrete Math: Jan 19, 2016: injective ZxZ->Z and surjective [-2,2]∩Q->Q: Discrete Math: Nov 2, 2015 Can someone please indicate to me why this also is the case? Show transcribed image text. The function is surjective because every point in the codomain is the value of f(x) for at least one point x in the domain. If y is in B, then g(y) is in A. and: f(g(y)) = (f o g)(y) = y. Thus, π A is a left inverse of ι b and ι b is a right inverse of π A. Secondly, Aluffi goes on to say the following: "Similarly, a surjective function in general will have many right inverses; they are often called sections." A function $g\colon B\to A$ is a pseudo-inverse of $f$ if for all $b\in R$, $g(b)$ is a preimage of $b$. Definition (Iden tit y map). Behavior under composition. T o define the inv erse function, w e will first need some preliminary definitions. Showing f is injective: Suppose a,a ′ ∈ A and f(a) = f(a′) ∈ B. Thus f is injective. Thread starter Showcase_22; Start date Nov 19, 2008; Tags function injective inverse; Home. id. Peter . Let A and B be non-empty sets and f: A → B a function. In other words, the function F maps X onto Y (Kubrusly, 2001). When A and B are subsets of the Real Numbers we can graph the relationship. This problem has been solved! Forums. unfold injective, left_inverse. De nition 2. A function is called to be bijective or bijection, if a function f: A → B satisfies both the injective (one-to-one function) and surjective function (onto function) properties. It means that each and every element “b” in the codomain B, there is exactly one element “a” in the domain A so that f(a) = b. here is another point of view: given a map f:X-->Y, another map g:Y-->X is a left inverse of f iff gf = id(Y), a right inverse iff fg = id(X), and a 2 sided inverse if both hold. Prove That: T Has A Right Inverse If And Only If T Is Surjective. a left inverse must be injective and a function with a right inverse must be surjective. Showcase_22. We will show f is surjective. Expert Answer . Implicit: v; t; e; A surjective function from domain X to codomain Y. What factors could lead to bishops establishing monastic armies? A surjective function, also called a surjection or an onto function, is a function where every point in the range is mapped to from a point in the domain. De nition. Proof. So let us see a few examples to understand what is going on. Introduction to the inverse of a function Proof: Invertibility implies a unique solution to f(x)=y Surjective (onto) and injective (one-to-one) functions Relating invertibility to being onto and one-to-one Determining whether a transformation is onto Simplifying conditions for invertibility Showing that inverses are linear. F or example, we will see that the inv erse function exists only. Let f: A !B be a function. (Note that these proofs are superfluous,-- given that Bijection is equivalent to Function.Inverse.Inverse.) Any function that is injective but not surjective su ces: e.g., f: f1g!f1;2g de ned by f(1) = 1. Function has left inverse iff is injective. De nition 1.1. ... Bijective functions have an inverse! Prove that: T has a right inverse if and only if T is surjective. For instance, if A is the set of non-negative real numbers, the inverse map of f: A → A, x → x 2 is called the square root map. If g is a left inverse for f, g f = id A, which is injective, so f is injective by problem 4(c). Showing g is surjective: let a ∈ a date Nov 19, 2008 ; Tags function injective inverse Home., w e will first need some preliminary definitions BA\ ) is something else to. Thread starter Showcase_22 ; Start date Nov 19, 2008 ; Tags function inverse! } } \ ) is something else surjective: let f: a A..: a → B a function defined by if f ( a ) a. I, MICHAELMAS 2016 1 an element a … is surjective group group... Non-Empty sets and f ( x ) = f ( a ) =b, then g ( B has! [ /math ] be a Bijection } \ ) is also surjective prove that: T has inverse! Is something else: suppose a, a ′ ∈ a and B are subsets of Real. ( it is not surjective ) please indicate to me why this also is the case define later, they... 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text:
# UTS ProgSoc ProgComp 2020

UTS Progsoc 2020 Competitive Programming Competition, 2020

Today, I participated in the ProgSocUTS 2020 Programming Competition, placing 3rd!

The programming competition, otherwise known as ProgComp was a 3-hour competitive programming event, open to all UTS students, alumni and the general public. This competition aimed to attempt to complete all of the programming competitions within the 3-hour time frame adhering to the questions requirements as well as time complexity requirements. Whilst we faced challenges, especially near the end, where we just managed to finish the last questions mere minutes before the deadline. It was an insightful event.

Throughout this event, we attempted a total of 7 challenges, however only in the end getting 5 of them within the prescribed time slot. The flags we attempted include; "ASCII Squares", "Mario's Big Jump", "Mark and Toys", "Extreme FizzBuzz", "Jess and Cookies", "Largest Rectangle" as well as "Helicopter Crash". Whilst all these challenges were difficult, all ranged with varying degrees of difficulty and varying locations of where we struggled. Overall this competition made us jump around a lot in our analytical and programmatical thinking and has in turn further highlighted sections where myself and my time need to improve on in the future with.

I'd like to thank my team "FAANG Or Bust", once again, for partaking with me, as without them I'd not be able to place anywhere as well without you guys.

## ASCII Squares

The first question of this competition was a simple one called "ASCII Squares" in ASCII squares as the name suggests, we had to make a program that given an integer input produced an ASCII square with the dimensions of the input as an output. This question wasn't too challenging and took us about 5 minutes to implement. Overall it was a fun but simple challenge that helped prime us into a more competitive mindset for the harder incoming questions.

### ASCII Squares Question

This simple task has therefore been offered to you as a sanity test: write a program that draws a square, using asterisks and SPACEs.

Input Format

A single integer, N, where 2 <= N <= 40. This will be the size of your square.

Constraints

N will be between 2 and 40

Output Format

There will be N horizontal lines in total.

The first and Nth lines will have asterisks (*) occupying every odd-numbered position on the line from positions 1 to (2N - 1). All even-numbered positions will be filled with SPACEs.

For all other lines (if any), only positions 1 and (2N - 1) will have asterisks. All other positions will be filled with SPACEs.

There are no SPACEs nor TABs following the (2N - 1)th position on each line.

Sample Input 0

5

Sample Output 0

* * * * *
*       *
*       *
*       *
* * * * *

### Solution

'use strict';

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

process.stdin.on('data', function (inputStdin) {
inputString += inputStdin;
});

process.stdin.on('end', function () {
inputString = inputString.split('\n');

main();
});

return inputString[currentLine++];
}

function main() {
const size = parseInt(readLine().trim(), 10);
if(size === 1){
console.log('*')
return;
}
const edge = '* '.repeat(size).trim();
const middle = '  '.repeat(size - 2);
const output = [];
output.push(edge);
for(let i = 0; i < size - 2; i++){
output.push(* ${middle}*); } output.push(edge); console.log(output.join('\n')); }  ## Mario's Big Jump This was the second challenge of the competition and wow! This was a lot more challenging. Whilst the first one was very trivial, boasting a 100% completion rate for all contestants, this one quickly shattered the idea that we could breeze through this competition in 30 or so minutes. In this challenge, which was shockingly similar to some HSC Physics questions I had to do last year, was to calculate how far Mario could jump given a running speed, jump speed and target distance. Whilst I may have been going with a more elegant approach initially, getting my pen and paper out and drafting up a general formula for it, my teammates went for a more "direct" approach, calculating the player's movement each "tick". Whilst I did draft up a formula for it in time, deriving it from some projectile motion equations: $v = u + at$ $v^2 = u^2 + 2as$ $s = ut + 1/2 at^2$ By the time I derived the equation, my teammates had already solved the problem and moved on to the next challenge. During this challenge, I came to realize, whilst my way would lead to a nicer and albeit faster solution. In a time-based competition, this was not worth doing, and the cheap but dirty solution was the more ideal solution. ### Mario's Big Jump Question In the Mushroom Kingdom, Mario is working out how far he can jump. The gravity in the kingdom will make him fall at 5 meters/second, and he will continue to move at a constant rate while he's in the air. For example,if Mario jumps up with a speed of 10 meters/second while running at 30 meters/second, the results are: As a result, Mario will jump 150 meters and stay in the air for 5 seconds. For simplicity, we will calculate Marios's position at the end of every second, even though more accurate calculations are possible by if shorter intervals are used. Input Format The input consists of a list of three numbers representing a jump to a target: 1. The speed Mario is running (in meters/second) 2. The speed Mario jumps up into the air (in meters/second) 3. The distance to the target (in meters) Constraints None Output Format Print the distance and height of Mario's jump after each second as a pair of numbers on a new line, starting at one second. Repeat each second until Mario lands (height reaches zero or below). Use integers for calculating results. For each target, output "yes" if Mario is within 3 meters of the target when he lands (height <= zero), else print "no". Sample Input 0 30 10 150  Sample Output 0 30 10 60 15 90 15 120 10 150 0 yes  ### Solution // input: function processData(input) { let [forwardSpeed, jumpSpeed, target] = input; let height = 0; let pos = 0; do { pos += forwardSpeed; height += jumpSpeed; jumpSpeed -= 5; console.log(pos, height); } while (height > 0) if (Math.abs(pos - target) < 3) { console.log('yes') } else { console.log('no') } } /* loopies untill dist to target === 0 input: [run speed, jump speed, disto target] INTS only output: print("dist height") after each second as number pair in new line */ process.stdin.resume(); process.stdin.setEncoding("ascii"); _input = ""; process.stdin.on("data", function (input) { _input += input; }); process.stdin.on("end", function () { let str = _input.replace(/\n/g, ''); processData(str.split(' ').map(i => parseInt(i, 10))); });  ## Mark and Toys Marks and Toys was another refreshing simple and quick question we completed in 10 to 15 minutes. In this one, we were given an array of items a mom and dad could buy for their first child and a specified budget. In this one, we just had to find the most items they could buy for a specified budget. This question was quite trivial and resulted in us more or less sorting the array list into an order and then iterating in a while loop checking how many items the couple could buy until the loop became less than 0. ### Mark and Toys Question Mark and Jane are very happy after having their first child. Their son loves toys, so Mark wants to buy some. There are a number of different toys lying in front of him, tagged with their prices. Mark has only a certain amount to spend, and he wants to maximize the number of toys he buys with this money. Given a list of prices and an amount to spend, what is the maximum number of toys Mark can buy? For example, if $prices = [1,2,3,4]$ and Mark has $k=7$ to spend, he can buy items $[1,2,3]$ for $6$, or $[3,4]$ for $7$ units of currency. He would choose the first group of $3$ items. Function Description Complete the function maximumToys in the editor below. It should return an integer representing the maximum number of toys Mark can purchase. maximumToys has the following parameter(s): • prices: an array of integers representing toy prices • k: an integer, Mark's budget Input Format The first line contains two integers, $n$ and $k$, the number of priced toys and the amount Mark has to spend. The next line contains $n$ space-separated integers $prices[i]$. Constraints $1 \leq n \leq 10^5$ $1 \leq k \leq 10^9$ $1 \leq prices[i] \leq 10^9$ A toy can't be bought multiple times. Output Format An integer that denotes the maximum number of toys Mark can buy for his son. Sample Input 7 50 1 12 5 111 200 1000 10  Sample Output 4  Explanation He can buy only $4$ toys at most. These toys have the following prices: $[1,12,5,10]$. ### Solution 'use strict'; const fs = require('fs'); process.stdin.resume(); process.stdin.setEncoding('utf-8'); let inputString = ''; let currentLine = 0; process.stdin.on('data', inputStdin => { inputString += inputStdin; }); process.stdin.on('end', _ => { inputString = inputString.replace(/\s*$/, '')
.split('\n')
.map(str => str.replace(/\s*\$/, ''));

main();
});

return inputString[currentLine++];
}

/*
input:
2 number: max toys num, marks budget
array: [prices]

i=0;
orderList
for(marks > 0) {
subtrack marks
i++
}

*/

// Complete the maximumToys function below.
function maximumToys(prices, k) {
// prices = [int]

const sorted = prices.sort((a, b) => a - b);

let count = 0;

let budget = k;

for(let i = 0; i < sorted.length; i++){
console.log(budget, sorted[i]);
if(budget - sorted[i] < 0){
return i;
}
budget -= sorted[i];
}

return sorted.length;

}

function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

const nk = readLine().split(' ');

const n = parseInt(nk[0], 10);

const k = parseInt(nk[1], 10);

const prices = readLine().split(' ').map(pricesTemp => parseInt(pricesTemp, 10));

let result = maximumToys(prices, k);

ws.write(result + "\n");

ws.end();
}

## Extreme FizzBuzz

This one also, was too was not too challenging, whilst it did sound somewhat complicated on paper, the actual implementation was quite quick. Thankfully a week or so previously I had watched a Tom Scott video regarding a more expandable and "scalable" FizzBuzz implementation, if that ever was a thing, so this solution was churned out quite quickly as well. The logic with this question was, having a normal loop which iterated through all the numbers and then having a second inner loop which looped through all the conditional operators, if any of them were true it be added to the response string, if multiple were true they would also be appended onwards. Whilst this may have been a dirty solution, I still can't think of a "clean" solution to this question and even in a non-competitional environment, I'd probably implement it the same way.

### Extreme FizzBuzz Question

The traditional FizzBuzz challenge is read a list of numbers, and for multiples of 3 print Fizz, multiples of 5 print Buzz, and print FizzBuzz if both (i.e. a multiple of 15).

We want a program where we can input a list of arbitrary rules that follow the same pattern for a range of numbers.

Input Format

The first line will have 3 numbers, the number of rules (rLen), the start of the range to print, and the end of the range to print.

The remaining rLen lines will each have a number and a word

Constraints

There will be at most 10 rules, and the program range print up to 1000 numbers

Output Format

A list of numbers from the start to the end of the range (inclusive), that follow the rules of the game and swap out the correct numbers with words, and join words for numbers that match multiple rules.

Sample Input 0

2 1 16
3 Fizz
5 Buzz

Sample Output 0

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16

Sample Input 1

3 10 24
2 Even
3 Triple

Sample Output 1

11
13
Even
Triple
17
EvenTriple
19
Triple
Even
23

### Solution

function processData(input) {
const lines = input.split('\n');
const [rc, start, end] = lines[0].split(' ').map(i => parseInt(i));
const rules = lines.slice(1)
.map(r => r.split(' '))
.map(r => ({
n: parseInt(r[0]),
txt: r[1]
}));
//   console.log(start, end, rules);

// loop
for (let i = start; i <= end; i++) {
let str = rules.map(r => {
if (i % r.n === 0) {
return r.txt;
}
return ''
}).join('')
if (str === '') {
str = i.toString();
}
console.log(str);
}

}

/*
input
[NumberOfRules, StartingIndex, EndIndex]

3 10 24
2 Even
3 Triple

output
Even
11
*/

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});

process.stdin.on("end", function () {
processData(_input);
});

## Jesse and Cookies

Now we get to the hard one, this question definitely made us toss and turn trying to get a solution and certainly made our heads bang at a couple of walls a handful of times. Whilst this question was easy to implement, we kept failing the hidden test cases, if not from random errors, from our programming from running too long. Initially, we thought we had a loop failing to terminate, but after reviewing the question we discovered that it had a time-complexity requirement of efficiency level requirement before passing, this one took us a good hour to solve as our team hasn't yet studied DSA and will most likely cover this content next year.

We initially assumed it was due to JavaScript being an interpreted language and just, in general, being unoptimized, so we opted to use... python... another interpreted language. Whilst it was certainly faster, during our benchmarks the only issue we failed to account for was that NumPY was an imported library, and this competition wouldn't allow it.

As a result, we had to jump to a third language at this point! After solving the question two times already we jumped to C#. Whilst this didn't immediately solve the problem, it did bring the time down dramatically, we were just a couple of hundred milliseconds away from passing now!

In the end, we stumbled around in libraries and discovered the best way to implement this solution was swapping from LinkedLists and implementing a priority queue solution using the stack and heap. Through this method, instead of knowing the entire order of every element, we instead only needed to know the top two biggest or smallest elements, saving a dramatic amount of time.

This question, whilst perhaps being trivial for someone strong in DSA was definitely a challenging and frustrating question for us. I can't wait to pick up DSA and see how our team goes next year with that additional knowledge, who knows maybe we'll become first next time :)

### Jesse and Cookies Question

Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value $K$. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with:

$sweetness = (1 * Least sweet cookie + 2 * 2nd Least sweet cookie)$.

He repeats this procedure until all the cookies in his collection have a sweetness $\geq K$.

You are given Jesse's cookies. Print the number of operations required to give the cookies a sweetness $\geq K$. Print $-1$ if this isn't possible.

Input Format

The first line consists of integers $N$, the number of cookies and $K$, the minimum required sweetness, separated by a space.
The next line contains $N$ integers describing the array $A$ where $A_i$ is the sweetness of the $i^{th}$ cookie in Jesse's collection.

Constraints

$1 \leq N \leq 10^6$

$0 \leq K \leq 10^9$

$0 \leq A_i \leq 10^6$

Output Format

Output the number of operations that are needed to increase the cookie's sweetness $\geq K$.
Output $-1$ if this isn't possible.

Sample Input

6 7
1 2 3 9 10 12

Sample Output

2

Explanation

Combine the first two cookies to create a cookie with sweetness = $1 * 1 + 2 * 2 = 5$
After this operation, the cookies are $[3,5,9,10,12]$.
Then, combine the cookies with sweetness $3$ and sweetness $5$, to create a cookie with resulting sweetness = $1 * 3 + 2 * 5 = 13$

Now, the cookies are $[9,10,12,13$.
All the cookies have a sweetness $\geq 7$.

Thus, $2$ operations are required to increase the sweetness.

### Solutions

'use strict';

const fs = require('fs');

process.stdin.resume();
process.stdin.setEncoding('utf-8');

let inputString = '';
let currentLine = 0;

process.stdin.on('data', inputStdin => {
inputString += inputStdin;
});

process.stdin.on('end', _ => {
inputString = inputString.trim().split('\n').map(str => str.trim());

main();
});

return inputString[currentLine++];
}

/*
* Complete the cookies function below.
*/
function cookies(k, A) {
/*
* Write your code here.

Input:

6 7
1 2 3 9 10 12

Output:
2

OrderArray
Get Least Two Sweetness

*/

//   let op = 0;

const check = () => {
return A.reduce((c, n) => c && n >= k, A, true)
}

let steps = 0
while (!check()) {
A = A.sort((a, b) => a - b);
if (A.length <= 1) {
return -1;
}

const [first, second] = A.splice(0, 2);
A.push(first + 2 * second);
steps++;
}

return steps

// return (op === 0)? -1 : op;
}

function main() {
const ws = fs.createWriteStream(process.env.OUTPUT_PATH);

const nk = readLine().split(' ');

const n = parseInt(nk[0], 10);

const k = parseInt(nk[1], 10);

const A = readLine().split(' ').map(ATemp => parseInt(ATemp, 10));

let result = cookies(k, A);

ws.write(result + "\n");

ws.end();
}

The solution in python goes as follows

### Python Solution

#!/bin/python3

import os
import sys
import numpy as np;

#
# Complete the cookies function below.
#

def check(A, k):
for a in A:
if a < k: return False
return True

A = np.array(A)

steps = 0
while not check(A, k):

#let steps = 0
#while (!check()) {
#  A = A.sort((a, b) => a - b);
#  if (A.length <= 1) {
#    return -1;
#  }

#  const [first, second] = A.splice(0, 2);
#  A.push(first + 2 * second);
#  steps++;
#}

return steps

if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')

nk = input().split()

n = int(nk[0])

k = int(nk[1])

A = list(map(int, input().rstrip().split()))

result = cookies(k, A)

fptr.write(str(result) + '\n')

fptr.close()

For our final, and more importingly working solution, we ended up using C#, with it finally passing this time. A key aspect we learnt though in this competition was the need of good data structures, whilst I don't have the original code we used for the solution on me as it was done by another team mate, our code followed something similar to this, but replaced with priority queues using stacks and heaps.

### C# Solution (Finally Working)

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;

class Solution
{

/*
* Complete the cookies function below.
*/

static bool check(IEnumerable<int> A, int k)
{
foreach(var a in A) {
if(a < k) return false;
}
return true;
}

static int cookies(int k, List<int> A)
{
/*
* Write your code here.
*/

A.Sort();
var L = new LinkedList<int>(A);

int steps = 0;
while (!check(L, k))
{
foreach(var node in A)
int first = -1;
int second = -1;
int firstVal = -1;
int secondVal = -1;
for(int i = 0; i < A.Count; i++){

}
}
return 0;

}

public static void Main(string[] args)
{
Console.WriteLine(1);
TextWriter textWriter = new StreamWriter(@System.Environment.GetEnvironmentVariable("OUTPUT_PATH"), true);

string[] nk = Console.ReadLine().Split(' ');

int n = Convert.ToInt32(nk[0]);

int k = Convert.ToInt32(nk[1]);

int[] A = Array.ConvertAll(Console.ReadLine().Split(' '), ATemp => Convert.ToInt32(ATemp))
;
int result = cookies(k, A.ToList());

textWriter.WriteLine(result);

textWriter.Flush();
textWriter.Close();
}
}

Solution.Main(new string[0]);

## Helicopter Crash

Unlike the cookies question, this one thankfully wasn't DSA heavy in regards to it wanting an efficient solution, however, it was hard in its own terms with it requiring some complex critical and abstract thinking. With this question we finished down to the hair, finishing mere minutes before the deadline. With this question whilst our critical thinking was there, or close we struggled to get the right answer initially from the question being somewhat vague.

Initially, we tried rounding, rounding up or rounding down for each value, however, through the 4th or 5th iteration of the loop we'd get discrepancies between the solution and the answer. This was the most frustrating part of the competition, next to the Cookies question. Every time we'd round up, round down or just in the general round, regardless of what we did, we'd be either 1 above or 1 below the intended answer.

After thinking, and clarifying the requirements with the event organizers, we discovered the actual solution. Instead of rounding the actual integer value, we had to round the outputted result! A simple, straightforward but overall overlooked answer was the solution to the challenge. In our panic, we resulted in trying to brute-forcing the question, instead of standing back, looking clearly at the question and thinking "What needs to be rounded" we instead opted to panic.

With this last flag solved, we concluded the competition, just in the nick of times, jumping up from 8th place way up to 3rd!

### Helicopter Crash Question

This was the last challenge we completed, also if not the most challenging question we faced. Whilst not DSA heavy on optimization like the Cookies question, it had its own fair share of issues and required some unique critical thinking skills that got me lost for a good portion of this question before landing the solution, unlike our poor old helicopter here :(

### Helicopter Crash Question

In the next scene, a helicopter will be crashing into the side of a building, and the studio has a few ideas of how to crash it. Given these potential impact points, they want to know what the explosion would look like. The parts of the building will stretch on impact, causing ripples that need to be accounted for.

To model the surface of a building, you can use a two-dimensional array of doubles, with a height of 7 and width of 5. The value of each point in the array indicates how far that part of the building surface has been stretched. You then apply an iterative finite-difference technique to model the ripple effects of the crash as follows:

1. Begin with all points at 0.0 (rest state) except the impact point, which is set to the impact value.
2. Calculate the new position of a point in the surface as the average of 1) current location and 2) average of neighbors.
3. Update all points at once.

The outermost positions of the building are assumed to always remain at 0. For simplicity, you may assume your building has a height of 7 and a width of 5.

Note that it is important to store changes to the array and apply them all at once at the end of the time step. Otherwise, algorithms utilize a mixture of old and new simulation data, which can yield undesirable results.

Input Format

The input will consist of four integer numbers. The first two numbers indicating the point of impact (height H, width W). The third number is the impact strength (S); The final number is how many time steps of the simulation should be displayed (T).

Constraints

H and W will fit within the buildings dimensions (7x5) S will be a value from 1-10000 T will be a value from 1-100

Output Format

For each time step of the simulation, the program should print "Time " (starting at time 0), then the state of the building at that timestep. The building state can be printed as a 7x5 array of numbers, with each floor of the building on a separate line. Even though the values are stored as doubles, for clarity the program should convert it and just print an integer values.

Sample Input 0

2 2 100 3

Sample Output 0

Time 0
0    0    0    0    0
0    0    0    0    0
0    0    100  0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
Time 1
0    0    0    0    0
0    6    6    6    0
0    6    50   6    0
0    6    6    6    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
Time 2
0    0    0    0    0
0    7    7    7    0
0    7    28   7    0
0    7    7    7    0
0    0    1    0    0
0    0    0    0    0
0    0    0    0    0

Sample Input 1

3 3 100 20

Sample Output 1

Time 0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    100  0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
Time 1
0    0    0    0    0
0    0    0    0    0
0    0    6    6    0
0    0    6    50   0
0    0    6    6    0
0    0    0    0    0
0    0    0    0    0
Time 2
0    0    0    0    0
0    0    0    0    0
0    0    7    7    0
0    1    7    26   0
0    0    7    7    0
0    0    0    0    0
0    0    0    0    0
Time 3
0    0    0    0    0
0    0    1    1    0
0    1    6    6    0
0    2    7    15    0
0    1    6    6    0
0    0    1    1    0
0    0    0    0    0
Time 4
0    0    0    0    0
0    0    1    1    0
0    1    5    5    0
0    2    6    9    0
0    1    5    5    0
0    0    1    1    0
0    0    0    0    0
Time 5
0    0    0    0    0
0    1    1    1    0
0    2    4    4    0
0    2    5    6    0
0    2    4    4    0
0    1    1    1    0
0    0    0    0    0
Time 6
0    0    0    0    0
0    1    1    1    0
0    1    3    3    0
0    2    4    4    0
0    1    3    3    0
0    1    1    1    0
0    0    0    0    0
Time 7
0    0    0    0    0
0    0    1    1    0
0    1    3    2    0
0    2    3    3    0
0    1    3    2    0
0    0    1    1    0
0    0    0    0    0
Time 8
0    0    0    0    0
0    0    1    1    0
0    1    2    2    0
0    2    3    2    0
0    1    2    2    0
0    0    1    1    0
0    0    0    0    0
Time 9
0    0    0    0    0
0    0    1    0    0
0    1    2    1    0
0    1    2    2    0
0    1    2    1    0
0    0    1    0    0
0    0    0    0    0
Time 10
0    0    0    0    0
0    0    1    0    0
0    1    2    1    0
0    1    2    1    0
0    1    2    1    0
0    0    1    0    0
0    0    0    0    0
Time 11
0    0    0    0    0
0    0    0    0    0
0    1    1    1    0
0    1    1    1    0
0    1    1    1    0
0    0    0    0    0
0    0    0    0    0
Time 12
0    0    0    0    0
0    0    0    0    0
0    0    1    1    0
0    1    1    1    0
0    0    1    1    0
0    0    0    0    0
0    0    0    0    0
Time 13
0    0    0    0    0
0    0    0    0    0
0    0    1    0    0
0    0    1    1    0
0    0    1    0    0
0    0    0    0    0
0    0    0    0    0
Time 14
0    0    0    0    0
0    0    0    0    0
0    0    1    0    0
0    0    1    0    0
0    0    1    0    0
0    0    0    0    0
0    0    0    0    0
Time 15
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    1    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
Time 16
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
Time 17
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
Time 18
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
Time 19
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0
0    0    0    0    0

### Solution

function processData(input) {
//Enter your code here
/*
Helicopter crash
Impact H, Width W, Third S, Final Steps T

Building is 7x5

Starts iyt 199
gives 6 units surrounding it val 50
goes up to 7 units surround val 28 now

*/

// let h, w, s, t = 0;
//Test case 0:
const [h, w, s, t] = input.split(' ').map(i => parseInt(i));
// h = 2;
// w = 2;
// s = 100;
// t = 3;

let arr = new Array(7).fill(0).map(() => new Array(5).fill(0));

arr[h][w] = s;

const isInside = (x, y) => {
if (x < 0 || y < 0) return false;
if (x >= 5 || y >= 7) return false;
return true;
}

for (var i = 0; i < t; i++) {
console.log("Time " + i);
console.log(arr.map(a => a.map(i => Math.floor(i)).join('\t')).join('\n'))

arr = new Array(7).fill(0).map((_, y) => new Array(5).fill(0).map((_, x) => {
if(x === 0 || y === 0 || x === 4 || y === 6) return 0;
sum = 0;
for (let xx = -1; xx <= 1; xx++) {
for (let yy = -1; yy <= 1; yy++) {
if(xx === 0 && yy === 0) continue;
if(isInside(xx + x, yy + y)){
sum += arr[yy + y][xx + x];
}
}
}
return (arr[y][x] + sum / 8) / 2;
}));
}

}

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
_input += input;
});

process.stdin.on("end", function () {
processData(_input);
});

## Conclusions

Whilst this competition, was somewhat stressful, challenging and fun all around. I'd especially like to thank my teammates from "FAANG Or Bust" for helping me throughout this competition, whilst we didn't ultimately win, we did come third, which was something I was not expecting, especially as a first-year!

With these competitions, whilst my honest reactions to them previously were boredom or just sheer frustration, I can finally see how fun they actually can end up being, with the right team and difficulty level behind you. Whilst this one was more opted for more hardcore university students, it was A LOT easier than the Google Kickstart competition, where I struggled to solve any questions within the given amount of time with even the ones I finished being a shallow victory, as they were just frustrating to complete.

For those reading, even if competitive programming doesn't sound appealing to yourself, as you think you wouldn't do well or it just sounds boring. Please give it a try! Whilst I hated or didn't like the first few I've done, this one has definitely shown me another side to it and have further sparked my interest of competing in another one again shortly!

format: plain


text:
# Solving the convection-diffusion equation using finite differences at high Peclet numbers

I am trying to solve the 2D convection-diffusion equation, which in non-dimensional variables can be expressed for my problem as:

$\frac{\partial c}{\partial t} = \lambda^2\frac{\partial^2c}{\partial x^2} + \frac{\partial^2c}{\partial y^2} - 6\, Pe\, y (1 - y)\frac{\partial c}{\partial x}$

Here, $\lambda$ is a geometrical parameter (an aspect ratio of sorts) and $Pe$ is the Peclet number. The $y\,(1 - y)$ appears due to a parabolic velocity profile at the inlet (left wall). The boundary conditions are:

• Left wall: Dirichlet type
• Top and Right walls: No flux, so $y$ and $x$ gradients are zero respectively
• Bottom wall: Robin type boundary conditions. Basically a surface reaction that follows Langmuir kinetics.

I have discretized the equation using central differences for space and am using the Crank-Nicholson scheme to implicitly march in time. My code blows up (extremely large values of c, ~ $10^{134}$) for $Pe > 100$. It works fine when $Pe$ is below that.

Why does this happen? What do I do?

• Do you have quantitative boundary conditions we could play around with? – dearN Apr 3 '14 at 19:47
• I'm sorry, but I don't understand your question. – user7847 Apr 5 '14 at 4:17

Without seeing your solution, or knowing $\lambda$, that's my best guess. I suspect that if you look at your first time step for $Pe$ greater than but near 100, the solution hasn't completely blown up, but it does appear oscillatory. If that's the case, you either need to stabilize, or you need more mesh in the boundary layer(s).
• Given that linear FEM and FVM methods can directly reduced to central-difference methods, you should expect that they should have similar $Pe$ restrictions. – Bill Barth Apr 4 '14 at 3:42


text:
nLab perfect module

Context

Higher algebra

higher algebra

universal algebra

Contents

Definition

In the generality of higher algebra:

Definition

Let $R$ be an A-∞ ring. Write $R Mod^{perf} \hookrightarrow R Mod$ for the smallest stable (∞,1)-category inside that of all ∞-modules which contains $R$ and is closed under retracts. An object in $R Mod$ is called a perfect $R$-module .

Properties

Relation to compact and dualizable objects

Propositon

Let $R$ be an A-∞ ring. The (∞,1)-category of ∞-modules $R Mod$ is a compactly generated (∞,1)-category and the compact objects coincide with the perfect modules, def. 1

If $R$ is commutative (E-∞) then the perfect modules (and hence the compact objects) also coincide with the dualizable objects.

The first statement is (HA, prop. 8.2.5.2), the second (HA, prop. 8.2.5.4). For perfect chain complexes this also appears as (BFN 08, lemma 3.5).

finite objects:

References

For perfect chain complexes see the references there.

In the general context of higher algebra perfect modules are discussed in

For the properties of perfect modules in derived algebraic geometry, see

Last revised on February 5, 2015 at 10:35:04. See the history of this page for a list of all contributions to it.


text:
# Inequality. What did I do wrong?

#### dukeham

##### New member
*English is not my native language. Sorry if i misspelled wrong.
Solve the inequality: (x+1)/sqrt(1-x)< 1

I know that the answer is: (-∞,0) , but i can´t get to it

First: 1-x ≥ 0
x ≤ 1 (-∞,1]

Then I resolve the exercise:

-multiply both sides by sqrt(1-x) and simplify x+1< sqrt(1-x)
-square both sides and cancel out square root (x+1)^2 < 1-x
-expand binomial x^2+2x+1<1-x
-subtract 1-x from both sides and simplify x^2+3x<0
-factor and find signs -3<x<0
-Combine ranges x≤1 and -3<x<0

What did I do wrong?

#### Subhotosh Khan

##### Super Moderator
Staff member
*English is not my native language. Sorry if i misspelled wrong.
Solve the inequality: (x+1)/sqrt(1-x)< 1
View attachment 11743

I know that the answer is: (-∞,0) , but i can´t get to it

First: 1-x ≥ 0
x ≤ 1 (-∞,1]

Then I resolve the exercise:

-multiply both sides by sqrt(1-x) and simplify x+1< sqrt(1-x)
-square both sides and cancel out square root (x+1)^2 < 1-x
-expand binomial x^2+2x+1<1-x
-subtract 1-x from both sides and simplify x^2+3x<0
-factor and find signs -3<x<0
-Combine ranges x≤1 and -3<x<0

What did I do wrong?
-subtract 1-x from both sides and simplify x^2+3x<0

x(x+3) < 0 .......................... Two solutions

Either x>0 and (x+3)<0 ............................impossible

Or

x<0 and x+3>0

then x > -3 .................. which implies

-3 < x < 0

#### Dr.Peterson

##### Elite Member
Squaring both sides of an inequality does not preserve the inequality. Your error is in assuming it does.

I would either break it into cases before squaring, or just graph the two sides. Using cases, note that if both sides of an inequality are positive, then squaring will preserve it; for example, 1 < 2 becomes 1 < 4. But if the left side is negative while the right is positive, e.g. -3 < 2, then it is always true, and you don't need to do anything further. Squaring would have produced 9 < 4, which is false.

The solution is not (-3, 0); just check the inequality at, say, -3 and -4 to see this.

#### lookagain

##### Senior Member
Solve the inequality: (x+1)/sqrt(1-x)< 1
View attachment 11743

I know the answer is: (-∞,0)
.
.
.
$$\displaystyle \dfrac{x + 1}{\sqrt{1 - x}} \ - \ 1 < 0$$

$$\displaystyle \dfrac{x + 1 }{\sqrt{1 - x}} \ - \ \dfrac{\sqrt{1 - x}}{\sqrt{1 - x}} < 0$$

$$\displaystyle \dfrac{x + 1 \ - \ \sqrt{1 - x}}{\sqrt{1 - x}} < 0$$

For potential critical numbers, set $$\displaystyle \ \ (x + 1 \ - \ \sqrt{1 - x}) \ \$$ equal to 0, and set $$\displaystyle \ \ \sqrt{1 - x} \ \$$ equal to 0.

x = 0 and x = 3 are candidates for solutions to the first equation, but x = 3 is extraneous and must be discarded..
x = 1 is the solution to the second equation.

Looking at test values on a number line:

To the right of x = 1: The expression is undefined.

Between x = 0 and x = 1, you could choose a convenient x-value because of the square root, say x = 3/4. The sign of the
expression is positive.

x = -3 turned out not to be a critical number. And, because it is to the left of the remaining critical number x = 0, and it is
a convenient choice because of the square root, then you may substitute it and see that the sign of the expression is negative.

So, the solution for the problem is $$\displaystyle \$$ x < 0, or $$\displaystyle \ \ (-\infty, 0)$$.

#### pka

##### Elite Member
Here is a slightly different approach.
After reducing the inequality to $$\displaystyle \dfrac{(x + 1) \ - \ \sqrt{1 - x}}{\sqrt{1 - x}} < 0$$
note that only the numerator determines the sign. The radical forces $$\displaystyle x<1$$.

If $$\displaystyle x<0$$ then $$\displaystyle (x+1)<1~\&~\sqrt{1-x}>1$$ so the numerator is negative.

We know that $$\displaystyle x=0$$ does not work so that leaves $$\displaystyle a\in(0,1)$$ to consider.
But then we would have $$\displaystyle (a+1)>1~\&~\sqrt{1-a}<1$$ so the numerator is positive and we don't want that.

Hence we have a solution: $$\displaystyle x<0$$.

#### dukeham

##### New member
Thank you all for the answers.


text:
# Why is dependence of radiation forcing on CO2 concentration logarithmic?

As a newbie on climate science and meteorology I'm trying to understand the almost logarithmic behavior of CO2 concentration on radiation forcing

$$F(c) = A \cdot \ln \frac{c}{c_0}$$

and why there is no "saturation" due to further increasing of CO2 even when the atmosphere is already totally absorbing as a whole.

So far I learned that the air layers which contribute most to outgoing radiation are placed where optical thickness, as measured from TOA, is about 1. Let's say in an undisturbed atmosphere this location is about a height of z=2 (arbitrary units), then a CO2 doubling leads to a shift of this layer to about z=3.

Since this higher layer has usually lower temperature, less energy is irradiated an, therefore, the planet will warm up.

On the other hand, other explanations, even mentioned in the cited publication, are based on spectral effects where the wings of the absorption band become more absorbing at higher CO2 concentrations:

Q1: Firstly, I don't understand why this last effect is, because CO2 is quite dilute in atmosphere and this effect cannot be due to pressure broadening. What is the reason for absorbing more CO2 at the wings when concentration rises up?

Q2: secondly, which of the two effects is more relevant to explain the observed log behavior? Are they related in any way or do they just happen to give the same overall result?

• Have a look at the 'curve of growth' of spectral line equivalent width: spiff.rit.edu/classes/phys440/lectures/curve/curve.html Even at high optical depths, 'saturation' doesn't mean that there is no increase in absorption anymore. Rather, absorption starts depending logarithmically, rather than linearly on the absorber density, as the line core is saturated, but the line doppler wings continue to contribute. The arguments for a single line then directly translate to the line forests of molecules, which is why the overall CO2 opacity also follows the log-behaviour. Nov 10, 2022 at 15:05
• Does it really mean that the absorption characteristics of a particular molecule, e.g. CO2 depend on the density of those molecules? Cannot imagine, because wit 300ppm CO2 molecules are very rare and they do not interact in any way. Most of the other molecules for a single CO2 are not CO2, but N2, O2, ... so if there is some modification because of collisions, most of the collisions come from other molecule types - not CO2. So I would assume some dependence on the total pressure, but not on partial pressure of CO2. Do we talk about the same thing? Nov 10, 2022 at 15:31
• The pressure broadening/collisional broadening does not matter on the collision partner when you are close to thermal equilibrium (we're far from the ionosphere here). Study the document I've linked carefully, it doesn't take a long time. The log(c) dependency of the total absorption depends on the number of excitations any single line gets hit with times the number of absorbers, which is just another way of encoding the optical depth and hence the partial pressure. Nov 10, 2022 at 17:45
• I would say that is correct. Nov 15, 2022 at 20:44

Your confusion, I think, comes from a subtlety, and I think I can only answer this from the perspective of an astrophysicist. We don't think in terms of "forcings", the vocabulary is still alien to me (I have yet to see a sensible, formal defition of it). But I will answer in terms of a certainly related concept.

In the optically thick part of an atmosphere (i.e. the one that potentially features a greenhouse effect), we know from analytic solutions that the Temperature $$T$$ as function of optical depth $$\tau$$ is approximately $$T^4=T^4_0(\frac{2}{3} + \tau)$$, where $$T_0$$ is a measure of the energy content of the optically thin radiation field.

The total optical depth $$\tau$$, itself is simply computed as a sum of all opacity carrier contributing species $$s$$, with number densities $$n_s$$, opacity function $$\kappa_s$$, and optical ray path $$ds$$ as $$\tau = \sum_s \int ds \, n_s \, \kappa_s.$$ This already answers the part of your question about the partial pressure dependency: $$\tau\propto n_s$$, no matter how the opacity is generated (I am using pressure and number density at constant temperature interchangeably, but to first order it is $$n_s$$ and not $$p_s$$ determining $$\tau$$).
Now collision-induced absorption, which gives rise to the broad Lorentzian line wings, which grow logarithmically at large optical depth, is dependent on all available collision partners, hence this is where the total pressure comes into play, via $$\kappa_{CO2} = \kappa_{CO2}(\sum_s n_s)$$. In the last sum, the $$CO_2$$ contribution would be negligible, but $$\kappa_{CO2}$$ still grows logarithmically.

So in my understanding we'd rather have something along the lines of $$\tau \propto n_s \times log(f(\sum_s n_s))$$

I would still like to better understand how the 'forcing' comes about from those quantities, but it does give you an answer as to why the temperature to which the system wants to get to (the above $$T^4$$) has the dependencies that you are confused about.

Just wondering about your statement above: "Since this higher layer has usually lower temperature, less energy is irradiated an, therefore, the planet will warm up". This doesn't sound right. For one thing, temperature at any layer is not a passive variable: if radiation (absorption and emission) is happening at a higher level, then the temperature at that level should rise too.


text:
# 2175. Cow Counting

Farmer John wants to label his N (1 <= N <= 1,000,000) cows, but cows don’t like having the digit L (0 <= L <= 9) written on them, so he avoids that.

If the cows are labeled with the smallest set of N positive integers that don’t have the digit L, what is the largest number that Farmer John has to write on any cow?

### 输入格式

• Line 1: Two space-separated integers: N and L.

### 输出格式

• Line 1: A single integer that is the largest number that Farmer John has to write on any cow.

### 样例

Input
10 1

Output
22

32 人解决，51 人已尝试。

37 份提交通过，共有 109 份提交。

4.9 EMB 奖励。


text:
# Difference between revisions of "2003 AMC 12A Problems/Problem 1"

The following problem is from both the 2003 AMC 12A #1 and 2003 AMC 10A #1, so both problems redirect to this page.

## Problem

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006$

## Solution 1

The first $2003$ even counting numbers are $2,4,6,...,4006$.

The first $2003$ odd counting numbers are $1,3,5,...,4005$.

Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$.

$(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$

$= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}$

## Solution 2

Using the sum of an arithmetic progression formula, we can write this as $\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}$.

## Solution 3

The formula for the sum of the first $n$ even numbers, is $S_E=n^{2}+n$, (E standing for even).

Sum of first $n$ odd numbers, is $S_O=n^{2}$, (O standing for odd).

Knowing this, plug $2003$ for $n$,

$S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow$ $\boxed{\mathrm{(D)}\ 2003}$.

## Solution 4

In the case that we don't know if $0$ is considered an even number, we note that it doesn't matter! The sum of odd numbers is $O=1+3+5+...+4005$. And the sum of even numbers is either $E_1=0+2+4...+4004$ or $E_2=2+4+6+...+4006$. When compared to the sum of odd numbers, we see that each of the $n$th term in the series of even numbers differ by $1$. For example, take series $O$ and $E_1$. The first terms are $1$ and $0$. Their difference is $|1-0|=1$. Similarly, take take series $O$ and $E_2$. The first terms are $1$ and $2$. Their difference is $|1-2|=1$. Since there are $2003$ terms in each set, the answer $\boxed{\mathrm{(D)}\ 2003}$.

Solution by franzliszt

## Solution 5 (Fastest method)

We can see that the difference of the first even number and the first odd number is one, the difference between the second even number and the second odd number is one and so on. Then, we get $1 * 2003$ which is $\boxed{\mathrm{(D)}\ 2003}$.

Solution by Penguin Spellcaster


text:
# Area Of A Rhombus

Here we will learn about finding the area of a rhombus, including compound area questions, questions with missing side lengths and questions involving unit conversion.

There are also area of a rhombus worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

## What is the area of a rhombus?

The area of a rhombus is the amount of space inside the rhombus. It is measured in units squared ( cm^2, \ m^2, \ mm^2 etc).

Area formula:

A=\frac{1}{2}\times{D}\times{d}

Where D is the long diagonal of the rhombus (the base) and d is the short diagonal in the rhombus (the height of the rhombus).

A rhombus is a polygon and a quadrilateral ( 4 sided shape) where opposite sides are parallel, all sides are equal length and opposite angles are equal. The interior angles of a rhombus add up to 360^{\circ} . A rhombus is a special type of a parallelogram.

The diagonals of a parallelogram bisect each other at a right angle as shown below:

To calculate the area of a rhombus, we need to know the length of each diagonal of the rhombus.

If we move the two triangles from half of the rhombus to the opposing side, we get a rectangle:

The width of the rectangle is half of the long diagonal of the rhombus ( D ), and the height of the rectangle is the length of the short diagonal in the rhombus ( d ).

This gives the area of the rhombus formula:

A=\frac{1}{2}\times{D}\times{d}

## How to find the area of a rhombus

In order to find the area of a rhombus:

1. Identify the length of the diagonals.
2. Write down the formula for the area of a rhombus.
3. Substitute the given values of the diagonals and solve.
4. Write down your final answer, including the units.

## Related lessons on area

Area of a rhombus is part of our series of lessons to support revision on area. You may find it helpful to start with the main area lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

## Area of a rhombus examples

### Example 1: finding area given the length of the diagonals

Find the area of the rhombus below:

1. Identify the length of the diagonals.

\begin{aligned} &AC = 2m \\\\ &BD = 9m \end{aligned}

These are the diagonals of a rhombus which correspond to the base and height of the corresponding rectangle.

2Write down the formula for the area of a rhombus.

A=\frac{1}{2}\times{D}\times{d}

3Substitute the given values of the diagonals and solve.

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{9}\times{2}\\\\ &=9 \end{aligned}

4Write down your final answer, including the units.

In this case we are working with metres so our final answer must be in square metres.

A=9m^2

### Example 2: finding the area of a rhombus requiring converting units

Calculate the area of the rhombus below:

\begin{aligned} &MO = 50cm \\\\ &NP= 1m \end{aligned}

We have two different units for the length and width so we must change the measures to a common unit.

In this case, let us change both units to centimetres.

As 1m=100cm .

\begin{aligned} &MO = 50cm \\\\ &NP = 100cm \end{aligned}

A=\frac{1}{2}\times{D}\times{d}

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{100}\times{50}\\\\ &=2500 \end{aligned}

As the area has been calculated using the lengths in centimetres, the area unit is square centimetres.

A=2500\text{ cm}^2

### Example 3: worded problem

A rhombus ABCD has the point O as the point of intersection of the two diagonals AC and BD . The length OB=3cm , and OC=5cm . Calculate the area of the rhombus.

As the point O is the intersection of the two diagonals, the lengths OB and OC are half of the length of each diagonal. This means that as OB=3cm, \ AB=3 \times 2=6cm and as OC=5cm, \ CD=5 \times 2=10cm.

A=\frac{1}{2}\times{D}\times{d}

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{10}\times{6}\\\\ &=30 \end{aligned}

As the diagonal lengths are given in centimetres, the area unit is square centimetres.

A=30\text{ cm}^2

### Example 4: calculating the diagonal given area and the other diagonal

Calculate the length of AC given the length of BD=14m and the area of the rhombus is 35m^2 . Use the diagram below to help you.

\begin{aligned} &BD = 14m \\\\ &AC = x \end{aligned}

In this case we are also given Area = 35m^2 .

A=\frac{1}{2}\times{D}\times{d}

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d} \\\\ 35&=\frac{1}{2}\times{14}\times{x} \\\\ 35&=7x \\\\ x&=5 \end{aligned}

In this case we are asked to find the length of a diagonal and not area so our answer should be written in metres:

AC=5\text{ m}

### Example 5: area given side length and one diagonal

ABCD is a rhombus where the edges AB and CD are parallel. Given that AB=20cm and BD=32cm, calculate the area of the rhombus.

\begin{aligned} &BD = 32cm \\\\ &AC = x \end{aligned}

We need to calculate the length of AC. In this question we are given one side of the rhombus which is AB=20cm .

Note: Since we are given the length of the side AB we now know the length of any side of the rhombus since all sides of a rhombus are the same length.

We can split the rhombus into 4 right-angled triangles. If we take triangle OAB , the hypotenuse AB=20cm and OB=16cm since it is half the length of BD . We now need to calculate the length of OA .

As OAB is a right-angled triangle, we can use Pythagoras’ theorem to calculate OA .

\begin{aligned} c^{2}&=a^{2}+b^{2}\\\\ 20^{2}&=16^{2}+x^{2}\\\\ 400&=256+x^{2}\\\\ x^{2}&=144\\\\ x&=\sqrt{144}\\\\ x&=12 \end{aligned}

As x=OA=12cm, \ AC=12 \times 2=24cm

Now that we have the length of two of the diagonals of the rhombus, we can calculate the area of the rhombus ABCD .

A=\frac{1}{2}\times{D}\times{d}

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{32}\times{24}\\\\ &=384 \end{aligned}

The area of the rhombus ABCD is 384cm^2 .

### Example 6: calculate the area using trigonometry (higher only)

A rhombus has a side length of 13cm . One pair of interior angles is double the other pair of interior angles. Calculate the area of the rhombus to 2 decimal places.

Sketching a diagram first, we have:

As the sum of angles in a rhombus is 360^{\circ} , we can quickly calculate the value of x by forming and solving an equation:

\begin{aligned} 2x+x+2x+x&=360 \\\\ 6x&=360 \\\\ x&=60^{\circ} \end{aligned}

So the angles in the rhombus are 60^{\circ} and 120^{\circ} . If we filled these in on the diagram, we have:

We can calculate the length AC by using the cosine rule c^{2}=a^{2}+b^{2}-2ab\cos(\theta) , looking at the top half of the rhombus, we have:

Labelling the sides and angles according to the formula ( \theta must be the included angle between the two sides a and b , we have:

Substituting the values of a, b, and \theta into the cosine rule, we have:

\begin{aligned} c^{2}&=a^{2}+b^{2}-2ab\cos(\theta) \\\\ c^{2}&=13^{2}+13^{2}-2\times{13}\times{13}\times\cos(120)\\\\ c^{2}&=507\\\\ c&=13\sqrt{3} \end{aligned}

So the long diagonal AC=13\sqrt{3}\text{ cm}

Repeating the same process for the diagonal BD, we have:

Substituting the values of a, b, and \theta into the cosine rule, we have:

\begin{aligned} c^{2}&=a^{2}+b^{2}-2ab\cos(\theta) \\\\ c^{2}&=13^{2}+13^{2}-2\times{13}\times{13}\times\cos(60) \\\\ c^{2}&=169 \\\\ c&=13 \end{aligned}

So the short diagonal BD=13\text{ cm} .

(For this example, BCD is an equilateral triangle!)

A=\frac{1}{2}\times{D}\times{d}

As the long diagonal AC=13\sqrt{3}\text{ cm} and the short diagonal BD=13\text{ cm}, we have:

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{13\sqrt{3}}\times{13}\\\\ &=146.3582932…\\\\ &=146.36\text{ cm}^2\text{ (2dp)} \end{aligned}

The area of the rhombus ABCD is 146.36cm^2 \ (2dp). .

### Common misconceptions

• Using incorrect units for the answer

A common error is to forget to include squared units when asked to calculate area.

• Forgetting to convert measures to a common unit

Before using the formula for calculating the area of a rectangle we need to ensure that units are the same. If different units are given (E.g. length = 4m and width = 3cm ) then you must convert them either both to cm or both to m.

• Using half of the diagonal

As the area of a rhombus is half the length of one diagonal multiplied by the length of the other diagonal, the other diagonal is also halved within the calculation, leaving a solution that is out by a factor of 2.

• Calculating the area of a square (using side lengths)

The area of a rhombus is calculated by using the side lengths, instead of the lengths of the diagonals.

### Practice area of a rhombus questions

1. Find the area of the rhombus below:

24m^2

48m^2

16m^2

48m
\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{12}\times{4}\\\\ &=24 \end{aligned}

A=24\text{ m}^2

2. The rhombus MNOP has the following information:

MP=250cm

NO=4.5m

Calculate the area of the rhombus in square centimetres.

11250cm^2

28125cm^2

56.25cm^2

56250cm^2
\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{250}\times{450}\\\\ &=56250 \end{aligned}

A=56250\text{ cm}^2

3. Mr. Perry owns a farm and is looking to make the chicken enclosure shown below. Each chicken needs 4 square metres to move around freely. What is the maximum number of chickens that can fit into the enclosure?

24

48

96

192

Calculating the area of the rhombus, we have

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{12}\times{16}\\\\ &=96 \end{aligned}

A=96\text{ m}^2

As the field is 96m^2 and each chicken requires 4m^2, \ 96\div{4}=24 chickens

4. The area of rhombus ABCD is 54m^2. The centre of the rhombus is located at point P . Given that AP=3m , calculate the length BD.

9m

18m

54m

324m

As A=54m^2 and AP=3m, \ AC=32=6m.

Using the formula for the area of a rhombus, we have

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ 54&=\frac{1}{2}\times{6}\times{x}\\\\ 54&=3x\\\\ x&=18 \end{aligned}

x=18\text{ m}

5. In the rhombus below AB = 5m, \ BD = 8m. Calculate the area of the rhombus.

24m^2

48m^2

6m^2

20m^2

We can calculate half the length of AC by using Pythagoras’ theorem. Let M be the centre of the rhombus. The distance BM is half the distance BD and so BM=4m.

Using Pythagoras’ theorem to calculate the length of AM , we have:

\begin{aligned} AM^{2}&=5^{2}-4^{2}\\\\ &=25-16\\\\ &=9\\\\ AM&=3 \end{aligned}

As AM=3m, \ AC=3 \times 2=6m.

The area of the rhombus is therefore:

\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{6}\times{8}\\\\ &=24 \end{aligned}

A=24\text{ m}^2

6. (Higher only) A rhombus PQRS has side length 5cm . The rhombus has two pairs of interior angles, equal to x and 3x-20 .

Use the cosine rule c^2=a^2+b^2-2ab \cos(\theta) to calculate the area of the rhombus to 1 decimal place.

19.2cm^2

12.5cm^2

25.0cm^2

20.0cm^2

First we need to calculate the value of the two pairs of angles in the rhombus:

\begin{aligned} x+3x-20+x+3x-20&=360\\\\ 8x-40&=360\\\\ 8x&=400\\\\ x&=50 \end{aligned}

x=50^{\circ}

3x-20=130^{\circ}

So we have a rhombus with a side length of 5cm and the interior angles are 50^{\circ} and 130^{\circ}.

Calculating the distance PR using the cosine rule, we have:

PR=\sqrt{5^{2}+5^{2}-2\times{5}\times{5}\times\cos(130)}=9.06307787

Calculating the distance QS using the cosine rule, we have:

QS=\sqrt{5^{2}+5^{2}-2\times{5}\times{5}\times\cos(50)}=4.226182617

The area of the rhombus is therefore:
\begin{aligned} A&=\frac{1}{2}\times{D}\times{d}\\\\ &=\frac{1}{2}\times{9.06307787}\times{4.226182617}\\\\ &=19.15111108 \end{aligned}

A=19.2\text{ cm}^2\text{ (1dp)}

### Area of a rhombus GCSE questions

1. Shown below is a rhombus where AM=6m , and BM=10m . Calculate the area of the rhombus.

(3 marks)

AC=12m, \ BD=20m

(1)

\frac{12 \times 20}{2}

(1)

120m^2

(1)

2. (a) A rectangular logo is composed of 8 congruent rhombuses. Calculate the area of one rhombus.

(b) Calculate the perimeter of the logo. Write your answer as a surd in its simplest form.

(6 marks)

(a)

12\div{3}=4\text{m and }4\div{2}=2\text{m}

(1)

\frac{4 \times 2}{2}

(1)

4\text{ m}^{2}

(1)

(b)

2^{2}+1^{2}=5

(1)

Side length of a rhombus =\sqrt{5}

(1)

16\sqrt{5}

(1)

3. Ms. Polly is looking to cover her garden with new grass. One roll of grass covers 6 square metres and costs £9. How much would it cost to cover the entire ground with grass?

(5 marks)

\sqrt{25^{2}-7^{2}}

(1)

24m

(1)

\frac{48 \times 14}{2}

(1)

(336 \div 6) \times 9

(1)

£504

(1)

## Learning checklist

You have now learned how to:

• Calculate the area of a rhombus and related composite shapes
• Calculate the area of parallelograms
• Recognise when it is possible to use formulae for area
• Derive and apply formulae to calculate and solve problems involving: perimeter and area of triangles and parallelograms

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#### FREE GCSE maths practice papers (Edexcel, AQA & OCR)

8 sets of free exam practice papers written by maths teachers and examiners for Edexcel, AQA and OCR.

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