---
title: "proofs"
date: "13/06/2025"
useKatex: true
useD3: true
---

import Pseudocode from '@components/Pseudocode.astro';

A computer science student attempting to learn proofs.

# 993

## A

Count-Pairs($n$):
1. Return $n-1$

---

*Proof.*

Suppose $(a,b)\in\mathbb{N}^2$. Because $a=n-b$ and $a\geq1$, it follows that $1\leq b\leq n-1$. Each choice of $b$ yields a unique $a=n-b$, so there are $n-1$ unique solutions.

$\blacksquare$

## B

Mirror-String($s$):
1. Reverse $s$
2. For each character $c$ in $s$:
  - If $c$ is "w": Print($c$)
  - If $c$ is "p": Print($q$)
  - If $c$ is "q": Print($p$)

---

*Proof.*

The string appears fipped on the y-axis from within the score due to the perspective
shifting. Structurally, it is read right-to-left. "p"/"q"/"w" appear as "q"/"p"/"w" when flipped on its y-axis.

$\blacksquare$

## C

Seat-Monkeys($a$, $b$, $c$, $m$):
1. Return $min(m, a)+min(m, b)+min(c, 2\cdot m-(min(m, a) + min(m, b)))$

---

*Proof.*

Consider an assignment of monkeys $S$ that sits the $a$ and $b$ monkeys in the first and second row and then fills the remaining seats with the $c$ monkeys.

Assume there exists a more optimal assignment of monkeys $S^{'}$. WLOG, assume $S^{'}$ sits $a$ and $b$ monkeys first in their respective rows.

$S^{'}$ can only differ from $S$ as follows:

1. Seats a $c$ monkey in row 1 instead of an $a$ monkey
  - $S^{'}$ leaves an $a$ monkey unseated. $S$ seats this monkey instead--the same number of monkeys are seated $S$.
2. Seats a $c$ monkey in row 2 instead of a $b$ monkey
  - $S^{'}$ leaves a $b$ monkey unseated. $S$ seats this monkey instead--the same number of monkeys are seated in $S$.
3. Does not seat a monkey where $S$ has
  - $S$ seats more than $S^{'}$.

In all cases, $S^{'}$ is no better than S, therefore $S$ is optimal.

$\blacksquare$

## D

Construct-B($a$):
1. Let $b$ be an array of size $n=\#a$ and $X$ be the set of numbers in $a$.
2. For each element $x$ of $a$ at index $i$:
- If $x\in X$:
  - $b[i]:=x$
  - $X:=X \backslash \{x\}$
3. Let $Y=\{1,2,\cdots,n\}\backslash X$
4. For each element $x$ of $b$ at index $i$:
- If $a[i]\in X$:
  - $b[i]:=\text{first-element}(Y)$
  - $Y:=Y\backslash\{\text{first-element}{(Y)}\}$
5. Return $b$

---

*Proof.*

Consider the array $b$ from Construct-B.

If there are $x$ unique elements in $a$, the algorithm assigns each of those to the first $x$ positions in $b$. There are $n-x$ duplicate elements in $a$ and $n-x$ remaining positions in $b$.

Since $\forall a\in A:1\leq a\leq n$, there are $n-x$ unused unique elements in $\{1,2,\cdots,n\}$, each of which is assigned a unique position in $b$.

Therefore, all elements of $b$ are unique and thus a mode. As every unique element in $a$ is assigned to an index no later in $b$, all $a_i$ is a mode in $b[i+1:]$.

$\blacksquare$

## E

Count-Pairs($l_1$, $l_2$, $r_1$, $r_2$, $k$):
1. Let $A:=\lfloor log_k(r_2/l_1)\rfloor$
2. Let $B:=\lfloor max(0, log_k(l_2/r_1))\rfloor$
3. Let $\text{total}:=0$
4. For each $A\leq i\leq B$:
- Let $r=\lfloor r_2/ k^n\rfloor$
- Let $l=\lfloor l_2/k^n\rfloor$
- $\text{total} := \text{total} + max(0, r - l + 1)$
5. Return $\text{total}$
---

*Proof.*

Each value of $n$ corresponds to a line with slope $k^n$ because $y/x=k^n\leftrightarrow y=x\cdot k^n$. The problem can be visualized as follows:

![graph of problem](/posts/proofs/graph.webp)

It is sufficient to count the number of ordered $(x,y)$ pairs for all valid $n$. Because $y=x\cdot k^n\leftrightarrow n=log_k(y/x)$, $n\in [log_k(l_2/r_1), log_k(r_2/l_1)]$.

For each $n_0$ in this range, the smallest $x$ satisfying $y=x\cdot k^n$ is $\lceil l_2/k^n\rceil$ and the largest $\lfloor r_2/k^n\rfloor$, so $n_0$ contributes $max(0, \lfloor r_2/k^n\rfloor - \lceil l_2/k^n\rceil + 1)$ ordered pairs.

## F

1. Let $A=\sum a$ and $B=\sum b$.
2. For each query with requested beauty $q$:
- If $\exists (i,j)\in(\{1,2,\cdots,n\},\{1,2,\cdots,m\}):(A-a[i])\cdot(B-b[j])=x$, print "YES"
- Otherwise, print "NO"

---

*Proof.*

The beauty of the grid equals $B=\sum_i \sum_j M_{i,j}=\sum_i\sum_j a_i\cdot b_j=\sum_i(a_i\cdot \sum_j b_j)=(\sum_i a_i)\cdot (\sum_j b_j)$.

Formulating setting row $i$ and column $j$ to zero, the new beauty is:

$q=B-(b_j\cdot(\sum a)+a_i\cdot(\sum b)-a_i\cdot b_j)$
$=((\sum a)-a_i)\cdot((\sum b)-b_j)$

If such $a_i$ and $b_j$ exist, the operation can be performed.

## G1

1. Let $G$ be the input graph
2. Let $\text{ans}=0$
3. For each component $C$ in $G$:
  - Let $\text{cycle}$ be the set of nodes in the cycle of $C$
  - Let $d=max_{u\in \text{cycle}}\text{Distance-To-Cycle(C, u)}$
  - $\text{ans}:=max(\text{ans}, d)$
4. Return $\text{ans}$

---

*Proof.*

WLOG, consider some $v_i\in V$. There must exist an edge $e=(v_i,v_j),i\neq j$. Following this path from $v_i$, the each edge must map to a previously seen node, forming a cycle, or a new node. Because the graph is finite, the path starting any $v_i$ must contain a cycle. Thus, G is a graph of components with one cycle each.

WLOG, consider some component $C\in G$. For every spider $v_i$ in $C$:

- If $v_i$ is in the cycle:
  - The cycle itself will always be stable. Every spider has a plushie and each spider will give and receive a plushie.
- Otherwise, $v_i$ is not in the cycle.
  - Let $v_j$ be the furthest spider on a path with a plushie containing $v_i$ to the cycle. When $v_j$ gives a plushie to its child, the graph is unstable because $v_j$ transitions from state $0$ to $1$. However, the path containing $v_j$ and its ancestors become stable the next year because they never receive or give a plushie again. The path takes $d_j$ years to become stable, the distance from $v_j$ to the path.

Therefore, $C$ becomes stable in $d_C:=max_{v_j\in C}(d_j)$ time.

The entire graph becomes stable when each component becomes stable, which is the longest time any component takes to become stable. Thus, the graph becomes stable in $max_{C\in G}d_C$ years.

$\blacksquare$

## G2

1. Let $G$ be the input graph
2. Let $\text{ans}=0$
3. For each component $C$ in $G$:
  - Let $U$ be the set of all nodes not in the cycle of $C$
  - Let $\text{count}$ be the number of plushies each spider has
  - Let $\text{par}$ be the set of parents for each spider
  - Run a multi-source BFS simulating the state transitions on $U$. For each iteration at year $y$:
    - $\text{count[u]}:=\text{count[u]}+\sum_{p\in \text{par[u]}}\text{count[p]}$
    - $\text{ans}:=max(\text{ans}, \text{count[u]} - $y$)$
    - $y:=y+1$
4. Return $\text{ans}$

---

*Proof.*

WLOG, consider some $v_i\in V$. There must exist an edge $e=(v_i,v_j),i\neq j$. Following this path from $v_i$, the each edge must map to a previously seen node, forming a cycle, or a new node. Because the graph is finite, the path starting any $v_i$ must contain a cycle. Thus, G is a graph of components with one cycle each.

WLOG, consider some component $C\in G$. For every spider $v_i$ in $C$:

- If $v_i$ is in the cycle:
  - The cycle itself will always be stable. If a spider has $x$ plushies in year $y$, it will give and receive one plushie and have $x$ in the next year as well.
- Otherwise, $v_i$ is not in the cycle.
  - Let $v_j$ be the furthest spider on a path containing $v_i$ to the cycle. When $v_j$ gives a plushie to its child, the graph is unstable because $v_j$ transitions from state $0$ to $1$. However, the path containing $v_j$ and its ancestors become stable the next year because they never receive or give a plushie again.
  - However, the child may have more than one plushie next year. If the child had $x_0$ plushies the on year $y$ received $x_1$ plushies the next, it must ultimately give $x_0+x_1$ plushies taking $x_0+x_1-y$ years. The path becomes stable in the maximum time it takes any spider on the path to give its plushies, $d_i$. The algorithm gathers the $x_1$ term by considering all parents of the child and propagates the plushie counts by simulation.

Therefore, $C$ becomes stable in $d_C:=max_{v_i\in C}(d_i)$ time.

The entire graph becomes stable when each component becomes stable, which is the longest time any component takes to become stable. Thus, the graph becomes stable in $max_{C\in G}d_C$ years.

$\blacksquare$

