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% Document Info
\author{AviMaths (AviMathPerson)}
\pdfinfo{
  /Title (discrete-math.pdf)
  /Creator (Cheatography)
  /Author (AviMaths (AviMathPerson))
  /Subject (Discrete Math Cheat Sheet)
}

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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
    \SetRowColor{DarkBackground}
    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
    }
\end{tabulary}
\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Discrete Math Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{AviMaths (AviMathPerson)} via \textcolor{DarkBackground}{\uline{cheatography.com/189338/cs/42865/}}}
\end{tabulary}
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}AviMaths (AviMathPerson) \\
  \uline{cheatography.com/avimathperson} \\
        \uline{\seqsplit{unitmeasure}.xyz}
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 27th March, 2024.\\
   Updated 27th March, 2024.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{1.29402 cm} x{3.68298 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Important relations}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Reflexive:} & Shorthand: Ia ⊆ R Meaning: Every element is related to itself. for all a ∈ A, aRa holds ( R ⊆ \{(a, a) | a ∈ A\} ) \tn 
% Row Count 5 (+ 5)
% Row 1
\SetRowColor{white}
\seqsplit{Transitive:} & Shorthand: ( R ∘ R = R2 ⊆ R ) Meaning: If ( (a, b) ∈ R ) and ( (b, c) ∈ R ), then ( (a, c) ∈ R ). (aRb and bRc) -\textgreater{} aRc \tn 
% Row Count 10 (+ 5)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Symmetric:} & Shorthand: ( R = R⁻$^{\textrm{1}}$ ) Meaning: If ( (a, b) ∈ R ), then ( (b, a) ∈ R ). When aRb \textless{}=\textgreater{} bRa \tn 
% Row Count 14 (+ 4)
% Row 3
\SetRowColor{white}
\seqsplit{Antisymmetric:} & Shorthand: ( R ∩ R⁻$^{\textrm{1}}$ ⊆ \{(a, a) | a ∈ A\} ) Meaning: If ( (a, b) ∈ R ) and ( (b, a) ∈ R ), then ( a = b ). (aRb and bRa) -\textgreater{} (a = b) - This does not mean not-symmetric \tn 
% Row Count 21 (+ 7)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Equivalence relation is one where Reflexivity, Transitivity, and Symmetry all hold}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Cardinality}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Cardinality} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Aleph 0} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Aleph} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Finite} \tn 
% Row Count 4 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.60195 cm} x{1.14425 cm} x{1.8308 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Combinatorics}}  \tn
% Row 0
\SetRowColor{LightBackground}
Case: & Order matters & Order doesn't matter \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
With repetition & n\textasciicircum{}k (case 1) & nCk (case 3) \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Without repetition & nPk (case  2) & (k+n-1)C(k) (case 4) \tn 
% Row Count 6 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Functions}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Let f,g be two functions, (f:A -\textgreater{} B) , (g:B -\textgreater{} A)} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Function f} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Onto} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{One to One} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Onto and One to One} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Identity Ia} \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{(g ∘ f)(a) = g(f(a)) means g composed with f}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Set Theory}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{A ∪ B = \{x ∈ A or x ∈ B or both\} \newline % Row Count 1 (+ 1)
A ∩ B = \{x ∈ A and x ∈ B\} \newline % Row Count 2 (+ 1)
A ⊕ B = (A - B) ∪ (B - A) = (A ∪ B) - (A ∩ B) \newline % Row Count 4 (+ 2)
A - B = A ∩ Bᶜ = \{x ∈ A and x ∉ B\} \newline % Row Count 5 (+ 1)
Demorgan's laws: \newline % Row Count 6 (+ 1)
(A ∪ B)ᶜ = Aᶜ ∩ Bᶜ \newline % Row Count 7 (+ 1)
(A ∩ B)ᶜ = Aᶜ ∪ Bᶜ \newline % Row Count 8 (+ 1)
Associativity: \newline % Row Count 9 (+ 1)
A ∪ (B ∪ C) = (A ∪ B) ∪ C = A ∪ B ∪ C \newline % Row Count 10 (+ 1)
Distributivity; \newline % Row Count 11 (+ 1)
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) \newline % Row Count 12 (+ 1)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)% Row Count 13 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
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\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Subsets}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{A ⊆ A ∪ B = B ∪ A \newline % Row Count 1 (+ 1)
B ⊆ A ∪ B \newline % Row Count 2 (+ 1)
If A ⊆ B, then A ∪ B = B \newline % Row Count 3 (+ 1)
A ∩ B ⊆ A ⊆ A ∪ B \newline % Row Count 4 (+ 1)
B ∩ A ⊆ B ⊆ A ∪ B% Row Count 5 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Cartesian product:}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{A × B = \{ (a, b) | a ∈ A and b ∈ B \} \newline % Row Count 1 (+ 1)
an unordered set of sets of ordered pairs where a is in A, b is in B \newline % Row Count 3 (+ 2)
if A = \{1,2\}, B = \{2,3\}, then A × B = \{(1,2), (1,3), (2,2), (2,3)\} \newline % Row Count 5 (+ 2)
A × B ≠ B × A (unless A = B) \newline % Row Count 6 (+ 1)
A ∩ (A × B) = ∅ \newline % Row Count 7 (+ 1)
|A × B| = |A| × |B| \newline % Row Count 8 (+ 1)
Distribution: \newline % Row Count 9 (+ 1)
A × (B ∪ C) = (A × B) ∪ (A × C) \newline % Row Count 10 (+ 1)
A × (B ∩ C) = (A × B) ∩ (A × C)% Row Count 11 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Order relation}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Partial order If and only if all (Reflexivity, Transitivity, and Anti-symmetry) hold \newline % Row Count 2 (+ 2)
clear hasse diagram can be drawn \newline % Row Count 3 (+ 1)
items for which the relation doesn't hold will be drawn but not connected to the others in the diagram% Row Count 6 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Total/Linear order: \newline Partial order holds \newline Totality: For any ( a, b ∈ A ), either ( (a, b) ∈ R ) or ( (b, a) ∈ R ). \newline In other words: For any two distinct elements a and b, either a is related to b (a ≤ b), or b is related to a (b ≤ a). \newline hasse diagram would be a straight line (all elements relate to one another in this set)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Universal Set}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Universal Set = U: \newline % Row Count 1 (+ 1)
for any finite set A \newline % Row Count 2 (+ 1)
U = \{ x ∈ U | x ∉ A \} \newline % Row Count 3 (+ 1)
Aᶜ = U - A \newline % Row Count 4 (+ 1)
(Aᶜ)ᶜ = A \newline % Row Count 5 (+ 1)
Uᶜ = ∅ \newline % Row Count 6 (+ 1)
∅ᶜ = U% Row Count 7 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Relations}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{ARB = (a,b) ∈ R \newline % Row Count 1 (+ 1)
Identity: Ia = (a,a) \newline % Row Count 2 (+ 1)
Ex: \{(1,1),(2,2),(3,3), … \} \newline % Row Count 3 (+ 1)
Relation on set itself: R ⊆ A × A \newline % Row Count 4 (+ 1)
ARA is another way to write it too. \newline % Row Count 5 (+ 1)
Empty relation when R = ∅ \newline % Row Count 6 (+ 1)
implies that the relation R is empty, meaning it does not hold between any two pairs. It's essentially a relation with no elements. \newline % Row Count 9 (+ 3)
Complete relation when R = A × B \newline % Row Count 10 (+ 1)
implies that the relation R contains all possible pairs that can be formed by taking one element from set A and one element from set B. It's a relation where every element of A is related to every element of B. \newline % Row Count 15 (+ 5)
Inverse relation is R⁻$^{\textrm{1}}$ = \{ (b, a) | (a, b) ∈ R \} \newline % Row Count 17 (+ 2)
R consists of all pairs in R but with their elements reversed. If (a,b) is in R, then (b,a) is in R⁻$^{\textrm{1}}$ \newline % Row Count 20 (+ 3)
Composition of relations: \newline % Row Count 21 (+ 1)
R ∘ S = \{ (a, c) | ∃ b : (a, b) ∈ R and (b, c) ∈ S \} \newline % Row Count 23 (+ 2)
Set of pairs (a,c) such that exists an element b for which both (a,b) is in R and (b,c) is in S \newline % Row Count 25 (+ 2)
R ∘ R = R2 is a relation composed with itself \newline % Row Count 26 (+ 1)
(R ∘ S) ∘ T = R ∘ (S ∘ T) i.e it is associative (but not communitive) \newline % Row Count 28 (+ 2)
Ia ∘ R = R% Row Count 29 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{p{0.4577 cm} x{2.01388 cm} x{2.10542 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{3}{x{5.377cm}}{\bf\textcolor{white}{Order relation terms}}  \tn
% Row 0
\SetRowColor{LightBackground}
\seqsplit{Minimal} & An element a is minimal if there is no b such that b precedes a. & Elements with nothing less than them (no predecessors) \tn 
% Row Count 4 (+ 4)
% Row 1
\SetRowColor{white}
\seqsplit{Minimum} & An element a is a minimum if for all b, a precedes b & Element that is less than everything else (either a set has 1 minimum or no minimum element) \tn 
% Row Count 10 (+ 6)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Maximal} & An element a is maximal if there is no b such that a precedes b & follows from minimal (with greater than) \tn 
% Row Count 14 (+ 4)
% Row 3
\SetRowColor{white}
\seqsplit{Maximum} & An element a is a maximum if for all b, b precedes a & follows from minimum (with greater than) \tn 
% Row Count 18 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}---}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Power sets}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The power set of A is denoted as P(A) or 2A \newline % Row Count 1 (+ 1)
A ∈ P(A), ∅ ∈ P(A) \newline % Row Count 2 (+ 1)
If |A| = n, then |P(A)| = 2n \newline % Row Count 3 (+ 1)
|P(A)| = 2|A| \newline % Row Count 4 (+ 1)
If A = ∅, then P(A) = \{∅\} \newline % Row Count 5 (+ 1)
If A = \{1,2\}, then P(A) = \{∅, \{1\}, \{2\}, \{1, 2\}\} \newline % Row Count 6 (+ 1)
|A| \textless{} |P(A)|% Row Count 7 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Power set proofs}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{P(A) ∩ P(B) = P(A ∩ B) \newline % Row Count 1 (+ 1)
Forward Inclusion: Let 𝑋 be an arbitrary element in 𝑃(𝐴) ∩ 𝑃(𝐵). By definition of intersection, 𝑋 belongs to both 𝑃(𝐴) and 𝑃(𝐵). This implies 𝑋 is a subset of both 𝐴 and 𝐵. Consequently, 𝑋 is also a subset of their intersection, 𝐴 ∩ 𝐵. Thus, 𝑋 is an element of 𝑃(𝐴 ∩ 𝐵). Therefore, 𝑃(𝐴) ∩ 𝑃(𝐵) ⊆ 𝑃(𝐴 ∩ 𝐵). \newline % Row Count 9 (+ 8)
Reverse Inclusion: Let 𝑌 be an arbitrary element in 𝑃(𝐴 ∩ 𝐵). By definition, 𝑌 is a subset of 𝐴 ∩ 𝐵, hence a subset of both 𝐴 and 𝐵. Consequently, 𝑌 belongs to both 𝑃(𝐴) and 𝑃(𝐵). Thus, 𝑌 is an element of 𝑃(𝐴) ∩ 𝑃(𝐵). Therefore, 𝑃(𝐴 ∩ 𝐵) ⊆ 𝑃(𝐴) ∩ 𝑃(𝐵). \newline % Row Count 16 (+ 7)
Conclusion: Combining both directions of inclusion, we've demonstrated that 𝑃(𝐴) ∩ 𝑃(𝐵) ⊆ 𝑃(𝐴 ∩ 𝐵) and 𝑃(𝐴 ∩ 𝐵) ⊆ 𝑃(𝐴) ∩ 𝑃(𝐵), implying 𝑃(𝐴) ∩ 𝑃(𝐵) = 𝑃(𝐴 ∩ 𝐵). Thus, the equality holds.% Row Count 22 (+ 6)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{P(A) ∪ P(B) ≠ P(A ∪ B) \newline  \newline Example of why these aren't equal: \newline A = \{1\}, B = \{2\}, A ∪ B = \{1,2\} =\textgreater{} P(A ∪ B) = \{∅, \{1\}, \{2\}, \{1,2\}\} \newline  \newline P(A) = \{∅, \{1\}\}, P(B) = \{∅, \{2\}\} =\textgreater{} P(A) ∪ P(B) = \{∅, \{1\}, \{2\}\}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{More combinatorics}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{number of ways to place k balls in n boxes. \newline % Row Count 1 (+ 1)
P = permutation \newline % Row Count 2 (+ 1)
C = combination \newline % Row Count 3 (+ 1)
Order matters = sequence of choices \newline % Row Count 4 (+ 1)
Order doesn't matter = if we picked ball 1 then ball 2… it would be equivalent to picking ball 2 then ball 1 \newline % Row Count 7 (+ 3)
With replacement = same item can be picked several times \newline % Row Count 9 (+ 2)
Without replacement= each item is chosen at most, 1 time \newline % Row Count 11 (+ 2)
{\bf{Case 1}} \newline % Row Count 12 (+ 1)
K times out of n objects \newline % Row Count 13 (+ 1)
Number of functions from A to B \newline % Row Count 14 (+ 1)
|A| = K, |B| = n \newline % Row Count 15 (+ 1)
if A has 3 elements, and B has 5… we would get 5\textasciicircum{}3 total functions that can be defined \newline % Row Count 17 (+ 2)
{\bf{Case 2}} \newline % Row Count 18 (+ 1)
K unique balls in n small boxes (can only fit 1 item in each box) \newline % Row Count 20 (+ 2)
number of one to one functions from A to B \newline % Row Count 21 (+ 1)
{\bf{Case 3}} \newline % Row Count 22 (+ 1)
K identical balls in n small boxes \newline % Row Count 23 (+ 1)
Binomial coefficients \newline % Row Count 24 (+ 1)
{\bf{Case 4}} \newline % Row Count 25 (+ 1)
Bars and stars \newline % Row Count 26 (+ 1)
K identical balls in n numbered boxes (but each box can hold \textgreater{}= 0 balls)% Row Count 28 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}