.

CS1029/notes-2023-11-15

@@ -0,0 +1 @@
+

CS1029/notes-2023-11-28

@@ -0,0 +1 @@
+

CS1029/practical-2023-11-17/combinatorics

@@ -0,0 +1,37 @@
+1. a) 5 8 7 * * = 280
+   b) 6 1 + 6 * 1 + 6 * = 258
+   c) 26 2 ^ 36 4 ^ * = 1 135 420 416
+   d) 17 3 P = 4080
+   e) 26 2 P 34 4 P * = 723 465 600
+   f) 8 5 C 11 7 C * = 18 480
+
+2. a) 8 2 C 5 2 C 3 2 C * * = 840
+   b) 8 2 C 14 4 C * = 28 028
+
+3. a) (sides + 1), or 7 for the likely indented d6
+   b) (sides_1 + sides_2), or 1 for the likely intended d6
+   c) 5
+
+4. (Assuming the unaccented latin alphabet only)
+   26 3 ^ = 17 576
+
+5. 26 2 ^ 10 4 ^ 10 2 ^ + 26 4 ^ * * = 3 120 049 337 600
+
+6. 99 50 * 1 + = 4 951
+
+7. a) 10 4 C = 210
+   b) 10 4 C 10 3 C 10 2 C 10 1 C + + + = 385
+   c) 10 4 C 10 5 C 10 6 C 10 7 C 10 8 C 10 9 C 1 + + + + + + = 848
+
+8. a) 100 4 P = 94 109 400
+   b) 99 3 P = 941 094
+   c) 99 4 P = 90 345 024
+   d) 97 4 ! * = 2 328
+
+9. a) 25 4 C = 12 650
+   b) 25 4 P = 303 600
+
+10. 3 5 ^ = 243
+
+11. 5 3 ^ 3 ! / = 20
+

CS1029/practical-2023-11-24/probability

@@ -0,0 +1,24 @@
+01. 4 52 / = 1/13
+02. 50 100 / = 1/2
+03. 1 1 52 5 C / - = 2598959/2598960
+04. 4 1 C 52 5 C / 4 2 C 52 5 C / 4 3 C 52 5 C / 1 52 5 C / sum =  1/173264
+05.
+ a) 1 50 5 C / = 1/2118760
+ b) 1 52 5 C / = 1/2598960
+ c) 1 56 5 C / = 1/3819816
+ d) 1 60 5 C / = 1/5461512
+06.
+ a) 1 200 3 P / = 1/7880400
+ b) 1 200 / 3 ^ = 1/8000000
+07.
+ a) 1 26 ! / = 1/403291461126605635584000000
+ b) 1 26 / = 1/26
+08. 4 3 C 1 2 / 4 ^ * = 1/4
+09. 6 1 C 1 2 / 6 ^ * = 3/32
+10. 2 5 / 1 2 / * 1 3 / / = 3/5
+11. P(U|P) = P(P|U)P(U) / P(P|U)P(U) + P(P|U')P(U')
+  ) 96 % 8 % * 96 % 8 % * 1 8 % - 9 % * + / = 64/133
+12.
+ a) 8 % 98 % * 8 % 98 % * 1 8 % - 3 % * + / = 196/265
+ b) 1 196 265 / - = 69/265
+ c) 2 % 8 % * 2 % 8 % * 92 % 97 % * + / = 4/2235

CS1032/lecture-2023-11-23/crypt.py

@@ -0,0 +1,17 @@
+from hashlib import sha256
+from itertools import cycle
+import base64
+
+def crypt(msg: bytes, key: str) -> bytes:
+	key_h = sha256()
+	key_h.update(key.encode('utf8'))
+	key = key_h.digest()
+
+	return bytes(p ^ k for (p, k) in zip(msg, cycle(key)))
+
+def encrypt(msg: str, key: str) -> str:
+	return base64.b64encode(crypt(bytes(msg, 'utf8'), key)).decode('utf8')
+
+def decrypt(msg: str, key: str) -> str:
+	return crypt(base64.b64decode(msg), key).decode('utf8')
+

CS1032/lecture-2023-11-23/nqueens.pl

@@ -0,0 +1,19 @@
+safe(Q1X-Q1Y, Q2X-Q2Y) :-
+	between(0, 7, Q1Y),
+	between(0, 7, Q2Y),
+	Q1X =\= Q2X,
+	Q1Y =\= Q2Y,
+	abs(Q1X - Q2X) =\= abs(Q1Y - Q2Y).
+safe(_, []).
+safe(QN, [Q | QS]) :- safe(QN, [Q | QS], 1).
+safe(_, [], _).
+safe(QN, [Q | QS], QX) :- safe(0-QN, QX-Q), safe(QN, QS, QX + 1).
+
+add(Qs, QsN) :-
+	QsN = [Q | Qs],
+	safe(Q, Qs).
+
+add(1, Qs, QsN) :- add(Qs, QsN).
+add(N, Qs, QsN) :- add(Qs, QA), add(M, QA, QsN), succ(M, N).
+
+solution(Qs) :- add(8, [], Qs).

CS1032/lecture-2023-11-23/nqueens.py

@@ -0,0 +1,18 @@
+def safe(q1, q2):
+    (q1x, q1y) = q1
+    (q2x, q2y) = q2
+    return q1x != q2x and q1y != q2y and abs(q1x - q2x) != abs(q1y - q2y)
+
+def possible_positions(current, n):
+    return ((*current, y) for y in range(n) if all(safe((len(current), y), q) for q in enumerate(current)))
+
+def solutions_r(current, n = 8):
+    if len(current) == n:
+        yield current
+        return
+
+    for pos in possible_positions(current, n):
+        yield from solutions_r(pos, n)
+
+def solutions(n = 8):
+    return solutions_r((), n)

CS1032/lecture-2023-11-23/pat.py

@@ -0,0 +1,7 @@
+def l(n=9):
+    for x in range(1, n):
+        print(" " * (x % 2), "*")
+
+def r(n=5):
+    for x in range(n, -1, -1):
+        print("*" * (x * 2 + 1))

CS1032/notes-2023-11-16

@@ -0,0 +1 @@
+

CS1032/practical-2023-11-15/decomment.py

@@ -0,0 +1,4 @@
+import sys
+
+for line in sys.stdin:
+    print(line.split('#')[0].rstrip())

CS1032/practical-2023-11-15/pass.py

@@ -0,0 +1,11 @@
+#! /usr/bin/env python3
+import sys, random
+
+with open(sys.argv[1]) as words_f:
+    words = [word for line in words_f if 3 <= len(word := line.strip()) < 10 and word.lower() == word]
+
+pair = []
+while not 8 <= sum(map(len, pair)) <= 10:
+    pair = random.choices(words, k=2)
+
+print(''.join(w.title() for w in pair))

CS1032/practical-2023-11-15/redact.py

@@ -0,0 +1,6 @@
+#! /usr/bin/env python3
+import sys, re
+redacted = re.compile(f"({'|'.join([s.strip() for s in open(sys.argv[1])])})")
+
+for line in sys.stdin:
+	print(redacted.sub(lambda m: '*' * len(m.group(0)), line), end='')

CS1032/practical-2023-11-15/sum.py

@@ -0,0 +1,9 @@
+#! /usr/bin/env python3
+
+sum = 0
+while True:
+    i = input(f"[{sum:g}]: ")
+    if not i: break
+    try:
+        sum += float(i)
+    except ValueError: print(f"\t'{i}' is not a number")

CS1032/practical-2023-11-22/search.py

@@ -0,0 +1,33 @@
+#! /usr/bin/env python3
+
+def sort(list):
+    if len(list) <= 1: return list
+    pivot = list.pop()
+    return sort([v for v in list if v < pivot]) + [v for v in list if v == pivot] + [pivot] + sort([v for v in list if v > pivot])
+
+def search(list, elem, start=0):
+    if list == [elem]:
+        return start
+    elif len(list) <= 1:
+        return None
+    i = len(list) // 2
+    pivot = list[i]
+    if elem > pivot:
+        return search(list[i:], elem, start = i + start)
+    elif elem < pivot:
+        return search(list[:i], elem, start=start)
+    elif elem == pivot:
+        return i + start
+
+    raise ValueError("Got NaN!")
+
+if __name__ == '__main__':
+    l = []
+    while line := input('> '):
+        l.append(line)
+
+    i = search(sort(l), input('Goal: '))
+    if i:
+        print(f"Found at {i}")
+    else:
+        print("Not present")

MA1006/linear.tex

@@ -45,7 +45,7 @@ Solutions invariant.
 \end{align*}
 \section*{As Matrices}
 \begin{align*}
-	\systeme{
+   & \systeme{
 		x + 2y = 1,
 		2x - y = 3
 	}
@@ -53,8 +53,8 @@ Solutions invariant.
 	\begin{pmatrix}[cc|c]
 		1 & 2  & 1 \\
 		2 & -1 & 3
-	\end{pmatrix}
+	\end{pmatrix}                                   \\
-	                               & \systeme{
+   & \systeme{
 		x - y + z = -2,
 		2x + 3y + z = 7,
 		x - 2y - z = -2
@@ -347,12 +347,107 @@ Inversion of a linear transformation is equivalent to inversion of its represent
 \[ \forall A: \text{$n$x$n$ matrix}.\quad P_A\paren{\lambda} = \det{\paren{A - \lambda I_n}} \tag{characteristic polynomial in $\lambda$}\]
 Eigenvalues of $A$ are the solutions of $P_A\paren{\lambda} = 0$
 \begin{align*}
-	& A\vec{x} = \lambda\vec{x} & x \neq 0\\
+	     & A\vec{x} = \lambda\vec{x}                                             & x \neq 0 \\
-	\iff & A\vec{x} - \lambda\vec{x} = 0 \\
+	\iff & A\vec{x} - \lambda\vec{x} = 0                                                    \\
-	\iff & (A - \lambda I_n)\vec{x} = 0 \\
+	\iff & (A - \lambda I_n)\vec{x} = 0                                                     \\
-	\iff & \det{\paren{A - \lambda I_n}} = 0 \\
+	\iff & \det{\paren{A - \lambda I_n}} = 0                                                \\
-	& \quad \text{ or $\paren{A - \lambda I_n}$ is invertible and $x = 0$ }
+	     & \quad \text{ or $\paren{A - \lambda I_n}$ is invertible and $x = 0$ }
 \end{align*}
-\[ P_{R\theta}(\lambda) = \frac{2\cos{\theta} \pm \sqrt{-4\lambda^2\sin^2{\theta}}}{2}\]
+\[ P_{R\theta}(\lambda) \text{ has roots } \frac{2\cos{\theta} \pm \sqrt{-4\lambda^2\sin^2{\theta}}}{2}\]
 \[ R_\theta \text{ has eigenvalues }\iff \sin{\theta} = 0 \]
-\end{document}
+\subsubsection*{Example}
+\begin{align*}
+	A                        & = \begin{pmatrix} 4 & 0 & 1 \\ -2 & 1	& 0 \\ -2 & 0 & 1 \end{pmatrix}                                                                                                                                                   \\
+	P_A(\lambda)             & = \det{A - \lambda I_3} = \det{\begin{pmatrix} 4 - \lambda & 0 & 1 \\ -2 & 1 - \lambda & 0 \\ -2 & 0 & 1 - \lambda \end{pmatrix}} = (1 - \lambda)\det{\begin{pmatrix}4 - \lambda & 1 \\ -2 & 1 - \lambda \end{pmatrix}} \\
+	                         & = (1 - \lambda)\paren{(4 - \lambda)(1 - \lambda) + 2} = (1 - \lambda)(\lambda^2 - 5\lambda + 6)                                                                                                                         \\
+	                         & = (1 - \lambda)(2 - \lambda)(3 - \lambda)                                                                                                                                                                               \\
+	\lambda                  & = 1, 2, 3                                                                                                                                                                                                               \\
+	A\vec{x}                 & = \lambda\vec{x}~\forall. \text{ eigenvectors } \vec{x}                                                                                                                                                                 \\
+	(A - \lambda I_n)\vec{x} & = 0                                                                                                                                                                                                                     \\
+	                         & \begin{pmatrix} 3 & 0 & 1 \\ -2 & 0 & 0 \\ -2 & 0 & 0 \end{pmatrix}\vec{x} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}                                                                                                  \\
+	                         & \text{ eigenvectors with eigenvalue 1 are } s\vec{e}_2~~\forall~s \in \R, \neq 0                                                                                                                                        \\
+	(A - 2I_n)\vec{x}        & = 0                                                                                                                                                                                                                     \\
+	                         & \begin{pmatrix} 2 & 0 & 1 \\ -2 & -1 & 0 \\ -2 & 0 & -1 \end{pmatrix}\vec{x} = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}                                                                                                 \\
+	                         & \systeme{
+		2x_1 + x_3 = 0,
+		-2x_1 - x_2 = 0,
+		-2x_1 -x_3 = 0
+	}: s\begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix}                                                                                                                                                                                                    \\ \\
+	B                        & = \begin{pmatrix} 5 & 3 & 3 \\ -3 & -1 & -3 \\ -1 & -3 & -1 \end{pmatrix}
+\end{align*}
+Repeated roots of the characteristic polynomial lead to multiple variables.
+\[ \forall \text{ matrices } A.~A \text{ is invertible } \iff 0 \text{ is not an eigenvalue }\]
+\begin{align*}
+	\text{If $0$ is an eigenvalue, } P_A(0) = 0 \therefore \det{\paren{A - 0I_n}} = 0
+\end{align*}
+\[ P_A(\lambda) = \det{\paren{(-I_n)(\lambda I_n - A)}} = \det{-I_n}\det{\lambda I_n - A}\]
+\[ = (-1)^n \lambda^n + c_{n-1}\lambda^{n-1} ... \]
+\begin{description}
+	\item[Trace] The sum of the diagonal of a matrix
+		\[ c_{n - 1} = (-1)^{n+1}\operatorname{tr}A \]
+	\item[Cayley-Hamilton Theorem:] \( P_A(A) = 0_n \)
+\end{description}
+\subsubsection*{Example}
+\begin{align*}
+	A            & = \begin{pmatrix} 1 & 4 \\ 3 & 2 \end{pmatrix} \\
+	P_A(\lambda) & = (1 - \lambda)(2 - \lambda) - 12              \\
+	             & = 2 - 3\lambda + \lambda^2 - 12                \\
+	             & = \lambda^2 - 3\lambda - 10                    \\
+	P_A          & (5 \text{ or } -2) = 0                         \\
+	P_A(A)       & = 0                                            \\
+	A^2          & = 3A - 10I_2                                   \\
+	A^3          & = (3A - 10I_2)A                                \\
+	A^{n + 2}    & = 3A^{n+1} + 10A^n
+\end{align*}
+\subsection*{Diagonalization}
+\begin{description}
+	\item[Similarity of matrices] $A$ and $B$ are \emph{similar} iff there exists an invertible matrix $P$ such that $B = P^{-1}AP$
+		\[ T_\alpha \text{ is similar to } T_0 (P = R_{-\alpha}) \]
+\end{description}
+
+For similar $A$, $B$:
+\begin{itemize}
+	\item $\det{A}$ = $\det{B}$
+	\item $P_A(\lambda) = P_B(\lambda)$
+	      \[\det{\paren{A - \lambda I_n}} = \det{\paren{PBP^{-1} - \lambda PP^{-1}}} = \det{\paren{P(B - \lambda I_n)P^{-1}}}\]
+	      \[ (A - \lambda I_n) \text{ and } (B - \lambda I_n) \text{ are similar }\]
+	\item eigenvalues are the same
+	\item trace is the same
+\end{itemize}
+
+\begin{description}
+	\item[Diagonalizable Matrix] a square matrix that is similar to a diagonal matrix
+		\[ P^{-1}AP = D \]
+		$P$ diagonalizes $A$ \quad
+		(P is not necessarily unique)
+\end{description}
+An $n$x$n$ matrix $A$ is dagonalizable iff there exists a matrix $P = (\vec{x}_1, \vec{x}_2, ...)$, where $\vec{x}_i$ is an eigenvector of $A$
+\[ P^{-1}AP = \begin{pmatrix} \lambda_1 & 0 & \cdots & \cdots \\ 0 & \lambda_2 & 0 & \cdots \\ \vdots & \vdots & \ddots & \cdots \end{pmatrix} \]
+i.e. it's diagonal, with the $ii^\text{th}$ element equal to the eigenvalue corresponding to $\vec{x}_i$
+\subsubsection*{Example}
+\begin{align*}
+	A         & = \begin{pmatrix} 4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1 \end{pmatrix} \\
+	\lambda_1 & = \begin{pmatrix} 0 \\ s \\ 0 \end{pmatrix}                           \\ \lambda_2 & = \begin{pmatrix} t \\ -2t \\ -2t \end{pmatrix} \\ \lambda_3 & = \begin{pmatrix} u \\ -u \\ -u \end{pmatrix} \\
+	P         & = \begin{pmatrix}
+		              0 & 1  & 1  \\
+		              1 & -2 & -1 \\
+		              0 & -2 & -1
+	              \end{pmatrix}                                                      \\
+	P^{-1}    & = \begin{pmatrix}
+		              0 & 1 & -1 \\ -1 & 0 & -1 \\ 2 & 0 & 1
+	              \end{pmatrix}                               \\
+	P^{-1}AP  & = \begin{pmatrix}
+		              1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3
+	              \end{pmatrix}
+\end{align*}
+\subsection*{Linear Independence}
+For any set of vectors $V = { v_i }~\forall_{i<n}$ in $R^n$, $v_i$ are linearly independent if
+\[ \sum k_i v_i = 0 \implies \forall i.~ k_i = 0 \quad (k_i \in \R) \]
+This implies that no vector in $V$ can be written as a sum of scalar multiples of the others. \\
+A square matrix is invertible iff its columns are linearly independent in $R^n$
+\\ An $n$x$n$ matrix $A$ is diagonalizable iff it admits $n$ linearly independent eigenvectors \\
+For $\vec{x}_i$ eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_i$, $\{ \vec{x}_i \}$ is linearly independent. \\
+Note: the inverse is \emph{not} true: \( \exists \text{ matrices } $A$\) with repeated eigenvalues and linearly independent eigenvectors.
+\[ \forall k \in \N, A = PDP^{-1}.~A^k = (PDP^{-1})^k = PD^kP^{-1} \]
+Exponentiation of a diagonal matrix is elementwise
+\end{document}