feat: 1522 l2 notes
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@ -215,9 +215,7 @@ nth(Idx, [_|Tail], Elem) :-
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Idx1 is Idx - 1,
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nth(Idx1, Tail, Elem).
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% Taufiq
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nth(0, [V | _], V) :- !.
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nth(I, [_|T], V) :- nth(I1, T, V), I is I1 + 1, I > 0.
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%%%%%%%%%%%%%%
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@ -258,9 +256,6 @@ pos_helper(Target, [_|Tail], Acc, Index) :-
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Acc1 is Acc + 1,
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pos_helper(Target, Tail, Acc1, Index).
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% Taufiq
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pos(X, [X | _], 0).
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pos(X, [_ | T], I) :- pos(X, T, I1), I is I1 + 1.
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%%%%%%%%%%%%%%
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@ -294,9 +289,6 @@ all_pos_helper(Target, [_|Tail], Acc, Rest) :-
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Acc1 is Acc + 1,
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all_pos_helper(Target, Tail, Acc1, Rest).
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% Taufiq
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all_pos(X, L, []) :- not(member(X, L)).
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all_pos(X, L, LI) :- bagof(I, pos(X, L, I), LI).
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%%%%%%%%%%%%%%
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@ -538,9 +538,8 @@ def average_increase_in_cases(n_cases_increase, n_adj_entries_avg=7):
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'''
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sd_win = np.lib.stride_tricks.sliding_window_view(n_cases_increase, 2 * n_adj_entries_avg + 1, axis=1)
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avg = np.mean(sd_win, axis=2)
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avg = np.floor(np.mean(sd_win, axis=2))
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res = np.pad(avg, ((0, 0), (n_adj_entries_avg, n_adj_entries_avg)), 'constant', constant_values=(np.nan, np.nan))
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print(res)
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return res
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@ -552,7 +551,7 @@ def test_26():
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assert(np.array_equal(actual, expected, equal_nan=True))
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test_26()
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# test_26()
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# Task 2.7
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def is_peak(n_cases_increase_avg, n_adj_entries_peak=7):
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@ -605,9 +604,23 @@ def is_peak(n_cases_increase_avg, n_adj_entries_peak=7):
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Hint: to determine `n_adj_entries_avg` from `n_cases_increase_avg`,
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`np.count_nonzero` and `np.isnan` may be helpful.
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'''
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# get the number of paddings to add (count nans and / 2)
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paddings = int(np.count_nonzero(np.isnan(n_cases_increase_avg[0])) / 2)
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nanless = n_cases_increase_avg[:, paddings: -paddings]
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# find peaks (get the sliding window)
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sd_win = np.lib.stride_tricks.sliding_window_view(nanless, 2 * n_adj_entries_peak + 1, axis=1)
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mids = sd_win[:, :, n_adj_entries_peak]
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left_vals = np.max(sd_win[:, :, :n_adj_entries_peak], axis=2)
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right_vals = np.max(sd_win[:, :, n_adj_entries_peak+1:], axis=2)
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peaks = (mids > left_vals) & (mids >= right_vals)
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# TODO: add your solution here and remove `raise NotImplementedError`
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raise NotImplementedError
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means = np.nanmean(nanless, axis=1)[:, np.newaxis] * 0.1
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significant = mids > means
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result = np.logical_and(peaks, significant)
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res = np.pad(result, ((0, 0), (paddings+n_adj_entries_peak, paddings+ n_adj_entries_peak)), 'constant', constant_values=(False, False))
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return res
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def test_27():
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@ -627,7 +640,7 @@ def test_27():
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[False, False, False, False, False, False, False, False, False]])
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assert np.all(actual2 == expected2)
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#test_27()
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test_27()
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def visualise_increase(n_cases_increase, n_cases_increase_avg=None):
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'''
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@ -166,9 +166,12 @@ To solve a linear system we perform operations:
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\item add a constant multiple of an equation to another
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\end{itemize}
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Likewise, for a augmented matrix, the operations are on the \textbf{rows} of the augmented matrix
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\begin{itemize}
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\item Multiply row by nonzero constant
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\item Interchange 2 rows
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\item add a constant multiple of a row to another row
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\end{itemize}
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To note: all these operations are revertible
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@ -0,0 +1,352 @@
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\subsection{Recap}
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Given the linear equation $a_1x_1 + a_2x_2 + ... + a_nx_n = b$
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\begin{enumerate}
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\item $a_1 = a_2 = ... = a_n = b = 0$ zero equation
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Solution: $x_1 = t_1, x_2 = t_2, ... = x_n = t_n$
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\item $a_1 = a_2 = ... = a_n = 0 \neq b$ inconsistent
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No Solution
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\item Not all $a_1 ... a_n$ are zero.
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Set $n-1$ of $x_i$ as params, solve for last variable
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\end{enumerate}
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\subsection{Elementary Row Operations Example}
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\begin{center}
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\begin{minipage}{.3\linewidth}
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\systeme{x+y+3z = 0, 2x-2y+2z=4, 3x+9y=3}
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\end{minipage}%
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\begin{minipage}{.3\linewidth}
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\begin{equation*}
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\begin{amatrix}{3}
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1 & 1 & 3 & 0 \\
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2 & 2 & 2 & 4 \\
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3 & 9 & 0 & 3
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\end{amatrix}
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\end{equation*}
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\end{minipage}
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\end{center}
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\subsection{Row Equivalent Matrices}
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2 Augmented Matrices are row equivalent if one can be obtained from the other by a series of elementary row operations
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Given a augmented matrix $A$, how to find a row equivalent augmented matrix B of which is of a \textbf{simple} form?
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\subsection{Row Echelon Form}
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\begin{defn}[Row Echelon Form (Simple)]
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Augmented Matrix is in row-echelon form if
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\begin{itemize}
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\item Zero rows are grouped together at the bottom
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\item For any 2 successive nonzero rows, The first nonzero number in the lower row appears to the right of the first nonzero number on the higher row
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$\begin{amatrix}{4}
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0 & 0 & 1 & 2 & 3 \\
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0 & 0 & 0 & 1 & 2 \\
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\end{amatrix}$
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\item Leading entry if a nonzero row is a \textbf{pivot point}
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\item Column of augmented matrix is called
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\begin{itemize}
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\item \textbf{Pivot Column} if it contains a pivot point
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\item \textbf{Non Pivot Column} if it contains no pivot point
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\end{itemize}
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\item Pivot Column contains exactly 1 pivot point
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\# of pivots = \# of leading entries = \# of nonzero rows
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\end{itemize}
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\end{defn}
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Examples of row echlon form:
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\begin{equation*}
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\begin{amatrix}{2}
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3 & 2 & 1 \\
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\end{amatrix}
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\begin{amatrix}{2}
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1 & -1 & 0 \\
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0 & 1 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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2 & 1 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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1 & 2 & 3 & 4 \\
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0 & 1 & 1 & 2 \\
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0 & 0 & 2 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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0 & 1 & 2 & 8 & 1 \\
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0 & 0 & 0 & 0 & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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Examples of NON row echlon form:
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\begin{equation*}
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\begin{amatrix}{2}
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0 & \textbf{1} & 0 \\
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\textbf{1} & 0 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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0 & 0 & \textbf{1} \\
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\textbf{1} & -1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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\textbf{1} & 0 & 2 & 1 \\
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0 & \textbf{1} & 0 & 2 \\
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0 & \textbf{1} & 1 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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\textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} & \textbf{0} \\
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1 & 0 & 2 & 0 & 1 \\
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0 & 0 & 0 & 1 & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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\subsection{Reduced Row-Echelon Form}
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\begin{defn}[Reduced Row-Echelon Form]
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Suppose an augmented matrix is in row-echelon form. It is in \textbf{reduced row-echelon form} if
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\begin{itemize}
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\item Leading entry of every nonzero row is 1
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\subitem Every pivot point is one
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\item In each pivot column, except the pivot point, all other entries are 0.
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\end{itemize}
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\end{defn}
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Examples of reduced row-echelon form:
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\begin{equation*}
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\begin{amatrix}{2}
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1 & 2 & 3 \\
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\end{amatrix}
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\begin{amatrix}{2}
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0 & 0 & 0 \\
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0 & 0 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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1 & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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1 & 0 & 0 & 1 \\
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0 & 1 & 0 & 2 \\
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0 & 0 & 1 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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0 & 1 & 2 & 0 & 1 \\
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0 & 0 & 0 & 1 & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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Examples of row-echelon form but not reduced: (pivot point is not 1 / all other elements \textbf{in pivot column} must be zero)
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\begin{equation*}
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\begin{amatrix}{2}
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\textbf{3} & 2 & 1 \\
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\end{amatrix}
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\begin{amatrix}{2}
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1 & \textbf{-1} & 0 \\
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0 & 1 & 0 \\
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\end{amatrix}
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\begin{amatrix}{2}
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\textbf{2} & 0 & 0 \\
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0 & 1 & 0 \\
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0 & 0 & 1 \\
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\end{amatrix}
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\begin{amatrix}{3}
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\textbf{-1} & 2 & 3 & 4 \\
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0 & 1 & 1 & 2 \\
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0 & 0 & 2 & 3 \\
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\end{amatrix}
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\begin{amatrix}{4}
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0 & 1 & 2 & \textbf{8} & 1 \\
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0 & 0 & 0 & \textbf{4} & 3 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}
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\end{equation*}
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To note: 2nd matrix has -1 in the pivot column, but 5th matrix has 2 in a non-pivot column so its fine
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\subsection{Solving Linear System}
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If Augmented Matrix is in reduced row-echelon form, then solving it is easy
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\begin{equation*}
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\begin{amatrix}{3}
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1 & 0 & 0 & 1 \\
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0 & 1 & 0 & 2 \\
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0 & 0 & 1 & 3 \\
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\end{amatrix}
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\text{then } x_1 = 1, x_2 = 2, x_3 = 3
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\end{equation*}
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\begin{note}
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\begin{itemize}
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\item If any equations in the system is inconsistent, the whole system is inconsistent
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\end{itemize}
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\end{note}
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\subsubsection{Examples}
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Augmented Matrix: $\begin{amatrix}{4}
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1 & -1 & 0 & 3 & -2 \\
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0 & 0 & 1 & 2 & 5 \\
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0 & 0 & 0 & 0 & 0 \\
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\end{amatrix}$
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\begin{itemize}
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\item The zero row can be ignored.
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\systeme{x_1 - x_2 + 3x_4 = -2, x_3 + 2x_4 = 5}
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\item Degree of freedom(\# cols): 4, number of restrictions (\# pivot cols): 2, arbitrary vars(\# non pivot cols): 4-2 = 2. Set this to the non-pivot cols
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\end{itemize}
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\begin{enumerate}
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\item Let $x_4 = t$ and sub into 2nd eqn
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\subitem $x_3 + 2t = 5 \Rightarrow x_3 = 5-2t$
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\item sub $x_4 = t$ into 1st eqn
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\subitem $x_1 - x_2 + 3t = -2$
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\subitem Let $x_2 = s$. Then $x_1 = -2 + s - 3t$
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\item Infinitely many sols with ($s$ and $t$ as arbitrary params)
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\subitem $x_1 = -2 + s - 3t, x_2 = s, x_3 = 5-2t, x_4 = t$
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\end{enumerate}
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Augmented Matrix: $\begin{amatrix}{5}
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0 & 2 & 2 & 1 & -2 & 2 \\
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0 & 0 & 1 & 1 & 1 & 3 \\
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0 & 0 & 0 & 0 & 2 & 4 \\
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\end{amatrix}$
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\begin{itemize}
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\item \systeme{0x_1 + 2x_2 + 2x_3 + 1x_4 -2x_5 = 2,x_3 + x_4 +x_5 = 3,2x_5 = 4}
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\item Degree of freedom: 5, number of restrictions: 3, arbitrary vars: 5-3 = 2
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\end{itemize}
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\begin{enumerate}
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\item by 3rd eqn, $2x_5 = 4 \Rightarrow x_5 = 2$
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\item sub $x_5 = 2$ into 2nd eqn
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\subitem $x_3 + x_4 + 2 = 3 \Rightarrow x_3 + x_4 = 1$
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\subitem let $x_4 = t$. Then $x_3 = 1-t$
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\item sub $x_5 = 2, x_3 = 1-t, x_4 = t$ into 1st eqn
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\subitem $2x_2 + 2(1-t) + t - 2(2) = 2 \Rightarrow 2x_2 -t = 4 \Rightarrow x_2 = \frac{t}{2} + 2$
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\item system has inf many solns: $x_1 = s, x_2 = \frac{t}{2} + 2, x_3 = 1-t, x_4 = t, x_5 = 2$ where $s$ and $t$ are arbitrary
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\end{enumerate}
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\subsubsection{Algorithm}
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Given the augmented matrix is in row-echelon form.
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\begin{enumerate}
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\item Set variables corresponding to non-pivot columns to be arbitrary parameters
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\item Solve variables corresponding to pivot columns by back substitution (from last eqn to first)
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\end{enumerate}
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\subsection{Gaussian Eliminiation}
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\begin{defn}[Gaussian Elimination]\ \\
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\begin{enumerate}
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\item Find the left most column which is not entirely zero
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\item If top entry of such column is 0, replace with nonzero number by swapping rows
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\item For each row below top row, add multiple of top row so that leading entry becomes 0
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\item Cover top row and repeat to remaining matrix
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\end{enumerate}
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\end{defn}
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\begin{note}[Algorithm with Example]\ \\
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$\begin{amatrix}{6}
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0 & 0 & 0 & 2 & 4 & 2 & 8 \\
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0 & 1 & 2 & 4 & 5 & 3 &-9 \\
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0 &-2 &-4 &-5 &-4 & 3 & 6 \\
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\end{amatrix}$
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\begin{enumerate}
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\item Find the left most column which is not all zero (2nd column)
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\item Check top entry of the selection. If its zero, replace it by a nonzero number by interchanging the top row with another row below
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\subitem $\begin{amatrix}{6}
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0 & 1 & 2 & 4 & 5 & 3 &-9 \\
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0 & 0 & 0 & 2 & 4 & 2 & 8 \\
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0 &-2 &-4 &-5 &-4 & 3 & 6 \\
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\end{amatrix}$
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\item For each row below the top row, adda suitable multiple of top row so that leading entry becomes 0.
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\subitem $2R_1 + R_3$ will ensure that the -2 turns to 0
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\subitem $\begin{amatrix}{6}
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0 & 1 & 2 & 4 & 5 & 3 &-9 \\
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0 & 0 & 0 & 2 & 4 & 2 & 8 \\
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0 & 0 & 0 & 3 & 6 & 9 &-12\\
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\end{amatrix}$
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\item Cover top row and repeat procedure to the remaining matrix
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\subitem $\begin{amatrix}{6}
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0 & 1 & 2 & 4 & 5 & 3 &-9 \\
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\hline
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0 & 0 & 0 & 2 & 4 & 2 & 8 \\
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0 & 0 & 0 & 3 & 6 & 9 &-12\\
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\end{amatrix}$
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\subitem Look at $C_4$. $R_3 \times -1.5R_2$ will set $R_3C_4$ to zero.
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\subitem $\begin{amatrix}{6}
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0 & 1 & 2 & 4 & 5 & 3 &-9 \\
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\hline
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0 & 0 & 0 & 2 & 4 & 2 & 8 \\
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0 & 0 & 0 & 0 & 0 & 6 & -24\\
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\end{amatrix}$
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\subitem This is now in row echelon form.
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\end{enumerate}
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Only use $R_i \Leftrightarrow R_j or R_i + CR_j$ in this method.
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\end{note}
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\subsection{Gauss-Jordan Elimination}
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\begin{defn}[Gauss Joran Elimination]\ \\
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\begin{enumerate}
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\item[1-4.] Use Gaussian Eliminiation to get row-echelon form
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\setcounter{enumi}{4}
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\item For each nonzero row, multiply a suitable constant so pivot point becomes 1
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\item Begin with last nonzero row and work backwords
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\subitem Add suitable multiple of each row to the rows above to introduce 0 above pivot point
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\end{enumerate}
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\begin{itemize}
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\item Every matrix has a unique reduced row-echelon form.
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\item Every nonzero matrix has infinitely many row-echelon ofrm
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\end{itemize}
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\end{defn}
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\begin{note}[Gauss Jordan Elimination Example] Suppose an augmented matrix is in row-echelon form.
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$\begin{amatrix}{5}
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1 & 2 & 4 & 5 & 3 & -9 \\
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0 & 0 & 2 & 4 & 2 & 8 \\
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0 & 0 & 0 & 0 & 6 & -24 \\
|
||||
\end{amatrix}$
|
||||
\begin{enumerate}
|
||||
\item All pivot points must be 1
|
||||
\subitem multiply $R_2$ by $\frac{1}{2}$ and $R_3$ by $\frac{1}{6}$
|
||||
\subitem $\begin{amatrix}{5}
|
||||
1 & 2 & 4 & 5 & 3 & -9 \\
|
||||
0 & 0 & 1 & 2 & 1 & 4 \\
|
||||
0 & 0 & 0 & 0 & 1 & -4 \\
|
||||
\end{amatrix}$
|
||||
\item In each pivot col, all entries other than pivot point must be 0. Work backwards
|
||||
\subitem $R_1 + -3R_1$, $R_2 + -R_1$
|
||||
\subitem $\begin{amatrix}{5}
|
||||
1 & 2 & 4 & 5 & 0 & 3 \\
|
||||
0 & 0 & 1 & 2 & 0 & 8 \\
|
||||
0 & 0 & 0 & 0 & 1 & -4 \\
|
||||
\end{amatrix}$
|
||||
\subitem $R_1 + -4R_2$
|
||||
\subitem $\begin{amatrix}{5}
|
||||
1 & 2 & 0 & -3 & 0 & -29 \\
|
||||
0 & 0 & 1 & 2 & 0 & 8 \\
|
||||
0 & 0 & 0 & 0 & 1 & -4 \\
|
||||
\end{amatrix}$
|
||||
|
||||
\end{enumerate}
|
||||
\end{note}
|
||||
|
||||
|
||||
Binary file not shown.
@ -69,6 +69,10 @@
|
||||
This is a note
|
||||
\end{note}
|
||||
|
||||
\begin{defn}[Some Term]
|
||||
This is a definition
|
||||
\end{defn}
|
||||
|
||||
% Maybe I need to add one more part: Examples.
|
||||
% Set style and colour later.
|
||||
|
||||
|
||||
@ -9,7 +9,10 @@
|
||||
\usepackage[english]{babel}
|
||||
\usepackage{framed}
|
||||
\usepackage[dvipsnames]{xcolor}
|
||||
\usepackage{soul}
|
||||
\usepackage{tcolorbox}
|
||||
\usepackage{systeme}
|
||||
\usepackage{multicol}
|
||||
|
||||
\colorlet{LightGray}{White!90!Periwinkle}
|
||||
\colorlet{LightOrange}{Orange!15}
|
||||
@ -34,6 +37,10 @@
|
||||
\declaretheorem[style=notesty,numbered=no]{note}
|
||||
\tcolorboxenvironment{note}{colback=LightRed}
|
||||
|
||||
\declaretheoremstyle[name=Definition,]{defnsty}
|
||||
\declaretheorem[style=defnsty,numbered=no]{defn}
|
||||
\tcolorboxenvironment{defn}{colback=LightGray}
|
||||
|
||||
\setstretch{1.2}
|
||||
\geometry{
|
||||
textheight=9in,
|
||||
|
||||
Loading…
Reference in New Issue
Block a user