La Fée Verte @Absinthe@qoto.org

Still trying to find a style i like for the double L... Thoughts on this one?

#calligraphy #handwriting #writing #copperplate

Valid

@Absinthe

Should be fairly easy to modify for other bracket types, just add them to the lists at the top and bottom (using the same index for an open bracket and its corresponding close bracket).

function valid = is_balanced(string)
open_brackets = '({[';
close_brackets = ')}]';
expecting = false(size(string));
valid = true;
expected_bracket = [];
for index = 1:numel(string)
if ismember(string(index), open_brackets)
expecting(index) = true;
expected_bracket = strchr(open_brackets, string(index));
elseif ismember(string(index), close_brackets)
if expected_bracket == strchr(close_brackets, string(index))
expecting(find(expecting, 1, 'last')) = false;
expected_bracket = strchr(open_brackets, string(find(expecting, 1, 'last')));
else
valid = false; end; end; end;
if any(expecting)
valid = false; end; end;

@Absinthe

Matlab normally represents strings as 1D arrays of chars. It's not usual to put chars in an 2D array because all the rows would have to be the same length, and that's pretty inflexible. Octave is happy to make such an array just as it would with any other type, though, and here that's exactly what we want.

function array = justify_text(string, width)
words = strtrunc(strsplit(string), width);
array = zeros(0, width);
line_length = 0;
start_index = 1;
for word_index = 1:numel(words)
this_word = words{word_index};
if line_length + numel(this_word) > width
array = [array; render_words(start_index, word_index - 1)];
start_index = word_index;
line_length = numel(this_word) + 1;
else
line_length += numel(this_word) + 1; end; end;
array = [array; render_words(start_index, numel(words))];
function this_line = render_words(start_index, end_index)
chars_raw = cellfun(@numel, words(start_index:end_index));
total_padding = width - sum(chars_raw);
if start_index == end_index
this_line = [words{start_index} repmat(' ', 1, total_padding)];
else
base_padding = floor(total_padding / (end_index - start_index));
extra_padding = rem(total_padding, end_index - start_index);
next_char = 1;
for index = start_index:end_index
this_padding = 0;
if index < end_index
this_padding = base_padding;
if index - start_index < extra_padding
this_padding++; end; end;
this_line(next_char:(next_char + chars_raw(index - start_index + 1) - 1)) = words{index};
next_char += chars_raw(index - start_index + 1);
this_line(next_char:(next_char + this_padding - 1)) = repmat(' ', 1, this_padding);
next_char += this_padding; end; end; end; end;

@Absinthe

This was super straightforward. Linear in time and space, plus whatever the sort algorithm needs (by default it's nlogn in Octave but you can implement other ones).

function room_count = min_rooms(class_times)
[~, indices] = sort(class_times'(1:numel(class_times)));
current = 0;
room_count = 0;
for index = indices
if mod(index, 2)
current++;
room_count = max(room_count, current);
else
current--; end; end; end;

@Absinthe@qoto.org I just did, pretty interesting.

#toyprogrammingchallenge

Okay, here's one. I just finished helping someone worth through this. So it's pretty fun.

Write a program in Python, that can accept a paragraph string and and page
width and return an array of left AND right justified strings. NOTE: No words
should be broken, the beginning and end of the line should be characters).

You should provide instructions on how to execute your program and provide a
sample output.

Example:

Sample input:

Paragraph = "This is a sample text but a complicated problem to be solved, so
we are adding more text to see that it actually works."

Page Width = 20

Output should look like this:

Array [1] = "This is a sample"
Array [2] = "text but a"
Array [3] = "complicated problem"
Array [4] = "to be solved, so we"
Array [5] = "are adding more text"
Array [6] = "to see that it"
Array [7] = "actually works.”

@Absinthe@qoto.org
def encode(input):
out = ""

let = input[0]
num = 1

for i in range(1, len(input)):
if input[i] == input[i-1]:
num = num + 1
else:
out = out + str(num) + input[i-1]
let = input[i]
num = 1

out = out + str(num) + input[len(input)-1]

return out

def decode(input):
out = ""

i = 0

while i != (len(input)):
for p in range(0, int(input[i])):
out = out + input[i+1]
i = i + 2

return out

Here is a #toyprogrammingchallenge which corresponds to the general case of a problem I ran up against recently.

Given a positive integer K and a directed graph G with weighted edges, return a new graph H satisfying all the following conditions, if such a graph exists:

1. G and H contain exactly the same set of vertices.
2. H contains only edges in G, but G may contain edges not in H.
3. A path exists in H of length at most K between each pair of vertices in each direction.
4. No edge can be removed from H while still satisfying condition 3.

Where more than one graph exists satsifying these conditions, return the one with the least total weight. You may assume G does not contain edges with negative weights.

Here is an example G, each triplet representing the <start, end, weight> of an edge:
<1, 2, 40>
<1, 3, 12>
<1, 4, 50>
<2, 1, 84>
<2, 3, 19>
<2, 4, 69>
<3, 1, 25>
<3, 2, 78>
<3, 4, 93>
<4, 1, 75>
<4, 2, 36>
<4, 3, 96>

Your program should produce the following H given the above G and K = 2:
<1, 2, 40>
<1, 4, 50>
<2, 1, 84>
<2, 3, 19>
<3, 1, 25>
<4, 2, 36>