Visual clear solution for English to Braille Translator by veky - python coding challenges

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Visual solution in Clear category for English to Braille Translator by veky

def braille_page(text: str):
    def rows(code):
        f, e, d, c, b, a = format(code, "06b")
        yield [a, d]
        yield [b, e]
        yield [c, f]
    R = " a,b*k*l*cif*msp*e-h*o!r*djg*ntq$***_u?v*****x****.**z******#yw*"
    coded = []
    for c in text:
        if c.isupper(): coded += ["$", c.lower()]
        elif c.isdigit(): coded += ["#", "jabcdefghi"[int(c)]]
        else: coded.append(c)
    grid = [[0] * (min(len(coded), 10) * 3 - 1) for line in range(3)]
    for index, point in enumerate(coded):
        offset = index % 10 * 3
        if offset == 0 < index: grid.extend([0] * 29 for line in range(4))
        for row, etch in zip(grid[index // 10 * 4 :], rows(R.index(point))):
            row[offset : offset + 2] = map(int, etch)
    return grid

Oct. 16, 2014

Comments:

SandeepManilal on Jan. 11, 2015, 11:28 a.m.
Hello, Its so short! Could you help me understand this please (whenever you can spare the time)? Thanks.

veky on Jan. 11, 2015, 12:25 p.m.
The secret is R. :-) Since binary values of patterns are just in range(64), it's much more profitable to put their characters (and two special chars, $ for shift and # for numbers) in a sequence of len 64, than in a general dict as default code seems to like. When you have R, you just need to .index it with your characters, and convert the result to visual pattern of 0s and 1s. `rows` function does the second thing, I guess it's visual enough, you can see the 3x2 matrix right in the code. :-) ("06b" is "six-bit binary notation") `coded` is preprocessed so capital letters are $-prefixed (T |-> $t) , and numbers are #-prefixed (3 |-> #c). Others are left alone. Then grid is initialized to 3 long enough lines of zeros, and we start filling it with our patterns. `offset` is an abscissa (horizontal offset) of the start of current char (called `point`). Whenever offset is 0, and we're not right at the beginning, we start a new row (extend grid with 4 lines of maximal length of zeros). The filling itself is accomplished in the lines 17 and 18. `etch` is a single horizontal "linette" of a letter pattern. map of int over it is needed since `rows` returns chars, and we need numbers 0 and 1 for filling. Anything else need explaining?

SandeepManilal on Jan. 11, 2015, 12:40 p.m.
OK, Wow. Thanks. Great stuff. This was a difficult problem I thought. Good solution. (to be honest I still don't get it, but never mind)

veky on Jan. 11, 2015, 1:15 p.m.
Well, what phase seems to be the problem? You get "R2t". Convert it to "$r#bt" ("$" means "shift to uppercase", "#" means "shift to numbers, where a means 1, b means 2, etc"). Convert these to indices in R, getting 32, 23, 60, 3, 30. Each of these is converted to a 3x2 matrix of binary numbers, giving <pre class='brush: python'>32: (0, 0), (0, 0), (0, 1) 23: (1, 0), (1, 1), (1, 0) 60: (0, 1), (0, 1), (1, 1) 3: (1, 0), (1, 0), (0, 0) 30: (0, 1), (1, 1), (1, 0) </pre> Then, the grid is prepared: since len is 5<10, one row of 5 letters is enough. It is of length 5(chars) * 2(bits per char) + (5-1)(spaces between chars) = 14(bits), and consists of 3 rows of bits. So, starting grid is <pre class='brush: python'>0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 </pre> Now the above patterns are written in the grid. First 00 00 01 is written, vertically, in the first two columns. Then 10 11 10 is written, vertically, in the fourth and fifth column (offset=index%10*3, so when index=0, offset=0, and when index=1, offset=3, and so on). After those writes, grid is <pre class='brush: python'>0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 ^ ^ ^ ^ those columns were written </pre> When all five patterns are written, grid is returned. Is these still anything problematic?

SandeepManilal on Jan. 12, 2015, 8:05 a.m.
Ah ok! Thank you. Very Clever!

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