3-liner: based on Tinus_Trotyl's creative solution for Morse Decoder by przemyslaw.daniel - python coding challenges

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3-liner: based on Tinus_Trotyl's solution in Creative category for Morse Decoder by przemyslaw.daniel

def morse_decoder(c,k="y7ac-it-oe4gw5qr1zx-jb-p-6kf0du9-s--m--n2lv-3h---8"):
 return ' '.join(''.join(k[int('1'+i.replace('.','3').replace('-','0'),6)%52]
  for i in j.split()) for j in c.split('   ')).capitalize()

June 5, 2018

Comments:

Tinus_Trotyl on June 10, 2018, 12:30 p.m.
Yeah . . . all right :-)

leggewie on June 1, 2021, 2:19 p.m.
now, this needs explanation, I'm afraid ;-)

Ekaterina_Hostinetska on Aug. 13, 2021, 11:31 a.m.
Wow, this definitely needs some comments for the newbies, looks very interesting but it's hard to read :)

toranzocalderonjs on Feb. 6, 2022, 1:26 a.m.
The function works as follows: 1) Consider c, and get the list of every substring not containing three contiguous spaces (meaning ' '). This should give you the list of substrings with every word of the secret message. 2) Consider j as any element of this list, and get the list of every substring not containing a space (i.e. ' '). This should give you the list of substrings with every Morse character of the word (being that word j). 3) Consider i as any element of a word j. Replace every dot with a '3' and every dash with a '0', and put a '1' in the front of the string. Produce an integer with this string, but in base 6. Then keep the remainder of dividing by 52. I don't know how, but apparently przemyslaw.daniel figured out this correspondence (the order is <Morse code, alphabet, replacements, rebase, remainder>): <ul> <li>.- a 130 54 2</li> <li>-... b 10333 1425 21</li> <li>-.-. c 10303 1407 3</li> <li>-.. d 1033 237 29</li> <li>. e 13 9 9</li> <li>..-. f 13303 2055 27</li> <li>--. g 1003 219 11</li> <li>.... h 13333 2073 45</li> <li>.. i 133 57 5</li> <li>.--- j 13000 1944 20</li> <li>-.- k 1030 234 26</li> <li>.-.. l 13033 1965 41</li> <li>-- m 100 36 36</li> <li>-. n 103 39 39</li> <li>--- o 1000 216 8</li> <li>.--. p 13003 1947 23</li> <li>--.- q 10030 1314 14</li> <li>.-. r 1303 327 15</li> <li>... s 1333 345 33</li> <li><ul> <li>t 10 6 6</li> </ul></li> <li>..- u 1330 342 30</li> <li>...- v 13330 2070 42</li> <li>.-- w 1300 324 12</li> <li>-..- x 10330 1422 18</li> <li>-.-- y 10300 1404 0</li> <li>--.. z 10033 1317 17</li> <li>----- 0 100000 7776 28</li> <li>.---- 1 130000 11664 16</li> <li>..--- 2 133000 12312 40</li> <li>...-- 3 133300 12420 44</li> <li>....- 4 133330 12438 10</li> <li>..... 5 133333 12441 13</li> <li>-.... 6 103333 8553 25</li> <li>--... 7 100333 7905 1</li> <li>---.. 8 100033 7797 49</li> <li>----. 9 100003 7779 31</li> </ul> 4) Now use this number (the remainder) as an index for k, and you have the corresponding character. 5) Join every i of every j, and join every j of every word (with the corresponding spaces between). Capitalize the string (the method knows that it has to take just the first element of the string and make it uppercase if it is a letter). (If someone knows how to rewrite this table, I would love to hear about, because I don't like how it looks. I think it could be prettier.)

Strings Theory

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