8
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Introduction

You must create a function to convert Arabic numerals into Greek numerals. The input will be an integer less than 1000 and more than 0.

Algorithm

  1. Split number into digits (e.g. 123 -> 1, 2, 3)
  2. Take each digit, and change to character found in table below, for letter symbol or letter name, (e.g. 123 -> 1, 2, 3 -> ΡΚΓ or RhoKappaGamma).

Specifications

  • No built-in number-system conversion
  • Result must be capitalized as in example.
  • Input will be in base 10.

Test Cases

123 -> RhoKappaGamma or ΡΚΓ
8 -> Eta or Η
777 -> PsiOmicronZeta or ΨΟΖ
100 -> Rho or Ρ
606 -> ChiDiGamma or ΧϜ

Table

Α = 1 = Alpha = 913 UTF-8
Β = 2 = Beta = 914 UTF-8
Γ = 3 = Gamma = 915 UTF-8
Δ = 4 = Delta = 916 UTF-8
Ε = 5 = Epsilon = 917 UTF-8
Ϝ = 6 = DiGamma = 988 UTF-8
Ζ = 7 = Zeta = 918 UTF-8
Η = 8 = Eta = 919 UTF-8
Θ = 9 = Theta = 920 UTF-8

Ι = 10 = Iota = 921 UTF-8
Κ = 20 = Kappa = 922 UTF-8
Λ = 30 = Lambda = 923 UTF-8
Μ = 40 = Mu = 924 UTF-8
Ν = 50 = Nu = 925 UTF-8
Ξ = 60 = Xi = 926 UTF-8
Ο = 70 = Omicron = 927 UTF-8
Π = 80 = Pi = 928 UTF-8
Ϙ = 90 = Koppa = 984 UTF-8

Ρ = 100 = Rho = 929 UTF-8   
Σ = 200 = Sigma = 931 UTF-8
Τ = 300 = Tau = 932 UTF-8
Υ = 400 = Upsilon = 933 UTF-8
Φ = 500 = Phi = 934 UTF-8
Χ = 600 = Chi = 935 UTF-8
Ψ = 700 = Psi = 936 UTF-8
Ω = 800 = Omega = 937 UTF-8
Ϡ = 900 = SamPi = 992 UTF-8
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  • \$\begingroup\$ Does that include base 10 number conversion? \$\endgroup\$ – Neil Apr 22 '16 at 0:34
  • 2
    \$\begingroup\$ I might be misunderstanding, but shouldn't 123 be "PKΓ" \$\endgroup\$ – Maltysen Apr 22 '16 at 1:03
  • 3
    \$\begingroup\$ You may want to add test cases such as 700 and 803 (with zeroes as digits). \$\endgroup\$ – Doorknob Apr 22 '16 at 1:16
  • 1
    \$\begingroup\$ what encodings are allowed for the greek letters? UTF 8? (please give codepoints.) Are codepages allowed, such as en.wikipedia.org/wiki/Code_page_737 ? it may be obsolete but greek letters are rendered with a single byte. \$\endgroup\$ – Level River St Apr 22 '16 at 1:39
  • 2
    \$\begingroup\$ The character codes don't check out for me. 930 appears to be ΢; 931 is Σ. \$\endgroup\$ – Dennis Apr 22 '16 at 3:14
2
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Jelly, 42 40 bytes

“#'nn(2jj33556;r”Or2/F+878Ọ
ṚĖ’ḅ9‘xṚQṚị¢

Try it online! or verify all test cases.

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11
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Pyth, 49 48 46 44 bytes

_s.b@+kNsYc3mC+913d-.iU25smXm17hT4Cd"KGO"17_

Try it here.

Still working on shortening the function to generate the alphabet... and yes, the generator is longer than hardcoding the alphabet by 4 characters, but not in bytes.

_s.b@+kNsYc3[...]_
                 _  get input (implicit) and reverse
            [...]   generate "ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ" (see below)
          c3        split into 3 equal chunks
  .b                map over each input digit (as Y) and chunk (as N)...
     +kN            prepend empty string to N (to support digits of 0)
    @   sY          get the Yth element of the empty-string-plus-digits array
 s                  concatenate all the digits
_                   reverse again to put digits back in the right order

Here's how the alphabet-generating function works. Let's take all the codepoints and subtract 913 (the smallest value):

0 1 2 3 4 75 5 6 7 8 9 10 11 12 13 14 15 71 16 18 19 20 21 22 23 24 79

Now let's pull out the ones that don't fit the pattern:

0 1 2 3 4    5 6 7 8 9 10 11 12 13 14 15    16 18 19 20 21 22 23 24
          75                             71                         79

Okay, so the strategy is forming: generate the range 0 to 24 (we'll deal with the missing 17 later), and then insert 75, 71, and 79 at the correct locations.

How can we insert the outliers? Pyth has an .interleave function that takes ex. [1,2,3] and [4,5,6] and results in [1,4,2,5,3,6]. We're going to have to add some placeholders, though, since obviously the outliers are far apart. Let's represent the placeholder with X:

0 1 2 3  4 5 6 7 8 9 10 11 12 13 14 15 16 18 19 20 21 22 23 24
X X X X 75 X X X X X  X  X  X  X  X 71  X  X  X  X  X  X  X 79

Right, so the bottom outlier array starts with four Xs. Can we group the outliers to each be preceded by four Xs and see what remains?

0 1 2 3  4 5 6 7 8 9 10 11 12 13 14 15 16 18 19 20 21 22 23 24
[four X]75 X X X X X  X [  four X  ]71  X  X  X [  four X  ]79

Now we make another key observation: since 79 comes at the very end, there can be any number of Xs preceding it, and any number of Xs following it as well! This is because interleaving [1,2,3] and [4,5,6,7,8,9] results in [1,4,2,5,3,6,7,8,9]—note that the extra elements in the second array end up at the end, and since they're going to be removed anyway, we can safely allow that.

That means we can normalize this such that every outlier is preceding by four Xs and succeeded by 6:

0 1 2 3  4 5 6 7 8 9 10 11 12 13 14 15 16 18 19 20 21 22 23 24
[four X]75[   six X   ] [  four X  ]71[     six X      ] [four X]79[six X]

Great! Now we can simply take the array [75, 71, 79], insert four placeholder values before each element and six after, and then interleave it with the range 0..24.

But wait, what happened to 17? Remember, the top line in all of these examples has been missing 17. We can simply remove the 17 after interleaving; but this leaves another delightfully evil possibility open. You guessed it—the placeholder value we've been referring to as X will be 17. This allows us to remove both the extraneous 17 and all the placeholder values in one fell swoop.

Finally! Here's the code used to implement this:

mC+913d-.iU25smXm17hT4Cd"KGO"17
              m         "KGO"    for each char in the string "KGO"...
                      Cd         take its ASCII value ("KGO" -> 75,71,79)
                m17hT            generate an 11-element array filled with 17
               X     4           replace the 5th element with the ASCII value ^^
             s                   concatenate all of these together
        .iU25                    interleave with range(25)
       -                     17  remove all the 17s
m     d                          for each value in the resulting array...
  +913                           add 913 (to get the real character value)
 C                               convert to character
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2
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JavaScript (ES6), 120 115 bytes

n=>(s=n+"").replace(/./g,(d,i)=>d--?"ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ"[d+9*(s.length+~i)]:"")

Byte count is using UTF-8. It's 88 characters long. Input is a number.

Test

var solution = n=>(s=n+"").replace(/./g,(d,i)=>d--?"ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ"[d+9*(s.length+~i)]:"")
var testCases = [ 123, 8, 777, 100, 606, 700, 803 ];
document.write("<pre>"+testCases.map(c=>c+": "+solution(c)).join("\n"));

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2
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Python 3, 118 bytes

f=lambda n,i=3:i*"_"and f(n//10,i-1)+"   ΡΙΑΣΚΒΤΛΓΥΜΔΦΝΕΧΞϜΨΟΖΩΠΗϠϘΘ"[i-1::3][n%10].strip()
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  • \$\begingroup\$ Nice work with switching the order of operands in i*"_"and to avoid the space. I wouldn't have found that. \$\endgroup\$ – Cyoce Apr 24 '16 at 0:57
2
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Retina, 57 bytes

  • 11 bytes saved thanks to @Martin Büttner.

Simple transliterate with appropriate character ranges and 0 deletion on each digit.

T`d`_ΡΣ-ΩϠ`.(?=..)
T`d`_Ι-ΠϘ`.\B
T`d`_Α-ΕϜΖ-Θ

Try it online. (First line added to allow all testcases to run together.)


Martin also suggested this one for the same score:

T`d`_J-R`.(?=..)
T`d`_L`.\B
T`dL`_Α-ΕϜΖ-ΠϘΡΣ-ΩϠ
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2
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Julia, 138 bytes

f(x,d=digits(x))=reverse(join([d[i]>0?[["ΑΒΓΔΕϜΖΗΘ","ΙΚΛΜΝΞΟΠϘ","ΡΣΤΥΦΧΨΩϠ"][i]...][d[i]]:""for i=1:endof(d)]))

This is a function that accepts an integer and returns a string.

For each digit in the input in reversed order, we get the character corresponding to that digit from the appropriate string then combine the characters and reverse the result. For any digits that are 0, we simply use the empty string.

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2
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Julia, 84 bytes

f(x,p=0)=x>0?f(x÷10,p+9)"$(x%10<1?"":"/0123z456789:;<=>v?ABCDEFG~"[p+x%10]+866)":""

Try it online!

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1
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JavaScript (ES6), 83 characters, 110 bytes

n=>`${1e3+n}`.replace(/./g,(d,i)=>i--&&d--?"ΡΣΤΥΦΧΨΩϠΙΚΛΜΝΞΟΠϘΑΒΓΔΕϜΖΗΘ"[d+9*i]:"")
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  • \$\begingroup\$ Ah, I almost tried this approach but somehow I forgot that 0s are replaced with nothing and thought they would break it! \$\endgroup\$ – user81655 Apr 22 '16 at 12:32
  • \$\begingroup\$ @user81655 The command-line interpreter I use adds a byte to the length for the newline character (unless the lambda is a block), but today I was using the browser console, which doesn't. Oops. \$\endgroup\$ – Neil Apr 22 '16 at 12:48
0
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Ruby, 134 bytes

Basic lookup dealie using modulus. It probably can be optimized somewhere, but Ruby won't coerce strings into integers the way Javascript can, haha

->n{((0<d=n/100)?"ΡΣΤΥΦΧΨΩϠ"[d-1]:"")+((0<d=n%100/10)?"ΙΚΛΜΝΞΟΠϘ"[d-1]:"")+((0<n%=10)?"ΑΒΓΔΕϜΖΗΘ"[n-1]:"")}
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  • \$\begingroup\$ Please see edit -- I changed the character for Koppa. \$\endgroup\$ – NoOneIsHere Apr 22 '16 at 14:17
  • \$\begingroup\$ All right, it's fixed. \$\endgroup\$ – Value Ink Apr 22 '16 at 17:54

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