8
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Introduction

You must create a function to convert Greek numerals into Arabic numerals. The input will be a Greek numeral less than 1000 and more than 0. This is the reverse of my previous challenge.

Algorithm

  1. Split input into letters (e.g. ΡΚΓ -> Ρ, Κ, Γ)
  2. Take each letter, and change to character found in table below, for letter symbol, (e.g. ΡΚΓ -> Ρ, Κ, Γ -> 100, 20, 3).
  3. Add (e.g. ΡΚΓ -> Ρ, Κ, Γ -> 100, 20, 3 -> 123)

Specifications

  • No built-in number-system conversion
  • Input will be capitalized as in example.
  • Output must be in base 10.
  • ΡΡΡΡ will never happen. It will be Υ.

Test Cases

ΡΚΓ -> 123
Η -> 8
ΨΟΖ -> 777
Ρ -> 100
ΧϜ -> 606
ΡϘ -> 190
ΜΒ -> 42
Ν -> 50

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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5
  • \$\begingroup\$ Will we ever have input like ΡΡΡΡ? If so, what would the result be? \$\endgroup\$ Apr 22, 2016 at 16:06
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ No. That would be Upsilon. \$\endgroup\$ Apr 22, 2016 at 16:18
  • \$\begingroup\$ Oh, I misunderstood the question, haha. \$\endgroup\$ Apr 22, 2016 at 16:21
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ I edited that spec in just now. You didn't miss it. \$\endgroup\$ Apr 22, 2016 at 16:22
  • 1
    \$\begingroup\$ I think your test cases should cover all possible patterns of zeroes, so at least add something like 180, 42 and 50. \$\endgroup\$ Apr 22, 2016 at 16:46

8 Answers 8

4
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Python 3, 112

Saved 4 bytes thanks to vaultah.

Booyah, beating JS!

lambda x,z="ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ".find:sum((z(c)%9+1)*10**(z(c)//9)for c in x)

With test cases:

assert(f("ΡΚΓ")==123)
assert(f("Η")==8)
assert(f("ΨΟΖ")==777)
assert(f("Ρ")==100)
assert(f("ΧϜ")==606)

Loops through the string and uses its index in the list of potentials chars to calculate how much it's worth.

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3
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JavaScript (ES7), 115 bytes

s=>[...s].map(c=>n+=((i="ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ".search(c))%9+1)*10**(i/9|0),n=0)|n
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3
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Haskell, 116 113 bytes

f x=sum[v|d<-x,(l,v)<-zip"ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ"$(*)<$>[1,10,100]<*>[1..9],d==l]

Usage example: map f ["ΡΚΓ","Η","ΨΟΖ","Ρ","ΧϜ","ΡϘ","ΜΒ","Ν"] -> [123,8,777,100,606,190,42,50].

Lookup the value of the greek letter from a list of pairs (letter, value) and sum. The list of values is build by (*)<$>[1,10,100]<*>[1..9], where (*)<$>[1,10,100] builds a list of functions [(*1),(*10),(*100)] (multiply by 1, 10 and 100) which are applied separately to the elements of [1..9] and concatenated into a single list.

Edit: 3 bytes with thanks to @xnor.

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3
  • \$\begingroup\$ It's shorter to take the product as (*)<$>[1,10,100]<*>[1..9]. \$\endgroup\$
    – xnor
    Apr 23, 2016 at 2:38
  • \$\begingroup\$ @xnor: <*> in list context, again. I never think of it myself. Thanks! \$\endgroup\$
    – nimi
    Apr 23, 2016 at 8:50
  • \$\begingroup\$ I wouldn't think of it either, I got it from here. \$\endgroup\$
    – xnor
    Apr 23, 2016 at 23:21
3
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Julia, 82 70 bytes

x->10.^((t=findin([1:5;76;6:16;72;17;19:25;80]+912,x)-1)÷9)⋅(t%9+1)

Try it online!

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2
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Jelly, 47 45 bytes

⁵*ב}
“#'nn(2jj33556;r”Or2/F+878Ọ
¢iЀ’d9ñ/€S

Try it online! or verify all test cases.

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3
  • \$\begingroup\$ What encoding are you using? Try It Online says it is 47 bytes. You seem to be missing a <newline>Ç€ at the end. \$\endgroup\$ Apr 22, 2016 at 16:34
  • 3
    \$\begingroup\$ Jelly has its own encoding (bytes link in the header). The verify all test cases link includes an extra ǀ that applies the function to all test cases. The first link shows to actual program, which is 44 bytes long. \$\endgroup\$
    – Dennis
    Apr 22, 2016 at 16:37
  • \$\begingroup\$ This seems to be the shortest so far... \$\endgroup\$ May 2, 2016 at 15:28
2
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JavaScript (ES6), 116 bytes

s=>[...s].map(c=>n+=+((i="ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ".search(c))%9+1+"e"+(i/9|0)),n=0)|n

Only 1 byte longer than ES7!

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3
  • \$\begingroup\$ I'm not familiar with JavaScript, what does the +"e" do? \$\endgroup\$ Apr 22, 2016 at 18:54
  • \$\begingroup\$ @MorganThrapp String concatenation. For example, with Ϡ you would get 9+"e"+2 and then the +("9e2") becomes 900. \$\endgroup\$
    – Neil
    Apr 22, 2016 at 18:56
  • 1
    \$\begingroup\$ Ah, that's really weird. Javascript always manages to ---terrify--- surprise me. \$\endgroup\$ Apr 22, 2016 at 18:57
1
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Python 3, 188 Bytes

def f(x,l=list,z=0):
 r=l(range(1,10));R=[a*10for a in r]
 for a,b in l(zip(l("ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ"),r+R+[a*10for a in R])):
  z+=(0,b)[a in x]
 return z

Try it out! (Test cases included)

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1
  • \$\begingroup\$ You can save 25 bytes by condensing it down to def f(x):r=list(range(1,10));R=[a*10for a in r];return sum(b*(a in x)for a,b in zip("ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ",r+R+[a*10for a in R])). \$\endgroup\$ Apr 22, 2016 at 17:57
1
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Retina, 72 bytes

T`_Ι-ΠϘ0ΡΣ-ΩϠ`dl
[a-j]
$0aa 
\d
$&0 
T`_Α-ΕϜΖ-Θl`dd
\d+
$*
1

Try it online.

Explanation

Basically - replace every Greek symbol with the number that it represents, then return the sum of all the resulting numbers:

Transliterate 10s digits to Arabic and 100s digits to the Latin alphabet (0-9 => a-j):

T`_Ι-ΠϘ0ΡΣ-ΩϠ`dl

Append "aa " to any 100s digits:

[a-j]
$0aa 

Append "0 " to any 10s digits:

\d
$&0 

Transliterate 1's digits and Latin alphabet to Arabic:

T`_Α-ΕϜΖ-Θl`dd

Convert all space-separated decimal numbers to unary:

\d+
$*

Count the total number of unary 1s:

1
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