# Reverse Greek Conversion Golf

### 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

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

# 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.

# JavaScript (ES7), 115 bytes

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


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.

• It's shorter to take the product as (*)<$>[1,10,100]<*>[1..9]. – xnor Apr 23, 2016 at 2:38 • @xnor: <*> in list context, again. I never think of it myself. Thanks! – nimi Apr 23, 2016 at 8:50 • I wouldn't think of it either, I got it from here. – xnor Apr 23, 2016 at 23:21 # 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! # Jelly, 47 45 bytes ⁵*×‘} “#'nn(2jj33556;r”Or2/F+878Ọ ¢iÐ€’d9ñ/€S  • What encoding are you using? Try It Online says it is 47 bytes. You seem to be missing a <newline>Ç€ at the end. Apr 22, 2016 at 16:34 • Jelly has its own encoding (bytes link in the header). The verify all test cases link includes an extra Ç€ that applies the function to all test cases. The first link shows to actual program, which is 44 bytes long. Apr 22, 2016 at 16:37 • This seems to be the shortest so far... May 2, 2016 at 15:28 ## 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! • I'm not familiar with JavaScript, what does the +"e" do? Apr 22, 2016 at 18:54 • @MorganThrapp String concatenation. For example, with Ϡ you would get 9+"e"+2 and then the +("9e2") becomes 900. – Neil Apr 22, 2016 at 18:56 • Ah, that's really weird. Javascript always manages to ---terrify--- surprise me. Apr 22, 2016 at 18:57 ## 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) • 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])). Apr 22, 2016 at 17:57 # Retina, 72 bytes T_Ι-ΠϘ0ΡΣ-ΩϠdl [a-j]$0aa
\d
$&0 T_Α-ΕϜΖ-Θldd \d+$*
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-ΩϠdl


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-Θldd


Convert all space-separated decimal numbers to unary:

\d+
\$*


Count the total number of unary 1s:

1