8
\$\begingroup\$

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
\$\endgroup\$
  • \$\begingroup\$ Will we ever have input like ΡΡΡΡ? If so, what would the result be? \$\endgroup\$ – Conor O'Brien Apr 22 '16 at 16:06
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ No. That would be Upsilon. \$\endgroup\$ – NoOneIsHere Apr 22 '16 at 16:18
  • \$\begingroup\$ Oh, I misunderstood the question, haha. \$\endgroup\$ – Conor O'Brien Apr 22 '16 at 16:21
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ I edited that spec in just now. You didn't miss it. \$\endgroup\$ – NoOneIsHere Apr 22 '16 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\$ – Martin Ender Apr 22 '16 at 16:46
2
\$\begingroup\$

Jelly, 47 45 bytes

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

Try it online! or verify all test cases.

\$\endgroup\$
  • \$\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\$ – NoOneIsHere Apr 22 '16 at 16:34
  • 3
    \$\begingroup\$ 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. \$\endgroup\$ – Dennis Apr 22 '16 at 16:37
  • \$\begingroup\$ This seems to be the shortest so far... \$\endgroup\$ – NoOneIsHere May 2 '16 at 15:28
4
\$\begingroup\$

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.

\$\endgroup\$
3
\$\begingroup\$

JavaScript (ES7), 115 bytes

s=>[...s].map(c=>n+=((i="ΑΒΓΔΕϜΖΗΘΙΚΛΜΝΞΟΠϘΡΣΤΥΦΧΨΩϠ".search(c))%9+1)*10**(i/9|0),n=0)|n
\$\endgroup\$
3
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ It's shorter to take the product as (*)<$>[1,10,100]<*>[1..9]. \$\endgroup\$ – xnor Apr 23 '16 at 2:38
  • \$\begingroup\$ @xnor: <*> in list context, again. I never think of it myself. Thanks! \$\endgroup\$ – nimi Apr 23 '16 at 8:50
  • \$\begingroup\$ I wouldn't think of it either, I got it from here. \$\endgroup\$ – xnor Apr 23 '16 at 23:21
3
\$\begingroup\$

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!

\$\endgroup\$
2
\$\begingroup\$

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!

\$\endgroup\$
  • \$\begingroup\$ I'm not familiar with JavaScript, what does the +"e" do? \$\endgroup\$ – Morgan Thrapp Apr 22 '16 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 '16 at 18:56
  • 1
    \$\begingroup\$ Ah, that's really weird. Javascript always manages to ---terrify--- surprise me. \$\endgroup\$ – Morgan Thrapp Apr 22 '16 at 18:57
1
\$\begingroup\$

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)

\$\endgroup\$
  • \$\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\$ – Morgan Thrapp Apr 22 '16 at 17:57
1
\$\begingroup\$

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
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.