Count Like a Babylonian

Question

Challenge

Given an ASCII representation of a Babylonian number as input, output the number in Western Arabic numerals.

Babylonian Numeral System

How did the Babylonians count? Interestingly, they used a Base 60 system with an element of a Base 10 system. Let's first consider the unit column of the system:

The Babylonians had only three symbols: T (or, if you can render it: 𒐕) which represented 1, and < (or, if you can render it: 𒌋) which represented 10, and \ (or, if you render it: 𒑊) which represented zero.

Note: Technically, \ (or 𒑊) isn't zero (because the Babylonians did not have a notion of 'zero'). 'Zero' was invented later, so \ was a placeholder symbol added later to prevent ambiguity. However, for the purposes of this challenge, it's enough to consider \ as zero

So, in each column you just add up the value of the symbols, e.g.:

<<< = 30
<<<<TTTTTT = 46
TTTTTTTTT = 9
\ = 0

There will never be more than five < or more than nine T in each column. \ will always appear alone in the column.

Now, we need to extend this to adding more columns. This works exactly the same as any other base sixty, where you multiply the value of the rightmost column by \$60^0\$, the one to the left by \$60^1\$, the one to the left by \$60^2\$ and so on. You then add up the value of each to get the value of the number.

Columns will be separated by spaces to prevent ambiguity.

Some examples:

<< <TT = 20*60 + 12*1 = 1212
<<<TT \ TTTT = 32*60^2 + 0*60 + 4*1 = 115204

Rules

• You are free to accept either ASCII input (T<\) or Unicode input (𒐕𒌋𒑊)
• The inputted number will always be under \$10^7\$
• The <s will always be to the left of the Ts in each column
• \ will always appear alone in a column

Winning

Shortest code in bytes wins.

Answer 1

JavaScript (ES6), 44 bytes

Takes input as an array of ASCII characters.

a=>a.map(c=>k+=c<1?k*59:c<'?'?10:c<{},k=0)|k

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How?

The Babylonian Numeral System can be seen as a 4-instruction language working with a single register -- let's call it the accumulator.

Starting with \$k=0\$, each character \$c\$ in the input array \$a\$ modifies the accumulator \$k\$ as follows:

• space: multiply \$k\$ by \$60\$ (implemented as: add \$59k\$ to \$k\$)
• <: add \$10\$ to \$k\$
• T: increment \$k\$
• \: do nothing; this is the NOP instruction of this language (implemented as: add \$0\$ to \$k\$)

Answer 2

C (gcc), 140 138 136 bytes

• Saved two bytes thanks to ceilingcat.

B,a,b,y;l(char*o){y=B=0,a=1;for(char*n=o;*n;n++,B++[o]=b,a*=60)for(b=0;*n&&*n-32;n++)b+=*n-84?*n-60?:10:1;for(B=a;B/=60;y+=*o++*B);B=y;}

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Answer 3

Perl 6, 39 bytes

-3 bytes thanks to nwellnhof

{:60[.words>>.&{sum .ords X%151 X%27}]}

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Uses the cuneiform characters.

Explanation:

{                                     }   # Anonymous code block
     .words  # Split input on spaces
           >>.&{                    }  # Convert each value to
                sum   # The sum of:
                    .ords # The codepoints
                          X%151 X%27   # Converted to 0,1 and 10 through modulo
 :60[                                ]  # Convert the list of values to base 60

Answer 4

Jelly,  13  12 bytes

ḲO%7C%13§ḅ60

A monadic link accepting a list of characters which yields an integer.

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How?

ḲO%7C%13§ḅ60 - Link: list of characters   e.g. "<<<TT \ TTTT"
Ḳ            - split at spaces                 ["<<<TT", "\", "TTTT"]
 O           - cast to ordinals                [[60,60,60,84,84],[92],[84,84,84,84]]
  %7         - modulo by seven (vectorises)    [[4,4,4,0,0],[1],[0,0,0,0]]
    C        - compliment (1-X)                [[-3,-3,-3,1,1],[0],[1,1,1,1]]
     %13     - modulo by thirteen              [[10,10,10,1,1],[0],[1,1,1,1]]
        §    - sum each                        [32,0,4]
         ḅ60 - convert from base sixty         115204

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Another 12: ḲO⁽¡€%:5§ḅ60 (⁽¡€ is 1013, so this modulos 1013 by the Ordinal values getting 53, 5, and 1 for <, T, \ respectively then performs integer division, : by 5 to get 10, 1 and 0)

Answer 5

05AB1E, 13 bytes

8740|Ç%4/O60β

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To make up for how lazy I've been with my Jelly answer, here is a submission in 05AB1E xD.

Answer 6

Python 2, 96 93 87 85 bytes

lambda s:sum(60**i*sum(8740%ord(c)/4for c in v)for i,v in enumerate(s.split()[::-1]))

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---

Saved:

• -1 byte, thanks to Mr. Xcoder
• -4 bytes, thanks to Poon Levi
• -2 bytes, thanks to Matthew Jensen

Answer 7

Excel VBA, 121 bytes

Restricted to 32-Bit Office as ^ serves as the LongLong type literal in 64-Bit versions

Takes input from cell A1 and outputs to the vbe immediate window.

a=Split([A1]):u=UBound(a):For i=0 To u:v=a(i):j=InStrRev(v,"<"):s=s+(j*10-(InStr(1,v,"T")>0)*(Len(v)-j))*60^(u-i):Next:?s

Ungolfed and Commented

a=Split([A1])       '' Split input into array
u=UBound(a)         '' Get length of array
For i=0 To u        '' Iter from 0 to length
v=a(i)              '' Get i'th column of input
j=InStrRev(v,"<")   '' Get count of <'s in input
                    '' Multiply count of <'s by 10; check for any T's, if present
                    ''   add count of T's
t=t+(j*10-(InStr(1,v,"T")>0)*(Len(v)-j))
    *60^(u-i)       '' Multiply by base
Next                '' Loop
?s                  '' Output to the VBE immediate window

Answer 8

Dyalog APL, 33 30 bytes

{+/(⌊10*⍵-3)×60*+\2=⍵}'\ T<'⍳⌽

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Edit: -3 bytes thanks to ngn

'\ T<'⍳ replaces the characters with numbers (their position in the string constant), and ⌽ reverses the input so most significant 'digits' are last. This allows +\2= to keep a running count of the desired power of 60 (applied by 60*) by counting the number of times a space (index 2 in the string constant) is encountered.

⌊10*⍵-3 gives the desired power of ten for each character. The order of characters in the string constant and the -3 offset cause '\' and space to go to negative numbers, resulting in fractions when those characters are raised to the power of 10, allowing them to be eliminated by ⌊.

All we have to do now is multiply the powers-of-10 digits by the powers-of-60 place values and sum the lot up with +/.

Answer 9

Python 2, 62 bytes

lambda s:reduce(lambda x,y:x+[10,0,59*x,1]["<\ ".find(y)],s,0)

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This uses the technique from Arnauld's answer.

Answer 10

Canvas, 20 17 16 bytes

Ｓ｛｛<≡ＡײT≡］∑］<ｃ┴

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Explanation:

E{          ]     map over input split on spaces
  {       ]         map over the characters
   <≡A×               (x=="<") * 10
       ²T≡            x=="T"
           ∑        sum all of the results
             <c┴  and encode from base (codepoint of "<") to 10

Answer 11

APL(NARS ⎕io←0), 28 chars, 56 bytes

{60⊥{+/⍵⍳⍨10⍴'\T'}¨⍵⊂⍨⍵≠' '}

some test with type check:

  q←{60⊥{+/⍵⍳⍨10⍴'\T'}¨⍵⊂⍨⍵≠' '}

  o←⎕fmt
  o q '<< <TT'
1212
~~~~
  o q '<<<TT \ TTTT'
115204
~~~~~~

Each type result is number.

Answer 12

JavaScript (Node.js), 122 114 107 106 83 bytes

a=>a.split` `.map(b=>[...b].map(c=>x+=c<'T'?10:c<'U',x=0)&&x).reduce((a,b)=>a*60+b)

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I'm a little obsessed with "functional-style" array operations, uses ASCII input, as far as I can tell, JS isn't very good at getting charcodes golfily

I'm keeping this for posterity's sake, but this is a naive/dumb solution, I suggest you check out Arnauld's answer which is far more interesting an implementation of the challenge

Answer 13

Retina, 29 26 23 bytes

<
10*T
+`^(.*)¶
60*$1
T

Try it online! Uses newline separation, but link includes header to use spaces instead for convenience. Edit: Saved 3 bytes with help from @KevinCruijssen. Saved a further 3 bytes thanks to @FryAmTheEggman. Explanation:

<
10*T

Replace each < with 10 Ts.

+`^(.*)¶
60*$1

Take the first line, multiply it by 60, and add the next line. Then repeat until there is only one line left.

T

Count the Ts.

Faster 51-byte version:

%`^(<*)(T*).*
$.(10*$1$2
+`^(.+)¶(.+)
$.($1*60*_$2*

Try it online! Uses newline separation, but link includes header to use spaces instead for convenience. Explanation:

%`^(<*)(T*).*
$.(10*$1$2

Match each line individually, and count the number of Ts and 10 times the number of <s. This converts each line into its base-60 "digit" value.

+`^(.+)¶(.+)
$.($1*60*_$2*

Base 60 conversion, running a line at a time. The computation is done in decimal for speed.

Answer 14

Bash (with sed and dc), 50 bytes

sed 's/</A+/g
s/T/1+/g
s/ /60*/g
s/\\//g'|dc -ez?p

Takes space-delimited input from stdin, outputs to stdout

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Explanation

Uses sed to transform the input with a bunch of regular expression matches until, for example, the input <<<TT \ TTTT has been transformed to A+A+A+1+1+60*60*1+1+1+1+. Then this input is fed to dc with the explicit input execution command ?, preceded by z (pushes the stack length (0) to the stack so that we have somewhere to ground the addition) and followed by p (print).

Answer 15

J, 34 30 bytes

60#.1#.^:2(10 1*'<T'=/])&>@cut

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Answer 16

Dyalog APL, 35 32 bytes

f←{⍵+('<T\ '⍳⍺)⌷10,1,0,59×⍵}/∘⌽0∘,

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31 in dzaima/APL

Answer 17

Noether, 55 bytes

I~sL(si/~c{"<"=}{k10+~k}c{"T"=}{!k}c{" "=}{k60*~k}!i)kP

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Same approach as @Arnauld.

Answer 18

Charcoal, 26 bytes

≔⁰θＦＳ«≡ι ≦×⁶⁰θ<≦⁺χθT≦⊕θ»Ｉθ

Try it online! Link is to verbose version of code. Explanation:

≔⁰θ

Clear the result.

ＦＳ«...»

Loop over the input characters. The ≡ command is wrapped in a block to prevent it from finding a "default" block.

≡ι

Switch over the current character...

 ≦×⁶⁰θ

if it's a space then multiply the result by 60...

<≦⁺χθ

if it's a < then add 10 to the result...

T≦⊕θ

if it's a T then increment the result.

Ｉθ

Print the result.

Answer 19

R, 98 81 bytes

(u=sapply(scan(,""),function(x,y=utf8ToInt(x))y%%3%*%(y%%6)))%*%60^(sum(u|1):1-1)

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Ridiculously long due to string parsing. Thanks Giusppe for shaving off 16 unnecessary bytes.

Define y the bytecode value of unicode input and R = y("T<\") = y("𒐕𒌋𒑊")

Observe that R%%3 = 1,2,0 and R%%6 = 1,5,0... so R%%3 * R%%6 = 1,10,0 !

The rest is easy: sum per column, then dot-product with decreasing powers of 60.

Answer 20

Ruby, 50 46 bytes

->a{x=0;a.bytes{|c|x+=[59*x,10,0,1][c%9%5]};x}

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A basic port of Arnauld's answer improved by G B for -4 bytes.

Answer 21

C (gcc), 65 64 63 bytes

f(s,o)char*s;{for(o=0;*s;s++)o+=*s>32?(93^*s)/9:o*59;return o;}

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Answer 22

Java 8, 64 60 bytes

a->{int r=0;for(int c:a)r+=c<33?r*59:c<63?10:84/c;return r;}

-4 bytes thanks to @ceilingcat.

Try it online. Explanation:

a->{            // Method with character-array parameter and integer return-type
  int r=0;      //  Result-integer, starting at 0
  for(int c:a)  //  Loop over each character `c` of the input-array
    r+=         //   Increase the result by:
       c<33?    //    Is the current character `c` a space:
        r*59    //     Increase it by 59 times itself
       :c<63?   //    Else-if it's a '<':
        10      //     Increase it by 10
       :c<85?   //    Else (it's a 'T' or '\'):
        84/c;   //     Increase it by 84 integer-divided by `c`,
                //     (which is 1 for 'T' and 0 for '\')
  return r;}    //  Return the result

Answer 23

APL (Dyalog Extended), 18 bytes^SBCS

Anonymous tacit prefix function.

60⊥10⊥¨≠'<T'∘⍧¨⍤⊆⊢

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                  ⊢  the argument; "<<<TT \ TTTT"
       ≠             mask where different from space; [1,1,1,1,1,0,1,0,1,1,1,1]
                ⊆    enclose runs of 1; ["<<<TT","\","TTTT"]
               ⍤     on that
              ¨      for each one
             ⍧       Count the occurrences In it of the elements
            ∘        of the entire list
        '<T'         ["<","T"]; [[3,2],[0,0],[0,4]]
      ¨              for each one
   10⊥               evaluate as base-10 digits
60⊥                  evaluate as base-60 digits

Answer 24

Perl -F// -E, 39 bytes

$w+=/</?10:/T/?1:/ /?59*$w:0for@F;say$w

This reads the to be converted number from STDIN.

This is essential the same solution as given by @Arnauld using JavaScript.

Answer 25

F#, 128 bytes

let s(v:string)=Seq.mapFoldBack(fun r i->i*Seq.sumBy(fun c->match c with|'<'->10|'T'->1|_->0)r,i*60)(v.Split ' ')1|>fst|>Seq.sum

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Ungolfed it would look like this:

let s (v:string) =
    Seq.mapFoldBack(fun r i ->
        i * Seq.sumBy(fun c ->
            match c with
                | '<' -> 10
                | 'T' ->1
                | _ -> 0
        ) r, 
        i * 60) (v.Split ' ') 1
    |> fst
    |> Seq.sum

Seq.mapFoldBack combines Seq.map and Seq.foldBack. Seq.mapFoldBack iterates through the sequence backwards, and threads an accumulator value through the sequence (in this case, i).

For each element in the sequence, the Babylonian number is computed (by Seq.sumBy, which maps each character to a number and totals the result) and then multiplied by i. i is then multiplied by 60, and this value is then passed to the next item in the sequence. The initial state for the accumulator is 1.

For example, the order of calls and results in Seq.mapFoldBack for input <<<TT \ TTTT would be:

(TTTT, 1)     -> (4, 60)
(\, 60)       -> (0, 3600)
(<<<TT, 3600) -> (115200, 216000)

The function will return a tuple of seq<int>, int. The fst function returns the first item in that tuple, and Seq.sum does the actual summing.

Why not use Seq.mapi or similar?

Seq.mapi maps each element in the sequence, and provides the index to the mapping function. From there you could do 60 ** index (where ** is the power operator in F#).

But ** requires floats, not ints, which means that you need to either initialise or cast all the values in the function as float. The entire function will return a float, which (in my opinion) is a little messy.

Using Seq.mapi it can be done like this for 139 bytes:

let f(v:string)=v.Split ' '|>Seq.rev|>Seq.mapi(fun i r->Seq.sumBy(fun c->match c with|'<'->10.0|'T'->1.0|_->0.0)r*(60.0**float i))|>Seq.sum

Answer 26

Tcl, 134 bytes

proc B l {regsub {\\} $l 0 l
lmap c [lreverse $l] {incr s [expr 60**([incr i]-1)*([regexp -all < $c]*10+[regexp -all T $c])]}
expr $s}

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In the reversed list, I loop incrementing the result in counting < and T (with -all regexp option) and incrementing a natural as power of 60.

Correct version (see comment)

Answer 27

05AB1E (legacy), 10 bytes

#Ç9%5BO60β

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#               # split input on spaces
 Ç              # convert each character to its codepoint
  9%            # modulo 9 (maps 𒌋 to 5, 𒐕 to 1, 𒑊 to 0)
    5B          # convert each to base 5 (5 becomes 10, 0 and 1 unchanged)
      O         # sum each column
       60β      # convert from base 60

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05AB1E, 11 bytes

#Ç9%5B€O60β

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Same algorithm, but in modern 05AB1E O doesn't work on lists of mixed ints and lists, so we need €O instead.