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EDIT: For those of you who think this question is off-topic, I asked it on a forum about linguistics because I expected the answer to come from historical linguistics. I expected the answer to be something along the lines of:

Because the mantra "English is spelt as it was pronounced around the invention of the printing press." is not true. The silent 'e' is way more common today than final 'e' ever was in spoken English. Look at the numbers 1-10: the silent 'e' in 'five' (Old English 'fif') and 'nine' (Old English 'nigon') was never pronounced.".

Or something like that...

In a paper I published in Valpovački Godišnjak and Regionalne Studije (it's basically this text, just edited differently), I measured the collision entropy of the consonants in five different languages: English, German, Italian, French and Croatian. The table with collision entropies In a long text, Croatian has the highest collision entropy, and it also has the shallowest orthography. French has a very deep orthography (though not as deep as English), and it has the lowest collision entropy in both a long text and an Aspell word list. And the correlation seems to be even higher once you group those languages by families. Italian and French are closely related but have a very different orthography, and French has a lower collision entropy. German and English are closely related, but have a very different orthography, and English has a lower collision entropy than German. Is this correlation real?

The obvious answer (as I've explained in my paper) seems to be no, because spelling represents how the words were pronounced centuries ago, and it seems absurd to suggest that the collision entropy of languages always increases with time. But has anybody done an actual study on that?

EDIT: In case it is relevant, here is the program I used to measure the collision entropy:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <time.h>
#include <string.h>

int main() {
    srand(time(0));
    FILE *input=fopen("text.txt","r");
    if (!input) {
        fprintf(stderr,"Can't open 'text.txt'!\n"),
        exit(1);
    }
    fseek(input,0,SEEK_END);
    int length=ftell(input),sum1=0,sum2=0;
    /*Variable "length" now contains the total number
    of characters in "text.txt".*/
    int alphabet[26]={0};
    const char *vowels="aeiou";
    for (int i=0; i<1000000; i++) {
        int tmp=rand()%length;
        fseek(input,tmp,SEEK_SET);
        char first=fgetc(input);
        tmp=rand()%length;
        fseek(input,tmp,SEEK_SET);
        char second=fgetc(input);
        if (isalpha(first) &&
            !strchr(vowels,tolower(first))
            && isalpha(second) &&
            !strchr(vowels,tolower(second))) {
                /*If both randomly chosen characters
                 from "text.txt" happen to be consonants...*/
                sum2++;
                if (first==second) {
                    sum1++;
                    alphabet[toupper(first)-'A']++;
                }
            }
    }
    printf("1/%f\n",(float)sum2/sum1);
    int max=0;
    for (int i=0; i<26; i++)
        if (alphabet[i]>alphabet[max]) max=i;
    printf("%c %f%%\n",max+'A',100.*alphabet[max]/sum1);
}
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  • 2
    How are you measuring the collision entropy? I don't speak Czech(?) so I'm afraid the paper doesn't help much with this.
    – Draconis
    Apr 18 at 20:20
  • 2
    Ah, so you're calculating the likelihood of collision with consonant letters, not consonant phonemes?
    – Draconis
    Apr 18 at 21:11
  • 2
    What does it mean for an orthography to be deep?
    – curiousdannii
    Apr 18 at 22:05
  • 1
    @curiousdannii To be very diverged from the spoken language. You know, like English or French or Tibetan. That's in contrast with shallow orthography, such as German, Italian or Croatian. Apr 18 at 22:12
  • 5
    There is a reason why this question has not been systematically studied, which is the lack of a metric for computing the depth of an orthography. Perhaps you devised some metric that measures how divergent English or German orthography are from pronunciation, but it surely depends on having data about pronunciation. Such data is only available for a very few languages. Your notion lacks generality and objectivity, in that it can't be applied to randomly-selected languages.
    – user6726
    Apr 19 at 17:14

1 Answer 1

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I'm not aware of any studies on this. If there are any, I expect them to be related to old-school cryptanalysis.

Intuitively, the collision entropy depends on the frequency distribution of different letters in a text. In a language with "shallow" orthography, this is pretty close to the frequency distribution of phonemes; in a language with "deep" orthography, it's not (e.g. H appears much more than /h/ in English because it's also used in the digraphs SH, TH, CH, AH, etc).

There have been studies on the frequency distribution of phonemes (here's the first one I found), and you could potentially compare a grapheme distribution against these to look for interesting patterns.

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