• Options
• Login
• Desktop Style

The Naibbe cipher

• Index

  • The Voynich Ninja
  • Voynich Research
  • Theories & Solutions
  • The Naibbe cipher
• The Naibbe cipher

• RE: The Naibbe cipher

  bi3mw > 13-08-2025, 05:50 PM

  (13-08-2025, 11:41 AM)magnesium Wrote: You are not allowed to view links. Register or Login to view.Very interesting, thanks! Out of curiosity, would you mind running the same analysis on just the subsections of ZL3a-n.txt assigned to Voynich/Currier B? The overrepresentation of qok within the Naibbe cipher is meant to mimic Voynich B specifically, so I'm curious to see how well that holds up.

  
  Yes, I could do that. However, I would need a prepared ZL3a-n.txt file. Does anyone have one, or can anyone tell me how to extract Currier B from it ?

• RE: The Naibbe cipher

  magnesium > 13-08-2025, 06:04 PM

  (13-08-2025, 05:50 PM)bi3mw Wrote: You are not allowed to view links. Register or Login to view.Yes, I could do that. However, I would need a prepared ZL3a-n.txt file. Does anyone have one, or can anyone tell me how to extract Currier B from it ?

  
  Thank you very much! If memory serves, it's in the metadata for each folio, under the value $L. Currier B is $L=B, Currier A is $L=A. For example, for f75r:
  
  <f75r>    <! $Q=M $P=A $F=a $B=1 $I=B $L=B $H=2 $C=2>

• RE: The Naibbe cipher

  bi3mw > 13-08-2025, 06:31 PM

  I was actually thinking of an IVTT command on the command line that does everything in one go. Maybe @ReneZ can help here. I know the following command that extracts the entire text:
  
  ivtt -x7 ZL3a-n.txt outputfile.txt
  
  I just need to know the parameter to extract Currier B.
  
  Edit: Is that correct? :
  ivtt -x7 -LA ZL3a-n.txt outputfile.txt

• RE: The Naibbe cipher

  magnesium > 13-08-2025, 06:38 PM

  (13-08-2025, 06:31 PM)bi3mw Wrote: You are not allowed to view links. Register or Login to view.I was actually thinking of an IVTT command on the command line that does everything in one go. Maybe @ReneZ can help here. I know the following command that extracts the entire text:
  
  ivtt -x7 ZL3a-n.txt outputfile.txt
  
  I just need to know the parameter to extract Currier B.

  
  I believe you just need to add "+LB," like in the example on page 6 here: You are not allowed to view links. Register or Login to view.

• RE: The Naibbe cipher

  RobGea > 13-08-2025, 06:45 PM

  Hi bi3mw,
  i would be interested to know which "language-independent morphological parser" you used. Web-based or software or whatever, thanks

• RE: The Naibbe cipher

  bi3mw > 13-08-2025, 08:16 PM

  OK, when I apply the parameter “+LB”, the output file is 146 KB in size, and when I use “-LA”, it is 163.8 KB. That's why I think the first option is the right choice for this comparison. The following was used: naibbe_Cleaned_52_01_10_word_lines.txt + naibbe_Cleaned_52_02_10_word_lines.txt - 116 kb = 145.5 kb.
  
  As you can see, ch / hy is disproportionately represented in the Voynich text. The difference to the other values should have been lower in relation to the previous test.
  
     
     

• RE: The Naibbe cipher

  bi3mw > 13-08-2025, 09:49 PM

  (13-08-2025, 06:45 PM)RobGea Wrote: You are not allowed to view links. Register or Login to view.i would be interested to know which "language-independent morphological parser" you used. Web-based or software or whatever, thanks

  
  First, you need to install Pandas for Python:
  pip3 install pandas seaborn matplotlib
  
  I wrote a Python script that does this job.
  Call example (the number is variable and specifies how many top n combinations are output):
  
  python parser.py naibbe_Cleaned_52_01_10_word_lines.txt parsed.txt 25
  
  

  Code:

  #!/usr/bin/env python3
import sys
import re
from collections import Counter

def find_affixes(words, min_prefix_len=2, min_suffix_len=2, top_n=None):
    prefixes = Counter()
    suffixes = Counter()

    for w in words:
        if len(w) >= min_prefix_len:
            for i in range(min_prefix_len, min(len(w), 6)):
                prefixes[w[:i]] += 1
        if len(w) >= min_suffix_len:
            for i in range(min_suffix_len, min(len(w), 6)):
                suffixes[w[-i:]] += 1

    if top_n is None:
        return prefixes.most_common(), suffixes.most_common()
    else:
        return prefixes.most_common(top_n), suffixes.most_common(top_n)

def segment_word_multi(word, prefix_set, suffix_set):
    parts = []
    stem = word

    # Mehrere Präfixe entfernen
    while True:
        match = None
        for p in sorted(prefix_set, key=len, reverse=True):
            if stem.startswith(p) and len(stem) > len(p) + 1:
                match = p
                break
        if match:
            parts.append(match)
            stem = stem[len(match):]
        else:
            break

    suffix_parts = []
    # Mehrere Suffixe entfernen
    while True:
        match = None
        for s in sorted(suffix_set, key=len, reverse=True):
            if stem.endswith(s) and len(stem) > len(s) + 1:
                match = s
                break
        if match:
            suffix_parts.insert(0, match)
            stem = stem[:-len(match)]
        else:
            break

    parts.append(stem)
    parts.extend(suffix_parts)
    return parts

def find_stem(parts, prefix_set, suffix_set):
    for p in parts:
        if p not in prefix_set and p not in suffix_set:
            return p
    return ""

def main():
    argc = len(sys.argv)
    if argc < 3 or argc > 4:
        print(f"Usage: {sys.argv[0]} <input_file> <output_file> [top_n|'all']")
        sys.exit(1)

    input_file = sys.argv[1]
    output_file = sys.argv[2]
    top_n = 10  # Standardwert

    if argc == 4:
        param = sys.argv[3].lower()
        if param == "all":
            top_n = None  # Alle ausgeben
        else:
            try:
                top_n = int(param)
                if top_n < 1:
                    raise ValueError()
            except ValueError:
                print("Der dritte Parameter muss eine positive ganze Zahl oder 'all' sein.")
                sys.exit(1)

    with open(input_file, "r", encoding="utf-8") as f:
        text = f.read()

    words = [w.lower() for w in re.findall(r"[A-Za-z]+", text)]
    top_prefixes, top_suffixes = find_affixes(words, top_n=top_n)
    prefix_set = {p for p, _ in top_prefixes}
    suffix_set = {s for s, _ in top_suffixes}

    stem_counter = Counter()

    def replacer(match):
        token = match.group(0)
        parts = segment_word_multi(token.lower(), prefix_set, suffix_set)
        stem = find_stem(parts, prefix_set, suffix_set)
        stem_counter[stem] += 1
        segmented = "/".join(parts)
        if token[0].isupper():
            segmented = segmented.capitalize()
        return segmented

    output_text = re.sub(r"[A-Za-z]+", replacer, text)

    with open(output_file, "w", encoding="utf-8") as f:
        f.write(output_text)
        f.write("\n\n=== Präfixe ===\n")
        for p, c in top_prefixes:
            f.write(f"{p}\t{c}\n")
        f.write("\n=== Stämme ===\n")
        for s, c in stem_counter.most_common():
            if s != "":
                f.write(f"{s}\t{c}\n")
        f.write("\n=== Suffixe ===\n")
        for s, c in top_suffixes:
            f.write(f"{s}\t{c}\n")

if __name__ == "__main__":
    main()

  
  For the heat map, the call is as follows:
  
  python heatmap_prefix_suffix.py parsed.txt 25
  
  

  Code:

  #!/usr/bin/env python3
import sys
from collections import Counter
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt

def extract_prefix_suffix(word):
    parts = word.split('/')
    if len(parts) < 3:
        return [], [], None
    stem_index = len(parts) // 2
    prefix = parts[:stem_index]
    suffix = parts[stem_index+1:]
    return prefix, suffix, parts[stem_index]

def main():
    if len(sys.argv) < 2:
        print("Usage: python heatmap_prefix_suffix.py output_segmented.txt [N]")
        sys.exit(1)

    filename = sys.argv[1]
    top_n = 20  # default number of top prefixes and suffixes
    if len(sys.argv) >= 3:
        try:
            top_n = int(sys.argv[2])
        except ValueError:
            print("Parameter N must be an integer.")
            sys.exit(1)

    prefix_suffix_counts = Counter()

    with open(filename, 'r', encoding='utf-8') as f:
        for line in f:
            words = line.strip().split()
            for w in words:
                prefix, suffix, stem = extract_prefix_suffix(w)
                for pre in prefix:
                    for suf in suffix:
                        prefix_suffix_counts[(pre, suf)] += 1

    if not prefix_suffix_counts:
        print("No prefix-suffix combinations found.")
        sys.exit(1)

    data = []
    for (pre, suf), count in prefix_suffix_counts.items():
        data.append({"prefix": pre, "suffix": suf, "count": count})

    df = pd.DataFrame(data)

    # Select top N prefixes and suffixes by total count
    top_prefixes = df.groupby('prefix')['count'].sum().nlargest(top_n).index
    top_suffixes = df.groupby('suffix')['count'].sum().nlargest(top_n).index

    pivot = df.pivot(index='prefix', columns='suffix', values='count').fillna(0)
    pivot_top = pivot.loc[top_prefixes, top_suffixes]

    plt.figure(figsize=(18, 12))
    sns.heatmap(pivot_top, annot=True, fmt=".0f", cmap="YlGnBu", cbar_kws={"shrink": 0.5})
    plt.title(f"Top {top_n} Prefix-Suffix Combinations")
    plt.xlabel("Suffix")
    plt.ylabel("Prefix")
    plt.xticks(rotation=45, ha='right', fontsize=12)
    plt.yticks(rotation=0, fontsize=12)  # Präfix-Beschriftungen waagerecht
    plt.subplots_adjust(bottom=0.2)  # make space for xlabel
    plt.show()

if __name__ == "__main__":
    main()

• RE: The Naibbe cipher

  quimqu > 13-08-2025, 10:40 PM

  This is very interesting. I find the cipher a bit complex (sort of dice and card game, and I don't see it very easy to cipher and not even very easy to decipher, but this is the trick! If it is cyphered, only the allowed people should be able to read it...).
  
  I did my analyse of the text by bigrams (reseting by word) and I talked about a bump in the Voynch entropy by n-grams 2-4, that could only be reproduced by a sort of multi-character substitution (sort of syllabic substitution). You can see my thread here: You are not allowed to view links. Register or Login to view.
  
  Well, having analyzed one of the first naibbe ciphered files (naibbe_Cleaned_52_01_10_word_lines.txt)... the bump appears!  This are good news!
  
  
  

• RE: The Naibbe cipher

  davidma > 14-08-2025, 12:25 AM

  (13-08-2025, 10:40 PM)quimqu Wrote: You are not allowed to view links. Register or Login to view.This is very interesting. I find the cipher a bit complex (sort of dice and card game, and I don't see it very easy to cipher and not even very easy to decipher, but this is the trick! If it is cyphered, only the allowed people should be able to read it...).
  

  
  To my understanding though the Naibbe cipher uses cards and dice only to give a plausible explanation to the frequency of substitution tables and the spacing using something that people in the 1400-1450s had readily on hand. As long as the frequency behaviour is replicated it could be anything, any sort of randomizer works as long as the probabilities of picking a given encryption table are respected. It could even be scribal preference, though there have been compelling arguments that the scribes didnt really know what they were writing so maybe not. For me the "worrying" part is not so much the complexity (the cipher in itself is not particularly complex and if you have the tables on hand it can be decrypted quite quickly) but that without the original tables we might not be able to decrypt the VM (if it is encrypted at all), or extremely complex.
  
  I do not recall seeing anything that came close to the VM on all statistical indicators (bar the line and paragraph phenomena) apart from the Naibbe cipher, which means that this surely is on the track to something. As i recall from your post quimqu no other cipher had that distinctive bump that the VM has, but the naibbe one does. The other worry I have is the lables, I can't see how they would fit in the naibbe cipher at all because being verbose implies they at most encrypt 4 or 5 characters which would not carry much meaning (unless everything in the VM is heavily abbreviated, which is possible)

• RE: The Naibbe cipher

  ReneZ > 14-08-2025, 02:24 AM

  On the ivtt-related question, note that the option +LB includes only those pages where the Currier language has been set to 'B'. The option -LA removes those pages where the language has been set to 'A'.
  The difference between the two cases are those pages where the variable has not been set to any value.
  
  This variable has been set according to Currier's values in his paper, not any later analyses by others.
  
  In many cases people may want to exclude things like labels, and this is achieved by adding the option +@P which only selects all text in running paragraphs.
  
  With -x7 , uncertain spaces are treated as spaces. With -x8 they are ignored.
  I often use -s3 in combination with these, which causes the output to appear with each word on a separate line.
• Next Oldest Next Newest