Today’s Cryptogram

[nikos1121]

Today’s Cryptogram

T NOPK TOCHHC KBED HGTZ T TODHOC DP CP, FLD

VDAPORHA DBEO EZZ WJ EYDHADBPLRBDV TV WJ YLAJ,

YLAJ DBED FATORV LIPO WPADEZV DBH RAHEDHVD HGTZV.

- HLATITCHV

Yesterday’s Cryptogram

OFWWLRMKK LK ULVM F VLKK. CIT QTKS KOFPM LS SI MRYIC LS. —NMPRFPE QMUSXMP

[Lonesome in Massachussets]

HLATITCHV, you menda deese.

[Ozymandias Ghost]

3. Hal’s Helper

[UpstateNYer]

Usually Hal's Helper but sometimes, when I'm feeling ambitious, I do them in my head.

I did today's in my head and it took 5+ minutes - pretty sure it would have been a quickie if I used Hal's.

[John4.11]

Hal’s Helper.

[nikos1121]

Oh boy, there’s always one... lol

[Parley Baer]

Pencil and paper.

[gundog]

Did pen and paper for a lot of years. Doing Hal’s, now. Some I’ll try in my head.

[goodnesswins]

2&3

[Grammy]

I always use Hal’s. Thanks for providing the link!

[nikos1121]

Can you share that with the group? I’m like not real smart so you’ll have do it step by step?

[Ingtar]

3.

[Mr Fuji]

Probably about 4-5 minutes today.

To answer your question: 3 & 4. It just depends on how mentally tired I am in the evening.

[doorgunner69]

Hal’s, this one took a long time for some reason. Trying them before morning coffee is usually a waste of time.

[Lonesome in Massachussets]

Here’s my vowel sniffer. I will post my formatter/histogram later. If you want my excel workbook give me contact information on freepmail, and I’ll send a copy.

If you uncomment the first line, and comment the line that hard wires the value of “string” it’s a function. As is, you have to edit it each time you run it to change the value of “string”. This example is from the last time I used it. In matlab and octave the apostrophe is a string delimiter, so you have to enter double apostrophes (’’) to represent a single apostrophe. In octave quote marks, “, can also be a string delimiter.

When you run it the results should be:

>> vowelsniffer

vowelChar =

D
F
G
K
L
M
U
Z

D 23
F 19
G 11
K 11
L 10
M 9
U 9
Z 8

Notice that D = A, F = D, G = T, K = E, L = M, M = G, U = O, Z = I, so that for small samples it does not work very well. Try it with a longer quote - it shouldn’t matter whether or not it’s encrypted, it simply performs statistical tests.

When I paste in the first three paragraphs of the Battle of Bennington from Wikipedia,unencrypted, this is the score:

A 325
E 268
G 251
I 209
K 198
O 187
U 183
Y 118

Not perfect, but better. These are the letters that “smell” like vowels when subjected to statistical tests.

---

%function [possibleVowels] = vowelsniffer(string)
%
% implements the algorithm in chapter 2 in
% Secret History: The Story of Cryptology (Discrete Mathematics
% and Its Applications) 1st Edition
% by Craig P. Bauer

% paste in the string you are analyzing here
string =’QKH BUIP QUH OKDFV GTK HUIOF’’V MIKDG EZGZKV ZQ GTK QSLYKI UW AKUAOK DIUSQF HTUL BUS VTUSOFQ’’G LDPK D VSFFKQ LURK. - FDRZF OKGGKILDQ’;

string = upper(string);
% pad the string with blanks. We look one character ahead and one back
string = [’ ‘,string,’ ‘];
contacts = zeros(26,26);

for rowValue = ‘A’:’Z’;
for columnValue = ‘A’:’Z’;
if columnValue == rowValue; continue;end
rowIndex = find(string == rowValue);
if isempty(rowIndex);continue;end
columnIndex = find(string(rowIndex+1)==columnValue);
if ~isempty(columnIndex)
contacts(abs(rowValue)-64, abs(columnValue)-64) = ...
contacts(abs(rowValue)-64, abs(columnValue)-64) + length(columnIndex);
end
columnIndex = find(string(rowIndex-1)==columnValue);
if ~isempty(columnIndex)
contacts(abs(rowValue)-64, abs(columnValue)-64) =...
contacts(abs(rowValue)-64, abs(columnValue)-64) + length(columnIndex);
end
end % for rowValue = ‘A’:’Z’;
end % for columnValue = ‘A’:’Z’;

workingMatrix = contacts;

rowSums = sum(workingMatrix,2);
isVowel = false(size(rowSums));

while(1);

rowSums = sum(workingMatrix,2);
[maxSum, ix] = max(rowSums);
if maxSum <= 0 ; break;end
ix = find(rowSums == maxSum); % allow for multiple values
isVowel(ix) = true;
%
workingMatrix(ix,:) = -1e6; % get these rows out of the way;
%
% subtract from every row not identified as a vowel twice the
% number of times it contacts the “vowel(s)” identified on this iteration,
% Which is the same as reversing the column value of the vowel contacts
workingMatrix(:,ix) = -workingMatrix(:,ix);

end

% print candidate “vowels” and their raw scores.
% raw scores are the number of times a letter contacts a different letter
% candidate vowels are letters that are not eliminated by subtracting from their
% score twice the number of times they contact a previously identified vowel.
rowSums = sum(contacts,2);
[~ , ix] = sort(rowSums, ‘descend’);
vowelChar = char(find(isVowel)+64)

for k = 1:sum(isVowel)
fprintf(’%c\t%d\n’, vowelChar(k), rowSums(ix(k)))
end

%endfunction

[Lonesome in Massachussets]

See also my 34

below is a little matlab/octave script for making histograms of a letters in cryptogram, and printing the cryptogram as tab delimited characters (easy to paste into Excel), along with a histogram.

It's an acquired taste.

-

CYPHER = 'T NOPK TOCHHC KBED HGTZ T TODHOC DP CP, FLD VDAPORHA DBEO EZZ WJ EYDHADBPLRBDV TV WJ YLAJ, YLAJ DBED FATORV LIPO WPADEZV DBH RAHEDHVD HGTZV. - HLATITCHV ';


fid = fopen('decoderRing.txt', 'w');

A = abs(upper(CYPHER));

B = unique(A);

N = hist( A, B); %
N = N(B>64 & B<91);
B = B(B>64 & B<91);
unusedLetters = setdiff( 'A':'Z', char(B));
N = [N, zeros(size(unusedLetters))];
B = [B, abs(unusedLetters)];

[N, ix] = sort(N, 'descend');
B = B(ix);


for k = 1:length(N)

if B(k)<65; continue;end
if B(k)>90; continue;end

fprintf( '%d\t%c\n', N(k), char(B(k)));
fprintf(fid, '%d\t%c\n', N(k), char(B(k)));
end

fprintf( '\n\n')
fprintf(fid, '\n\n');

lineNo = 1;
for k = 1:length(CYPHER)

% char(39) = ' the apostrophe
if k> 30*lineNo && ~ismember(CYPHER(k), ['A':'Z', char(39), '.', ':',';'])
fprintf(fid, '\n\n\n');
fprintf( '\n\n\n');
lineNo = lineNo+1;
else

fprintf('%c\t',CYPHER(k))
fprintf(fid, '%c\t',CYPHER(k));
end
end

fclose (fid);
fprintf('\n\n')