[QUIZ] Cryptogram II (#206)

[Daniel Moore]

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## Cryptogram II (#206)

YRFFJ OSLXA PQPQY APVRR DPNSA ZAPIK
OXMQJ BOIMY XMSZZ FRIHE AUXJS IOROR
IEAHB QYAPQ YRHXJ STRIF ORIEX KOIKD
REAQK JHBOI QSFIQ AJHPA QPKIF FREKO
XMQJB OIGAA LRKIS PRNSA Z13VI PIKOX
MQJBO IGNSA ZIPVR FFYJM RXJSY IEUSH

 

[Robert Dober]

-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-= =3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-=3D-

The three rules of Ruby Quiz:

1. =A0Please do not post any solutions or spoiler discussion for this
quiz until 48 hours have elapsed from the time this message was
sent.

2. =A0Support Ruby Quiz by submitting ideas and responses
as often as you can!
Visit: http://rubyquiz.strd6.com/suggestions

3. =A0Enjoy!

Suggestion: =A0A [QUIZ] in the subject of emails about the problem
helps everyone on Ruby Talk follow the discussion. =A0Please reply to
the original quiz message, if you can.

RSS Feed: http://rubyquiz.strd6.com/quizzes.rss

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## Cryptogram II (#206)

YRFFJ OSLXA PQPQY APVRR DPNSA ZAPIK
OXMQJ BOIMY XMSZZ FRIHE AUXJS IOROR
IEAHB QYAPQ YRHXJ STRIF ORIEX KOIKD
REAQK JHBOI QSFIQ AJHPA QPKIF FREKO
XMQJB OIGAA LRKIS PRNSA Z13VI PIKOX
MQJBO IGNSA ZIPVR FFYJM RXJSY IEUSH

This is a strange way to spell
MARY HADA LITT LELA MBDA

R.



--=20
Toutes les grandes personnes ont d=92abord =E9t=E9 des enfants, mais peu
d=92entre elles s=92en souviennent.

All adults have been children first, but not many remember.

[Antoine de Saint-Exup=E9ry]

 

[lith]

1. =A0Please do not post any solutions or spoiler discussion for this

quiz until 48 hours have elapsed from the time this message was
sent.

YRFFJ OSLXA PQPQY APVRR DPNSA ZAPIK

Since 48 hours have passed, maybe somebody could at least explain what
this quiz was about. Or maybe simply post a solution.

Cheers.

 

[Martin DeMello]

Since 48 hours have passed, maybe somebody could at least explain what
this quiz was about. Or maybe simply post a solution.

About: Write a ruby program to convert that into English text.

martin

 

[Joshua Ball]

Check out this link to understand what a cryptogram is:
http://en.wikipedia.org/wiki/Cryptogram




-----Original Message-----
From: Martin DeMello [mailto:[email protected]]=20
Sent: Monday, May 25, 2009 8:30 AM
To: ruby-talk ML
Subject: Re: Cryptogram II (#206)

Since 48 hours have passed, maybe somebody could at least explain what
this quiz was about. Or maybe simply post a solution.

About: Write a ruby program to convert that into English text.

martin

 

[lith]

Check out this link to understand what a cryptogram is:http://en.wikipedi=

a.org/wiki/Cryptogram

I was wondering if the blanks represent actual word breaks or not. If
this were the case and if we were dealing with a normal cryptogram,
the word "FFYJM" would begin with two times the same letter. English
isn't my mother tongue but this seems rather unlikely to me which is
why I have concluded that is isn't a normal cryptogram, which is about
the point where I stopped.

Also, there already was a rubyquiz about solving cryptograms, which
made me suspect that we are dealing with something else here. Was I
wrong?

 

[Luke Cowell]

I was wondering if the blanks represent actual word breaks or not. If
this were the case and if we were dealing with a normal cryptogram,
the word "FFYJM" would begin with two times the same letter. English
isn't my mother tongue but this seems rather unlikely to me which is
why I have concluded that is isn't a normal cryptogram, which is about
the point where I stopped.

I was thinking that each tuple represents a single letter. It seems
unlikely that we have 36 words that all have exactly 5 letters in them.

Luke

 

[Martin Boese]

I was starting something like this. But it doesn't seem to work out:

crypt = <<-ENDC
YRFFJ OSLXA PQPQY APVRR DPNSA ZAPIK
OXMQJ BOIMY XMSZZ FRIHE AUXJS IOROR
IEAHB QYAPQ YRHXJ STRIF ORIEX KOIKD
REAQK JHBOI QSFIQ AJHPA QPKIF FREKO
XMQJB OIGAA LRKIS PRNSA Z13VI PIKOX
MQJBO IGNSA ZIPVR FFYJM RXJSY IEUSH
ENDC

('A'[0]..'Z'[0]).each { |c| puts "#{c.chr} -> #{crypt.count(c.chr)}" }
A -> 14
B -> 5
C -> 0
D -> 2
E -> 6
F -> 9
G -> 2
H -> 6
I -> 18
J -> 10
K -> 8
L -> 2
M -> 6
N -> 3
O -> 12
P -> 11
Q -> 11
R -> 15
S -> 11
T -> 1
U -> 2
V -> 3
W -> 0
X -> 9
Y -> 7
Z -> 5


english = File.read('/usr/share/dict/words').upcase

('A'[0]..'Z'[0]).each { |c| puts "#{c.chr} ->#{english.count(c.chr)}" }
A -> 64123
B -> 15524
C -> 31569
D -> 28877
E -> 88441
F -> 10551
G -> 23073
H -> 19313
I -> 66892
J -> 1915
K -> 8456
L -> 40826
M -> 22433
N -> 57227
O -> 48779
P -> 21891
Q -> 1510
R -> 57035
S -> 88135
T -> 52446
U -> 25927
V -> 7903
W -> 7431
X -> 2124
Y -> 12507
Z -> 3238

 

[Harry Kakueki]

I was wondering if the blanks represent actual word breaks or not. If
this were the case and if we were dealing with a normal cryptogram,
the word "FFYJM" would begin with two times the same letter. English
isn't my mother tongue but this seems rather unlikely to me which is
why I have concluded that is isn't a normal cryptogram, which is about
the point where I stopped.

Also, there already was a rubyquiz about solving cryptograms, which
made me suspect that we are dealing with something else here. Was I
wrong?

"FFYJM" could be ["oomph", "oozed", "oozes"] but "PQPQY" could be
["cacao", "cocoa", "dodos", "lulus", "mamas", "mimic", "tutus",
"vivid"].

Looking at these, there is no common letter for "Y".
So, I don't think these are just a bunch of five letter words.

Harry

 

[Rob Biedenharn]

I was wondering if the blanks represent actual word breaks or not. If
this were the case and if we were dealing with a normal cryptogram,
the word "FFYJM" would begin with two times the same letter. English
isn't my mother tongue but this seems rather unlikely to me which is
why I have concluded that is isn't a normal cryptogram, which is
about
the point where I stopped.

Also, there already was a rubyquiz about solving cryptograms, which
made me suspect that we are dealing with something else here. Was I
wrong?

"FFYJM" could be ["oomph", "oozed", "oozes"] but "PQPQY" could be
["cacao", "cocoa", "dodos", "lulus", "mamas", "mimic", "tutus",
"vivid"].

Looking at these, there is no common letter for "Y".
So, I don't think these are just a bunch of five letter words.

Harry

Come on people! This is a standard technique to prevent things like
lone letters or letter pairs providing too much help. The whole
cryptogram is likely a simple substitution (one letter stands for
another) with possible "filler" letters at the end (to make a multiple
of 5). You need to find word breaks after you apply a substitution.

Martin is probably on the right track with comparing frequency counts
between English and the cryptogram, but that's only a start. Once you
have a candidate substitution, you have to see if you can pull letters
off the cryptogram to form words. If you get a series of words that
leaves fewer than 5 letters at the end, then you very likely have a
solution.

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)

 

[Harry Kakueki]

Check out this link to understand what a cryptogram
is:http://en.wikipedia.org/wiki/Cryptogram

I was wondering if the blanks represent actual word breaks or not. If
this were the case and if we were dealing with a normal cryptogram,
the word "FFYJM" would begin with two times the same letter. English
isn't my mother tongue but this seems rather unlikely to me which is
why I have concluded that is isn't a normal cryptogram, which is about
the point where I stopped.

Also, there already was a rubyquiz about solving cryptograms, which
made me suspect that we are dealing with something else here. Was I
wrong?

"FFYJM" could be ["oomph", "oozed", "oozes"] but "PQPQY" could be
["cacao", "cocoa", "dodos", "lulus", "mamas", "mimic", "tutus",
"vivid"].

Looking at these, there is no common letter for "Y".
So, I don't think these are just a bunch of five letter words.

Harry

Come on people! This is a standard technique to prevent things like lone
letters or letter pairs providing too much help. The whole cryptogram is
likely a simple substitution (one letter stands for another) with possible
"filler" letters at the end (to make a multiple of 5). You need to find word
breaks after you apply a substitution.

Martin is probably on the right track with comparing frequency counts
between English and the cryptogram, but that's only a start. Once you have a
candidate substitution, you have to see if you can pull letters off the
cryptogram to form words. If you get a series of words that leaves fewer
than 5 letters at the end, then you very likely have a solution.

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)

I think most people are aware of everything you said.
And it will probably be easy to understand after seeing the result.
But there are many tricks that could be used here and I suspect that
most people only have a few hours to work on this, not days.
Also, it is easy to criticize others without actually having a
solution of your own.

Harry

 

[Ken Bloom]

I was starting something like this. But it doesn't seem to work out:

crypt = <<-ENDC
YRFFJ OSLXA PQPQY APVRR DPNSA ZAPIK
OXMQJ BOIMY XMSZZ FRIHE AUXJS IOROR
IEAHB QYAPQ YRHXJ STRIF ORIEX KOIKD
REAQK JHBOI QSFIQ AJHPA QPKIF FREKO
XMQJB OIGAA LRKIS PRNSA Z13VI PIKOX
MQJBO IGNSA ZIPVR FFYJM RXJSY IEUSH
ENDC

('A'[0]..'Z'[0]).each { |c| puts "#{c.chr} -> #{crypt.count(c.chr)}" }

cipherside=('A'[0]..'Z'[0]).map { |c| crypt.count(c.chr) }

english = File.read('/usr/share/dict/words').upcase

('A'[0]..'Z'[0]).each { |c| puts "#{c.chr} ->#{english.count(c.chr)}" }

englishside = ('A'[0]..'Z'[0]).map { |c| english.count(c.chr) }

class Array
def rotate num
self[num..-1]+self[0...num]
end
end

(0...26).each do |rot|
dotprod=cipherside.zip(englishside.rotate(rot)).inject(0) do |a,(b,c)|
a+b*c
end
printf "%d -> %d\n", rot, dotprod
end

0 -> 6418143
1 -> 5568695
2 -> 5549352
3 -> 6915018
4 -> 6262198
5 -> 5279474
6 -> 4898071
7 -> 4508113
8 -> 5015397
9 -> 5761530
10 -> 6017183
11 -> 5443382
12 -> 5440166
13 -> 5944635
14 -> 5750203
15 -> 4540453
16 -> 5307004
17 -> 5184543
18 -> 5899138
19 -> 5232100
20 -> 5903220
21 -> 5759249
22 -> 6291068
23 -> 4976438
24 -> 4537099
25 -> 5448116

def decrypt msg, num
msg.each_byte do |x|
next if x>?Z or x<?A
x+=num
x-=26 if x>?Z
x+=26 if x<?A
print x.chr
end
puts
end

(0...26).each do |rot|
decrypt crypt, rot
puts
end

I'll let you run this to see the output, but clearly this isn't a
rotation cipher.

 

[Todd Benson]

Check out this link to understand what a cryptogram
is:http://en.wikipedia.org/wiki/Cryptogram

I was wondering if the blanks represent actual word breaks or not. If
this were the case and if we were dealing with a normal cryptogram,
the word "FFYJM" would begin with two times the same letter. English
isn't my mother tongue but this seems rather unlikely to me which is
why I have concluded that is isn't a normal cryptogram, which is about
the point where I stopped.

Also, there already was a rubyquiz about solving cryptograms, which
made me suspect that we are dealing with something else here. Was I
wrong?

"FFYJM" could be ["oomph", "oozed", "oozes"] but "PQPQY" could be
["cacao", "cocoa", "dodos", "lulus", "mamas", "mimic", "tutus",
"vivid"].

Looking at these, there is no common letter for "Y".
So, I don't think these are just a bunch of five letter words.

Harry

Come on people! This is a standard technique to prevent things like lone
letters or letter pairs providing too much help. The whole cryptogram is
likely a simple substitution (one letter stands for another) with possible
"filler" letters at the end (to make a multiple of 5). You need to find word
breaks after you apply a substitution.

Martin is probably on the right track with comparing frequency counts
between English and the cryptogram, but that's only a start. Once you have a
candidate substitution, you have to see if you can pull letters off the
cryptogram to form words. If you get a series of words that leaves fewer
than 5 letters at the end, then you very likely have a solution.

-Rob

Rob Biedenharn http://agileconsultingllc.com
(e-mail address removed)

I think most people are aware of everything you said.
And it will probably be easy to understand after seeing the result.
But there are many tricks that could be used here and I suspect that
most people only have a few hours to work on this, not days.
Also, it is easy to criticize others without actually having a
solution of your own.

I love this quiz! I'll throw it at my friend, though. In real life,
these cryptograms can be tough, because you don't know if it's
english, and if it is, you don't know the order because of the 5 by 6
representation. That's the way I'd change it up, anyway.

I'll assume it's left-top to right-bottom, and see if he can come up
with something that works using frequency analysis.

Todd

 

[Martin Boese]

Got it ;-)

Below the result.


How? Instead of counting characters look for repeating sequences. Then
think carefully about the long word that it finds. I used this (ugly)
code:



crypted = <<HERE
YRFFJ OSLXA PQPQY APVRR DPNSA ZAPIK
OXMQJ BOIMY XMSZZ FRIHE AUXJS IOROR
IEAHB QYAPQ YRHXJ STRIF ORIEX KOIKD
REAQK JHBOI QSFIQ AJHPA QPKIF FREKO
XMQJB OIGAA LRKIS PRNSA Z13VI PIKOX
MQJBO IGNSA ZIPVR FFYJM RXJSY IEUSH
HERE
crypted = crypted.gsub(/[^A-Z]/, '')


class String
def count(ss)
r = 0
self.split(//).each_index do |idx|
r+=1 if self[idx..-1][0..(ss.length-1)] == ss
end
r
end
end

cfound = {}
(1..11).each do |len|
found = {}
crypted.split(//).each_index do |idx|
next if crypted[idx+len].nil?
search = crypted[idx..(idx+len)]
nfound = crypted.count(search)
found[search] = nfound if nfound > 1
end
cfound[len] = found.sort { |a,b| a[1]<=>b[1] }
end

(1..11).each do |idx|
k = idx
v = cfound[idx]
n = cfound[k+1] || []
v.each do |f|
puts "#{f[0]} => #{f[1]}" unless n.find { |e|
e[0].include?(f[0]) }
end
end


This website then helps:


http://tomacorp.com/codes/workbench.html



























Result:






sub = [['A', 'I'],
['B', 'G'],
['C', '.'] ,
['D', 'K'],
['E', 'D'],
['F', 'L'],
['G', '.'],
['H', 'N'],
['I', 'A'],
['J', 'O'],
['K', 'C'],
['L', 'B'],
['M', 'P'],
['N', 'Q'],
['O', 'R'],
['P', 'S'],
['Q', 'T'],
['R', 'E'],
['S', 'U'],
['T', 'F'],
['U', '.'],
['V', 'W'],
['W', '.'],
['X', 'Y'],
['Y', 'H'],
['Z', 'Z']
]



Now back to work...

Martin

 

[Martin DeMello]

Got it ;-)

Below the result.


How? Instead of counting characters look for repeating sequences. Then
think carefully about the long word that it finds. I used this (ugly)
code:

Bravo Very nicely done indeed.

martin