How to know the difference between two ciphertexts without key stream in stream ciphers
$begingroup$
If I have two cipher texts lets say $C_1$ and $C_2$ of the same length encrypted through stream cipher technique using the same keystream. Let's say they are:
$$C_1: texttt{96 C6 A1 08 E7 F2 33 3B 3F 5C AB}$$
$$C_2: texttt{90 C6 A1 1E E6 F3 31 2B 37 4A B6}$$
$C_1$ is encrypted as ($P_1 oplus text{Keystream}$) and $C_2$ by ($P_2 oplus text{Keystream}$) where $P_1$ and $P_2$ are corresponding plaintexts.
- I am asked to tell how can I differentiate between corresponding plain text $P_1$ and plain text $P_2$ from $C_1$ and $C_2$ as an attacker without knowing the keystream?
So, I think the answer would be since both ciphers are encrypted through the same key stream, they would have similarities where the same plain text and keystream value exists. In this way, I can differentiate the other parts of the plain text. Is there anything more to it?
Thanks.
encryption stream-cipher
edited 13 hours ago
kelalaka
8,70522351
asked 14 hours ago
TahirTahir
83
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
If I have two cipher texts lets say $C_1$ and $C_2$ of the same length encrypted through stream cipher technique using the same keystream. Let's say they are:
$$C_1: texttt{96 C6 A1 08 E7 F2 33 3B 3F 5C AB}$$
$$C_2: texttt{90 C6 A1 1E E6 F3 31 2B 37 4A B6}$$
$C_1$ is encrypted as ($P_1 oplus text{Keystream}$) and $C_2$ by ($P_2 oplus text{Keystream}$) where $P_1$ and $P_2$ are corresponding plaintexts.
- I am asked to tell how can I differentiate between corresponding plain text $P_1$ and plain text $P_2$ from $C_1$ and $C_2$ as an attacker without knowing the keystream?
So, I think the answer would be since both ciphers are encrypted through the same key stream, they would have similarities where the same plain text and keystream value exists. In this way, I can differentiate the other parts of the plain text. Is there anything more to it?
Thanks.
encryption stream-cipher
edited 13 hours ago
kelalaka
8,70522351
asked 14 hours ago
TahirTahir
83
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If I have two cipher texts lets say $C_1$ and $C_2$ of the same length encrypted through stream cipher technique using the same keystream. Let's say they are:
$$C_1: texttt{96 C6 A1 08 E7 F2 33 3B 3F 5C AB}$$
$$C_2: texttt{90 C6 A1 1E E6 F3 31 2B 37 4A B6}$$
$C_1$ is encrypted as ($P_1 oplus text{Keystream}$) and $C_2$ by ($P_2 oplus text{Keystream}$) where $P_1$ and $P_2$ are corresponding plaintexts.
- I am asked to tell how can I differentiate between corresponding plain text $P_1$ and plain text $P_2$ from $C_1$ and $C_2$ as an attacker without knowing the keystream?
So, I think the answer would be since both ciphers are encrypted through the same key stream, they would have similarities where the same plain text and keystream value exists. In this way, I can differentiate the other parts of the plain text. Is there anything more to it?
Thanks.
encryption stream-cipher
edited 13 hours ago
kelalaka
8,70522351
asked 14 hours ago
TahirTahir
83
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If I have two cipher texts lets say $C_1$ and $C_2$ of the same length encrypted through stream cipher technique using the same keystream. Let's say they are:
$$C_1: texttt{96 C6 A1 08 E7 F2 33 3B 3F 5C AB}$$
$$C_2: texttt{90 C6 A1 1E E6 F3 31 2B 37 4A B6}$$
$C_1$ is encrypted as ($P_1 oplus text{Keystream}$) and $C_2$ by ($P_2 oplus text{Keystream}$) where $P_1$ and $P_2$ are corresponding plaintexts.
- I am asked to tell how can I differentiate between corresponding plain text $P_1$ and plain text $P_2$ from $C_1$ and $C_2$ as an attacker without knowing the keystream?
So, I think the answer would be since both ciphers are encrypted through the same key stream, they would have similarities where the same plain text and keystream value exists. In this way, I can differentiate the other parts of the plain text. Is there anything more to it?
Thanks.
encryption stream-cipher
encryption stream-cipher
edited 13 hours ago
kelalaka
8,70522351
asked 14 hours ago
TahirTahir
83
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
kelalaka
8,70522351
asked 14 hours ago
TahirTahir
83
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
kelalaka
8,70522351
edited 13 hours ago
kelalaka
8,70522351
edited 13 hours ago
kelalaka
8,70522351
8,70522351
asked 14 hours ago
TahirTahir
83
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 hours ago
TahirTahir
83
asked 14 hours ago
TahirTahir
83
83
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tahir is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's say $C_1 = P_1 oplus K$ and $C_2 = P_2 oplus K$ where $P$ is a plaintext, $K$ is the key stream and $C$ is the ciphertext.
Then if you XOR the two ciphertext together you get:
$$C_1 oplus C_2 =\
P_1 oplus K oplus P2 oplus K =\
P_1 oplus P_2$$
There are all kinds of interesting properties of the XOR of two plaintext together. For instance, one of the most common characters is the space, so you can easily guess many characters by just flipping a bit (space is 0x20 or 0b0010_0000 after all). You can see that a lot of combinations are not possible or unlikely and you can perform frequency analysis.
This becomes even more powerful if you have 3 or more ciphertexts, as you can compare each and every pair, and if there are $n$ ciphertext then there are ${n cdot (n - 1)} over 2$ combinations to be made.
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
$endgroup$
add a comment |
$begingroup$
In the stream-ciphers, same key-stream is not used two times, I mean that when you encrypt P1 with a Keystream (P1⊕Keystream), the same key-stream should never used for encrypting P2 (P2⊕Keystream). if you use same key-stream for two different encryption, then you cipher-texts are susceptible to "two time pad Attack". In this attack, Attacker captures C1 and C2 which they are encrypted in this way:
C1 = (P1⊕Keystream)
C2 = (P2⊕Keystream)
Then attacker works out C1 ⊕ C2; which leads to P1 ⊕ P2. We know that (Keystream ⊕ Keystream = 1).
Now attacker bases on some characteristics of plaintext (P1,P2) such as redundancy of ASCII codes, we can get the original plaintext.
But we should remember that we use IV beside the Key for preventing of producing the same keystream.
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Let's say $C_1 = P_1 oplus K$ and $C_2 = P_2 oplus K$ where $P$ is a plaintext, $K$ is the key stream and $C$ is the ciphertext.
Then if you XOR the two ciphertext together you get:
$$C_1 oplus C_2 =\
P_1 oplus K oplus P2 oplus K =\
P_1 oplus P_2$$
There are all kinds of interesting properties of the XOR of two plaintext together. For instance, one of the most common characters is the space, so you can easily guess many characters by just flipping a bit (space is 0x20 or 0b0010_0000 after all). You can see that a lot of combinations are not possible or unlikely and you can perform frequency analysis.
This becomes even more powerful if you have 3 or more ciphertexts, as you can compare each and every pair, and if there are $n$ ciphertext then there are ${n cdot (n - 1)} over 2$ combinations to be made.
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
$endgroup$
add a comment |
$begingroup$
Let's say $C_1 = P_1 oplus K$ and $C_2 = P_2 oplus K$ where $P$ is a plaintext, $K$ is the key stream and $C$ is the ciphertext.
Then if you XOR the two ciphertext together you get:
$$C_1 oplus C_2 =\
P_1 oplus K oplus P2 oplus K =\
P_1 oplus P_2$$
There are all kinds of interesting properties of the XOR of two plaintext together. For instance, one of the most common characters is the space, so you can easily guess many characters by just flipping a bit (space is 0x20 or 0b0010_0000 after all). You can see that a lot of combinations are not possible or unlikely and you can perform frequency analysis.
This becomes even more powerful if you have 3 or more ciphertexts, as you can compare each and every pair, and if there are $n$ ciphertext then there are ${n cdot (n - 1)} over 2$ combinations to be made.
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
$endgroup$
add a comment |
$begingroup$
Let's say $C_1 = P_1 oplus K$ and $C_2 = P_2 oplus K$ where $P$ is a plaintext, $K$ is the key stream and $C$ is the ciphertext.
Then if you XOR the two ciphertext together you get:
$$C_1 oplus C_2 =\
P_1 oplus K oplus P2 oplus K =\
P_1 oplus P_2$$
There are all kinds of interesting properties of the XOR of two plaintext together. For instance, one of the most common characters is the space, so you can easily guess many characters by just flipping a bit (space is 0x20 or 0b0010_0000 after all). You can see that a lot of combinations are not possible or unlikely and you can perform frequency analysis.
This becomes even more powerful if you have 3 or more ciphertexts, as you can compare each and every pair, and if there are $n$ ciphertext then there are ${n cdot (n - 1)} over 2$ combinations to be made.
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
$endgroup$
Let's say $C_1 = P_1 oplus K$ and $C_2 = P_2 oplus K$ where $P$ is a plaintext, $K$ is the key stream and $C$ is the ciphertext.
Then if you XOR the two ciphertext together you get:
$$C_1 oplus C_2 =\
P_1 oplus K oplus P2 oplus K =\
P_1 oplus P_2$$
There are all kinds of interesting properties of the XOR of two plaintext together. For instance, one of the most common characters is the space, so you can easily guess many characters by just flipping a bit (space is 0x20 or 0b0010_0000 after all). You can see that a lot of combinations are not possible or unlikely and you can perform frequency analysis.
This becomes even more powerful if you have 3 or more ciphertexts, as you can compare each and every pair, and if there are $n$ ciphertext then there are ${n cdot (n - 1)} over 2$ combinations to be made.
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
edited 11 hours ago
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
answered 13 hours ago
Maarten Bodewes♦Maarten Bodewes
55.7k679196
55.7k679196
add a comment |
add a comment |
$begingroup$
In the stream-ciphers, same key-stream is not used two times, I mean that when you encrypt P1 with a Keystream (P1⊕Keystream), the same key-stream should never used for encrypting P2 (P2⊕Keystream). if you use same key-stream for two different encryption, then you cipher-texts are susceptible to "two time pad Attack". In this attack, Attacker captures C1 and C2 which they are encrypted in this way:
C1 = (P1⊕Keystream)
C2 = (P2⊕Keystream)
Then attacker works out C1 ⊕ C2; which leads to P1 ⊕ P2. We know that (Keystream ⊕ Keystream = 1).
Now attacker bases on some characteristics of plaintext (P1,P2) such as redundancy of ASCII codes, we can get the original plaintext.
But we should remember that we use IV beside the Key for preventing of producing the same keystream.
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
$endgroup$
add a comment |
$begingroup$
In the stream-ciphers, same key-stream is not used two times, I mean that when you encrypt P1 with a Keystream (P1⊕Keystream), the same key-stream should never used for encrypting P2 (P2⊕Keystream). if you use same key-stream for two different encryption, then you cipher-texts are susceptible to "two time pad Attack". In this attack, Attacker captures C1 and C2 which they are encrypted in this way:
C1 = (P1⊕Keystream)
C2 = (P2⊕Keystream)
Then attacker works out C1 ⊕ C2; which leads to P1 ⊕ P2. We know that (Keystream ⊕ Keystream = 1).
Now attacker bases on some characteristics of plaintext (P1,P2) such as redundancy of ASCII codes, we can get the original plaintext.
But we should remember that we use IV beside the Key for preventing of producing the same keystream.
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
$endgroup$
add a comment |
$begingroup$
In the stream-ciphers, same key-stream is not used two times, I mean that when you encrypt P1 with a Keystream (P1⊕Keystream), the same key-stream should never used for encrypting P2 (P2⊕Keystream). if you use same key-stream for two different encryption, then you cipher-texts are susceptible to "two time pad Attack". In this attack, Attacker captures C1 and C2 which they are encrypted in this way:
C1 = (P1⊕Keystream)
C2 = (P2⊕Keystream)
Then attacker works out C1 ⊕ C2; which leads to P1 ⊕ P2. We know that (Keystream ⊕ Keystream = 1).
Now attacker bases on some characteristics of plaintext (P1,P2) such as redundancy of ASCII codes, we can get the original plaintext.
But we should remember that we use IV beside the Key for preventing of producing the same keystream.
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
$endgroup$
In the stream-ciphers, same key-stream is not used two times, I mean that when you encrypt P1 with a Keystream (P1⊕Keystream), the same key-stream should never used for encrypting P2 (P2⊕Keystream). if you use same key-stream for two different encryption, then you cipher-texts are susceptible to "two time pad Attack". In this attack, Attacker captures C1 and C2 which they are encrypted in this way:
C1 = (P1⊕Keystream)
C2 = (P2⊕Keystream)
Then attacker works out C1 ⊕ C2; which leads to P1 ⊕ P2. We know that (Keystream ⊕ Keystream = 1).
Now attacker bases on some characteristics of plaintext (P1,P2) such as redundancy of ASCII codes, we can get the original plaintext.
But we should remember that we use IV beside the Key for preventing of producing the same keystream.
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
answered 12 hours ago
Arsalan VahiArsalan Vahi
917
917
add a comment |
add a comment |
Tahir is a new contributor. Be nice, and check out our Code of Conduct.
Tahir is a new contributor. Be nice, and check out our Code of Conduct.
Tahir is a new contributor. Be nice, and check out our Code of Conduct.
Tahir is a new contributor. Be nice, and check out our Code of Conduct.
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