What does this paraphrase of the birthday problem mean? The Next CEO of Stack OverflowWhat are the differences between collision attack and birthday attack?k(k-1)/2: Combinations and the Birthday boundIs there any function that does not suffers birthday problem?Why k-lists generalized birthday problem when $k=2$ is classical birthday problem?Time complexity of birthday attack type problemHow does hashing twice protect against birthday attacks?What is a wide block cipher and why does it avoid birthday bound problems?Can the birthday attack be extended in this case?On a lower bound for the birthday problemWhat's the Error in This Birthday-Problem Estimation?
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What does this paraphrase of the birthday problem mean?
The Next CEO of Stack OverflowWhat are the differences between collision attack and birthday attack?k(k-1)/2: Combinations and the Birthday boundIs there any function that does not suffers birthday problem?Why k-lists generalized birthday problem when $k=2$ is classical birthday problem?Time complexity of birthday attack type problemHow does hashing twice protect against birthday attacks?What is a wide block cipher and why does it avoid birthday bound problems?Can the birthday attack be extended in this case?On a lower bound for the birthday problemWhat's the Error in This Birthday-Problem Estimation?
$begingroup$
The following is an excerpt from A Generalized Birthday Problem - David Wagner:
One of the best-known combinatorial tools in cryptology is the birthday problem:
Problem 1. Given two lists $L_1, space L_2$ of elements drawn uniformly and independently at random from $0, 1^n$, find $x_1 in L_1$ and $x_2 in L_2$ such that $x_1 oplus x_2 = 0$.
It's not so intuitive for me to understand. In my understanding, the birthday problem is about the probability that at least 2 people in a room have the same birthday. How does the birthday problem transfers to this? Please give me some hints.
birthday-attack
$endgroup$
add a comment |
$begingroup$
The following is an excerpt from A Generalized Birthday Problem - David Wagner:
One of the best-known combinatorial tools in cryptology is the birthday problem:
Problem 1. Given two lists $L_1, space L_2$ of elements drawn uniformly and independently at random from $0, 1^n$, find $x_1 in L_1$ and $x_2 in L_2$ such that $x_1 oplus x_2 = 0$.
It's not so intuitive for me to understand. In my understanding, the birthday problem is about the probability that at least 2 people in a room have the same birthday. How does the birthday problem transfers to this? Please give me some hints.
birthday-attack
$endgroup$
add a comment |
$begingroup$
The following is an excerpt from A Generalized Birthday Problem - David Wagner:
One of the best-known combinatorial tools in cryptology is the birthday problem:
Problem 1. Given two lists $L_1, space L_2$ of elements drawn uniformly and independently at random from $0, 1^n$, find $x_1 in L_1$ and $x_2 in L_2$ such that $x_1 oplus x_2 = 0$.
It's not so intuitive for me to understand. In my understanding, the birthday problem is about the probability that at least 2 people in a room have the same birthday. How does the birthday problem transfers to this? Please give me some hints.
birthday-attack
$endgroup$
The following is an excerpt from A Generalized Birthday Problem - David Wagner:
One of the best-known combinatorial tools in cryptology is the birthday problem:
Problem 1. Given two lists $L_1, space L_2$ of elements drawn uniformly and independently at random from $0, 1^n$, find $x_1 in L_1$ and $x_2 in L_2$ such that $x_1 oplus x_2 = 0$.
It's not so intuitive for me to understand. In my understanding, the birthday problem is about the probability that at least 2 people in a room have the same birthday. How does the birthday problem transfers to this? Please give me some hints.
birthday-attack
birthday-attack
edited Mar 22 at 13:51
Squeamish Ossifrage
22k132100
22k132100
asked Mar 22 at 3:20
Cedric SunCedric Sun
314
314
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
$x_1 oplus x_2 = 0$ is equivalent to $x_1=x_2$ (because $oplus$ is bitwise XOR, and that equivalence stands for bits, and multibit quantities being equal in all their respective bits is equivalent to these quantities being equal).
Now assume that $x_i$ is the birthday of person $i$ in the room, expressed as days since the first day of the year, in binary, with a year of $2^n$ days, and what's meant should be clear.
Notice that the problem studied in the quote is about two lists/rooms, rather than one in the standard birthday problem.
$endgroup$
add a comment |
$begingroup$
Note that $x_1=x_2$, i.e., there is a birthday collision in $0,1^n$ if and only if $x_1oplus x_2=0.$
In a general additive group $G$, $x_1=x_2$, i.e., there is a birthday collision in $G$ if and only if $x_1-x_2=0.$
If you have two lists $L_1,L_2,$ then with probability roughly
$$expleft-frac^22^n+1right$$
there will be no collisions.
In the birthday paradox, for $N=2^n$ bins, the probability of no collisions after $m$ balls is roughly
$$expleft-fracm^22Nright$$
while here we have $|L_1||L_2|$ pairs to consider so $m=|L_1||L_2|.$
Wagner's paper is about finding efficient algorithms for vectors adding to zero for higher numbers (e.g., 4) of lists.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$x_1 oplus x_2 = 0$ is equivalent to $x_1=x_2$ (because $oplus$ is bitwise XOR, and that equivalence stands for bits, and multibit quantities being equal in all their respective bits is equivalent to these quantities being equal).
Now assume that $x_i$ is the birthday of person $i$ in the room, expressed as days since the first day of the year, in binary, with a year of $2^n$ days, and what's meant should be clear.
Notice that the problem studied in the quote is about two lists/rooms, rather than one in the standard birthday problem.
$endgroup$
add a comment |
$begingroup$
$x_1 oplus x_2 = 0$ is equivalent to $x_1=x_2$ (because $oplus$ is bitwise XOR, and that equivalence stands for bits, and multibit quantities being equal in all their respective bits is equivalent to these quantities being equal).
Now assume that $x_i$ is the birthday of person $i$ in the room, expressed as days since the first day of the year, in binary, with a year of $2^n$ days, and what's meant should be clear.
Notice that the problem studied in the quote is about two lists/rooms, rather than one in the standard birthday problem.
$endgroup$
add a comment |
$begingroup$
$x_1 oplus x_2 = 0$ is equivalent to $x_1=x_2$ (because $oplus$ is bitwise XOR, and that equivalence stands for bits, and multibit quantities being equal in all their respective bits is equivalent to these quantities being equal).
Now assume that $x_i$ is the birthday of person $i$ in the room, expressed as days since the first day of the year, in binary, with a year of $2^n$ days, and what's meant should be clear.
Notice that the problem studied in the quote is about two lists/rooms, rather than one in the standard birthday problem.
$endgroup$
$x_1 oplus x_2 = 0$ is equivalent to $x_1=x_2$ (because $oplus$ is bitwise XOR, and that equivalence stands for bits, and multibit quantities being equal in all their respective bits is equivalent to these quantities being equal).
Now assume that $x_i$ is the birthday of person $i$ in the room, expressed as days since the first day of the year, in binary, with a year of $2^n$ days, and what's meant should be clear.
Notice that the problem studied in the quote is about two lists/rooms, rather than one in the standard birthday problem.
edited Mar 23 at 11:31
answered Mar 22 at 7:22
fgrieufgrieu
81.9k7178349
81.9k7178349
add a comment |
add a comment |
$begingroup$
Note that $x_1=x_2$, i.e., there is a birthday collision in $0,1^n$ if and only if $x_1oplus x_2=0.$
In a general additive group $G$, $x_1=x_2$, i.e., there is a birthday collision in $G$ if and only if $x_1-x_2=0.$
If you have two lists $L_1,L_2,$ then with probability roughly
$$expleft-frac^22^n+1right$$
there will be no collisions.
In the birthday paradox, for $N=2^n$ bins, the probability of no collisions after $m$ balls is roughly
$$expleft-fracm^22Nright$$
while here we have $|L_1||L_2|$ pairs to consider so $m=|L_1||L_2|.$
Wagner's paper is about finding efficient algorithms for vectors adding to zero for higher numbers (e.g., 4) of lists.
$endgroup$
add a comment |
$begingroup$
Note that $x_1=x_2$, i.e., there is a birthday collision in $0,1^n$ if and only if $x_1oplus x_2=0.$
In a general additive group $G$, $x_1=x_2$, i.e., there is a birthday collision in $G$ if and only if $x_1-x_2=0.$
If you have two lists $L_1,L_2,$ then with probability roughly
$$expleft-frac^22^n+1right$$
there will be no collisions.
In the birthday paradox, for $N=2^n$ bins, the probability of no collisions after $m$ balls is roughly
$$expleft-fracm^22Nright$$
while here we have $|L_1||L_2|$ pairs to consider so $m=|L_1||L_2|.$
Wagner's paper is about finding efficient algorithms for vectors adding to zero for higher numbers (e.g., 4) of lists.
$endgroup$
add a comment |
$begingroup$
Note that $x_1=x_2$, i.e., there is a birthday collision in $0,1^n$ if and only if $x_1oplus x_2=0.$
In a general additive group $G$, $x_1=x_2$, i.e., there is a birthday collision in $G$ if and only if $x_1-x_2=0.$
If you have two lists $L_1,L_2,$ then with probability roughly
$$expleft-frac^22^n+1right$$
there will be no collisions.
In the birthday paradox, for $N=2^n$ bins, the probability of no collisions after $m$ balls is roughly
$$expleft-fracm^22Nright$$
while here we have $|L_1||L_2|$ pairs to consider so $m=|L_1||L_2|.$
Wagner's paper is about finding efficient algorithms for vectors adding to zero for higher numbers (e.g., 4) of lists.
$endgroup$
Note that $x_1=x_2$, i.e., there is a birthday collision in $0,1^n$ if and only if $x_1oplus x_2=0.$
In a general additive group $G$, $x_1=x_2$, i.e., there is a birthday collision in $G$ if and only if $x_1-x_2=0.$
If you have two lists $L_1,L_2,$ then with probability roughly
$$expleft-frac^22^n+1right$$
there will be no collisions.
In the birthday paradox, for $N=2^n$ bins, the probability of no collisions after $m$ balls is roughly
$$expleft-fracm^22Nright$$
while here we have $|L_1||L_2|$ pairs to consider so $m=|L_1||L_2|.$
Wagner's paper is about finding efficient algorithms for vectors adding to zero for higher numbers (e.g., 4) of lists.
edited Mar 22 at 6:19
answered Mar 22 at 4:14
kodlukodlu
9,24311331
9,24311331
add a comment |
add a comment |
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