Most efficient set of characters for cracking passwords (Parallelized)?
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For an assignment, we are required to crack a password from a hash given a salt.
The password will always be 4 characters that are case sensitive (ex: CMPS, cmps, CAMP, LIST).
We are to parallelize this process. What is the most efficient way to parallelize this process if we are to split it into 24 processes?
In other words, we are to split up the search for the password into 24 sub-searches.
EX:
Process 1 = [AAAA - BBBB]
Process 2 = [CCCC - DDDD]
Process n = [8888 - 9999]
and so on. This is likely not to work or be efficient. What would be a better way to split the process?
What I was thinking:
Make each set start with the most used letter in the English language, such that:
Process 1 = [EAAA - EZZZ]
Process 2 = [TAAA - TZZZ]
Thoughts?
probability sequences-and-series algorithms multisets
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
$endgroup$
add a comment |
$begingroup$
For an assignment, we are required to crack a password from a hash given a salt.
The password will always be 4 characters that are case sensitive (ex: CMPS, cmps, CAMP, LIST).
We are to parallelize this process. What is the most efficient way to parallelize this process if we are to split it into 24 processes?
In other words, we are to split up the search for the password into 24 sub-searches.
EX:
Process 1 = [AAAA - BBBB]
Process 2 = [CCCC - DDDD]
Process n = [8888 - 9999]
and so on. This is likely not to work or be efficient. What would be a better way to split the process?
What I was thinking:
Make each set start with the most used letter in the English language, such that:
Process 1 = [EAAA - EZZZ]
Process 2 = [TAAA - TZZZ]
Thoughts?
probability sequences-and-series algorithms multisets
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
$endgroup$
$begingroup$
This depends on how you define efficiency. In a cracking situation, it's likely that you are going to think about the worst-case scenario, in which case any equal-sized 24-way split without overlaps is efficient (presuming all hashes take constant time).
$endgroup$
– obscurans
Dec 3 '18 at 1:36
$begingroup$
@obscurans By efficiency I simply mean fast. It does not need to be the fastest, it simply needs to work relatively fast. No specific time complexity.
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:39
add a comment |
$begingroup$
For an assignment, we are required to crack a password from a hash given a salt.
The password will always be 4 characters that are case sensitive (ex: CMPS, cmps, CAMP, LIST).
We are to parallelize this process. What is the most efficient way to parallelize this process if we are to split it into 24 processes?
In other words, we are to split up the search for the password into 24 sub-searches.
EX:
Process 1 = [AAAA - BBBB]
Process 2 = [CCCC - DDDD]
Process n = [8888 - 9999]
and so on. This is likely not to work or be efficient. What would be a better way to split the process?
What I was thinking:
Make each set start with the most used letter in the English language, such that:
Process 1 = [EAAA - EZZZ]
Process 2 = [TAAA - TZZZ]
Thoughts?
probability sequences-and-series algorithms multisets
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
$endgroup$
For an assignment, we are required to crack a password from a hash given a salt.
The password will always be 4 characters that are case sensitive (ex: CMPS, cmps, CAMP, LIST).
We are to parallelize this process. What is the most efficient way to parallelize this process if we are to split it into 24 processes?
In other words, we are to split up the search for the password into 24 sub-searches.
EX:
Process 1 = [AAAA - BBBB]
Process 2 = [CCCC - DDDD]
Process n = [8888 - 9999]
and so on. This is likely not to work or be efficient. What would be a better way to split the process?
What I was thinking:
Make each set start with the most used letter in the English language, such that:
Process 1 = [EAAA - EZZZ]
Process 2 = [TAAA - TZZZ]
Thoughts?
probability sequences-and-series algorithms multisets
probability sequences-and-series algorithms multisets
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
asked Dec 3 '18 at 1:33
Gabriel GarciaGabriel Garcia
31
31
$begingroup$
This depends on how you define efficiency. In a cracking situation, it's likely that you are going to think about the worst-case scenario, in which case any equal-sized 24-way split without overlaps is efficient (presuming all hashes take constant time).
$endgroup$
– obscurans
Dec 3 '18 at 1:36
$begingroup$
@obscurans By efficiency I simply mean fast. It does not need to be the fastest, it simply needs to work relatively fast. No specific time complexity.
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:39
add a comment |
$begingroup$
This depends on how you define efficiency. In a cracking situation, it's likely that you are going to think about the worst-case scenario, in which case any equal-sized 24-way split without overlaps is efficient (presuming all hashes take constant time).
$endgroup$
– obscurans
Dec 3 '18 at 1:36
$begingroup$
@obscurans By efficiency I simply mean fast. It does not need to be the fastest, it simply needs to work relatively fast. No specific time complexity.
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:39
$begingroup$
This depends on how you define efficiency. In a cracking situation, it's likely that you are going to think about the worst-case scenario, in which case any equal-sized 24-way split without overlaps is efficient (presuming all hashes take constant time).
$endgroup$
– obscurans
Dec 3 '18 at 1:36
$begingroup$
This depends on how you define efficiency. In a cracking situation, it's likely that you are going to think about the worst-case scenario, in which case any equal-sized 24-way split without overlaps is efficient (presuming all hashes take constant time).
$endgroup$
– obscurans
Dec 3 '18 at 1:36
$begingroup$
@obscurans By efficiency I simply mean fast. It does not need to be the fastest, it simply needs to work relatively fast. No specific time complexity.
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:39
$begingroup$
@obscurans By efficiency I simply mean fast. It does not need to be the fastest, it simply needs to work relatively fast. No specific time complexity.
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:39
add a comment |
1 Answer
1
active
oldest
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$begingroup$
If you don't presume any particular definition of "efficiency" other than generally being fast, then we can just take the worst-case scenario where you only succeed when hashing the very last possible password.
To optimize for that, you simply want the subsearches to cover the entire space of possible passwords, and divide them equally. Any scheme that splits the space into 24 mostly-equal sized pieces without overlap is efficient (presuming hashes are constant-time).
This gives you a 24x speedup relative to single-process brute force (worst-case vs worst-case).
One particular implementation: number all possible passwords. Each process knows its ID from 0-23. Process #i hashes password i, i+24, i+48, ... until done.
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
$endgroup$
$begingroup$
Thank you for your answer @obscurans ! What do the increments mean? (i+24, i+28) Are you trying to say that process 1 would start at letter 1(a) and go through all letters (a-Z and 0-9)? Therefore, process 2 would start at letter 2(b) and would go all the way until loopback to a? such that process 2 = [BBBB] - [AAAA]?
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:53
$begingroup$
Make something that numbers the passwords. Say AAAA is 0, AAAB is 1, ..., perhaps AABA is 26 (make sure it's a unique number for each password, and all numbers in range map to some password). (i+24) is the numbering for some password, which process #i will next try and hash.
$endgroup$
– obscurans
Dec 3 '18 at 1:56
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you don't presume any particular definition of "efficiency" other than generally being fast, then we can just take the worst-case scenario where you only succeed when hashing the very last possible password.
To optimize for that, you simply want the subsearches to cover the entire space of possible passwords, and divide them equally. Any scheme that splits the space into 24 mostly-equal sized pieces without overlap is efficient (presuming hashes are constant-time).
This gives you a 24x speedup relative to single-process brute force (worst-case vs worst-case).
One particular implementation: number all possible passwords. Each process knows its ID from 0-23. Process #i hashes password i, i+24, i+48, ... until done.
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
$endgroup$
$begingroup$
Thank you for your answer @obscurans ! What do the increments mean? (i+24, i+28) Are you trying to say that process 1 would start at letter 1(a) and go through all letters (a-Z and 0-9)? Therefore, process 2 would start at letter 2(b) and would go all the way until loopback to a? such that process 2 = [BBBB] - [AAAA]?
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:53
$begingroup$
Make something that numbers the passwords. Say AAAA is 0, AAAB is 1, ..., perhaps AABA is 26 (make sure it's a unique number for each password, and all numbers in range map to some password). (i+24) is the numbering for some password, which process #i will next try and hash.
$endgroup$
– obscurans
Dec 3 '18 at 1:56
add a comment |
$begingroup$
If you don't presume any particular definition of "efficiency" other than generally being fast, then we can just take the worst-case scenario where you only succeed when hashing the very last possible password.
To optimize for that, you simply want the subsearches to cover the entire space of possible passwords, and divide them equally. Any scheme that splits the space into 24 mostly-equal sized pieces without overlap is efficient (presuming hashes are constant-time).
This gives you a 24x speedup relative to single-process brute force (worst-case vs worst-case).
One particular implementation: number all possible passwords. Each process knows its ID from 0-23. Process #i hashes password i, i+24, i+48, ... until done.
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
$endgroup$
$begingroup$
Thank you for your answer @obscurans ! What do the increments mean? (i+24, i+28) Are you trying to say that process 1 would start at letter 1(a) and go through all letters (a-Z and 0-9)? Therefore, process 2 would start at letter 2(b) and would go all the way until loopback to a? such that process 2 = [BBBB] - [AAAA]?
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:53
$begingroup$
Make something that numbers the passwords. Say AAAA is 0, AAAB is 1, ..., perhaps AABA is 26 (make sure it's a unique number for each password, and all numbers in range map to some password). (i+24) is the numbering for some password, which process #i will next try and hash.
$endgroup$
– obscurans
Dec 3 '18 at 1:56
add a comment |
$begingroup$
If you don't presume any particular definition of "efficiency" other than generally being fast, then we can just take the worst-case scenario where you only succeed when hashing the very last possible password.
To optimize for that, you simply want the subsearches to cover the entire space of possible passwords, and divide them equally. Any scheme that splits the space into 24 mostly-equal sized pieces without overlap is efficient (presuming hashes are constant-time).
This gives you a 24x speedup relative to single-process brute force (worst-case vs worst-case).
One particular implementation: number all possible passwords. Each process knows its ID from 0-23. Process #i hashes password i, i+24, i+48, ... until done.
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
$endgroup$
If you don't presume any particular definition of "efficiency" other than generally being fast, then we can just take the worst-case scenario where you only succeed when hashing the very last possible password.
To optimize for that, you simply want the subsearches to cover the entire space of possible passwords, and divide them equally. Any scheme that splits the space into 24 mostly-equal sized pieces without overlap is efficient (presuming hashes are constant-time).
This gives you a 24x speedup relative to single-process brute force (worst-case vs worst-case).
One particular implementation: number all possible passwords. Each process knows its ID from 0-23. Process #i hashes password i, i+24, i+48, ... until done.
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
answered Dec 3 '18 at 1:44
obscuransobscurans
908311
908311
$begingroup$
Thank you for your answer @obscurans ! What do the increments mean? (i+24, i+28) Are you trying to say that process 1 would start at letter 1(a) and go through all letters (a-Z and 0-9)? Therefore, process 2 would start at letter 2(b) and would go all the way until loopback to a? such that process 2 = [BBBB] - [AAAA]?
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:53
$begingroup$
Make something that numbers the passwords. Say AAAA is 0, AAAB is 1, ..., perhaps AABA is 26 (make sure it's a unique number for each password, and all numbers in range map to some password). (i+24) is the numbering for some password, which process #i will next try and hash.
$endgroup$
– obscurans
Dec 3 '18 at 1:56
add a comment |
$begingroup$
Thank you for your answer @obscurans ! What do the increments mean? (i+24, i+28) Are you trying to say that process 1 would start at letter 1(a) and go through all letters (a-Z and 0-9)? Therefore, process 2 would start at letter 2(b) and would go all the way until loopback to a? such that process 2 = [BBBB] - [AAAA]?
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:53
$begingroup$
Make something that numbers the passwords. Say AAAA is 0, AAAB is 1, ..., perhaps AABA is 26 (make sure it's a unique number for each password, and all numbers in range map to some password). (i+24) is the numbering for some password, which process #i will next try and hash.
$endgroup$
– obscurans
Dec 3 '18 at 1:56
$begingroup$
Thank you for your answer @obscurans ! What do the increments mean? (i+24, i+28) Are you trying to say that process 1 would start at letter 1(a) and go through all letters (a-Z and 0-9)? Therefore, process 2 would start at letter 2(b) and would go all the way until loopback to a? such that process 2 = [BBBB] - [AAAA]?
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:53
$begingroup$
Thank you for your answer @obscurans ! What do the increments mean? (i+24, i+28) Are you trying to say that process 1 would start at letter 1(a) and go through all letters (a-Z and 0-9)? Therefore, process 2 would start at letter 2(b) and would go all the way until loopback to a? such that process 2 = [BBBB] - [AAAA]?
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:53
$begingroup$
Make something that numbers the passwords. Say AAAA is 0, AAAB is 1, ..., perhaps AABA is 26 (make sure it's a unique number for each password, and all numbers in range map to some password). (i+24) is the numbering for some password, which process #i will next try and hash.
$endgroup$
– obscurans
Dec 3 '18 at 1:56
$begingroup$
Make something that numbers the passwords. Say AAAA is 0, AAAB is 1, ..., perhaps AABA is 26 (make sure it's a unique number for each password, and all numbers in range map to some password). (i+24) is the numbering for some password, which process #i will next try and hash.
$endgroup$
– obscurans
Dec 3 '18 at 1:56
add a comment |
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$begingroup$
This depends on how you define efficiency. In a cracking situation, it's likely that you are going to think about the worst-case scenario, in which case any equal-sized 24-way split without overlaps is efficient (presuming all hashes take constant time).
$endgroup$
– obscurans
Dec 3 '18 at 1:36
$begingroup$
@obscurans By efficiency I simply mean fast. It does not need to be the fastest, it simply needs to work relatively fast. No specific time complexity.
$endgroup$
– Gabriel Garcia
Dec 3 '18 at 1:39