MD5 Round Operations RFC1321
According to this documentation of md5 function the Step 4 involves rounds of functions where
Let [abcd k s i] denote the operation
a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s).
Do the following 16 operations.
I am unsure what the "+" symbolises in this operation, be it addition or bitwise AND. Earlier in the paper it states
Let the symbol "+" denote addition of words (i.e., modulo-2^32
addition). Let X <<< s denote the 32-bit value obtained by circularly
shifting (rotating) X left by s bit positions.
I know that [abcd] are 32 bit 'Words' and T[i] is a float (ive converted into 32 bit word and X[k] is a single bit. By doing this function with bitwise AND the result at the end of the function is always a 32 bitstring of FALSE which leads me to think somewhere is going wrong.
This is the python code responsible:
# parse to bitarrays
xb = bitarray(str(x))
tb = bitarray(str(t))
value = b & ((a & F(b,c,d) & xb & tb)) # value always turns false
value = leftshift(value, s)
return value
And a screenshot of variables watch that may be useful.
Thanks in advance SO community
python cryptography md5
asked Jan 2 at 14:49
euan joneseuan jones
336
add a comment |
According to this documentation of md5 function the Step 4 involves rounds of functions where
Let [abcd k s i] denote the operation
a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s).
Do the following 16 operations.
I am unsure what the "+" symbolises in this operation, be it addition or bitwise AND. Earlier in the paper it states
Let the symbol "+" denote addition of words (i.e., modulo-2^32
addition). Let X <<< s denote the 32-bit value obtained by circularly
shifting (rotating) X left by s bit positions.
I know that [abcd] are 32 bit 'Words' and T[i] is a float (ive converted into 32 bit word and X[k] is a single bit. By doing this function with bitwise AND the result at the end of the function is always a 32 bitstring of FALSE which leads me to think somewhere is going wrong.
This is the python code responsible:
# parse to bitarrays
xb = bitarray(str(x))
tb = bitarray(str(t))
value = b & ((a & F(b,c,d) & xb & tb)) # value always turns false
value = leftshift(value, s)
return value
And a screenshot of variables watch that may be useful.
Thanks in advance SO community
python cryptography md5
asked Jan 2 at 14:49
euan joneseuan jones
336
2
Since the document itself defines+as modulo-2^32 addition I think it is safe to assume that+does not mean bitwise AND, but rather that you add the leftmost 31 bits of the two words together and discard any resulting overflow.
– BoarGules
Jan 2 at 15:07
@BoarGules: leftmost 32 bits.
– James K Polk
Jan 2 at 22:09
I don't understand where the confusion lies. "+" means addition, not bitwise "and". X[k] is not a single bit, it is a 32-bit value. You really need to re-read the RFC.
– James K Polk
Jan 2 at 22:12
@JamesKPolk Quite correct. A careless error on my part.
– BoarGules
Jan 3 at 8:30
add a comment |
According to this documentation of md5 function the Step 4 involves rounds of functions where
Let [abcd k s i] denote the operation
a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s).
Do the following 16 operations.
I am unsure what the "+" symbolises in this operation, be it addition or bitwise AND. Earlier in the paper it states
Let the symbol "+" denote addition of words (i.e., modulo-2^32
addition). Let X <<< s denote the 32-bit value obtained by circularly
shifting (rotating) X left by s bit positions.
I know that [abcd] are 32 bit 'Words' and T[i] is a float (ive converted into 32 bit word and X[k] is a single bit. By doing this function with bitwise AND the result at the end of the function is always a 32 bitstring of FALSE which leads me to think somewhere is going wrong.
This is the python code responsible:
# parse to bitarrays
xb = bitarray(str(x))
tb = bitarray(str(t))
value = b & ((a & F(b,c,d) & xb & tb)) # value always turns false
value = leftshift(value, s)
return value
And a screenshot of variables watch that may be useful.
Thanks in advance SO community
python cryptography md5
asked Jan 2 at 14:49
euan joneseuan jones
336
According to this documentation of md5 function the Step 4 involves rounds of functions where
Let [abcd k s i] denote the operation
a = b + ((a + F(b,c,d) + X[k] + T[i]) <<< s).
Do the following 16 operations.
I am unsure what the "+" symbolises in this operation, be it addition or bitwise AND. Earlier in the paper it states
Let the symbol "+" denote addition of words (i.e., modulo-2^32
addition). Let X <<< s denote the 32-bit value obtained by circularly
shifting (rotating) X left by s bit positions.
I know that [abcd] are 32 bit 'Words' and T[i] is a float (ive converted into 32 bit word and X[k] is a single bit. By doing this function with bitwise AND the result at the end of the function is always a 32 bitstring of FALSE which leads me to think somewhere is going wrong.
This is the python code responsible:
# parse to bitarrays
xb = bitarray(str(x))
tb = bitarray(str(t))
value = b & ((a & F(b,c,d) & xb & tb)) # value always turns false
value = leftshift(value, s)
return value
And a screenshot of variables watch that may be useful.
Thanks in advance SO community
python cryptography md5
python cryptography md5
asked Jan 2 at 14:49
euan joneseuan jones
336
asked Jan 2 at 14:49
euan joneseuan jones
336
asked Jan 2 at 14:49
euan joneseuan jones
336
asked Jan 2 at 14:49
euan joneseuan jones
336
asked Jan 2 at 14:49
euan joneseuan jones
336
336
2
Since the document itself defines+as modulo-2^32 addition I think it is safe to assume that+does not mean bitwise AND, but rather that you add the leftmost 31 bits of the two words together and discard any resulting overflow.
– BoarGules
Jan 2 at 15:07
@BoarGules: leftmost 32 bits.
– James K Polk
Jan 2 at 22:09
I don't understand where the confusion lies. "+" means addition, not bitwise "and". X[k] is not a single bit, it is a 32-bit value. You really need to re-read the RFC.
– James K Polk
Jan 2 at 22:12
@JamesKPolk Quite correct. A careless error on my part.
– BoarGules
Jan 3 at 8:30
add a comment |
2
Since the document itself defines+as modulo-2^32 addition I think it is safe to assume that+does not mean bitwise AND, but rather that you add the leftmost 31 bits of the two words together and discard any resulting overflow.
– BoarGules
Jan 2 at 15:07
@BoarGules: leftmost 32 bits.
– James K Polk
Jan 2 at 22:09
I don't understand where the confusion lies. "+" means addition, not bitwise "and". X[k] is not a single bit, it is a 32-bit value. You really need to re-read the RFC.
– James K Polk
Jan 2 at 22:12
@JamesKPolk Quite correct. A careless error on my part.
– BoarGules
Jan 3 at 8:30
2
2
Since the document itself defines
+ as modulo-2^32 addition I think it is safe to assume that + does not mean bitwise AND, but rather that you add the leftmost 31 bits of the two words together and discard any resulting overflow.– BoarGules
Jan 2 at 15:07
Since the document itself defines
+ as modulo-2^32 addition I think it is safe to assume that + does not mean bitwise AND, but rather that you add the leftmost 31 bits of the two words together and discard any resulting overflow.– BoarGules
Jan 2 at 15:07
@BoarGules: leftmost 32 bits.
– James K Polk
Jan 2 at 22:09
@BoarGules: leftmost 32 bits.
– James K Polk
Jan 2 at 22:09
I don't understand where the confusion lies. "+" means addition, not bitwise "and". X[k] is not a single bit, it is a 32-bit value. You really need to re-read the RFC.
– James K Polk
Jan 2 at 22:12
I don't understand where the confusion lies. "+" means addition, not bitwise "and". X[k] is not a single bit, it is a 32-bit value. You really need to re-read the RFC.
– James K Polk
Jan 2 at 22:12
@JamesKPolk Quite correct. A careless error on my part.
– BoarGules
Jan 3 at 8:30
@JamesKPolk Quite correct. A careless error on my part.
– BoarGules
Jan 3 at 8:30
add a comment |
1 Answer
1
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oldest
votes
When the specification says + means 32 bit addition implement + using 32 bit addition instead of bitwise and.
When in doubt you can see other MD5 implementations.
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
add a comment |
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1 Answer
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When the specification says + means 32 bit addition implement + using 32 bit addition instead of bitwise and.
When in doubt you can see other MD5 implementations.
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
add a comment |
When the specification says + means 32 bit addition implement + using 32 bit addition instead of bitwise and.
When in doubt you can see other MD5 implementations.
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
add a comment |
When the specification says + means 32 bit addition implement + using 32 bit addition instead of bitwise and.
When in doubt you can see other MD5 implementations.
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
When the specification says + means 32 bit addition implement + using 32 bit addition instead of bitwise and.
When in doubt you can see other MD5 implementations.
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
answered Jan 2 at 16:55
Thomas M. DuBuissonThomas M. DuBuisson
55.9k690153
55.9k690153
add a comment |
add a comment |
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2
Since the document itself defines
+as modulo-2^32 addition I think it is safe to assume that+does not mean bitwise AND, but rather that you add the leftmost 31 bits of the two words together and discard any resulting overflow.– BoarGules
Jan 2 at 15:07
@BoarGules: leftmost 32 bits.
– James K Polk
Jan 2 at 22:09
I don't understand where the confusion lies. "+" means addition, not bitwise "and". X[k] is not a single bit, it is a 32-bit value. You really need to re-read the RFC.
– James K Polk
Jan 2 at 22:12
@JamesKPolk Quite correct. A careless error on my part.
– BoarGules
Jan 3 at 8:30