Great video! Btw, in the beginning you said that whatever bit length your input has, the padded message length would be the next 512 multiple. That is actually not fully true. That would only be true if the length of the input is smaller or equal to n*512 - 65 (if equal then k = 0). so if e.g. the input length is 960, the padded message will not be 1024 bits long, bc we would then have negative solution for k. But we cannot have a negative solution for k, so that’s why we would need to go with 1536.
Thank you so much for this video, for going through the trouble of understanding how it works and breaking it down into something meaningful! Having the example is SUPER helpful. Exactly what I was looking for.
Oh my dear god, thanks for going through this insanity :D You're making it so much better than having to go through the whole paper itself. Same goes for Chaum's video! And I do enjoy the music and the humor! :)
Thanks for making this video. It was nice to see all the steps actually laid out and what a crazy series of mathematical knots the algorithm ties the input data up in. (Also, for future videos, it would be worth normalizing audio volume up, mastering up, or running some different compression).
Thanks very much for the video!! I had to pause several times but I now understand better the algorithm. I also like that you show screenshots of the standard, it is a gentle introduction. I have some suggestions: - Use namespaced fonts. This avoid the misalignment of 10:34 - Do you make the animations using Manim? In this case would be great that you share the source code :)
Great explanation. The easiest to understand from what Ive seen on TH-cam so far. But I dont quite get how the looping works... Are all the iterations added to eachother at the end?
Why wouldn't you say modulo to 32nd? How would modulo prevent it from being greater than 32 bits? That is the only reason for the modulo portion, correct?
what if some one uses the final hash and applies these all same operation in a reverse manner and get to the initial input?? (i know doing these operation in reverse will require hell lot of brains but this might be possible ?)
Thanks for the explainations. However the music is very distracting. If I want to listen to music while learning I would just have a video in the background.
Question: at 7.23, how can you perform an xor on three inputs? There are only two bits available. I tried doing an xor function on the first two 32 byte numbers and then that output and the final input, and it still did not yield a correct answer??
7:24 when making addition modulo 2 for sigma0: Is the result shown correct? ("00000001 10001111 11101001 00000101") In my calculation it should be ("00000011 10001111 11101001 10110000")
Yes, as a hashing function is many-to-one relationship, and involves a finite number of outputs given an infinite domain - if you are interested in this more, you might want to read up on hashing functions, hash maps and most importantly, how a function deals with "collisions" - which are what you outlined, hashes that are the same but given different outputs. :)
If you do the bitwise addition of Sigma0 shown at 07:24 then the result I found is 00000011100011111110100110110000. Would you please explain if there is sth wrong ?
Yep, you're correct. I believe I used the correct sum going forward though so the error doesn't carry forward through the rest of the steps. Good spot.
Do you mean: Given a 256 bit output, find the input? No this is not possible, SHA256 (as with all cryptographic hash functions) is a 'one-way' function' meaning its impossible to invert (go google those terms and look at wiki if you're unfamiliar). To convince yourself this is true just look at the inner workings of the algorithm: it's composed of several 'mini' one-way functions such as bit-wise addition mod 2 (i.e. we pass the bits through an XOR gate) if the output is 1 then the input is either 01 or 10.. we can't even invert this function! Same goes for 'shift' function in the video, SHIFT3(11011100) = 00011011... those first 3 bits of the input could be anything and still give same output (b/c they're set to zero regardless) so we cannot invert this mini function either
@@redblockblue thank you for the response. I had a challenge to find a string that returns something with the substring b00da when passed through the Sha256. I'm no criptographer, I had to brute force it
3:59 we still doing stuff allways same. we can reverse. we not even need calculate those if it allwats start 2 then 3 then 5. we allways have those "hashes"
What will happen if the text we are trying to hash is 504 bits long? You don't have 64 bits to represent the length of the text in binary to be able to add to the end of the binary for it to be 512 bits in length. Your hash will instead be 568 bits long.
Yep, although that may sound dodgy it’s not an issue, the algorithm is completely described in detail and any ‘random’ numbers are generated in a natural way (fractional part of sqrt(2), etc.) and not just ‘hey, use *this* ‘random’ number. They have planted back doors into publicly used algorithms before though, watch the vid in description from computerphile
When it says (t-15) or (t-7) t is the loop from 16-63. t is going through the loop, when solving w16, t is 16 so w16 -15 = w1, if its w38 for example then its w38-15 = w23 if u get it?
10:18 well using prime number those are allways same and function ch result same thing or is efg change every step. hope fully. make no sende otherwise LOL
If the n is number of blocks youre talking about then you have more than one block, follow the same steps just in the final when youre adding h0 = h0 +a Then you take h0 from the previous block and use the new a from the next block
I didn't skip anything, I gave example of going through entire loop once and then you repeat.. why would I make the video longer by going through same loop again?
5:31 real messy LOL bcoz you explain it wrong. could just say ROR 7 times and make animation it roll 7 and not high light this and this if all they shift LOL now 18 ROR is easy. then 3 is easy but zero 3 after thats like without carry AND this is first time when math backward become cracking part we not know what those 3 bits was when they erased LOL so we need do 10 diff all plausable combo but from this point not matter LOL. what comes next i not yet reverse think lol
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Great video!
Btw, in the beginning you said that whatever bit length your input has, the padded message length would be the next 512 multiple. That is actually not fully true. That would only be true if the length of the input is smaller or equal to n*512 - 65 (if equal then k = 0). so if e.g. the input length is 960, the padded message will not be 1024 bits long, bc we would then have negative solution for k. But we cannot have a negative solution for k, so that’s why we would need to go with 1536.
Ah yes ofc, I'll pin your comment so others don't miss that. Thanks!
@@redblockblue
Alright, great. But really bro..your video was fantastic, good job brother! Keep it up!👍🔥
Btw, do you have Twitter or tg?
0:36
▶️⏯️🎦🆚◀️
@@redblockblue3:25
I searched this video thinking maybe I can reverse engineer this algorithm but turns out I have to give up on it😂😂
Love this type of video that goes into the meat of the algorithms instead of speaking abstractly. Thank you for this and well done!
Yep, that’s the style I’m going for with this channel, there are far too many channels using only analogies and never going deeper than that
Thank you so much for this video, for going through the trouble of understanding how it works and breaking it down into something meaningful! Having the example is SUPER helpful. Exactly what I was looking for.
Oh my dear god, thanks for going through this insanity :D You're making it so much better than having to go through the whole paper itself. Same goes for Chaum's video! And I do enjoy the music and the humor! :)
Only Video of Real SHA 256 algorithm implementation. Thank you for this.❤
this is the best explanation you will find on youtube.
Great video, you're really underrated for the amount of effort you put into this.
Amazing video. I was able to fully code the algorithm with the help of this explanation. Thank you!
Best one untill NOW..... Hats off
you teach in a better way i have seen anyone else, thank you so much when is the video on the ethereum white paper coming out?
Great video for sure! Very nice, simple explanation.
Good job!
This guy deserves more subscribers!!
Amazing video, exactly what I was looking for.
insane explanations thanks man!
Wow, Thank you so much. Truly no bs.
Thanks for making this video. It was nice to see all the steps actually laid out and what a crazy series of mathematical knots the algorithm ties the input data up in. (Also, for future videos, it would be worth normalizing audio volume up, mastering up, or running some different compression).
Thanks very much for the video!! I had to pause several times but I now understand better the algorithm.
I also like that you show screenshots of the standard, it is a gentle introduction.
I have some suggestions:
- Use namespaced fonts. This avoid the misalignment of 10:34
- Do you make the animations using Manim? In this case would be great that you share the source code :)
Best explanation ever !!
very interesting, cant wait to implement it 😋
Great video!!!
Great explanation. The easiest to understand from what Ive seen on TH-cam so far. But I dont quite get how the looping works... Are all the iterations added to eachother at the end?
Love it, thank you for video!
nice explanation i love it
OMG
this was the best
really nice video. Thank you for creating it
Great video. Thank you for the content. Can you give the link to the paper you are referencing in the video?
Great Video!
amazing video
the best explanation
This out of this world
First 2 min of the video is already better than me reading for 30 min on wiki
Nice vidéo !
hey why you stopped making videos ?
hmm I don't get it, why creating initial hash values at 3:28 introduces randomness? if we know, which prime numbers we are taking?
brain usage 100%, alert death imminent😂😂
Amazing video.
Your pedagogy skills are off the charts.
... but the music man ..............
Would you provide a resource to the pdf you keep referencing in the video?
Why wouldn't you say modulo to 32nd? How would modulo prevent it from being greater than 32 bits? That is the only reason for the modulo portion, correct?
What is the source of the document you show screenshots of?
what if some one uses the final hash and applies these all same operation in a reverse manner and get to the initial input?? (i know doing these operation in reverse will require hell lot of brains but this might be possible ?)
What book or site is that in the beginning?
8:05 are those calculated all original values or with answer we got so order is important LOL. i bet original numbers
Knowing all the steps why can't people work backwords to get the original string?
Where is the document that is referenced in this video?
Can you tell me the name of the book from which you have been explaining in the video?
Did you find it? I'm also looking for it
8:50 - why do we only go through that for loop once? What does t=0 to 63 repreent what is the 't' in this case?
when you're a not a programmer and tries to understand sha256
Thanks for the explainations. However the music is very distracting. If I want to listen to music while learning I would just have a video in the background.
bạn có tài liệu như ở trong video không, tôi hiện đang rất cần nó, cảm ơn
Can u do one for ripemd160?
man the audio is very low, but amazing video
can u do the sha3 please?
Question: at 7.23, how can you perform an xor on three inputs? There are only two bits available. I tried doing an xor function on the first two 32 byte numbers and then that output and the final input, and it still did not yield a correct answer??
I found you could do xor in python using the ^ symbol, but it produced the answer: 00000011100011111110100110110000, which is incorrect
@@EwarRobI think the content in the video is wrong, I get the same result as you in a different programming language!!!
7:24 when making addition modulo 2 for sigma0:
Is the result shown correct? ("00000001 10001111 11101001 00000101")
In my calculation it should be ("00000011 10001111 11101001 10110000")
I agree, I also got 00000011100011111110100110110000
One thing I dont get is, since every oitpit is the same lenght, can there be 2 strings with the same output?
Yes, as a hashing function is many-to-one relationship, and involves a finite number of outputs given an infinite domain - if you are interested in this more, you might want to read up on hashing functions, hash maps and most importantly, how a function deals with "collisions" - which are what you outlined, hashes that are the same but given different outputs. :)
yeah but there are so many gajillions of outputs, good luck finding two strings that give the same output
If you do the bitwise addition of Sigma0 shown at 07:24 then the result I found is 00000011100011111110100110110000. Would you please explain if there is sth wrong ?
Yep, you're correct. I believe I used the correct sum going forward though so the error doesn't carry forward through the rest of the steps. Good spot.
I was having the same issue, appreciate this comment!
Encountered the same thing. Glad I'm not crazy
I have a question, Can you explain this to me? I have the character "耀". What are the steps to encode this character? It's not in the ascii encoding.
yes it is
if it wasnt then we couldnt see it
Can you do the reverse example? getting a string to get a determined value from the function?
Do you mean:
Given a 256 bit output, find the input?
No this is not possible, SHA256 (as with all cryptographic hash functions) is a 'one-way' function' meaning its impossible to invert (go google those terms and look at wiki if you're unfamiliar).
To convince yourself this is true just look at the inner workings of the algorithm: it's composed of several 'mini' one-way functions such as bit-wise addition mod 2 (i.e. we pass the bits through an XOR gate) if the output is 1 then the input is either 01 or 10.. we can't even invert this function! Same goes for 'shift' function in the video, SHIFT3(11011100) = 00011011... those first 3 bits of the input could be anything and still give same output (b/c they're set to zero regardless) so we cannot invert this mini function either
@@redblockblue thank you for the response. I had a challenge to find a string that returns something with the substring b00da when passed through the Sha256. I'm no criptographer, I had to brute force it
10:50 how is he calculating this binary number. I still don’t quite understand
3:59 we still doing stuff allways same. we can reverse. we not even need calculate those if it allwats start 2 then 3 then 5. we allways have those "hashes"
What will happen if the text we are trying to hash is 504 bits long? You don't have 64 bits to represent the length of the text in binary to be able to add to the end of the binary for it to be 512 bits in length. Your hash will instead be 568 bits long.
you will then need to extend it it 1024 I guess
@@kyrylopisariev9791yes exactly
Have you tried convert the text into shingles/ tokens? In that way, you avoid having text longer than 504 bits
@@diogocardoso265 How do you do that?
SHA-256 algorithm designed by the US National Security Agency, is this correct??
Yep, although that may sound dodgy it’s not an issue, the algorithm is completely described in detail and any ‘random’ numbers are generated in a natural way (fractional part of sqrt(2), etc.) and not just ‘hey, use *this* ‘random’ number. They have planted back doors into publicly used algorithms before though, watch the vid in description from computerphile
3:25 ok i found myself solving Rubic's cube and thinking about what I can implement on my website 😅
If I am calculating W(16), on which bit I will have to apply the sigma0 and sigma1?
Can anyone tell me?
When it says (t-15) or (t-7) t is the loop from 16-63. t is going through the loop, when solving w16, t is 16 so w16 -15 = w1, if its w38 for example then its w38-15 = w23 if u get it?
holy hell
10:18 well using prime number those are allways same and function ch result same thing or is efg change every step. hope fully. make no sende otherwise LOL
3:46, why are the online decimal to binary tools inaccurate? They give me 0.00000010110100011001
What do you mean? Perhaps you calculated incorrectly.
Correct algorithm is: (**Psuedo code)
n = 2
nSqrt = sqrt(n) = 1.4142135623730950488016887242097 ..
nDecimals = 0.4142135623730950488016887242097..
nDecimals32Bits = nDecimals * (2^32)
hexValue = hex(nDecimals32Bits) = "0x6a09e667"
binaryValue = 01101010000010011110011001100111
Wonderful video my dear friend
I urgently need the presentation file you used please
..
nvlpubs.nist.gov/nistpubs/FIPS/NIST.FIPS.180-4.pdf
Can i use this for completing my school project
You can use it for whatever you’d like 👍
@@redblockblue thanks mate
Impossible, even for quantum computers to crack SHA-256??
Encryption built by the government in the NSA but Bitcoin totally wasn't sure
This video is good, but the music is so distracting.I had to stop after a bit
only understood first 3 mins
i now understand why its impossible to decode the output of hash256
Only problem this video has is the low volume even with headphone
I guess you need to change your headphones
you are probably going deaf
The frequencies carrying this recording are pretty low, could have used more mid/high, you should in any case try a better pair of headphones!
what would we do if our N was more than 1
If the n is number of blocks youre talking about then you have more than one block, follow the same steps just in the final when youre adding h0 = h0 +a
Then you take h0 from the previous block and use the new a from the next block
thanks@@hyxl3r254
8:33 prime number that allways same
Can you link me to the paper you read from?
nvlpubs.nist.gov/nistpubs/FIPS/NIST.FIPS.180-4.pdf
@@redblockblue thank you
"that is it".
Ya skipped so many parts and still it's tedious. 😅
I didn't skip anything, I gave example of going through entire loop once and then you repeat.. why would I make the video longer by going through same loop again?
kya hai ye kyo hai ye
5:31 real messy LOL bcoz you explain it wrong. could just say ROR 7 times and make animation it roll 7 and not high light this and this if all they shift LOL now 18 ROR is easy. then 3 is easy but zero 3 after thats like without carry AND this is first time when math backward become cracking part we not know what those 3 bits was when they erased LOL so we need do 10 diff all plausable combo but from this point not matter LOL. what comes next i not yet reverse think lol
the video is great but the audio is awful
Underaged thanks if I solve this some how lol u get half (‘if’)
I meant underrated
Thanks Bon, good luck solving whatever it is you’re trying to solve 👍
WTF..
nice video. but that is some math bro..
TL;DR - its a lot of complicated math.
Great video!