What does the symbol $W_t$ mean in the SHA-256 specification?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I'm really confused about the computation about SHA 256.
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
The $W_t$ variable: What's the value of $W_t$? How do I get that value?
I'm still confused because of the explanation using the English language. I'm a beginner guys, any help from you will be a plus point for me about hashing.
hash sha-256
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
asked Mar 8 at 10:00
Onta SsOnta Ss
416
$endgroup$
add a comment |
$begingroup$
I'm really confused about the computation about SHA 256.
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
The $W_t$ variable: What's the value of $W_t$? How do I get that value?
I'm still confused because of the explanation using the English language. I'm a beginner guys, any help from you will be a plus point for me about hashing.
hash sha-256
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
asked Mar 8 at 10:00
Onta SsOnta Ss
416
$endgroup$
add a comment |
$begingroup$
I'm really confused about the computation about SHA 256.
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
The $W_t$ variable: What's the value of $W_t$? How do I get that value?
I'm still confused because of the explanation using the English language. I'm a beginner guys, any help from you will be a plus point for me about hashing.
hash sha-256
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
asked Mar 8 at 10:00
Onta SsOnta Ss
416
$endgroup$
I'm really confused about the computation about SHA 256.
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
The $W_t$ variable: What's the value of $W_t$? How do I get that value?
I'm still confused because of the explanation using the English language. I'm a beginner guys, any help from you will be a plus point for me about hashing.
hash sha-256
hash sha-256
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
asked Mar 8 at 10:00
Onta SsOnta Ss
416
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
asked Mar 8 at 10:00
Onta SsOnta Ss
416
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
edited Mar 9 at 15:57
Squeamish Ossifrage
22.1k132100
22.1k132100
asked Mar 8 at 10:00
Onta SsOnta Ss
416
asked Mar 8 at 10:00
Onta SsOnta Ss
416
asked Mar 8 at 10:00
Onta SsOnta Ss
416
416
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is in SHA-256 message schedule (NIST-FIPS 180-4);
The message $M$ with length $l$ is first padded as the usual way;
- append 1 to the end of the message,
- then, add $k$ zero bits such that $$l+1+k equiv 448 mod 512$$
- finally, add the length of the message in 64-bit. Now, the total padded length is divisible by 512.
After padding, the padded message parsed into $M^1,ldots,M^N$ where each has size 512-bit.
The sub index $t$ represents 32-bits in 512 bits. Thus, The $M_t^(i)$ is the $t$-th 32-bit in the $M^(i)$ for $0 leq t leq 15$
The $W_t$ is defined as your equation;
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
Where
$$sigma_1^256(x) = operatornameROTR^17(x) oplus operatornameROTR^19(x) oplus operatornameSHR^10(x)$$
$$sigma_0^256(x) = operatornameROTR^7(x) oplus operatornameROTR^18(x) oplus operatornameSHR^3(x)$$
$sigma_1^256(x)$ and $sigma_0^256(x)$ operate on 32 bits and produce 32 bits.
Note: the 256 above the $sigma$ represents the 256 in SHA-256. Similarly, there is $sigma_0^512(x)$ and $sigma_1^512(x)$ for SHA-512.
edited Mar 9 at 4:19
Richie Frame
10.1k11530
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
$endgroup$
$begingroup$
so you mean Wt = M16.....Mn ?
$endgroup$
– Onta Ss
Mar 8 at 10:44
1
$begingroup$
No. $W_1 = M_1^(i), W_2 = M_2^(i), ldots, W_15 = M_15^(i)$
$endgroup$
– kelalaka
Mar 8 at 11:00
$begingroup$
@OntaSs there is no $M_16$, the 16 words of the input block are numbered 0 to 15 only
$endgroup$
– Richie Frame
Mar 9 at 0:50
$begingroup$
@kelalaka Wt = M1 u mean (Wt-2) = (M1-2) ? sorry bro i learn self-taught about this. i got little confused if read mathematic formulas
$endgroup$
– Onta Ss
Mar 9 at 2:25
$begingroup$
@RichieFrame yeah i got it. How about W16....W63 ? what is the value ?
$endgroup$
– Onta Ss
Mar 9 at 3:27
|
show 3 more comments
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "281"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f67873%2fwhat-does-the-symbol-w-t-mean-in-the-sha-256-specification%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is in SHA-256 message schedule (NIST-FIPS 180-4);
The message $M$ with length $l$ is first padded as the usual way;
- append 1 to the end of the message,
- then, add $k$ zero bits such that $$l+1+k equiv 448 mod 512$$
- finally, add the length of the message in 64-bit. Now, the total padded length is divisible by 512.
After padding, the padded message parsed into $M^1,ldots,M^N$ where each has size 512-bit.
The sub index $t$ represents 32-bits in 512 bits. Thus, The $M_t^(i)$ is the $t$-th 32-bit in the $M^(i)$ for $0 leq t leq 15$
The $W_t$ is defined as your equation;
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
Where
$$sigma_1^256(x) = operatornameROTR^17(x) oplus operatornameROTR^19(x) oplus operatornameSHR^10(x)$$
$$sigma_0^256(x) = operatornameROTR^7(x) oplus operatornameROTR^18(x) oplus operatornameSHR^3(x)$$
$sigma_1^256(x)$ and $sigma_0^256(x)$ operate on 32 bits and produce 32 bits.
Note: the 256 above the $sigma$ represents the 256 in SHA-256. Similarly, there is $sigma_0^512(x)$ and $sigma_1^512(x)$ for SHA-512.
edited Mar 9 at 4:19
Richie Frame
10.1k11530
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
$endgroup$
$begingroup$
so you mean Wt = M16.....Mn ?
$endgroup$
– Onta Ss
Mar 8 at 10:44
1
$begingroup$
No. $W_1 = M_1^(i), W_2 = M_2^(i), ldots, W_15 = M_15^(i)$
$endgroup$
– kelalaka
Mar 8 at 11:00
$begingroup$
@OntaSs there is no $M_16$, the 16 words of the input block are numbered 0 to 15 only
$endgroup$
– Richie Frame
Mar 9 at 0:50
$begingroup$
@kelalaka Wt = M1 u mean (Wt-2) = (M1-2) ? sorry bro i learn self-taught about this. i got little confused if read mathematic formulas
$endgroup$
– Onta Ss
Mar 9 at 2:25
$begingroup$
@RichieFrame yeah i got it. How about W16....W63 ? what is the value ?
$endgroup$
– Onta Ss
Mar 9 at 3:27
|
show 3 more comments
$begingroup$
It is in SHA-256 message schedule (NIST-FIPS 180-4);
The message $M$ with length $l$ is first padded as the usual way;
- append 1 to the end of the message,
- then, add $k$ zero bits such that $$l+1+k equiv 448 mod 512$$
- finally, add the length of the message in 64-bit. Now, the total padded length is divisible by 512.
After padding, the padded message parsed into $M^1,ldots,M^N$ where each has size 512-bit.
The sub index $t$ represents 32-bits in 512 bits. Thus, The $M_t^(i)$ is the $t$-th 32-bit in the $M^(i)$ for $0 leq t leq 15$
The $W_t$ is defined as your equation;
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
Where
$$sigma_1^256(x) = operatornameROTR^17(x) oplus operatornameROTR^19(x) oplus operatornameSHR^10(x)$$
$$sigma_0^256(x) = operatornameROTR^7(x) oplus operatornameROTR^18(x) oplus operatornameSHR^3(x)$$
$sigma_1^256(x)$ and $sigma_0^256(x)$ operate on 32 bits and produce 32 bits.
Note: the 256 above the $sigma$ represents the 256 in SHA-256. Similarly, there is $sigma_0^512(x)$ and $sigma_1^512(x)$ for SHA-512.
edited Mar 9 at 4:19
Richie Frame
10.1k11530
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
$endgroup$
$begingroup$
so you mean Wt = M16.....Mn ?
$endgroup$
– Onta Ss
Mar 8 at 10:44
1
$begingroup$
No. $W_1 = M_1^(i), W_2 = M_2^(i), ldots, W_15 = M_15^(i)$
$endgroup$
– kelalaka
Mar 8 at 11:00
$begingroup$
@OntaSs there is no $M_16$, the 16 words of the input block are numbered 0 to 15 only
$endgroup$
– Richie Frame
Mar 9 at 0:50
$begingroup$
@kelalaka Wt = M1 u mean (Wt-2) = (M1-2) ? sorry bro i learn self-taught about this. i got little confused if read mathematic formulas
$endgroup$
– Onta Ss
Mar 9 at 2:25
$begingroup$
@RichieFrame yeah i got it. How about W16....W63 ? what is the value ?
$endgroup$
– Onta Ss
Mar 9 at 3:27
|
show 3 more comments
$begingroup$
It is in SHA-256 message schedule (NIST-FIPS 180-4);
The message $M$ with length $l$ is first padded as the usual way;
- append 1 to the end of the message,
- then, add $k$ zero bits such that $$l+1+k equiv 448 mod 512$$
- finally, add the length of the message in 64-bit. Now, the total padded length is divisible by 512.
After padding, the padded message parsed into $M^1,ldots,M^N$ where each has size 512-bit.
The sub index $t$ represents 32-bits in 512 bits. Thus, The $M_t^(i)$ is the $t$-th 32-bit in the $M^(i)$ for $0 leq t leq 15$
The $W_t$ is defined as your equation;
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
Where
$$sigma_1^256(x) = operatornameROTR^17(x) oplus operatornameROTR^19(x) oplus operatornameSHR^10(x)$$
$$sigma_0^256(x) = operatornameROTR^7(x) oplus operatornameROTR^18(x) oplus operatornameSHR^3(x)$$
$sigma_1^256(x)$ and $sigma_0^256(x)$ operate on 32 bits and produce 32 bits.
Note: the 256 above the $sigma$ represents the 256 in SHA-256. Similarly, there is $sigma_0^512(x)$ and $sigma_1^512(x)$ for SHA-512.
edited Mar 9 at 4:19
Richie Frame
10.1k11530
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
$endgroup$
It is in SHA-256 message schedule (NIST-FIPS 180-4);
The message $M$ with length $l$ is first padded as the usual way;
- append 1 to the end of the message,
- then, add $k$ zero bits such that $$l+1+k equiv 448 mod 512$$
- finally, add the length of the message in 64-bit. Now, the total padded length is divisible by 512.
After padding, the padded message parsed into $M^1,ldots,M^N$ where each has size 512-bit.
The sub index $t$ represents 32-bits in 512 bits. Thus, The $M_t^(i)$ is the $t$-th 32-bit in the $M^(i)$ for $0 leq t leq 15$
The $W_t$ is defined as your equation;
$$W_t = begincases
M_t^(i) & 0 leq t leq 15 \
&\
sigma_1^256(W_t-2) + W_t-7 + sigma_0^256(W_t-15) + W_t-16 & 16 leq t leq 63
endcases$$
Where
$$sigma_1^256(x) = operatornameROTR^17(x) oplus operatornameROTR^19(x) oplus operatornameSHR^10(x)$$
$$sigma_0^256(x) = operatornameROTR^7(x) oplus operatornameROTR^18(x) oplus operatornameSHR^3(x)$$
$sigma_1^256(x)$ and $sigma_0^256(x)$ operate on 32 bits and produce 32 bits.
Note: the 256 above the $sigma$ represents the 256 in SHA-256. Similarly, there is $sigma_0^512(x)$ and $sigma_1^512(x)$ for SHA-512.
edited Mar 9 at 4:19
Richie Frame
10.1k11530
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
edited Mar 9 at 4:19
Richie Frame
10.1k11530
edited Mar 9 at 4:19
Richie Frame
10.1k11530
edited Mar 9 at 4:19
Richie Frame
10.1k11530
10.1k11530
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
answered Mar 8 at 10:11
kelalakakelalaka
8,68022351
8,68022351
$begingroup$
so you mean Wt = M16.....Mn ?
$endgroup$
– Onta Ss
Mar 8 at 10:44
1
$begingroup$
No. $W_1 = M_1^(i), W_2 = M_2^(i), ldots, W_15 = M_15^(i)$
$endgroup$
– kelalaka
Mar 8 at 11:00
$begingroup$
@OntaSs there is no $M_16$, the 16 words of the input block are numbered 0 to 15 only
$endgroup$
– Richie Frame
Mar 9 at 0:50
$begingroup$
@kelalaka Wt = M1 u mean (Wt-2) = (M1-2) ? sorry bro i learn self-taught about this. i got little confused if read mathematic formulas
$endgroup$
– Onta Ss
Mar 9 at 2:25
$begingroup$
@RichieFrame yeah i got it. How about W16....W63 ? what is the value ?
$endgroup$
– Onta Ss
Mar 9 at 3:27
|
show 3 more comments
$begingroup$
so you mean Wt = M16.....Mn ?
$endgroup$
– Onta Ss
Mar 8 at 10:44
1
$begingroup$
No. $W_1 = M_1^(i), W_2 = M_2^(i), ldots, W_15 = M_15^(i)$
$endgroup$
– kelalaka
Mar 8 at 11:00
$begingroup$
@OntaSs there is no $M_16$, the 16 words of the input block are numbered 0 to 15 only
$endgroup$
– Richie Frame
Mar 9 at 0:50
$begingroup$
@kelalaka Wt = M1 u mean (Wt-2) = (M1-2) ? sorry bro i learn self-taught about this. i got little confused if read mathematic formulas
$endgroup$
– Onta Ss
Mar 9 at 2:25
$begingroup$
@RichieFrame yeah i got it. How about W16....W63 ? what is the value ?
$endgroup$
– Onta Ss
Mar 9 at 3:27
$begingroup$
so you mean Wt = M16.....Mn ?
$endgroup$
– Onta Ss
Mar 8 at 10:44
$begingroup$
so you mean Wt = M16.....Mn ?
$endgroup$
– Onta Ss
Mar 8 at 10:44
1
1
$begingroup$
No. $W_1 = M_1^(i), W_2 = M_2^(i), ldots, W_15 = M_15^(i)$
$endgroup$
– kelalaka
Mar 8 at 11:00
$begingroup$
No. $W_1 = M_1^(i), W_2 = M_2^(i), ldots, W_15 = M_15^(i)$
$endgroup$
– kelalaka
Mar 8 at 11:00
$begingroup$
@OntaSs there is no $M_16$, the 16 words of the input block are numbered 0 to 15 only
$endgroup$
– Richie Frame
Mar 9 at 0:50
$begingroup$
@OntaSs there is no $M_16$, the 16 words of the input block are numbered 0 to 15 only
$endgroup$
– Richie Frame
Mar 9 at 0:50
$begingroup$
@kelalaka Wt = M1 u mean (Wt-2) = (M1-2) ? sorry bro i learn self-taught about this. i got little confused if read mathematic formulas
$endgroup$
– Onta Ss
Mar 9 at 2:25
$begingroup$
@kelalaka Wt = M1 u mean (Wt-2) = (M1-2) ? sorry bro i learn self-taught about this. i got little confused if read mathematic formulas
$endgroup$
– Onta Ss
Mar 9 at 2:25
$begingroup$
@RichieFrame yeah i got it. How about W16....W63 ? what is the value ?
$endgroup$
– Onta Ss
Mar 9 at 3:27
$begingroup$
@RichieFrame yeah i got it. How about W16....W63 ? what is the value ?
$endgroup$
– Onta Ss
Mar 9 at 3:27
|
show 3 more comments
Thanks for contributing an answer to Cryptography Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcrypto.stackexchange.com%2fquestions%2f67873%2fwhat-does-the-symbol-w-t-mean-in-the-sha-256-specification%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
