あなたが持って考えてみましょうハッシュ関数 $\mathcal{H}$の長さの文字列を取る$2n$の長さの戻り文字列$n$、それがあることを素敵なプロパティがある耐衝突、すなわち、見つけるのは難しいです二つの異なる文字列$s \neq s'$同じハッシュを持つ$\mathcal{H}(s) = \mathcal{H}(s')$。

ここで、任意の長さの文字列を取得し、それらを長さnの文字列にマッピングする新しいハッシュ関数$\mathcal{H'}$を作成します。$n$

幸運なことに、すでに1979年に、Merkle–Damgård構造として知られる方法が公開されました。

この課題の課題は、このアルゴリズムを実装することです。そのため、アプローチがより簡単であることを示すステップバイステップの例に進む前に、最初にマークル・ダムガード構造の形式的な説明を見ていきます。最初に表示される場合があります。

いくつかの整数$n > 0$、上記のハッシュ関数$\mathcal{H}$および任意の長さ$s$入力文字列sが与えられると、新しいハッシュ関数$\mathcal{H'}$は次のことを行います。

• 設定$l = |s|$、$s$の長さ、および長さnのチャンクに$s$を分割し、必要に応じて最後のチャンクを末尾のゼロで埋めます。この収率は、mは= ⌈ L$n$$m = \lceil \frac{l}{n} \rceil$標識されている多くのチャンク$c_1, c_2, \dots, c_m$。
• リーディングを追加して、後続チャンク$c_0$と$c_{m+1}$、ここで、$c_0$からなる文字列であり、$n$ゼロと$c_{m+1}$であり、$n$バイナリで、で埋め主要長さにゼロ$n$。
• 今反復適用$\mathcal{H}$現在のチャンクに$c_i$、前の結果に添付$r_{i-1}$：$r_i = \mathcal{H}(r_{i-1}c_i)$、$r_0 = c_0$。（このステップは、以下の例を見るとより明確になるかもしれません。）
• H ′の出力$\mathcal{H'}$は、最終結果$r_{m+1}$です。

タスク

入力として正の整数かかるプログラムや関数書き込み$n$、ハッシュ関数$\mathcal{H}$としてブラックボックスおよび空でない文字列$s$、H 'と同じ結果を返す$\mathcal{H'}$、同じ入力で。

これはcode-golfであるため、各言語の最短の回答が優先されます。

例

レッツ・発言$n = 5$、そう私たちの与えられたハッシュ関数$\mathcal{H}$長さ5の長さ10とリターンの文字列の文字列を取ります。

• 入力所与$s = \texttt{"Programming Puzzles"}$、我々は、次のチャンクを取得：$s_1 = \texttt{"Progr"}$、$s_2 = \texttt{"ammin"}$、$s_3 = \texttt{"g Puz"}$と$s_4 = \texttt{"zles0"}$。$s_4$は、長さが5になるように、末尾にゼロが1つ埋め込まれる必要があることに注意してください。
• $c_0 = \texttt{"00000"}$は5つのゼロのストリングであり、$c_5 = \texttt{"00101"}$は5のバイナリ（$\texttt{101}$）であり、2つの先行ゼロが埋め込まれます。
• チャンクは$\mathcal{H}$と結合されます：
  $r_0 = c_0 = \texttt{"00000"}$
  $r_1 = \mathcal{H}(r_0c_1) = \mathcal{H}(\texttt{"00000Progr"})$
  $r_2 = \mathcal{H}(r_1c_2) = \mathcal{H}(\mathcal{H}(\texttt{"00000Progr"})\texttt{"ammin"})$ $r_3 = \mathcal{H}(r_2c_3) = \mathcal{H}(\mathcal{H}(\mathcal{H}(\texttt{"00000Progr"})\texttt{"ammin"})\texttt{"g Puz"})$
  $r_4 = \mathcal{H}(r_3c_4) = \mathcal{H}(\mathcal{H}(\mathcal{H}(\mathcal{H}(\texttt{"00000Progr"})\texttt{"ammin"})\texttt{"g Puz"})\texttt{"zles0"})$
  $r_5 = \mathcal{H}(r_4c_5) = \mathcal{H}(\mathcal{H}(\mathcal{H}(\mathcal{H}(\mathcal{H}(\texttt{"00000Progr"})\texttt{"ammin"})\texttt{"g Puz"})\texttt{"zles0"})\texttt{"00101"})$
• $r_5$は出力です。

のは、この出力は、いくつかの選択肢に応じて、どのように見えるかを見てみましょう¹のための$\mathcal{H}$：

• 場合は$\mathcal{H}(\texttt{"0123456789"}) = \texttt{"13579"}$、すなわち$\mathcal{H}$ちょうど毎秒文字を返す、我々 GET：
  $r_1 = \mathcal{H}(\texttt{"00000Progr"}) = \texttt{"00Por"}$
  $r_2 = \mathcal{H}(\texttt{"00Porammin"}) = \texttt{"0oamn"}$
  $r_3 = \mathcal{H}(\texttt{"0oamng Puz"}) = \texttt{"omgPz"}$
  $r_4 = \mathcal{H}(\texttt{"omgPzzles0"}) = \texttt{"mPze0"}$
  $r_5 = \mathcal{H}(\texttt{"mPze000101"}) = \texttt{"Pe011"}$
  So $\texttt{"Pe011"}$ needs to be the output if such a $\mathcal{H}$ is given as black box function.
• If $\mathcal{H}$ simply returns the first 5 chars of its input, the output of $\mathcal{H'}$ is $\texttt{"00000"}$. Similarly if $\mathcal{H}$ returns the last 5 chars, the output is $\texttt{"00101"}$.
• If $\mathcal{H}$ multiplies the character codes of its input and returns the first five digits of this number, e.g. $\mathcal{H}(\texttt{"PPCG123456"}) = \texttt{"56613"}$, then $\mathcal{H}'(\texttt{"Programming Puzzles"}) = \texttt{"91579"}$.

^{1 For simplicity, those $\mathcal{H}$ are actually not collision resistant, though this does not matter for testing your submission.}

Haskell, 91 90 86 bytes

• -1 byte thanks to Laikoni
• -4 bytes thanks to xnor

n!h|let a='0'<$[1..n];c?""=c;c?z=h(c++take n(z++a))?drop n z=h.(++mapM(:"1")a!!n).(a?)

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Explanation

a='0'<$[1..n]

Just assigns the string "00...0" ('0' $n$ times) to a

c?""=c
c?z=h(c++take n(z++a))?drop n z

The function ? implements the recursive application of h: c is the hash we have obtained so far (length $n$), z is the rest of the string. If z is empty then we simply return c, otherwise we take the first $n$ characters of z (possibly padding with zeros from a), prepend c and apply h. This gives the new hash, and then we call ? recursively on this hash and the remaining characters of z.

n!h=h.(++mapM(:"1")a!!n).(a?)

The function ! is the one actually solving the challenge. It takes n, h and s (implicit) as inputs. We compute a?s, and all we have to do is append n in binary and apply h once more. mapM(:"1")a!!n returns the binary representation of $n$.

R, 159 154 bytes

function(n,H,s,`?`=paste0,`*`=strrep,`/`=Reduce,`+`=nchar,S=0*n?s?0*-(+s%%-n)?"?"/n%/%2^(n:1-1)%%2)(function(x,y)H(x?y))/substring(S,s<-seq(,+S,n),s--n-1)

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Yuck! Answering string challenges in R is never pretty, but this is horrible. This is an instructive answer on how not to write "normal" R code...

Thanks to nwellnhof for fixing a bug, at a cost of 0 bytes!

Thanks to J.Doe for swapping the operator aliasing to change the precedence, good for -4 bytes.

The explanation below is for the previous version of the code, but the principles remain the same.

function(n,H,s,               # harmless-looking function arguments with horrible default arguments 
                              # to prevent the use of {} and save two bytes
                              # then come the default arguments,
                              # replacing operators as aliases for commonly used functions:
 `+`=paste0,                  # paste0 with binary +
 `*`=strrep,                  # strrep for binary *
 `/`=Reduce,                  # Reduce with binary /
 `?`=nchar,                   # nchar with unary ?
 S=                           # final default argument S, the padded string:
  0*n+                        # rep 0 n times
  s+                          # the original string
  0*-((?s)%%-n)+              # 0 padding as a multiple of n
  "+"/n%/%2^(n:1-1)%%2)       # n as an n-bit number
                              # finally, the function body:
 (function(x,y)H(x+y)) /      # Reduce/Fold (/) by H operating on x + y
  substring(S,seq(1,?S,n),seq(n,?S,n))  # operating on the n-length substrings of S

C (gcc), 251 bytes

#define P sprintf(R,
b(_){_=_>1?10*b(_/2)+_%2:_;}f(H,n,x)void(*H)(char*);char*x;{char R[2*n+1],c[n+1],*X=x;P"%0*d",n,0);while(strlen(x)>n){strncpy(c,x,n);x+=n;strcat(R,c);H(R);}P"%s%s%0*d",R,x,n-strlen(x),0);H(R);P"%s%0*d",R,n,b(n));H(R);strcpy(X,R);}

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Not as clean as the bash solution, and highly improvable.

The function is f taking H as a function that replaces its string input with that string's hash, n as in the description, and x the input string and output buffer.

Description:

#define P sprintf(R,     // Replace P with sprintf(R, leading to unbalanced parenthesis
                         // This is replaced and expanded for the rest of the description
b(_){                    // Define b(x). It will return the integer binary expansion of _
                         // e.g. 5 -> 101 (still as integer)
  _=_>1?                 // If _ is greater than 1
    10*b(_/2)+_%2        // return 10*binary expansion of _/2 + last binary digit
    :_;}                 // otherwise just _
f(H,n,x)                 // Define f(H,n,x)
  void(*H)(char*);       // H is a function taking a string
  char*x; {              // x is a string
  char R[2*n+1],c[n+1],  // Declare R as a 2n-length string and c as a n-length string
  *X=x;                  // save x so we can overwrite it later
  sprintf(R,"%0*d",n,0); // print 'n' 0's into R
  while(strlen(x)>n){    // while x has at least n characters
    strncpy(c,x,n);x+=n; // 'move' the first n characters of x into c
    strcat(R,c);         // concatenate c and R
    H(R);}               // Hash R
  sprintf(R,"%s%s%0*d"   // set R to two strings concatenated followed by some zeroes
    R,x,                 // the two strings being R and (what's left of) x
    n-strlen(x),0);      // and n-len(x) zeroes
  H(R);                  // Hash R
  sprintf(R,"%s%*d",R,n, // append to R the decimal number, 0 padded to width n
    b(n));               // The binary expansion of n as a decimal number
  H(R);strcpy(X,R);}     // Hash R and copy it into where x used to be

Ruby, 78 bytes

->n,s,g{(([?0*n]*2*s).chop.scan(/.{#{n}}/)+["%0#{n}b"%n]).reduce{|s,x|g[s+x]}}

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How it works:

([?0*n]*2*s).chop    # Padding: add leading and trailing 
                     # zeros, then remove the last one
.scan(/.{#{n}}/)     # Split the string into chunks
                     # of length n
+["%0#{n}b"%n]       # Add the trailing block
.reduce{|s,x|g[s+x]} # Apply the hashing function
                     # repeatedly

Jelly, 23 bytes

0Ṿ;s;BṾ€ṚWƲ}z”0ZU0¦;Ç¥/

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Accepts $\mathcal H$ at the line above it, $s$ as its left argument, and $n$ as its right argument.

Bash, 127-ε bytes

Z=`printf %0*d $1` R=$Z
while IFS= read -rn$1 c;do R=$R$c$Z;R=`H<<<${R::2*$1}`;done
H< <(printf $R%0*d $1 `bc <<<"obase=2;$1"`)

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This works as a program/function/script/snippet. H must be resolveable to a program or function that will perform the hashing. N is the argument. Example call:

$ H() {
>   sed 's/.\(.\)/\1/g'
> }
$ ./wherever_you_put_the_script.sh 5 <<< "Programming Puzzles"  # if you add a shebang
Pe011

Description:

Z=`printf %0*d $1`

This creates a string of $1 zeroes. This works by calling printf and telling it to print an integer padded to extra argument width. That extra argument we pass is $1, the argument to the program/function/script which stores n.

R=$Z

This merely copies Z, our zero string, to R, our result string, in preparation for the hashing loop.

while IFS= read -rn$1 c; do

This loops over the input every $1 (n) characters loading the read characters into c. If the input ends then c merely ends up too short. The r option ensures that any special characters in the input don't get bash-interpreted. This is the -ε in the title - that r isn't strictly necessary, but makes the function more accurately match the input.

R=$R$c$Z

This concatenates the n characters read from input to R along with zeroes for padding (too many zeroes for now).

R=`H<<<${R::2*$1}`;done

This uses a here string as input to the hash function. The contents ${R::2*$1} are a somewhat esoteric bash parameter substitution which reads: R, starting from 0, only 2n characters.

Here the loop ends and we finish with:

H< <(printf $R%0*d $1 `bc <<<"obase=2;$1"`)

Here the same format string trick is used to 0 pad the number. bc is used to convert it to binary by setting the output base (obase) to 2. The result is passed to the hash function/program whose output is not captured and thus is shown to the user.

Pyth, 24 bytes

Since Pyth doesn't allow H to be used for a function name, I use y instead.

uy+GH+c.[E=`ZQQ.[ZQ.BQ*Z

Try it online! Example is with the "every second character" version of H.

Perl 6, 79 68 bytes

{reduce &^h o&[~],comb 0 x$^n~$^s~$n.fmt("%.{$n-$s.comb%-$n}b"): $n}

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Explanation

{
  reduce         # Reduce with
    &^h o&[~],   # composition of string concat and hash function
    comb         # Split string
      0 x$^n     # Zero repeated n times
      ~$^s       # Append input string s
      ~$n.fmt("  # Append n formatted
        %.       # with leading zeroes,
        {$n             # field width n for final chunk
         -$s.comb%-$n}  # -(len(s)%-n) for padding,
        b")      # as binary number
      :          # Method call with colon syntax
      $n         # Split into substrings of length n
}

Clean, 143 bytes

import StdEnv
r=['0':r]
$n h s=foldl(\a b=h(a++b))(r%(1,n))([(s++r)%(i,i+n-1)\\i<-[0,n..length s]]++[['0'+toChar((n>>(n-p))rem 2)\\p<-[1..n]]])

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Python 2, 126 113 bytes

lambda n,H,s:reduce(lambda x,y:H(x+y),re.findall('.'*n,'0'*n+s+'0'*(n-len(s)%n))+[bin(n)[2:].zfill(n)])
import re

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-13 thanks to Triggernometry.

Yeah, this is an abomination, why can't I just use a built-in to split a string into chunks...? :-(

Python 2, 106 102 bytes

For once, the function outgolfs the lambda. -4 bytes for simple syntax manipulation, thanks to Jo King.

def f(n,H,s):
 x='0'*n;s+='0'*(n-len(s)%n)+bin(n)[2:].zfill(n)
 while s:x=H(x+s[:n]);s=s[n:]
 return x

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Japt, 27 bytes

òV ú'0 pV¤ùTV)rÈ+Y gOvW}VçT

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I haven't found any capability for Japt to take functions directly as an input, so this takes a string which is interpreted as Japt code and expects it to define a function. Specifically, OvW takes the third input and interprets it as Japt, then g calls it. Replacing that with OxW allows input as a Javascript function instead, or if the function were (somehow) already stored in W it could just be W and save 2 bytes. The link above has the worked example of $\mathcal{H}$ that takes characters at odd indexes, while this one is the "multiply char-codes and take the 5 highest digits" example.

Due to the way Japt takes inputs, $s$ will be U, $n$ will be V, and $\mathcal{H}$ will be W

Explanation:

òV                             Split U into segments of length V
   ú'0                         Right-pad the short segment with "0" to the same length as the others
       p     )                 Add an extra element:
        V¤                       V as a base-2 string
          ùTV                    Left-pad with "0" until it is V digits long
              r                Reduce...
                        VçT          ...Starting with "0" repeated V times...
               È       }                                                  ...By applying:
                +Y               Combine with the previous result
                   gOvW          And run W as Japt code

GolfScript, 47 bytes

~:π'0'*:s\+s+π/);[sπ2base{48+}%+0π->]+{+1$~}*\;

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oK, 41 bytes

{(x#48)(y@,)/(0N,x)#z,,/$((x+x!-#z)#2)\x}

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{                                       } /x is n, y is H, z is s.
                          (x+x!-#z)       /number of padding 0's needed + x
                         (         #2)\x  /binary(x) with this length
                      ,/$                 /to string
                    z,                    /append to z
             (0N,x)#                      /split into groups of length x
       (y@,)/                             /foldl of y(concat(left, right))...
 (x#48)                                   /...with "0"*x as the first left string