あなたが持って考えてみましょうハッシュ関数 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ss同じハッシュを持つH(s)=H(s)





  • 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= Lnm=ln標識されている多くのチャンクc1,c2,,cm
  • リーディングを追加して、後続チャンクc0cm+1、ここで、c0からなる文字列であり、nゼロとcm+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H現在のチャンクにci、前の結果に添付ri1ri=H(ri1ci)r0=c0(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Hは、最終結果rm+1です。


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Hとしてブラックボックスおよび空でない文字列sH 'と同じ結果を返すH、同じ入力で。



  • 入力所与s="Programming Puzzles"、我々は、次のチャンクを取得:s1="Progr"s2="ammin"s3="g Puz"s4="zles0"s4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。
  • c0="00000"は5つのゼロのストリングであり、c5="00101"は5のバイナリ(101)であり、2つの先行ゼロが埋め込まれます。
  • チャンクはHと結合されます:
    r2=H(r1c2)=H(H("00000Progr")"ammin") r3=H(r2c3)=H(H(H("00000Progr")"ammin")"g Puz")
    r4=H(r3c4)=H(H(H(H("00000Progr")"ammin")"g Puz")"zles0")
    r5=H(r4c5)=H(H(H(H(H("00000Progr")"ammin")"g Puz")"zles0")"00101")
  • r5は出力です。


  • 場合はH("0123456789")="13579"、すなわちHちょうど毎秒文字を返す、我々 GET:
    r3=H("0oamng Puz")="omgPz"
    So "Pe011" needs to be the output if such a H is given as black box function.
  • If H simply returns the first 5 chars of its input, the output of H is "00000". Similarly if H returns the last 5 chars, the output is "00101".
  • If H multiplies the character codes of its input and returns the first five digits of this number, e.g. H("PPCG123456")="56613", then H("Programming Puzzles")="91579".

1 For simplicity, those H are actually not collision resistant, though this does not matter for testing your submission.

I must say it's fun that the example given has the last 'full' hash be of "OMG Puzzles!" effectively omgPzzles0. Well chosen example input!

Can we assume some flexibility on the input format for H (e.g. it takes two strings of length n, or a longer string of which it only considers the first 2n characters)?

Are space characters, e.g., between "g P" valid output?

@guest271314 If the space is part of the resulting hash, it needs to be outputted. If the hash is actually "gP", you may not output a space inbetween.



Haskell, 91 90 86 bytes

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?)

Try it online!



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

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.


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.

let in a guard is shorter than using where: Try it online!

It looks like mapM(\_->"01")a can be mapM(:"1")a.


R, 159 154 bytes


Try it online!

Yuck! Answering 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

I think 0*(n-(?s)%%n) doesn't work if n divides s evenly. But 0*-((?s)%%-n) should work.

@nwellnhof ah, of course, thank you, fixed.

Minor changes, 155 bytes

@J.Doe nice! I saved another byte since seq has 1 as its from argument by default.


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);}

Try it online!

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.


#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

I think: 227 bytes (going off of ceilingcat's comment)


Ruby, 78 bytes


Try it online!

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


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"`)

Try it online!

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


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.


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 in the title - that r isn't strictly necessary, but makes the function more accurately match the input.


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


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.

Why "127-ε"? Why not just "127"?
Solomon Ucko

I don't know. I was on the fence about the necessity of the r flag. I figured 1 byte doesn't really matter, but if pushed I could shave it.

For the read command?
Solomon Ucko

Because without it a `` in the input will be interpreted instead of ignored, so they'd have to be escaped.

Maybe add a note about that?
Solomon Ucko


Pyth, 24 bytes

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


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}

Try it online!


  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
$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]]])

Try it online!


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

Try it online!

-13 thanks to Triggernometry.

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

codegolf.stackexchange.com/a/173952/55696 A while loop is the best builtin I could hope for. 104 bytes
Steven H.

@StevenH. Yeah, especially if you're actually focusing on the golfing itself. >_>
Erik the Outgolfer

'0'*~-n instead of '0'*(len(s)%n) is shorter (and actually correct for shorter inputs).

@nwellnhof Yeah, but it's definitely not the same thing.
Erik the Outgolfer

Maybe I wasn't clear enough. Your solution gives the wrong answer for strings like Programming Puzz (16 chars). Replacing '0'*(len(s)%n) with '0'*~-n fixes that and saves 7 bytes.


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):
 while s:x=H(x+s[:n]);s=s[n:]
 return x

Try it online!

Shouldn't the result be 'Pe011', not 'e011'?

That it should. Fixed!
Steven H.

Use semi-colons instead of newlines. -4 bytes
Jo King

I didn't realize that worked for while loops as well, thanks!
Steven H.


Japt, 27 bytes

òV ú'0 pV¤ùTV)rÈ+Y gOvW}VçT

Try it!

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 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 H will be W


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


oK, 41 bytes


Try it online!

{                                       } /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
Licensed under cc by-sa 3.0 with attribution required.