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<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"  
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<head><title>Theory used by libgfshare</title> 
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>
   <div class="maketitle">



<h2 class="titleHead">Theory used by libgfshare</h2>
<div class="author" ><span 
class="cmr-12">Simon McVittie</span></div><br />
<div class="date" ><span 
class="cmr-12">23rd April 2006</span></div>
   </div>
   <h3 class="sectionHead"><span class="titlemark">1   </span> <a 
 id="x1-10001"></a>Introduction</h3>
<!--l. 13--><p class="noindent" >libgfshare implements Shamir secret sharing [SHAMIR] over the field <span 
class="cmmi-10">GF</span>(2<sup><span 
class="cmr-7">8</span></sup>), instead of <span 
class="cmmi-10">GF</span>(<span 
class="cmmi-10">p</span>) for a prime <span 
class="cmmi-10">p </span>as suggested
by Shamir&#8217;s paper. This document aims to prove the security and integrity of this scheme.
<!--l. 17--><p class="indent" >   Note that while I believe this document to be correct, I accept no responsibility for loss or damage caused by relying on
the correctness of my proof.
<!--l. 21--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">2   </span> <a 
 id="x1-20002"></a>Definitions</h3>
<!--l. 23--><p class="noindent" >Let <span 
class="cmmi-10">F </span>be a field with multiplicative identity 1 and additive identity 0.
<!--l. 25--><p class="indent" >   If <span 
class="cmmi-10">A </span>= <span 
class="cmsy-10">{</span>(<span 
class="cmmi-10">a</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,b</span><sub><span 
class="cmr-7">1</span></sub>)<span 
class="cmmi-10">,</span><img 
src="theory0x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,</span>(<span 
class="cmmi-10">a</span><sub><span 
class="cmmi-7">n</span></sub><span 
class="cmmi-10">,b</span><sub><span 
class="cmmi-7">n</span></sub>)<span 
class="cmsy-10">}</span>, with the <span 
class="cmmi-10">a</span><sub><span 
class="cmmi-7">i</span></sub> distinct nonzero elements of F and the <span 
class="cmmi-10">b</span><sub><span 
class="cmmi-7">i</span></sub> elements of <span 
class="cmmi-10">F</span>, indexed by
<span 
class="cmmi-10">I </span>= <span 
class="cmsy-10">{</span>1<span 
class="cmmi-10">,</span><img 
src="theory1x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,n</span><span 
class="cmsy-10">}</span>, then define
   <center class="par-math-display" >
<img 
src="theory2x.png" alt="       &sum;      &prod;
PA(x) =    bj       (x &minus; ak)(aj &minus; ak)&minus;1
        j&isin;I  k&isin;I,k&frasl;=j
" class="par-math-display" ></center>
<!--l. 31--><p class="nopar" >
<!--l. 33--><p class="indent" >   a polynomial of degree at most <span 
class="cmmi-10">n </span><span 
class="cmsy-10">&minus; </span>1. (By distinctness of the <span 
class="cmmi-10">a</span><sub><span 
class="cmmi-7">i</span></sub>, the inverses required exist.) This is the Lagrange
interpolating polynomial for the points in <span 
class="cmmi-10">A</span>.
<!--l. 37--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">3   </span> <a 
 id="x1-30003"></a>Lemma 1</h3>
<!--l. 39--><p class="noindent" >Let <span 
class="cmmi-10">a</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory3x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,a</span><sub><span 
class="cmmi-7">t</span></sub> <span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F </span>be distinct and nonzero; let <span 
class="cmmi-10">b</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory4x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,b</span><sub><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,c </span><span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F </span>be arbitrary. Then there exists <span 
class="cmmi-10">b</span><sub><span 
class="cmmi-7">t</span></sub> <span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F </span>such that if
<span 
class="cmmi-10">A </span>= <span 
class="cmsy-10">{</span>(<span 
class="cmmi-10">a</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,b</span><sub><span 
class="cmr-7">1</span></sub>)<span 
class="cmmi-10">,</span><img 
src="theory5x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,</span>(<span 
class="cmmi-10">a</span><sub><span 
class="cmmi-7">t</span></sub><span 
class="cmmi-10">,b</span><sub><span 
class="cmmi-7">t</span></sub>)<span 
class="cmsy-10">} </span>then <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">A</span></sub>(0) = <span 
class="cmmi-10">c</span>.
<!--l. 43--><p class="noindent" >
   <h4 class="subsectionHead"><span class="titlemark">3.1   </span> <a 
 id="x1-40003.1"></a>Proof</h4>
<!--l. 45--><p class="noindent" >Let <span 
class="cmmi-10">I </span>= <span 
class="cmsy-10">{</span>1<span 
class="cmmi-10">,</span><img 
src="theory6x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,t</span><span 
class="cmsy-10">}</span>. We have
   <center class="par-math-display" >
<img 
src="theory7x.png" alt="       &sum;      &prod;               &minus;1  &sum;      &prod;             &minus;1
PA(0) =   bj       &minus; ak(aj &minus; ak) =   yj      ak(ak &minus; aj)
        j&isin;I  k&isin;I,k&frasl;=j                j&isin;I  k&isin;I,k&frasl;=j
" class="par-math-display" ></center>
<!--l. 50--><p class="nopar" >
<!--l. 52--><p class="indent" >   Let
   <center class="par-math-display" >
<img 
src="theory8x.png" alt="    &lfloor;                             &rfloor; &lfloor;                &rfloor;
          &sum;       &prod;                    &prod;
bt = &lceil;c +     bj      ak(aj &minus; ak)&minus;1&rceil; &lceil;     a&minus;k1(ak &minus; at)&rceil;
         j&isin;I,j&frasl;=t  k&isin;I,k&frasl;=j               k&isin;I,k&frasl;=t
" class="par-math-display" ></center>
<!--l. 57--><p class="nopar" >
<!--l. 59--><p class="indent" >   Then
   <center class="par-math-display" >
<img 
src="theory9x.png" alt="         &sum;       &prod;                     &prod;
PA (0) =      bj       ak(ak &minus; aj)&minus;1 + bt    ak(ak &minus; at)&minus;1
        j&isin;I,j&frasl;=t  k&isin;I,k&frasl;=j                k&isin;I,k&frasl;=t
" class="par-math-display" ></center>
<!--l. 63--><p class="nopar" >
   <center class="math-display" >
<img 
src="theory10x.png" alt="    &sum;       &prod;             &minus;1   &sum;       &prod;             &minus; 1
=       bj      ak(ak &minus; aj) &minus;       bj      ak(ak &minus; aj) + c
  j&isin;I,j&frasl;=t  k&isin;I,k&frasl;=j              j&isin;I,j&frasl;=t  k&isin;I,k&frasl;=j
" class="math-display" ></center>
<!--l. 65--><p class="nopar" >
   <center class="math-display" >
<img 
src="theory11x.png" alt="= c
" class="math-display" ></center>
<!--l. 67--><p class="nopar" >
<!--l. 69--><p class="indent" >   as required.
<!--l. 71--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">4   </span> <a 
 id="x1-50004"></a>Lemma 2</h3>
<!--l. 73--><p class="noindent" >For any <span 
class="cmmi-10">x</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory12x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,x</span><sub><span 
class="cmmi-7">t</span></sub> distinct and nonzero elements of <span 
class="cmmi-10">F</span>, and any <span 
class="cmmi-10">y</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory13x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,y</span><sub><span 
class="cmmi-7">t</span></sub><span 
class="cmmi-10">,u </span>arbitrary elements of <span 
class="cmmi-10">F</span>, let
   <center class="par-math-display" >
<img 
src="theory14x.png" alt="X  = {(x1,y1),&sdot;&sdot;&sdot;,(xt,yt)}
" class="par-math-display" ></center>
<!--l. 76--><p class="nopar" >
<!--l. 78--><p class="indent" >   and
   <center class="par-math-display" >
<img 
src="theory15x.png" alt="U = {(x1,y1),&sdot;&sdot;&sdot;,(xt&minus; 1,yt&minus;1),(u,PX (u))}
" class="par-math-display" ></center>
<!--l. 80--><p class="nopar" >
<!--l. 82--><p class="indent" >   Then <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">X</span></sub> = <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">U</span></sub><span 
class="cmmi-10">,i.e.P</span><sub><span 
class="cmmi-7">X</span></sub>(<span 
class="cmmi-10">x</span>) = <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">U</span></sub>(<span 
class="cmmi-10">x</span>) for all <span 
class="cmmi-10">x </span><span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F</span>.
<!--l. 84--><p class="noindent" >
   <h4 class="subsectionHead"><span class="titlemark">4.1   </span> <a 
 id="x1-60004.1"></a>Proof</h4>
<!--l. 86--><p class="noindent" >Let <span 
class="cmmi-10">S</span><sub><span 
class="cmmi-7">a,b</span></sub> = <img 
src="theory16x.png" alt="{(x1,y1),&sdot;&sdot;&sdot;,(xt&minus; 1,yt&minus;1),(a,b)}"  class="left" align="middle">. Then
   <center class="par-math-display" >
<img 
src="theory17x.png" alt="         &sum;                    &prod;                        &prod;
PSa,b(x) =   yj(x&minus; a)(xj &minus; a)&minus; 1      (x &minus; xk)(xj &minus; xk)&minus;1 + b (x &minus; xk)(a&minus; xk)&minus;1
         j&#x003C;t                 k&frasl;=j,k&#x003C;t                    k&#x003C;t
" class="par-math-display" ></center>
<!--l. 93--><p class="nopar" >
<!--l. 95--><p class="indent" >   Hence if we let <span 
class="cmmi-10">d</span><sub><span 
class="cmmi-7">i,j</span></sub> = <span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">i</span></sub> <span 
class="cmsy-10">&minus;</span><span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">j</span></sub> and <span 
class="cmmi-10">e</span><sub><span 
class="cmmi-7">i</span></sub> = <span 
class="cmmi-10">u</span><span 
class="cmsy-10">&minus;</span><span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">i</span></sub> (both of which are necessarily nonzero, by distinctness of the <span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">i</span></sub> and <span 
class="cmmi-10">u</span>) we
have
   <center class="par-math-display" >
<img 
src="theory18x.png" alt="        &sum;      &minus; 1 &prod;       &minus;1    &prod;     &minus;1
PX (u) =   yjetdj,t       ekdj,k + yt  ekdt,k
        j&#x003C;t       k&frasl;=j,k&#x003C;t         k&#x003C;t
" class="par-math-display" ></center>
<!--l. 102--><p class="nopar" >
<!--l. 104--><p class="indent" >   and if we also let <span 
class="cmmi-10">f</span><sub><span 
class="cmmi-7">i</span></sub> = <span 
class="cmmi-10">x </span><span 
class="cmsy-10">&minus; </span><span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">i</span></sub>,
   <center class="par-math-display" >
<img 
src="theory19x.png" alt="P  (x) = &sum; y (u&minus; x)e&minus;1  &prod;   f d&minus;1+ P  (u)&prod;  f e&minus; 1
  U     j&#x003C;t j       j k&frasl;=j,k&#x003C;t k j,k   X    k&#x003C;t kk
" class="par-math-display" ></center>
<!--l. 109--><p class="nopar" >
   <center class="math-display" >
<img 
src="theory20x.png" alt="                                              (                                )
        &sum;              &prod;           { &prod;      } { &sum;          &prod;            &prod;      }
PU (x) =    yj(u &minus; x)e&minus;j 1      fkd&minus;j,1k +    fke&minus;k1       yjetd&minus;j1,t      ekd&minus;j,1l + yt  eld&minus;t1,l
        j&#x003C;t           k&frasl;=j,k&#x003C;t        k&#x003C;t       ( j&#x003C;t      l&frasl;=j,l&#x003C;t         l&#x003C;t     )
" class="math-display" ></center>
<!--l. 115--><p class="nopar" >
<!--l. 117--><p class="indent" >   Expanding,
   <center class="par-math-display" >
<img 
src="theory21x.png" alt="           &sum;              { &prod;            }
PU (x ) =     j&#x003C;tyj(u&minus; x)e&minus;j1    k&frasl;=j,k&#x003C;t fkd&minus;j,1k
           +&sum;    y e d&minus;1{&prod;      e d&minus;1} {&prod;    f e&minus;1}
              j{&#x003C;&prod;t j t j,t   l&frasl;=j,l&#x003C;}t k j,l     k&#x003C;t k k
           +yt   k&#x003C;tekd&minus;t,1kfke&minus;k1
" class="par-math-display" ></center>
<!--l. 127--><p class="nopar" >
   <center class="par-math-display" >
<img 
src="theory22x.png" alt="            &lfloor;          ({           )}             ({                )} &rfloor;
PU(x) = &sum; yj&lceil; (u &minus; x)e&minus; 1   &prod;   fkd&minus; 1 + etd&minus;1fje&minus;1   &prod;    ekd&minus; 1fke&minus;1  &rceil; + yt&prod; d&minus; 1fk
       j&#x003C;t          j  (k&frasl;=j,k&#x003C;t   j,k)      j,t   j ( k&frasl;=j,k&#x003C;t  j,k  k )       k&#x003C;t t,k
" class="par-math-display" ></center>
<!--l. 137--><p class="nopar" >
   <center class="math-display" >
<img 
src="theory23x.png" alt="     &lfloor;             &rfloor;
  &sum;       &prod;          [                    ]    &prod;
=    &lceil;yj      fkd&minus;j1,k&rceil;  (u &minus; x)e&minus;j 1+ ete&minus;j1d&minus;j,t1fj + yt   d&minus;t1,kfk
  j&#x003C;t   k&frasl;=j,k&#x003C;t                                 k&#x003C;t
" class="math-display" ></center>
<!--l. 141--><p class="nopar" >
<!--l. 143--><p class="indent" >   Now
   <center class="par-math-display" >
<img 
src="theory24x.png" alt="(u &minus; x )e&minus;j1 + ete&minus;j1d&minus;j1,tfj = (e&minus;j1d&minus;j1,t)[(u &minus; x)dj,t + etfj]
" class="par-math-display" ></center>
<!--l. 148--><p class="nopar" >
   <center class="math-display" >
<img 
src="theory25x.png" alt="    &minus;1 &minus;1
= (ej dj,t)[(u &minus; x)(xj &minus; xt) +(u &minus; xt)(x&minus; xj)]
" class="math-display" ></center>
<!--l. 150--><p class="nopar" >
   <center class="math-display" >
<img 
src="theory26x.png" alt="    &minus;1 &minus;1
= (ej dj,t)(x &minus; xt)(u&minus; xj)
" class="math-display" ></center>
<!--l. 152--><p class="nopar" >
   <center class="math-display" >
<img 
src="theory27x.png" alt="   &minus;1
= dj,tft
" class="math-display" ></center>
<!--l. 154--><p class="nopar" >
<!--l. 156--><p class="indent" >   Hence
   <center class="par-math-display" >
<img 
src="theory28x.png" alt="           &lfloor;             &rfloor;
P  (x) = &sum;  &lceil;y   &prod;   f d&minus;1&rceil; [f d&minus;1]+ y &prod;  d&minus;1f =  P (x)
 U      j&#x003C;t  jk&frasl;=j,k&#x003C;t k j,k   t j,t    tk&#x003C;t t,k k    X
" class="par-math-display" ></center>
<!--l. 163--><p class="nopar" >
<!--l. 165--><p class="indent" >   as required.
<!--l. 167--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">5   </span> <a 
 id="x1-70005"></a>Construction</h3>
<!--l. 169--><p class="noindent" >Let <span 
class="cmmi-10">s </span>be the number of &#8220;shares&#8221; and <span 
class="cmmi-10">t </span>be the required threshold to recover the shared secret (i.e. we construct a &#8220;<span 
class="cmmi-10">t </span>of <span 
class="cmmi-10">s</span>&#8221;
share).
<!--l. 172--><p class="indent" >   Given a secret <span 
class="cmmi-10">f </span><span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F </span>we may construct a Lagrange interpolating polynomial <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">X</span></sub> of degree no more than <span 
class="cmmi-10">t </span><span 
class="cmsy-10">&minus; </span>1, with
<span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">X</span></sub>(0) = <span 
class="cmmi-10">f</span>, as follows:
<!--l. 176--><p class="indent" >   - choose distinct nonzero <span 
class="cmmi-10">x</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory29x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,x</span><sub><span 
class="cmmi-7">s</span></sub> <span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F</span>
<!--l. 178--><p class="indent" >   - choose arbitrary (and unpredictable) <span 
class="cmmi-10">y</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory30x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,y</span><sub><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sub> <span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F</span>
<!--l. 180--><p class="indent" >   - use Lemma 1 to select <span 
class="cmmi-10">y</span><sub><span 
class="cmmi-7">t</span></sub> such that <span 
class="cmmi-10">X </span>= <span 
class="cmsy-10">{</span>(<span 
class="cmmi-10">x</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,y</span><sub><span 
class="cmr-7">1</span></sub>)<span 
class="cmmi-10">,</span><img 
src="theory31x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,</span>(<span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">t</span></sub><span 
class="cmmi-10">,y</span><sub><span 
class="cmmi-7">t</span></sub>)<span 
class="cmsy-10">} </span>has the desired intercept <span 
class="cmmi-10">f</span>
<!--l. 183--><p class="indent" >   To obtain additional shares, calculate <span 
class="cmmi-10">y</span><sub><span 
class="cmmi-7">t</span><span 
class="cmr-7">+1</span></sub> = <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">X</span></sub>(<span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">t</span><span 
class="cmr-7">+1</span></sub>)<span 
class="cmmi-10">,</span><img 
src="theory32x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,y</span><sub><span 
class="cmmi-7">s</span></sub> = <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">X</span></sub>(<span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">s</span></sub>).
<!--l. 186--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">6   </span> <a 
 id="x1-80006"></a>Alternate construction, as used in libgfshare</h3>
<!--l. 188--><p class="noindent" >In libgfshare the construction used is as follows:
<!--l. 190--><p class="indent" >   - construct a polynomial <span 
class="cmmi-10">P </span>by choosing arbitrary and unpredictable coefficients of <span 
class="cmmi-10">x,</span><img 
src="theory33x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,x</span><sup><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sup> from <span 
class="cmmi-10">F</span>, and setting the
coefficient of <span 
class="cmmi-10">x</span><sup><span 
class="cmr-7">0</span></sup> to <span 
class="cmmi-10">f</span>: this therefore has the desired intercept <span 
class="cmmi-10">f</span>
<!--l. 194--><p class="indent" >   - choose distinct nonzero <span 
class="cmmi-10">x</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory34x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,x</span><sub><span 
class="cmmi-7">s</span></sub> <span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F </span>and evaluate <span 
class="cmmi-10">y</span><sub><span 
class="cmr-7">1</span></sub> = <span 
class="cmmi-10">P</span>(<span 
class="cmmi-10">x</span><sub><span 
class="cmr-7">1</span></sub>)<span 
class="cmmi-10">,</span><img 
src="theory35x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,y</span><sub><span 
class="cmmi-7">s</span></sub> = <span 
class="cmmi-10">P</span>(<span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">s</span></sub>)
<!--l. 197--><p class="noindent" >
   <h4 class="subsectionHead"><span class="titlemark">6.1   </span> <a 
 id="x1-90006.1"></a>Proof of equivalence in a finite field <span 
class="cmmi-10">F</span></h4>
<!--l. 199--><p class="noindent" >Suppose <span 
class="cmmi-10">F </span>is finite, as is the case in libgfshare, and that in each construction, arbitrary choices are made from among all
possible values in <span 
class="cmmi-10">F</span>.
<!--l. 203--><p class="indent" >   In the alternate construction, given <span 
class="cmmi-10">x</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory36x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,x</span><sub><span 
class="cmmi-7">t</span></sub><span 
class="cmmi-10">,f </span>we choose a polynomial <span 
class="cmmi-10">P</span>(<span 
class="cmmi-10">x</span>) = <span 
class="cmmi-10">f </span>+ <span 
class="cmmi-10">m</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">x </span>+ <img 
src="theory37x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" > + <span 
class="cmmi-10">m</span><sub><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">x</span><sup><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sup> by choosing
arbitrary coefficients <span 
class="cmmi-10">m</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory38x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,m</span><sub><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sub> <span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F</span>, i.e. choosing arbitrarily from among the <img 
src="theory39x.png" alt="|F|"  class="left" align="middle"><sup><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sup> distinct polynomials of degree no
more than <span 
class="cmmi-10">t </span><span 
class="cmsy-10">&minus; </span>1 with intercept <span 
class="cmmi-10">f</span>.
<!--l. 209--><p class="indent" >   In the first construction, given <span 
class="cmmi-10">x</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory40x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,x</span><sub><span 
class="cmmi-7">t</span></sub><span 
class="cmmi-10">,f </span>we obtain a polynomial by choosing arbitrary <span 
class="cmmi-10">y</span><sub><span 
class="cmr-7">1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory41x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,y</span><sub><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sub> <span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F</span>. The
polynomials chosen are necessarily distinct since no polynomial can pass through both (<span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">i</span></sub><span 
class="cmmi-10">,p</span>) and (<span 
class="cmmi-10">x</span><sub><span 
class="cmmi-7">i</span></sub><span 
class="cmmi-10">,q</span>) for any <span 
class="cmmi-10">p</span><span 
class="cmmi-10">&ne;</span><span 
class="cmmi-10">q</span>, so by
choosing each <span 
class="cmmi-10">y</span><sub><span 
class="cmmi-7">i</span></sub> from among the <img 
src="theory42x.png" alt="|F |"  class="left" align="middle"> elements of <span 
class="cmmi-10">F</span>, we choose arbitrarily from a set of <img 
src="theory43x.png" alt="|F |"  class="left" align="middle"><sup><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sup> distinct polynomials whose
intercepts are all <span 
class="cmmi-10">f</span>.
<!--l. 216--><p class="indent" >   Since there are only <img 
src="theory44x.png" alt="|F |"  class="left" align="middle"><sup><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sup> such polynomials, each construction chooses arbitrarily from among the same set, and by
the pigeonhole principle there exists a bijective mapping between sets of arbitrary <span 
class="cmmi-10">y </span>values in the first construction and sets
of arbitrary coefficients in the second.
<!--l. 222--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">7   </span> <a 
 id="x1-100007"></a>Theorem: With at least <span 
class="cmmi-10">t </span>pieces the secret is recoverable</h3>
<!--l. 224--><p class="noindent" >Let <span 
class="cmmi-10">B </span><span 
class="cmsy-10">&sub;</span><img 
src="theory45x.png" alt="{(x ,y),&sdot;&sdot;&sdot;,(x ,y )}
   1 1       s  s"  class="left" align="middle"> with <span 
class="cmsy-10">|</span><span 
class="cmmi-10">B</span><span 
class="cmsy-10">| </span>= <span 
class="cmmi-10">t</span>. Then <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">B</span></sub>(0) = <span 
class="cmmi-10">c</span>.
<!--l. 227--><p class="indent" >   Further, if <span 
class="cmmi-10">B</span><sup><span 
class="cmsy-7">&prime;</span></sup><span 
class="cmsy-10">&sub;</span><img 
src="theory46x.png" alt="{(x ,y ),&sdot;&sdot;&sdot;,(x ,y )}
   1  1      s  s"  class="left" align="middle"> with <span 
class="cmsy-10">|</span><span 
class="cmmi-10">B</span><sup><span 
class="cmsy-7">&prime;</span></sup><span 
class="cmsy-10">| </span><span 
class="cmmi-10">&#x003E; t</span>, then for every subset <span 
class="cmmi-10">B </span>of <span 
class="cmmi-10">B</span><sup><span 
class="cmsy-7">&prime;</span></sup> with <span 
class="cmsy-10">|</span><span 
class="cmmi-10">B</span><span 
class="cmsy-10">| </span>= <span 
class="cmmi-10">t</span>, <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">B</span></sub>(0) = <span 
class="cmmi-10">f</span>.
<!--l. 231--><p class="noindent" >
   <h4 class="subsectionHead"><span class="titlemark">7.1   </span> <a 
 id="x1-110007.1"></a>Proof</h4>
<!--l. 233--><p class="noindent" >The second part is trivially implied by the first.
<!--l. 235--><p class="indent" >   Recall that <span 
class="cmmi-10">X </span>= <img 
src="theory47x.png" alt="{(x1,y1),&sdot;&sdot;&sdot;,(xt,yt)}"  class="left" align="middle"> and that <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">X</span></sub>(0) = <span 
class="cmmi-10">f</span>. If <span 
class="cmmi-10">B </span>= <span 
class="cmmi-10">X </span>the result is true. If not, repeatedly
apply Lemma 2 to replace an element of <span 
class="cmmi-10">X </span>not in <span 
class="cmmi-10">B </span>with an element of <span 
class="cmmi-10">B </span>not in <span 
class="cmmi-10">X</span>, preserving the value of
<span 
class="cmmi-10">P</span>(0).
<!--l. 240--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">8   </span> <a 
 id="x1-120008"></a>Theorem: With fewer than <span 
class="cmmi-10">t </span>pieces no information is gained</h3>
<!--l. 242--><p class="noindent" >Let <span 
class="cmmi-10">C </span><span 
class="cmsy-10">&sub;</span><img 
src="theory48x.png" alt="{(x1,y1),&sdot;&sdot;&sdot;,(xs,ys)}"  class="left" align="middle"> with <span 
class="cmsy-10">|</span><span 
class="cmmi-10">C</span><span 
class="cmsy-10">| </span><span 
class="cmmi-10">&#x003C; t</span>. Then for each <span 
class="cmmi-10">d </span><span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F</span>, there exists <span 
class="cmmi-10">D </span><span 
class="cmsy-10">&sup; </span><span 
class="cmmi-10">C</span>, <span 
class="cmsy-10">|</span><span 
class="cmmi-10">D</span><span 
class="cmsy-10">| </span>= <span 
class="cmmi-10">t</span>, such that
<span 
class="cmmi-10">d </span>= <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">D</span></sub>(0).
<!--l. 246--><p class="indent" >   (In other words, any <span 
class="cmmi-10">d </span><span 
class="cmsy-10">&isin; </span><span 
class="cmmi-10">F </span>remains a possible value for the secret, so an attacker with fewer than <span 
class="cmmi-10">t </span>shares has gained no
information.)
<!--l. 249--><p class="noindent" >
   <h4 class="subsectionHead"><span class="titlemark">8.1   </span> <a 
 id="x1-130008.1"></a>Proof</h4>
<!--l. 251--><p class="noindent" >Let <span 
class="cmmi-10">a</span><sub><span 
class="cmmi-7">i</span></sub>, <span 
class="cmmi-10">b</span><sub><span 
class="cmmi-7">i</span></sub> be such that <span 
class="cmmi-10">C </span>= <img 
src="theory49x.png" alt="{(a1,b1),&sdot;&sdot;&sdot;,(an,bn)}"  class="left" align="middle">, some <span 
class="cmmi-10">n &#x003C; t</span>. Choose arbitrary <span 
class="cmmi-10">a</span><sub><span 
class="cmmi-7">n</span><span 
class="cmr-7">+1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory50x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,a</span><sub><span 
class="cmmi-7">t</span></sub> and arbitrary <span 
class="cmmi-10">b</span><sub><span 
class="cmmi-7">n</span><span 
class="cmr-7">+1</span></sub><span 
class="cmmi-10">,</span><img 
src="theory51x.png" alt="&sdot;&sdot;&sdot;"  class="cdots" ><span 
class="cmmi-10">,b</span><sub><span 
class="cmmi-7">t</span><span 
class="cmsy-7">&minus;</span><span 
class="cmr-7">1</span></sub>.
Let <span 
class="cmmi-10">b</span><sub><span 
class="cmmi-7">t</span></sub> be chosen by applying Lemma 1 with <span 
class="cmmi-10">c </span>:= <span 
class="cmmi-10">d</span>. Then by choice of <span 
class="cmmi-10">b</span><sub><span 
class="cmmi-7">t</span></sub>, <span 
class="cmmi-10">P</span><sub><span 
class="cmmi-7">C</span></sub>(0) = <span 
class="cmmi-10">d </span>as required.
<!--l. 256--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">9   </span> <a 
 id="x1-140009"></a>Implementation in <span 
class="cmmi-10">GF</span>(2<sup><span 
class="cmr-7">8</span></sup>)</h3>
<!--l. 258--><p class="noindent" >The program <span 
class="cmtt-10">test</span><span 
class="cmtt-10">_gfshare</span><span 
class="cmtt-10">_isfield</span>, compiled and run by <span 
class="cmtt-10">make check</span>, demonstrates that the calculations done by
libgfshare are indeed performed in a field.
<!--l. 262--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">10   </span> <a 
 id="x1-1500010"></a>Attacks not addressed</h3>
<!--l. 264--><p class="noindent" >This document has not addressed the following:
<!--l. 266--><p class="indent" >   - Attacks based on the use of a predictable or partially predictable pseudorandom number generator might be
possible.
<!--l. 269--><p class="indent" >   - In the implementation used in libgfshare, the field <span 
class="cmmi-10">F </span>is the field of byte values, with addition being bitwise exclusive-or,
and multiplication as usual; each byte of the secret is shared separately by applying this algorithm separately.
This means that when a secret file is shared, the length in bytes of each share equals the length in bytes of
the secret. If the length of the secret is itself secret, it should be padded to some standard length before
sharing.
<!--l. 277--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">11   </span> <a 
 id="x1-1600011"></a>References</h3>
<!--l. 279--><p class="noindent" >[SHAMIR] Adi Shamir, &#8221;How to share a secret&#8221;, Communications of the ACM, 22(1), pp612&#8211;613, 1979. Available at
<a 
href="http://www.cs.tau.ac.il/~bchor/Shamir.html" class="url" ><span 
class="cmtt-10">http://www.cs.tau.ac.il/</span><span 
class="cmtt-10">~</span><span 
class="cmtt-10">bchor/Shamir.html</span></a>
<!--l. 281--><p class="noindent" >
   <h3 class="sectionHead"><span class="titlemark">12   </span> <a 
 id="x1-1700012"></a>Copyright and disclaimer</h3>
<!--l. 283--><p class="noindent" >Copyright 2006 Simon McVittie, <a 
href="http://smcv.pseudorandom.co.uk/" class="url" ><span 
class="cmtt-10">http://smcv.pseudorandom.co.uk/</span></a>
<!--l. 285--><p class="indent" >   Permission is hereby granted, free of charge, to any person obtaining a copy of this software and associated
documentation files (the &#8221;Software&#8221;), to deal in the Software without restriction, including without limitation the rights to
use, copy, modify, merge, publish, distribute, sublicense, and/or sell copies of the Software, and to permit persons to whom
the Software is furnished to do so, subject to the following conditions:
<!--l. 292--><p class="indent" >   The above copyright notice and this permission notice shall be included in all copies or substantial portions of the
Software.
<!--l. 295--><p class="indent" >   THE SOFTWARE IS PROVIDED &#8221;AS IS&#8221;, WITHOUT WARRANTY OF ANY KIND, EXPRESS OR IMPLIED,
INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, FITNESS FOR A
PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT
HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF
CONTRACT, TORT OR OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE
OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
    
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