Program: 21009
Project: 21009
Buggy source:
/** 
 * Encodes a given field element in its 32 byte representation. This is done in two steps: <ol> <li>Reduce the value of the field element modulo $p$. <li>Convert the field element to the 32 byte representation. </ol><p> The idea for the modulo $p$ reduction algorithm is as follows: <p> <h2>Assumption:</h2> <p><ul> <li>$p = 2^{255} - 19$ <li>$h = h_0 + 2^{25} * h_1 + 2^{(26+25)} * h_2 + \dots + 2^{230} * h_9$ where $0 \le |h_i| \lt 2^{27}$ for all $i=0,\dots,9$. <li>$h \cong r \mod p$, i.e. $h = r + q * p$ for some suitable $0 \le r \lt p$ and an integer $q$. </ul><p> Then $q = [2^{-255} * (h + 19 * 2^{-25} * h_9 + 1/2)]$ where $[x] = floor(x)$. <p> <h2>Proof:</h2> <p> We begin with some very raw estimation for the bounds of some expressions: <p> $$ \begin{equation} |h| \lt 2^{230} * 2^{30} = 2^{260} \Rightarrow |r + q * p| \lt 2^{260} \Rightarrow |q| \lt 2^{10}. \\ \Rightarrow -1/4 \le a := 19^2 * 2^{-255} * q \lt 1/4. \\ |h - 2^{230} * h_9| = |h_0 + \dots + 2^{204} * h_8| \lt 2^{204} * 2^{30} = 2^{234}. \\ \Rightarrow -1/4 \le b := 19 * 2^{-255} * (h - 2^{230} * h_9) \lt 1/4 \end{equation} $$ <p> Therefore $0 \lt 1/2 - a - b \lt 1$. <p> Set $x := r + 19 * 2^{-255} * r + 1/2 - a - b$. Then: <p> $$ 0 \le x \lt 255 - 20 + 19 + 1 = 2^{255} \\ \Rightarrow 0 \le 2^{-255} * x \lt 1. $$ <p> Since $q$ is an integer we have <p> $$ [q + 2^{-255} * x] = q \quad (1) $$ <p> Have a closer look at $x$: <p> $$ \begin{align} x &amp;= h - q * (2^{255} - 19) + 19 * 2^{-255} * (h - q * (2^{255} - 19)) + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * (h - 2^{230} * h_9) \\ &amp;= h - q * 2^{255} + 19 * q + 19 * 2^{-255} * h - 19 * q + 19^2 * 2^{-255} * q + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * h + 19 * 2^{-25} * h_9 \\ &amp;= h + 19 * 2^{-25} * h_9 + 1/2 - q^{255}. \end{align} $$ <p> Inserting the expression for $x$ into $(1)$ we get the desired expression for $q$.
 */
public byte[] encode(FieldElement x){
  int[] h=((Ed25519FieldElement)x).t;
  int h0=h[0];
  int h1=h[1];
  int h2=h[2];
  int h3=h[3];
  int h4=h[4];
  int h5=h[5];
  int h6=h[6];
  int h7=h[7];
  int h8=h[8];
  int h9=h[9];
  int q;
  int carry0;
  int carry1;
  int carry2;
  int carry3;
  int carry4;
  int carry5;
  int carry6;
  int carry7;
  int carry8;
  int carry9;
  q=(19 * h9 + (1 << 24)) >> 25;
  q=(h0 + q) >> 26;
  q=(h1 + q) >> 25;
  q=(h2 + q) >> 26;
  q=(h3 + q) >> 25;
  q=(h4 + q) >> 26;
  q=(h5 + q) >> 25;
  q=(h6 + q) >> 26;
  q=(h7 + q) >> 25;
  q=(h8 + q) >> 26;
  q=(h9 + q) >> 25;
  h0+=19 * q;
  carry0=h0 >> 26;
  h1+=carry0;
  h0-=carry0 << 26;
  carry1=h1 >> 25;
  h2+=carry1;
  h1-=carry1 << 25;
  carry2=h2 >> 26;
  h3+=carry2;
  h2-=carry2 << 26;
  carry3=h3 >> 25;
  h4+=carry3;
  h3-=carry3 << 25;
  carry4=h4 >> 26;
  h5+=carry4;
  h4-=carry4 << 26;
  carry5=h5 >> 25;
  h6+=carry5;
  h5-=carry5 << 25;
  carry6=h6 >> 26;
  h7+=carry6;
  h6-=carry6 << 26;
  carry7=h7 >> 25;
  h8+=carry7;
  h7-=carry7 << 25;
  carry8=h8 >> 26;
  h9+=carry8;
  h8-=carry8 << 26;
  carry9=h9 >> 25;
  h9-=carry9 << 25;
  byte[] s=new byte[32];
  s[0]=(byte)h0;
  s[1]=(byte)(h0 >> 8);
  s[2]=(byte)(h0 >> 16);
  s[3]=(byte)((h0 >> 24) | (h1 << 2));
  s[4]=(byte)(h1 >> 6);
  s[5]=(byte)(h1 >> 14);
  s[6]=(byte)((h1 >> 22) | (h2 << 3));
  s[7]=(byte)(h2 >> 5);
  s[8]=(byte)(h2 >> 13);
  s[9]=(byte)((h2 >> 21) | (h3 << 5));
  s[10]=(byte)(h3 >> 3);
  s[11]=(byte)(h3 >> 11);
  s[12]=(byte)((h3 >> 19) | (h4 << 6));
  s[13]=(byte)(h4 >> 2);
  s[14]=(byte)(h4 >> 10);
  s[15]=(byte)(h4 >> 18);
  s[16]=(byte)h5;
  s[17]=(byte)(h5 >> 8);
  s[18]=(byte)(h5 >> 16);
  s[19]=(byte)((h5 >> 24) | (h6 << 1));
  s[20]=(byte)(h6 >> 7);
  s[21]=(byte)(h6 >> 15);
  s[22]=(byte)((h6 >> 23) | (h7 << 3));
  s[23]=(byte)(h7 >> 5);
  s[24]=(byte)(h7 >> 13);
  s[25]=(byte)((h7 >> 21) | (h8 << 4));
  s[26]=(byte)(h8 >> 4);
  s[27]=(byte)(h8 >> 12);
  s[28]=(byte)((h8 >> 20) | (h9 << 6));
  s[29]=(byte)(h9 >> 2);
  s[30]=(byte)(h9 >> 10);
  s[31]=(byte)(h9 >> 18);
  return s;
}

Fixed version (ground truth):
/** 
 * Encodes a given field element in its 32 byte representation. This is done in two steps: <ol> <li>Reduce the value of the field element modulo $p$. <li>Convert the field element to the 32 byte representation. </ol><p> The idea for the modulo $p$ reduction algorithm is as follows: </p> <h2>Assumption:</h2> <ul> <li>$p = 2^{255} - 19$ <li>$h = h_0 + 2^{25} * h_1 + 2^{(26+25)} * h_2 + \dots + 2^{230} * h_9$ where $0 \le |h_i| \lt 2^{27}$ for all $i=0,\dots,9$. <li>$h \cong r \mod p$, i.e. $h = r + q * p$ for some suitable $0 \le r \lt p$ and an integer $q$. </ul><p> Then $q = [2^{-255} * (h + 19 * 2^{-25} * h_9 + 1/2)]$ where $[x] = floor(x)$. </p> <h2>Proof:</h2> <p> We begin with some very raw estimation for the bounds of some expressions: <p> $$ \begin{equation} |h| \lt 2^{230} * 2^{30} = 2^{260} \Rightarrow |r + q * p| \lt 2^{260} \Rightarrow |q| \lt 2^{10}. \\ \Rightarrow -1/4 \le a := 19^2 * 2^{-255} * q \lt 1/4. \\ |h - 2^{230} * h_9| = |h_0 + \dots + 2^{204} * h_8| \lt 2^{204} * 2^{30} = 2^{234}. \\ \Rightarrow -1/4 \le b := 19 * 2^{-255} * (h - 2^{230} * h_9) \lt 1/4 \end{equation} $$ <p> Therefore $0 \lt 1/2 - a - b \lt 1$. <p> Set $x := r + 19 * 2^{-255} * r + 1/2 - a - b$. Then: <p> $$ 0 \le x \lt 255 - 20 + 19 + 1 = 2^{255} \\ \Rightarrow 0 \le 2^{-255} * x \lt 1. $$ <p> Since $q$ is an integer we have <p> $$ [q + 2^{-255} * x] = q \quad (1) $$ <p> Have a closer look at $x$: <p> $$ \begin{align} x &amp;= h - q * (2^{255} - 19) + 19 * 2^{-255} * (h - q * (2^{255} - 19)) + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * (h - 2^{230} * h_9) \\ &amp;= h - q * 2^{255} + 19 * q + 19 * 2^{-255} * h - 19 * q + 19^2 * 2^{-255} * q + 1/2 - 19^2 * 2^{-255} * q - 19 * 2^{-255} * h + 19 * 2^{-25} * h_9 \\ &amp;= h + 19 * 2^{-25} * h_9 + 1/2 - q^{255}. \end{align} $$ <p> Inserting the expression for $x$ into $(1)$ we get the desired expression for $q$.
 */
public byte[] encode(FieldElement x){
  int[] h=((Ed25519FieldElement)x).t;
  int h0=h[0];
  int h1=h[1];
  int h2=h[2];
  int h3=h[3];
  int h4=h[4];
  int h5=h[5];
  int h6=h[6];
  int h7=h[7];
  int h8=h[8];
  int h9=h[9];
  int q;
  int carry0;
  int carry1;
  int carry2;
  int carry3;
  int carry4;
  int carry5;
  int carry6;
  int carry7;
  int carry8;
  int carry9;
  q=(19 * h9 + (1 << 24)) >> 25;
  q=(h0 + q) >> 26;
  q=(h1 + q) >> 25;
  q=(h2 + q) >> 26;
  q=(h3 + q) >> 25;
  q=(h4 + q) >> 26;
  q=(h5 + q) >> 25;
  q=(h6 + q) >> 26;
  q=(h7 + q) >> 25;
  q=(h8 + q) >> 26;
  q=(h9 + q) >> 25;
  h0+=19 * q;
  carry0=h0 >> 26;
  h1+=carry0;
  h0-=carry0 << 26;
  carry1=h1 >> 25;
  h2+=carry1;
  h1-=carry1 << 25;
  carry2=h2 >> 26;
  h3+=carry2;
  h2-=carry2 << 26;
  carry3=h3 >> 25;
  h4+=carry3;
  h3-=carry3 << 25;
  carry4=h4 >> 26;
  h5+=carry4;
  h4-=carry4 << 26;
  carry5=h5 >> 25;
  h6+=carry5;
  h5-=carry5 << 25;
  carry6=h6 >> 26;
  h7+=carry6;
  h6-=carry6 << 26;
  carry7=h7 >> 25;
  h8+=carry7;
  h7-=carry7 << 25;
  carry8=h8 >> 26;
  h9+=carry8;
  h8-=carry8 << 26;
  carry9=h9 >> 25;
  h9-=carry9 << 25;
  byte[] s=new byte[32];
  s[0]=(byte)h0;
  s[1]=(byte)(h0 >> 8);
  s[2]=(byte)(h0 >> 16);
  s[3]=(byte)((h0 >> 24) | (h1 << 2));
  s[4]=(byte)(h1 >> 6);
  s[5]=(byte)(h1 >> 14);
  s[6]=(byte)((h1 >> 22) | (h2 << 3));
  s[7]=(byte)(h2 >> 5);
  s[8]=(byte)(h2 >> 13);
  s[9]=(byte)((h2 >> 21) | (h3 << 5));
  s[10]=(byte)(h3 >> 3);
  s[11]=(byte)(h3 >> 11);
  s[12]=(byte)((h3 >> 19) | (h4 << 6));
  s[13]=(byte)(h4 >> 2);
  s[14]=(byte)(h4 >> 10);
  s[15]=(byte)(h4 >> 18);
  s[16]=(byte)h5;
  s[17]=(byte)(h5 >> 8);
  s[18]=(byte)(h5 >> 16);
  s[19]=(byte)((h5 >> 24) | (h6 << 1));
  s[20]=(byte)(h6 >> 7);
  s[21]=(byte)(h6 >> 15);
  s[22]=(byte)((h6 >> 23) | (h7 << 3));
  s[23]=(byte)(h7 >> 5);
  s[24]=(byte)(h7 >> 13);
  s[25]=(byte)((h7 >> 21) | (h8 << 4));
  s[26]=(byte)(h8 >> 4);
  s[27]=(byte)(h8 >> 12);
  s[28]=(byte)((h8 >> 20) | (h9 << 6));
  s[29]=(byte)(h9 >> 2);
  s[30]=(byte)(h9 >> 10);
  s[31]=(byte)(h9 >> 18);
  return s;
}