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/* Can we get down to 8 registers? Possible strategy:
*
* Do what's needed to compute y3. At this point, we need to keep
* inputs, t01, t04, y3, so we only have one temporary left (and we
* discard t02, recomputing it frpm x3 when needed later.
*
* Compute y1: Discard t04. Use one register to compute t08 and t10.
* Use the other to compute t05 (then but t15 in x2), and t09. We can
* then get to y1, and have one spare register.
*
* Put t13 in x1. Use the spare register to compute t04 (again), t11,
* t16 and y2. The rest should be easy, y0 can be done in parallel
* with y2.
*/
strict digraph sbox6 {
x0 [shape=box];
x1 [shape=box];
x2 [shape=box];
x3 [shape=box];
t01 [label="\N\ny0\n1"];
t04 [label="\N\ny2\n2"];
t06 [label="\N\ny3\n3"];
t02 [label="\N\ny1,x3\n4,18"];
t03 [label="\N\ny1\n5"];
y3 [label="\N\n6", shape=box];
t11 [label="\N\ny2\n7"];
t08 [label="\N\ny1\n8"];
t10 [label="\N\ny1\n9"];
t05 [label="\N\nt\n10"];
t15 [label="\N\nx2\n11"];
t09 [label="\N\nt\n12"];
y1 [label="\N\n13", shape=box];
t13 [label="\N\nx1\n14"];
t16 [label="\N\ny2\n15"];
y2 [label="\N\n16", shape=box];
t14 [label="\N\ny0\n17"];
t17 [label="\N\ny0\n18"];
y0 [label="\N\n19", shape=box];
x0 -> t01; x2 -> t01; // t01 = x0 & x2;
x3 -> t02; // t02 = ~ x3;
x0 -> t03; t02 -> t03; // t03 = x0 & t02;
x1 -> t04; t01 -> t04; // t04 = x1 | t01;
x0 -> t05; x1 -> t05; // t05 = x0 & x1;
x2 -> t06; t04 -> t06; // t06 = x2 ^ t04;
t03 -> y3; t06 -> y3; // y3 = t03 ^ t06;
x2 -> t08; y3 -> t08; // t08 = x2 | y3;
x3 -> t09; t05 -> t09; // t09 = x3 | t05;
x0 -> t10; t08 -> t10; // t10 = x0 ^ t08;
t04 -> t11; y3 -> t11; // t11 = t04 & y3;
t09 -> y1; t10 -> y1; // y1 = t09 ^ t10;
x1 -> t13; y1 -> t13; // t13 = x1 ^ y1;
t01 -> t14; y1 -> t14; // t14 = t01 ^ y1;
x2 -> t15; t05 -> t15; // t15 = x2 ^ t05;
t11 -> t16; t13 -> t16; // t16 = t11 | t13;
t02 -> t17; t14 -> t17; // t17 = t02 | t14;
t15 -> y0; t17 -> y0; // y0 = t15 ^ t17;
x0 -> y2; t16 -> y2; // y2 = x0 ^ t16;
}
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