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fix(examples): replace switch statement with std::popcount in sudoku example #2251

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fix(examples): replace switch statement with std::popcount in sudoku example #2251

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fix(examples): replace switch statement with std::popcount in sudoku …
alwaysprince05 Sep 4, 2026
9fdf92e
fix(examples): simplify popcount check in sudoku examine_potentials
alwaysprince05 Sep 4, 2026
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65 changes: 6 additions & 59 deletions examples/sudoku.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -24,6 +24,7 @@
#include <cstdlib>

#include <atomic>
#include <bit>
#include <cassert>
#include <chrono>
#include <string_view>
Expand Down Expand Up @@ -221,67 +222,13 @@ bool examine_potentials(board_element *b, bool *progress)
{
if (b[i].solved_element == 0 && b[i].potential_set == 0) // empty set
return false;
switch (b[i].potential_set)
// Check if potential_set is a power of 2 (only one bit set)
if (std::popcount(b[i].potential_set) == 1)
{
case 1:
{
b[i].solved_element = 1;
singletons = true;
break;
}
case 2:
{
b[i].solved_element = 2;
singletons = true;
break;
}
case 4:
{
b[i].solved_element = 3;
singletons = true;
break;
}
case 8:
{
b[i].solved_element = 4;
singletons = true;
break;
}
case 16:
{
b[i].solved_element = 5;
singletons = true;
break;
}
case 32:
{
b[i].solved_element = 6;
singletons = true;
break;
}
case 64:
{
b[i].solved_element = 7;
singletons = true;
break;
}
case 128:
{
b[i].solved_element = 8;
singletons = true;
break;
}
case 256:
{
b[i].solved_element = 9;
// It's a singleton - solve it
// countr_zero gives the bit position (0-indexed), add 1 to get the element value
b[i].solved_element = std::countr_zero(b[i].potential_set) + 1;
singletons = true;
break;
}
default:
{
assert(!"potential set is not a power of 2");
break;
}
}
}
*progress = singletons;
Expand Down