From f7055d26925e515c8a3ffd139e9791f6b763b546 Mon Sep 17 00:00:00 2001
From: alwaysprince05 <prince.p24@medhaviskillsuniversity.edu.in>
Date: Sat, 5 Sep 2026 00:46:51 +0530
Subject: [PATCH 1/2] fix(examples): replace switch statement with
 std::popcount in sudoku example
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Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

The examine_potentials function in sudoku.cpp used a switch statement that
only handled powers of 2 (1, 2, 4, 8, 16, 32, 64, 128, 256). If the
potential set was not a power of 2, it hit the default case which asserted.

This caused the sudoku example to crash in Debug mode with:
  Assertion failed: (!"potential set is not a power of 2")

The fix uses std::popcount and std::countr_zero to handle non-power-of-2
values gracefully. When popcount == 1, it's a singleton and can be solved.
When popcount > 1, it's normal during solving and continues.

Fixes #2105

🤖 Generated with Codebuff
Co-Authored-By: Codebuff <noreply@codebuff.com>
---
 examples/sudoku.cpp | 65 +++++----------------------------------------
 1 file changed, 6 insertions(+), 59 deletions(-)

diff --git a/examples/sudoku.cpp b/examples/sudoku.cpp
index 36f028f54..4bddd7c8c 100644
--- a/examples/sudoku.cpp
+++ b/examples/sudoku.cpp
@@ -24,6 +24,7 @@
 #include <cstdlib>
 
 #include <atomic>
+#include <bit>
 #include <cassert>
 #include <chrono>
 #include <string_view>
@@ -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 (b[i].potential_set != 0 && 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;

From 9fdf92ec13bd01ae4d9a67b8459447d6ded1cde8 Mon Sep 17 00:00:00 2001
From: alwaysprince05 <prince.p24@medhaviskillsuniversity.edu.in>
Date: Sat, 5 Sep 2026 04:59:50 +0530
Subject: [PATCH 2/2] fix(examples): simplify popcount check in sudoku
 examine_potentials
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit

Remove redundant `potential_set != 0` guard — std::popcount(0) == 0
which is never == 1, so the check was always false when potential_set
was zero. Addresses review feedback from @ericniebler.

🤖 Generated with Codebuff
Co-Authored-By: Codebuff <noreply@codebuff.com>
---
 examples/sudoku.cpp | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/examples/sudoku.cpp b/examples/sudoku.cpp
index 4bddd7c8c..a6a295198 100644
--- a/examples/sudoku.cpp
+++ b/examples/sudoku.cpp
@@ -223,7 +223,7 @@ bool examine_potentials(board_element *b, bool *progress)
     if (b[i].solved_element == 0 && b[i].potential_set == 0)  // empty set
       return false;
     // Check if potential_set is a power of 2 (only one bit set)
-    if (b[i].potential_set != 0 && std::popcount(b[i].potential_set) == 1)
+    if (std::popcount(b[i].potential_set) == 1)
     {
       // It's a singleton - solve it
       // countr_zero gives the bit position (0-indexed), add 1 to get the element value