[numpy_vs_numba_vs_jax] Fix issue in parallel Numba implementation

The previous implementation had a race condition where multiple threads
simultaneously updated the shared variable 'm', causing incorrect results
(returning -inf instead of the actual maximum).

The fix computes per-row maximums in parallel (each thread writes to a
unique index), then reduces to find the global maximum.

Changes:
- Replace shared variable 'm' with thread-safe 'row_maxes' array
- Each parallel iteration computes a thread-local 'row_max'
- Final np.max(row_maxes) combines partial results
- Remove broken nested prange version (same race condition)
- Add explanatory text about avoiding race conditions in reductions

Author: mmcky

lectures/numpy_vs_numba_vs_jax.md

@@ -200,23 +200,28 @@ working with a single one-dimensional grid.
 
 ### Parallelized Numba
 
-Now let's try parallelization with Numba using `prange`:
+Now let's try parallelization with Numba using `prange`.
 
-First we parallelize just the outer loop.
+When parallelizing a reduction (like finding a maximum), we need to avoid race conditions
+where multiple threads try to update the same variable simultaneously.
+
+The solution is to compute partial results (row maximums) in parallel, then combine them.
 
 ```{code-cell} ipython3
 @numba.jit(parallel=True)
 def compute_max_numba_parallel(grid):
     n = len(grid)
-    m = -np.inf
+    row_maxes = np.empty(n)
     for i in numba.prange(n):
+        row_max = -np.inf
         for j in range(n):
             x = grid[i]
             y = grid[j]
             z = np.cos(x**2 + y**2) / (1 + x**2 + y**2)
-            if z > m:
-                m = z
-    return m
+            if z > row_max:
+                row_max = z
+        row_maxes[i] = row_max
+    return np.max(row_maxes)
 
 with qe.Timer(precision=8):
     compute_max_numba_parallel(grid)
@@ -228,32 +233,6 @@ with qe.Timer(precision=8):
     compute_max_numba_parallel(grid)
 ```
 
-Next we parallelize both loops.
-
-```{code-cell} ipython3
-@numba.jit(parallel=True)
-def compute_max_numba_parallel_nested(grid):
-    n = len(grid)
-    m = -np.inf
-    for i in numba.prange(n):
-        for j in numba.prange(n):
-            x = grid[i]
-            y = grid[j]
-            z = np.cos(x**2 + y**2) / (1 + x**2 + y**2)
-            if z > m:
-                m = z
-    return m
-
-with qe.Timer(precision=8):
-    compute_max_numba_parallel_nested(grid)
-```
-
-```{code-cell} ipython3
-with qe.Timer(precision=8):
-    compute_max_numba_parallel_nested(grid)
-```
-
-
 Depending on your machine, you might or might not see large benefits from parallelization here.
 
 If you have a small number of cores, the overhead of thread management and synchronization can