[PATCH] Better PLL frequency matching for tdfxfb driver

[PATCH] Better PLL frequency matching for tdfxfb driver

[Richard Drummond]

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Attached is a patch against 2.6.11.7 which improves the PLL frequency matching 
in the tdfxfb driver. Instead of requiring 64260 iterations to obtain the 
closest supported PLL frequency, this code does it with the same degree of 
accuracy in at most 768 iterations.

Signed-off-by: Richard Drummond <evilrich@rcdrummond.net>

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--- drivers/video/tdfxfb.c_orig	2005-04-20 15:04:23.000000000 -0500
+++ drivers/video/tdfxfb.c	2005-04-21 09:26:18.000000000 -0500
@@ -320,30 +320,40 @@
 
 static u32 do_calc_pll(int freq, int* freq_out) 
 {
-	int m, n, k, best_m, best_n, best_k, f_cur, best_error;
+	int m, n, k, best_m, best_n, best_k, best_error;
 	int fref = 14318;
   
-	/* this really could be done with more intelligence --
-	   255*63*4 = 64260 iterations is silly */
 	best_error = freq;
 	best_n = best_m = best_k = 0;
-	for (n = 1; n < 256; n++) {
-		for (m = 1; m < 64; m++) {
-			for (k = 0; k < 4; k++) {
-				f_cur = fref*(n + 2)/(m + 2)/(1 << k);
-				if (abs(f_cur - freq) < best_error) {
-					best_error = abs(f_cur-freq);
-					best_n = n;
-					best_m = m;
-					best_k = k;
-				}
+   
+	for (k = 3; k >= 0; k--) {	
+		for (m = 63; m >= 0; m--) {
+			/* Estimate value of n that produces target frequency with current m and k */
+			int n_estimated = (freq * (m + 2) * (1 << k) / fref) - 2;
+	     
+			/* Search neighborhood of estimated n */
+			for (n = max(0, n_estimated - 1); n <= min(255, n_estimated + 1); n++) {
+				/* Calculate PLL freqency with current m, k and estimated n */
+				int f     = fref * (n + 2) / (m + 2) / (1 << k);
+				int error = abs (f - freq);
+		  
+				/* If this is the closest we've come to the target frequency
+				 * then remember n, m and k */
+				if (error  < best_error) {
+					best_error = error;
+					best_n     = n;
+					best_m     = m;
+					best_k     = k;
+				}  
 			}
 		}
 	}
+   
 	n = best_n;
 	m = best_m;
 	k = best_k;
 	*freq_out = fref*(n + 2)/(m + 2)/(1 << k);
+   
 	return (n << 8) | (m << 2) | k;
 }