Repatch (from linux) of OSX IRAF

1 files changed, 63 insertions, 0 deletions

diff --git a/sys/vops/lz/asoks.x b/sys/vops/lz/asoks.x
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+# Copyright(c) 1986 Association of Universities for Research in Astronomy Inc.
+
+include <mach.h>
+
+# ASOK -- Select the Kth smallest element from a vector.  The algorithm used
+# is selection by tail recursion (Gonnet 1984).  In each iteration a pivot key
+# is selected (somewhat arbitrarily) from the array.  The array is then split
+# into two subarrays, those with key values less than or equal to the pivot key
+# and those with values greater than the pivot.  The size of the two subarrays
+# determines which contains the median value, and the process is repeated
+# on that subarray, and so on until all of the elements of the subarray
+# are equal, e.g., there is only one element left in the subarray.  For a
+# randomly ordered array the expected running time is O(3.38N).  The selection
+# is carried out in place, leaving the array in a partially ordered state.
+#
+# N.B.: Behaviour is O(N) if the input array is sorted.
+# N.B.: The cases ksel=1 and ksel=npix, i.e., selection of the minimum and
+# maximum values, are more efficiently handled by ALIM which is O(2N).
+#
+# Jul99 - The above algorithm was found to be pathologically slow in cases
+# where many or all elements of the array are equal.  The version of the
+# algorithm below, from Wirth, appears to avoid this problem.
+
+short procedure asoks (a, npix, ksel)
+
+short	a[ARB]			# input array
+int	npix			# number of pixels
+int	ksel			# element to be selected
+
+int	lo, up, i, j, k, dummy
+short	temp, wtemp
+
+begin
+	lo = 1
+	up = npix
+	k  = max (lo, min (up, ksel))
+
+	# while (lo < up)
+	do dummy = 1, MAX_INT {
+	    if (! (lo < up))
+		break
+
+	    temp = a[k];  i = lo;  j = up
+
+	    repeat {
+		while (a[i] < temp)
+		    i = i + 1
+		while (temp < a[j])
+		    j = j - 1
+		if (i <= j) {
+		    wtemp = a[i];  a[i] = a[j];  a[j] = wtemp
+		    i = i + 1;  j = j - 1
+		}
+	    } until (i > j)
+
+	    if (j < k)
+		lo = i
+	    if (k < i)
+		up = j
+	}
+
+	return (a[k])
+end