问题
Note: Please don't interpret this as "homework question." This is just a thing I curious to know :)
The median of five is sometimes used as an exercise in algorithm design and is known to be computable using only 6 comparisons.
What is the best way to implement this "median of five using 6 comparisons" in C# ? All of my attempts seem to result in awkward code :( I need nice and readable code while still using only 6 comparisons.
public double medianOfFive(double a, double b, double c, double d, double e){
//
// return median
//
return c;
}
Note: I think I should provide the "algorithm" here too:
I found myself not able to explain the algorithm clearly as Azereal did in his forum post. So I will reference his post here. From http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1061827085
Well I was posed this problem in one of my assignments and I turned to this forum for help but no help was here. I eventually found out how to do it.
Start a mergesort with the first 4 elements and order each pair (2 comparisons)
Compare the two lower ones of each pair and eliminate the lowest one from the possibilities (3 comparisons)
Add in the 5th number set aside to the number without a pair and compare the two (4 comparisons)
Compare the two lowest of the two new pairs and eliminate the lower one (5 comparisons)
Compare the one by itself and the lower of the last pair and the lower number is the median
The possible median is within the parentesis
(54321)
5:4 3:2 2 comparisons
(4<5 2<3 1)
4:2 3 comparisons
2(4<5 3 1)
1:3 4 comparisons
2(4<5 1<3)
4:1 5 comparisons
1,2(4<5 3)
4:3 6 comparisons
1,2(3)4,5
Three is the median
EDIT: As your request and to prevent myself from getting more downvotes, this are C++ code I wrote to find median of five. Don't mind it's awkwardness:
double StageGenerator::MedianOfFive(double n1, double n2, double n3, double n4, double n5){
double *a = &n1, *b = &n2, *c = &n3, *d = &n4, *e = &n5;
double *tmp;
// makes a < b and b < d
if(*b < *a){
tmp = a; a = b; b = tmp;
}
if(*d < *c){
tmp = c; c = d; d = tmp;
}
// eleminate the lowest
if(*c < *a){
tmp = b; b = d; d = tmp;
c = a;
}
// gets e in
a = e;
// makes a < b and b < d
if(*b < *a){
tmp = a; a = b; b = tmp;
}
// eliminate another lowest
// remaing: a,b,d
if(*a < *c){
tmp = b; b = d; d = tmp;
a = c;
}
if(*d < *a)
return *d;
else
return *a;
}
It should be more compact, isn't it ?
EDIT:
As @pablito pointed out in his answer. The built-in List.Sort() cannot fulfill this requirement since it uses up to 13 comparisons :]
回答1:
This is basically just factoring out the swapping and sorting code from your C++ example:
private static void Swap(ref double a, ref double b) {
double t = a;
a = b;
b = t;
}
private static void Sort(ref double a, ref double b) {
if (a > b) {
double t = a;
a = b;
b = t;
}
}
private static double MedianOfFive(double a, double b, double c, double d, double e){
// makes a < b and c < d
Sort(ref a, ref b);
Sort(ref c, ref d);
// eleminate the lowest
if (c < a) {
Swap(ref b, ref d);
c = a;
}
// gets e in
a = e;
// makes a < b
Sort(ref a, ref b);
// eliminate another lowest
// remaing: a,b,d
if (a < c) {
Swap(ref b, ref d);
a = c;
}
return Math.Min(d, a);
}
回答2:
I found this post interesting and as an exercise I created this which ONLY does 6 comparisons and NOTHING else:
static double MedianOfFive(double a, double b, double c, double d, double e)
{
return b < a ? d < c ? b < d ? a < e ? a < d ? e < d ? e : d
: c < a ? c : a
: e < d ? a < d ? a : d
: c < e ? c : e
: c < e ? b < c ? a < c ? a : c
: e < b ? e : b
: b < e ? a < e ? a : e
: c < b ? c : b
: b < c ? a < e ? a < c ? e < c ? e : c
: d < a ? d : a
: e < c ? a < c ? a : c
: d < e ? d : e
: d < e ? b < d ? a < d ? a : d
: e < b ? e : b
: b < e ? a < e ? a : e
: d < b ? d : b
: d < c ? a < d ? b < e ? b < d ? e < d ? e : d
: c < b ? c : b
: e < d ? b < d ? b : d
: c < e ? c : e
: c < e ? a < c ? b < c ? b : c
: e < a ? e : a
: a < e ? b < e ? b : e
: c < a ? c : a
: a < c ? b < e ? b < c ? e < c ? e : c
: d < b ? d : b
: e < c ? b < c ? b : c
: d < e ? d : e
: d < e ? a < d ? b < d ? b : d
: e < a ? e : a
: a < e ? b < e ? b : e
: d < a ? d : a;
}
回答3:
Thanks. I know your posts are quite old, but it was useful for my issue.
I needed a way to compute the median of 5 SSE/AVX registers (4 floats / 8 floats at once, or 2 doubles/4 doubles at once):
without any conditional jumps
only with min/max instructions
If the min/max functions are programmed for scalar registers with conditional jumps, my code is not optimal in term of comparisons. But if the min/max functions are coded with corresponding CPU instructions, my code is very effective because no conditional jump is done by the CPU when executing.
template<class V>
inline V median(const V &a, const V &b, const V &c)
{
return max(min(a,b),min(c,max(a,b)));
}
template<class V>
inline V median(const V &a, const V &b, const V &c, const V &d, const V &e)
{
V f=max(min(a,b),min(c,d)); // discards lowest from first 4
V g=min(max(a,b),max(c,d)); // discards biggest from first 4
return median(e,f,g);
}
回答4:
This is pretty ugly and could use some refactoring, but it explicitly walks through all the comparisons and swaps so you can see what's going on.
public double medianOfFive(double a, double b, double c, double d, double e){
double median;
// sort a and b
if(a > b) // comparison # 1
{
double temp = a;
a = b;
b = temp;
}
// sort c and d
if(c > d) // comparison # 2
{
double temp = c;
c = d;
d = temp;
}
// replace the lower of a and c with e
// because the lowest of the first four cannot be the median
if(a < c) // comparison # 3
{
a = e;
// re-sort a and b
if(a > b) // comparison # 4
{
double temp = a;
a = b;
b = temp;
}
}
else
{
c = e;
// re-sort c and d
if(c > d) // comparison # 4
{
double temp = c;
c = d;
d = temp;
}
}
// eliminate a or c, because the lowest
// of the remaining four can't be the median either
if(a < c) // comparison #5
{
if(b < c) // comparison #6
{
median = c;
}
else
{
median = b;
}
}
else
{
if(d < a) // comparison #6
{
median = a;
}
else
{
median = d;
}
}
return median;
}
回答5:
An interesting thread here:
http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_cs;action=display;num=1061827085
Quote from thread:
Put the numbers in an array.
Use three comparisons and shuffle around the numbers so that a[1] < a[2], a[4] < a[5], and a[1] < a[4].
If a[3] > a[2], then the problem is fairly easy. If a[2] < a[4], the median value is the smaller of a[3] and a[4]. If not, the median value is the smaller of a[2] and a[5].
So a[3] < a[2]. If a[3] > a[4], then the solution is the smaller of a[3] and a[5]. Otherwise, the solution is the smaller of a[2] and a[4].
回答6:
Just to check how many comparisons:
class MyComparable : IComparable
{
public static int NumberOfComparisons = 0;
public int NumPart { get; set; }
#region IComparable Members
public int CompareTo(object obj)
{
NumberOfComparisons++; //I know, not thread safe but just for the sample
MyComparable mc = obj as MyComparable;
if (mc == null)
return -1;
else
return NumPart.CompareTo(mc.NumPart);
}
#endregion
}
class Program
{
static void Main(string[] args)
{
List<MyComparable> list = new List<MyComparable>();
list.Add(new MyComparable() { NumPart = 5 });
list.Add(new MyComparable() { NumPart = 4 });
list.Add(new MyComparable() { NumPart = 3 });
list.Add(new MyComparable() { NumPart = 2 });
list.Add(new MyComparable() { NumPart = 1 });
list.Sort();
Console.WriteLine(MyComparable.NumberOfComparisons);
}
}
the result is 13.
回答7:
For completeness, the question is a specific case of a sorting network, which Knuth (Art of Computer Programming, vol 3) covers in great detail. The classic paper by K.E. Batcher on the subject is brief and worth reading.
回答8:
This should do it
private Double medianofFive(double[] input)
{
Double temp;
if (input[0] > input[1])//#1 - sort First and Second
{
temp = input[0];
input[0] = input[1];
input[1] = temp;
}
if (input[2] > input[3])//#2 sort Third and Fourth
{
temp = input[2];
input[2] = input[3];
input[3] = temp;
}
// replace the smaller of first and third with 5th, then sort
int smallerIndex = input[0] < input[2] ? 0 : 2;//#3
input[smallerIndex] = input[4];
//sort the new pair
if(input[smallerIndex]>input[smallerIndex+1])//#4
{
temp = input[smallerIndex];
input[smallerIndex] = input[smallerIndex+1];
input[smallerIndex+1] = temp;
}
//compare the two smaller numbers
// then compare the smaller of the two's partner with larger of the two
// the smaller of THOSE two is the median
if (input[2] > input[0])
//#5
{
temp = input[2] > input[1] ? input[1] : input[2];//#6
}
else
{
temp = input[0] > input[3] ? input[3] : input[0];//#6
}
return temp;
}
回答9:
-- In Haskell the solution could look like
import Data.Function
median5By pred [a,b,c,d,e] = fst $ merge2 c' d' where
merge2 = merge2By pred
merge2By pred x y = if x `pred` y then (x,y) else (y,x)
((_,b'), de ) = merge2By (pred `on` fst) (merge2 a b) (merge2 d e)
((_,c'),(d',_)) = merge2By (pred `on` fst) (merge2 b' c) de
main = print $ median5By (<) [2,5,4,1,3]
回答10:
Interesting how many comparisons in MSDN sample...
public static double Median(this IEnumerable<double> source) {
if (source.Count() == 0) throw new InvalidOperationException("Cannot compute median for an empty set.");
var sortedList = from number in source
orderby number
select number;
int itemIndex = (int)sortedList.Count() / 2;
if (sortedList.Count() % 2 == 0) {
// Even number of items.
return (sortedList.ElementAt(itemIndex) + sortedList.ElementAt(itemIndex - 1)) / 2; } else {
// Odd number of items.
return sortedList.ElementAt(itemIndex); }
}
来源：https://stackoverflow.com/questions/480960/code-to-calculate-median-of-five-in-c-sharp