Algorithm L
Algorithm L (List merge sort). Records R1, … , RN are assumed to contain
keys K1,…, KN, together with link fields L1, …, LN capable of holding the
numbers -(N + 1) through (N + 1). There are two auxiliary link fields L0 and
LN+1 in artificial records R0 and RN+1 at the beginning and end of the file. This
algorithm is a “list sort” that sets the link fields so that the records are linked
together in ascending order. After sorting is complete, L0 will be the index of
the record with the smallest key; and Lk, for 1 <= k <= N, will be the index of the
record that follows Rk, or Lk = 0 if Rk is the record with the largest key. (See
Eq. 5.2.1-(13).)
During the course of this algorithm, R0 and RN+1 serve as list heads for two
linear lists whose sublists are being merged. A negative link denotes the end of
a sublist known to be ordered; a zero link denotes the end of the entire list. We
assume that N >=2.
The notation “|LS| <– p” means “Set Ls to p or -p, retaining the previous
sign of Ls.” This operation is well-suited to MIX, but unfortunately not to most
computers; it is possible to modify the algorithm in straightforward ways to
obtain an equally efficient method for most other machines.
L1. [Prepare two lists.] Set L0 <– 1, LN+1 <– 2, Li <– -(i + 2) for 1 <= i <= N-2,
and LN-1 <– LN <– 0. (We have created two lists containing R1, R3, R5, …
and R2, R4, R6, … respectively; the negative links indicate that each or-
dered sublist consists of one element only. For another way to do this step,
taking advantage of ordering that may be present in the initial data, see
exercise 12.)
L2. [Begin new pass.] Set s <– 0, t <– N + 1, p <– Ls, q <– Lt. If q = 0, the
algorithm terminates. (During each pass, p and q traverse the lists being
merged; s usually points to the most recently processed record of the current
sublist, while t points to the end of the previously output sublist.)
L3. [Compare Kp : Kq] If Kp > Kq, go to L6.
L4. [Advance p.] Set |LS| <– p, s <– p, p <– Lp. If p > 0, return to L3.
L5. [Complete the sublist.] Set Ls <– q, s <– t. Then set t <– q and q <– Lq, one
or more times, until q <= 0. Finally go to L8.
L6. [Advance q] (Steps L6 and L7 are dual to L4 and L5.) Set |LS| <– q, s <– q,
q <– Lq. If q > 0, return to L3.
L7. [Complete the sublist.] Set Ls <– p, s <– t. Then set t <– p and p <– Lp, one
or more times, until p <= 0.
L8. [End of pass?] (At this point, p <= 0 and q <= 0, since both pointers have
moved to the end of their respective sublists.) Set p <– -p, q <– -q. If
q = 0, set |LS| <– p, |Lt| <– 0 and return to L2. Otherwise return to L3. |
Data table
Java program
In this program, R1,…,RN were simplified to K1,…,KN.
/**
* Created with IntelliJ IDEA.
* User: 1O1O
* Date: 12/4/13
* Time: 10:01 PM
* :)~
* List Merge Sort:Sorting by Merging:Internal Sorting
*/
public class Main {
public static void main(String[] args) {
int N = 16;
int[] K = new int[17];
int[] L = new int[18];
/*Prepare the input data*/
K[1] = 503;
K[2] = 87;
K[3] = 512;
K[4] = 61;
K[5] = 908;
K[6] = 170;
K[7] = 897;
K[8] = 275;
K[9] = 653;
K[10] = 426;
K[11] = 154;
K[12] = 509;
K[13] = 612;
K[14] = 677;
K[15] = 765;
K[16] = 703;
/*Output unsorted Ks*/
System.out.println("Unsorted Ks:");
for(int i=1; i<=N; i++){
System.out.println(i+":"+K[i]);
}
System.out.println();
/*Kernel of the Algorithm!*/
int s;
int t;
int p;
int q;
L[0] = 1; /*L1*/
L[N+1] = 2;
for(int i=1; i<=N-2; i++){
L[i] = -(i+2);
}
L[N-1] = L[N] = 0;
do{
s = 0; /*L2*/
t = N+1;
p = L[s];
q = L[t];
if(q == 0){
break;
}
do{
if(K[p] > K[q]){ /*L3*/
if(L[s] >= 0){ /*L6*/
L[s] = Math.abs(q);
}else {
L[s] = -Math.abs(q);
}
s = q;
q = L[q];
if(q > 0){
continue;
}
L[s] = p; /*L7*/
s = t;
do{
t = p;
p = L[p];
}while (p > 0);
} else {
if(L[s] >= 0){ /*L4*/
L[s] = Math.abs(p);
}else {
L[s] = -Math.abs(p);
}
s = p;
p = L[p];
if(p > 0){
continue;
}
L[s] = q; /*L5*/
s = t;
do{
t = q;
q = L[q];
}while (q > 0);
}
p = -p; /*L8*/
q = -q;
if(q == 0){
if(L[s] >= 0){
L[s] = Math.abs(p);
}else {
L[s] = -Math.abs(p);
}
L[t] = 0;
break;
}
}while (true);
}while (true);
/*Output sorted Ks*/
System.out.println("Sorted Ks:");
int index = 0;
int i = 1;
while (L[index] > 0){
System.out.println(i+":"+K[L[index]]);
index = L[index];
i++;
}
}
}Outputs
Unsorted Ks:
1:503
2:87
3:512
4:61
5:908
6:170
7:897
8:275
9:653
10:426
11:154
12:509
13:612
14:677
15:765
16:703
Sorted Ks:
1:61
2:87
3:154
4:170
5:275
6:426
7:503
8:509
9:512
10:612
11:653
12:677
13:703
14:765
15:897
16:908Reference
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