Algorithm R
Algorithm R (Radix exchange sort). Records R1,… ,RN are rearranged in
place; after sorting is complete, their keys will be in order, K1 <= … <= KN. Each
key is assumed to be a nonnegative m-bit binary number, (a1 a2 … am)2; the ith
most significant bit, ai, is called “bit i” of the key. An auxiliary stack with
room for at most m-1 entries is needed for temporary storage. This algorithm
essentially follows the radix exchange partitioning procedure described in the
text above; certain improvements in its efficiency are possible, as described in
the text and exercises below.
R1. [Initialize.] Set the stack empty, and set I <– 1, r <– N, b <– 1.
R2. [Begin new stage.] (We now wish to sort the subfile Rl … Rr on bit b;
from the nature of the algorithm, we have I <= r.) If I = r, go to step R10
(since a one-word file is already sorted). Otherwise set i <– I, j <– r.
R3. [Inspect Ki for 1.] Examine bit b of Ki. If it is a 1, go to step R6.
R4. [Increase i.] Increase i by 1. If i <= j, return to step R3; otherwise go to
step R8.
R5. [Inspect Kj+1 for 0.] Examine bit b of Kj+1. If it is a 0, go to step R7.
R6. [Decrease j.] Decrease j by 1. If i <= j, go to step R5; otherwise go to
step R8.
R7. [Exchange Ri, Rj+1] Interchange records Ri <–> Rj+1; then go to step R4.
R8. [Test special cases.] (At this point a partitioning stage has been completed;
i = j+1, bit b of keys Kl, … , Kj is 0, and bit b of keys Ki, … ,Kr is 1.)
Increase b by 1. If b > m, where m is the total number of bits in the keys,
go to step R10. (In such a case, the subfile Rl … Rr has been sorted. This
test need not be made if there is no chance of having equal keys present in
the file.) Otherwise if j < I or j = r, go back to step R2 (all bits examined
were 1 or 0, respectively). Otherwise if j = I, increase I by 1 and go to
step R2 (there was only one 0 bit).
R9. [Put on stack.] Insert the entry (r, b) on top of the stack; then set r <– j
and go to step R2.
R10. [Take off stack.] If the stack is empty, we are done sorting; otherwise set
I <– r+1, remove the top entry (r’, b’) of the stack, set r <– r’, b <– b’, and
return to step R2. |
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
* :)~
* Radix exchange sort:Sorting by Exchanging:Internal Sorting
*/
public class Main {
public static void main(String[] args) {
int N = 16;
String[] K = new String[17];
int m = 10; /*Total number of bits in the Ks*/
int[] R = new int[10]; /*At most m-1*/
int[] B = new int[10]; /*At most m-1*/
/*Prepare the input data*/
K[1] = Integer.toBinaryString(503);
K[2] = Integer.toBinaryString(87);
K[3] = Integer.toBinaryString(512);
K[4] = Integer.toBinaryString(61);
K[5] = Integer.toBinaryString(908);
K[6] = Integer.toBinaryString(170);
K[7] = Integer.toBinaryString(897);
K[8] = Integer.toBinaryString(275);
K[9] = Integer.toBinaryString(653);
K[10] = Integer.toBinaryString(426);
K[11] = Integer.toBinaryString(154);
K[12] = Integer.toBinaryString(509);
K[13] = Integer.toBinaryString(612);
K[14] = Integer.toBinaryString(677);
K[15] = Integer.toBinaryString(765);
K[16] = Integer.toBinaryString(703);
/*Transform Decimal to Binary(all 10bits in this example) and output unsorted Ks*/
System.out.println("Unsorted Ks:");
for(int i=1; i<=N; i++){
if(K[i].length() < 10){
switch (10 - K[i].length()){
case 1: K[i] = "0" + K[i]; break;
case 2: K[i] = "00" + K[i]; break;
case 3: K[i] = "000" + K[i]; break;
case 4: K[i] = "0000" + K[i]; break;
case 5: K[i] = "00000" + K[i]; break;
case 6: K[i] = "000000" + K[i]; break;
case 7: K[i] = "0000000" + K[i]; break;
case 8: K[i] = "00000000" + K[i]; break;
case 9: K[i] = "000000000" + K[i]; break;
}
}
System.out.println(String.format("%3s", i+":")+String.format("%11s", K[i])+":"+Integer.parseInt(K[i], 2));
}
System.out.println();
/*Kernel of the Algorithm!*/
int l = 1; /*R1*/
int r = N;
int b = 1;
int len = 0;
int i;
int j;
String temp;
do{
do{
if(l == r){ /*R2*/
break;
}else {
i = l;
j = r;
}
do{
if(K[i].charAt(b-1) == '1'){ /*R3*/
do{
j--; /*R6*/
if(i <= j){
if(K[j+1].charAt(b-1) == '0'){ /*R5*/
temp = K[i]; /*R7*/
K[i] = K[j+1];
K[j+1] = temp;
break;
}
}else {
break;
}
}while (true);
if(i > j){
break;
}
}
i++; /*R4*/
if(i <= j){
/*Go to R3*/
}else {
break;/*Go to R8*/
}
}while (true);
b++; /*R8*/
if(b > m){
break;
}else if(j < l || j == r){
/*Go to R2*/
}else if(j == l){
l++;
/*Go to R2*/
}else {
len++;
R[len] = r; /*R9*/
B[len] = b;
r = j;
}
}while (true);
if(len == 0){ /*R10*/
break;
}else {
l = r+1;
r = R[len];
b = B[len];
len--;
}
}while (true);
/*Output sorted Ks*/
System.out.println("Sorted Ks:");
for(i=1; i<=N; i++){
System.out.println(String.format("%3s", i+":")+String.format("%11s", K[i])+":"+Integer.parseInt(K[i], 2)); /*Transform Binary to Decimal*/
}
}
}Outputs
Unsorted Ks:
1: 0111110111:503
2: 0001010111:87
3: 1000000000:512
4: 0000111101:61
5: 1110001100:908
6: 0010101010:170
7: 1110000001:897
8: 0100010011:275
9: 1010001101:653
10: 0110101010:426
11: 0010011010:154
12: 0111111101:509
13: 1001100100:612
14: 1010100101:677
15: 1011111101:765
16: 1010111111:703
Sorted Ks:
1: 0000111101:61
2: 0001010111:87
3: 0010011010:154
4: 0010101010:170
5: 0100010011:275
6: 0110101010:426
7: 0111110111:503
8: 0111111101:509
9: 1000000000:512
10: 1001100100:612
11: 1010001101:653
12: 1010100101:677
13: 1010111111:703
14: 1011111101:765
15: 1110000001:897
16: 1110001100:908Reference
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