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two arrays and random permutations
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$begingroup$
First time poster here, very new to programming. Its for a Homework for a Java class i'm taking, I can not figure it out for the life of me. I understand there are probably a lot of errors in the code, but my primary focus is to get the program working without incorperating alot of high level knowledge The problem im trying to solve is as follows
" Write a program that displays a random permutation of the numbers 1 to 10.
The resulting permutation is stored in an array of integers. To generate a random permutation, you need
Hint: Use two arrays:
- One filled with the numbers in sequence 1-10 (this is your Array1)
- Then create a second array (Array2) initialized with 0 that you fill replace with the 0-10 numbers
of Array1 using the following algorithm:
Generate a random number in the range 0-9 that you use to represent the index (position) in Array1.
Copy the number found at that position in Array2.
You must keep track of the values already entered in Array2 since duplicates are not permitted
(this means that every time you try to enter a new number in Array2,
you must traverse Array2 to check if the number is already present. If it is present you try again
using a while loop. It it is not present, you append it to the values already entered.
Repeat 10 times.".
Here is the code I have thus far, please excuse typos that may be in comments.
//method for generating range of int numbers. Max is included in range.
public static int getRandomIntRange(int min, int max) {
int x = (int)(Math.random() * ((max + 1 - min) + 1)) + min;
if (x > (max + 1))
x = max;
return x;
}
public static void main(String[] args) {
//loop for storing 1-10 int number withing A1[].
int[] A1 = new int[10];
int count = 1;
for (int k = 0; k<A1.length; k++) {
A1[k] =count;
count++;;
}
// loop to test A1 is in sequence.
//for (int k = 0; k < 10 ; k++) {
//System.out.println(A1[k]);
//}
int [] A2 = new int[10];
int ranNum;
//loop to give A1 value to A2 with no repeats.
for(int k=0;k<A2.length; k++){
ranNum=getRandomIntRange(0,9);
if ((A2[0]!=A1[ranNum] )&&(A2[1] !=A1[ranNum])&&
(A2[2]!=A1[ranNum])&&(A2[3]!=A1[ranNum])&&
(A2[4]!=A1[ranNum])&&(A2[5]!=A1[ranNum])&&
(A2[6]!=A1[ranNum])&&(A2[7]!=A1[ranNum])&&(A2[8]!=A1[ranNum])&&(A2[9]!=A1[ranNum])) {
A2[k]=A1[ranNum];
}
}
for(int k =0; k<A2.length; k++) {
System.out.println(A2[k]);
}
}
}
java
asked 5 mins ago
PRDPRD
1
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
First time poster here, very new to programming. Its for a Homework for a Java class i'm taking, I can not figure it out for the life of me. I understand there are probably a lot of errors in the code, but my primary focus is to get the program working without incorperating alot of high level knowledge The problem im trying to solve is as follows
" Write a program that displays a random permutation of the numbers 1 to 10.
The resulting permutation is stored in an array of integers. To generate a random permutation, you need
Hint: Use two arrays:
- One filled with the numbers in sequence 1-10 (this is your Array1)
- Then create a second array (Array2) initialized with 0 that you fill replace with the 0-10 numbers
of Array1 using the following algorithm:
Generate a random number in the range 0-9 that you use to represent the index (position) in Array1.
Copy the number found at that position in Array2.
You must keep track of the values already entered in Array2 since duplicates are not permitted
(this means that every time you try to enter a new number in Array2,
you must traverse Array2 to check if the number is already present. If it is present you try again
using a while loop. It it is not present, you append it to the values already entered.
Repeat 10 times.".
Here is the code I have thus far, please excuse typos that may be in comments.
//method for generating range of int numbers. Max is included in range.
public static int getRandomIntRange(int min, int max) {
int x = (int)(Math.random() * ((max + 1 - min) + 1)) + min;
if (x > (max + 1))
x = max;
return x;
}
public static void main(String[] args) {
//loop for storing 1-10 int number withing A1[].
int[] A1 = new int[10];
int count = 1;
for (int k = 0; k<A1.length; k++) {
A1[k] =count;
count++;;
}
// loop to test A1 is in sequence.
//for (int k = 0; k < 10 ; k++) {
//System.out.println(A1[k]);
//}
int [] A2 = new int[10];
int ranNum;
//loop to give A1 value to A2 with no repeats.
for(int k=0;k<A2.length; k++){
ranNum=getRandomIntRange(0,9);
if ((A2[0]!=A1[ranNum] )&&(A2[1] !=A1[ranNum])&&
(A2[2]!=A1[ranNum])&&(A2[3]!=A1[ranNum])&&
(A2[4]!=A1[ranNum])&&(A2[5]!=A1[ranNum])&&
(A2[6]!=A1[ranNum])&&(A2[7]!=A1[ranNum])&&(A2[8]!=A1[ranNum])&&(A2[9]!=A1[ranNum])) {
A2[k]=A1[ranNum];
}
}
for(int k =0; k<A2.length; k++) {
System.out.println(A2[k]);
}
}
}
java
asked 5 mins ago
PRDPRD
1
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
First time poster here, very new to programming. Its for a Homework for a Java class i'm taking, I can not figure it out for the life of me. I understand there are probably a lot of errors in the code, but my primary focus is to get the program working without incorperating alot of high level knowledge The problem im trying to solve is as follows
" Write a program that displays a random permutation of the numbers 1 to 10.
The resulting permutation is stored in an array of integers. To generate a random permutation, you need
Hint: Use two arrays:
- One filled with the numbers in sequence 1-10 (this is your Array1)
- Then create a second array (Array2) initialized with 0 that you fill replace with the 0-10 numbers
of Array1 using the following algorithm:
Generate a random number in the range 0-9 that you use to represent the index (position) in Array1.
Copy the number found at that position in Array2.
You must keep track of the values already entered in Array2 since duplicates are not permitted
(this means that every time you try to enter a new number in Array2,
you must traverse Array2 to check if the number is already present. If it is present you try again
using a while loop. It it is not present, you append it to the values already entered.
Repeat 10 times.".
Here is the code I have thus far, please excuse typos that may be in comments.
//method for generating range of int numbers. Max is included in range.
public static int getRandomIntRange(int min, int max) {
int x = (int)(Math.random() * ((max + 1 - min) + 1)) + min;
if (x > (max + 1))
x = max;
return x;
}
public static void main(String[] args) {
//loop for storing 1-10 int number withing A1[].
int[] A1 = new int[10];
int count = 1;
for (int k = 0; k<A1.length; k++) {
A1[k] =count;
count++;;
}
// loop to test A1 is in sequence.
//for (int k = 0; k < 10 ; k++) {
//System.out.println(A1[k]);
//}
int [] A2 = new int[10];
int ranNum;
//loop to give A1 value to A2 with no repeats.
for(int k=0;k<A2.length; k++){
ranNum=getRandomIntRange(0,9);
if ((A2[0]!=A1[ranNum] )&&(A2[1] !=A1[ranNum])&&
(A2[2]!=A1[ranNum])&&(A2[3]!=A1[ranNum])&&
(A2[4]!=A1[ranNum])&&(A2[5]!=A1[ranNum])&&
(A2[6]!=A1[ranNum])&&(A2[7]!=A1[ranNum])&&(A2[8]!=A1[ranNum])&&(A2[9]!=A1[ranNum])) {
A2[k]=A1[ranNum];
}
}
for(int k =0; k<A2.length; k++) {
System.out.println(A2[k]);
}
}
}
java
asked 5 mins ago
PRDPRD
1
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
First time poster here, very new to programming. Its for a Homework for a Java class i'm taking, I can not figure it out for the life of me. I understand there are probably a lot of errors in the code, but my primary focus is to get the program working without incorperating alot of high level knowledge The problem im trying to solve is as follows
" Write a program that displays a random permutation of the numbers 1 to 10.
The resulting permutation is stored in an array of integers. To generate a random permutation, you need
Hint: Use two arrays:
- One filled with the numbers in sequence 1-10 (this is your Array1)
- Then create a second array (Array2) initialized with 0 that you fill replace with the 0-10 numbers
of Array1 using the following algorithm:
Generate a random number in the range 0-9 that you use to represent the index (position) in Array1.
Copy the number found at that position in Array2.
You must keep track of the values already entered in Array2 since duplicates are not permitted
(this means that every time you try to enter a new number in Array2,
you must traverse Array2 to check if the number is already present. If it is present you try again
using a while loop. It it is not present, you append it to the values already entered.
Repeat 10 times.".
Here is the code I have thus far, please excuse typos that may be in comments.
//method for generating range of int numbers. Max is included in range.
public static int getRandomIntRange(int min, int max) {
int x = (int)(Math.random() * ((max + 1 - min) + 1)) + min;
if (x > (max + 1))
x = max;
return x;
}
public static void main(String[] args) {
//loop for storing 1-10 int number withing A1[].
int[] A1 = new int[10];
int count = 1;
for (int k = 0; k<A1.length; k++) {
A1[k] =count;
count++;;
}
// loop to test A1 is in sequence.
//for (int k = 0; k < 10 ; k++) {
//System.out.println(A1[k]);
//}
int [] A2 = new int[10];
int ranNum;
//loop to give A1 value to A2 with no repeats.
for(int k=0;k<A2.length; k++){
ranNum=getRandomIntRange(0,9);
if ((A2[0]!=A1[ranNum] )&&(A2[1] !=A1[ranNum])&&
(A2[2]!=A1[ranNum])&&(A2[3]!=A1[ranNum])&&
(A2[4]!=A1[ranNum])&&(A2[5]!=A1[ranNum])&&
(A2[6]!=A1[ranNum])&&(A2[7]!=A1[ranNum])&&(A2[8]!=A1[ranNum])&&(A2[9]!=A1[ranNum])) {
A2[k]=A1[ranNum];
}
}
for(int k =0; k<A2.length; k++) {
System.out.println(A2[k]);
}
}
}
java
java
asked 5 mins ago
PRDPRD
1
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
PRDPRD
1
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
PRDPRD
1
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
PRDPRD
1
asked 5 mins ago
PRDPRD
1
1
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
PRD is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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