Custom shuffle algorithm
I know there is an efficient shuffling algorithm called Fisher Yates I believe but I try to build a custom one (out of stubbornness).
I am trying to generate a new random number in the right range, than use it to determine a new random place in a copied array. And then store this random number to NOT use it twice.
My method does not work. Would someone care to give me some enlightment?
public int shuffled(int numbers) {
int rand;
int copyNumbers = new int[numbers.length];
int usedRand = new int[numbers.length];
for (int i = 0; i < numbers.length; i++) {
rand = new Random().nextInt(numbers.length + 1);
if ((i != rand) && (usedRand[i]!=rand)) {
copyNumbers[rand] = numbers[i];
usedRand[i]=rand;
}
}
numbers = copyNumbers;
return numbers;
}
arrays shuffle
add a comment |
I know there is an efficient shuffling algorithm called Fisher Yates I believe but I try to build a custom one (out of stubbornness).
I am trying to generate a new random number in the right range, than use it to determine a new random place in a copied array. And then store this random number to NOT use it twice.
My method does not work. Would someone care to give me some enlightment?
public int shuffled(int numbers) {
int rand;
int copyNumbers = new int[numbers.length];
int usedRand = new int[numbers.length];
for (int i = 0; i < numbers.length; i++) {
rand = new Random().nextInt(numbers.length + 1);
if ((i != rand) && (usedRand[i]!=rand)) {
copyNumbers[rand] = numbers[i];
usedRand[i]=rand;
}
}
numbers = copyNumbers;
return numbers;
}
arrays shuffle
What does not work? evenRandom
should follow a certain structure. If you really want to use a random value, you should mix in some real random variables, like time, date etc.
– dnsiv
Nov 21 '18 at 10:08
I don't get a properly shuffled array. This is a school project so they do not look that close if our shuffling is "real" random. But of course I take any avice.
– Dovendyr
Nov 21 '18 at 10:29
1
Right now I see 2 possible errors. The first one isrand = new Random().nextInt(numbers.length+1);
You need a random number from 0 to numbers.length-1, since arrays start with indexing at 0, but right now you will get a random number from 1 to numbers.length. You should get anIndexOutOfBoundsException
this way.
– dnsiv
Nov 21 '18 at 10:50
The second possible error lies in this phrase(usedRand[i] != rand)
. You're just comparing the current entry (i
) in usedRand with the current random value, if that entry already used this random number. But this way, you wont be able to access one entry a second time. You need to create a second for loop, which iterates through all entries of usedRand and compares their values with the current random number
– dnsiv
Nov 21 '18 at 10:52
add a comment |
I know there is an efficient shuffling algorithm called Fisher Yates I believe but I try to build a custom one (out of stubbornness).
I am trying to generate a new random number in the right range, than use it to determine a new random place in a copied array. And then store this random number to NOT use it twice.
My method does not work. Would someone care to give me some enlightment?
public int shuffled(int numbers) {
int rand;
int copyNumbers = new int[numbers.length];
int usedRand = new int[numbers.length];
for (int i = 0; i < numbers.length; i++) {
rand = new Random().nextInt(numbers.length + 1);
if ((i != rand) && (usedRand[i]!=rand)) {
copyNumbers[rand] = numbers[i];
usedRand[i]=rand;
}
}
numbers = copyNumbers;
return numbers;
}
arrays shuffle
I know there is an efficient shuffling algorithm called Fisher Yates I believe but I try to build a custom one (out of stubbornness).
I am trying to generate a new random number in the right range, than use it to determine a new random place in a copied array. And then store this random number to NOT use it twice.
My method does not work. Would someone care to give me some enlightment?
public int shuffled(int numbers) {
int rand;
int copyNumbers = new int[numbers.length];
int usedRand = new int[numbers.length];
for (int i = 0; i < numbers.length; i++) {
rand = new Random().nextInt(numbers.length + 1);
if ((i != rand) && (usedRand[i]!=rand)) {
copyNumbers[rand] = numbers[i];
usedRand[i]=rand;
}
}
numbers = copyNumbers;
return numbers;
}
arrays shuffle
arrays shuffle
edited Dec 4 '18 at 5:37
Cœur
19k9113155
19k9113155
asked Nov 21 '18 at 10:03
DovendyrDovendyr
52
52
What does not work? evenRandom
should follow a certain structure. If you really want to use a random value, you should mix in some real random variables, like time, date etc.
– dnsiv
Nov 21 '18 at 10:08
I don't get a properly shuffled array. This is a school project so they do not look that close if our shuffling is "real" random. But of course I take any avice.
– Dovendyr
Nov 21 '18 at 10:29
1
Right now I see 2 possible errors. The first one isrand = new Random().nextInt(numbers.length+1);
You need a random number from 0 to numbers.length-1, since arrays start with indexing at 0, but right now you will get a random number from 1 to numbers.length. You should get anIndexOutOfBoundsException
this way.
– dnsiv
Nov 21 '18 at 10:50
The second possible error lies in this phrase(usedRand[i] != rand)
. You're just comparing the current entry (i
) in usedRand with the current random value, if that entry already used this random number. But this way, you wont be able to access one entry a second time. You need to create a second for loop, which iterates through all entries of usedRand and compares their values with the current random number
– dnsiv
Nov 21 '18 at 10:52
add a comment |
What does not work? evenRandom
should follow a certain structure. If you really want to use a random value, you should mix in some real random variables, like time, date etc.
– dnsiv
Nov 21 '18 at 10:08
I don't get a properly shuffled array. This is a school project so they do not look that close if our shuffling is "real" random. But of course I take any avice.
– Dovendyr
Nov 21 '18 at 10:29
1
Right now I see 2 possible errors. The first one isrand = new Random().nextInt(numbers.length+1);
You need a random number from 0 to numbers.length-1, since arrays start with indexing at 0, but right now you will get a random number from 1 to numbers.length. You should get anIndexOutOfBoundsException
this way.
– dnsiv
Nov 21 '18 at 10:50
The second possible error lies in this phrase(usedRand[i] != rand)
. You're just comparing the current entry (i
) in usedRand with the current random value, if that entry already used this random number. But this way, you wont be able to access one entry a second time. You need to create a second for loop, which iterates through all entries of usedRand and compares their values with the current random number
– dnsiv
Nov 21 '18 at 10:52
What does not work? even
Random
should follow a certain structure. If you really want to use a random value, you should mix in some real random variables, like time, date etc.– dnsiv
Nov 21 '18 at 10:08
What does not work? even
Random
should follow a certain structure. If you really want to use a random value, you should mix in some real random variables, like time, date etc.– dnsiv
Nov 21 '18 at 10:08
I don't get a properly shuffled array. This is a school project so they do not look that close if our shuffling is "real" random. But of course I take any avice.
– Dovendyr
Nov 21 '18 at 10:29
I don't get a properly shuffled array. This is a school project so they do not look that close if our shuffling is "real" random. But of course I take any avice.
– Dovendyr
Nov 21 '18 at 10:29
1
1
Right now I see 2 possible errors. The first one is
rand = new Random().nextInt(numbers.length+1);
You need a random number from 0 to numbers.length-1, since arrays start with indexing at 0, but right now you will get a random number from 1 to numbers.length. You should get an IndexOutOfBoundsException
this way.– dnsiv
Nov 21 '18 at 10:50
Right now I see 2 possible errors. The first one is
rand = new Random().nextInt(numbers.length+1);
You need a random number from 0 to numbers.length-1, since arrays start with indexing at 0, but right now you will get a random number from 1 to numbers.length. You should get an IndexOutOfBoundsException
this way.– dnsiv
Nov 21 '18 at 10:50
The second possible error lies in this phrase
(usedRand[i] != rand)
. You're just comparing the current entry (i
) in usedRand with the current random value, if that entry already used this random number. But this way, you wont be able to access one entry a second time. You need to create a second for loop, which iterates through all entries of usedRand and compares their values with the current random number– dnsiv
Nov 21 '18 at 10:52
The second possible error lies in this phrase
(usedRand[i] != rand)
. You're just comparing the current entry (i
) in usedRand with the current random value, if that entry already used this random number. But this way, you wont be able to access one entry a second time. You need to create a second for loop, which iterates through all entries of usedRand and compares their values with the current random number– dnsiv
Nov 21 '18 at 10:52
add a comment |
1 Answer
1
active
oldest
votes
As I already pointed the errors out in the comments. Here is possible implementation of shuffle(). I commented the code based on my modifications.
If you happen to have questions, please feel free to ask!
public static void main(String args) {
//sample values
int numbers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18};
//First call of recursive function
int x = shuffled(numbers, null, null, false, 0);
//Print out values
for (int i = 0; i < x.length; i++) {
System.out.print(x[i] + "|");
}
}
public static int shuffled(int numbers, int usedRand, int copyNumbers, boolean newRandomValue, int current) {
//Initialisation
int rand = -1;
int start = 0;
//If function hasn't been called before, initialise.
if (usedRand == null) {
usedRand = new int[numbers.length];
Arrays.fill(usedRand, -1);
}
if (copyNumbers == null) {
copyNumbers = new int[numbers.length];
}
//If function has been called recursively, reassign the start-Integer for for-loop
if (newRandomValue) {
start = current;
}
//Iterate over numbers-array with <start> as start-index
for (int i = start; i < numbers.length; i++) {
/*
new Random().nextInt(numbers.length+1 ) gets a random number between 0 and numbers.length + 1
In our example, numbers.length+1 is 20. So we will get a random number between 0 and 20.
But we want to get random values between -1 and 19, so
new Random().nextInt(numbers.length+1 ) --> values between 0 to 20
rand = new Random().nextInt(numbers.length+1 ) -1 --> values between -1 to 19
*/
rand = new Random().nextInt(numbers.length+1 ) -1;
//Iterate through usedRand array and compare if current rand value has already been used yet.
for (int j = 0; j < usedRand.length; j++) {
if (usedRand[j] == rand) {
/*
If current rand value has already been used, recursively call shuffle again with the current index i
as new start-index
*/
return shuffled(numbers, usedRand, copyNumbers, true, i);
}
}
copyNumbers[rand] = numbers[i];
usedRand[i] = rand;
}
numbers = copyNumbers;
return numbers;
}
add a comment |
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1 Answer
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1 Answer
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active
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active
oldest
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active
oldest
votes
As I already pointed the errors out in the comments. Here is possible implementation of shuffle(). I commented the code based on my modifications.
If you happen to have questions, please feel free to ask!
public static void main(String args) {
//sample values
int numbers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18};
//First call of recursive function
int x = shuffled(numbers, null, null, false, 0);
//Print out values
for (int i = 0; i < x.length; i++) {
System.out.print(x[i] + "|");
}
}
public static int shuffled(int numbers, int usedRand, int copyNumbers, boolean newRandomValue, int current) {
//Initialisation
int rand = -1;
int start = 0;
//If function hasn't been called before, initialise.
if (usedRand == null) {
usedRand = new int[numbers.length];
Arrays.fill(usedRand, -1);
}
if (copyNumbers == null) {
copyNumbers = new int[numbers.length];
}
//If function has been called recursively, reassign the start-Integer for for-loop
if (newRandomValue) {
start = current;
}
//Iterate over numbers-array with <start> as start-index
for (int i = start; i < numbers.length; i++) {
/*
new Random().nextInt(numbers.length+1 ) gets a random number between 0 and numbers.length + 1
In our example, numbers.length+1 is 20. So we will get a random number between 0 and 20.
But we want to get random values between -1 and 19, so
new Random().nextInt(numbers.length+1 ) --> values between 0 to 20
rand = new Random().nextInt(numbers.length+1 ) -1 --> values between -1 to 19
*/
rand = new Random().nextInt(numbers.length+1 ) -1;
//Iterate through usedRand array and compare if current rand value has already been used yet.
for (int j = 0; j < usedRand.length; j++) {
if (usedRand[j] == rand) {
/*
If current rand value has already been used, recursively call shuffle again with the current index i
as new start-index
*/
return shuffled(numbers, usedRand, copyNumbers, true, i);
}
}
copyNumbers[rand] = numbers[i];
usedRand[i] = rand;
}
numbers = copyNumbers;
return numbers;
}
add a comment |
As I already pointed the errors out in the comments. Here is possible implementation of shuffle(). I commented the code based on my modifications.
If you happen to have questions, please feel free to ask!
public static void main(String args) {
//sample values
int numbers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18};
//First call of recursive function
int x = shuffled(numbers, null, null, false, 0);
//Print out values
for (int i = 0; i < x.length; i++) {
System.out.print(x[i] + "|");
}
}
public static int shuffled(int numbers, int usedRand, int copyNumbers, boolean newRandomValue, int current) {
//Initialisation
int rand = -1;
int start = 0;
//If function hasn't been called before, initialise.
if (usedRand == null) {
usedRand = new int[numbers.length];
Arrays.fill(usedRand, -1);
}
if (copyNumbers == null) {
copyNumbers = new int[numbers.length];
}
//If function has been called recursively, reassign the start-Integer for for-loop
if (newRandomValue) {
start = current;
}
//Iterate over numbers-array with <start> as start-index
for (int i = start; i < numbers.length; i++) {
/*
new Random().nextInt(numbers.length+1 ) gets a random number between 0 and numbers.length + 1
In our example, numbers.length+1 is 20. So we will get a random number between 0 and 20.
But we want to get random values between -1 and 19, so
new Random().nextInt(numbers.length+1 ) --> values between 0 to 20
rand = new Random().nextInt(numbers.length+1 ) -1 --> values between -1 to 19
*/
rand = new Random().nextInt(numbers.length+1 ) -1;
//Iterate through usedRand array and compare if current rand value has already been used yet.
for (int j = 0; j < usedRand.length; j++) {
if (usedRand[j] == rand) {
/*
If current rand value has already been used, recursively call shuffle again with the current index i
as new start-index
*/
return shuffled(numbers, usedRand, copyNumbers, true, i);
}
}
copyNumbers[rand] = numbers[i];
usedRand[i] = rand;
}
numbers = copyNumbers;
return numbers;
}
add a comment |
As I already pointed the errors out in the comments. Here is possible implementation of shuffle(). I commented the code based on my modifications.
If you happen to have questions, please feel free to ask!
public static void main(String args) {
//sample values
int numbers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18};
//First call of recursive function
int x = shuffled(numbers, null, null, false, 0);
//Print out values
for (int i = 0; i < x.length; i++) {
System.out.print(x[i] + "|");
}
}
public static int shuffled(int numbers, int usedRand, int copyNumbers, boolean newRandomValue, int current) {
//Initialisation
int rand = -1;
int start = 0;
//If function hasn't been called before, initialise.
if (usedRand == null) {
usedRand = new int[numbers.length];
Arrays.fill(usedRand, -1);
}
if (copyNumbers == null) {
copyNumbers = new int[numbers.length];
}
//If function has been called recursively, reassign the start-Integer for for-loop
if (newRandomValue) {
start = current;
}
//Iterate over numbers-array with <start> as start-index
for (int i = start; i < numbers.length; i++) {
/*
new Random().nextInt(numbers.length+1 ) gets a random number between 0 and numbers.length + 1
In our example, numbers.length+1 is 20. So we will get a random number between 0 and 20.
But we want to get random values between -1 and 19, so
new Random().nextInt(numbers.length+1 ) --> values between 0 to 20
rand = new Random().nextInt(numbers.length+1 ) -1 --> values between -1 to 19
*/
rand = new Random().nextInt(numbers.length+1 ) -1;
//Iterate through usedRand array and compare if current rand value has already been used yet.
for (int j = 0; j < usedRand.length; j++) {
if (usedRand[j] == rand) {
/*
If current rand value has already been used, recursively call shuffle again with the current index i
as new start-index
*/
return shuffled(numbers, usedRand, copyNumbers, true, i);
}
}
copyNumbers[rand] = numbers[i];
usedRand[i] = rand;
}
numbers = copyNumbers;
return numbers;
}
As I already pointed the errors out in the comments. Here is possible implementation of shuffle(). I commented the code based on my modifications.
If you happen to have questions, please feel free to ask!
public static void main(String args) {
//sample values
int numbers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18};
//First call of recursive function
int x = shuffled(numbers, null, null, false, 0);
//Print out values
for (int i = 0; i < x.length; i++) {
System.out.print(x[i] + "|");
}
}
public static int shuffled(int numbers, int usedRand, int copyNumbers, boolean newRandomValue, int current) {
//Initialisation
int rand = -1;
int start = 0;
//If function hasn't been called before, initialise.
if (usedRand == null) {
usedRand = new int[numbers.length];
Arrays.fill(usedRand, -1);
}
if (copyNumbers == null) {
copyNumbers = new int[numbers.length];
}
//If function has been called recursively, reassign the start-Integer for for-loop
if (newRandomValue) {
start = current;
}
//Iterate over numbers-array with <start> as start-index
for (int i = start; i < numbers.length; i++) {
/*
new Random().nextInt(numbers.length+1 ) gets a random number between 0 and numbers.length + 1
In our example, numbers.length+1 is 20. So we will get a random number between 0 and 20.
But we want to get random values between -1 and 19, so
new Random().nextInt(numbers.length+1 ) --> values between 0 to 20
rand = new Random().nextInt(numbers.length+1 ) -1 --> values between -1 to 19
*/
rand = new Random().nextInt(numbers.length+1 ) -1;
//Iterate through usedRand array and compare if current rand value has already been used yet.
for (int j = 0; j < usedRand.length; j++) {
if (usedRand[j] == rand) {
/*
If current rand value has already been used, recursively call shuffle again with the current index i
as new start-index
*/
return shuffled(numbers, usedRand, copyNumbers, true, i);
}
}
copyNumbers[rand] = numbers[i];
usedRand[i] = rand;
}
numbers = copyNumbers;
return numbers;
}
answered Nov 21 '18 at 11:35
dnsivdnsiv
347117
347117
add a comment |
add a comment |
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What does not work? even
Random
should follow a certain structure. If you really want to use a random value, you should mix in some real random variables, like time, date etc.– dnsiv
Nov 21 '18 at 10:08
I don't get a properly shuffled array. This is a school project so they do not look that close if our shuffling is "real" random. But of course I take any avice.
– Dovendyr
Nov 21 '18 at 10:29
1
Right now I see 2 possible errors. The first one is
rand = new Random().nextInt(numbers.length+1);
You need a random number from 0 to numbers.length-1, since arrays start with indexing at 0, but right now you will get a random number from 1 to numbers.length. You should get anIndexOutOfBoundsException
this way.– dnsiv
Nov 21 '18 at 10:50
The second possible error lies in this phrase
(usedRand[i] != rand)
. You're just comparing the current entry (i
) in usedRand with the current random value, if that entry already used this random number. But this way, you wont be able to access one entry a second time. You need to create a second for loop, which iterates through all entries of usedRand and compares their values with the current random number– dnsiv
Nov 21 '18 at 10:52