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Not anymore (note: a different answer was edited 4 hours ago, so I'm not bumping the challenge)
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The Fifth Marshal
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Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Stackylogic, 226 bytes

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

replaced http://codegolf.stackexchange.com/ with https://codegolf.stackexchange.com/
Source Link

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found herehere. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

corrected an error; NOR not NAND
Source Link

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NANDNOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NAND of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Stackylogic, 226 bytes (non-competing)

Yes, that is right. I made a program in Stackylogic (non-TC), which was invented by Helka Homba, for the challenge found here. This is made after the challenge, so non-competing.

Stackylogic has only binary input, so 10 (or more, any more digits will be ignored) bit binary must be used (least significant bit inputted first). Any dates outside the specified range might fail, as it simply checks what the inputted number is: it doesn't cover unnecessary dates

Not only is this my first challenge with stackylogic, but the first challenge with stackylogic at all.

Get ready for this mess:

1
0
1?
010
1?0
010
1?10
?1010
001010
?1010
?1010
?010
?10
?0
0
?
110
?10
11010
?010
11010
?1010
001010
?1010
?1010
?1010
?1010
?010
?0
110
?10
11010
?010
1010
01010
01010
?010
?0
110
?0
110
?0
110
1?0
?10
0?10
?10
?0
01
?
?<
0

This took me so long to make, because Stackylogic is the most confusing language I have encountered, and extremely unreadable: you have to know how the rest of the program has executed before you can read the current section being edited. I even had to add spaces for readability while creating it.

Meagre explanation

This is a simple explanation of what it does.

Stackylogic does not have any mathematical functions, so that made this harder. I had to hardcode most of it, to check if it was a specific number.

First, this program will do a NOR of the least significant bits, discarding them in the process. this means that if it is divisible by 4, it will proceed to the main part of the program, otherwise output 0.

Second, the pointer is carried over to the labyrinth of stackylogic, from here, if the next two bits are zero, it will instantly output 1 (as then it is divisible by 16, and so a leap year despite any other conditions), other wise it will check if it is not any of the numbers that is divisible by 4 but not a leap year, between 1801 and 2400.

To explain in detail, would involve making this post many times longer than it already is

Source Link
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