Write an algorithm accepts a positive integer n as input, when executed, prints out a list of all self-aware arrays of length n
Self-Aware Arrays.
An integer array int A is self-aware if for each i < A.length, A[i] is the exact number of occurrences of i in A.
For example, [2, 0, 2, 0] is self-aware.
Write an algorithm accepts a positive integer n as input, when executed, prints out a list of all self-aware arrays of that length.
Here are some Self-aware arrays have been found. But I haven't found an algorithm.
N = 4
2, 0, 2, 0
1, 2, 1, 0
N = 5
2, 1, 2, 0, 0
N = 6
None
N = 7
3, 2, 1, 1, 0, 0, 0
N = 8
4, 2, 1, 0, 1, 0, 0, 0
N = 9
5, 2, 1, 0, 0, 1, 0, 0, 0
N = 10
6, 2, 1, 0, 0, 0, 1, 0, 0, 0
N = 11
7, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0
N = a
a-4, 2, 1, <a-7 0s>, 1, 0, 0, 0
Seems we have the properties
sum(A[i]) = n = sum(i * A[i])
A[0] = sum( (i-1) * A[i] ) i>=2
arrays algorithm
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
add a comment |
Self-Aware Arrays.
An integer array int A is self-aware if for each i < A.length, A[i] is the exact number of occurrences of i in A.
For example, [2, 0, 2, 0] is self-aware.
Write an algorithm accepts a positive integer n as input, when executed, prints out a list of all self-aware arrays of that length.
Here are some Self-aware arrays have been found. But I haven't found an algorithm.
N = 4
2, 0, 2, 0
1, 2, 1, 0
N = 5
2, 1, 2, 0, 0
N = 6
None
N = 7
3, 2, 1, 1, 0, 0, 0
N = 8
4, 2, 1, 0, 1, 0, 0, 0
N = 9
5, 2, 1, 0, 0, 1, 0, 0, 0
N = 10
6, 2, 1, 0, 0, 0, 1, 0, 0, 0
N = 11
7, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0
N = a
a-4, 2, 1, <a-7 0s>, 1, 0, 0, 0
Seems we have the properties
sum(A[i]) = n = sum(i * A[i])
A[0] = sum( (i-1) * A[i] ) i>=2
arrays algorithm
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
What have you tried so far?
– jrook
Nov 14 '18 at 22:02
See Self-descriptive number for a reference. en.m.wikipedia.org/wiki/Self-descriptive_number
– Shang Gao
Nov 15 '18 at 15:51
add a comment |
Self-Aware Arrays.
An integer array int A is self-aware if for each i < A.length, A[i] is the exact number of occurrences of i in A.
For example, [2, 0, 2, 0] is self-aware.
Write an algorithm accepts a positive integer n as input, when executed, prints out a list of all self-aware arrays of that length.
Here are some Self-aware arrays have been found. But I haven't found an algorithm.
N = 4
2, 0, 2, 0
1, 2, 1, 0
N = 5
2, 1, 2, 0, 0
N = 6
None
N = 7
3, 2, 1, 1, 0, 0, 0
N = 8
4, 2, 1, 0, 1, 0, 0, 0
N = 9
5, 2, 1, 0, 0, 1, 0, 0, 0
N = 10
6, 2, 1, 0, 0, 0, 1, 0, 0, 0
N = 11
7, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0
N = a
a-4, 2, 1, <a-7 0s>, 1, 0, 0, 0
Seems we have the properties
sum(A[i]) = n = sum(i * A[i])
A[0] = sum( (i-1) * A[i] ) i>=2
arrays algorithm
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
Self-Aware Arrays.
An integer array int A is self-aware if for each i < A.length, A[i] is the exact number of occurrences of i in A.
For example, [2, 0, 2, 0] is self-aware.
Write an algorithm accepts a positive integer n as input, when executed, prints out a list of all self-aware arrays of that length.
Here are some Self-aware arrays have been found. But I haven't found an algorithm.
N = 4
2, 0, 2, 0
1, 2, 1, 0
N = 5
2, 1, 2, 0, 0
N = 6
None
N = 7
3, 2, 1, 1, 0, 0, 0
N = 8
4, 2, 1, 0, 1, 0, 0, 0
N = 9
5, 2, 1, 0, 0, 1, 0, 0, 0
N = 10
6, 2, 1, 0, 0, 0, 1, 0, 0, 0
N = 11
7, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0
N = a
a-4, 2, 1, <a-7 0s>, 1, 0, 0, 0
Seems we have the properties
sum(A[i]) = n = sum(i * A[i])
A[0] = sum( (i-1) * A[i] ) i>=2
arrays algorithm
arrays algorithm
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
edited Nov 15 '18 at 15:53
Shang Gao
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
asked Nov 14 '18 at 21:59
Shang GaoShang Gao
212
212
What have you tried so far?
– jrook
Nov 14 '18 at 22:02
See Self-descriptive number for a reference. en.m.wikipedia.org/wiki/Self-descriptive_number
– Shang Gao
Nov 15 '18 at 15:51
add a comment |
What have you tried so far?
– jrook
Nov 14 '18 at 22:02
See Self-descriptive number for a reference. en.m.wikipedia.org/wiki/Self-descriptive_number
– Shang Gao
Nov 15 '18 at 15:51
What have you tried so far?
– jrook
Nov 14 '18 at 22:02
What have you tried so far?
– jrook
Nov 14 '18 at 22:02
See Self-descriptive number for a reference. en.m.wikipedia.org/wiki/Self-descriptive_number
– Shang Gao
Nov 15 '18 at 15:51
See Self-descriptive number for a reference. en.m.wikipedia.org/wiki/Self-descriptive_number
– Shang Gao
Nov 15 '18 at 15:51
add a comment |
1 Answer
1
active
oldest
votes
Outer loop: Generate all candidates of length n. This can be done with any convenient set product software; you need n elements in the range 0:n-1. For each candidate, filter and check (below).
Filter:
- The sum of the candidates elements must equal
n. Note that if you're writing your own candidate generator, you can use this to greatly reduce the quantity of candidates. - The first element can never be 0.
Check:
Count the quantity of each number 0:n-1 in order. The resulting list is your check count. If this list is equal to the original candidate, you have a solution: print it.
Shortcuts:
Are you allowed to use known results? It's long proved that for n >= 7, the formula you give is the only solution: you can directly generate the one solution and return. For n < 7, brute force will produce all valid solutions quite quickly.
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
How can we prove for n >= 7, the formul is the only solution ?
– Shang Gao
Nov 15 '18 at 1:51
The first place I saw it proved was, I think, in Hofstadter's "MetaMagical Themas" column in Scientific American. I've seen it in other places. You can either research the known solutions, or work through the combinatorics. There are some early restrictions that make it easier: the sum of the array must ben, and the first element cannot be0. Once you get up to 7 and 8, I recall that there's an inductive proof for the others.
– Prune
Nov 15 '18 at 16:44
add a comment |
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Outer loop: Generate all candidates of length n. This can be done with any convenient set product software; you need n elements in the range 0:n-1. For each candidate, filter and check (below).
Filter:
- The sum of the candidates elements must equal
n. Note that if you're writing your own candidate generator, you can use this to greatly reduce the quantity of candidates. - The first element can never be 0.
Check:
Count the quantity of each number 0:n-1 in order. The resulting list is your check count. If this list is equal to the original candidate, you have a solution: print it.
Shortcuts:
Are you allowed to use known results? It's long proved that for n >= 7, the formula you give is the only solution: you can directly generate the one solution and return. For n < 7, brute force will produce all valid solutions quite quickly.
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
How can we prove for n >= 7, the formul is the only solution ?
– Shang Gao
Nov 15 '18 at 1:51
The first place I saw it proved was, I think, in Hofstadter's "MetaMagical Themas" column in Scientific American. I've seen it in other places. You can either research the known solutions, or work through the combinatorics. There are some early restrictions that make it easier: the sum of the array must ben, and the first element cannot be0. Once you get up to 7 and 8, I recall that there's an inductive proof for the others.
– Prune
Nov 15 '18 at 16:44
add a comment |
Outer loop: Generate all candidates of length n. This can be done with any convenient set product software; you need n elements in the range 0:n-1. For each candidate, filter and check (below).
Filter:
- The sum of the candidates elements must equal
n. Note that if you're writing your own candidate generator, you can use this to greatly reduce the quantity of candidates. - The first element can never be 0.
Check:
Count the quantity of each number 0:n-1 in order. The resulting list is your check count. If this list is equal to the original candidate, you have a solution: print it.
Shortcuts:
Are you allowed to use known results? It's long proved that for n >= 7, the formula you give is the only solution: you can directly generate the one solution and return. For n < 7, brute force will produce all valid solutions quite quickly.
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
How can we prove for n >= 7, the formul is the only solution ?
– Shang Gao
Nov 15 '18 at 1:51
The first place I saw it proved was, I think, in Hofstadter's "MetaMagical Themas" column in Scientific American. I've seen it in other places. You can either research the known solutions, or work through the combinatorics. There are some early restrictions that make it easier: the sum of the array must ben, and the first element cannot be0. Once you get up to 7 and 8, I recall that there's an inductive proof for the others.
– Prune
Nov 15 '18 at 16:44
add a comment |
Outer loop: Generate all candidates of length n. This can be done with any convenient set product software; you need n elements in the range 0:n-1. For each candidate, filter and check (below).
Filter:
- The sum of the candidates elements must equal
n. Note that if you're writing your own candidate generator, you can use this to greatly reduce the quantity of candidates. - The first element can never be 0.
Check:
Count the quantity of each number 0:n-1 in order. The resulting list is your check count. If this list is equal to the original candidate, you have a solution: print it.
Shortcuts:
Are you allowed to use known results? It's long proved that for n >= 7, the formula you give is the only solution: you can directly generate the one solution and return. For n < 7, brute force will produce all valid solutions quite quickly.
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
Outer loop: Generate all candidates of length n. This can be done with any convenient set product software; you need n elements in the range 0:n-1. For each candidate, filter and check (below).
Filter:
- The sum of the candidates elements must equal
n. Note that if you're writing your own candidate generator, you can use this to greatly reduce the quantity of candidates. - The first element can never be 0.
Check:
Count the quantity of each number 0:n-1 in order. The resulting list is your check count. If this list is equal to the original candidate, you have a solution: print it.
Shortcuts:
Are you allowed to use known results? It's long proved that for n >= 7, the formula you give is the only solution: you can directly generate the one solution and return. For n < 7, brute force will produce all valid solutions quite quickly.
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
answered Nov 14 '18 at 23:20
PrunePrune
44.7k143457
44.7k143457
How can we prove for n >= 7, the formul is the only solution ?
– Shang Gao
Nov 15 '18 at 1:51
The first place I saw it proved was, I think, in Hofstadter's "MetaMagical Themas" column in Scientific American. I've seen it in other places. You can either research the known solutions, or work through the combinatorics. There are some early restrictions that make it easier: the sum of the array must ben, and the first element cannot be0. Once you get up to 7 and 8, I recall that there's an inductive proof for the others.
– Prune
Nov 15 '18 at 16:44
add a comment |
How can we prove for n >= 7, the formul is the only solution ?
– Shang Gao
Nov 15 '18 at 1:51
The first place I saw it proved was, I think, in Hofstadter's "MetaMagical Themas" column in Scientific American. I've seen it in other places. You can either research the known solutions, or work through the combinatorics. There are some early restrictions that make it easier: the sum of the array must ben, and the first element cannot be0. Once you get up to 7 and 8, I recall that there's an inductive proof for the others.
– Prune
Nov 15 '18 at 16:44
How can we prove for n >= 7, the formul is the only solution ?
– Shang Gao
Nov 15 '18 at 1:51
How can we prove for n >= 7, the formul is the only solution ?
– Shang Gao
Nov 15 '18 at 1:51
The first place I saw it proved was, I think, in Hofstadter's "MetaMagical Themas" column in Scientific American. I've seen it in other places. You can either research the known solutions, or work through the combinatorics. There are some early restrictions that make it easier: the sum of the array must be
n, and the first element cannot be 0. Once you get up to 7 and 8, I recall that there's an inductive proof for the others.– Prune
Nov 15 '18 at 16:44
The first place I saw it proved was, I think, in Hofstadter's "MetaMagical Themas" column in Scientific American. I've seen it in other places. You can either research the known solutions, or work through the combinatorics. There are some early restrictions that make it easier: the sum of the array must be
n, and the first element cannot be 0. Once you get up to 7 and 8, I recall that there's an inductive proof for the others.– Prune
Nov 15 '18 at 16:44
add a comment |
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What have you tried so far?
– jrook
Nov 14 '18 at 22:02
See Self-descriptive number for a reference. en.m.wikipedia.org/wiki/Self-descriptive_number
– Shang Gao
Nov 15 '18 at 15:51