Expand and Contract
$begingroup$
Take a positive integer $k$ as input. Start with $n := 1$ and repeatedly increase $n$ by the largest integer power of ten $i$ such that $i le n$ and $i + n le k$.
Repeat until $n = k$ and return a list of all intermediate values of $n$, including both the initial $1$ and the final $k$.
During this process, growth will initially be limited by the former inequality, and only afterwards by the latter; the growth will take the form of an initial "expansion" period, during which $n$ is increased by ever-larger powers, followed by a "contract" period, during which $n$ is increased by ever-smaller powers in order to "zoom in" on the correct number.
Test Cases
1 => [1]
10 => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
321 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 310, 320, 321]
1002 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 400, 500, 600, 700, 800, 900,
1000, 1001, 1002]
This is code-golf, so the shortest answer (in bytes) wins.
code-golf number decimal
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
$endgroup$
add a comment |
$begingroup$
Take a positive integer $k$ as input. Start with $n := 1$ and repeatedly increase $n$ by the largest integer power of ten $i$ such that $i le n$ and $i + n le k$.
Repeat until $n = k$ and return a list of all intermediate values of $n$, including both the initial $1$ and the final $k$.
During this process, growth will initially be limited by the former inequality, and only afterwards by the latter; the growth will take the form of an initial "expansion" period, during which $n$ is increased by ever-larger powers, followed by a "contract" period, during which $n$ is increased by ever-smaller powers in order to "zoom in" on the correct number.
Test Cases
1 => [1]
10 => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
321 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 310, 320, 321]
1002 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 400, 500, 600, 700, 800, 900,
1000, 1001, 1002]
This is code-golf, so the shortest answer (in bytes) wins.
code-golf number decimal
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
$endgroup$
add a comment |
$begingroup$
Take a positive integer $k$ as input. Start with $n := 1$ and repeatedly increase $n$ by the largest integer power of ten $i$ such that $i le n$ and $i + n le k$.
Repeat until $n = k$ and return a list of all intermediate values of $n$, including both the initial $1$ and the final $k$.
During this process, growth will initially be limited by the former inequality, and only afterwards by the latter; the growth will take the form of an initial "expansion" period, during which $n$ is increased by ever-larger powers, followed by a "contract" period, during which $n$ is increased by ever-smaller powers in order to "zoom in" on the correct number.
Test Cases
1 => [1]
10 => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
321 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 310, 320, 321]
1002 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 400, 500, 600, 700, 800, 900,
1000, 1001, 1002]
This is code-golf, so the shortest answer (in bytes) wins.
code-golf number decimal
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
$endgroup$
Take a positive integer $k$ as input. Start with $n := 1$ and repeatedly increase $n$ by the largest integer power of ten $i$ such that $i le n$ and $i + n le k$.
Repeat until $n = k$ and return a list of all intermediate values of $n$, including both the initial $1$ and the final $k$.
During this process, growth will initially be limited by the former inequality, and only afterwards by the latter; the growth will take the form of an initial "expansion" period, during which $n$ is increased by ever-larger powers, followed by a "contract" period, during which $n$ is increased by ever-smaller powers in order to "zoom in" on the correct number.
Test Cases
1 => [1]
10 => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
321 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 310, 320, 321]
1002 => [1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 20, 30, 40, 50, 60, 70, 80, 90,
100, 200, 300, 400, 500, 600, 700, 800, 900,
1000, 1001, 1002]
This is code-golf, so the shortest answer (in bytes) wins.
code-golf number decimal
code-golf number decimal
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
asked 1 hour ago
Esolanging FruitEsolanging Fruit
8,65932774
8,65932774
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Python 2, 61 bytes
f=lambda k,n=1:n<k and[n]+f(k,n+10**~-len(`min(n,k-n)`))or[n]
Try it online!
answered 37 mins ago
Chas BrownChas Brown
5,0891523
$endgroup$
add a comment |
$begingroup$
Perl 6, 48 41 bytes
->k{1,{$_+10**min($_,k-$_).comb/10}...k}
Try it online!
Explanation:
->k{ } # Anonymous code block taking k
1, ...k # Start a sequence from 1 to k
{ } # Where each element is
$_+ # The previous element plus
10** # 10 to the power of
.comb # The length of
min($_,k-$_) # The min of the current count and the remainder
/10 # Minus one
answered 49 mins ago
Jo KingJo King
26.3k364129
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Python 2, 61 bytes
f=lambda k,n=1:n<k and[n]+f(k,n+10**~-len(`min(n,k-n)`))or[n]
Try it online!
answered 37 mins ago
Chas BrownChas Brown
5,0891523
$endgroup$
add a comment |
$begingroup$
Python 2, 61 bytes
f=lambda k,n=1:n<k and[n]+f(k,n+10**~-len(`min(n,k-n)`))or[n]
Try it online!
answered 37 mins ago
Chas BrownChas Brown
5,0891523
$endgroup$
add a comment |
$begingroup$
Python 2, 61 bytes
f=lambda k,n=1:n<k and[n]+f(k,n+10**~-len(`min(n,k-n)`))or[n]
Try it online!
answered 37 mins ago
Chas BrownChas Brown
5,0891523
$endgroup$
Python 2, 61 bytes
f=lambda k,n=1:n<k and[n]+f(k,n+10**~-len(`min(n,k-n)`))or[n]
Try it online!
answered 37 mins ago
Chas BrownChas Brown
5,0891523
edited 31 mins ago
answered 37 mins ago
Chas BrownChas Brown
5,0891523
answered 37 mins ago
Chas BrownChas Brown
5,0891523
answered 37 mins ago
Chas BrownChas Brown
5,0891523
5,0891523
add a comment |
add a comment |
$begingroup$
Perl 6, 48 41 bytes
->k{1,{$_+10**min($_,k-$_).comb/10}...k}
Try it online!
Explanation:
->k{ } # Anonymous code block taking k
1, ...k # Start a sequence from 1 to k
{ } # Where each element is
$_+ # The previous element plus
10** # 10 to the power of
.comb # The length of
min($_,k-$_) # The min of the current count and the remainder
/10 # Minus one
answered 49 mins ago
Jo KingJo King
26.3k364129
$endgroup$
add a comment |
$begingroup$
Perl 6, 48 41 bytes
->k{1,{$_+10**min($_,k-$_).comb/10}...k}
Try it online!
Explanation:
->k{ } # Anonymous code block taking k
1, ...k # Start a sequence from 1 to k
{ } # Where each element is
$_+ # The previous element plus
10** # 10 to the power of
.comb # The length of
min($_,k-$_) # The min of the current count and the remainder
/10 # Minus one
answered 49 mins ago
Jo KingJo King
26.3k364129
$endgroup$
add a comment |
$begingroup$
Perl 6, 48 41 bytes
->k{1,{$_+10**min($_,k-$_).comb/10}...k}
Try it online!
Explanation:
->k{ } # Anonymous code block taking k
1, ...k # Start a sequence from 1 to k
{ } # Where each element is
$_+ # The previous element plus
10** # 10 to the power of
.comb # The length of
min($_,k-$_) # The min of the current count and the remainder
/10 # Minus one
answered 49 mins ago
Jo KingJo King
26.3k364129
$endgroup$
Perl 6, 48 41 bytes
->k{1,{$_+10**min($_,k-$_).comb/10}...k}
Try it online!
Explanation:
->k{ } # Anonymous code block taking k
1, ...k # Start a sequence from 1 to k
{ } # Where each element is
$_+ # The previous element plus
10** # 10 to the power of
.comb # The length of
min($_,k-$_) # The min of the current count and the remainder
/10 # Minus one
answered 49 mins ago
Jo KingJo King
26.3k364129
edited 25 mins ago
answered 49 mins ago
Jo KingJo King
26.3k364129
answered 49 mins ago
Jo KingJo King
26.3k364129
answered 49 mins ago
Jo KingJo King
26.3k364129
26.3k364129
add a comment |
add a comment |
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…Please make sure to answer the question and provide sufficient detail.
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