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6

$begingroup$

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

share|improve this question

asked 1 hour ago

Esolanging FruitEsolanging Fruit

8,65932774

$endgroup$

add a comment | 

6

$begingroup$

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

share|improve this question

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

share|improve this question

asked 1 hour ago

Esolanging FruitEsolanging Fruit

8,65932774

$endgroup$

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]

share|improve this question

asked 1 hour ago

Esolanging FruitEsolanging Fruit

8,65932774

share|improve this question

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

2 Answers
2

active

oldest

votes

3

$begingroup$

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!

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Chas BrownChas Brown

5,0991523

$endgroup$

add a comment | 

2

$begingroup$

Perl 6, 48 41 bytes

->k1,$_+10**min($_,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

share|improve this answer

edited 33 mins ago

answered 57 mins ago

Jo KingJo King

26.3k364129

$endgroup$

add a comment | 

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3

$begingroup$

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!

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Chas BrownChas Brown

5,0991523

$endgroup$

add a comment | 

3

$begingroup$

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!

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Chas BrownChas Brown

5,0991523

$endgroup$

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!

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Chas BrownChas Brown

5,0991523

$endgroup$

f=lambda k,n=1:n<k and[n]+f(k,n+10**~-len(`min(n,k-n)`))or[n]

share|improve this answer

edited 38 mins ago

answered 44 mins ago

Chas BrownChas Brown

5,0991523

answered 44 mins ago

Chas BrownChas Brown

5,0991523

answered 44 mins ago

Chas BrownChas Brown

5,0991523

2

$begingroup$

Perl 6, 48 41 bytes

->k1,$_+10**min($_,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

share|improve this answer

edited 33 mins ago

answered 57 mins ago

Jo KingJo King

26.3k364129

$endgroup$

add a comment | 

2

$begingroup$

Perl 6, 48 41 bytes

->k1,$_+10**min($_,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

share|improve this answer

edited 33 mins ago

answered 57 mins ago

Jo KingJo King

26.3k364129

$endgroup$

add a comment | 

$begingroup$

Perl 6, 48 41 bytes

->k1,$_+10**min($_,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

share|improve this answer

edited 33 mins ago

answered 57 mins ago

Jo KingJo King

26.3k364129

$endgroup$

->k1,$_+10**min($_,k-$_).comb/10...k

->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

share|improve this answer

edited 33 mins ago

answered 57 mins ago

Jo KingJo King

26.3k364129

answered 57 mins ago

Jo KingJo King

26.3k364129

answered 57 mins ago

Jo KingJo King

26.3k364129