Find Array Runs

Clash Royale CLAN TAG#URR8PPP

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Find the runs inside an array

A run is defined as three or more numbers that increment from the previous with a constant step. For example [1,2,3] would be a run with step 1, [1,3,5,7] would be a run with step 2, and [1,2,4,5] is not a run.

We can express these runs by the notation "i to j by s" where i is the first number of the run, j is the last number of the run, and s is the step. However, runs of step 1 will be expressed "i to j".

So using the arrays before, we get:

• [1,2,3] -> "1to3"




• [1,3,5,7] -> "1to7by2"




• [1,2,4,5] -> "1 2 4 5"

In this challenge, it is your task to do this for arrays that may have multiple runs.

Example Python code with recursion:

def arr_comp_rec(a, start_index):
 # Early exit and recursion end point
 if start_index == len(a)-1:
 return str(a[-1])
 elif start_index == len(a):
 return ''

 # Keep track of first delta to compare while searching
 first_delta = a[start_index+1] - a[start_index]
 last = True
 for i in range(start_index, len(a)-1):
 delta = a[i+1] - a[i]
 if delta != first_delta:
 last = False
 break
 # If it ran through the for loop, we need to make sure it gets the last value
 if last: i += 1

 if i - start_index > 1:
 # There is more than 2 numbers between the indexes
 if first_delta == 1:
 # We don't need by if step = 1
 return "to ".format(a[start_index], a[i]) + arr_comp_rec(a, i+1)
 else:
 return "toby ".format(a[start_index], a[i], first_delta) + arr_comp_rec(a, i+1)
 else:
 # There is only one number we can return
 return " ".format(a[start_index]) + arr_comp_rec(a, i)

IO is flexible

Input

Array of sorted positive ints (no duplicates)

Output

String of the runs separated by a space, or a string array of the runs

Does not need to be greedy in a particular direction

Can have trailing whitespace

Test Cases

In: [1000, 1002, 1004, 1006, 1008, 1010]
Out: "1000to1010by2"

In: [1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233]
Out: "1to3 5 8 13 21 34 55 89 144 233"

In: [10, 20, 30, 40, 60]
Out: "10to40by10 60"

In: [5, 6, 8, 11, 15, 16, 17]
Out: "5 6 8 11 15to17"

In: [1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 30, 45, 50, 60, 70, 80, 90, 91, 93]
Out: "1to7 9to15by2 30 45 50to90by10 91 93"

This is code-golf so least number of bytes wins.

code-golf array-manipulation

share|improve this question

edited Aug 27 at 3:20

asked Aug 26 at 23:19

WretchedLout

15016

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  related
  – Luis felipe De jesus Munoz
  Aug 26 at 23:23
  

  
  
  
  


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  2
  

  
  
  
  
  
  

  
  Must it be greedy left-to-right? (i.e. can [4, 5, 6, 7, 9, 11, 13, 15] not be 4to6 7to15by2?)
  – Jonathan Allan
  Aug 26 at 23:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan No, it does not necessarily need to be left greedy.
  – WretchedLout
  Aug 27 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose there will be no duplicate entries?
  – user71546
  Aug 27 at 1:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Only Positive Integers. @Οurous Trailing whitespace acceptable.
  – WretchedLout
  Aug 27 at 3:19
  

  
  
  
  

 | 
show 4 more comments

up vote
14
down vote

favorite

Find the runs inside an array

A run is defined as three or more numbers that increment from the previous with a constant step. For example [1,2,3] would be a run with step 1, [1,3,5,7] would be a run with step 2, and [1,2,4,5] is not a run.

We can express these runs by the notation "i to j by s" where i is the first number of the run, j is the last number of the run, and s is the step. However, runs of step 1 will be expressed "i to j".

So using the arrays before, we get:

• [1,2,3] -> "1to3"




• [1,3,5,7] -> "1to7by2"




• [1,2,4,5] -> "1 2 4 5"

In this challenge, it is your task to do this for arrays that may have multiple runs.

Example Python code with recursion:

def arr_comp_rec(a, start_index):
 # Early exit and recursion end point
 if start_index == len(a)-1:
 return str(a[-1])
 elif start_index == len(a):
 return ''

 # Keep track of first delta to compare while searching
 first_delta = a[start_index+1] - a[start_index]
 last = True
 for i in range(start_index, len(a)-1):
 delta = a[i+1] - a[i]
 if delta != first_delta:
 last = False
 break
 # If it ran through the for loop, we need to make sure it gets the last value
 if last: i += 1

 if i - start_index > 1:
 # There is more than 2 numbers between the indexes
 if first_delta == 1:
 # We don't need by if step = 1
 return "to ".format(a[start_index], a[i]) + arr_comp_rec(a, i+1)
 else:
 return "toby ".format(a[start_index], a[i], first_delta) + arr_comp_rec(a, i+1)
 else:
 # There is only one number we can return
 return " ".format(a[start_index]) + arr_comp_rec(a, i)

IO is flexible

Input

Array of sorted positive ints (no duplicates)

Output

String of the runs separated by a space, or a string array of the runs

Does not need to be greedy in a particular direction

Can have trailing whitespace

Test Cases

In: [1000, 1002, 1004, 1006, 1008, 1010]
Out: "1000to1010by2"

In: [1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233]
Out: "1to3 5 8 13 21 34 55 89 144 233"

In: [10, 20, 30, 40, 60]
Out: "10to40by10 60"

In: [5, 6, 8, 11, 15, 16, 17]
Out: "5 6 8 11 15to17"

In: [1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 30, 45, 50, 60, 70, 80, 90, 91, 93]
Out: "1to7 9to15by2 30 45 50to90by10 91 93"

This is code-golf so least number of bytes wins.

code-golf array-manipulation

share|improve this question

edited Aug 27 at 3:20

asked Aug 26 at 23:19

WretchedLout

15016

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  related
  – Luis felipe De jesus Munoz
  Aug 26 at 23:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Must it be greedy left-to-right? (i.e. can [4, 5, 6, 7, 9, 11, 13, 15] not be 4to6 7to15by2?)
  – Jonathan Allan
  Aug 26 at 23:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan No, it does not necessarily need to be left greedy.
  – WretchedLout
  Aug 27 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose there will be no duplicate entries?
  – user71546
  Aug 27 at 1:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Only Positive Integers. @Οurous Trailing whitespace acceptable.
  – WretchedLout
  Aug 27 at 3:19
  

  
  
  
  

 | 
show 4 more comments

up vote
14
down vote

favorite

up vote
14
down vote

favorite

Find the runs inside an array

A run is defined as three or more numbers that increment from the previous with a constant step. For example [1,2,3] would be a run with step 1, [1,3,5,7] would be a run with step 2, and [1,2,4,5] is not a run.

We can express these runs by the notation "i to j by s" where i is the first number of the run, j is the last number of the run, and s is the step. However, runs of step 1 will be expressed "i to j".

So using the arrays before, we get:

• [1,2,3] -> "1to3"




• [1,3,5,7] -> "1to7by2"




• [1,2,4,5] -> "1 2 4 5"

In this challenge, it is your task to do this for arrays that may have multiple runs.

Example Python code with recursion:

def arr_comp_rec(a, start_index):
 # Early exit and recursion end point
 if start_index == len(a)-1:
 return str(a[-1])
 elif start_index == len(a):
 return ''

 # Keep track of first delta to compare while searching
 first_delta = a[start_index+1] - a[start_index]
 last = True
 for i in range(start_index, len(a)-1):
 delta = a[i+1] - a[i]
 if delta != first_delta:
 last = False
 break
 # If it ran through the for loop, we need to make sure it gets the last value
 if last: i += 1

 if i - start_index > 1:
 # There is more than 2 numbers between the indexes
 if first_delta == 1:
 # We don't need by if step = 1
 return "to ".format(a[start_index], a[i]) + arr_comp_rec(a, i+1)
 else:
 return "toby ".format(a[start_index], a[i], first_delta) + arr_comp_rec(a, i+1)
 else:
 # There is only one number we can return
 return " ".format(a[start_index]) + arr_comp_rec(a, i)

IO is flexible

Input

Array of sorted positive ints (no duplicates)

Output

String of the runs separated by a space, or a string array of the runs

Does not need to be greedy in a particular direction

Can have trailing whitespace

Test Cases

In: [1000, 1002, 1004, 1006, 1008, 1010]
Out: "1000to1010by2"

In: [1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233]
Out: "1to3 5 8 13 21 34 55 89 144 233"

In: [10, 20, 30, 40, 60]
Out: "10to40by10 60"

In: [5, 6, 8, 11, 15, 16, 17]
Out: "5 6 8 11 15to17"

In: [1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 30, 45, 50, 60, 70, 80, 90, 91, 93]
Out: "1to7 9to15by2 30 45 50to90by10 91 93"

This is code-golf so least number of bytes wins.

code-golf array-manipulation

share|improve this question

edited Aug 27 at 3:20

asked Aug 26 at 23:19

WretchedLout

15016

Find the runs inside an array

A run is defined as three or more numbers that increment from the previous with a constant step. For example [1,2,3] would be a run with step 1, [1,3,5,7] would be a run with step 2, and [1,2,4,5] is not a run.

We can express these runs by the notation "i to j by s" where i is the first number of the run, j is the last number of the run, and s is the step. However, runs of step 1 will be expressed "i to j".

So using the arrays before, we get:

• [1,2,3] -> "1to3"




• [1,3,5,7] -> "1to7by2"




• [1,2,4,5] -> "1 2 4 5"

In this challenge, it is your task to do this for arrays that may have multiple runs.

Example Python code with recursion:

def arr_comp_rec(a, start_index):
 # Early exit and recursion end point
 if start_index == len(a)-1:
 return str(a[-1])
 elif start_index == len(a):
 return ''

 # Keep track of first delta to compare while searching
 first_delta = a[start_index+1] - a[start_index]
 last = True
 for i in range(start_index, len(a)-1):
 delta = a[i+1] - a[i]
 if delta != first_delta:
 last = False
 break
 # If it ran through the for loop, we need to make sure it gets the last value
 if last: i += 1

 if i - start_index > 1:
 # There is more than 2 numbers between the indexes
 if first_delta == 1:
 # We don't need by if step = 1
 return "to ".format(a[start_index], a[i]) + arr_comp_rec(a, i+1)
 else:
 return "toby ".format(a[start_index], a[i], first_delta) + arr_comp_rec(a, i+1)
 else:
 # There is only one number we can return
 return " ".format(a[start_index]) + arr_comp_rec(a, i)

IO is flexible

Input

Array of sorted positive ints (no duplicates)

Output

String of the runs separated by a space, or a string array of the runs

Does not need to be greedy in a particular direction

Can have trailing whitespace

Test Cases

In: [1000, 1002, 1004, 1006, 1008, 1010]
Out: "1000to1010by2"

In: [1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233]
Out: "1to3 5 8 13 21 34 55 89 144 233"

In: [10, 20, 30, 40, 60]
Out: "10to40by10 60"

In: [5, 6, 8, 11, 15, 16, 17]
Out: "5 6 8 11 15to17"

In: [1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 30, 45, 50, 60, 70, 80, 90, 91, 93]
Out: "1to7 9to15by2 30 45 50to90by10 91 93"

This is code-golf so least number of bytes wins.

code-golf array-manipulation

share|improve this question

edited Aug 27 at 3:20

asked Aug 26 at 23:19

WretchedLout

15016

share|improve this question

share|improve this question

edited Aug 27 at 3:20

edited Aug 27 at 3:20

edited Aug 27 at 3:20

asked Aug 26 at 23:19

WretchedLout

15016

asked Aug 26 at 23:19

WretchedLout

15016

asked Aug 26 at 23:19

WretchedLout

15016

15016

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  related
  – Luis felipe De jesus Munoz
  Aug 26 at 23:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Must it be greedy left-to-right? (i.e. can [4, 5, 6, 7, 9, 11, 13, 15] not be 4to6 7to15by2?)
  – Jonathan Allan
  Aug 26 at 23:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan No, it does not necessarily need to be left greedy.
  – WretchedLout
  Aug 27 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose there will be no duplicate entries?
  – user71546
  Aug 27 at 1:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Only Positive Integers. @Οurous Trailing whitespace acceptable.
  – WretchedLout
  Aug 27 at 3:19
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  related
  – Luis felipe De jesus Munoz
  Aug 26 at 23:23
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Must it be greedy left-to-right? (i.e. can [4, 5, 6, 7, 9, 11, 13, 15] not be 4to6 7to15by2?)
  – Jonathan Allan
  Aug 26 at 23:44
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @JonathanAllan No, it does not necessarily need to be left greedy.
  – WretchedLout
  Aug 27 at 0:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I suppose there will be no duplicate entries?
  – user71546
  Aug 27 at 1:18
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Only Positive Integers. @Οurous Trailing whitespace acceptable.
  – WretchedLout
  Aug 27 at 3:19
  

  
  
  
  

1

1

related
– Luis felipe De jesus Munoz
Aug 26 at 23:23

related
– Luis felipe De jesus Munoz
Aug 26 at 23:23

2

2

Must it be greedy left-to-right? (i.e. can [4, 5, 6, 7, 9, 11, 13, 15] not be 4to6 7to15by2?)
– Jonathan Allan
Aug 26 at 23:44

Must it be greedy left-to-right? (i.e. can [4, 5, 6, 7, 9, 11, 13, 15] not be 4to6 7to15by2?)
– Jonathan Allan
Aug 26 at 23:44

1

1

@JonathanAllan No, it does not necessarily need to be left greedy.
– WretchedLout
Aug 27 at 0:01

@JonathanAllan No, it does not necessarily need to be left greedy.
– WretchedLout
Aug 27 at 0:01

I suppose there will be no duplicate entries?
– user71546
Aug 27 at 1:18

I suppose there will be no duplicate entries?
– user71546
Aug 27 at 1:18

1

1

Only Positive Integers. @Οurous Trailing whitespace acceptable.
– WretchedLout
Aug 27 at 3:19

Only Positive Integers. @Οurous Trailing whitespace acceptable.
– WretchedLout
Aug 27 at 3:19

 | 
show 4 more comments

16 Answers
16

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oldest

votes

up vote
5
down vote



accepted

Jelly,  42  40 bytes

-2 thanks to Kevin Cruijssen (filter out twos, ḟ2, rather than replacing twos with zeros, 2,0y)

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K

A full program printing the result.

(As a monadic Link a list containing a mixture of integers and characters would be yielded)

Try it online!

(Too inefficient for the largest test-case to complete within 60s, so I removed [1,2,3,4].)

How?

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K - Main Link: list of numbers
ŒṖ - all partitions
 ƊÞ - sort by last three links as a monad:
 Ʋ€ - last four links as a monad for €ach:
 I - incremental differences (of the part)
 E - all equal?
 L - length (of the part)
 × - multiply
 ḟ2 - filter discard twos
 S - sum
 Ṫ - tail (gets us the desired partition of the input)
 µ ) - perform this monadic chain for €ach:
 . - literal 0.5
 ị - index into (the part) - getting [tail,head]
 U - upend - getting [head,tail]
 Ɗ - last three links as a monad:
 I - incremental differences (of the part)
 1 - literal one
 ḟ - filter discard (remove the ones)
 , - pair -> [[head,tail],[deltasWithoutOnes]]
 Q€ - de-duplicate €ach -> [[head,tail],[delta]] or [[head,tail],] or [[loneValue],]
 F - flatten -> [head,tail,delta] or [head,tail] or [loneValue]
 $ - last two links as a monad:
 Ɗ - last three links as a monad:
 “to“by” - literal list [['t', 'o'], ['b', 'y']]
 Ṗ - pop (get flattened result without rightmost entry)
 ṁ - mould ["to","by"] like that (i.e. ["to","by"] or ["to"] or )
 ż - zip together 
 K - join with spaces
 - implicit print

share|improve this answer

edited Aug 28 at 17:19

answered Aug 27 at 3:07

Jonathan Allan

48k534158

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  'by' should not be used if step = 1
  – WretchedLout
  Aug 27 at 3:25
  

  
  
  
  


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  oh Gawwwd :p Thanks for the heads-up!
  – Jonathan Allan
  Aug 27 at 3:30
  

  
  
  
  


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  2,0ySƲÞ can be golfed to ḟ2SƊÞ for -2 bytes.
  – Kevin Cruijssen
  Aug 28 at 9:42
  
  

  
  
  
  


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  @KevinCruijssen very true, nice golf, thanks
  – Jonathan Allan
  Aug 28 at 17:02
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  @KevinCruijssen FYI I remember why that was there now - originally I was trying to do my sorting with a product, P, rather than a sum S and would have needed the zeros.
  – Jonathan Allan
  Aug 28 at 17:21
  

  
  
  
  

 | 
show 1 more comment

up vote
6
down vote

Swift, 246 bytes

func f(_ a:[Int])
var c=a.count,m=a+a,z=0,r:String=""
for i in 1..<cm[i]-=a[i-1];if m[i]==m[i-1]m[i-1]=0;m[0]=1
for i in 0..<cif m[i]==0 z=1else if z==0r+=" (a[i])"elser+="to(a[i])"+(m[i]>1 ? "by(m[i])":"");z=0;m[i+1]=1
print(r)

Try it online!

share|improve this answer

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

answered Aug 27 at 8:14

drawnonward

1611

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  Welcome to Programming Puzzles & Code Golf! Very nice first answer. I hope you'll stay around because the Swift community here is... really small!
  – Mr. Xcoder
  Aug 27 at 8:23
  

  
  
  
  


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  This saves you a couple of bytes
  – Mr. Xcoder
  Aug 27 at 8:26
  

  
  
  
  

add a comment | 

up vote
3
down vote

K (ngn/k), 102 bytes

1_,//$H:" ",h:*x;l:+/&x=h+(!#x)*d:--/2#x;$[l>2;(H;"to";x l-1;(~d=1)#,"by",d;o l_x);l;H,o 1_x;""]x

Try it online!

share|improve this answer

edited Aug 27 at 1:13

answered Aug 27 at 0:52

ngn

6,10812256

add a comment | 

up vote
3
down vote

JavaScript (ES6), 129 bytes

Returns an array of strings.

a=>(a.map((n,i)=>n+[n-a[i+1]||''])+'').replace(/(d+)(-d+)(?:,d+2)+,(d+)/g,(_,a,b,c)=>a+'to'+(~b?c+'by'+-b:c)).split(/-d+,/)

Try it online!

How?

Step #1

We first append to each number a suffix consisting of a leading '-' followed by the difference with the next number, except for the last entry which is left unchanged. This new array is coerced to a string.

(a.map((n, i) => n + [n - a[i + 1] || '']) + '')

Example:

Input : [ 1, 2, 3, 5, 9, 11, 13, 20 ]
Output: "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"

Step #2

We identify all runs in the resulting string and replace them with the appropriate notation.

.replace(
 /(d+)(-d+)(?:,d+2)+,(d+)/g,
 (_, a, b, c) => a + 'to' + (~b ? c + 'by' + -b : c)
)

Example:

Input : "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"
Output: "1to3-2,5-4,9to13by2-7,20"

Step #3

Finally, we split the string on the remaining suffixes, including the trailing commas.

.split(/-d+,/)

Example:

Input : "1to3-2,5-4,9to13by2-7,20"
Output: [ '1to3', '5', '9to13by2', '20' ]

share|improve this answer

edited Aug 28 at 11:37

answered Aug 27 at 11:49

Arnauld

63.5k580268

add a comment | 

up vote
2
down vote

Ruby, 125 118 bytes

->ai=y=0;a.chunkx.map2-i<(s=w.size)?"#w[i*s]to#w[~i=0]"+"by#z"*(z<=>1)+' ':' '*i+w*' '*i=1*''

Try it online!

Explanation

Ruby's Enumerable has a useful chunk method that does precisely what we need here - it groups items by consecutive runs of the same return value from the block, in our case - the difference between the current (x) and previous (y) value.

The caveat is that such strategy won't capture the first element of the run, e.g. here only the two last elements are grouped together:

Input: [5, 6, 8, 11, 15, 16, 17]
Grouped: [[5, [5]], [1, [6]], [2, [8]], [3, [11]], [4, [15]], [1, [16, 17]]]

Therefore, while mapping to the correctly formatted strings, when we encounter a new potential run (chunk with > 1 item), we must track if the previous item was single (i=1) or already used in another run (i=0). If there is an unused single item, it becomes the starting point of the run, and lowers the chunk size threshold from 3 to 2.

share|improve this answer

edited Aug 27 at 20:30

answered Aug 27 at 11:06

Kirill L.

2,4061116

add a comment | 

up vote
2
down vote

R, 238 bytes

Almost certainly not the easiest way to do it.

function(x,r=rle(diff(x)),n=0,L=length)
while(n<L(x))
m=r$l[1]
cat(`if`(L(r$l)&&m>1,c(x[n+1],"to",x[n+m+1],ifelse(r$v[1]>1,paste0("by",r$v[1]),"")),x[n+1])," ",sep="")	
if(L(r$l)==0) break;
n=n+m+1*(m>1)
r=rle(diff(c(tail(x,L(x)-n))))

Try it online!

share|improve this answer

answered Aug 28 at 19:18

J.Doe

5218

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use F instead of n as it's already initialized to 0, which should save a few bytes, I think.
  – Giuseppe
  Aug 28 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This one is frustrating. You're nearly there by just using split, diff and rle. Unfortunately, the greedy search for runs means a lot of fiddling.
  – J.Doe
  Aug 29 at 10:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  214 bytes Almost there, the only problem is the uncorrect spacing sometimes...
  – digEmAll
  Aug 29 at 12:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That 'by'[D>1] is a good trick.
  – J.Doe
  Aug 29 at 12:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This seems to work fine 217 bytes
  – digEmAll
  Aug 29 at 18:46
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote

R, 180 175 bytes

r=rle(c(0,diff(a<-scan())));for(j in 1:sum(1|r$l))l=r$l[j];v=r$v[j];i=T+l-1;cat("if"(l>2-F,paste0(a[T][!F],"to",a[i],"by"[v>1],v[v>1]," "),c("",a[T:i])[-3^F]));T=i+1;F=l<3-F

Try it online!

Conceptually, this is a port of my Ruby answer, though obviously quite a bit different technically.

5 bytes saved by JayCe.

share|improve this answer

edited Aug 30 at 16:29

answered Aug 30 at 8:41

Kirill L.

2,4061116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wanted to to something with rle but was too lazy... you can save 1 byte doing sum(1|x) in place of length(x): TIO
  – JayCe
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually you can have just 1 cat for 175 bytes: TIO
  – JayCe
  Aug 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, great, thanks!
  – Kirill L.
  Aug 30 at 16:30
  

  
  
  
  

add a comment | 

up vote
1
down vote

JavaScript (Node.js), 177 173 bytes

a=>for(h=t=p=d=0,o=;(c=a[t])

Try it online!

share|improve this answer

edited Aug 27 at 4:53

answered Aug 27 at 4:43

user71546

1,31529

add a comment | 

up vote
1
down vote

Clean, 208 ... 185 bytes

import StdEnv,Data.List,Text
l=length
$=""
$k=last[u<+(if(v>)("to"<+last v<+if(u<v!!0-1)("by"<+(v!!0-u))"")"")+" "+ $t\i=:[u:v]<-inits k&t<-tails k|l(nub(zipWith(-)i v))<2&&l i<>2]

Try it online!

share|improve this answer

edited Aug 27 at 10:49

answered Aug 27 at 2:45

Οurous

5,1931931

add a comment | 

up vote
1
down vote

Jelly, 41 bytes

ŒṖIE€Ạ$>Ẉ2eƊƲƇṪµ.ịṚj⁾to;IḢ⁾by;ẋ⁻1$ƲƊµḊ¡€K

Try it online!

Full program.

share|improve this answer

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
1
down vote

Python 2, 170 166 bytes

def f(a):
 j=len(a)
 while j>2:
	l,r=a[:j:j-1];s=(r-l)/~-j
	if a[:j]==range(l,r+1,s):return[`l`+'to%d'%r+'by%d'%s*(s>1)]+f(a[j:])
	j-=1
 return a and[`a[0]`]+f(a[1:])

Try it online!

share|improve this answer

edited Aug 27 at 12:00

answered Aug 27 at 9:07

TFeld

11.2k2833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if this output format is valid; the runs should all be strings, no?
  – Erik the Outgolfer
  Aug 27 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EriktheOutgolfer Fixed
  – TFeld
  Aug 27 at 12:01
  

  
  
  
  

add a comment | 

up vote
1
down vote

Python 2, 138 136 bytes

-2 bytes thanks to Erik the Outgolfer.

r=;d=0
for n in input()+[0]:
 a=n-([n]+r)[-1]
 if(a-d)*d:
	if r[2:]:r=["%dto"%r[0]+`r[-1]`+"by%s"%d*(1%d)]
	print r.pop(0),
 r+=n,;d=a

Try it online!

share|improve this answer

edited Aug 27 at 12:24

answered Aug 27 at 12:05

ovs

17.2k21056

add a comment | 

up vote
1
down vote

gvm (commit 2612106) bytecode, 108 bytes

Expects the size of the array in one line, then the members each in one line.

Hexdump:

> hexdump -C findruns.bin
00000000 e1 0a 00 10 00 e1 0b 00 ff c8 92 00 f4 f7 10 00 |................|
00000010 01 b0 20 03 00 ff 0a 01 a2 01 c8 92 00 f4 01 c0 |.. .............|
00000020 03 00 ff 0a 02 d0 72 01 0a 03 c8 92 00 f4 05 b0 |......r.........|
00000030 20 a2 02 c0 02 02 6a 03 8b 00 ff f6 06 b0 20 a2 | .....j....... .|
00000040 02 f4 ce 0a 02 c8 92 00 f6 07 6a 03 8b 00 ff f6 |..........j.....|
00000050 f2 b9 66 01 a2 02 00 01 8a 03 f6 05 b9 69 01 a2 |..f..........i..|
00000060 03 92 00 f4 ac c0 74 6f 00 62 79 00 |......to.by.|
0000006c

Test runs:

> echo -e "7n5n6n8n11n15n16n17n" | ./gvm findruns.bin
5 6 8 11 15to17
> echo -e "20n1n2n3n4n5n6n7n9n11n13n15n30n45n50n60n70n80n90n91n93n" | ./gvm findruns.bin
1to7 9to15by2 30 45 50to90by10 91 93

Manually assembled from this:

0100 e1 rud ; read length of array
0101 0a 00 sta $00 ; -> to $00
0103 10 00 ldx #$00 ; loop counter
 readloop:
0105 e1 rud ; read unsigned number
0106 0b 00 ff sta $ff00,x ; store in array at ff00
0109 c8 inx ; next index
010a 92 00 cpx $00 ; length reached?
010c f4 f7 bne readloop ; no -> read next number
010e 10 00 ldx #$00 ; loop counter
0110 01 ; 'lda $20b0', to skip next instruction
 runloop:
0111 b0 20 wch #' ' ; write space character
0113 03 00 ff lda $ff00,x ; load next number from array
0116 0a 01 sta $01 ; -> to $01
0118 a2 01 wud $01 ; and output
011a c8 inx ; next index
011b 92 00 cpx $00 ; length reached?
011d f4 01 bne compare ; if not calculate difference
011f c0 hlt ; done
 compare:
0120 03 00 ff lda $ff00,x ; load next number from array
0123 0a 02 sta $02 ; -> to $01
0125 d0 sec ; calculate ...
0126 72 01 sbc $01 ; ... difference ...
0128 0a 03 sta $03 ; ... to $03
012a c8 inx ; next index
012b 92 00 cpx $00 ; length reached?
012d f4 05 bne checkrun ; if not check whether we have a run
012f b0 20 wch #' ' ; output space
0131 a2 02 wud $02 ; output number
0133 c0 hlt ; done
 checkrun:
0134 02 02 lda $02 ; calculate next ...
0136 6a 03 adc $03 ; ... expected number in run
0138 8b 00 ff cmp $ff00,x ; compare with real next number
013b f6 06 beq haverun ; ok -> found a run
013d b0 20 wch #' ' ; otherwise output space ...
013f a2 02 wud $02 ; ... and number
0141 f4 ce bne runloop ; and repeat searching for runs
 haverun:
0143 0a 02 sta $02 ; store number to $02
0145 c8 inx ; next index
0146 92 00 cpx $00 ; length reached?
0148 f6 07 beq outputrun ; yes -> output this run
014a 6a 03 adc $03 ; calculate next expected number
014c 8b 00 ff cmp $ff00,x ; compare with real next number
014f f6 f2 beq haverun ; ok -> continue parsing run
 outputrun:
0151 b9 66 01 wtx str_to ; write "to"
0154 a2 02 wud $02 ; write end number of run
0156 00 01 lda #$01 ; compare #1 with ...
0158 8a 03 cmp $03 ; ... step size
015a f6 05 beq skip_by ; equal, then skip output of "by"
015c b9 69 01 wtx str_by ; output "by"
015f a2 03 wud $03 ; output step size
 skip_by:
0161 92 00 cpx $00 ; length of array reached?
0163 f4 ac bne runloop ; no -> repeat searching for runs
0165 c0 hlt ; done
 str_to:
0166 74 6f 00 ; "to"
 str_by:
0169 62 79 00 ; "by"
016c

share|improve this answer

edited Aug 27 at 21:20

answered Aug 27 at 11:54

Felix Palmen

3,026423

add a comment | 

up vote
1
down vote

05AB1E (legacy), 49 50 bytes

.œʒε¥Ë}P}Σ€g2KO>}¤εD©g≠i¬s¤„toý¬®¥0èDU≠i„byX««]˜ðý

Way too long, but I'm already glad it's working. This challenge is a lot harder than it looks imo.. Can without a doubt be golfed further.
Σ€g2KO>}¤ is a port of 2,0ySƲÞṪ from @JonathanAllan's Jelly answer (thanks!).

Try it online. (NOTE: Times out for the big test cases.)

+1 byte as bug-fix because 0 is always put at a trailing position when sorting.

Explanation:

.œ # Get all partions of the (implicit) input-list
 # i.e. [1,2,3,11,18,20,22,24,32,33,34]
 # → [[[1],[2],[3],[11],[18],[20],[22],[24],[32],[33],[34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32],[33,34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32,33],[34]],
 # ...]
 ʒ } # Filter this list by:
 ε } # Map the current sub-list by:
 ¥Ë # Take the deltas, and check if all are equal
 # i.e. [1,2,3] → [1,1] → 1
 # i.e. [1,2,3,11] → [1,1,8] → 0
 P # Check if all sub-list have equal deltas
 Σ } # Now that we've filtered the list, sort it by:
 €g # Take the length of each sub-list
 # i.e. [[1,2,3],[11,18],[20,22,24],[32,33],[34]]
 # → [3,2,3,2,1]
 # i.e. [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 # → [3,1,4,3]
 2K # Remove all 2s
 # i.e. [3,2,3,2,1] → ['3','3','1']
 O # And take the sum
 # i.e. ['3','3','1'] → 7
 # i.e. [3,1,4,3] → 11
 > # And increase the sum by 1 (0 is always trailing when sorting)
 ¤ # And then take the last item of this sorted list
 # i.e. for input [1,2,3,11,18,20,22,24,32,33,34]
 # → [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 ε # Now map each of the sub-lists to:
 D© # Save the current sub-list in the register
 g≠i # If its length is not 1:
 # i.e. [11] → 1 → 0 (falsey)
 # i.e. [18,20,22,24] → 4 → 1 (truthy)
 ¬s¤ # Take the head and tail of the sub-list
 # i.e. [18,20,22,24] → 18 and 24
 „toý # And join them with "to"
 # i.e. 18 and 24 → ['18to24', '18to20to22to24']
 ¬ # (head to remove some access waste we no longer need)
 # i.e. ['18to24', '18to20to22to24'] → '18to24'
 ® # Get the sub-list from the register again
 ¥ # Take its deltas
 # i.e. [18,20,22,24] → [2,2,2]
 0è # Get the first item (should all be the same delta)
 # i.e. [2,2,2] → 2
 DU # Save it in variable `X`
 ≠i # If the delta is not 1:
 # i.e. 2 → 1 (truthy)
 „byX« # Merge "by" with `X`
 # i.e. 2 → 'by2'
 « # And merge it with the earlier "#to#"
 # i.e. '18to24' and 'by2' → '18to24by2'
 ] # Close the mapping and both if-statements
˜ # Flatten the list
 # i.e. ['1to3',[11],'18to24by2','32to34']
 # → ['1to3',11,'18to24by2','32to34']
 ðý # And join by spaces (which we implicitly output as result)
 # i.e. ['1to3',11,'18to24by2','32to34']
 # → '1to3 11 18to24by2 32to34'

share|improve this answer

edited Aug 28 at 9:32

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

add a comment | 

up vote
0
down vote

Perl 5, 154 bytes

$$r[1]-$$r[0]==$_-$$r[-1])?push@$r,$_:push@r,$r=[$_]for@_;join" ",map@$_<3?@$_:("$$_[0]to$$_[-1]by".($$_[1]-$$_[0]))=~s/by1$//r,@r

Same with spaces, newlines, #comments and sub by:

sub by 
 my(@r,$r);
 @r && # if at least one run candidate exists and...
 ( @$r<2 # ...just one elem so far
 

Try it online!

...for passing tests from OP.

share|improve this answer

edited Aug 27 at 12:51

answered Aug 27 at 11:47

Kjetil S.

51115

add a comment | 

up vote
0
down vote

Retina 0.8.2, 77 bytes

d+
$*
(1+)(?= 1(1+))
$1:$2
1:(1+) (1+:1 )+(1+)
1to$3by$1
:1+|by1b

1+
$.&

Try it online! Link includes test cases. Explanation:

d+
$*

Convert to unary.

(1+)(?= 1(1+))
$1:$2

Compute consecutive differences.

1:(1+) (1+:1 )+(1+)
1to$3by$1

Convert runs to to...by syntax.

:1+|by1b

Remove unconverted differences and by1.

1+
$.&

Convert to decimal.

share|improve this answer

answered Aug 29 at 12:12

Neil

75k744170

add a comment | 

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16 Answers
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up vote
5
down vote



accepted

Jelly,  42  40 bytes

-2 thanks to Kevin Cruijssen (filter out twos, ḟ2, rather than replacing twos with zeros, 2,0y)

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K

A full program printing the result.

(As a monadic Link a list containing a mixture of integers and characters would be yielded)

Try it online!

(Too inefficient for the largest test-case to complete within 60s, so I removed [1,2,3,4].)

How?

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K - Main Link: list of numbers
ŒṖ - all partitions
 ƊÞ - sort by last three links as a monad:
 Ʋ€ - last four links as a monad for €ach:
 I - incremental differences (of the part)
 E - all equal?
 L - length (of the part)
 × - multiply
 ḟ2 - filter discard twos
 S - sum
 Ṫ - tail (gets us the desired partition of the input)
 µ ) - perform this monadic chain for €ach:
 . - literal 0.5
 ị - index into (the part) - getting [tail,head]
 U - upend - getting [head,tail]
 Ɗ - last three links as a monad:
 I - incremental differences (of the part)
 1 - literal one
 ḟ - filter discard (remove the ones)
 , - pair -> [[head,tail],[deltasWithoutOnes]]
 Q€ - de-duplicate €ach -> [[head,tail],[delta]] or [[head,tail],] or [[loneValue],]
 F - flatten -> [head,tail,delta] or [head,tail] or [loneValue]
 $ - last two links as a monad:
 Ɗ - last three links as a monad:
 “to“by” - literal list [['t', 'o'], ['b', 'y']]
 Ṗ - pop (get flattened result without rightmost entry)
 ṁ - mould ["to","by"] like that (i.e. ["to","by"] or ["to"] or )
 ż - zip together 
 K - join with spaces
 - implicit print

share|improve this answer

edited Aug 28 at 17:19

answered Aug 27 at 3:07

Jonathan Allan

48k534158

• 
  
  
  

  
  

  
  
  
  
  
  

  
  'by' should not be used if step = 1
  – WretchedLout
  Aug 27 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh Gawwwd :p Thanks for the heads-up!
  – Jonathan Allan
  Aug 27 at 3:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2,0ySƲÞ can be golfed to ḟ2SƊÞ for -2 bytes.
  – Kevin Cruijssen
  Aug 28 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen very true, nice golf, thanks
  – Jonathan Allan
  Aug 28 at 17:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen FYI I remember why that was there now - originally I was trying to do my sorting with a product, P, rather than a sum S and would have needed the zeros.
  – Jonathan Allan
  Aug 28 at 17:21
  

  
  
  
  

 | 
show 1 more comment

up vote
5
down vote



accepted

Jelly,  42  40 bytes

-2 thanks to Kevin Cruijssen (filter out twos, ḟ2, rather than replacing twos with zeros, 2,0y)

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K

A full program printing the result.

(As a monadic Link a list containing a mixture of integers and characters would be yielded)

Try it online!

(Too inefficient for the largest test-case to complete within 60s, so I removed [1,2,3,4].)

How?

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K - Main Link: list of numbers
ŒṖ - all partitions
 ƊÞ - sort by last three links as a monad:
 Ʋ€ - last four links as a monad for €ach:
 I - incremental differences (of the part)
 E - all equal?
 L - length (of the part)
 × - multiply
 ḟ2 - filter discard twos
 S - sum
 Ṫ - tail (gets us the desired partition of the input)
 µ ) - perform this monadic chain for €ach:
 . - literal 0.5
 ị - index into (the part) - getting [tail,head]
 U - upend - getting [head,tail]
 Ɗ - last three links as a monad:
 I - incremental differences (of the part)
 1 - literal one
 ḟ - filter discard (remove the ones)
 , - pair -> [[head,tail],[deltasWithoutOnes]]
 Q€ - de-duplicate €ach -> [[head,tail],[delta]] or [[head,tail],] or [[loneValue],]
 F - flatten -> [head,tail,delta] or [head,tail] or [loneValue]
 $ - last two links as a monad:
 Ɗ - last three links as a monad:
 “to“by” - literal list [['t', 'o'], ['b', 'y']]
 Ṗ - pop (get flattened result without rightmost entry)
 ṁ - mould ["to","by"] like that (i.e. ["to","by"] or ["to"] or )
 ż - zip together 
 K - join with spaces
 - implicit print

share|improve this answer

edited Aug 28 at 17:19

answered Aug 27 at 3:07

Jonathan Allan

48k534158

• 
  
  
  

  
  

  
  
  
  
  
  

  
  'by' should not be used if step = 1
  – WretchedLout
  Aug 27 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh Gawwwd :p Thanks for the heads-up!
  – Jonathan Allan
  Aug 27 at 3:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2,0ySƲÞ can be golfed to ḟ2SƊÞ for -2 bytes.
  – Kevin Cruijssen
  Aug 28 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen very true, nice golf, thanks
  – Jonathan Allan
  Aug 28 at 17:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen FYI I remember why that was there now - originally I was trying to do my sorting with a product, P, rather than a sum S and would have needed the zeros.
  – Jonathan Allan
  Aug 28 at 17:21
  

  
  
  
  

 | 
show 1 more comment

up vote
5
down vote



accepted

up vote
5
down vote



accepted

Jelly,  42  40 bytes

-2 thanks to Kevin Cruijssen (filter out twos, ḟ2, rather than replacing twos with zeros, 2,0y)

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K

A full program printing the result.

(As a monadic Link a list containing a mixture of integers and characters would be yielded)

Try it online!

(Too inefficient for the largest test-case to complete within 60s, so I removed [1,2,3,4].)

How?

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K - Main Link: list of numbers
ŒṖ - all partitions
 ƊÞ - sort by last three links as a monad:
 Ʋ€ - last four links as a monad for €ach:
 I - incremental differences (of the part)
 E - all equal?
 L - length (of the part)
 × - multiply
 ḟ2 - filter discard twos
 S - sum
 Ṫ - tail (gets us the desired partition of the input)
 µ ) - perform this monadic chain for €ach:
 . - literal 0.5
 ị - index into (the part) - getting [tail,head]
 U - upend - getting [head,tail]
 Ɗ - last three links as a monad:
 I - incremental differences (of the part)
 1 - literal one
 ḟ - filter discard (remove the ones)
 , - pair -> [[head,tail],[deltasWithoutOnes]]
 Q€ - de-duplicate €ach -> [[head,tail],[delta]] or [[head,tail],] or [[loneValue],]
 F - flatten -> [head,tail,delta] or [head,tail] or [loneValue]
 $ - last two links as a monad:
 Ɗ - last three links as a monad:
 “to“by” - literal list [['t', 'o'], ['b', 'y']]
 Ṗ - pop (get flattened result without rightmost entry)
 ṁ - mould ["to","by"] like that (i.e. ["to","by"] or ["to"] or )
 ż - zip together 
 K - join with spaces
 - implicit print

share|improve this answer

edited Aug 28 at 17:19

answered Aug 27 at 3:07

Jonathan Allan

48k534158

Jelly,  42  40 bytes

-2 thanks to Kevin Cruijssen (filter out twos, ḟ2, rather than replacing twos with zeros, 2,0y)

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K

A full program printing the result.

(As a monadic Link a list containing a mixture of integers and characters would be yielded)

Try it online!

(Too inefficient for the largest test-case to complete within 60s, so I removed [1,2,3,4].)

How?

ŒṖIE×LƲ€ḟ2SƊÞṪµ.ịU,Iḟ1ƊQ€Fż“to“by”ṁṖƊ$)K - Main Link: list of numbers
ŒṖ - all partitions
 ƊÞ - sort by last three links as a monad:
 Ʋ€ - last four links as a monad for €ach:
 I - incremental differences (of the part)
 E - all equal?
 L - length (of the part)
 × - multiply
 ḟ2 - filter discard twos
 S - sum
 Ṫ - tail (gets us the desired partition of the input)
 µ ) - perform this monadic chain for €ach:
 . - literal 0.5
 ị - index into (the part) - getting [tail,head]
 U - upend - getting [head,tail]
 Ɗ - last three links as a monad:
 I - incremental differences (of the part)
 1 - literal one
 ḟ - filter discard (remove the ones)
 , - pair -> [[head,tail],[deltasWithoutOnes]]
 Q€ - de-duplicate €ach -> [[head,tail],[delta]] or [[head,tail],] or [[loneValue],]
 F - flatten -> [head,tail,delta] or [head,tail] or [loneValue]
 $ - last two links as a monad:
 Ɗ - last three links as a monad:
 “to“by” - literal list [['t', 'o'], ['b', 'y']]
 Ṗ - pop (get flattened result without rightmost entry)
 ṁ - mould ["to","by"] like that (i.e. ["to","by"] or ["to"] or )
 ż - zip together 
 K - join with spaces
 - implicit print

share|improve this answer

edited Aug 28 at 17:19

answered Aug 27 at 3:07

Jonathan Allan

48k534158

share|improve this answer

share|improve this answer

edited Aug 28 at 17:19

edited Aug 28 at 17:19

edited Aug 28 at 17:19

answered Aug 27 at 3:07

Jonathan Allan

48k534158

answered Aug 27 at 3:07

Jonathan Allan

48k534158

answered Aug 27 at 3:07

Jonathan Allan

48k534158

48k534158

• 
  
  
  

  
  

  
  
  
  
  
  

  
  'by' should not be used if step = 1
  – WretchedLout
  Aug 27 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh Gawwwd :p Thanks for the heads-up!
  – Jonathan Allan
  Aug 27 at 3:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2,0ySƲÞ can be golfed to ḟ2SƊÞ for -2 bytes.
  – Kevin Cruijssen
  Aug 28 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen very true, nice golf, thanks
  – Jonathan Allan
  Aug 28 at 17:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen FYI I remember why that was there now - originally I was trying to do my sorting with a product, P, rather than a sum S and would have needed the zeros.
  – Jonathan Allan
  Aug 28 at 17:21
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  'by' should not be used if step = 1
  – WretchedLout
  Aug 27 at 3:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  oh Gawwwd :p Thanks for the heads-up!
  – Jonathan Allan
  Aug 27 at 3:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  2,0ySƲÞ can be golfed to ḟ2SƊÞ for -2 bytes.
  – Kevin Cruijssen
  Aug 28 at 9:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KevinCruijssen very true, nice golf, thanks
  – Jonathan Allan
  Aug 28 at 17:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @KevinCruijssen FYI I remember why that was there now - originally I was trying to do my sorting with a product, P, rather than a sum S and would have needed the zeros.
  – Jonathan Allan
  Aug 28 at 17:21
  

  
  
  
  

'by' should not be used if step = 1
– WretchedLout
Aug 27 at 3:25

'by' should not be used if step = 1
– WretchedLout
Aug 27 at 3:25

oh Gawwwd :p Thanks for the heads-up!
– Jonathan Allan
Aug 27 at 3:30

oh Gawwwd :p Thanks for the heads-up!
– Jonathan Allan
Aug 27 at 3:30

2,0ySƲÞ can be golfed to ḟ2SƊÞ for -2 bytes.
– Kevin Cruijssen
Aug 28 at 9:42

2,0ySƲÞ can be golfed to ḟ2SƊÞ for -2 bytes.
– Kevin Cruijssen
Aug 28 at 9:42

@KevinCruijssen very true, nice golf, thanks
– Jonathan Allan
Aug 28 at 17:02

@KevinCruijssen very true, nice golf, thanks
– Jonathan Allan
Aug 28 at 17:02

1

1

@KevinCruijssen FYI I remember why that was there now - originally I was trying to do my sorting with a product, P, rather than a sum S and would have needed the zeros.
– Jonathan Allan
Aug 28 at 17:21

@KevinCruijssen FYI I remember why that was there now - originally I was trying to do my sorting with a product, P, rather than a sum S and would have needed the zeros.
– Jonathan Allan
Aug 28 at 17:21

 | 
show 1 more comment

up vote
6
down vote

Swift, 246 bytes

func f(_ a:[Int])
var c=a.count,m=a+a,z=0,r:String=""
for i in 1..<cm[i]-=a[i-1];if m[i]==m[i-1]m[i-1]=0;m[0]=1
for i in 0..<cif m[i]==0 z=1else if z==0r+=" (a[i])"elser+="to(a[i])"+(m[i]>1 ? "by(m[i])":"");z=0;m[i+1]=1
print(r)

Try it online!

share|improve this answer

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

answered Aug 27 at 8:14

drawnonward

1611

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Welcome to Programming Puzzles & Code Golf! Very nice first answer. I hope you'll stay around because the Swift community here is... really small!
  – Mr. Xcoder
  Aug 27 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This saves you a couple of bytes
  – Mr. Xcoder
  Aug 27 at 8:26
  

  
  
  
  

add a comment | 

up vote
6
down vote

Swift, 246 bytes

func f(_ a:[Int])
var c=a.count,m=a+a,z=0,r:String=""
for i in 1..<cm[i]-=a[i-1];if m[i]==m[i-1]m[i-1]=0;m[0]=1
for i in 0..<cif m[i]==0 z=1else if z==0r+=" (a[i])"elser+="to(a[i])"+(m[i]>1 ? "by(m[i])":"");z=0;m[i+1]=1
print(r)

Try it online!

share|improve this answer

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

answered Aug 27 at 8:14

drawnonward

1611

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Welcome to Programming Puzzles & Code Golf! Very nice first answer. I hope you'll stay around because the Swift community here is... really small!
  – Mr. Xcoder
  Aug 27 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This saves you a couple of bytes
  – Mr. Xcoder
  Aug 27 at 8:26
  

  
  
  
  

add a comment | 

up vote
6
down vote

up vote
6
down vote

Swift, 246 bytes

func f(_ a:[Int])
var c=a.count,m=a+a,z=0,r:String=""
for i in 1..<cm[i]-=a[i-1];if m[i]==m[i-1]m[i-1]=0;m[0]=1
for i in 0..<cif m[i]==0 z=1else if z==0r+=" (a[i])"elser+="to(a[i])"+(m[i]>1 ? "by(m[i])":"");z=0;m[i+1]=1
print(r)

Try it online!

share|improve this answer

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

answered Aug 27 at 8:14

drawnonward

1611

Swift, 246 bytes

func f(_ a:[Int])
var c=a.count,m=a+a,z=0,r:String=""
for i in 1..<cm[i]-=a[i-1];if m[i]==m[i-1]m[i-1]=0;m[0]=1
for i in 0..<cif m[i]==0 z=1else if z==0r+=" (a[i])"elser+="to(a[i])"+(m[i]>1 ? "by(m[i])":"");z=0;m[i+1]=1
print(r)

Try it online!

share|improve this answer

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

answered Aug 27 at 8:14

drawnonward

1611

share|improve this answer

share|improve this answer

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

edited Aug 27 at 8:21

Mr. Xcoder

30.3k757193

30.3k757193

answered Aug 27 at 8:14

drawnonward

1611

answered Aug 27 at 8:14

drawnonward

1611

answered Aug 27 at 8:14

drawnonward

1611

1611

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Welcome to Programming Puzzles & Code Golf! Very nice first answer. I hope you'll stay around because the Swift community here is... really small!
  – Mr. Xcoder
  Aug 27 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This saves you a couple of bytes
  – Mr. Xcoder
  Aug 27 at 8:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  Welcome to Programming Puzzles & Code Golf! Very nice first answer. I hope you'll stay around because the Swift community here is... really small!
  – Mr. Xcoder
  Aug 27 at 8:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This saves you a couple of bytes
  – Mr. Xcoder
  Aug 27 at 8:26
  

  
  
  
  

4

4

Welcome to Programming Puzzles & Code Golf! Very nice first answer. I hope you'll stay around because the Swift community here is... really small!
– Mr. Xcoder
Aug 27 at 8:23

Welcome to Programming Puzzles & Code Golf! Very nice first answer. I hope you'll stay around because the Swift community here is... really small!
– Mr. Xcoder
Aug 27 at 8:23

This saves you a couple of bytes
– Mr. Xcoder
Aug 27 at 8:26

This saves you a couple of bytes
– Mr. Xcoder
Aug 27 at 8:26

add a comment | 

up vote
3
down vote

K (ngn/k), 102 bytes

1_,//$H:" ",h:*x;l:+/&x=h+(!#x)*d:--/2#x;$[l>2;(H;"to";x l-1;(~d=1)#,"by",d;o l_x);l;H,o 1_x;""]x

Try it online!

share|improve this answer

edited Aug 27 at 1:13

answered Aug 27 at 0:52

ngn

6,10812256

add a comment | 

up vote
3
down vote

K (ngn/k), 102 bytes

1_,//$H:" ",h:*x;l:+/&x=h+(!#x)*d:--/2#x;$[l>2;(H;"to";x l-1;(~d=1)#,"by",d;o l_x);l;H,o 1_x;""]x

Try it online!

share|improve this answer

edited Aug 27 at 1:13

answered Aug 27 at 0:52

ngn

6,10812256

add a comment | 

up vote
3
down vote

up vote
3
down vote

K (ngn/k), 102 bytes

1_,//$H:" ",h:*x;l:+/&x=h+(!#x)*d:--/2#x;$[l>2;(H;"to";x l-1;(~d=1)#,"by",d;o l_x);l;H,o 1_x;""]x

Try it online!

share|improve this answer

edited Aug 27 at 1:13

answered Aug 27 at 0:52

ngn

6,10812256

K (ngn/k), 102 bytes

1_,//$H:" ",h:*x;l:+/&x=h+(!#x)*d:--/2#x;$[l>2;(H;"to";x l-1;(~d=1)#,"by",d;o l_x);l;H,o 1_x;""]x

Try it online!

share|improve this answer

edited Aug 27 at 1:13

answered Aug 27 at 0:52

ngn

6,10812256

share|improve this answer

share|improve this answer

edited Aug 27 at 1:13

edited Aug 27 at 1:13

edited Aug 27 at 1:13

answered Aug 27 at 0:52

ngn

6,10812256

answered Aug 27 at 0:52

ngn

6,10812256

answered Aug 27 at 0:52

ngn

6,10812256

6,10812256

add a comment | 

add a comment | 

up vote
3
down vote

JavaScript (ES6), 129 bytes

Returns an array of strings.

a=>(a.map((n,i)=>n+[n-a[i+1]||''])+'').replace(/(d+)(-d+)(?:,d+2)+,(d+)/g,(_,a,b,c)=>a+'to'+(~b?c+'by'+-b:c)).split(/-d+,/)

Try it online!

How?

Step #1

We first append to each number a suffix consisting of a leading '-' followed by the difference with the next number, except for the last entry which is left unchanged. This new array is coerced to a string.

(a.map((n, i) => n + [n - a[i + 1] || '']) + '')

Example:

Input : [ 1, 2, 3, 5, 9, 11, 13, 20 ]
Output: "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"

Step #2

We identify all runs in the resulting string and replace them with the appropriate notation.

.replace(
 /(d+)(-d+)(?:,d+2)+,(d+)/g,
 (_, a, b, c) => a + 'to' + (~b ? c + 'by' + -b : c)
)

Example:

Input : "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"
Output: "1to3-2,5-4,9to13by2-7,20"

Step #3

Finally, we split the string on the remaining suffixes, including the trailing commas.

.split(/-d+,/)

Example:

Input : "1to3-2,5-4,9to13by2-7,20"
Output: [ '1to3', '5', '9to13by2', '20' ]

share|improve this answer

edited Aug 28 at 11:37

answered Aug 27 at 11:49

Arnauld

63.5k580268

add a comment | 

up vote
3
down vote

JavaScript (ES6), 129 bytes

Returns an array of strings.

a=>(a.map((n,i)=>n+[n-a[i+1]||''])+'').replace(/(d+)(-d+)(?:,d+2)+,(d+)/g,(_,a,b,c)=>a+'to'+(~b?c+'by'+-b:c)).split(/-d+,/)

Try it online!

How?

Step #1

We first append to each number a suffix consisting of a leading '-' followed by the difference with the next number, except for the last entry which is left unchanged. This new array is coerced to a string.

(a.map((n, i) => n + [n - a[i + 1] || '']) + '')

Example:

Input : [ 1, 2, 3, 5, 9, 11, 13, 20 ]
Output: "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"

Step #2

We identify all runs in the resulting string and replace them with the appropriate notation.

.replace(
 /(d+)(-d+)(?:,d+2)+,(d+)/g,
 (_, a, b, c) => a + 'to' + (~b ? c + 'by' + -b : c)
)

Example:

Input : "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"
Output: "1to3-2,5-4,9to13by2-7,20"

Step #3

Finally, we split the string on the remaining suffixes, including the trailing commas.

.split(/-d+,/)

Example:

Input : "1to3-2,5-4,9to13by2-7,20"
Output: [ '1to3', '5', '9to13by2', '20' ]

share|improve this answer

edited Aug 28 at 11:37

answered Aug 27 at 11:49

Arnauld

63.5k580268

add a comment | 

up vote
3
down vote

up vote
3
down vote

JavaScript (ES6), 129 bytes

Returns an array of strings.

a=>(a.map((n,i)=>n+[n-a[i+1]||''])+'').replace(/(d+)(-d+)(?:,d+2)+,(d+)/g,(_,a,b,c)=>a+'to'+(~b?c+'by'+-b:c)).split(/-d+,/)

Try it online!

How?

Step #1

We first append to each number a suffix consisting of a leading '-' followed by the difference with the next number, except for the last entry which is left unchanged. This new array is coerced to a string.

(a.map((n, i) => n + [n - a[i + 1] || '']) + '')

Example:

Input : [ 1, 2, 3, 5, 9, 11, 13, 20 ]
Output: "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"

Step #2

We identify all runs in the resulting string and replace them with the appropriate notation.

.replace(
 /(d+)(-d+)(?:,d+2)+,(d+)/g,
 (_, a, b, c) => a + 'to' + (~b ? c + 'by' + -b : c)
)

Example:

Input : "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"
Output: "1to3-2,5-4,9to13by2-7,20"

Step #3

Finally, we split the string on the remaining suffixes, including the trailing commas.

.split(/-d+,/)

Example:

Input : "1to3-2,5-4,9to13by2-7,20"
Output: [ '1to3', '5', '9to13by2', '20' ]

share|improve this answer

edited Aug 28 at 11:37

answered Aug 27 at 11:49

Arnauld

63.5k580268

JavaScript (ES6), 129 bytes

Returns an array of strings.

a=>(a.map((n,i)=>n+[n-a[i+1]||''])+'').replace(/(d+)(-d+)(?:,d+2)+,(d+)/g,(_,a,b,c)=>a+'to'+(~b?c+'by'+-b:c)).split(/-d+,/)

Try it online!

How?

Step #1

We first append to each number a suffix consisting of a leading '-' followed by the difference with the next number, except for the last entry which is left unchanged. This new array is coerced to a string.

(a.map((n, i) => n + [n - a[i + 1] || '']) + '')

Example:

Input : [ 1, 2, 3, 5, 9, 11, 13, 20 ]
Output: "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"

Step #2

We identify all runs in the resulting string and replace them with the appropriate notation.

.replace(
 /(d+)(-d+)(?:,d+2)+,(d+)/g,
 (_, a, b, c) => a + 'to' + (~b ? c + 'by' + -b : c)
)

Example:

Input : "1-1,2-1,3-2,5-4,9-2,11-2,13-7,20"
Output: "1to3-2,5-4,9to13by2-7,20"

Step #3

Finally, we split the string on the remaining suffixes, including the trailing commas.

.split(/-d+,/)

Example:

Input : "1to3-2,5-4,9to13by2-7,20"
Output: [ '1to3', '5', '9to13by2', '20' ]

share|improve this answer

edited Aug 28 at 11:37

answered Aug 27 at 11:49

Arnauld

63.5k580268

share|improve this answer

share|improve this answer

edited Aug 28 at 11:37

edited Aug 28 at 11:37

edited Aug 28 at 11:37

answered Aug 27 at 11:49

Arnauld

63.5k580268

answered Aug 27 at 11:49

Arnauld

63.5k580268

answered Aug 27 at 11:49

Arnauld

63.5k580268

63.5k580268

add a comment | 

add a comment | 

up vote
2
down vote

Ruby, 125 118 bytes

->ai=y=0;a.chunkx.map2-i<(s=w.size)?"#w[i*s]to#w[~i=0]"+"by#z"*(z<=>1)+' ':' '*i+w*' '*i=1*''

Try it online!

Explanation

Ruby's Enumerable has a useful chunk method that does precisely what we need here - it groups items by consecutive runs of the same return value from the block, in our case - the difference between the current (x) and previous (y) value.

The caveat is that such strategy won't capture the first element of the run, e.g. here only the two last elements are grouped together:

Input: [5, 6, 8, 11, 15, 16, 17]
Grouped: [[5, [5]], [1, [6]], [2, [8]], [3, [11]], [4, [15]], [1, [16, 17]]]

Therefore, while mapping to the correctly formatted strings, when we encounter a new potential run (chunk with > 1 item), we must track if the previous item was single (i=1) or already used in another run (i=0). If there is an unused single item, it becomes the starting point of the run, and lowers the chunk size threshold from 3 to 2.

share|improve this answer

edited Aug 27 at 20:30

answered Aug 27 at 11:06

Kirill L.

2,4061116

add a comment | 

up vote
2
down vote

Ruby, 125 118 bytes

->ai=y=0;a.chunkx.map2-i<(s=w.size)?"#w[i*s]to#w[~i=0]"+"by#z"*(z<=>1)+' ':' '*i+w*' '*i=1*''

Try it online!

Explanation

Ruby's Enumerable has a useful chunk method that does precisely what we need here - it groups items by consecutive runs of the same return value from the block, in our case - the difference between the current (x) and previous (y) value.

The caveat is that such strategy won't capture the first element of the run, e.g. here only the two last elements are grouped together:

Input: [5, 6, 8, 11, 15, 16, 17]
Grouped: [[5, [5]], [1, [6]], [2, [8]], [3, [11]], [4, [15]], [1, [16, 17]]]

Therefore, while mapping to the correctly formatted strings, when we encounter a new potential run (chunk with > 1 item), we must track if the previous item was single (i=1) or already used in another run (i=0). If there is an unused single item, it becomes the starting point of the run, and lowers the chunk size threshold from 3 to 2.

share|improve this answer

edited Aug 27 at 20:30

answered Aug 27 at 11:06

Kirill L.

2,4061116

add a comment | 

up vote
2
down vote

up vote
2
down vote

Ruby, 125 118 bytes

->ai=y=0;a.chunkx.map2-i<(s=w.size)?"#w[i*s]to#w[~i=0]"+"by#z"*(z<=>1)+' ':' '*i+w*' '*i=1*''

Try it online!

Explanation

Ruby's Enumerable has a useful chunk method that does precisely what we need here - it groups items by consecutive runs of the same return value from the block, in our case - the difference between the current (x) and previous (y) value.

The caveat is that such strategy won't capture the first element of the run, e.g. here only the two last elements are grouped together:

Input: [5, 6, 8, 11, 15, 16, 17]
Grouped: [[5, [5]], [1, [6]], [2, [8]], [3, [11]], [4, [15]], [1, [16, 17]]]

Therefore, while mapping to the correctly formatted strings, when we encounter a new potential run (chunk with > 1 item), we must track if the previous item was single (i=1) or already used in another run (i=0). If there is an unused single item, it becomes the starting point of the run, and lowers the chunk size threshold from 3 to 2.

share|improve this answer

edited Aug 27 at 20:30

answered Aug 27 at 11:06

Kirill L.

2,4061116

Ruby, 125 118 bytes

->ai=y=0;a.chunkx.map2-i<(s=w.size)?"#w[i*s]to#w[~i=0]"+"by#z"*(z<=>1)+' ':' '*i+w*' '*i=1*''

Try it online!

Explanation

Ruby's Enumerable has a useful chunk method that does precisely what we need here - it groups items by consecutive runs of the same return value from the block, in our case - the difference between the current (x) and previous (y) value.

The caveat is that such strategy won't capture the first element of the run, e.g. here only the two last elements are grouped together:

Input: [5, 6, 8, 11, 15, 16, 17]
Grouped: [[5, [5]], [1, [6]], [2, [8]], [3, [11]], [4, [15]], [1, [16, 17]]]

Therefore, while mapping to the correctly formatted strings, when we encounter a new potential run (chunk with > 1 item), we must track if the previous item was single (i=1) or already used in another run (i=0). If there is an unused single item, it becomes the starting point of the run, and lowers the chunk size threshold from 3 to 2.

share|improve this answer

edited Aug 27 at 20:30

answered Aug 27 at 11:06

Kirill L.

2,4061116

share|improve this answer

share|improve this answer

edited Aug 27 at 20:30

edited Aug 27 at 20:30

edited Aug 27 at 20:30

answered Aug 27 at 11:06

Kirill L.

2,4061116

answered Aug 27 at 11:06

Kirill L.

2,4061116

answered Aug 27 at 11:06

Kirill L.

2,4061116

2,4061116

add a comment | 

add a comment | 

up vote
2
down vote

R, 238 bytes

Almost certainly not the easiest way to do it.

function(x,r=rle(diff(x)),n=0,L=length)
while(n<L(x))
m=r$l[1]
cat(`if`(L(r$l)&&m>1,c(x[n+1],"to",x[n+m+1],ifelse(r$v[1]>1,paste0("by",r$v[1]),"")),x[n+1])," ",sep="")	
if(L(r$l)==0) break;
n=n+m+1*(m>1)
r=rle(diff(c(tail(x,L(x)-n))))

Try it online!

share|improve this answer

answered Aug 28 at 19:18

J.Doe

5218

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use F instead of n as it's already initialized to 0, which should save a few bytes, I think.
  – Giuseppe
  Aug 28 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This one is frustrating. You're nearly there by just using split, diff and rle. Unfortunately, the greedy search for runs means a lot of fiddling.
  – J.Doe
  Aug 29 at 10:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  214 bytes Almost there, the only problem is the uncorrect spacing sometimes...
  – digEmAll
  Aug 29 at 12:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That 'by'[D>1] is a good trick.
  – J.Doe
  Aug 29 at 12:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This seems to work fine 217 bytes
  – digEmAll
  Aug 29 at 18:46
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote

R, 238 bytes

Almost certainly not the easiest way to do it.

function(x,r=rle(diff(x)),n=0,L=length)
while(n<L(x))
m=r$l[1]
cat(`if`(L(r$l)&&m>1,c(x[n+1],"to",x[n+m+1],ifelse(r$v[1]>1,paste0("by",r$v[1]),"")),x[n+1])," ",sep="")	
if(L(r$l)==0) break;
n=n+m+1*(m>1)
r=rle(diff(c(tail(x,L(x)-n))))

Try it online!

share|improve this answer

answered Aug 28 at 19:18

J.Doe

5218

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use F instead of n as it's already initialized to 0, which should save a few bytes, I think.
  – Giuseppe
  Aug 28 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This one is frustrating. You're nearly there by just using split, diff and rle. Unfortunately, the greedy search for runs means a lot of fiddling.
  – J.Doe
  Aug 29 at 10:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  214 bytes Almost there, the only problem is the uncorrect spacing sometimes...
  – digEmAll
  Aug 29 at 12:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That 'by'[D>1] is a good trick.
  – J.Doe
  Aug 29 at 12:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This seems to work fine 217 bytes
  – digEmAll
  Aug 29 at 18:46
  

  
  
  
  

 | 
show 2 more comments

up vote
2
down vote

up vote
2
down vote

R, 238 bytes

Almost certainly not the easiest way to do it.

function(x,r=rle(diff(x)),n=0,L=length)
while(n<L(x))
m=r$l[1]
cat(`if`(L(r$l)&&m>1,c(x[n+1],"to",x[n+m+1],ifelse(r$v[1]>1,paste0("by",r$v[1]),"")),x[n+1])," ",sep="")	
if(L(r$l)==0) break;
n=n+m+1*(m>1)
r=rle(diff(c(tail(x,L(x)-n))))

Try it online!

share|improve this answer

answered Aug 28 at 19:18

J.Doe

5218

R, 238 bytes

Almost certainly not the easiest way to do it.

function(x,r=rle(diff(x)),n=0,L=length)
while(n<L(x))
m=r$l[1]
cat(`if`(L(r$l)&&m>1,c(x[n+1],"to",x[n+m+1],ifelse(r$v[1]>1,paste0("by",r$v[1]),"")),x[n+1])," ",sep="")	
if(L(r$l)==0) break;
n=n+m+1*(m>1)
r=rle(diff(c(tail(x,L(x)-n))))

Try it online!

share|improve this answer

answered Aug 28 at 19:18

J.Doe

5218

share|improve this answer

share|improve this answer

answered Aug 28 at 19:18

J.Doe

5218

answered Aug 28 at 19:18

J.Doe

5218

answered Aug 28 at 19:18

J.Doe

5218

5218

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use F instead of n as it's already initialized to 0, which should save a few bytes, I think.
  – Giuseppe
  Aug 28 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This one is frustrating. You're nearly there by just using split, diff and rle. Unfortunately, the greedy search for runs means a lot of fiddling.
  – J.Doe
  Aug 29 at 10:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  214 bytes Almost there, the only problem is the uncorrect spacing sometimes...
  – digEmAll
  Aug 29 at 12:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That 'by'[D>1] is a good trick.
  – J.Doe
  Aug 29 at 12:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This seems to work fine 217 bytes
  – digEmAll
  Aug 29 at 18:46
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use F instead of n as it's already initialized to 0, which should save a few bytes, I think.
  – Giuseppe
  Aug 28 at 19:43
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This one is frustrating. You're nearly there by just using split, diff and rle. Unfortunately, the greedy search for runs means a lot of fiddling.
  – J.Doe
  Aug 29 at 10:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  214 bytes Almost there, the only problem is the uncorrect spacing sometimes...
  – digEmAll
  Aug 29 at 12:13
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  That 'by'[D>1] is a good trick.
  – J.Doe
  Aug 29 at 12:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This seems to work fine 217 bytes
  – digEmAll
  Aug 29 at 18:46
  

  
  
  
  

use F instead of n as it's already initialized to 0, which should save a few bytes, I think.
– Giuseppe
Aug 28 at 19:43

use F instead of n as it's already initialized to 0, which should save a few bytes, I think.
– Giuseppe
Aug 28 at 19:43

This one is frustrating. You're nearly there by just using split, diff and rle. Unfortunately, the greedy search for runs means a lot of fiddling.
– J.Doe
Aug 29 at 10:08

This one is frustrating. You're nearly there by just using split, diff and rle. Unfortunately, the greedy search for runs means a lot of fiddling.
– J.Doe
Aug 29 at 10:08

214 bytes Almost there, the only problem is the uncorrect spacing sometimes...
– digEmAll
Aug 29 at 12:13

214 bytes Almost there, the only problem is the uncorrect spacing sometimes...
– digEmAll
Aug 29 at 12:13

That 'by'[D>1] is a good trick.
– J.Doe
Aug 29 at 12:54

That 'by'[D>1] is a good trick.
– J.Doe
Aug 29 at 12:54

This seems to work fine 217 bytes
– digEmAll
Aug 29 at 18:46

This seems to work fine 217 bytes
– digEmAll
Aug 29 at 18:46

 | 
show 2 more comments

up vote
2
down vote

R, 180 175 bytes

r=rle(c(0,diff(a<-scan())));for(j in 1:sum(1|r$l))l=r$l[j];v=r$v[j];i=T+l-1;cat("if"(l>2-F,paste0(a[T][!F],"to",a[i],"by"[v>1],v[v>1]," "),c("",a[T:i])[-3^F]));T=i+1;F=l<3-F

Try it online!

Conceptually, this is a port of my Ruby answer, though obviously quite a bit different technically.

5 bytes saved by JayCe.

share|improve this answer

edited Aug 30 at 16:29

answered Aug 30 at 8:41

Kirill L.

2,4061116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wanted to to something with rle but was too lazy... you can save 1 byte doing sum(1|x) in place of length(x): TIO
  – JayCe
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually you can have just 1 cat for 175 bytes: TIO
  – JayCe
  Aug 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, great, thanks!
  – Kirill L.
  Aug 30 at 16:30
  

  
  
  
  

add a comment | 

up vote
2
down vote

R, 180 175 bytes

r=rle(c(0,diff(a<-scan())));for(j in 1:sum(1|r$l))l=r$l[j];v=r$v[j];i=T+l-1;cat("if"(l>2-F,paste0(a[T][!F],"to",a[i],"by"[v>1],v[v>1]," "),c("",a[T:i])[-3^F]));T=i+1;F=l<3-F

Try it online!

Conceptually, this is a port of my Ruby answer, though obviously quite a bit different technically.

5 bytes saved by JayCe.

share|improve this answer

edited Aug 30 at 16:29

answered Aug 30 at 8:41

Kirill L.

2,4061116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wanted to to something with rle but was too lazy... you can save 1 byte doing sum(1|x) in place of length(x): TIO
  – JayCe
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually you can have just 1 cat for 175 bytes: TIO
  – JayCe
  Aug 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, great, thanks!
  – Kirill L.
  Aug 30 at 16:30
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

R, 180 175 bytes

r=rle(c(0,diff(a<-scan())));for(j in 1:sum(1|r$l))l=r$l[j];v=r$v[j];i=T+l-1;cat("if"(l>2-F,paste0(a[T][!F],"to",a[i],"by"[v>1],v[v>1]," "),c("",a[T:i])[-3^F]));T=i+1;F=l<3-F

Try it online!

Conceptually, this is a port of my Ruby answer, though obviously quite a bit different technically.

5 bytes saved by JayCe.

share|improve this answer

edited Aug 30 at 16:29

answered Aug 30 at 8:41

Kirill L.

2,4061116

R, 180 175 bytes

r=rle(c(0,diff(a<-scan())));for(j in 1:sum(1|r$l))l=r$l[j];v=r$v[j];i=T+l-1;cat("if"(l>2-F,paste0(a[T][!F],"to",a[i],"by"[v>1],v[v>1]," "),c("",a[T:i])[-3^F]));T=i+1;F=l<3-F

Try it online!

Conceptually, this is a port of my Ruby answer, though obviously quite a bit different technically.

5 bytes saved by JayCe.

share|improve this answer

edited Aug 30 at 16:29

answered Aug 30 at 8:41

Kirill L.

2,4061116

share|improve this answer

share|improve this answer

edited Aug 30 at 16:29

edited Aug 30 at 16:29

edited Aug 30 at 16:29

answered Aug 30 at 8:41

Kirill L.

2,4061116

answered Aug 30 at 8:41

Kirill L.

2,4061116

answered Aug 30 at 8:41

Kirill L.

2,4061116

2,4061116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wanted to to something with rle but was too lazy... you can save 1 byte doing sum(1|x) in place of length(x): TIO
  – JayCe
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually you can have just 1 cat for 175 bytes: TIO
  – JayCe
  Aug 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, great, thanks!
  – Kirill L.
  Aug 30 at 16:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I wanted to to something with rle but was too lazy... you can save 1 byte doing sum(1|x) in place of length(x): TIO
  – JayCe
  Aug 30 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Actually you can have just 1 cat for 175 bytes: TIO
  – JayCe
  Aug 30 at 13:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ah, great, thanks!
  – Kirill L.
  Aug 30 at 16:30
  

  
  
  
  

I wanted to to something with rle but was too lazy... you can save 1 byte doing sum(1|x) in place of length(x): TIO
– JayCe
Aug 30 at 13:51

I wanted to to something with rle but was too lazy... you can save 1 byte doing sum(1|x) in place of length(x): TIO
– JayCe
Aug 30 at 13:51

Actually you can have just 1 cat for 175 bytes: TIO
– JayCe
Aug 30 at 13:59

Actually you can have just 1 cat for 175 bytes: TIO
– JayCe
Aug 30 at 13:59

Ah, great, thanks!
– Kirill L.
Aug 30 at 16:30

Ah, great, thanks!
– Kirill L.
Aug 30 at 16:30

add a comment | 

up vote
1
down vote

JavaScript (Node.js), 177 173 bytes

a=>for(h=t=p=d=0,o=;(c=a[t])

Try it online!

share|improve this answer

edited Aug 27 at 4:53

answered Aug 27 at 4:43

user71546

1,31529

add a comment | 

up vote
1
down vote

JavaScript (Node.js), 177 173 bytes

a=>for(h=t=p=d=0,o=;(c=a[t])

Try it online!

share|improve this answer

edited Aug 27 at 4:53

answered Aug 27 at 4:43

user71546

1,31529

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript (Node.js), 177 173 bytes

a=>for(h=t=p=d=0,o=;(c=a[t])

Try it online!

share|improve this answer

edited Aug 27 at 4:53

answered Aug 27 at 4:43

user71546

1,31529

JavaScript (Node.js), 177 173 bytes

a=>for(h=t=p=d=0,o=;(c=a[t])

Try it online!

share|improve this answer

edited Aug 27 at 4:53

answered Aug 27 at 4:43

user71546

1,31529

share|improve this answer

share|improve this answer

edited Aug 27 at 4:53

edited Aug 27 at 4:53

edited Aug 27 at 4:53

answered Aug 27 at 4:43

user71546

1,31529

answered Aug 27 at 4:43

user71546

1,31529

answered Aug 27 at 4:43

user71546

1,31529

1,31529

add a comment | 

add a comment | 

up vote
1
down vote

Clean, 208 ... 185 bytes

import StdEnv,Data.List,Text
l=length
$=""
$k=last[u<+(if(v>)("to"<+last v<+if(u<v!!0-1)("by"<+(v!!0-u))"")"")+" "+ $t\i=:[u:v]<-inits k&t<-tails k|l(nub(zipWith(-)i v))<2&&l i<>2]

Try it online!

share|improve this answer

edited Aug 27 at 10:49

answered Aug 27 at 2:45

Οurous

5,1931931

add a comment | 

up vote
1
down vote

Clean, 208 ... 185 bytes

import StdEnv,Data.List,Text
l=length
$=""
$k=last[u<+(if(v>)("to"<+last v<+if(u<v!!0-1)("by"<+(v!!0-u))"")"")+" "+ $t\i=:[u:v]<-inits k&t<-tails k|l(nub(zipWith(-)i v))<2&&l i<>2]

Try it online!

share|improve this answer

edited Aug 27 at 10:49

answered Aug 27 at 2:45

Οurous

5,1931931

add a comment | 

up vote
1
down vote

up vote
1
down vote

Clean, 208 ... 185 bytes

import StdEnv,Data.List,Text
l=length
$=""
$k=last[u<+(if(v>)("to"<+last v<+if(u<v!!0-1)("by"<+(v!!0-u))"")"")+" "+ $t\i=:[u:v]<-inits k&t<-tails k|l(nub(zipWith(-)i v))<2&&l i<>2]

Try it online!

share|improve this answer

edited Aug 27 at 10:49

answered Aug 27 at 2:45

Οurous

5,1931931

Clean, 208 ... 185 bytes

import StdEnv,Data.List,Text
l=length
$=""
$k=last[u<+(if(v>)("to"<+last v<+if(u<v!!0-1)("by"<+(v!!0-u))"")"")+" "+ $t\i=:[u:v]<-inits k&t<-tails k|l(nub(zipWith(-)i v))<2&&l i<>2]

Try it online!

share|improve this answer

edited Aug 27 at 10:49

answered Aug 27 at 2:45

Οurous

5,1931931

share|improve this answer

share|improve this answer

edited Aug 27 at 10:49

edited Aug 27 at 10:49

edited Aug 27 at 10:49

answered Aug 27 at 2:45

Οurous

5,1931931

answered Aug 27 at 2:45

Οurous

5,1931931

answered Aug 27 at 2:45

Οurous

5,1931931

5,1931931

add a comment | 

add a comment | 

up vote
1
down vote

Jelly, 41 bytes

ŒṖIE€Ạ$>Ẉ2eƊƲƇṪµ.ịṚj⁾to;IḢ⁾by;ẋ⁻1$ƲƊµḊ¡€K

Try it online!

Full program.

share|improve this answer

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
1
down vote

Jelly, 41 bytes

ŒṖIE€Ạ$>Ẉ2eƊƲƇṪµ.ịṚj⁾to;IḢ⁾by;ẋ⁻1$ƲƊµḊ¡€K

Try it online!

Full program.

share|improve this answer

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

add a comment | 

up vote
1
down vote

up vote
1
down vote

Jelly, 41 bytes

ŒṖIE€Ạ$>Ẉ2eƊƲƇṪµ.ịṚj⁾to;IḢ⁾by;ẋ⁻1$ƲƊµḊ¡€K

Try it online!

Full program.

share|improve this answer

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

Jelly, 41 bytes

ŒṖIE€Ạ$>Ẉ2eƊƲƇṪµ.ịṚj⁾to;IḢ⁾by;ẋ⁻1$ƲƊµḊ¡€K

Try it online!

Full program.

share|improve this answer

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

share|improve this answer

share|improve this answer

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

answered Aug 27 at 11:47

Erik the Outgolfer

29.4k42698

29.4k42698

add a comment | 

add a comment | 

up vote
1
down vote

Python 2, 170 166 bytes

def f(a):
 j=len(a)
 while j>2:
	l,r=a[:j:j-1];s=(r-l)/~-j
	if a[:j]==range(l,r+1,s):return[`l`+'to%d'%r+'by%d'%s*(s>1)]+f(a[j:])
	j-=1
 return a and[`a[0]`]+f(a[1:])

Try it online!

share|improve this answer

edited Aug 27 at 12:00

answered Aug 27 at 9:07

TFeld

11.2k2833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if this output format is valid; the runs should all be strings, no?
  – Erik the Outgolfer
  Aug 27 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EriktheOutgolfer Fixed
  – TFeld
  Aug 27 at 12:01
  

  
  
  
  

add a comment | 

up vote
1
down vote

Python 2, 170 166 bytes

def f(a):
 j=len(a)
 while j>2:
	l,r=a[:j:j-1];s=(r-l)/~-j
	if a[:j]==range(l,r+1,s):return[`l`+'to%d'%r+'by%d'%s*(s>1)]+f(a[j:])
	j-=1
 return a and[`a[0]`]+f(a[1:])

Try it online!

share|improve this answer

edited Aug 27 at 12:00

answered Aug 27 at 9:07

TFeld

11.2k2833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if this output format is valid; the runs should all be strings, no?
  – Erik the Outgolfer
  Aug 27 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EriktheOutgolfer Fixed
  – TFeld
  Aug 27 at 12:01
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Python 2, 170 166 bytes

def f(a):
 j=len(a)
 while j>2:
	l,r=a[:j:j-1];s=(r-l)/~-j
	if a[:j]==range(l,r+1,s):return[`l`+'to%d'%r+'by%d'%s*(s>1)]+f(a[j:])
	j-=1
 return a and[`a[0]`]+f(a[1:])

Try it online!

share|improve this answer

edited Aug 27 at 12:00

answered Aug 27 at 9:07

TFeld

11.2k2833

Python 2, 170 166 bytes

def f(a):
 j=len(a)
 while j>2:
	l,r=a[:j:j-1];s=(r-l)/~-j
	if a[:j]==range(l,r+1,s):return[`l`+'to%d'%r+'by%d'%s*(s>1)]+f(a[j:])
	j-=1
 return a and[`a[0]`]+f(a[1:])

Try it online!

share|improve this answer

edited Aug 27 at 12:00

answered Aug 27 at 9:07

TFeld

11.2k2833

share|improve this answer

share|improve this answer

edited Aug 27 at 12:00

edited Aug 27 at 12:00

edited Aug 27 at 12:00

answered Aug 27 at 9:07

TFeld

11.2k2833

answered Aug 27 at 9:07

TFeld

11.2k2833

answered Aug 27 at 9:07

TFeld

11.2k2833

11.2k2833

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if this output format is valid; the runs should all be strings, no?
  – Erik the Outgolfer
  Aug 27 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EriktheOutgolfer Fixed
  – TFeld
  Aug 27 at 12:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I'm not sure if this output format is valid; the runs should all be strings, no?
  – Erik the Outgolfer
  Aug 27 at 11:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @EriktheOutgolfer Fixed
  – TFeld
  Aug 27 at 12:01
  

  
  
  
  

I'm not sure if this output format is valid; the runs should all be strings, no?
– Erik the Outgolfer
Aug 27 at 11:56

I'm not sure if this output format is valid; the runs should all be strings, no?
– Erik the Outgolfer
Aug 27 at 11:56

@EriktheOutgolfer Fixed
– TFeld
Aug 27 at 12:01

@EriktheOutgolfer Fixed
– TFeld
Aug 27 at 12:01

add a comment | 

up vote
1
down vote

Python 2, 138 136 bytes

-2 bytes thanks to Erik the Outgolfer.

r=;d=0
for n in input()+[0]:
 a=n-([n]+r)[-1]
 if(a-d)*d:
	if r[2:]:r=["%dto"%r[0]+`r[-1]`+"by%s"%d*(1%d)]
	print r.pop(0),
 r+=n,;d=a

Try it online!

share|improve this answer

edited Aug 27 at 12:24

answered Aug 27 at 12:05

ovs

17.2k21056

add a comment | 

up vote
1
down vote

Python 2, 138 136 bytes

-2 bytes thanks to Erik the Outgolfer.

r=;d=0
for n in input()+[0]:
 a=n-([n]+r)[-1]
 if(a-d)*d:
	if r[2:]:r=["%dto"%r[0]+`r[-1]`+"by%s"%d*(1%d)]
	print r.pop(0),
 r+=n,;d=a

Try it online!

share|improve this answer

edited Aug 27 at 12:24

answered Aug 27 at 12:05

ovs

17.2k21056

add a comment | 

up vote
1
down vote

up vote
1
down vote

Python 2, 138 136 bytes

-2 bytes thanks to Erik the Outgolfer.

r=;d=0
for n in input()+[0]:
 a=n-([n]+r)[-1]
 if(a-d)*d:
	if r[2:]:r=["%dto"%r[0]+`r[-1]`+"by%s"%d*(1%d)]
	print r.pop(0),
 r+=n,;d=a

Try it online!

share|improve this answer

edited Aug 27 at 12:24

answered Aug 27 at 12:05

ovs

17.2k21056

Python 2, 138 136 bytes

-2 bytes thanks to Erik the Outgolfer.

r=;d=0
for n in input()+[0]:
 a=n-([n]+r)[-1]
 if(a-d)*d:
	if r[2:]:r=["%dto"%r[0]+`r[-1]`+"by%s"%d*(1%d)]
	print r.pop(0),
 r+=n,;d=a

Try it online!

share|improve this answer

edited Aug 27 at 12:24

answered Aug 27 at 12:05

ovs

17.2k21056

share|improve this answer

share|improve this answer

edited Aug 27 at 12:24

edited Aug 27 at 12:24

edited Aug 27 at 12:24

answered Aug 27 at 12:05

ovs

17.2k21056

answered Aug 27 at 12:05

ovs

17.2k21056

answered Aug 27 at 12:05

ovs

17.2k21056

17.2k21056

add a comment | 

add a comment | 

up vote
1
down vote

gvm (commit 2612106) bytecode, 108 bytes

Expects the size of the array in one line, then the members each in one line.

Hexdump:

> hexdump -C findruns.bin
00000000 e1 0a 00 10 00 e1 0b 00 ff c8 92 00 f4 f7 10 00 |................|
00000010 01 b0 20 03 00 ff 0a 01 a2 01 c8 92 00 f4 01 c0 |.. .............|
00000020 03 00 ff 0a 02 d0 72 01 0a 03 c8 92 00 f4 05 b0 |......r.........|
00000030 20 a2 02 c0 02 02 6a 03 8b 00 ff f6 06 b0 20 a2 | .....j....... .|
00000040 02 f4 ce 0a 02 c8 92 00 f6 07 6a 03 8b 00 ff f6 |..........j.....|
00000050 f2 b9 66 01 a2 02 00 01 8a 03 f6 05 b9 69 01 a2 |..f..........i..|
00000060 03 92 00 f4 ac c0 74 6f 00 62 79 00 |......to.by.|
0000006c

Test runs:

> echo -e "7n5n6n8n11n15n16n17n" | ./gvm findruns.bin
5 6 8 11 15to17
> echo -e "20n1n2n3n4n5n6n7n9n11n13n15n30n45n50n60n70n80n90n91n93n" | ./gvm findruns.bin
1to7 9to15by2 30 45 50to90by10 91 93

Manually assembled from this:

0100 e1 rud ; read length of array
0101 0a 00 sta $00 ; -> to $00
0103 10 00 ldx #$00 ; loop counter
 readloop:
0105 e1 rud ; read unsigned number
0106 0b 00 ff sta $ff00,x ; store in array at ff00
0109 c8 inx ; next index
010a 92 00 cpx $00 ; length reached?
010c f4 f7 bne readloop ; no -> read next number
010e 10 00 ldx #$00 ; loop counter
0110 01 ; 'lda $20b0', to skip next instruction
 runloop:
0111 b0 20 wch #' ' ; write space character
0113 03 00 ff lda $ff00,x ; load next number from array
0116 0a 01 sta $01 ; -> to $01
0118 a2 01 wud $01 ; and output
011a c8 inx ; next index
011b 92 00 cpx $00 ; length reached?
011d f4 01 bne compare ; if not calculate difference
011f c0 hlt ; done
 compare:
0120 03 00 ff lda $ff00,x ; load next number from array
0123 0a 02 sta $02 ; -> to $01
0125 d0 sec ; calculate ...
0126 72 01 sbc $01 ; ... difference ...
0128 0a 03 sta $03 ; ... to $03
012a c8 inx ; next index
012b 92 00 cpx $00 ; length reached?
012d f4 05 bne checkrun ; if not check whether we have a run
012f b0 20 wch #' ' ; output space
0131 a2 02 wud $02 ; output number
0133 c0 hlt ; done
 checkrun:
0134 02 02 lda $02 ; calculate next ...
0136 6a 03 adc $03 ; ... expected number in run
0138 8b 00 ff cmp $ff00,x ; compare with real next number
013b f6 06 beq haverun ; ok -> found a run
013d b0 20 wch #' ' ; otherwise output space ...
013f a2 02 wud $02 ; ... and number
0141 f4 ce bne runloop ; and repeat searching for runs
 haverun:
0143 0a 02 sta $02 ; store number to $02
0145 c8 inx ; next index
0146 92 00 cpx $00 ; length reached?
0148 f6 07 beq outputrun ; yes -> output this run
014a 6a 03 adc $03 ; calculate next expected number
014c 8b 00 ff cmp $ff00,x ; compare with real next number
014f f6 f2 beq haverun ; ok -> continue parsing run
 outputrun:
0151 b9 66 01 wtx str_to ; write "to"
0154 a2 02 wud $02 ; write end number of run
0156 00 01 lda #$01 ; compare #1 with ...
0158 8a 03 cmp $03 ; ... step size
015a f6 05 beq skip_by ; equal, then skip output of "by"
015c b9 69 01 wtx str_by ; output "by"
015f a2 03 wud $03 ; output step size
 skip_by:
0161 92 00 cpx $00 ; length of array reached?
0163 f4 ac bne runloop ; no -> repeat searching for runs
0165 c0 hlt ; done
 str_to:
0166 74 6f 00 ; "to"
 str_by:
0169 62 79 00 ; "by"
016c

share|improve this answer

edited Aug 27 at 21:20

answered Aug 27 at 11:54

Felix Palmen

3,026423

add a comment | 

up vote
1
down vote

gvm (commit 2612106) bytecode, 108 bytes

Expects the size of the array in one line, then the members each in one line.

Hexdump:

> hexdump -C findruns.bin
00000000 e1 0a 00 10 00 e1 0b 00 ff c8 92 00 f4 f7 10 00 |................|
00000010 01 b0 20 03 00 ff 0a 01 a2 01 c8 92 00 f4 01 c0 |.. .............|
00000020 03 00 ff 0a 02 d0 72 01 0a 03 c8 92 00 f4 05 b0 |......r.........|
00000030 20 a2 02 c0 02 02 6a 03 8b 00 ff f6 06 b0 20 a2 | .....j....... .|
00000040 02 f4 ce 0a 02 c8 92 00 f6 07 6a 03 8b 00 ff f6 |..........j.....|
00000050 f2 b9 66 01 a2 02 00 01 8a 03 f6 05 b9 69 01 a2 |..f..........i..|
00000060 03 92 00 f4 ac c0 74 6f 00 62 79 00 |......to.by.|
0000006c

Test runs:

> echo -e "7n5n6n8n11n15n16n17n" | ./gvm findruns.bin
5 6 8 11 15to17
> echo -e "20n1n2n3n4n5n6n7n9n11n13n15n30n45n50n60n70n80n90n91n93n" | ./gvm findruns.bin
1to7 9to15by2 30 45 50to90by10 91 93

Manually assembled from this:

0100 e1 rud ; read length of array
0101 0a 00 sta $00 ; -> to $00
0103 10 00 ldx #$00 ; loop counter
 readloop:
0105 e1 rud ; read unsigned number
0106 0b 00 ff sta $ff00,x ; store in array at ff00
0109 c8 inx ; next index
010a 92 00 cpx $00 ; length reached?
010c f4 f7 bne readloop ; no -> read next number
010e 10 00 ldx #$00 ; loop counter
0110 01 ; 'lda $20b0', to skip next instruction
 runloop:
0111 b0 20 wch #' ' ; write space character
0113 03 00 ff lda $ff00,x ; load next number from array
0116 0a 01 sta $01 ; -> to $01
0118 a2 01 wud $01 ; and output
011a c8 inx ; next index
011b 92 00 cpx $00 ; length reached?
011d f4 01 bne compare ; if not calculate difference
011f c0 hlt ; done
 compare:
0120 03 00 ff lda $ff00,x ; load next number from array
0123 0a 02 sta $02 ; -> to $01
0125 d0 sec ; calculate ...
0126 72 01 sbc $01 ; ... difference ...
0128 0a 03 sta $03 ; ... to $03
012a c8 inx ; next index
012b 92 00 cpx $00 ; length reached?
012d f4 05 bne checkrun ; if not check whether we have a run
012f b0 20 wch #' ' ; output space
0131 a2 02 wud $02 ; output number
0133 c0 hlt ; done
 checkrun:
0134 02 02 lda $02 ; calculate next ...
0136 6a 03 adc $03 ; ... expected number in run
0138 8b 00 ff cmp $ff00,x ; compare with real next number
013b f6 06 beq haverun ; ok -> found a run
013d b0 20 wch #' ' ; otherwise output space ...
013f a2 02 wud $02 ; ... and number
0141 f4 ce bne runloop ; and repeat searching for runs
 haverun:
0143 0a 02 sta $02 ; store number to $02
0145 c8 inx ; next index
0146 92 00 cpx $00 ; length reached?
0148 f6 07 beq outputrun ; yes -> output this run
014a 6a 03 adc $03 ; calculate next expected number
014c 8b 00 ff cmp $ff00,x ; compare with real next number
014f f6 f2 beq haverun ; ok -> continue parsing run
 outputrun:
0151 b9 66 01 wtx str_to ; write "to"
0154 a2 02 wud $02 ; write end number of run
0156 00 01 lda #$01 ; compare #1 with ...
0158 8a 03 cmp $03 ; ... step size
015a f6 05 beq skip_by ; equal, then skip output of "by"
015c b9 69 01 wtx str_by ; output "by"
015f a2 03 wud $03 ; output step size
 skip_by:
0161 92 00 cpx $00 ; length of array reached?
0163 f4 ac bne runloop ; no -> repeat searching for runs
0165 c0 hlt ; done
 str_to:
0166 74 6f 00 ; "to"
 str_by:
0169 62 79 00 ; "by"
016c

share|improve this answer

edited Aug 27 at 21:20

answered Aug 27 at 11:54

Felix Palmen

3,026423

add a comment | 

up vote
1
down vote

up vote
1
down vote

gvm (commit 2612106) bytecode, 108 bytes

Expects the size of the array in one line, then the members each in one line.

Hexdump:

> hexdump -C findruns.bin
00000000 e1 0a 00 10 00 e1 0b 00 ff c8 92 00 f4 f7 10 00 |................|
00000010 01 b0 20 03 00 ff 0a 01 a2 01 c8 92 00 f4 01 c0 |.. .............|
00000020 03 00 ff 0a 02 d0 72 01 0a 03 c8 92 00 f4 05 b0 |......r.........|
00000030 20 a2 02 c0 02 02 6a 03 8b 00 ff f6 06 b0 20 a2 | .....j....... .|
00000040 02 f4 ce 0a 02 c8 92 00 f6 07 6a 03 8b 00 ff f6 |..........j.....|
00000050 f2 b9 66 01 a2 02 00 01 8a 03 f6 05 b9 69 01 a2 |..f..........i..|
00000060 03 92 00 f4 ac c0 74 6f 00 62 79 00 |......to.by.|
0000006c

Test runs:

> echo -e "7n5n6n8n11n15n16n17n" | ./gvm findruns.bin
5 6 8 11 15to17
> echo -e "20n1n2n3n4n5n6n7n9n11n13n15n30n45n50n60n70n80n90n91n93n" | ./gvm findruns.bin
1to7 9to15by2 30 45 50to90by10 91 93

Manually assembled from this:

0100 e1 rud ; read length of array
0101 0a 00 sta $00 ; -> to $00
0103 10 00 ldx #$00 ; loop counter
 readloop:
0105 e1 rud ; read unsigned number
0106 0b 00 ff sta $ff00,x ; store in array at ff00
0109 c8 inx ; next index
010a 92 00 cpx $00 ; length reached?
010c f4 f7 bne readloop ; no -> read next number
010e 10 00 ldx #$00 ; loop counter
0110 01 ; 'lda $20b0', to skip next instruction
 runloop:
0111 b0 20 wch #' ' ; write space character
0113 03 00 ff lda $ff00,x ; load next number from array
0116 0a 01 sta $01 ; -> to $01
0118 a2 01 wud $01 ; and output
011a c8 inx ; next index
011b 92 00 cpx $00 ; length reached?
011d f4 01 bne compare ; if not calculate difference
011f c0 hlt ; done
 compare:
0120 03 00 ff lda $ff00,x ; load next number from array
0123 0a 02 sta $02 ; -> to $01
0125 d0 sec ; calculate ...
0126 72 01 sbc $01 ; ... difference ...
0128 0a 03 sta $03 ; ... to $03
012a c8 inx ; next index
012b 92 00 cpx $00 ; length reached?
012d f4 05 bne checkrun ; if not check whether we have a run
012f b0 20 wch #' ' ; output space
0131 a2 02 wud $02 ; output number
0133 c0 hlt ; done
 checkrun:
0134 02 02 lda $02 ; calculate next ...
0136 6a 03 adc $03 ; ... expected number in run
0138 8b 00 ff cmp $ff00,x ; compare with real next number
013b f6 06 beq haverun ; ok -> found a run
013d b0 20 wch #' ' ; otherwise output space ...
013f a2 02 wud $02 ; ... and number
0141 f4 ce bne runloop ; and repeat searching for runs
 haverun:
0143 0a 02 sta $02 ; store number to $02
0145 c8 inx ; next index
0146 92 00 cpx $00 ; length reached?
0148 f6 07 beq outputrun ; yes -> output this run
014a 6a 03 adc $03 ; calculate next expected number
014c 8b 00 ff cmp $ff00,x ; compare with real next number
014f f6 f2 beq haverun ; ok -> continue parsing run
 outputrun:
0151 b9 66 01 wtx str_to ; write "to"
0154 a2 02 wud $02 ; write end number of run
0156 00 01 lda #$01 ; compare #1 with ...
0158 8a 03 cmp $03 ; ... step size
015a f6 05 beq skip_by ; equal, then skip output of "by"
015c b9 69 01 wtx str_by ; output "by"
015f a2 03 wud $03 ; output step size
 skip_by:
0161 92 00 cpx $00 ; length of array reached?
0163 f4 ac bne runloop ; no -> repeat searching for runs
0165 c0 hlt ; done
 str_to:
0166 74 6f 00 ; "to"
 str_by:
0169 62 79 00 ; "by"
016c

share|improve this answer

edited Aug 27 at 21:20

answered Aug 27 at 11:54

Felix Palmen

3,026423

gvm (commit 2612106) bytecode, 108 bytes

Expects the size of the array in one line, then the members each in one line.

Hexdump:

> hexdump -C findruns.bin
00000000 e1 0a 00 10 00 e1 0b 00 ff c8 92 00 f4 f7 10 00 |................|
00000010 01 b0 20 03 00 ff 0a 01 a2 01 c8 92 00 f4 01 c0 |.. .............|
00000020 03 00 ff 0a 02 d0 72 01 0a 03 c8 92 00 f4 05 b0 |......r.........|
00000030 20 a2 02 c0 02 02 6a 03 8b 00 ff f6 06 b0 20 a2 | .....j....... .|
00000040 02 f4 ce 0a 02 c8 92 00 f6 07 6a 03 8b 00 ff f6 |..........j.....|
00000050 f2 b9 66 01 a2 02 00 01 8a 03 f6 05 b9 69 01 a2 |..f..........i..|
00000060 03 92 00 f4 ac c0 74 6f 00 62 79 00 |......to.by.|
0000006c

Test runs:

> echo -e "7n5n6n8n11n15n16n17n" | ./gvm findruns.bin
5 6 8 11 15to17
> echo -e "20n1n2n3n4n5n6n7n9n11n13n15n30n45n50n60n70n80n90n91n93n" | ./gvm findruns.bin
1to7 9to15by2 30 45 50to90by10 91 93

Manually assembled from this:

0100 e1 rud ; read length of array
0101 0a 00 sta $00 ; -> to $00
0103 10 00 ldx #$00 ; loop counter
 readloop:
0105 e1 rud ; read unsigned number
0106 0b 00 ff sta $ff00,x ; store in array at ff00
0109 c8 inx ; next index
010a 92 00 cpx $00 ; length reached?
010c f4 f7 bne readloop ; no -> read next number
010e 10 00 ldx #$00 ; loop counter
0110 01 ; 'lda $20b0', to skip next instruction
 runloop:
0111 b0 20 wch #' ' ; write space character
0113 03 00 ff lda $ff00,x ; load next number from array
0116 0a 01 sta $01 ; -> to $01
0118 a2 01 wud $01 ; and output
011a c8 inx ; next index
011b 92 00 cpx $00 ; length reached?
011d f4 01 bne compare ; if not calculate difference
011f c0 hlt ; done
 compare:
0120 03 00 ff lda $ff00,x ; load next number from array
0123 0a 02 sta $02 ; -> to $01
0125 d0 sec ; calculate ...
0126 72 01 sbc $01 ; ... difference ...
0128 0a 03 sta $03 ; ... to $03
012a c8 inx ; next index
012b 92 00 cpx $00 ; length reached?
012d f4 05 bne checkrun ; if not check whether we have a run
012f b0 20 wch #' ' ; output space
0131 a2 02 wud $02 ; output number
0133 c0 hlt ; done
 checkrun:
0134 02 02 lda $02 ; calculate next ...
0136 6a 03 adc $03 ; ... expected number in run
0138 8b 00 ff cmp $ff00,x ; compare with real next number
013b f6 06 beq haverun ; ok -> found a run
013d b0 20 wch #' ' ; otherwise output space ...
013f a2 02 wud $02 ; ... and number
0141 f4 ce bne runloop ; and repeat searching for runs
 haverun:
0143 0a 02 sta $02 ; store number to $02
0145 c8 inx ; next index
0146 92 00 cpx $00 ; length reached?
0148 f6 07 beq outputrun ; yes -> output this run
014a 6a 03 adc $03 ; calculate next expected number
014c 8b 00 ff cmp $ff00,x ; compare with real next number
014f f6 f2 beq haverun ; ok -> continue parsing run
 outputrun:
0151 b9 66 01 wtx str_to ; write "to"
0154 a2 02 wud $02 ; write end number of run
0156 00 01 lda #$01 ; compare #1 with ...
0158 8a 03 cmp $03 ; ... step size
015a f6 05 beq skip_by ; equal, then skip output of "by"
015c b9 69 01 wtx str_by ; output "by"
015f a2 03 wud $03 ; output step size
 skip_by:
0161 92 00 cpx $00 ; length of array reached?
0163 f4 ac bne runloop ; no -> repeat searching for runs
0165 c0 hlt ; done
 str_to:
0166 74 6f 00 ; "to"
 str_by:
0169 62 79 00 ; "by"
016c

share|improve this answer

edited Aug 27 at 21:20

answered Aug 27 at 11:54

Felix Palmen

3,026423

share|improve this answer

share|improve this answer

edited Aug 27 at 21:20

edited Aug 27 at 21:20

edited Aug 27 at 21:20

answered Aug 27 at 11:54

Felix Palmen

3,026423

answered Aug 27 at 11:54

Felix Palmen

3,026423

answered Aug 27 at 11:54

Felix Palmen

3,026423

3,026423

add a comment | 

add a comment | 

up vote
1
down vote

05AB1E (legacy), 49 50 bytes

.œʒε¥Ë}P}Σ€g2KO>}¤εD©g≠i¬s¤„toý¬®¥0èDU≠i„byX««]˜ðý

Way too long, but I'm already glad it's working. This challenge is a lot harder than it looks imo.. Can without a doubt be golfed further.
Σ€g2KO>}¤ is a port of 2,0ySƲÞṪ from @JonathanAllan's Jelly answer (thanks!).

Try it online. (NOTE: Times out for the big test cases.)

+1 byte as bug-fix because 0 is always put at a trailing position when sorting.

Explanation:

.œ # Get all partions of the (implicit) input-list
 # i.e. [1,2,3,11,18,20,22,24,32,33,34]
 # → [[[1],[2],[3],[11],[18],[20],[22],[24],[32],[33],[34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32],[33,34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32,33],[34]],
 # ...]
 ʒ } # Filter this list by:
 ε } # Map the current sub-list by:
 ¥Ë # Take the deltas, and check if all are equal
 # i.e. [1,2,3] → [1,1] → 1
 # i.e. [1,2,3,11] → [1,1,8] → 0
 P # Check if all sub-list have equal deltas
 Σ } # Now that we've filtered the list, sort it by:
 €g # Take the length of each sub-list
 # i.e. [[1,2,3],[11,18],[20,22,24],[32,33],[34]]
 # → [3,2,3,2,1]
 # i.e. [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 # → [3,1,4,3]
 2K # Remove all 2s
 # i.e. [3,2,3,2,1] → ['3','3','1']
 O # And take the sum
 # i.e. ['3','3','1'] → 7
 # i.e. [3,1,4,3] → 11
 > # And increase the sum by 1 (0 is always trailing when sorting)
 ¤ # And then take the last item of this sorted list
 # i.e. for input [1,2,3,11,18,20,22,24,32,33,34]
 # → [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 ε # Now map each of the sub-lists to:
 D© # Save the current sub-list in the register
 g≠i # If its length is not 1:
 # i.e. [11] → 1 → 0 (falsey)
 # i.e. [18,20,22,24] → 4 → 1 (truthy)
 ¬s¤ # Take the head and tail of the sub-list
 # i.e. [18,20,22,24] → 18 and 24
 „toý # And join them with "to"
 # i.e. 18 and 24 → ['18to24', '18to20to22to24']
 ¬ # (head to remove some access waste we no longer need)
 # i.e. ['18to24', '18to20to22to24'] → '18to24'
 ® # Get the sub-list from the register again
 ¥ # Take its deltas
 # i.e. [18,20,22,24] → [2,2,2]
 0è # Get the first item (should all be the same delta)
 # i.e. [2,2,2] → 2
 DU # Save it in variable `X`
 ≠i # If the delta is not 1:
 # i.e. 2 → 1 (truthy)
 „byX« # Merge "by" with `X`
 # i.e. 2 → 'by2'
 « # And merge it with the earlier "#to#"
 # i.e. '18to24' and 'by2' → '18to24by2'
 ] # Close the mapping and both if-statements
˜ # Flatten the list
 # i.e. ['1to3',[11],'18to24by2','32to34']
 # → ['1to3',11,'18to24by2','32to34']
 ðý # And join by spaces (which we implicitly output as result)
 # i.e. ['1to3',11,'18to24by2','32to34']
 # → '1to3 11 18to24by2 32to34'

share|improve this answer

edited Aug 28 at 9:32

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

add a comment | 

up vote
1
down vote

05AB1E (legacy), 49 50 bytes

.œʒε¥Ë}P}Σ€g2KO>}¤εD©g≠i¬s¤„toý¬®¥0èDU≠i„byX««]˜ðý

Way too long, but I'm already glad it's working. This challenge is a lot harder than it looks imo.. Can without a doubt be golfed further.
Σ€g2KO>}¤ is a port of 2,0ySƲÞṪ from @JonathanAllan's Jelly answer (thanks!).

Try it online. (NOTE: Times out for the big test cases.)

+1 byte as bug-fix because 0 is always put at a trailing position when sorting.

Explanation:

.œ # Get all partions of the (implicit) input-list
 # i.e. [1,2,3,11,18,20,22,24,32,33,34]
 # → [[[1],[2],[3],[11],[18],[20],[22],[24],[32],[33],[34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32],[33,34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32,33],[34]],
 # ...]
 ʒ } # Filter this list by:
 ε } # Map the current sub-list by:
 ¥Ë # Take the deltas, and check if all are equal
 # i.e. [1,2,3] → [1,1] → 1
 # i.e. [1,2,3,11] → [1,1,8] → 0
 P # Check if all sub-list have equal deltas
 Σ } # Now that we've filtered the list, sort it by:
 €g # Take the length of each sub-list
 # i.e. [[1,2,3],[11,18],[20,22,24],[32,33],[34]]
 # → [3,2,3,2,1]
 # i.e. [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 # → [3,1,4,3]
 2K # Remove all 2s
 # i.e. [3,2,3,2,1] → ['3','3','1']
 O # And take the sum
 # i.e. ['3','3','1'] → 7
 # i.e. [3,1,4,3] → 11
 > # And increase the sum by 1 (0 is always trailing when sorting)
 ¤ # And then take the last item of this sorted list
 # i.e. for input [1,2,3,11,18,20,22,24,32,33,34]
 # → [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 ε # Now map each of the sub-lists to:
 D© # Save the current sub-list in the register
 g≠i # If its length is not 1:
 # i.e. [11] → 1 → 0 (falsey)
 # i.e. [18,20,22,24] → 4 → 1 (truthy)
 ¬s¤ # Take the head and tail of the sub-list
 # i.e. [18,20,22,24] → 18 and 24
 „toý # And join them with "to"
 # i.e. 18 and 24 → ['18to24', '18to20to22to24']
 ¬ # (head to remove some access waste we no longer need)
 # i.e. ['18to24', '18to20to22to24'] → '18to24'
 ® # Get the sub-list from the register again
 ¥ # Take its deltas
 # i.e. [18,20,22,24] → [2,2,2]
 0è # Get the first item (should all be the same delta)
 # i.e. [2,2,2] → 2
 DU # Save it in variable `X`
 ≠i # If the delta is not 1:
 # i.e. 2 → 1 (truthy)
 „byX« # Merge "by" with `X`
 # i.e. 2 → 'by2'
 « # And merge it with the earlier "#to#"
 # i.e. '18to24' and 'by2' → '18to24by2'
 ] # Close the mapping and both if-statements
˜ # Flatten the list
 # i.e. ['1to3',[11],'18to24by2','32to34']
 # → ['1to3',11,'18to24by2','32to34']
 ðý # And join by spaces (which we implicitly output as result)
 # i.e. ['1to3',11,'18to24by2','32to34']
 # → '1to3 11 18to24by2 32to34'

share|improve this answer

edited Aug 28 at 9:32

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

add a comment | 

up vote
1
down vote

up vote
1
down vote

05AB1E (legacy), 49 50 bytes

.œʒε¥Ë}P}Σ€g2KO>}¤εD©g≠i¬s¤„toý¬®¥0èDU≠i„byX««]˜ðý

Way too long, but I'm already glad it's working. This challenge is a lot harder than it looks imo.. Can without a doubt be golfed further.
Σ€g2KO>}¤ is a port of 2,0ySƲÞṪ from @JonathanAllan's Jelly answer (thanks!).

Try it online. (NOTE: Times out for the big test cases.)

+1 byte as bug-fix because 0 is always put at a trailing position when sorting.

Explanation:

.œ # Get all partions of the (implicit) input-list
 # i.e. [1,2,3,11,18,20,22,24,32,33,34]
 # → [[[1],[2],[3],[11],[18],[20],[22],[24],[32],[33],[34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32],[33,34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32,33],[34]],
 # ...]
 ʒ } # Filter this list by:
 ε } # Map the current sub-list by:
 ¥Ë # Take the deltas, and check if all are equal
 # i.e. [1,2,3] → [1,1] → 1
 # i.e. [1,2,3,11] → [1,1,8] → 0
 P # Check if all sub-list have equal deltas
 Σ } # Now that we've filtered the list, sort it by:
 €g # Take the length of each sub-list
 # i.e. [[1,2,3],[11,18],[20,22,24],[32,33],[34]]
 # → [3,2,3,2,1]
 # i.e. [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 # → [3,1,4,3]
 2K # Remove all 2s
 # i.e. [3,2,3,2,1] → ['3','3','1']
 O # And take the sum
 # i.e. ['3','3','1'] → 7
 # i.e. [3,1,4,3] → 11
 > # And increase the sum by 1 (0 is always trailing when sorting)
 ¤ # And then take the last item of this sorted list
 # i.e. for input [1,2,3,11,18,20,22,24,32,33,34]
 # → [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 ε # Now map each of the sub-lists to:
 D© # Save the current sub-list in the register
 g≠i # If its length is not 1:
 # i.e. [11] → 1 → 0 (falsey)
 # i.e. [18,20,22,24] → 4 → 1 (truthy)
 ¬s¤ # Take the head and tail of the sub-list
 # i.e. [18,20,22,24] → 18 and 24
 „toý # And join them with "to"
 # i.e. 18 and 24 → ['18to24', '18to20to22to24']
 ¬ # (head to remove some access waste we no longer need)
 # i.e. ['18to24', '18to20to22to24'] → '18to24'
 ® # Get the sub-list from the register again
 ¥ # Take its deltas
 # i.e. [18,20,22,24] → [2,2,2]
 0è # Get the first item (should all be the same delta)
 # i.e. [2,2,2] → 2
 DU # Save it in variable `X`
 ≠i # If the delta is not 1:
 # i.e. 2 → 1 (truthy)
 „byX« # Merge "by" with `X`
 # i.e. 2 → 'by2'
 « # And merge it with the earlier "#to#"
 # i.e. '18to24' and 'by2' → '18to24by2'
 ] # Close the mapping and both if-statements
˜ # Flatten the list
 # i.e. ['1to3',[11],'18to24by2','32to34']
 # → ['1to3',11,'18to24by2','32to34']
 ðý # And join by spaces (which we implicitly output as result)
 # i.e. ['1to3',11,'18to24by2','32to34']
 # → '1to3 11 18to24by2 32to34'

share|improve this answer

edited Aug 28 at 9:32

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

05AB1E (legacy), 49 50 bytes

.œʒε¥Ë}P}Σ€g2KO>}¤εD©g≠i¬s¤„toý¬®¥0èDU≠i„byX««]˜ðý

Way too long, but I'm already glad it's working. This challenge is a lot harder than it looks imo.. Can without a doubt be golfed further.
Σ€g2KO>}¤ is a port of 2,0ySƲÞṪ from @JonathanAllan's Jelly answer (thanks!).

Try it online. (NOTE: Times out for the big test cases.)

+1 byte as bug-fix because 0 is always put at a trailing position when sorting.

Explanation:

.œ # Get all partions of the (implicit) input-list
 # i.e. [1,2,3,11,18,20,22,24,32,33,34]
 # → [[[1],[2],[3],[11],[18],[20],[22],[24],[32],[33],[34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32],[33,34]],
 # [[1],[2],[3],[11],[18],[20],[22],[24],[32,33],[34]],
 # ...]
 ʒ } # Filter this list by:
 ε } # Map the current sub-list by:
 ¥Ë # Take the deltas, and check if all are equal
 # i.e. [1,2,3] → [1,1] → 1
 # i.e. [1,2,3,11] → [1,1,8] → 0
 P # Check if all sub-list have equal deltas
 Σ } # Now that we've filtered the list, sort it by:
 €g # Take the length of each sub-list
 # i.e. [[1,2,3],[11,18],[20,22,24],[32,33],[34]]
 # → [3,2,3,2,1]
 # i.e. [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 # → [3,1,4,3]
 2K # Remove all 2s
 # i.e. [3,2,3,2,1] → ['3','3','1']
 O # And take the sum
 # i.e. ['3','3','1'] → 7
 # i.e. [3,1,4,3] → 11
 > # And increase the sum by 1 (0 is always trailing when sorting)
 ¤ # And then take the last item of this sorted list
 # i.e. for input [1,2,3,11,18,20,22,24,32,33,34]
 # → [[1,2,3],[11],[18,20,22,24],[32,33,34]]
 ε # Now map each of the sub-lists to:
 D© # Save the current sub-list in the register
 g≠i # If its length is not 1:
 # i.e. [11] → 1 → 0 (falsey)
 # i.e. [18,20,22,24] → 4 → 1 (truthy)
 ¬s¤ # Take the head and tail of the sub-list
 # i.e. [18,20,22,24] → 18 and 24
 „toý # And join them with "to"
 # i.e. 18 and 24 → ['18to24', '18to20to22to24']
 ¬ # (head to remove some access waste we no longer need)
 # i.e. ['18to24', '18to20to22to24'] → '18to24'
 ® # Get the sub-list from the register again
 ¥ # Take its deltas
 # i.e. [18,20,22,24] → [2,2,2]
 0è # Get the first item (should all be the same delta)
 # i.e. [2,2,2] → 2
 DU # Save it in variable `X`
 ≠i # If the delta is not 1:
 # i.e. 2 → 1 (truthy)
 „byX« # Merge "by" with `X`
 # i.e. 2 → 'by2'
 « # And merge it with the earlier "#to#"
 # i.e. '18to24' and 'by2' → '18to24by2'
 ] # Close the mapping and both if-statements
˜ # Flatten the list
 # i.e. ['1to3',[11],'18to24by2','32to34']
 # → ['1to3',11,'18to24by2','32to34']
 ðý # And join by spaces (which we implicitly output as result)
 # i.e. ['1to3',11,'18to24by2','32to34']
 # → '1to3 11 18to24by2 32to34'

share|improve this answer

edited Aug 28 at 9:32

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

share|improve this answer

share|improve this answer

edited Aug 28 at 9:32

edited Aug 28 at 9:32

edited Aug 28 at 9:32

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

answered Aug 27 at 17:27

Kevin Cruijssen

29.5k550162

29.5k550162

add a comment | 

add a comment | 

up vote
0
down vote

Perl 5, 154 bytes

$$r[1]-$$r[0]==$_-$$r[-1])?push@$r,$_:push@r,$r=[$_]for@_;join" ",map@$_<3?@$_:("$$_[0]to$$_[-1]by".($$_[1]-$$_[0]))=~s/by1$//r,@r

Same with spaces, newlines, #comments and sub by:

sub by 
 my(@r,$r);
 @r && # if at least one run candidate exists and...
 ( @$r<2 # ...just one elem so far
 

Try it online!

...for passing tests from OP.

share|improve this answer

edited Aug 27 at 12:51

answered Aug 27 at 11:47

Kjetil S.

51115

add a comment | 

up vote
0
down vote

Perl 5, 154 bytes

$$r[1]-$$r[0]==$_-$$r[-1])?push@$r,$_:push@r,$r=[$_]for@_;join" ",map@$_<3?@$_:("$$_[0]to$$_[-1]by".($$_[1]-$$_[0]))=~s/by1$//r,@r

Same with spaces, newlines, #comments and sub by:

sub by 
 my(@r,$r);
 @r && # if at least one run candidate exists and...
 ( @$r<2 # ...just one elem so far
 

Try it online!

...for passing tests from OP.

share|improve this answer

edited Aug 27 at 12:51

answered Aug 27 at 11:47

Kjetil S.

51115

add a comment | 

up vote
0
down vote

up vote
0
down vote

Perl 5, 154 bytes

$$r[1]-$$r[0]==$_-$$r[-1])?push@$r,$_:push@r,$r=[$_]for@_;join" ",map@$_<3?@$_:("$$_[0]to$$_[-1]by".($$_[1]-$$_[0]))=~s/by1$//r,@r

Same with spaces, newlines, #comments and sub by:

sub by 
 my(@r,$r);
 @r && # if at least one run candidate exists and...
 ( @$r<2 # ...just one elem so far
 

Try it online!

...for passing tests from OP.

share|improve this answer

edited Aug 27 at 12:51

answered Aug 27 at 11:47

Kjetil S.

51115

Perl 5, 154 bytes

$$r[1]-$$r[0]==$_-$$r[-1])?push@$r,$_:push@r,$r=[$_]for@_;join" ",map@$_<3?@$_:("$$_[0]to$$_[-1]by".($$_[1]-$$_[0]))=~s/by1$//r,@r

Same with spaces, newlines, #comments and sub by:

sub by 
 my(@r,$r);
 @r && # if at least one run candidate exists and...
 ( @$r<2 # ...just one elem so far
 

Try it online!

...for passing tests from OP.

share|improve this answer

edited Aug 27 at 12:51

answered Aug 27 at 11:47

Kjetil S.

51115

share|improve this answer

share|improve this answer

edited Aug 27 at 12:51

edited Aug 27 at 12:51

edited Aug 27 at 12:51

answered Aug 27 at 11:47

Kjetil S.

51115

answered Aug 27 at 11:47

Kjetil S.

51115

answered Aug 27 at 11:47

Kjetil S.

51115

51115

add a comment | 

add a comment | 

up vote
0
down vote

Retina 0.8.2, 77 bytes

d+
$*
(1+)(?= 1(1+))
$1:$2
1:(1+) (1+:1 )+(1+)
1to$3by$1
:1+|by1b

1+
$.&

Try it online! Link includes test cases. Explanation:

d+
$*

Convert to unary.

(1+)(?= 1(1+))
$1:$2

Compute consecutive differences.

1:(1+) (1+:1 )+(1+)
1to$3by$1

Convert runs to to...by syntax.

:1+|by1b

Remove unconverted differences and by1.

1+
$.&

Convert to decimal.

share|improve this answer

answered Aug 29 at 12:12

Neil

75k744170

add a comment | 

up vote
0
down vote

Retina 0.8.2, 77 bytes

d+
$*
(1+)(?= 1(1+))
$1:$2
1:(1+) (1+:1 )+(1+)
1to$3by$1
:1+|by1b

1+
$.&

Try it online! Link includes test cases. Explanation:

d+
$*

Convert to unary.

(1+)(?= 1(1+))
$1:$2

Compute consecutive differences.

1:(1+) (1+:1 )+(1+)
1to$3by$1

Convert runs to to...by syntax.

:1+|by1b

Remove unconverted differences and by1.

1+
$.&

Convert to decimal.

share|improve this answer

answered Aug 29 at 12:12

Neil

75k744170

add a comment | 

up vote
0
down vote

up vote
0
down vote

Retina 0.8.2, 77 bytes

d+
$*
(1+)(?= 1(1+))
$1:$2
1:(1+) (1+:1 )+(1+)
1to$3by$1
:1+|by1b

1+
$.&

Try it online! Link includes test cases. Explanation:

d+
$*

Convert to unary.

(1+)(?= 1(1+))
$1:$2

Compute consecutive differences.

1:(1+) (1+:1 )+(1+)
1to$3by$1

Convert runs to to...by syntax.

:1+|by1b

Remove unconverted differences and by1.

1+
$.&

Convert to decimal.

share|improve this answer

answered Aug 29 at 12:12

Neil

75k744170

Retina 0.8.2, 77 bytes

d+
$*
(1+)(?= 1(1+))
$1:$2
1:(1+) (1+:1 )+(1+)
1to$3by$1
:1+|by1b

1+
$.&

Try it online! Link includes test cases. Explanation:

d+
$*

Convert to unary.

(1+)(?= 1(1+))
$1:$2

Compute consecutive differences.

1:(1+) (1+:1 )+(1+)
1to$3by$1

Convert runs to to...by syntax.

:1+|by1b

Remove unconverted differences and by1.

1+
$.&

Convert to decimal.

share|improve this answer

answered Aug 29 at 12:12

Neil

75k744170

share|improve this answer

share|improve this answer

answered Aug 29 at 12:12

Neil

75k744170

answered Aug 29 at 12:12

Neil

75k744170

answered Aug 29 at 12:12

Neil

75k744170

75k744170

add a comment | 

add a comment | 

 

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