Find palindrome dates in the Linux command line - explained

By Lazarus Lazaridis

Yesterday I saw a great command that finds the palindrome dates between now and x days ago.

The command was posted on Twitter by @climagic (a great account to follow if you want to find awesome command line stuff).

In this post we are going to break it down and explain how it works.

What is a palindrome

Palindrome is a word/number/phrase or other sequence of characters that reads the same backward as forward. Palindrome - Wikipedia

Few days ago, the date was 02/02/2020 which was a palindrome:

Image showing the palindrome 02022020

The command

printf "now - %d days\n" {1..332044} |date -f- +%Y%_m%_d$'\n'%Y%m%d |tr -d ' ' > alldates.txt; rev alldates.txt >revdates.txt; paste alldates.txt revdates.txt |awk '$1==$2{print $1}'

And here’s a more readable version using new lines after each pipe.

printf "now - %d days\n" {1..332044} |
date -f- +%Y%_m%_d$'\n'%Y%m%d |
tr -d ' ' > alldates.txt; rev alldates.txt >revdates.txt; paste alldates.txt revdates.txt |
awk '$1==$2{print $1}'

I am going to explain each part of the command separately.

Note that the output of each part is the input of the next part (following the pipe |).

printf "now - %d days\n" {1..332044}

The first part seems pretty straight forward but it does have an interesting part, the brace expansion mechanism, which I didn’t know of.

Open a terminal and type the following commands:

# From 1 to 10
$ echo {1..10}
1 2 3 4 5 6 7 8 9 10 # From a to z
$ echo {a..z}
a b c d e f g h i j k l m n o p q r s t u v w x y z # From 10 to 20
$ echo Before-{10..20}-After
Before-10-After Before-11-After Before-12-After Before-13-After Before-14-After Before-15-After Before-16-After Before-17-After Before-18-After Before-19-After Before-20-After

As you can see, the brace expansion allows the generation of values based on a “sequence” we set.

Learn more about the Brace expansion mechanism here.

Part explained

  • we print the phrase “now - x days” so many times as the number of values expanded from the sequence
  • {1..332044} will generate 332044 values starting from 1 and increasing by 1
  • we add a new line at the end of each produced string

Output

now - 1 days
now - 2 days
now - 3 days
now - 4 days
now - 5 days
...
now - 332038 days
now - 332039 days
now - 332040 days
now - 332041 days
now - 332042 days
now - 332043 days
now - 332044 days

In this part we use the date command to parse the previously piped output and print two lines for each date, one with padding zeros and one without zeros for single digit days and months.

Familiarize yourself with the command in your terminal:

# Show current date = now
$ date
Thu Feb 6 07:30:10 EET 2020 # Show yesterday
$ date --date="now - 1 days"
Wed Feb 5 07:31:22 EET 2020 # Show yesterday as dd/mm/yyyy (zero padding)
$ date --date="now - 1 days" +%d/%m/%Y
05/02/2020 # Show yesterday as dd/mm/yyyy (space padded)
$ date --date="now - 1 days" +%_d/%_m/%Y 5/ 2/2020 # Show yesterday as dd/mm/yyyy (without padding)
$ date --date="now - 1 days" +%-d/%-m/%Y
5/2/2020

Part explained

  • we use the -f option to define the input of the date to be the lines of a file, followed by a hyphen - to set this file to be the stdin
  • we set the output format (what follows the plus + symbol) to be the numeric representation of the year (%Y), the month (padded with spaces: %_m, padded with zeros: %m) and the day of the month (padded with spaces: %_d, padded with zeros: %d)
  • the space padded format is %Y%\_m%\_d and the zero padded format is %Y%m%d
  • we combine these two formats for each date and we separate them with a new line $'\n' thus, for each date we parse, we produce two lines, one padded with zeros and one padded with spaces

Note: instead of padding with spaces (using the underscore _ in the format), we could use the hyphen - instead which would skip the padding of the field. This way wouldn’t have to remove the spaces as described in the next part.

Output

2020 2 5
20200205
2020 2 4
20200204
2020 2 3
20200203
2020 2 2
20200202
2020 2 1
20200201
2020 131
20200131
2020 130
20200130

In this part we remove the spaces produced in the previous step and we create two files, one with all the dates and one with all the dates reversed.

Then we combine these two files by merging each line of the first file with the corresponding line of the second file separated with a tab.

Familiarize your self with the commands of this part in your terminal:

# Remove empty spaces
$ echo "Good night " | tr -d " "
Goodnight # Remove hyphens
$ echo "Good-night" | tr -d "-"
Goodnight # Print a-x-z where x is a number between 1 and 9
$ printf "a-%d-z\n" {1..9}
a-1-z
a-2-z
a-3-z
a-4-z
a-5-z
a-6-z
a-7-z
a-8-z
a-9-z # and then reverse the order of the characters of each line
$ printf "a-%d-z\n" {1..9} | rev
z-1-a
z-2-a
z-3-a
z-4-a
z-5-a
z-6-a
z-7-a
z-8-a
z-9-a

Part explained

  • tr -d '': we remove the spaces of the input (which is the output of the previous part) and we save the output in a file named alldates.txt
  • rev: we reverse the order of the characters of each line of the alldates.txt file and we save the output in a new file named revdates.txt
  • paste: we merge the two files; each line of the alldates.txt file is merged with the corresponding line of the file revdates.txt separated by a tab (the two files have the exact number of lines).

Output

202025	520202
20200205	50200202
202024	420202
20200204	40200202
202023	320202
20200203	30200202
202022	220202
20200202	20200202
202021	120202
20200201	10200202

awk '$1==$2{print $1}'

In this final part, we use awk to print the palindromes.

Familiarize yourself with the usage of awk related to this post, in your terminal:

# Print the whole input (awk's default behavior) if the fields are equal
$ echo "a a" | awk '$1==$2'
a a # Print the whole input (awk's default behavior) if the fields are not equal
$ echo "a a" | awk '$1!=$2'
# nothing here # Print the whole input (awk's default behavior) if the fields are not equal
$ echo "a b" | awk '$1!=$2'
a b # Print the first field (awk's default behavior) if the fields are not equal
$ echo "a b" | awk '$1!=$2 {print $1}'
a # Print the second field (awk's default behavior) if the fields are not equal
$ echo "a b" | awk '$1!=$2 {print $2}'
b

Part explained

  • the previous output that is now the input of this part consists of lines and each line has two values separated by a TAB. With this kind of input, awk resolves two fields in each line. We can access these two fields using the $1 and $2 variables.
  • as described in a previous post, awk statements consist of a pattern-expression and an associated action.
    <pattern/expression> { <action> }
    

    If the pattern succeeds then the associated action is being executed. In this part the expression is translated as “if the two fields are equal” ($1==$2) and the action as “print the first field” (printf $1).

  • if the fields are equal we do have a palindrome since each pair of fields is actually a date and its reversed value.

Output

20200202
2019102
2018102
2017102
2016102
2015102
2014102
2013102
2012102
...
11111111
11111111
1111111
1111111
111111

Cat photo

That’s all, thanks for reading.

Image with image reflection of my cat, Irida

References