JavaScript (ES6), 112 bytes

All credit for this shorter version goes to @nwellnhof.

Takes input as an integer. Returns an array of three decimal digits.

n=>[(x=n>>4,y=x>>3,q=n/2&55,p=q%8)>5&&q-39?8|y&1:y,(p&5^5?x&6:q-23?8:y&6)|x&1,(p<5?p*2:p<6?x&6:p%q<7?y&6:8)|n&1]

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JavaScript (ES6), 118 117 bytes

Takes input as an integer. Returns an array of three decimal digits.

n=>[(x=n>>4&7,y=n>>7,p=n/2&7)>5&&p<7|x/2^2?8|y&1:y,(p<7?p-5?x:8:x/2^1?8:y&6)|x&1,(p<5?p*2:p<6?x&6:p<7|x<2?y&6:8)|n&1]

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How?

Instead of trying to apply the 'official' algorithm, this code is based on some kind of reverse-engineering of the patterns that can be found in the expected results.

Given the input integer $n$, we compute:

$$begin{align}&x=leftlfloorfrac{n}{16}rightrfloor bmod 8\
&y=leftlfloorfrac{n}{128}rightrfloor\
&p=leftlfloorfrac{n}{2}rightrfloor bmod 8end{align}
$$

Example: first digit (hundreds)

x     | 0                | 1                | 2                | 3               

n & 1 | 0101010101010101 | 0101010101010101 | 0101010101010101 | 0101010101010101

p     | 0011223344556677 | 0011223344556677 | 0011223344556677 | 0011223344556677

------+------------------+------------------+------------------+-----------------

y = 0 | 0000000000008888 | 0000000000008888 | 0000000000008888 | 0000000000008888

y = 1 | 1111111111119999 | 1111111111119999 | 1111111111119999 | 1111111111119999

y = 2 | 2222222222228888 | 2222222222228888 | 2222222222228888 | 2222222222228888

y = 3 | 3333333333339999 | 3333333333339999 | 3333333333339999 | 3333333333339999

y = 4 | 4444444444448888 | 4444444444448888 | 4444444444448888 | 4444444444448888

y = 5 | 5555555555559999 | 5555555555559999 | 5555555555559999 | 5555555555559999

y = 6 | 6666666666668888 | 6666666666668888 | 6666666666668888 | 6666666666668888

y = 7 | 7777777777779999 | 7777777777779999 | 7777777777779999 | 7777777777779999



x     | 4                | 5                | 6                | 7               

n & 1 | 0101010101010101 | 0101010101010101 | 0101010101010101 | 0101010101010101

p     | 0011223344556677 | 0011223344556677 | 0011223344556677 | 0011223344556677

------+------------------+------------------+------------------+-----------------

y = 0 | 0000000000008800 | 0000000000008800 | 0000000000008888 | 0000000000008888

y = 1 | 1111111111119911 | 1111111111119911 | 1111111111119999 | 1111111111119999

y = 2 | 2222222222228822 | 2222222222228822 | 2222222222228888 | 2222222222228888

y = 3 | 3333333333339933 | 3333333333339933 | 3333333333339999 | 3333333333339999

y = 4 | 4444444444448844 | 4444444444448844 | 4444444444448888 | 4444444444448888

y = 5 | 5555555555559955 | 5555555555559955 | 5555555555559999 | 5555555555559999

y = 6 | 6666666666668866 | 6666666666668866 | 6666666666668888 | 6666666666668888

y = 7 | 7777777777779977 | 7777777777779977 | 7777777777779999 | 7777777777779999

Algorithm:

• If $p<6$, we have $d=y$
  


• If $p=6$, we have $d=8+(ybmod2)$
  


• If $p=7text{ AND }(x<4text{ OR }x>5)$, we have $d=8+(ybmod2)$
  


• If $p=7text{ AND }(x=4text{ OR }x=5)$, we have $d=y$
  

As JS code:

p > 5 && p < 7 | x / 2 ^ 2 ? 8 | y & 1 : y

Python 3, 229 ... 97 96 bytes

lambda a:[[a&6,a>>4&6,a>>7&6,8][b"  eW7B]Oys"[~a&8or~a&6or~6|a>>4]%x&3]|a>>x%9&1for x in[7,4,9]]

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-4 bytes by @xnor

-6 bytes by @nwellnhof

Formatted:

h = lambda a:[

    [a&6, a>>4&6, a>>7&6, 8][              List to take high bits from

        b"  eW7B]Oys"[                     10 char string; where to get high bits for

                                             indicator values 1-8. 0th,1st chars not used.

            ~a&8 or ~a&6 or ~6|a>>4]       Compute indicator (by @nwellnhof)

        %x&3]                              High bits of each digit

    | a >> x%9 & 1                         bitwise OR with low bit of each digit

    for x in [7,4,9]]

Explanation

Because I originally wanted to implement this in Jelly, I take a different approach from most answers here, which is simple and perhaps suited to a golfing language. Though the golfed function takes an integer, let the input as a bit list be [a0,a1,...,a9]. Then we can derive three values from the input

• The low bits [a2,a5,a9]: These will always be the low bits of [d0,d1,d2] respectively.


• The high bits [2*a0a1,2*a3a4,2*a7a8,8]: The high bits of each digit will be one of these.


• The indicator bits, [a3,a4,a5,a7,a8], determining how to get the high bits of each digit. We compute the indicator (between 1 and 8) as follows:
  
  
  
  
  • If a5 == 0, the indicator is 8 (originally 0, but using 8 instead saves a byte)
  
  
  • If a3 nand a4, the indicator is 6 - 2*a3a4
  
  
  • Otherwise the indicator is 2*a7a8 + 1 (actually calculated as a negative number).
  
  
  
  

Then the nth digit can be elegantly computed as high_bits[arr[indicator][n]] | low_bits[n] by the table below, which is compressed into a string.

arr = [

    [0,1,2],

    [3,1,2],

    [1,3,2],

    [2,1,3],

    [2,3,3],

    [3,2,3],

    [3,3,2],

    [3,3,3]

]

JavaScript (Node.js), 126 119 117 112 111 bytes

(a,b,c,d,e,f,g,h,i,j)=>[(g&h&i+(b+=a*4+b,e+=d*4+e)!=5?8:b)+c,(g&i?h+e-3?8:b:e)+f,(g?h<i?e:h>i*e?b:8:h*4+i*2)+j]

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-5 bytes thanks @tsh (and 2 by myself) So l can make more effort than I expected.

-2 more bytes using @tsh's technique!

-5 bytes thanks @Arnauld

-1 byte thanks @Neil

Input as a list of 10 bits (as 10 arguments), output as a list of 3 digits.

C (gcc), 138 129 bytes

f(w){int t=w/2&55,s=t%8,v=w/16,u=v/8;w=((s<6|t==39?u:8|u%2)*10+v%2+(s&5^5?v&6:t-23?8:u&6))*10+w%2+(s<5?s*2:s<6?v&6:s%t<7?u&6:8);}

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First extracts some bits into variables s and t, so that the eight rows of the conversion table can be identified by:

1.  s < 4              u v w¹

2.  s = 4              u v 8¹

3.  s = 5              u 8 v

4.  s = 6              8 v u

5.  s = 7, t =  7      8 8 u

6.  s = 7, t = 23      8 u 8

7.  s = 7, t = 39      u 8 8

8.  s = 7, t = 55      8 8 8



¹ Can be computed with s*2

Then sets up u and v with divisions (right shifts), so that u, v and the input w contain the lower three BCD bits in positions 0-2. The rest is bit shuffling depending on s and t. Two notable tricks are:

s&5^5  // Rows 1, 2 and 4.

s%t<7  // Rows 1-5.

A port of Shieru Asakoto's Javascript solution is only 124 bytes:

f(a,b,c,d,e,f,g,h,i,j){a=(((g&h&i+(b+=a*4+b,e+=d*4+e)!=5?8:b)+c)*10+(g&i?h+e-3?8:b:e)+f)*10+(g?h-i|h&!e?h?b:e:8:h*4+i*2)+j;}

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Ruby, 153 ... 119 117 bytes

->n{n+=n&896;a,b,c=n&1536,n&96,n&14;"%x"%n+=c<9?0:2036+[b/16-b-1918,r=a>>8,[r+126,a/16-26,a-1978,38][b/32]-a][c/2-5]}

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How it works:

->n{n+=n&896;

This is the starting point: convert to BCD by shifting 3 bits to the left, which works for most of the patterns.

a,b,c=n&1536,n&96,n&14;

Get the middle bits of each nibble (and one extra bit of the third nibble, but mask the least significant bit).

"%x"%n+=c<9?0

If the third digit is less than 10 (less than 9 because we never cared for the LSB anyway), we're set: this is plain BCD, we can output the hex without changing anything

:2036+[b/16-b-1918,r=a>>8,[r+126,a/16-26,a-1978,38][b/32]-a][c/2-5]}

Otherwise do some black magic by shifting bits around and adding magic numbers until we get the result we want.

Retina 0.8.2, 191 181 bytes

(...)(...)

:$1,$2;

..(.),11(.);111

100$1,100$2;100

(10|(..)(.,)01)(.);111

100$3$2$4;100

(..)(.),(00.);111

100$2,1$3;0$1

(..)((.{5});110|(.);101)

100$3$4;$1

1

01

+`10

011

.0+(1*)

$.1

Try it online! Link includes test cases. Edit: Saved 10 byte by not padding digits to 4 bits except where necessary. Explanation:

(...)(...)

:$1,$2;

Insert separators so that each digit can be converted to decimal separately. This effectively handles the first two cases in the conversion table.

..(.),11(.);111

100$1,100$2;100

Handle the last (eighth) case in the conversion table.

(10|(..)(.,)01)(.);111

100$3$2$4;100

Handle the sixth and seventh cases in the conversion table.

(..)(.),(00.);111

100$2,1$3;0$1

Handle the fifth case in the conversion table.

(..)((.{5});110|(.);101)

100$3$4;$1

Handle the third and fourth cases in the conversion table.

1

01

+`10

011

.0+(1*)

$.1

Perform binary to decimal conversion.

Python 2, 157 bytes

lambda a,b,c,d,e,f,g,h,i,j:[c+[a*4+b*2,8][g*h*~(d*~e*i)],f+[d*4+e*2,8,a*4+b*2][g*i+(d<e)*g*i*h],j+[h*4+i*2,[8,[a*4+b*2,d*4+e*2][h<i]][h^i or(h&i-(d|e))]][g]]

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Clean, 238 ... 189 bytes

-2 bytes thanks to Neil

import StdEnv

$a b c d e f g h i j=100*(c+2*b+4*a)+10*(f+2*e+4*d)+j+2*i+4*h-2*(h*(99*b+198*a-394)+i*(9*e+18*d+h*(e+2*d-4+(b+2*a-4)*(1-10*e-100*d+110*e*d))-35)-4)*g+0^(e+d)*(2*b+4*a-8*i*h*g)

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Takes a 'list' of 10 bits in the form of 10 arguments, using a direct formula to compute the result.

Perl 5, 195 bytes

sub f{$n=shift;@p=((map{($n>>$_&3)*2}(8,5,1)),8);for(16390,28935,29005,227791,29108,225788,226803,228863){return 2*$n&256|$n&17|$p[$_>>4&3]<<8|$p[$_/4&3]<<4|$p[$_&3]if($_>>12&$n/2)==($_>>6&63);}}

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I know 195 bytes is far too much for this contest, but I had no idea how to further compress the Perl code. Suggestions?

Explanation of the code

In a more readable version, the code intention should become apparent:

sub dpd {

  my $n = shift;

  my $v=2*($n&128)|$n&17;

  my @p=((map{($n>>$_&3)*2}(8,5,1)),8);

  for (16390,28935,29005,227791,29108,225788,226803,228863) {

    return $v |$p[$_>>4&3]<<8|$p[$_>>2&3]<<4|$p[$_&3]

      if(($_>>12&$n/2)==($_>>6&63));

  }

}

In the rules for DPD encoding, each line is encoded into a 18 bit value, segmentation into (6,6,(2,2,2)) bits.

• The first 6 bits are an appropriate bit mask for the bits 1 (=h) to 6 (=d) of the input (bit 4 = f is redundant, but it simplifies the evaluation code to have it included).


• The next 6 bits are the value bits for this bit mask. The values are checked on all places where the bit mask has a 1 value.


• The following 3*2 bits contain the indices for the array @p for the 3-bit sequences which are to spliced into bits 11-9, 7-5 and 3-1 of the result.


• The array @p is constructed from bits 9-8, 6-5, 3-2 of the input, and the number 8 as fourth member


• The bits at position 7,4 and 0 of the input are transferred directly into bits 8,4 and 0 of the result.

For example, the first number in the list, 16390, which is 100000000000110 as a bit field, carries the following information:

000100 : bit mask says: only consider bit 3 of the input

000000 : bit values say: bit 3 should be 0

00     : use '0ab' as higher bits of first digit

01     : use '0de' as higher bits of second digit

10     : use '0gh' as higher bits of third digit

05AB1E, 104 103 bytes

•3γã•S£©4èUXтÌ‹X111QVY®2èDˆTQ*~i0®нëт}®1èY¯`*i0®нëY_Xт>Ê*i0¯`ëт]®3èY¯`_*X110Q~i0®нëXт›iYiтë0¯`ëX]®θJ4ôC

Definitely not the right language for this kind of challenge, but ah well..

Input as string, output as list of three digits.

Try it online or verify all test cases.

Explanation:

We have the following eight scenarios to consider:

     1st 2nd 3rd 4th 5th 6th                          1st digit    2nd digit    3rd digit

1.   ab  c   de  f   0gh i   →   0abc 0def 0ghi   →   '0'+1st 2nd  '0'+3rd 4th  5th     6th

2.   ab  c   de  f   100 i   →   0abc 0def 100i   →   '0'+1st 2nd  '0'+3rd 4th  5th     6th

3.   ab  c   gh  f   101 i   →   0abc 100f 0ghi   →   '0'+1st 2nd  '100'   4th  '0'+3rd 6th

4.   gh  c   de  f   110 i   →   100c 0def 0ghi   →   '100'   2nd  '0'+3rd 4th  '0'+1st 6th

5.   gh  c   00  f   111 i   →   100c 100f 0ghi   →   '100'   2nd  '100'   4th  '0'+1st 6th

6.   de  c   01  f   111 i   →   100c 0def 100i   →   '100'   2nd  '0'+1st 4th  '100'   6th

7.   ab  c   10  f   111 i   →   0abc 100f 100i   →   '0'+1st 2nd  '100'   4th  '100'   6th

8.   xx  c   11  f   111 i   →   100c 100f 100i   →   '100'   2nd  '100'   4th  '100'   6th

I first split the (implicit) input into chunks of size [2,1,2,1,3,1] and store that list in the register:

•3γã•     # Push compressed integer 212131

     S    # Convert it to a list of digits

      £   # Split the (implicit) input in chunks of that size

       ©  # Store it in the register (without popping)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •3γã• is 212131

Now we're first going to built the 0s and 1s for the first digit of the output. Scenarios 1,2,3,7 use '0'+1st+2nd; and scenarios 4,5,6,8 use '100'+2nd:

4è                    # Take the 5th item of the list

  U                   # Pop and store it in variable `X`

XтÌ‹                  #  Check if `X` is below 102

                  ~   # OR

   X111Q              #  `X` is equal to 111

        VY            #  And store that result in variable `Y`

                 *    #  and

          ®2è         #  Get the 3rd item from the list of the register

             Dˆ       #  Push it to the global array

               TQ     #  And check if it's equal to 10

i                     # If the combined check above is truthy (exactly 1):

 0                    #  Push 0 to the stack

 ®н                   #  Push the 1st item of the list to the stack

ë                     # Else:

 т                    #  Push 100 to the stack

}                     # Close the if-else

®1è                   # And push the 2nd item of the list to the stack

Then we're going to built the 0s and 1s for the second digit of the output. Scenarios 1,2,4 use '0'+3rd+4th; scenarios 3,5,7,8 use '100'+4th; and scenario 6 uses '0'+1st+4th:

Y                # Push `Y` (check if `X` equals 111)

   *             # and

 ¯`              # Push the item from the global array (3rd item of the list)

i                # If both checks above are truthy (exactly 1):

 0               #  Push 0 to the stack

 ®н              #  Push the 1st item of the list to the stack

ë                # Else:

 Y_              #  Push inverted `Y` (check if `X` does NOT equal 111)

       *         #  and

   Xт>Ê          #  Check if `X` (5th item of the list) does NOT equal 101

 i               #  If both checks above are truthy (exactly 1):

  0              #   Push 0 to the stack

  ¯`             #   Push the item from the global array (3rd item of the list)

 ë               #  Else:

  т              #   Push 100 to the stack

]                # Close both if-else cases

®3è              # And push the 4th item of the list to the stack

Then we're going to built the 0s and 1s for the third digit of the output. Scenarios 1,2 use 5th+6th; scenario 3 uses '0'+3rd+6th; scenarios 4,5 use '0'+1st+6th; and scenarios 6,7,8 use '100'+6th:

Y           #  Push `Y` (check if `X` equals 111)

    *       #  and

 ¯`_        #  Check if the item from the global array (3rd item of the list) is exactly 0

         ~  # OR

    X110Q   #  Check if `X` (5th item of the list) equals 110

i           # If the combined check above is truthy (exactly 1):

 0          #  Push 0 to the stack

 ®н         #  Push the 1st item of the list to the stack

ë           # Else:

 Xт›i       #  If `X` (5th item of the list) is larger than 100 (so 101/110/111):

     Yi     #   If `Y` (if `X` equals 111):

       т    #    Push 100 to the stack

      ë     #   Else:

       0    #    Push 0 to the stack

       ¯`   #    Push the item from the global array (3rd item of the list)

    ë       #  Else:

     X      #   Push `X` (5th item of the list) to the stack

]           # Close all if-else cases

®θ          # And push the last (6th) item of the list to the stack

Now we have all 0s and 1s on the stack, so we can convert it to the three output digits:

J     # Join the entire stack together

 4ô   # Split it into parts of size 4

   C  # Convert each part from binary to an integer (and output implicitly)

Jelly, 51 bytes

“¬Y-‘D¤ịȦƤS4x;“€-Y‘D¤ị$Ḅ;0,-4,5s3ḣ2SAœ?ɗ@/Ḥ+“¤©½‘ị$

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