Why are there two different truth tables for BCD to Gray?

[Differentiate1]

Having a bit of trouble understanding why there are two different truth tables for BCD to Gray.

Using ABCD as the BCD and WXYZ as the Gray,
The way I learned it is:
1) MSB of BCD (A) = MSB of Gray (W)
2) A ⊕ B = X
3) B ⊕ C = Y
4) C ⊕ D = Z
5) Since BCD only ranges from 0 - 9, 10 - 15 become "don't cares"

Following the rules, chart prints:
http://i.imgur.com/IPsPoAS.png

However, while browsing through examples of converting BCD to Gray, I found these:
Example Problem
Example Problem 2

The rules I posted don't work in the Example Problems.

Are there some sort of exceptions that may alter the Gray Code?

 

[Hesch]

There are at least 3 different BCD-codes:

Maybe that's the cause to confusion?

But why not choose the one you like, practise some Karnough mapping and make your own truth table ?

Observe: BCD₄₂₂₁(5) is wrong: It must be 1001 or 0111

 

[Differentiate1]

Thanks for the reply,

The problem I'm facing requires me to,
"implement a BCD to Gray Code converter using a 4x16x4 PAL where each output collects 4 different product terms."

I completed the problem using the 8421 BCD code to Gray, but the solutions solved it the way shown here: Example Problem
Would using the 8421 BCD to Gray functions where
W = A
X = A'B
Y = A'BC' + B'C
Z = A'BC'D + B'C'D + AD + BCD + B'CD'

be considered correct in this problem still? I'm skeptical regarding my answer since the question requested "output collects 4 different product terms." I'm assuming that part just means the OR gates in the PAL needs to collect 4 connections from the AND gates.

 

[Hesch]

The rules here:

1) MSB of BCD (A) = MSB of Gray (W)
2) A ⊕ B = X
3) B ⊕ C = Y
4) C ⊕ D = Z

matches the BCD and gray-codes here:

In your examples in #1, there are other gray-codes, and of course your rules will not match these.

For example, how have you found: Z = A'BC'D + B'C'D + AD + BCD + B'CD' in #3 ? That's not right.

 

[Differentiate1]

Yeah that's not right.
Re-did the K-Map and found Z = A'C'D + B'C'D + A'CD'

But how would you solve the problem in the example using their BCD-Gray Code instead of the 8421 choice?
Certainly the rules will not work in that.

 

[Hesch]

Re-did the K-Map and found Z = A'C'D + B'C'D + A'CD'

I cannot find out what you are doing. Which BCD-code and which gray-code ar you using?

As for the BCD- and gray-codes shown in #4, the result is:

Z = C ⊕ D = ( C * !D ) + ( !C * D )

( ! means "not" )