2 bit binary counter karnaugh diagram

[hadoque]

Hi
I'm supposed to design a 2 bit binary counter with an extra input, x, to decide whether it should count up or down (x=1 counts up). The design uses two D-flip-flops.
To help I got a couple of truth tables to fill in. The first is for both flip-flops, D1 and D2. Then this one is split to two separate tables, for each flip-flop.
q1 is the output of the D1 flip-flop and so forth.
This is how I filled the tables:

q1q2|x=0|x=1
____|___|___
00  | 11|01
____|___|___
01  | 00|10
____|___|___
11  | 10|00
____|___|___
10  | 01|11
____|___|___

     D1 D2

q1q2|x=0|x=1
____|___|___
00  |  1|  1
____|___|___
01  |  0| 0
____|___|___
11  |  0| 0
____|___|___
10  |  1| 1
____|___|___

     D1

q1q2|x=0|x=1
____|___|___
00  | 1 |0
____|___|___
01  | 0 |1
____|___|___
11  | 1 |0
____|___|___
10  | 0 |1
____|___|___

     D2

When I collect the ones to make the boolean functions, I get:

D1 = q1' + q2'
D2 = x'q1'q2' + xq1'q2 + x'q1q2 + xq1q2'

But D2 equals 1, so I must have made an error somewhere. Please help.

[Thav]

How does D2 equal 1? It looks correct to me. Your equation for D1 is off, however.

[hadoque]

Originally Posted by Thav

How does D2 equal 1?

Because I suck at Boolean algebra..
The correct answer should be:
D1 = q2'
D2 = (x(XOR)q1(XOR)q2)'
but I also separated the tables wrong, so D1=D2, and vice versa.