Puzzle: DONALD + GERALD = ROBERT
  Each letter represents a distinct digit.  What is the mapping

Solution (not necessarily unique):
   
   c1 c2 c3 c4 c5    carries
   D  O  N  A  L  D
   G  E  R  A  L  D
   ----------------
   R  O  B  E  R  T

Steps:
 1. T is even ( D + D)

 2. D >= 5 <=> c5 == 1
    D <= 4 <=> c5 == 0

 3. L >= 5 <=> c4 == 1
    L <= 4 <=> c4 == 0

 4. A >= 5 <=> c3 == 1
    A <= 4 <=> c3 == 0

 5. c5 == 1 <=> R is odd
    c5 == 0 <=> R is even ( L + L)
    R not 0 (because R and L cannot both be 0)

 6. c4 == 1 <=> E is odd
    c4 == 0 <=> E is even ( A + A)
    E not 0 (because E and A cannot both be 0)

 7. c2 == 0  <=> E == 0
    c2 == 1  <=> E == 9 and c3 + N + R > 9

 8. c1 + D + G <= 9 (no carry)
    R <= D
    R <= G

 9. By (6) and (7), E must odd and hence 9.
 	By (7) it is either 0 or 9, but if it is 0, then it is even
	and by (6) it cannot be even and 0.
    ==> c4 = 1, E = 9, c2 = 1
        O is the last variable to be solved for - it can be anything

10. By (9): E == 9, c2 == 1
    ==> c1 = 1

11. By (9) and (8),
    ==> D + G <= 8

12. By (9): c4 == 1
    ==> L >= 5

13. By (9): c4 == 1, E == 9
    ==> A = 4 (A cannot be 9 because E is already 9)
    ==> c3 = 0

14. Three instance of R and of D.  Make assumption.
	Aside: simplified version of this puzzle gives clue "R is odd".
	14a. R is odd ==> c5 == 1, D >= 5
    Choices sorted for this additional information.
    c5 == 1 :
      T == 0 <=> D == 5
      T == 2 <=> D == 6
      T == 4 <=> D == 7	: unavailable: (13): D == 4
      T == 6 <=> D == 8
      T == 8 <=> D == 9 : unavailable:  (9): E == 9
    c5 == 0 (not possibilities with the additional clue):
      T == 2 <=> D == 1
      T == 4 <=> D == 2 : unavailable: (13): D == 4
      T == 6 <=> D == 3
      T == 8 <=> D == 4 : unavailable: (13): D == 4

    Iterate through the possbilities:
    ==> D=5, T=0, c5=1

15. By (14): D == 5
  and (10): c1 == 1
    ==> R == c1 + D + G == 1 + 5 + G
    ==> R == G + 6

16. By (15), 
	R == 6 <=> G == 0
	R == 7 <=> G == 1
	R == 8 <=> G == 2
	R == 9 <=> G == 3

17. Unavailable values:
     By (14): 0
     By (9):  9
	R == 7 <=> G == 1
	R == 8 <=> G == 2	(invalidated by additional clue that "R is odd")
    
18. c3 + N + R == B with carry c2
    By (9): c2 == 1
    By (13): c3 == 0
    	mod10(N + R) == 1
	rem10(N + R) == B

19. Assume two possible values for R from (17).
    First, assume R == 7, and hence G == 1.
    By (18), N >= 3.
	N == 3 <=> B == 0 : unavailable value (14)
	N == 4 unavailable value (13)
	N == 5 unavailable value (14)
	N == 6 <=> B == 3
	N == 7 unavailable value (19) (value assumed here for R)
	N == 8 <=> B == 5 : unavailable value (14)
	N == 9 unavailable value (9)
    ==> N = 6, B = 3

20. By (12): L >= 5,
    Unavailable values: 5, 6, 7, 9
    ==> L = 8

21. By unavailable values:
    ==> O = 2

Done.  Two assumptions (14) and (19) that led to a consistent solution.
  Alternate branches not explored.

======================================================    
Adding restrictions (sorted by variable):

A:
 13). 4

B:
 19). 3

D:
 14). 5

E:
  7). 0 or 9
  9). 9

G:
 19). 1

L:
 12). >= 5
 20). 8

N:
 19). 6

O:
 21). 2

R:
  5). not 0
 19). 7

T:
 14). 0
 
c1:
  10). 1

c2:
   9). 1

c3:
  13). 0

c4: 
   9). 1

c5:
  14). 1


======================================================    
Final assignment:
   0  1  2  3  4  5  6  7  8  9
   T  G  O  B  A  D  N  R  L  E

   A  B  D  E  G  L  N  O  R  T
   4  3  5  9  1  8  6  2  7  0