Since the bases and operators are finite, this puzzle can be solved by a

brute force method.  According to the question, the basis is 4.  With unitary

operations, i.e., n!, and SQR(n), we now have v1= < 4, 24, 2 > as our basis.

By using binary operations, i.e., +, -, *, /,and ^ we can generate 

the following matrix:


 +      4       24      2

 4      8       28      6     

24  +/-20       48      26

 2  +/- 2    +/-22      4

			   0

 Similarly,

 *      4       24      2

 4     16       96      8 

24      6      576     48 

 2      2       12      4

			   1

 And so on...                               

 

 So by using two 4's, we have v2=<-22, -20, -2, 0, 1/12, 1/6, 1/2, 1

 2 4 6 8 12 16 20 22 26 24 26 28 (44) 48 96 256, ... 24^24> as our basis.

 (After sorting, and ignoring SQR(v2) and (v2)! for simplicity.)

 By using v1 and v2, we can proceed to generate v3 with the same 

 method. v3 should consist all the possible numbers that can generate

 from three 4's with the operators mentioned.  We then can generate all

 the possible number that can form from four 4's and the operators by

 v4=SORT(operator(v3,v1)/operator(v2,v2)). (/=logical AND)

 (ignoring (v4)!, SQR(v4), ((v4)!)!, SQR(SQR(v4)), ... etc)

 To save time, I found SORT(+(v2,v2)/-(v2,v2)/*(v2,v2)) will generate

 all the number between 0-> 100 except:

 11 15 19 23 31 33 35 37 39 41 43 51 53 55 57 58

 59 61 62 63 65 67 69 71 73 75 77 78 79 81 82 83

 84 85 86 87 89 90 91 93 95 and 99. (That's more than 50 generated.)

 62 and 90 belong to operator(v3,v1):

 4*4*4-2=62 and (4!/4)/4/SQR(4)=90 and other numbers in the 'miss-list' 

 is also belong to operator(v3,v1).   Those numbers that do not appear

 in v4 and unitary operations on v4 can NEVER be generate by four 4's

 and the operators.  Mr. Kelly don't think anyone is awesome. ;)