> 
> End-of-chapter exercises, p 272:
> 1	convert binary to decimal
> 2	convert decimal to binary
> 5	binary numbers that are all 1's: eg 1111, 11111111111
> 6	Adding a 0 to a binary number: 1101 -> 11010, etc
> 10	PIPPIN tasks: (a) swap, (b) compare
> 11	PIPPIN: X = X/2
> 14revised: Python terminates def, while, for, if lines with ":". 
> What would happen if we eliminated these colons from python?
> 15	Draw parse trees
> 16	describe code generation
> 
> 
> End-of-chapter exercises, p 308:
> 6	find logic table. that is, for inputs 
> 	(a,b) = (0,0), (0,1), (1,0), (1,1), find respective output x
> 8	three-input AND, three-input OR
> 9	majority circuit: 3 inputs: output is 1 <===> at least 2 inputs are 1
> 15	pipelining
> 
> =======================================================================
> 
> 1. What is the value of s after execution of the following python code:
> 
> n=1
> s=1
> while n<7:
>    s = s + n*n
>    n = n+2
 
We go through the loop for n=1,3,5; s = 1 + 1*1 + 3*3 + 5*5 = 36
 
 
> 2. What is the value of L after execution of the following python code:
> Note L is a list.
> 
> n = 0
> L = []
> while n < 5:
>     L = L + [2*n]
>     n = n + 1
 
We go through with n = 0,1,2,3,4, each time appending 2n to the end:

	L = [0,2,4,6,8]
 
 
> 3. What is printed by the following:
> 
> for i in range(1,5):
>     print range(0,i)
 
Recall that range(0,i) is simply the list [0,1,2,...,i-1].
The four lists are printed one per line:

	[0]
	[0,1]
	[0,1,2]
	[0,1,2,3]
 

> 4. What is string t after the following?
> t = ""
> s = "foobar"
> for i in range(len(s)):
>     t = t + s[i]

t contains the reverse of s: "raboof"
    
> Pippin review deleted
 
> 5. Give a parse tree for the following expression, and write PIPPIN instructions 
> to leave the value of the expression in the accumulator;
> 
>     (X+Y)/2

	LOD X
	ADD Y
	DIV #2
	
> 6  Give a parse tree for the following expression, and write PIPPIN instructions 
> to leave the value of the following expression in the accumulator. 
> Hint: Store (X+Y) in the temporary location T1.
> Unlike the previous example, this one cannot be done without temporary storage.
> 
>     12/(X+Y)	

LOD X
ADD Y
STO T1		# store in temp loc
LOD #12		# Load Acc with 12
DIV T1		# now divide 12 by what we worked out before

> 7. Give parse trees for each of the following expressions.
>     (X+Y)*(X+2)
>     X*Y + 3*Z*(W+1)
> 
> 8. Assume that the location P is initially 1, and N is initially 5.
> Give the value in P at the end.
> 
> 	LOD N
> START:	JMZ END		# jump to END if Acc == 0
> 	MUL P		# multiply Acc by P
> 	STO P
> 	LOD N		# load the original N again
> 	SUB #1
> 	STO N
> 	JMP START
> END:	HLT

Each time through the loop, N is decremented by one.
Before that, we multiply P by N. So, the final value in N is 0,
and the final value in P is 5*4*3*2*1 = 120

> 9. Give the output of the following for all inputs.
> 
> x----+------+ 
>      | NAND |--+ w
> y----+------+  |
>                +----+------+
>                     | NAND |--output
> z-------------------+------+
> 
> 
> x  y  z  |  output
> 0  0  0  |  
> 0  0  1  |  
> 0  1  0  |  
> 0  1  1  |  
> 1  0  0  |  
> 1  0  1  |  
> 1  1  0  |  
> 1  1  1  |  
> 
> The table for one NAND gate is as follows:
> 
> x  y  output
> 0  0   1
> 0  1   1
> 1  0   1
> 1  1   0


I add a column "w" representing the output of the first NAND gate.
Then we can use the NAND table to fill in the w column.
To get the final output, use the NAND table and input values w and z.
NAND(w,z)=1 whenever EITHER w=0 or z=0.

 x  y  w  z  |  output
 0  0  1  0  |  1
 0  0  1  1  |  0
 0  1  1  0  |  1
 0  1  1  1  |  0
 1  0  1  0  |  1
 1  0  1  1  |  0
 1  1  0  0  |  1
 1  1  0  1  |  1