using previous/linked value in equation - z3

Question

It is possible to resolve this kind of equation :
const = [0x50, 0xe8, 0bcb, 0x9f, 0xa1]
data = IntVector('data', len(const))
for i in range(0, len(const)-1):
s.add(data[i] >= 32, data[i] <= 126)
s.add(data[i+1] >= 32, data[i+1] <= 126)
s.add(data[i] + data[i+1] == const[i]
or I misuse Z3 library ?

Answer 1

Yes. After fixing the obvious syntax errors, my Z3 reports it as unsatisfiable.

Answer 2

Supposing that 0bcb means something like 0xcb, then the list of constants is [80, 232, 203, 159, 161].
Then you ask for 5 variables, lets call them d0, d1, d2, d3, d4, which are each between 32 and 126. And where d0 + d1 == 80 (so d1 <= 48) and d1 + d2 == 232 (so d1 >= 106). This clearly contradicts, in which Z3 agrees. (Note that your constraints don't use the last element of const.)
Here is a slightly more pythonic version of the code which includes each of the 5 constants (with the second constant lowered to obtain a solvable system of constraints):
from Z3 import IntVector, Solver, sat
const = [0x50, 0xa9, 0xcb, 0x9f, 0xa1]
s = Solver()
data = IntVector('data', len(const)+1)
for d in data:
s.add(d >= 32, d <= 126)
for d0, d1, c0 in zip(data, data[1:], const):
s.add(d0 + d1 == c0)
result = s.check()
if result == sat:
print("Here is a solution: ")
m = s.model()
values = [m[d].as_long() for d in data]
print(values, " sums:", [hex(v0 + v1) for v0, v1 in zip(values, values[1:])])
elif result == unsat:
print("There is no solution")
else:
print("Z3 could not solve the constraints")
Output:
Here is a solution:
[37, 43, 126, 77, 82, 79] sums: ['0x50', '0xa9', '0xcb', '0x9f', '0xa1']