Modulo arithmetic solving

That took more fiddling than I'd hoped to get all the modulo operators in the right place, but exactly the same system works for modulo equations:

from pylogram import constrain, default_vars as vars

constrain(   vars.a + 5*vars.b == 22 , mod=17 )
constrain( 2*vars.a +   vars.b == -5 , mod=17 )

print( " >> a , b =", vars.a, ",", vars.b )

#  >> a , b = 8, 13

assert (  8 + 5*13)%17 , (22)%17
assert (2*8 +   13)%17 , (-6)%17

Pylogram arrays

I've also added arrays, so you can do:

# Warning, untested
fib = Arr(10,Var)
for a,b,c in fib.adj_objs(3):
 c = a + b
fib[0] = fib[1] = 1
assert fib == (1, 1, 2, 3, 5, 8, 13, 21, 34, 55)

graph = Arr(10,Point)
for x,(pt,f) in enumerate(zip(graph,fib)):
 pt.x = 10 * x
 pt.y = 150 - 2 * f

But I really wanted to give an alternative to the "for ... in" syntax, something like:

 fib_pts = Arr(10,Point)
 fib.each() = fib.prev() + fib.prev(2)

or maybe:

 fib[pyl.idx] = fib[pyl.idx-1] + fib[pyl.idx-2]

Bottle tops

For people in Cambridgeshire, we now can recycle plastic bottle tops. Apparently.

The fact that they changed the advice hopefully suggests they have some reason for it better that guessing.