def instances(GD, N, MD, p):
"""
The formula from "Optimizing the oracle under a depth limit". Assuming single-target, t = 1.
:params GD: Grover's oracle depth
:params N: keyspace size
:params MD: MAXDEPTH
:params p: target success probability
Assuming p = 1
In depth MD can fit j = floor(MD/GD) iterations.
These give probability 1 for a search space of size M.
p(j) = sin((2j+1)theta)**2
1 = sin((2j+1)theta)**2
1 = sin((2j+1)theta)
(2j+1)theta = pi/2
theta = pi/(2(2j+1)) = sqrt(t/M) = 1/sqrt(M).
sqrt(M) = 2(2j+1)/pi
M = (2(2j+1)/pi)**2
Hence need S = ceil(N/M) machines.
S = ceil(N/(2(2*floor(MD/GD)+1)/pi)**2)
Else
Could either lower each individual computer's success prob, since the target is inside only one computer's state.
Then given a requested p, we have
p = sin((2j+1)theta)**2
arcsine(sqrt(p)) = (2j+1)theta = (2j+1)/sqrt(M)
M = (2j+1)**2/arcsine(sqrt(p))**2
S = ceil(N*(arcsine(sqrt(p))/(2j+1))**2)
Or could just run full depth but have less machines.
For a target p, one would choose to have ceil(p*S) machines, where S is chosen as in the p = 1 case.
Then look at which of both strategies gives lower cost.
"""
# compute the p=1 case first
S1 = ceil(N / (2 * (2 * floor(MD / GD) + 1) / pi)**2)
# An alternative reasoning giving the same result for p == 1 (up to a very small difference):
# Inner parallelisation gives sqrt(S) speedup without loosing success prob.
# Set ceil(sqrt(N) * pi/4) * GD/sqrt(S) = MAXDEPTH
# S1 = float(ceil(((pi*sqrt(N)/4) * GD / MD)**2))
if p == 1:
return S1
else:
Sp = ceil(N * (arcsin(sqrt(R(p))) / (2 * floor(MD / GD) + 1))**2)
if ceil(p * S1) == Sp:
print(
"NOTE: for (GD, log2(N), log2(MD), p) == (%d, %.2f, %.2f, %.2f) naive reduction of parallel machines is not worse!"
% (GD, log(N, 2).n(), log(MD, 2).n(), p))
elif ceil(p * S1) < Sp:
print(
"NOTE: for (GD, log2(N), log2(MD), p) == (%d, %.2f, %.2f, %.2f) naive reduction of parallel machines is better!"
% (GD, log(N, 2).n(), log(MD, 2).n(), p))
res = min(Sp, ceil(p * S1))
return res
def iterations(p, N, t):
"""
Grover succeeds with probability p(j) = sin^2((2j+1)theta).
This is the inverse function for obtaining the number of iterations j from theta and p(j).
:params p: Grover's success prob
:params N: Grover's search space size
:params t: targets. Right now, assumed to be 1.
"""
return (R(arcsin(sqrt(p)))/theta(N,t) - 1)/2
def FGRooDhFkch():
pspict, fig = SinglePicture("FGRooDhFkch")
pspict.dilatation_X(3)
pspict.dilatation_Y(1)
x = var('x')
f = phyFunction(arcsin(x)).graph(-1, 1)
pspict.DrawGraphs(f)
pspict.axes.single_axeY.axes_unit = AxesUnit(pi / 2, '')
pspict.grid.Dy = pi / 2
pspict.DrawDefaultGrid()
pspict.DrawDefaultAxes()
#fig.no_figure()
fig.conclude()
fig.write_the_file()
def theta(N, t):
"""
Given search space of size N with t solutions, returns angle theta that each Grover iteration advances.
"""
return R(arcsin(sqrt(R(t)/N)))