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index.jl
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index.jl
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### A Pluto.jl notebook ###
# v0.19.19
using Markdown
using InteractiveUtils
# This Pluto notebook uses @bind for interactivity. When running this notebook outside of Pluto, the following 'mock version' of @bind gives bound variables a default value (instead of an error).
macro bind(def, element)
quote
local iv = try Base.loaded_modules[Base.PkgId(Base.UUID("6e696c72-6542-2067-7265-42206c756150"), "AbstractPlutoDingetjes")].Bonds.initial_value catch; b -> missing; end
local el = $(esc(element))
global $(esc(def)) = Core.applicable(Base.get, el) ? Base.get(el) : iv(el)
el
end
end
# ╔═╡ 0eaaa803-9bd3-4efe-9766-1f7bc7993854
begin
using Plots
using Latexify, LaTeXStrings
using PlutoUI
using Markdown
using Test
using Distributions: Bernoulli
using Polynomials
end
# ╔═╡ 1446d3db-7da7-4e3d-b7cd-390b69957bf3
md"""
*This document is an interactive Julia notebook (using `Pluto.jl`), or a document exported from it. The original notebook is completely self-contained. Expand cells in the notebook to see the code (in particular, LaTeX equation are often pretty-printed Julia code). All plots can be regenerated with any of the parameters changed and all calculations are included in the notebook file.*
"""
# ╔═╡ 8560ed6c-55c2-483c-8cfb-9d9508aeb0c4
md"""
### Assymptotically optimal limit
In the limit of many detection events to be compressed together, the optimal compression rate is given by [Shannon's source coding theorem](https://en.wikipedia.org/wiki/Shannon%27s_source_coding_theorem). Note that in this limit, a realization of the Poisson process is in one-to-one correspondence to a sequence of waiting times. Thus, both have the same optimal compression limit.
The number of bits (on average) required to compress a single difference between time frame indices is given by the Shannon entropy of the [Geometric distribution](https://en.wikipedia.org/wiki/Geometric_distribution):
"""
# ╔═╡ d5f7ee81-2d0a-42f5-89b3-81bfd191a7b7
md"""
Where ``h_2`` is the Binary Shannon entropy function
"""
# ╔═╡ 5b24a2fc-a4a2-4fba-835a-99a3ddc5da1d
@latexrun h₂(p) = -p * log2(p) - (1 - p) * log2(1 - p)
# ╔═╡ 71962a4a-c5ad-4b25-89a4-964dde13f638
@latexrun H_ΔT(p_det) = h₂(p_det) / p_det
# ╔═╡ 6f8a0039-953e-4ddd-8eef-5a35e1a0dec9
md"""
### Realistic compression (Golomb-Rice)
The Shannon limit is not achievable in practice. However, given our large numbers of detections and small probabilities of detection in a given time frame, we can come very close.
The [Golomb-Rice code](https://en.wikipedia.org/wiki/Golomb_coding) is a near optimal data compression algorithm for the Geometric distribution that is straightforward to implement. It encodes each frame index separately into a sequence of bits, which makes it easier to use and analyze. (By contrast, most efficient compression algorithms, compress a large file as a whole).
"""
# ╔═╡ dfdad6a0-a928-4dd7-bc5a-51233e236e07
md"""
The following plot shows that, under conditions relevant to our application, the Golomb-Rice method comes very close to the optimum as given by the Shannon entropy if the parameter (exponent b) is chosen carefully.
"""
# ╔═╡ 6afcf2d8-f617-4d43-b918-75a282538509
md"""
## Estimating ``\ p_{det}`` (static)
"""
# ╔═╡ 643b0ebb-008e-45f5-afa8-6672d6900899
md"""
To estimate the probability of getting a detection in a given time frame, a number of parameters are important, several of which we do not know yet.
To start, these parameters are fairly certain. They are taken from [[Rusca et al., 2018](https://doi.org/10.48550/arXiv.1801.03443)] and the supplementary material (for the one-decoy-protocol; we do not consider different protocols in this document).
"""
# ╔═╡ 4d688d01-bc55-4859-a7bd-67dc796de559
"Signal state average photon number"
μ₁ = 0.5
# ╔═╡ 31be7bc7-7dfd-4014-8b74-748d6f1ff7c6
"Decoy state average photon number"
μ₂ = 0.15
# ╔═╡ bc7693c2-d503-4b8d-b21f-bf8a2cee289d
"Probability to send a signal state"
Pμ₁ = 0.7
# ╔═╡ b89c31fc-9786-47a2-900c-884eb282d05d
md"""
##### The following parameters we control precisely.
"""
# ╔═╡ f206f0fc-054e-4898-b309-3360fd7833d2
"Repetition frequency of QKD signals (pulses):"
f_rep = 100e6 # Hz
# ╔═╡ ec9b70e5-f647-4a89-be02-1035c54e59d4
md"##### The following paremeters should be measured or simulated"
# ╔═╡ a9f40c72-86de-4a3a-87e7-7591091126ef
begin
η_trans = 1e-4 # optical channel loss
η_receiver = 0.7 # transmissivity of receiver
η_spectral = 0.7 # loss due to spectral filter
η_detector = 0.6 # quantum efficiency of detector
η_safety = 1. # safety margin.
@latexrun η = η_trans * η_receiver * η_spectral * η_detector * η_safety
end
# ╔═╡ 5e91af77-79d4-4afd-b200-7223d3ef64db
begin
# note: use this to make slider:
#md""" $(@bind R_ni Slider(0:300, default = 100, show_value=true)) Hz"""
R_ni = 500 # stray light (counts per second), spectrally filtered
f_dark = 80 # dark counts from one detector (per second)
t_dark = 1e-9 # time filter window (seconds)
@latexrun PDC = (
(R_ni * η_receiver * η_spectral * η_detector + 4 * f_dark) * t_dark
)
end
# ╔═╡ 2a893d6c-1e60-407a-a16e-1f8c16089681
md"""
By adding up the equations (B2) in [Rusca et al., 2018](https://doi.org/10.48550/arXiv.1801.03443) supplementary material and ignoring the dark time correction (negligible for channels with high loss), we get an equation for the probability to get a detection in a given time frame. The dark count term ``PDC`` could also be neglected and is included for completeness.
The 'one minus exponential' factor arises from the Poisson distribution, assuming a detection happens as soon as there is at least one photon in the pulse. The detector efficiency is taken into account inside the total channel transmission η.
"""
# ╔═╡ 3df7671a-5d7a-4d81-a8f6-a9227046bfd9
@latexrun p_det(;η, Pμ₁, μ₁, μ₂, PDC) = (
PDC + Pμ₁*(1 - exp(-μ₁ * η)) + (1 - Pμ₁)*(1 - exp(-μ₂ * η))
)
# ╔═╡ 15f5daf5-44ba-41ce-a963-e23e20d7af36
md"""
## Estimating ``\ p_{det}`` (time dependent loss)
"""
# ╔═╡ b46a64ec-532b-4ae3-8c2a-d3f0bbce4625
"""
Example time dependent channel transmission. Should model a satellite scenario reasonably well, which has a best-case transmission of -31.8dB.
"""
function eta_trans_of_t(t)
p = Polynomial([-31.8, -1.5e-17, -1.01e-3, 1.8e-20, 5.96e-8, 0., -2.4e-12, 0., 4.9e-17, 0., -3.9e-22])
half_time = 205. # acceptable interval: -205 ... 205 seconds.
if abs(t) < half_time
return exp10(1/10 * p(t))
else
return NaN
end
end
# ╔═╡ 462a1f28-a08a-4a5f-b8a0-a0f35042b3da
Markdown.parse("""
We multiply it with the factors from above to obtain total transmission as a function of time
""")
# ╔═╡ 2c3dc8f1-8fb1-4ccd-a1c7-ccadb5068e65
eta_of_t(t) = (
eta_trans_of_t(t) * η_receiver * η_spectral * η_detector * η_safety
)
# ╔═╡ 289a7553-733d-4a63-83a4-086722e9832e
Markdown.parse("""
And detection probability per frame as a function of time
""")
# ╔═╡ eea0f925-705d-4dba-984c-e55a82755c80
p_det_t(t) = p_det(; η=eta_of_t(t), Pμ₁, μ₁, μ₂, PDC)
# ╔═╡ 0567991b-cf8e-41c0-8d80-f0a2123ab885
let
link_time_interval = -250.:250.
plt = plot(eta_trans_of_t, link_time_interval,
yaxis=:log,
yticks=exp10.(-7.:0.2:-2),
xlabel="t (seconds)",
label=L"channel transmission $η_{trans}$",
legend=:topleft,
ylims=(-Inf, 1.5e-3), # to make legend fit
)
plot!(plt,
eta_of_t,
link_time_interval,
label="total transmission η",
)
plot!(plt,
p_det_t,
link_time_interval,
label=latexify(Symbol(p_det))
)
end
# ╔═╡ 68bb0e85-cf67-4d34-a459-d85d0b9d648a
md"""We can use the detection probability to estimate the total number of detection events by integrating over the curve:
"""
# ╔═╡ b8055769-0cb1-4b48-853e-eb85e40369fe
let
δt = 1e-2 # seconds
t_range_integrate = -200:δt:200
curve_avg_detections_per_time_frame = p_det_t.(t_range_integrate)
integral_total_detections = sum(δt .* curve_avg_detections_per_time_frame * f_rep)
Markdown.parse("""
### Total expected detections: $(latexify(integral_total_detections, fmt=FancyNumberFormatter(2),))
""")
end
# ╔═╡ 4a8f68d7-19cd-4121-8562-91bc8db271b3
md"""
Using this detection probability, we estimate the total number of bits needed to perform time sifting on data collected during the entire time.
To compute the total number of bits needed we treat the detection probability and the 'avg. bits per timestamp' as smooth functions of time and integrate their product.
"""
# ╔═╡ 29f93574-b9bf-457c-bae1-42bf1ab29a50
md"""
.
.
.
# Helper functions"""
# ╔═╡ 9129c148-91d6-46a2-b6bc-88a3b26b5b02
TableOfContents() # generates an interactive table of contents for the notebook.
# ╔═╡ 9bbf3207-ea72-434f-8406-84525d0eddf1
"""
Golomb-Rice encoding of a vector of time frame indices (is converted do differences)
"""
function encode_golomb_rice(input::Vector{Int}, b::Int)
result = Bool[] # vector of bits
current = 0
for i in eachindex(input)
next = input[i]
diff = next - current
current = next
factor = div(diff, 2^b) # floor division
remainder = diff % (2^b)
# unary encoding of factor
for f = factor:-1:1
push!(result, true)
end
push!(result, false) # marks end of factor encoding
# binary encoding of remainder
for bit_idx in 0:(b-1)
push!(result, (remainder >> bit_idx) & 1)
end
end
return result
end
# ╔═╡ 84b60a6f-48e7-4319-86b4-af98fbfdb496
"""
Decodes differences between timestamps that correspond to input to `encode_golomb_rice`
"""
function decode_golomb_rice(input::Vector{Bool}, b::Int)
result = Int[]
current_factor = 0
current_remainder = 0
i = 1
while i <= length(input)
if input[i]
current_factor += 1
i += 1
continue
else
i += 1 # skip the zero-bit marking end of factor.
for j in 0:(b-1)
current_remainder += input[i + j] * 2^j
end
i += (b - 1)
end
push!(result, 2^b * current_factor + current_remainder)
current_factor = 0
current_remainder = 0
i += 1
end
return result
end
# ╔═╡ 784ced0b-fd1a-42e7-9df2-f7b4a6615819
let # unit test to test the compression functions
v = [1, 25, 56, 123]
b = 4;
encoded = encode_golomb_rice(v, b)
# encodes 0, 1, 24, 31, 67 (differences of v)
expected = Bool[
0, 1, 0, 0, 0, # 1
1, 0, 0, 0, 0, 1, # 24
1, 0, 1, 1, 1, 1, # 31
1, 1, 1, 1, 0, 1, 1, 0, 0 # 67
]
@test encoded == expected
end
# ╔═╡ 2eac341f-cd24-4ae9-8582-c87019589063
let # unit test to test the compression functions
timestamps = [1, 25, 56, 123]
encoded = encode_golomb_rice(timestamps, 4)
decoded = decode_golomb_rice(encoded, 4)
expected_diffs = [1, 24, 31, 67]
@test expected_diffs == [timestamps[1], diff(timestamps)...]
@test decoded == expected_diffs
end
# ╔═╡ c87460cb-792e-41d5-9fa7-9dc54ba44710
"""
Computes compression efficiency of Golomb-Rice code
(Assume p_det < 0.5)
`n_samples` is actually the number of possible frames.
E.g., for a frame duration of 10ns (1/100MHz) and a link duration of 7 minutes,
n_samples = 4.2e10. Such large sample sizes become computationally demanding.
The parameter b specifies the exponent that is used for the Golomb-Rice code. A good (but not provably optimal) choice is given by default.
"""
function golombrice_avg_bits_per_timestamp(
p_det;
b=round(Int, -log2(-log2(1-p_det))), # known to be a good value
n_samples=1_000_000,
)
dist = Bernoulli(p_det)
frame_indices = findall(rand(dist, n_samples))
compressed = encode_golomb_rice(frame_indices, b)
avg_bits_per_detection = length(compressed) / length(frame_indices)
return avg_bits_per_detection
end
# ╔═╡ 9c290708-e0ee-42cf-bdc8-ac39b9120a1a
let
# warning: this cell both plots and reruns the monte-carlo simulation every time it is executed.
# Beware the computation time for large sample sizes.
p_range = exp10.(-6:0.2:-3)
n_samples = Int(1e8)
plt = plot(H_ΔT, p_range,
title="""Time stamp compression
($(1. *n_samples) frames simulated)
""",
xaxis=:log,
xflip=true,
xlabel=L"p_{det}",
ylabel="avg. bits per time frame index",
xticks=exp10.(-6:1:-3),
yticks=10:2:26,
label=L"\mathrm{Shannon\ entropy } \ H_{ΔT} \ (optimal)",
legend=:bottomright
)
plot!(plt,
p -> golombrice_avg_bits_per_timestamp(p; n_samples=n_samples),
p_range,
label="Golomb-Rice (b chosen as shown)",
)
__b = 17
plot!(plt,
p -> golombrice_avg_bits_per_timestamp(p;
n_samples=n_samples, b=__b),
p_range,
label="Golomb-Rice (b=$__b)",
)
plot!(plt,
p -> round(Int, -log2(-log2(1-p))),
exp10.(-6:0.01:-3),
label=L"b = \mathrm{round}(-\log_2(-\log_2(1-p_{det})))",
)
plt
end
# ╔═╡ 89e67b55-10b4-4a5e-9892-4193fab77300
# curve to integrate to get total transmitted bits
begin
b_const_integrate = 14
δt = 2 # seconds
t_range_integrate = -200:δt:200
avg_bits_per_time_frame(t) = golombrice_avg_bits_per_timestamp(p_det_t(t);
n_samples=Int(1e8), b=b_const_integrate) * p_det_t(t)
curve_avg_bits_per_time_frame = avg_bits_per_time_frame.(t_range_integrate)
integral_total_bits = sum(δt .* curve_avg_bits_per_time_frame * f_rep)
plt_bits_per_timeframe = plot(
t_range_integrate, curve_avg_bits_per_time_frame,
xlabel="t (seconds)",
ylabel="avg. bits per time frame",
label="Golomb-Rice (b = $b_const_integrate)",
)
opt_avg_bits_per_time_frame(t) = p_det_t(t) * H_ΔT(p_det_t(t))
opt_curve_avg_bits_per_time_frame = opt_avg_bits_per_time_frame.(t_range_integrate
)
opt_integral_total_bits = sum(δt .* opt_curve_avg_bits_per_time_frame * f_rep)
plot!(plt_bits_per_timeframe,
t_range_integrate, opt_curve_avg_bits_per_time_frame,
label="Shannon Limit",
)
plt_bits_per_timeframe
end
# ╔═╡ 03f9ab84-59aa-4118-83e0-6ff6824a8e0e
Markdown.parse("""The integral of the realistic (Golomb-Rice) curve yields the
### Total bits transmitted for time-sifting: $(latexify(integral_total_bits, fmt=FancyNumberFormatter(3),))
for comparison: the Shannon limit gives the lower bound for total bits transmitted of $(latexify(opt_integral_total_bits, fmt=FancyNumberFormatter(3),)).
""")
# ╔═╡ 52e511ac-3025-4b9b-9380-5fa06c569f09
"""
Like @show but returns a LaTeXString instead of printing to stdout.
The expression is not evaluated!!! Don't use to define things!
"""
macro show_latex(ex)
quote
$(latexify)($(string(ex)) * " = " * $string($(esc(ex))),
env=:raw,
fmt=FancyNumberFormatter(),
)
end
end
# ╔═╡ 9b97a6b0-57c9-4b2d-8257-5a7dfb68a2e3
Markdown.parse("""
# Classical data for QKD time-sifting
This notebook describes the amount of classical data transmission required to perform **time-sifting**, which is the first step in QKD post-processing (often omitted in theoretical analyses). Time-sifting can take up to 90% of the total data transmission required for QKD post-processing.
In time-sifting, the QKD detector (Bob) informs (via classical communication) the sender (Alice) at which times quantum detection events were recorded. Assume a BB84 scenario where the sender sends pulses with a constant repetition rate (e.g. ``$(@show_latex f_rep)`` Hz). Assuming that Alice and Bob have synchronized clocks and agree on a start time, to perform time sifting, Bob only needs to inform Alice in which **time frames** (intervals of length ``1/f_{rep}``) detection events happened. This is to say, for each detection event Bob transmits via the classical link one number (**index**) indicating the time frame where it happend.
Note that for 10 nanosecond long time frames and 7 minutes of link duration, there are ``4.2\\cdot 10^{10}`` frames. If time sifting during the QKD link is not possible, the QKD sender has to store all random numbers used to prepare each of the ``4.2\\cdot 10^{10}`` quantum states until the time sifting is completed. If each state takes up one byte, this amounts to 42 Gigabytes, which can be a challenge for the sender hardware.
""")
# ╔═╡ e500d4a6-84e4-4689-9dc5-7cdb1eddb70c
begin
display_p_det = latexify(
p_det(; η, Pμ₁, μ₁, μ₂, PDC)
; fmt = FancyNumberFormatter(2)
) # these are default values defined below
Markdown.parse("""
## How many bits per detection? (Stationary case)
Assume that the optical transmission of the channel (e.g. ``$(@show_latex η)`` or ``$(round(10*log10(η); sigdigits=3))`` dB) is constant over time (e.g. for a free-space optical link between buildings). To a good approximation, there is a constant and independent probability (e.g. ``p_{det} = `` $display_p_det ) to get a detection in any given time interval. (This is an approximation, since fluctuations in the optical losses will be correlated in time even for a stationary link.)
The task of transmitting time sifting information thus becomes transmitting a number of i.i.d. samples from a Bernoulli distribution, i.e., a finite part of a [Bernoulli process](https://en.wikipedia.org/wiki/Bernoulli_process). We assume that there are many detection events and subsequently consider the process itself, rather than a finite part.
To minimize the amount of classical communication needed for time sifting, we want to compress this data as much as possible. While these indices keep increasing, the waiting times (difference between indices) of consecutive detections for a Bernoulli process follow a stationary distribution: the geometric distribution (discrete analog of the exponential distribution, just like a Bernoulli process is the discrete analog of a Poisson process). More precisely, let ``\\Delta T`` be the random variable of waiting times (time frame index differences) between consecutive detection events. Then
```math
P(\\Delta T = N) = (1 - p_{det})^{N-1} p_{det}, \\qquad N=1,2,\\dots
```
""")
end
# ╔═╡ d60402c6-6527-44c7-8145-95387b5820cf
Markdown.parse("""
Channel attentuation
See [Gruneisen et al.](https://doi.org/10.1364/OE.23.023924)
(e.g. ``$(@show_latex η)``)
""")
# ╔═╡ 77f1ddd1-aba6-452f-87b7-96944e73a71a
Markdown.parse("""
The probability to get a dark count inside one frame of length ``1/f_{rep}``
Includes contributions from sky radiance and detector dark counts.
(e.g. ``$(@show_latex PDC)``)
See [Gruneisen et al.](https://doi.org/10.1364/OE.23.023924)
""")
# ╔═╡ f5e3cf21-4347-4452-8b50-4938268fbba1
Markdown.parse("""
For the example values considered here, we obtain
```math
$(@show_latex p_det(; η, Pμ₁, μ₁, μ₂, PDC))
```
""")
# ╔═╡ 7733a685-ef5f-48bd-8aa7-44d02ac4aa14
Markdown.parse("""
Consider the following time dependent channel transmission (example for around 500km). This corresponds to the variable ``$(@show_latex η_trans)`` above.
Using this, we can calculate the time dependent 'detection probabilities in a single frame'.
""")
# ╔═╡ 00000000-0000-0000-0000-000000000001
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