Copyright By PowCoder代写 加微信 powcoder
x = Int16[10, 5, 3, 11]
# we can access memory directly
x_1 = Base.unsafe_load(pointer(x))
# memory is laid out in bytes (8 bits)
x_h = Base.unsafe_load(pointer(x) + 1)
# Advanced: bits combine half of bits of 4 and bits of 5
# but not as expected since bits are stored “backward” (little-Endian)
sz = sizeof(eltype(x))
x_3 = Base.unsafe_load(pointer(x) + 2sz)
Base.unsafe_load(pointer(x) + 4sz)
# WARNING: May crash. High-performance languages are NOT safe
4-element Vector{Int16}:
x = [1.3, 1.2, 1.5]
3-element Vector{Float64}:
A = [1 2 3; 4 5 6; 7 8 9]
3×3 Matrix{Int64}:
sz = sizeof(Int)
Base.unsafe_load(pointer(A) + 3sz)
9-element Vector{Int64}:
function mul_rows(A, x)
m,n = size(A)
ret = zeros(m)
for k = 1:m
for j = 1:n
ret[k] += A[k,j] * x[j]
function mul(A, x)
m,n = size(A)
ret = zeros(m)
for j = 1:n
for k = 1:m
ret[k] += A[k,j] * x[j]
mul (generic function with 1 method)
A = randn(n,n)
x = randn(n)
mul_rows(A,x) ≈ A*x
] add BenchmarkTools
Resolving package versions…
No Changes to `~/Documents/Lectures/MATH50003NumericalAnalysis/Project.toml`
No Changes to `~/Documents/Lectures/MATH50003NumericalAnalysis/Manifest.toml`
using BenchmarkTools
@btime mul_rows(A, x)
@btime mul(A, x)
@btime A*x
18.228 μs (1 allocation: 896 bytes)
6.637 μs (1 allocation: 896 bytes)
1.883 μs (1 allocation: 896 bytes)
100-element Vector{Float64}:
-11.091082683943975
15.375023837352812
1.7863169866347457
3.7581125078770223
-11.382201133296679
-12.391297354798967
-9.666072728450594
17.852819140415107
-6.526323923317621
2.3809426807502265
-26.6358042787717
0.3943165121789224
-2.6864548720750006
-8.39162579412428
12.791452114985063
-7.941425454384663
12.516025757522588
-14.57435214893141
-7.375001104828355
4.095507806518102
4.643869749457856
-7.70968610579241
9.519075191195832
10.605253221605262
-7.2323433646541035
We can estimate algorithm complexity by considering n and 2n at same time. E.g. if algorithm takes ~Cn^p operations on n then it will take ~2^pCn^poperations on 2n, in other words time on 2n / time on n -> 2^p
A = randn(n,n)
x = randn(n)
B = randn(2n,2n)
y = randn(2n);
mul_rows(A,x))
4.411648351648352
3.8463392588128955
2.187420284925831
程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com