Pairs in lua programming - lua

I have two set of values defined:
local A1 = {100, 200, 300, 400}
local A2 = {500, 600, 700, 800}
I want to iterate a loop assigning values for another variable B1 and B2 as pairs from A1 and A2 as follows:
B1 = 100 and B2 = 500 (first iteration)
B1 =200 and B2 = 600 (second iteration)
B1 = 300 and B2 = 700 (third iteration)
B1=400 and B2 = 800 (fourth iteration)
I tried to use ipairs as follows:
for i, f1 in ipairs(A1) do
for j, f2 in ipairs(A2) do
B1 = f1
B2 = f2
end
end
but this gave me
B1 = 100 and B2 = 500 (first iteration)
B1 =100 and B2 = 600 (second iteration)
B1 = 100 and B2 = 700 (third iteration)
B1=100 and B2 = 800 (fourth iteration)
B1 = 200 and B2 = 500 (fifth iteration)
B1 =200 and B2 = 600 (sixth iteration)
B1 =200 and B2 = 700 (seventh iteration)
....
...
...
so on...
can anyone help me to code in the right way?

You can easily do this with a numerical loop:
for i = 1, 4 do
local a, b = A1[i], B1[i]
--- use them
end
How you go about determining the number of iterations you'll need is the tricky part. If the sizes are variant, but each table is the same length as the others you can instead use the length operator (#A1).
Alternatively, you might want a function that returns the largest length of a given set of tables.
local function max_table_len (...)
local tabs = { ... }
local len = 0
for i = 1, #tabs do
local l = #tabs[i]
if l > len then
len = l
end
end
return len
end
And maybe even a helper function to get each value.
local function get_from_tables (index, ...)
local values = { ... }
local len = #values
for i = 1, len do
values[i] = values[i][index]
end
return table.unpack(values, 1, len)
end
Ending up with something like:
for index = 1, max_table_len(A1, B1) do
local a, b = get_from_tables(index, A1, B1)
end

You can build on the ipairs example from Programming in Lua. For instance this version iterates over 2 sequences in parallel:
-- iterator function
local function iter_ipairs2(tablePair, i)
i = i + 1
local v1 = tablePair[1][i]
local v2 = tablePair[2][i]
-- if you use 'and' here the iteration stops after finishing
-- the shortest sequence. If you use 'or' the iteration
-- will stop after it finishes the longest sequence.
if v1 and v2 then
return i, v1, v2
end
end
-- this is the function you'll call from your other code:
local function ipairs2(t1, t2)
return iter_ipairs2, {t1, t2}, 0
end
-- usage:
local A1 = {100, 200, 300, 400, 500}
local A2 = {500, 600, 700, 800}
for i, v1, v2 in ipairs2(A1, A2) do
print(i, v1, v2)
end

The previous answers are more detailed and provide a more general and better answer.
This one is for someone very new to Lua. Not only does it show two loops, it reinforces that there is usually more than one way to get where you want to go.
local A1 = {100, 200, 300, 400}
local A2 = {500, 600, 700, 800}
print("simplest answer:")
-- doesn't use ipairs and assumes A1 and A2 are the same size
for i = 1, #A1 do
B1 = A1[i]
B2 = A2[i]
print(B1, B2, "(iteration #"..i..")")
end
print()
print("answer that uses ipairs:")
-- again, assumes A1 and A2 are the same size
for i, v in ipairs(A1) do
B1 = A1[i] -- i steps through A1 and A2
B2 = A2[i] -- this works because A1 and A2 are same size
print(B1, B2, "(iteration #"..i..")")
end
Gives this output:
simplest answer:
100 500 (iteration #1)
200 600 (iteration #2)
300 700 (iteration #3)
400 800 (iteration #4)
answer that uses ipairs:
100 500 (iteration #1)
200 600 (iteration #2)
300 700 (iteration #3)
400 800 (iteration #4)

Related

finding minimum values from a cut table Lua 5.1.5

I have a Lua script that turns a table into segments:
function tablecut(t, n)
local result = {}
local j = 0
for i = 1, #t do
if (i-1) % n == 0 then
j = j + 1
result[j] = {}
end
result[j][#result[j]+1] = t[i]
end
return result
end
output = tablecut({'15', '62', '14', '91', '33', '55', '29', '4'}, 4)
for i = 1, #output do
for j = 1, #output[i] do
io.write(tostring(output[i][j])..' ')
end
print()
end
output:
15 62 14 91
33 55 29 4
And I am trying to find the minima from the cut lists so the output would look like this:
15 62 14 91
min = 14
33 55 29 4
min = 4
Edit: If its of any importance this is how I got it to work on Lua 5.3 but there is no table.move function on Lua 5.1. I can't remember how my thought function worked when I wrote this code.
function indexOf(array, value)
for i, v in ipairs(array) do
if v == value then
return i
end
end
return nil
end
Indicies = {}
Answers = {}
function chunks(lst, size)
local i = 1
local count = 0
return function()
if i > #lst then return end
local chunk = table.move(lst, i, i + size -1, 1, {})
i = i + size
count = count + 1
return count, chunk
end
end
local a = {91,52,19,59,38,29,58,11,717,91,456,49,30,62,43,8,17,15,26,22,13,10,2,23} --Test list
for i, chunk in chunks(a, 4) do
x=math.min(a)
print(string.format("#%d: %s", i, table.concat(chunk, ",")))
table.sort(chunk)
print(math.min(chunk[1]))
table.insert(Answers, chunk[1])
table.insert(Indicies, (indexOf(a, chunk[1])))
Output:
#1: 91,52,19,59
19
#2: 38,29,58,11
11
#3: 717,91,456,49
49
your table cut function could be simplified, and your output for loop needs you use an iterator if you want to get an output simply like you do in your 5.3 script.
function cuttable(t,n)
local binned = {}
for i=1,#t,n do
local bin = {}
for j=1,n do
table.insert(bin, t[i + ((j - 1) % n)])
end
table.insert(binned, bin)
end
return binned
end
For the for loop, we can use ipairs on the output of cuttable keeping things pretty simple, then we just do the same steps of concat then sort and print out our results.
for k, bin in ipairs(cuttable(a,4)) do
local output = "#" .. k .. ":" .. table.concat(bin, ",")
table.sort(bin)
print(output)
print(bin[1])
end
Output
#1:91,52,19,59
19
#2:38,29,58,11
11
#3:717,91,456,49
49
#4:30,62,43,8
8
#5:17,15,26,22
15
#6:13,10,2,23
2
One way to implement the cutting would be using a for loop & unpack. I have handled the case of the length not being divisible by 4 after the for loop to (1) maximize performance (check doesn't need to be done every iteration) and (2) be able to directly pass the values to math.min, which doesn't accept nils.
for i = 1, math.floor(#t / 4), 4 do
print(unpack(t, i, i+4))
print("min = " .. math.min(unpack(t, i, i+4)))
end
-- If #t is not divisible by 4, deal with the remaining elements
local remaining = #t % 4
if remaining > 0 then
print(unpack(t, #t - remaining, remaining))
print("min = " .. math.min(unpack(t, #t - remaining, remaining)))
end

multi key tuple to map a multi value tuple in Lua

Is there a lib in Lua that supports a map from a tuple to a tuple? I have a key {a,b,c} to map to a value {c,d,e}
There are libs such as, http://lua-users.org/wiki/MultipleKeyIndexing for multikey but not where the value is a tuple.
Here's one way to use Egor's suggestion for making a key through string concatenation. Make your own simple insert and get methods for a table, t.
local a, b, c = 10, 20, 30
local d, e, f = 100, 200, 300
local t = {}
t.key = function (k)
local key = ""
for _,v in ipairs(k) do
key = key .. tostring(v) .. ";"
end
return key
end
t.set = function (k, v)
local key = t.key(k)
t[key] = v
end
t.get = function (k)
local key = t.key(k)
return t[key]
end
t.set ({a, b, c}, {d, e, f}) -- using variables
t.set ({40, 50, 60}, {400, 500, 600}) -- using constants
local w = t.get ({a, b, c}) -- using variables
local x = t.get ({40, 50, 60}) -- using constants
print(w[1], w[2], w[3]) -- 100 200 300
print(x[1], x[2], x[3]) -- 400 500 600

Lua Musical Notes to Numbers

I need to convert some musical note inputs representing a chord to numbers above it's root note 0 using Lua.
So from the midi data we get the notes of a C13 Chord
input: C, E, G, A#, D, F, A
as the root note 0 is C we start on the C note,
below we have 2 octaves of a piano keyboard, 12 notes on each where chords are played
0C 1C# 2D 3D# 4E 5F 6F# 7G 8G# 9A 10A# 11B 12C 13C# 14D 15D# 16E 17F 18F# 19G 20G# 21A 22A# 23B
so C is the root note 0
D,F,A are played on the next octave
result: 0,4,7,10,14,17,21
so if we have a D chord
input: D,F#,A
D is the root note 0
all notes played on the first octave
0D 1D# 2E 3F 4F# 5G 6G# 7A 8A# 9B 10C 11C# 12D 13D# 14E 15F 16F# 17G 18G# 19A 20A# 21B 22C 23C#
result: 0,4,7
G#m7#9 Chord
input: G#,B,D#,F#,B
0G# 1A 2A# 3B 4C 5C# 6D 7D# 8E 9F 10F# 11G 12G# 13A 14A# 15B 16C 17C# 18D 19D# 20E 21F 22F# 23G
result: 0,3,7,10,15
Something like this may work:
local function notes2nums(input)
local map = {A = 9, ["A#"] = 10, B = 11, C = 0, ["C#"] = 1, D = 2, ["D#"] = 3, E = 4, F = 5, ["F#"] = 6, G = 7, ["G#"] = 8}
local base, prev
return (input:gsub("([^,]+)", function(note)
local num = map[note] or error(("Unexpected note value '%s'"):format(note))
base = base or num
num = num - base
if prev and num < prev then num = num + 12 end
prev = num
return tostring(num)
end))
end
print(notes2nums("D,F#,A"))
print(notes2nums("C,E,G,A#,D,F,A"))
print(notes2nums("G#,B,D#,F#,B"))
This prints:
0,4,7
0,4,7,10,14,17,21
0,3,7,10,15

Adaptation of SHA2 512 gives incorrect results

I am trying to adapt the pure Lua implementation of the SecureHashAlgorithm found here for SHA2 512 instead of SHA2 256. When I try to use the adaptation, it does not give the correct answer.
Here is the adaptation:
--
-- UTILITY FUNCTIONS
--
-- transform a string of bytes in a string of hexadecimal digits
local function str2hexa (s)
local h = string.gsub(s, ".", function(c)
return string.format("%02x", string.byte(c))
end)
return h
end
-- transforms number 'l' into a big-endian sequence of 'n' bytes
--(coded as a string)
local function num2string(l, n)
local s = ""
for i = 1, n do
--most significant byte of l
local remainder = l % 256
s = string.char(remainder) .. s
--remove from l the bits we have already transformed
l = (l-remainder) / 256
end
return s
end
-- transform the big-endian sequence of eight bytes starting at
-- index 'i' in 's' into a number
local function s264num (s, i)
local n = 0
for i = i, i + 7 do
n = n*256 + string.byte(s, i)
end
return n
end
--
-- MAIN SECTION
--
-- FIRST STEP: INITIALIZE HASH VALUES
--(second 32 bits of the fractional parts of the square roots of the first 9th through 16th primes 23..53)
local HH = {}
local function initH512(H)
H = {0x6a09e667f3bcc908, 0xbb67ae8584caa73b, 0x3c6ef372fe94f82b, 0xa54ff53a5f1d36f1, 0x510e527fade682d1, 0x9b05688c2b3e6c1f, 0x1f83d9abfb41bd6b, 0x5be0cd19137e2179}
return H
end
-- SECOND STEP: INITIALIZE ROUND CONSTANTS
--(first 80 bits of the fractional parts of the cube roots of the first 80 primes 2..409)
local k = {
0x428a2f98d728ae22, 0x7137449123ef65cd, 0xb5c0fbcfec4d3b2f, 0xe9b5dba58189dbbc, 0x3956c25bf348b538,
0x59f111f1b605d019, 0x923f82a4af194f9b, 0xab1c5ed5da6d8118, 0xd807aa98a3030242, 0x12835b0145706fbe,
0x243185be4ee4b28c, 0x550c7dc3d5ffb4e2, 0x72be5d74f27b896f, 0x80deb1fe3b1696b1, 0x9bdc06a725c71235,
0xc19bf174cf692694, 0xe49b69c19ef14ad2, 0xefbe4786384f25e3, 0x0fc19dc68b8cd5b5, 0x240ca1cc77ac9c65,
0x2de92c6f592b0275, 0x4a7484aa6ea6e483, 0x5cb0a9dcbd41fbd4, 0x76f988da831153b5, 0x983e5152ee66dfab,
0xa831c66d2db43210, 0xb00327c898fb213f, 0xbf597fc7beef0ee4, 0xc6e00bf33da88fc2, 0xd5a79147930aa725,
0x06ca6351e003826f, 0x142929670a0e6e70, 0x27b70a8546d22ffc, 0x2e1b21385c26c926, 0x4d2c6dfc5ac42aed,
0x53380d139d95b3df, 0x650a73548baf63de, 0x766a0abb3c77b2a8, 0x81c2c92e47edaee6, 0x92722c851482353b,
0xa2bfe8a14cf10364, 0xa81a664bbc423001, 0xc24b8b70d0f89791, 0xc76c51a30654be30, 0xd192e819d6ef5218,
0xd69906245565a910, 0xf40e35855771202a, 0x106aa07032bbd1b8, 0x19a4c116b8d2d0c8, 0x1e376c085141ab53,
0x2748774cdf8eeb99, 0x34b0bcb5e19b48a8, 0x391c0cb3c5c95a63, 0x4ed8aa4ae3418acb, 0x5b9cca4f7763e373,
0x682e6ff3d6b2b8a3, 0x748f82ee5defb2fc, 0x78a5636f43172f60, 0x84c87814a1f0ab72, 0x8cc702081a6439ec,
0x90befffa23631e28, 0xa4506cebde82bde9, 0xbef9a3f7b2c67915, 0xc67178f2e372532b, 0xca273eceea26619c,
0xd186b8c721c0c207, 0xeada7dd6cde0eb1e, 0xf57d4f7fee6ed178, 0x06f067aa72176fba, 0x0a637dc5a2c898a6,
0x113f9804bef90dae, 0x1b710b35131c471b, 0x28db77f523047d84, 0x32caab7b40c72493, 0x3c9ebe0a15c9bebc,
0x431d67c49c100d4c, 0x4cc5d4becb3e42b6, 0x597f299cfc657e2a, 0x5fcb6fab3ad6faec, 0x6c44198c4a475817
}
-- THIRD STEP: PRE-PROCESSING (padding)
local function preprocess(toProcess, len)
--append a single '1' bit
--append K '0' bits, where K is the minimum number >= 0 such that L + 1 + K = 896mod1024
local extra = 128 - (len + 9) % 128
len = num2string(8 * len, 8)
toProcess = toProcess .. "\128" .. string.rep("\0", extra) .. len
assert(#toProcess % 128 == 0)
return toProcess
end
local function rrotate(rot, n)
return (rot >> n) | ((rot << 64 - n))
end
local function digestblock(msg, i, H)
local w = {}
for j = 1, 16 do w[j] = s264num(msg, i + (j - 1)*4) end
for j = 17, 80 do
local v = w[j - 15]
local s0 = rrotate(v, 1) ~ rrotate(v, 8) ~ (v >> 7)
v = w[j - 2]
w[j] = w[j - 16] + s0 + w[j - 7] + ((rrotate(v, 19) ~ rrotate(v, 61)) ~ (v >> 6))
end
local a, b, c, d, e, f, g, h = H[1], H[2], H[3], H[4], H[5], H[6], H[7], H[8]
for i = 1, 80 do
a, b, c, d, e, f, g, h = a , b , c , d , e , f , g , h
local s0 = rrotate(a, 28) ~ (rrotate(a, 34) ~ rrotate(a, 39))
local maj = ((a & b) ~ (a & c)) ~ (b & c)
local t2 = s0 + maj
local s1 = rrotate(e, 14) ~ (rrotate(e, 18) ~ rrotate(e, 41))
local ch = (e & f) ~ (~e & g)
local t1 = h + s1 + ch + k[i] + w[i]
h, g, f, e, d, c, b, a = g, f, e, d + t1, c, b, a, t1 + t2
end
H[1] = (H[1] + a)
H[2] = (H[2] + b)
H[3] = (H[3] + c)
H[4] = (H[4] + d)
H[5] = (H[5] + e)
H[6] = (H[6] + f)
H[7] = (H[7] + g)
H[8] = (H[8] + h)
end
local function finalresult512 (H)
-- Produce the final hash value:
return
str2hexa(num2string(H[1], 8)..num2string(H[2], 8)..num2string(H[3], 8)..num2string(H[4], 8)..
num2string(H[5], 8)..num2string(H[6], 8)..num2string(H[7], 8)..num2string(H[8], 8))
end
-- Returns the hash512 for the given string.
local function hash512 (msg)
msg = preprocess(msg, #msg)
local H = initH512(HH)
-- Process the message in successive 1024-bit (128 bytes) chunks:
for i = 1, #msg, 128 do
digestblock(msg, i, H)
end
return finalresult512(H)
end
Given hash512("a"):
Expect: 1f40fc92da241694750979ee6cf582f2d5d7d28e18335de05abc54d0560e0f5302860c652bf08d560252aa5e74210546f369fbbbce8c12cfc7957b2652fe9a75
Actual: e0b9623f2194cb81f2a62616a183edbe390be0d0b20430cadc3371efc237fa6bf7f8b48311f2fa249131c347fee3e8cde6acfdab286d648054541f92102cfc9c
I know that I am creating a message of the correct bit size (1024 bits) and also working in 1024-bit chunks, or at least I believe I am.
I am not sure if it has to do with the handling of the integers (the standard requires unsigned integers) or whether I made a mistake in one of the utility functions, or both. If it is indeed an issue with the handling of the integers, how would I go about taking care of the problem. I was able to resolve this when working on the 256-bit version of the adaptation by using mod 2^32 when working with numbers in the digestblock method. I attempted to do mod 2^64 and 2^63 with the 512-bit version and it does not correct the problem. I am stumped.
I should mention that I cannot use one of the many library implementations as I am using a sandboxed Lua that does not provide this access, which is why I need a pure lua implementation. Thanks in advance.
Unfortunately, after introducing integers in Lua 5.3 writing scripts for Lua becomes a more complicated task.
You must always think about transformations between integers and floating point numbers.
ALWAYS. Yes, that's boring.
One of your mistakes is an excellent example of this "dark corner of Lua".
local remainder = l % 256
s = string.char(remainder) .. s
--remove from l the bits we have already transformed
l = (l-remainder) / 256
Your value l is initially a 64-bit integer.
After cutting off its first byte l contains (64-8) = 56 bits, but now it's a floating point-number (with 53-bit precision, of course).
Possible solution: use l = l >> 8 or l = l // 256 instead of l = (l-remainder) / 256
Another mistake is using s264num(msg, i + (j - 1) * 4) instead of s264num(msg, i + (j - 1) * 8)
One more mistake is in the following line:
local extra = 128 - (len + 9) % 128
The correct code is
local extra = - (len + 17) % 128 + 8
(Please note that -a%m+b is not the same as b-a%m due to operator precedence)
After fixing these 3 mistakes your code works correctly.

Lua 3 point crossover help to start

I want to implement a 3 point crossover for genetic programming but I don't know how to do it and where to start.
My input is:
a = {(first pair), (second pair), ... etc.}
For example a = {(12345,67890), (09876,54321)} (those are numbers, not strings)
Output: Something like this:
Example: a_1 = {(12895), (67340)} also numbers.
Thanks for reply and sorry for my bad English.
Here is my quick implementation of k-point crossover for integers using mostly integer arithmetic. Starting with this, you can extend it to crossover your chromosomes of many pairs of integers using a loop.
math.randomseed(111)
-- math.randomseed(os.time())
a = 12345
b = 67890
len = 5 -- number of digits
function split(mum, dad, len, base)
local split = math.pow(base, math.random(len))
local son = math.floor(dad / split) * split + mum % split
local daughter = math.floor(mum / split) * split + dad % split
return son, daughter
end
function kpoint(mum, dad, len, base, k)
for i=1, k do
mum, dad = split(mum, dad, len, base)
end
return mum, dad
end
s, d = kpoint(a, b, len, 10, 3) -- 3 point crossover in base 10
print(s) -- 67395
print(d) -- 12840
-- binary, (crossover binary representation)
s, d = kpoint(tonumber("10001", 2), tonumber("10110", 2), 5, 2, 3)
print(s) -- 23 which is (10111) in base 2
print(d) -- 16 which is (10000) in base 2
-- binary, (crossover base 10, but interpret as binary)
s, d = kpoint(1101, 1010, 4, 10, 3)
print(s) -- 1001
print(d) -- 1110

Resources