Creating Set in F# with elements from 1111 to 6666 - f#

How do i create a Set in F# with elements from 1111 to 6666 without any values being 0, 7 or higher.
E.g. [1111,1112,1113,1114,1115,1116,1121]
I'd like to make it a set.
Thanks in advance


You can use a sequence comprehension:
let values = seq {
for i in 1110 .. 10 .. 6660 do
for j in 1 .. 6 do
yield i + j
}
and create a set using Set.ofSeq e.g.
let s = Set.ofSeq values


There must be an easier way than:
let values = seq {
for a in 1000 .. 1000 .. 6000 do
for b in 100 .. 100 .. 600 do
for c in 10 .. 10 .. 60 do
for d in 1 .. 6 do
yield a + b + c + d
}

Related

Why does SCAN/LAMBDA give unexpected results?

It is very possible that I dont understand the lambda logic or do I? I have dataset A2:A5 like:
1
3
6
10
If I do: =SCAN(0, A2:A5, LAMBDA(aa, bb, aa+bb)) i get:
1
4
10
20
If I do: =SCAN(0, A2:A5, LAMBDA(aa, bb, ROW(bb)-1)) I get
1
2
3
4
if I run: =SCAN(0, A2:A5, LAMBDA(aa, bb, (aa+bb)*(ROW(bb)-1))) the result is
1
8
42
208
Why there is 42 and 208 ? How this results in such values? How can it be 42 and 208 ?
Expected result is
1
8
30
80
And I can get it with:
=ArrayFormula(SCAN(0, A2:A5, LAMBDA(aa, bb, aa+bb))*(ROW(A2:A5)-1))
But not with
=SCAN(0, A2:A5, LAMBDA(aa, bb, (aa+bb)*(ROW(bb)-1)))

SCAN is a great intermediate results function. To understand how SCAN operates, you need to understand how REDUCE operates. The syntax is:
=REDUCE(initial_value, array, LAMBDA(accumulator, current_value, some_function()))
Going through =SCAN(0, A2:A5, LAMBDA(aa, bb, (aa+bb)*(ROW(bb)-1))) step by step,
A2:A5 is 1,3,6,10
Step 1:
aa = 0(initial_value)
bb = 1(current_value:A2)
Result((aa+bb)*(ROW(bb)-1)): (0+1)*(2-1)=1
Step 2:
aa = 1(accumulator(previous return value))
bb = 3(current_value:A3)
Result((aa+bb)*(ROW(bb)-1)): (1+3)*(3-1)=8
Step 3:
aa = 8(accumulator(previous return value))
bb = 6(current_value:A4)
Result((aa+bb)*(ROW(bb)-1)): (8+6)*(4-1)=42
Step 4:
aa = 42(accumulator(previous return value))
bb = 10(current_value:A5)
Result((aa+bb)*(ROW(bb)-1)): (42+10)*(5-1)=52*4=208

aa stores the result of the previous calculation, so you have:

above answers pretty much contain all so I will add only this:
you probably expected that by doing (aa+bb)*(ROW(bb)-1) you will get:
(aa+bb)
*
(ROW(bb)-1)
1
*
1
=
1
4
*
2
=
8
10
*
3
=
30
20
*
4
=
80
but that's not how it works. to get your expected result and by not using your formula where ROW is outside of SCAN:
=ArrayFormula(SCAN(0, A2:A5, LAMBDA(aa, bb, aa+bb))*(ROW(A2:A5)-1))
you would need to do:
=INDEX(MAP(SCAN(0, A2:A5, LAMBDA(aa, bb, (aa+bb))), ROW(A2:A5)-1, LAMBDA(cc, dd, cc*dd)))
where cc is the entire SCAN and dd is ROW(A2:A5)-1 eg. first do the running total and then multiplication, which is not so feasible length-wise.
or shorter but with SEQUENCE:
=MAP(SCAN(0, A2:A5, LAMBDA(aa, bb, (aa+bb))), SEQUENCE(4), LAMBDA(cc, dd, cc*dd))

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

formatting strings in lua in a pattern

I want to make a script that takes any number, counts up to them and returns them in a format.
so like this
for i = 1,9 do
print(i)
end
will return
1
2
3
4
5
6
7
8
9
however I want it to print like this
1 2 3
4 5 6
7 8 9
and I want it to work even with things more than 9 so things like 20 would be like this
1 2 3
4 5 6
7 8 9
10 11 12
13 14 15
16 17 18
19 20
I'm sure it can be done using the string library in lua but I am not sure how to use that library.
Any help?

function f(n,per_line)
per_line = per_line or 3
for i = 1,n do
io.write(i,'\t')
if i % per_line == 0 then io.write('\n') end
end
end
f(9)
f(20)

The for loop takes an optional third step:
for i = 1, 9, 3 do
print(string.format("%d %d %d", i, i + 1, i + 2))
end

I can think of 2 ways to do this:
local NUMBER = 20
local str = {}
for i=1,NUMBER-3,3 do
table.insert(str,i.." "..i+1 .." "..i+2)
end
local left = {}
for i=NUMBER-NUMBER%3+1,NUMBER do
table.insert(left,i)
end
str = table.concat(str,"\n").."\n"..table.concat(left," ")
And another one using gsub:
local NUMBER = 20
local str = {}
for i=1,NUMBER do
str[i] = i
end
-- Makes "1 2 3 4 ..."
str = table.concat(str," ")
-- Divides it per 3 numbers
-- "%d+ %d+ %d+" matches 3 numbers divided by spaces
-- (You can replace the spaces (including in concat) with "\t")
-- The (...) capture allows us to get those numbers as %1
-- The "%s?" at the end is to remove any trailing whitespace
-- (Else each line would be "N N N " instead of "N N N")
-- (Using the '?' as the last triplet might not have a space)
-- ^ e.g. NUMBER = 6 would make it end with "4 5 6"
-- The "%1\n" just gets us our numbers back and adds a newline
str = str:gsub("(%d+ %d+ %d+)%s?","%1\n")
print(str)
I've benchmarked both code snippets. The upper one is a tiny bit faster, although the difference is almost nothing:
Benchmarked using 10000 interations
NUMBER 20 20 20 100 100
Upper 256 ms 276 ms 260 ms 1129 ms 1114 ms
Lower 284 ms 280 ms 282 ms 1266 ms 1228 ms

Use a temporary table to contain the values until you print them:
local temp = {}
local cols = 3
for i = 1,9 do
if #temp == cols then
print(table.unpack(temp))
temp = {}
end
temp[#temp + 1] = i
end
--Last minute check for leftovers
if #temp > 0 then
print(table.unpack(temp))
end
temp = nil

"Bitwise AND" in Lua

I'm trying to translate a code from C to Lua and I'm facing a problem.
How can I translate a Bitwise AND in Lua?
The source C code contains:
if ((command&0x80)==0)
...
How can this be done in Lua?
I am using Lua 5.1.4-8

Implementation of bitwise operations in Lua 5.1 for non-negative 32-bit integers
OR, XOR, AND = 1, 3, 4
function bitoper(a, b, oper)
local r, m, s = 0, 2^31
repeat
s,a,b = a+b+m, a%m, b%m
r,m = r + m*oper%(s-a-b), m/2
until m < 1
return r
end
print(bitoper(6,3,OR)) --> 7
print(bitoper(6,3,XOR)) --> 5
print(bitoper(6,3,AND)) --> 2

Here is a basic, isolated bitwise-and implementation in pure Lua 5.1:
function bitand(a, b)
local result = 0
local bitval = 1
while a > 0 and b > 0 do
if a % 2 == 1 and b % 2 == 1 then -- test the rightmost bits
result = result + bitval -- set the current bit
end
bitval = bitval * 2 -- shift left
a = math.floor(a/2) -- shift right
b = math.floor(b/2)
end
return result
end
usage:
print(bitand(tonumber("1101", 2), tonumber("1001", 2))) -- prints 9 (1001)

Here's an example of how i bitwise-and a value with a constant 0x8000:
result = (value % 65536) - (value % 32768) -- bitwise and 0x8000

In case you use Adobe Lightroom Lua, Lightroom SDK contains LrMath.bitAnd() method for "bitwise AND" operation:
-- x = a AND b
local a = 11
local b = 6
local x = import 'LrMath'.bitAnd(a, b)
-- x is 2
And there are also LrMath.bitOr(a, b) and LrMath.bitXor(a, b) methods for "bitwise OR" and "biwise XOR" operations.

This answer is specifically for Lua 5.1.X
you can use
if( (bit.band(command,0x80)) == 0) then
...
in Lua 5.3.X and onwards it's very straight forward...
print(5 & 6)
hope that helped 😉

Calculated nCr mod m (n choose r) for large values of n (10^9)

Now that CodeSprint 3 is over, I've been wondering how to solve this problem. We need to simply calculate nCr mod 142857 for large values of r and n (0<=n<=10^9 ; 0<=r<=n). I used a recursive method which goes through min(r, n-r) iterations to calculate the combination. Turns out this wasn't efficient enough. I've tried a few different methods, but they all seem to not be efficient enough. Any suggestions?

For non-prime mod, factor it (142857 = 3^3 * 11 * 13 * 37) and compute C(n,k) mod p^q for each prime factor of the mod using the general Lucas theorem, and combine them using Chinese remainder theorem.
For example, C(234, 44) mod 142857 = 6084, then
C(234, 44) mod 3^3 = 9
C(234, 44) mod 11 = 1
C(234, 44) mod 13 = 0
C(234, 44) mod 37 = 16
The Chinese Remainder theorem involves finding x such that
x = 9 mod 3^3
x = 1 mod 11
x = 0 mod 13
x = 16 mod 37
The result is x = 6084.
Example
C(234, 44) mod 3^3
First convert n, k, and n-k to base p
n = 234_10 = 22200_3
k = 44_10 = 1122_3
r = n-k = 190_10 = 21001_3
Next find the number of carries
e[i] = number of carries from i to end
e 4 3 2 1 0
1 1
r 2 1 0 0 1
k 1 1 2 2
n 2 2 2 0 0
Now create the factorial function needed for general Lucas
def f(n, p):
r = 1
for i in range(1, n+1):
if i % p != 0:
r *= i
return r
Since q = 3, you will consider only three digits of the base p representation at a time
So
f(222_3, 3)/[f(210_3, 3) * f(011_3, 3)] *
f(220_3, 3)/[f(100_3, 3) * f(112_3, 3)] *
f(200_3, 3)/[f(001_3, 3) * f(122_3, 3)] = 6719344775 / 7
Now
s = 1 if p = 2 and q >= 3 else -1
Then
p^e[0] * s * 6719344775 / 7 mod 3^3
e[0] = 2
p^e[0] = 3^2 = 9
s = -1
p^e[0] * s * 6719344775 = -60474102975
Now you have
-60474102975 / 7 mod 3^3
This is a linear congruence and can be solved with
ModularInverse(7, 3^3) = 4
4 * -60474102975 mod 27 = 9
Hence C(234, 44) mod 3^3 = 9

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