max = 0
value = 0
LOOP
INPUT value
IF (value == 0)
EXIT LOOP
ENDIF
IF (value > max)
max = value
ENDIF
ENDLOOP
PRINT max
STOP
I am using https://marie.js.org/ but i'm having a lot of trouble trying to figure out how to do an if statement. I've attempted to use skipcond. I'm also struggling a bit with the endless loop. Any help to get me started off would be really appreciated.
First, convert the pseudo code to the if-goto style of assembly language & machine code.
if a then
b
endif
translates into
if !a then goto endif1
b
endif1,
Second, translate your pseudo code variables into Marie assembly language/machine code variables.
For example, you have an integer max in the pseudo code, so in the data area put:
max, dec 0
Finally, translate each line of if-goto code into assembly.
Conditional tests if a < b goto are done by comparison using subtraction. So, load a into the accumulator, subtract b, which sets the condition codes, and then do a SkipCond and goto to skip or not skip code you want to execute.
Marie.js has a number of simple examples. Look at the multiplication example, to see data/variable declarations, conditional branches, loops, input, output.
Related
I've tried searching for help but I haven't found a solution yet, I'm trying to repeat math.random.
current code:
local ok = ""
for i = 0,10 do
local ok = ok..math.random(0,10)
end
print(ok)
no clue why it doesn't work, please help
Long answer
Even if the preferable answer is already given, just copying it will probably not lead to the solution you may expect or less future mistakes. So I decided to explain why your code fails and to fix it and also help better understand how DarkWiiPlayer's answer works (except for string.rep and string.gsub).
Issues
There are at least three issues in your code:
the math.random(m, n) function includes lower and the upper values
local declarations hide a same-name objects in outer scopes
math.random gives the same number sequence unless you set its seed with math.randomseed
See Detailed explanation section below for more.
Another point seems at least worth mentioning or suspicious to me, as I assume you might be puzzled by the result (it seems to me to reflect exactly the perspective of the C programmer, from which I also got to know Lua): the Lua for loop specifies start and end value, so both of these values are included.
Attempt to repair
Here I show how a version of your code that yields the same results as the answer you accepted: a sequence of 10 percent-encoded decimal digits.
-- this will change the seed value (but mind that its resolution is seconds)
math.randomseed(os.time())
-- initiate the (only) local variable we are working on later
local ok = ""
-- encode 10 random decimals (Lua's for-loop is one-based and inclusive)
for i = 1, 10 do
ok = ok ..
-- add fixed part
'%3' ..
-- concatenation operator implicitly converts number to string
math.random(0, 9) -- a random number in range [0..9]
end
print(ok)
Detailed explanation
This explanation makes heavily use of the assert function instead of adding print calls or comment what the output should be. In my opinion assert is the superior choice for illustrating expected behavior: The function guides us from one true statement - assert(true) - to the next, at the first miss - assert(false) - the program is exited.
Random ranges
The math library in Lua provides actually three random functions depending on the count of arguments you pass to it. Without arguments, the result is in the interval [0,1):
assert(math.random() >= 0)
assert(math.random() < 1)
the one-argument version returns a value between 1 and the argument:
assert(math.random(1) == 1)
assert(math.random(10) >= 1)
assert(math.random(10) <= 10)
the two-argument version explicitly specifies min and max values:
assert(math.random(2,2) == 2)
assert(math.random(0, 9) >= 0)
assert(math.random(0, 9) <= 9)
Hidden outer variable
In this example, we have two variables x of different type, the outer x is not accessible from the inner scope.
local x = ''
assert(type(x) == 'string')
do
local x = 0
assert(type(x) == 'number')
-- inner x changes type
x = x .. x
assert(x == '00')
end
assert(type(x) == 'string')
Predictable randomness
The first call to math.random() in a Lua program will return always the same number because the pseudorandom number generator (PRNG) starts at seed 1. So if you call math.randomseed(1), you'll reset the PRNG to its initial state.
r0 = math.random()
math.randomseed(1)
r1 = math.random()
assert(r0 == r1)
After calling math.randomseed(os.time()) calls to math.random() will return different sequences presuming that subsequent program starts differ at least by one second. See question Current time in milliseconds and its answers for more information about the resolutions of several Lua functions.
string.rep(".", 10):gsub(".", function() return "%3" .. math.random(0, 9) end)
That should give you what you want
I was just wondering if there is a special way of saying when something equals something. For example in python, if you declare something equals 2, you say something = 2, whereas when you check if something equals something else, you would say:
if something == somethingelse:
So my question is in pseudocode for algorithms if I'm checking to see if a entered password equals a stored password in an IF THEN ELSE ENDIF loop, would I use one or two equal signs:
WHILE attempts < 3
Get EnteredPassword
**IF EnteredPassword = StoredPassword THEN**
Validated = TRUE
ELSE
attempts = attempts + 1
ENDIF
ENDWHILE
Usually, pseudocode is very broad and every author has their own way of expressing it. As
Aziz has noted, usually x <- 1 is used for an assignment and x := x + 1 for an update. Read ':=' as 'becomes' instead of 'equals', however, they are interchangeably used. As for your question, both = and == are accepted answers, as long as it is clear to your reader what your intention is.
To express equals you use the equal mark symbol once, unlike in python where you use the symbol twice to compare two values (eg if variable == 'one'). An example syntax is:
variable = 'one'
WHILE variable = 'one' DO
SEND "hi" TO DISPLAY
I'm writing a checkpoint function in my Monte Carlo simulation in Fortran 90/95, the compiler I'm using is ifort 18.0.2, before going through detail just to clarify the version of pseudo-random generator I'm using:
A C-program for MT19937, with initialization, improved 2002/1/26.
Coded by Takuji Nishimura and Makoto Matsumoto.
Code converted to Fortran 95 by Josi Rui Faustino de Sousa
Date: 2002-02-01
See mt19937 for the source code.
The general structure of my Monte Carlo simulation code is given below:
program montecarlo
call read_iseed(...)
call mc_subroutine(...)
end
Within the read_iseed
subroutine read_iseed(...)
use mt19937
if (Restart == 'n') then
call system('od -vAn -N4 -td4 < /dev/urandom > '//trim(IN_ISEED)
open(unit=7,file=trim(IN_ISEED),status='old')
read(7,*) i
close(7)
!This is only used to initialise the PRNG sequence
iseed = abs(i)
else if (Restart == 'y') then
!Taking seed value from the latest iteration of previous simulation
iseed = RestartSeed
endif
call init_genrand(iseed)
print *, 'first pseudo-random value ',genrand_real3(), 'iseed ',iseed
return
end subroutine
Based on my understanding, if the seed value holds a constant, the PRNG should be able to reproduce the pseudo-random sequence every time?
In order to prove this is the case, I ran two individual simulations by using the same seed value, they are able to reproduce the exact sequence. So far so good!
Based on the previous test, I'd further assume that regardless the number of times init_genrand() being called within one individual simulation, the PRNG should also be able to reproduce the pseudo-random value sequence? So I did a little modification to my read_iseed() subroutine
subroutine read_iseed(...)
use mt19937
if (Restart == 'n') then
call system('od -vAn -N4 -td4 < /dev/urandom > '//trim(IN_ISEED)
open(unit=7,file=trim(IN_ISEED),status='old')
read(7,*) i
close(7)
!This is only used to initialise the PRNG sequence
iseed = abs(i)
else if (Restart == 'y') then
!Taking seed value from the latest iteration of the previous simulation
iseed = RestartSeed
endif
call init_genrand(iseed)
print *, 'first time initialisation ',genrand_real3(), 'iseed ',iseed
call init_genrand(iseed)
print *, 'second time initialisation ',genrand_real3(), 'iseed ',iseed
return
end subroutine
The output is surprisingly not the case I thought would be, by all means iseed outputs are identical in between two initializations, however, genrand_real3() outputs are not identical.
Because of this unexpected result, I struggled with resuming the simulation at an arbitrary state of the system since the simulation is not reproducing the latest configuration state of the system I'm simulating.
I'm not sure if I've provided enough information, please let me know if any part of this question needs to be more specific?
From the source code you've provided (See [mt19937]{http://web.mst.edu/~vojtat/class_5403/mt19937/mt19937ar.f90} for the source code.), the init_genrand does not clear the whole state.
There are 3 critical state variables:
integer( kind = wi ) :: mt(n) ! the array for the state vector
logical( kind = wi ) :: mtinit = .false._wi ! means mt[N] is not initialized
integer( kind = wi ) :: mti = n + 1_wi ! mti==N+1 means mt[N] is not initialized
The first one is the "array for the state vector", second one is a flag that ensures we don't start with uninitialized array, and the third one is some position marker, as I guess from the condition stated in the comment.
Looking at subroutine init_genrand( s ), it sets mtinit flag, and fills the mt() array from 1 upto n. Alright.
Looking at genrand_real3 it's based on genrand_int32.
Looking at genrand_int32, it starts up with
if ( mti > n ) then ! generate N words at one time
! if init_genrand() has not been called, a default initial seed is used
if ( .not. mtinit ) call init_genrand( seed_d )
and does its arithmetic magic and then starts getting the result:
y = mt(mti)
mti = mti + 1_wi
so.. mti is a positional index in the 'state array', and it is incremented by 1 after each integer read from the generator.
Back to init_genrand - remember? it have been resetting the array mt() but it has not resetted the MTI back to its starting mti = n + 1_wi.
I bet this is the cause of the phenomenon you've observed, since after re-initializing with the same seed, the array would be filled with the same set of values, but later the int32 generator would read from a different starting point. I doubt it was intended, so it's probably a tiny bug easy to overlook.
Lets take this code for example;
If (Random) <> (0 to Variable) then
Its very simple, I just want it to do something if Random is different than 0 to another number set in that variable, im not sure how to do this tho
if you want to check if the variable is not in range from 0 to Variable, it's better to use the code:
if (Random1 < 0) or (Random1 > Variable) then ...
I'm not sure that the code
if not (Random1 in [0..Variable]) then ...
will work for Variable values out of range [0..255]
Sorry, I should of explained better, they are both integer, and Random cant be used since its an instruction, it has to be Random1.
But the answer is
"if not (Random1 in [0..Variable]) then. [0..Variable]"
Thank you very much #lurker.
I've heard that it's been proven theoretically possible to express any control flow in a Turing-complete language using only structured programming constructs, (conditionals, loops and loop-breaks, and subroutine calls,) without any arbitrary GOTO statements. Is there any way to use that theory to automate refactoring of code that contains GOTOs into code that does not?
Let's say I have an arbitrary single subroutine in a simple imperative language, such as C or Pascal. I also have a parser that can verify that this subroutine is valid, and produce an Abstract Syntax Tree from it. But the code contains GOTOs and Labels, which could jump forwards or backwards to any arbitrary point, including into or out of conditional or loop blocks, but not outside of the subroutine itself.
Is there an algorithm that could take this AST and rework it into new code which is semantically identical, but does not contain any Labels or GOTO statements?
In principle, it is always possible to do this, though the results might not be pretty.
One way to always eliminate gotos is to transform the program in the following way. Start off by numbering all the instructions in the original program. For example, given this program:
start:
while (true) {
if (x < 5) goto start;
x++
}
You could number the statements like this:
0 start:
1 while (x < 3) {
2 if (x < 5) goto start;
3 x++
}
To eliminate all gotos, you can simulate the flow of the control through this function by using a while loop, an explicit variable holding the program counter, and a bunch of if statements. For example, you might translate the above code like this:
int PC = 0;
while (PC <= 3) {
if (PC == 0) {
PC = 1; // Label has no effect
} else if (PC == 1) {
if (x < 3) PC = 4; // Skip loop, which ends this function.
else PC = 2; // Enter loop.
} else if (PC == 2) {
if (x < 5) PC = 0; // Simulate goto
else PC = 3; // Simulate if-statement fall-through
} else if (PC == 3) {
x++;
PC = 1; // Simulate jump back up to the top of the loop.
}
}
This is a really, really bad way to do the translation, but it shows that in theory it is always possible to do this. Actually implementing this would be very messy - you'd probably number the basic blocks of the function, then generate code that puts the basic blocks into a loop, tracks which basic block is currently executing, then simulates the effect of running a basic block and the transition from that basic block to the appropriate next basic block.
Hope this helps!
I think you want to read Taming Control Flow by Erosa and Hendren, 1994. (Earlier link on Google scholar).
By the way, loop-breaks are also easy to eliminate. There is a simple mechanical procedure involving the creating of a boolean state variable and the restructuring of nested conditionals to create straight-line control flow. It does not produce pretty code :)
If your target language has tail-call optimization (and, ideally, inlining), you can mechanically remove both break and continue by turning the loop into a tail-recursive function. (If the index variable is modified by the loop body, you need to work harder at this. I'll just show the simplest case.) Here's the transformation of a simple loop:
for (Type Index = Start; function loop(Index: Type):
Condition(Index); if (Condition)
Index = Advance(Index)){ return // break
Body Body
} return loop(Advance(Index)) // continue
loop(Start)
The return statements labeled "continue" and "break" are precisely the transformation of continue and break. Indeed, the first step in the procedure might have been to rewrite the loop into its equivalent form in the original language:
{
Type Index = Start;
while (true) {
if (!Condition(Index))
break;
Body;
continue;
}
}
I use either/both Polyhedron's spag and vast's 77to90 to begin the process of refactoring fortran and then converting it to matlab source. However, these tools always leave 1/4 to 1/2 of the goto's in the program.
I wrote up a goto remover which accomplishes something similar to what you were describing: it takes fortran code and refactors all the remaining goto's from a program and replacing them with conditionals and do/cycle/exit's which can then be converted into other languages like matlab. You can read more about the process I use here:
http://engineering.dartmouth.edu/~d30574x/consulting/consulting_gotorefactor.html
This program could be adapted to work with other languages, but I have not gotten than far yet.