I am new to Prolog and am having some difficulty fixing the errors of my first program.
The program requirement is that it divides the 2 inputs using recursion, returning 0 if the dividend is larger than the divisor, and ignores remainders.
%Author: Justin Taylor
testquotient :-
repeat,
var(Divident), var(Divisor), var(Answer), var(End),
write('Enter Divident: '),
read(Divident),
write('Enter Divisor: '),
read(Divisor),
quotient(Divident, Divisor, Answer),
nl,
write('Quotient is = '),
write(Answer),
nl,
write('Enter 0 to quit, 1 to continue: '),
read(End),
(End =:= 0),!.
quotient(_, 0, 'Undefined').
quotient(0, _, 0).
quotient(Divisor == Divident -> Answer = 1).
quotient(Divisor < Divident -> Answer = 0).
quotient(Divident, Divisor, Answer) :-
(Divisor > Divident -> Divisor = Divisor - Divident,
quotient(Divident, Divisor, Answer + 1);
Answer = Answer).
First, read up on is. Type help(is). at the SWI-Prolog's prompt. Read the whole section about "Arithmetic" carefully. Second, your first few clauses for quotient are completely off-base, invalid syntax. I'll show you how to rewrite one of them, you'll have to do the other yourself:
%% WRONG: quotient(Divisor == Divident -> Answer = 1).
quotient(Divisor, Divident, Answer) :-
Divisor =:= Divident -> Answer = 1.
%% WRONG: quotient(Divisor < Divident -> Answer = 0).
....
Note the use of =:= instead of ==.
Your last clause for quotient looks almost right at the first glance, save for the major faux pas: prolog's unification, =, is not, repeat not, an assignment operator! We don't change values assigned to logical variables (if X is 5, what's there to change about it? It is what it is). No, instead we define new logical variable, like this
( Divisor > Divident -> NewDivisor = Divisor - Divident,
and we use it in the recursive call,
%% WRONG: quotient(Divident, NewDivisor, Answer + 1) ;
but this is wrong too, w.r.t. the new Answer. If you add 1 on your way down (as you subtract Divident from your Divisor - btw shouldn't it be the other way around?? check your logic or at least swap your names, "divisor" is what you divide by ) that means you should've supplied the initial value. But you seem to supply the terminal value as 0, and that means that you should build your result on your way back up from the depths of recursion:
%%not quite right yet
quotient(Divident, NewDivisor, NewAnswer), Answer = NewAnswer + 1 ;
Next, Answer = Answer succeeds always. We just write true in such cases.
Lastly, you really supposed to use is on each recursion step, and not just in the very end:
( Divisor > Divident -> NewDivisor is Divisor - Divident, %% use "is"
quotient(Divident, NewDivisor, NewAnswer), Answer is NewAnswer+1 %% use "is"
; true ). %% is this really necessary?
Your 'Undefined' will cause an error on 0, but leave it at that, for now. Also, you don't need to "declare" your vars in Prolog. The line var(Divident), ..., var(End), serves no purpose.
Related
I'm new to z3 and trying to use it to solve logic puzzles. The puzzle type I'm working on, Skyscrapers, includes given constraints on the number of times that a new maximum value is found while reading a series of integers.
For example, if the constraint given was 3, then the series [2,3,1,5,4] would satisfy the constraint as we'd detect the maximums '2', '3', '5'.
I've implemented a recursive solution, but the rule does not apply correctly and the resulting solutions are invalid.
for (int i = 0; i < clues.Length; ++i)
{
IntExpr clue = c.MkInt(clues[i].count);
IntExpr[] orderedCells = GetCells(clues[i].x, clues[i].y, clues[i].direction, cells, size);
IntExpr numCells = c.MkInt(orderedCells.Length);
ArrayExpr localCells = c.MkArrayConst(string.Format("clue_{0}", i), c.MkIntSort(), c.MkIntSort());
for (int j = 0; j < orderedCells.Length; ++j)
{
c.MkStore(localCells, c.MkInt(j), orderedCells[j]);
}
// numSeen counter_i(index, localMax)
FuncDecl counter = c.MkFuncDecl(String.Format("counter_{0}", i), new Sort[] { c.MkIntSort(), c.MkIntSort()}, c.MkIntSort());
IntExpr index = c.MkIntConst(String.Format("index_{0}", i));
IntExpr localMax = c.MkIntConst(String.Format("localMax_{0}", i));
s.Assert(c.MkForall(new Expr[] { index, localMax }, c.MkImplies(
c.MkAnd(c.MkAnd(index >= 0, index < numCells), c.MkAnd(localMax >= 0, localMax <= numCells)), c.MkEq(c.MkApp(counter, index, localMax),
c.MkITE(c.MkOr(c.MkGe(index, numCells), c.MkLt(index, c.MkInt(0))),
c.MkInt(0),
c.MkITE(c.MkOr(c.MkEq(localMax, c.MkInt(0)), (IntExpr)localCells[index] >= localMax),
1 + (IntExpr)c.MkApp(counter, index + 1, (IntExpr)localCells[index]),
c.MkApp(counter, index + 1, localMax)))))));
s.Assert(c.MkEq(clue, c.MkApp(counter, c.MkInt(0), c.MkInt(0))));
Or as an example of how the first assertion is stored:
(forall ((index_3 Int) (localMax_3 Int))
(let ((a!1 (ite (or (= localMax_3 0) (>= (select clue_3 index_3) localMax_3))
(+ 1 (counter_3 (+ index_3 1) (select clue_3 index_3)))
(counter_3 (+ index_3 1) localMax_3))))
(let ((a!2 (= (counter_3 index_3 localMax_3)
(ite (or (>= index_3 5) (< index_3 0)) 0 a!1))))
(=> (and (>= index_3 0) (< index_3 5) (>= localMax_3 0) (<= localMax_3 5))
a!2))))
From reading questions here, I get the sense that defining functions via Assert should work. However, I didn't see any examples where the function had two arguments. Any ideas what is going wrong? I realize that I could define all primitive assertions and avoid recursion, but I want a general solver not dependent on the size of the puzzle.
Stack-overflow works the best if you post entire code segments that can be independently run to debug. Unfortunately posting chosen parts makes it really difficult for people to understand what might be the problem.
Having said that, I wonder why you are coding this in C/C# to start with? Programming z3 using these lower level interfaces, while certainly possible, is a terrible idea unless you've some other integration requirement. For personal projects and learning purposes, it's much better to use a higher level API. The API you are using is extremely low-level and you end up dealing with API-centric issues instead of your original problem.
In Python
Based on this, I'd strongly recommend using a higher-level API, such as from Python or Haskell. (There are bindings available in many languages; but I think Python and Haskell ones are the easiest to use. But of course, this is my personal bias.)
The "skyscraper" constraint can easily be coded in the Python API as follows:
from z3 import *
def skyscraper(clue, xs):
# If list is empty, clue has to be 0
if not xs:
return clue == 0;
# Otherwise count the visible ones:
visible = 1 # First one is always visible!
curMax = xs[0]
for i in xs[1:]:
visible = visible + If(i > curMax, 1, 0)
curMax = If(i > curMax, i, curMax)
# Clue must equal number of visibles
return clue == visible
To use this, let's create a row of skyscrapers. We'll make the size based on a constant you can set, which I'll call N:
s = Solver()
N = 5 # configure size
row = [Int("v%d" % i) for i in range(N)]
# Make sure row is distinct and each element is between 1-N
s.add(Distinct(row))
for i in row:
s.add(And(1 <= i, i <= N))
# Add the clue, let's say we want 3 for this row:
s.add(skyscraper(3, row))
# solve
if s.check() == sat:
m = s.model()
print([m[i] for i in row])
else:
print("Not satisfiable")
When I run this, I get:
[3, 1, 2, 4, 5]
which indeed has 3 skyscrapers visible.
To solve the entire grid, you'd create NxN variables and add all the skyscraper assertions for all rows/columns. This is a bit of coding, but you can see that it's quite high-level and a lot easier to use than the C-encoding you're attempting.
In Haskell
For reference, here's the same problem encoded using the Haskell SBV library, which is built on top of z3:
import Data.SBV
skyscraper :: SInteger -> [SInteger] -> SBool
skyscraper clue [] = clue .== 0
skyscraper clue (x:xs) = clue .== visible xs x 1
where visible [] _ sofar = sofar
visible (x:xs) curMax sofar = ite (x .> curMax)
(visible xs x (1+sofar))
(visible xs curMax sofar)
row :: Integer -> Integer -> IO SatResult
row clue n = sat $ do xs <- mapM (const free_) [1..n]
constrain $ distinct xs
constrain $ sAll (`inRange` (1, literal n)) xs
constrain $ skyscraper (literal clue) xs
Note that this is even shorter than the Python encoding (about 15 lines of code, as opposed to Python's 30 or so), and if you're familiar with Haskell quite a natural description of the problem without getting lost in low-level details. When I run this, I get:
*Main> row 3 5
Satisfiable. Model:
s0 = 1 :: Integer
s1 = 4 :: Integer
s2 = 5 :: Integer
s3 = 3 :: Integer
s4 = 2 :: Integer
which tells me the heights should be 1 4 5 3 2, again giving a row with 3 visible skyscrapers.
Summary
Once you're familiar with the Python/Haskell APIs and have a good idea on how to solve your problem, you can code it in C# if you like. I'd advise against it though, unless you've a really good reason to do so. Sticking the Python or Haskell is your best bet not to get lost in the details of the API.
I'm very new to Julia but I've got a some background in Scheme/Rust/F#.
Today I wanted to make yesterday's AoC nicer without an explicit number of nested loops.
I arrived at this working solution, but I don't like the last if. In the languages mentioned above I would call a function (or use a computation expression) that gives me the first result that is not None. For Julia, I expected something to do that. It does, but unexpectedly in an eager fashion.
So When I tried return something(find(r, n, start + 1, which), find(r, n - 1, start + 1, extended)), that also evaluated the second argument when the first already had a result—and thus crashed.
Is there a macro/lazy version or something that I didn't find? How are you supposed to handle a case like that?
I also thought about (short-circuited) or'ing them together, but I guess Julia's strictness in that matter spoils that.
using DataStructures
function find(r::Array{Int}, n, start = 1, which = nil())::Union{Int,Nothing}
if start <= length(r)
extended = cons(start, which)
with_current = sum(i -> r[i], extended)
if with_current == 2020 && n == 1
return prod(i -> r[i], extended)
else
# Unfortunately no :(
#return something(find(r, n, start + 1, which), find(r, n - 1, start + 1, extended))
re = find(r, n, start + 1, which)
if isnothing(re)
return find(r, n - 1, start + 1, extended)
else
re
end
end
end
end
Let me comment more on it why it is not possible given the discussion in the comments.
In Julia function arguments are evaluated eagerly, so Julia evaluates both find(r, n, start + 1, which) and find(r, n - 1, start + 1, extended) before passing them to something function.
Now, with macros you have (I am not writing in a fully general case for simplicity and I hope I got the hygiene right :)):
julia> macro something(x, y)
quote
local vx = $(esc(x))
isnothing(vx) ? $(esc(y)) : vx
end
end
#something (macro with 1 method)
julia> #something 1 2
1
julia> #something nothing 2
2
julia> #something 1 sqrt(-1)
1
julia> #something nothing sqrt(-1)
ERROR: DomainError with -1.0:
sqrt will only return a complex result if called with a complex argument. Try sqrt(Complex(x)).
(in a full-blown version of the macro varargs and Some should be handled to replicate something exactly)
Piqued by Bogumił's answer I wanted to write my first Julia macro. It took some time and numerous attempts to figure out syntax, hygiene and escaping but I'm quite happy now.
I thought it might be worth sharing and provide opportunity for suggestions/improvements.
A lazy #something analog to Base.something
function _something_impl(thing)
:(something($(esc(thing))))
end
function _something_impl(thing, rest...)
quote
local evalued = $(esc(thing))
if isnothing(evalued)
$(_something_impl(rest...))
else
something(evalued)
end
end
end
macro something(things...)
_something_impl(things...)
end
Version without exceptions
As I found exceptions raised from a macro like this not quite suitable, I also made a version that falls back to nothing.
function _something_nothing_impl(thing)
quote
local evaluated = $(esc(thing))
if isa(evaluated, Some)
evaluated.value
else
evaluated
end
end
end
function _something_nothing_impl(thing, rest...)
quote
local evalued = $(esc(thing))
if isnothing(evalued)
$(_something_nothing_impl(rest...))
else
something(evalued)
end
end
end
macro something_nothing(things...)
_something_nothing_impl(things...)
end
Now I guess the recursive middle function could also generated by a macro. :)
After doing some Prolog in uni and doing some exercises I decided to go along somewhat further although I got to admit I don't understand recursion that well, I get the concept and idea but how to code it, is still a question for me. So that's why I was curious if anyone knows how to help tackle this problem.
The idea is given a number e.g. 45, check whether it is possible to make a list starting with 1 going n+1 into the list and if the sum of the list is the same as the given number.
So for 45, [1,2,3,4,5,6,7,8,9] would be correct.
So far I tried looking at the [sum_list/2][1] implemented in Prolog itself but that only checks whether a list is the same as the number it follows.
So given a predicate lijstSom(L,S) (dutch for listSum), given
?- lijstSom(L, 45)
L = [1,2,3,4,5,6,7,8,9];
False
My Idea was something along the line of for example if S = 45, doing steps of the numbers (increasing by 1) and subtracting it of S, if 0 is the remainder, return the list, else return false.
But for that you need counters and I find it rather hard to grasp that in recursion.
EDIT:
Steps in recursion.
Base case empty list, 0 (counter nr, that is minus S), 45 (S, the remainder)
[1], 1, 44
[1,2], 2, 42
[1,2,3], 3, 39
I'm not sure how to read the example
?- lijstSom(L, 45)
L = [1,2,3,4,5,6,7,8,9],
False
...but think of the predicate lijstSom(List, Sum) as relating certain lists of integers to their sum, as opposed to computing the sum of lists of integers. Why "certain lists"? Because we have the constraint that the integers in the list of integers must be monotonically increasing in increments of 1, starting from 1.
You can thus ask the Prolog Processor the following:
"Say something about the relationship between the first argument of lijstSom/2 and the second argument lijstSom/2 (assuming the first is a list of monotonically increasing integers, and the second an integer):
lijstSom([1,2,3], Sum)
... should return true (because yes, there is at least one solution) and give Sum = 6 (because it constructs the solution, too ... we are some corner of Construtivism here.
lijstSom(L, 6)
... should return true (because yes, there is at least one solution) and give the solution [1,2,3].
lijstSom([1,2,3], 6)
... should return true (because yes, [1,2,3] has a sum 6); no further information is needed.
lijstSom(L, S)
... should an infinite series of true and pairs of solution ("generate the solutions").
L = [1], S = 1;
L = [1,2], S = 3;
L = [1,2,3], S = 6;
...
lijstSom([1,2,3], 7)
...should return false ("fail") because 7 is not in a relation lijstSom with [1,2,3] as 7 =/= 1+2+3.
One might even want things to have Prolog Processor say something interesting about:
lijstSom([1,2,X], 6)
X = 3
or even
lijstSom([1,2,X], S)
X = 3
S = 6
In fact, lijstSom/2 as near to mathematically magical as physically possible, which is to say:
Have unrestricted access to the full table of list<->sum relationships floating somewhere in Platonic Math Space.
Be able to find the correct entry in seriously less than infinite number of steps.
And output it.
Of course we are restricted to polynomial algorithms of low exponent and finite number of dstinguishable symbols for eminently practical reasons. Sucks!
So, first define lijstSom(L,S) using an inductive definition:
lijstSom([a list with final value N],S) ... is true if ... lijstSom([a list],S-N and
lijstSom([],0) because the empty list has sum 0.
This is nice because it gives the recipe to reduce a list of arbitrary length down to a list of size 0 eventually while keeping full knowledge its sum!
Prolog is not good at working with the tail of lists, but good with working with the head, so we cheat & change our definition of lijstSom/2 to state that the list is given in reverse order:
lijstSom([3,2,1], 6)
Now some code.
#= is the "constain to be equal" operator from library(clpfd). To employ it, we need to issue use_module(library(clpfd)). command first.
lijstSom([],0).
lijstSom([K|Rest],N) :- lijstSom([Rest],T), T+K #= N.
The above follows the mathematical desiderate of lijstSom and allows the Prolog Processor to perform its computation: in the second clause, it can compute the values for a list of size A from the values of a list of size A-1, "falling down" the staircase of always decreasing list length until it reaches the terminating case of lijstSom([],0)..
But we haven't said anything about the monotonically decreasing-by-1 list.
Let's be more precise:
lijstSom([],0) :- !.
lijstSom([1],1) :- ! .
lijstSom([K,V|Rest],N) :- K #= V+1, T+K #= N, lijstSom([V|Rest],T).
Better!
(We have also added '!' to tell the Prolog Processor to not look for alternate solutions past this point, because we know more about the algorithm than it will ever do. Additionally, the 3rd line works, but only because I got it right after running the tests below and having them pass.)
If the checks fail, the Prolog Processor will says "false" - no solution for your input. This is exactly what we want.
But does it work? How far can we go in the "mathematic-ness" of this eminently physical machine?
Load library(clpfd) for constraints and use library(plunit) for unit tests:
Put this into a file x.pl that you can load with [x] alias consult('x') or reload with make on the Prolog REPL:
:- use_module(library(clpfd)).
lijstSom([],0) :-
format("Hit case ([],0)\n"),!.
lijstSom([1],1) :-
format("Hit case ([1],1)\n"),!.
lijstSom([K,V|Rest],N) :-
format("Called with K=~w, V=~w, Rest=~w, N=~w\n", [K,V,Rest,N]),
K #= V+1,
T+K #= N,
T #> 0, V #> 0, % needed to avoid infinite descent
lijstSom([V|Rest],T).
:- begin_tests(listsom).
test("0 verify") :- lijstSom([],0).
test("1 verify") :- lijstSom([1],1).
test("3 verify") :- lijstSom([2,1],3).
test("6 verify") :- lijstSom([3,2,1],6).
test("0 construct") :- lijstSom(L,0) , L = [].
test("1 construct") :- lijstSom(L,1) , L = [1].
test("3 construct") :- lijstSom(L,3) , L = [2,1].
test("6 construct") :- lijstSom(L,6) , L = [3,2,1].
test("0 sum") :- lijstSom([],S) , S = 0.
test("1 sum") :- lijstSom([1],S) , S = 1.
test("3 sum") :- lijstSom([2,1],S) , S = 3.
test("6 sum") :- lijstSom([3,2,1],S) , S = 6.
test("1 partial") :- lijstSom([X],1) , X = 1.
test("3 partial") :- lijstSom([X,1],3) , X = 2.
test("6 partial") :- lijstSom([X,2,1],6) , X = 3.
test("1 extreme partial") :- lijstSom([X],S) , X = 1, S = 1.
test("3 extreme partial") :- lijstSom([X,1],S) , X = 2, S = 3.
test("6 extreme partial") :- lijstSom([X,2,1],S) , X = 3, S = 6.
test("6 partial list") :- lijstSom([X|L],6) , X = 3, L = [2,1].
% Important to test the NOPES
test("bad list", fail) :- lijstSom([3,1],_).
test("bad sum", fail) :- lijstSom([3,2,1],5).
test("reversed list", fail) :- lijstSom([1,2,3],6).
test("infinite descent from 2", fail) :- lijstSom(_,2).
test("infinite descent from 9", fail) :- lijstSom(_,9).
:- end_tests(listsom).
Then
?- run_tests(listsom).
% PL-Unit: listsom ...................... done
% All 22 tests passed
What would Dijkstra say? Yeah, he would probably bitch about something.
To split a number into digits in a given base, Julia has the digits() function:
julia> digits(36, base = 4)
3-element Array{Int64,1}:
0
1
2
What's the reverse operation? If you have an array of digits and the base, is there a built-in way to convert that to a number? I could print the array to a string and use parse(), but that sounds inefficient, and also wouldn't work for bases > 10.
The previous answers are correct, but there is also the matter of efficiency:
sum([x[k]*base^(k-1) for k=1:length(x)])
collects the numbers into an array before summing, which causes unnecessary allocations. Skip the brackets to get better performance:
sum(x[k]*base^(k-1) for k in 1:length(x))
This also allocates an array before summing: sum(d.*4 .^(0:(length(d)-1)))
If you really want good performance, though, write a loop and avoid repeated exponentiation:
function undigit(d; base=10)
s = zero(eltype(d))
mult = one(eltype(d))
for val in d
s += val * mult
mult *= base
end
return s
end
This has one extra unnecessary multiplication, you could try to figure out some way of skipping that. But the performance is 10-15x better than the other approaches in my tests, and has zero allocations.
Edit: There's actually a slight risk to the type handling above. If the input vector and base have different integer types, you can get a type instability. This code should behave better:
function undigits(d; base=10)
(s, b) = promote(zero(eltype(d)), base)
mult = one(s)
for val in d
s += val * mult
mult *= b
end
return s
end
The answer seems to be written directly within the documentation of digits:
help?> digits
search: digits digits! ndigits isdigit isxdigit disable_sigint
digits([T<:Integer], n::Integer; base::T = 10, pad::Integer = 1)
Return an array with element type T (default Int) of the digits of n in the given base,
optionally padded with zeros to a specified size. More significant digits are at higher
indices, such that n == sum([digits[k]*base^(k-1) for k=1:length(digits)]).
So for your case this will work:
julia> d = digits(36, base = 4);
julia> sum([d[k]*4^(k-1) for k=1:length(d)])
36
And the above code can be shortened with the dot operator:
julia> sum(d.*4 .^(0:(length(d)-1)))
36
Using foldr and muladd for maximum conciseness and efficiency
undigits(d; base = 10) = foldr((a, b) -> muladd(base, b, a), d, init=0)
I am brand new to prolog and I feel like there is a concept that I am failing to understand, which is preventing me from grasping the concept of recursion in prolog. I am trying to return S, which is the sum of the square of each digit, taken as a list from an integer that is entered by the user in a query. E.g The user enters 12345, I must return S = (1^2)+(2^2)+(3^2)+(4^2)+(5^2) = 55.
In my program below, I understand why the each segment of the calculation of S is printed multiple time as it is part of the recursive rule. However, I do not understand how I would be able to print S as the final result. I figured that I could set a variable = to the result from sos in the second rule and add it as a parameter for intToList but can't seem to figure this one out. The compiler warns that S is a singleton variable in the intToList rule.
sos([],0).
sos([H|T],S) :-
sos(T, S1),
S is (S1 + (H * H)),
write('S is: '),write(S),nl.
intToList(0,[]).
intToList(N,[H|T]) :-
N1 is floor(N/10),
H is N mod 10,
intToList(N1,T),
sos([H|T],S).
The issue with your original code is that you're trying to handle your call to sos/2 within your recursive clause for intToList/2. Break it out (and rename intToList/2 to something more meaningful):
sosDigits(Number, SoS) :-
number_digits(Number, Digits),
sos(Digits, SoS).
Here's your original sos/2 without the write, which seems to work fine:
sos([], 0).
sos([H|T], S) :-
sos(T, S1),
S is (S1 + (H * H)).
Or better, use an accumulator for tail recursion:
sos(Numbers, SoS) :-
sos(Numbers, 0, SoS).
sos([], SoS, SoS).
sos([X|Xs], A, SoS) :-
A1 is A + X*X,
sos(Xs, A1, SoS).
You can also implement sos/2 using maplist/3 and sumlist/2:
square(X, S) :- S is X * X.
sos(Numbers, SoS) :- maplist(square, Numbers, Squares), sumlist(Squares, SoS).
Your intToList/2 needs to be refactored using an accumulator to maintain correct digit order and to get rid of the call to sos/2. Renamed as explained above:
number_digits(Number, Digits) :-
number_digits(Number, [], Digits).
number_digits(Number, DigitsSoFar, [Number | DigitsSoFar]) :-
Number < 10.
number_digits(Number, DigitsSoFar, Digits) :-
Number >= 10,
NumberPrefix is Number div 10,
ThisDigit is Number mod 10,
number_digits(NumberPrefix, [ThisDigit | DigitsSoFar], Digits).
The above number_digits/2 also handles 0 correctly, so that number_digits(0, Digits) yields Digit = [0] rather than Digits = [].
You can rewrite the above implementation of number_digits/3 using the -> ; construct:
number_digits(Number, DigitsSoFar, Digits) :-
( Number < 10
-> Digits = [Number | DigitsSoFar]
; NumberPrefix is Number div 10,
ThisDigit is Number mod 10,
number_digits(NumberPrefix, [ThisDigit | DigitsSoFar], Digits)
).
Then it won't leave a choice point.
Try this:
sos([],Accumulator,Accumulator).
sos([H|T],Accumulator,Result_out) :-
Square is H * H,
Accumulator1 is Accumulator + Square,
sos(T,Accumulator1,Result_out).
int_to_list(N,R) :-
atom_chars(N,Digit_Chars),
int_to_list1(Digit_Chars,Digits),
sos(Digits,0,R).
int_to_list1([],[]).
int_to_list1([Digit_Char|Digit_Chars],[Digit|Digits]) :-
atom_number(Digit_Char,Digit),
int_to_list1(Digit_Chars,Digits).
For int_to_list I used atom_chars which is built-in e.g.
?- atom_chars(12345,R).
R = ['1', '2', '3', '4', '5'].
And then used a typical loop to convert each character to a number using atom_number e.g.
?- atom_number('2',R).
R = 2.
For sos I used an accumulator to accumulate the answer, and then once the list was empty moved the value in the accumulator to the result with
sos([],Accumulator,Accumulator).
Notice that there are to different variables for the accumulator e.g.
Accumulator1 is Accumulator + Square,
sos(T,Accumulator1,Result_out).
this is because in Prolog variables are immutable, so one can not keep assigning new values to the same variable.
Here are some example runs
?- int_to_list(1234,R).
R = 30.
?- int_to_list(12345,R).
R = 55.
?- int_to_list(123456,R).
R = 91.
If you have any questions just ask in the comments under this answer.