# RabbitFarm

### 2021-03-07

The Weekly Challenge 102 (Prolog Solutions)

*The examples used here are from the weekly challenge problem statement and demonstrate the working solution. Also, the challenge statements are given in a Perl context although for Prolog solutions some adjustments are made to account for the differing semantics between the two languages.*

## Part 1

*You are given a positive integer $N. Write a script to generate all Rare Numbers of size $N if any exist.*

### Solution

```
:-initialization(main).
perfect_square(N):-
A is floor(sqrt(N)),
N is A * A.
rare(N, N, Rares, Rares).
rare(Lower, Upper, RareAccum, Rares):-
number_codes(Lower, C),
reverse(C, CR),
number_codes(R1, CR),
X0 is Lower + R1,
X1 is Lower - R1,
perfect_square(X0),
perfect_square(X1),
Next is Lower + 1,
rare(Next, Upper, [Lower|RareAccum], Rares).
rare(Lower, Upper, RareAccum, Rares):-
Next is Lower + 1,
rare(Next, Upper, RareAccum, Rares).
rare_numbers(N, Rares):-
Lower is 10 ^ (N - 1),
Upper is (10 ^ N) - 1,
rare(Lower, Upper, [], Rares).
main:-
rare_numbers(2, Rares2),
write(Rares2), nl,
rare_numbers(6, Rares6),
write(Rares6), nl,
rare_numbers(9, Rares9),
write(Rares9), nl,
halt.
```

### Sample Run

```
$ gplc prolog/ch-1.p
$ prolog/ch-1
[65]
[621770]
[281089082]
```

### Notes

This is maybe straightforward bit of Prolog with arithmetic. One might say that I missed an opportunity to use `between/3`

but, in fact, I was happy to terminate the recursion of `rare/4`

by identifying the case where the first two arguments (`Lower`

and `Upper`

) are equal.

## Part 2

*You are given a positive integer $N. Write a script to produce hash counting string of that length.*

### Solution

```
:-initialization(main).
hcs(0, String, String).
hcs(1, StringAccum, String):-
hcs(0, [35|StringAccum], String).
hcs(N, StringAccum, String):-
number_codes(N, C),
append(C, "#", Accum),
length(Accum, L),
N0 is N - L,
append(Accum, StringAccum, StringAccum0),
hcs(N0, StringAccum0, String).
hash_counting_string(N, String):-
hcs(N, [], S),
atom_codes(String, S).
main:-
hash_counting_string(1, String1),
write(String1), nl,
hash_counting_string(2, String2),
write(String2), nl,
hash_counting_string(3, String3),
write(String3), nl,
hash_counting_string(10, String10),
write(String10), nl,
hash_counting_string(14, String14),
write(String14), nl,
halt.
```

### Sample Run

```
$ gplc ch-2.p
$ prolog/ch-2
#
2#
#3#
#3#5#7#10#
2#4#6#8#11#14#
```

### Notes

As mentioned elsewhere I first wrote a solution to this in Perl. This code, follows pretty directly from that, a clean bit of recursion. Probably we should prefer a more *prological* piece of code versus this which is more …what should we call it…*procursive*? That is, recursive fairly functional style code which does not make much use of the power of Prolog backtracking.

Time is running out to submit solutions for Weekly Challenge 102, but I can imagine a solution which uses DCGs perhaps? The hash counting string being assembled by processing a list of success numbers which in turn generate the appropriate characters. I’ll update this article if I get a chance.

## References

posted at: 16:27 by: Adam Russell | path: /prolog | permanent link to this entry

### 2021-02-21

The Weekly Challenge 100 (Prolog Solutions)

*The examples used here are from the weekly challenge problem statement and demonstrate the working solution. Also, the challenge statements are given in a Perl context although for Prolog solutions some adjustments are made to account for the differing semantics between the two languages.*

## Part 1

*You are given a time (12 hour / 24 hour). Write a one-liner to convert the given time from 12 hour format to 24 hour format and vice versa.*

### Solution

```
:-initialization(main).
hour_to_12(H12, H24):-
append(H, [58|R], H24),
number_codes(N0, H),
N is N0 - 12,
number_codes(N, C0),
flatten([C0, 58, R], C),
atom_codes(A, C),
H12 = A.
hour_to_24(H12, H24):-
append(H, [58|R], H12),
number_codes(N0, H),
N is N0 + 12,
number_codes(N, C0),
flatten([C0, 58, R], C),
atom_codes(A, C),
H24 = A.
twenty_four_hour(H12, H24):-
nonvar(H12),
hour_to_24(H12, H24).
twenty_four_hour(H12, H24):-
nonvar(H24),
hour_to_12(H12, H24).
main:-
twenty_four_hour("05:15 pm", HOUR_24),
write(HOUR_24), nl,
twenty_four_hour(HOUR_12, "17:15 pm"),
write(HOUR_12), nl,
halt.
```

### Sample Run

```
$ gplc ch-1.p
$ ch-1
17:15 pm
5:15 pm
```

### Notes

Keep in mind that any two points determine a line. Therefore to consider all possible non-trivial lines we need to review all triples of points.

In determining collinearity I calculate the area of a triangle using the triple of points. If the area is zero we know that all the points lay on the same line.

## Part 2

*You are given triangle array. Write a script to find the minimum path sum from top to bottom. When you are on index i on the current row then you may move to either index i or index i + 1 on the next row.*

### Solution

```
:-initialization(main).
minimum_sum(Triangle, Sum):-
minimum_sum(Triangle, 1, 0, Sum).
minimum_sum([H|[]], Index, PartialSum, Sum):-
nth(Index, H, N),
Sum is PartialSum + N.
minimum_sum([H0, H1|T], Index, PartialSum, Sum):-
nth(Index, H0, N0),
PartialSum0 is PartialSum + N0,
I0 is Index + 1,
nth(I0, H1, N1),
nth(Index, H1, N2),
N1 > N2,
minimum_sum([H1|T], Index, PartialSum0, Sum).
minimum_sum([H0, H1|T], Index, PartialSum, Sum):-
nth(Index, H0, N0),
PartialSum0 is PartialSum + N0,
I0 is Index + 1,
nth(I0, H1, N1),
nth(Index, H1, N2),
N1 =< N2,
minimum_sum([H1|T], I0, PartialSum0, Sum).
main:-
minimum_sum([[1], [2, 4], [6, 4, 9], [5, 1, 7, 2]], Sum0),
write(Sum0), nl,
minimum_sum([[3], [3, 1], [5, 2, 3], [4, 3, 1, 3]], Sum1),
write(Sum1), nl,
halt.
```

### Sample Run

```
$ gplc ch-2.p
$ prolog/ch-2
8
7
```

### Notes

This code is more *functional* than *logical*. Code is not written in a vacuum! The night before working this problem I was reading about Functional Programming in Java and it seems to have slightly warped my brain. Well, or at least I decided it would be fun to do things this way.

A more idiomatically Prolog solution would surely make use of Constraint Programming and just solve the more general case without the triangle restriction.

## References

posted at: 17:08 by: Adam Russell | path: /prolog | permanent link to this entry

### 2021-02-07

The Weekly Challenge 098 (Prolog Solutions)

*The examples used here are from the weekly challenge problem statement and demonstrate the working solution. Also, the challenge statements are given in a Perl context although for Prolog solutions some adjustments are made to account for the differing semantics between the two languages.*

## Part 1

*You are given file $FILE. Create subroutine readN($FILE, $number) returns the first n-characters and moves the pointer to the (n+1)th character.*

### Solution

```
:-dynamic(position/1).
:-initialization(main).
position(0).
read_n_chars(Stream, N, Chars):-
read_n_chars(Stream, N, [], Chars).
read_n_chars(_, 0, ByteAccum, Chars):-
atom_codes(Chars, ByteAccum).
read_n_chars(Stream, N, ByteAccum, Chars):-
\+ at_end_of_stream(Stream),
get_byte(Stream, C),
N0 is N - 1,
append(ByteAccum, [C], ByteAccum0),
read_n_chars(Stream, N0, ByteAccum0, Chars).
read_n_chars(Stream, _, ByteAccum, Chars):-
at_end_of_stream(Stream),
read_n_chars(Stream, 0, ByteAccum, Chars).
read_n(File, N, Chars):-
open(File, read, Stream,[type(binary),reposition(true)]),
position(Position),
seek(Stream, bof, Position, _),
read_n_chars(Stream, N, Chars),
X is N + Position,
retract(position(Position)),
asserta(position(X)),
close(Stream).
main:-
read_n('../ch-1.dat', 4, Chars0),
write(Chars0), nl,
read_n('../ch-1.dat', 4, Chars1),
write(Chars1), nl,
read_n('../ch-1.dat', 4, Chars2),
write(Chars2), nl,
halt.
```

### Sample Run

```
$ gplc ch-1.p
$ ./ch-1
1234
5678
90
```

### Notes

Clearly given the semantics of Prolog we need to adjust the challenge specification slightly. For the Prolog version we’ll write a predicate `read_n/3`

which will instantiate the third argument each time it is called in the way described, resuming the file reads after the last read position. Because we want to initialize the position to 0 to start and then update with `retract/1`

and `asserta/1`

we use the `dynamic/1`

specification to allow for this. If not specified dynamic we’d see an error such as `system_error(cannot_catch_throw(error(permission_error(modify,static_procedure,position/1),retract/1)))`

when trying to make alterations.

The last read position is stored as a Prolog fact after each read. Each `read_n/3`

call will use that last position to do a `seek/4`

and then read in the characters. While some Prologs will allow seeking within text files Gnu Prolog disallows this. This is easily worked around, however, by opening the file in binary mode, reading in the characters using `get_byte/2`

and then using `atom_codes/2`

to convert. To make this work we need to specify `Options`

of `[type(binary),reposition(true)]`

to `open/4`

.

## Part 2

*You are given a sorted array of distinct integers @N and a target `$N``. Write a script to return the index of the given target if found otherwise place the target in the sorted array and return the index.*

### Solution

```
:-initialization(main).
needle_haystack([H|T], N, Index):-
((N < H, Index is 0);
(last(T, Last), N > Last,
length([H|T], Length), Index is Length)).
needle_haystack([H|T], N, Index):-
needle_haystack([H|T], N, 0, Index).
needle_haystack([], _, Index, Index).
needle_haystack([_|[]], _, Index, Index).
needle_haystack([H0, _|_], N, Counter, Index):-
H0 == N,
Index is Counter.
needle_haystack([_, H1|_], N, Counter, Index):-
H1 == N,
Index is Counter + 1.
needle_haystack([H0, H1|T], N, Counter, Index):-
H0 \== N,
H1 \== N,
\+ between(H0, H1, N),
C is Counter + 2,
needle_haystack(T, N, C, Index).
needle_haystack([H0, H1|_], N, Counter, Index):-
H0 \== N,
H1 \== N,
between(H0, H1, N),
Index is Counter + 1.
main:-
needle_haystack([1, 2, 3, 4], 3, Index0),
write(Index0), nl,
needle_haystack([1, 3, 5, 7], 6, Index1),
write(Index1), nl,
needle_haystack([12, 14, 16, 18], 10, Index2),
write(Index2), nl,
needle_haystack([11, 13, 15, 17], 19, Index3),
write(Index3), nl,
halt.
```

### Sample Run

```
$ gplc prolog/ch-2.p
$ ./ch-2
2
3
0
4
```

### Notes

This is a somewhat convoluted sounding problem at first but it actually ends up being straight forward to solve, albeit slightly tedious. The code is largely an examination of the possible cases where `N`

may be. If it would occur before the head of the list or after the last element we are done immediately. Otherwise we evaluate the list in search of `N`

, returning the index if we find it. If we find two successive elements in which `N`

would be between, but is not in the list, we have identified the point in which `N`

would be inserted.

While the challenge statement indicates `N`

should be inserted we only really seem to care about the index and so I don’t actually provide an updated list. To do so we could add an additional argument to `needle_haystack/3`

which would be instantiated to an updated list.

## References

posted at: 14:12 by: Adam Russell | path: /prolog | permanent link to this entry

### 2021-01-31

The Weekly Challenge 097 (Prolog Solutions)

*The examples used here are from the weekly challenge problem statement and demonstrate the working solution.*

## Part 1

*You are given string $S containing alphabets A..Z only and a number $N. Write a script to encrypt the given string $S using Caesar Cipher with left shift of size $N.*

### Solution

```
:-initialization(main).
caesar([], [], _).
caesar([H|T], [C|Cypher], N):-
C is H - N,
C >= 65,
caesar(T, Cypher, N).
caesar([H|T], [C|Cypher], N):-
C0 is H - N,
C0 < 65,
C is C0 + 26,
caesar(T, Cypher, N).
main:-
caesar("ABCDEFGHIJKLMNOPQRSTUVWXYZ", C, 3),
atom_codes(CypherText, C),
write(CypherText), nl,
halt.
```

### Sample Run

```
$ gplc ch-1.p
$ ./ch-1
XYZABCDEFGHIJKLMNOPQRSTUVW
```

## Part 2

*You are given a binary string $B and an integer $S. Write a script to split the binary string $B of size $S and then find the minimum number of flips required to make it all the same.*

### Solution

```
:-initialization(main).
substrings(BinaryString, N, SubStrings):-
substrings(BinaryString, N, [], SubStrings).
substrings([], _, SubStrings, SubStrings).
substrings(BinaryString, N, SubStringAccum, SubStrings):-
length(L, N),
append(L, X, BinaryString),
substrings(X, N, [L|SubStringAccum], SubStrings).
count_flips(B, [H|T], Flips):-
count_flips(B, [H|T], 0, Flips).
count_flips(_, [], Flips, Flips).
count_flips(B, [H|T], FlipsSum, Flips):-
number_codes(B0, B),
number_codes(H0, H),
X is xor(B0, H0),
Flips0 is X + FlipsSum,
count_flips(B, T, Flips0, Flips).
min_flips(SubStrings, MinFlips):-
min_flips(SubStrings, [], Flips),
sort(Flips,[MinFlips|_]).
min_flips([_], Flips, Flips).
min_flips([H|T], FlipsAccum, Flips):-
count_flips(H, T, Flips0),
min_flips(T, [Flips0|FlipsAccum], Flips).
main:-
substrings("101100101", 3, SubStrings),
min_flips(SubStrings, Flips),
write(Flips),nl,
halt.
```

### Sample Run

```
$ gplc prolog/ch-2.p
$ ./ch-2
1
```

## References

posted at: 18:48 by: Adam Russell | path: /prolog | permanent link to this entry

### 2021-01-18

Making Prolog Code Bimodal

## Palindromic Numbers

Suppose we are asked the following, as was the case for The Weekly Challenge 095.

*Write a script to figure out if the given number is Palindrome. Print 1 if true otherwise 0.*

Here is my original solution to this.

### Solution

```
:-initialization(main).
palindrome_even([]).
palindrome_even([H|T]):-
last(T, Last),
H == Last,
append(L, [Last], T),
palindrome_even(L).
palindrome_odd([_|[]]).
palindrome_odd([H|T]):-
last(T, Last),
H == Last,
append(L, [Last], T),
palindrome_odd(L).
palindrome(N):-
N > 0,
number_chars(N, C),
length(C, Length),
M is Length mod 2,
M == 0,
palindrome_even(C).
palindrome(N):-
N > 0,
number_chars(N, C),
length(C, Length),
M is Length mod 2,
M == 1,
palindrome_odd(C).
is_palindrome(N):-
palindrome(N),
write(1), nl.
is_palindrome(N):-
\+ palindrome(N),
write(0), nl.
main:-
is_palindrome(1221),
is_palindrome(-101),
is_palindrome(90),
halt.
```

### Sample Run

```
$ gplc prolog/ch-1.p
$ prolog/ch-1
1
0
0
```

### Notes

I wrote a short description of this code before.

To repeat a little here: this is a Prolog version of Perl code I wrote for the same problem. The basic approach is roughly the same as the Perl implementation: starting at both ends of the number work inwards comparing one pair of digits at a time. Here we define all negative numbers as being non-palindromic due to the inherent asymmetry of the ‘-’.

It is slightly odd in Prolog to print out a 0 or 1 as asked in this problem statement. In languages like Perl doing so is really just as simple as printing out a boolean value returned from some comparison or function call. Although odd it is not unheard of, however, and the best advice I have seen on the matter is to define a small predicate to wrap the predicate(s) doing the real work and print whatever is needed. That job here is done by `is_palindrome/1`

.

Let’s ignore `is_palindrome/1`

, though, and consider something else.

This code works fine, as is shown, but it is unsatisfying from the perspective in that it only works one way. If we have defined the validation of palindromic numbers correctly then in fact Prolog should also be able to generate palindromic numbers as well! That is,

a *bimodal predicate* that used in one mode will validate palindromic numbers and in another mode create them.

Trying the existing code out with an uninstantiated variable this is the result

```
?- palindrome(N).
uncaught exception: error(instantiation_error,(>)/2)
```

Of course it does not take a Prolog genius to see pretty quickly why this does not work. A couple of things that jump out right away:

- We are assuming that
`palindrome/1`

is being given a valid integer `N > 0`

requires an instantiated number variable`N`

, as shown in the error message.- The list of digits we evaluate to be a palindrome is created by
`number_chars/2`

which, for our purposes, also requires an instantiated number variable.

To adjust this code to work both ways need to handle whether `N`

is in fact instantiated or not. But here we need to be careful, we need to create one in such a way that when a non-palindromic number is created a new one is created upon backtracking. We want to be as general as possible.

Also, we already have two instances of `palindrome/1`

predicates to handle the two cases of even/odd number of digits. Let’s take care to not write more code than necessary when modifying this code.

So, taking all this into account, what is to be done? Well, it turns out we can make this code work both ways by adding adding two lines (and removing 1 existing line) from `palindrome/1`

.

The changes are:

- Check to see if N is between 1 and the largest possible integer available using
`between/3`

. Here I am using Gnu Prolog on a 32-bit platform (NetBSD/macppc) and this value is`268435455`

. - When the third argument to
`between/3`

is uninstantiated it will be instantiated to be an integer in the range given by the first two arguments. - The check if
`N`

is positive is no longer necessary and can be removed.

The updated code looks like this.

```
palindrome(N):-
current_prolog_flag(max_integer, MAX_INTEGER),
between(1, MAX_INTEGER, N),
number_chars(N, C),
length(C, Length),
M is Length mod 2,
M == 0,
palindrome_even(C).
```

From the Gnu Prolog top level we can see this in action. The output is truncated to just show some of the solutions generated upon backtracking.

```
| ?- palindrome(1200021).
true ?
yes
| ?- palindrome(X).
X = 11 ? ;
X = 22 ? ;
.
.
.
```

How many palindromic numbers are there between `1`

and `MAX_INTEGER`

?

```
| ?- findall(X, palindrome(X), Xs), length(Xs, L).
L = 36842.
```

36842 palindromic numbers is about 0.01% of all integers in this range.

The full code for this is below. I’ve left the `main/0`

predicate alone as it demonstrates what was necessary for the original problem. Output showing the *bimodal* functionality was run in the Gnu Prolog top level. If this code interests you the `palindrome/1`

predicate can be used directly as shown.

```
:-initialization(main).
palindrome_even([]).
palindrome_even([H|T]):-
last(T, Last),
H == Last,
append(L, [Last], T),
palindrome_even(L).
palindrome_odd([_|[]]).
palindrome_odd([H|T]):-
last(T, Last),
H == Last,
append(L, [Last], T),
palindrome_odd(L).
palindrome(N):-
current_prolog_flag(max_integer, MAX_INTEGER),
between(1, MAX_INTEGER, N),
number_chars(N, C),
length(C, Length),
M is Length mod 2,
M == 0,
palindrome_even(C).
palindrome(N):-
current_prolog_flag(max_integer, MAX_INTEGER),
between(1, MAX_INTEGER, N),
number_chars(N, C),
length(C, Length),
M is Length mod 2,
M == 1,
palindrome_odd(C).
is_palindrome(N):-
palindrome(N),
write(1), nl.
is_palindrome(N):-
\+ palindrome(N),
write(0), nl.
main:-
is_palindrome(1221),
is_palindrome(-101),
is_palindrome(90).
```

## References

posted at: 17:46 by: Adam Russell | path: /prolog | permanent link to this entry

### 2021-01-17

The Weekly Challenge 095 (Prolog Solutions)

*The examples used here are from the weekly challenge problem statement and demonstrate the working solution.*

## Part 1

*Write a script to figure out if the given number is Palindrome. Print 1 if true otherwise 0.*

### Solution

```
:-initialization(main).
palindrome_even([]).
palindrome_even([H|T]):-
last(T, Last),
H == Last,
append(L, [Last], T),
palindrome_even(L).
palindrome_odd([_|[]]).
palindrome_odd([H|T]):-
last(T, Last),
H == Last,
append(L, [Last], T),
palindrome_odd(L).
palindrome(N):-
N > 0,
number_chars(N, C),
length(C, Length),
M is Length mod 2,
M == 0,
palindrome_even(C).
palindrome(N):-
N > 0,
number_chars(N, C),
length(C, Length),
M is Length mod 2,
M == 1,
palindrome_odd(C).
is_palindrome(N):-
palindrome(N),
write(1), nl.
is_palindrome(N):-
\+ palindrome(N),
write(0), nl.
main:-
is_palindrome(1221),
is_palindrome(-101),
is_palindrome(90),
halt.
```

### Sample Run

```
$ gplc prolog/ch-1.p
$ prolog/ch-1
1
0
0
```

### Notes

This is a Prolog version of Perl code I wrote for the same problem. The basic approach is roughly the same as the Perl implementation: starting at both ends of the number work inwards comparing one pair of digits at a time. Here we define all negative numbers as being non-palindromic due to the inherent asymmetry of the ‘-’.

It is slightly odd in Prolog to print out a 0 or 1 as asked in this problem statement. In languages like Perl doing so is really just as simple as printing out a boolean value returned from some comparison or function call. Although odd it is not unheard of, however, and the best advice I have seen on the matter is to define a small predicate to wrap the predicate(s) doing the real work and print whatever is needed. That job here is done by `is_palindrome/1`

## Part 2

*Demonstrate Stack operations.*

### Solution

```
:-initialization(main).
new_stack([]).
push(Stack, Value, [Value|Stack]).
pop([H|T], H, T).
top([H|_], H).
min(Stack, Min):-
min_list(Stack, Min).
main:-
new_stack(Stack),
push(Stack, 2, NewStack0),
push(NewStack0, -1, NewStack1),
push(NewStack1, 0, NewStack2),
pop(NewStack2, Top0, NewStack3),
write(Top0), nl,
top(NewStack3, Top1),
write(Top1), nl,
push(NewStack3, 0, NewStack4),
min(NewStack4, Min),
write(Min), nl,
halt.
```

### Sample Run

```
$ gplc prolog/ch-2.p
$ prolog/ch-2
0
-1
-1
```

### Notes

Just writing this code made me feel like I was committing a crime against Prolog! Implementing this sort of data structure in Prolog is really just writing some small predicates which wrap otherwise ordinary list operations. Then again, implementing Stacks in Prolog is not entirely unheard of. This sort of thing is done as an exercise when studying data structures. The excellent text by Luger and Stubblefield describes a Stack implementation, among other Abstract Data Types, in Prolog.

## References

AI Algorithms, Data Structures, and Idioms in Prolog, Lisp, and Java

posted at: 16:03 by: Adam Russell | path: /prolog | permanent link to this entry

### 2021-01-10

The Weekly Challenge 094 (Prolog Solutions)

*The examples used here are from the weekly challenge problem statement and demonstrate the working solution.*

## Part 1

*You are given a list of strings S. Write a script to group Anagrams together in any random order.*

### Solution

```
:-initialization(main).
letter_factor(e, 2).
letter_factor(t, 3).
letter_factor(a, 5).
letter_factor(o, 7).
letter_factor(i, 11).
letter_factor(n, 13).
letter_factor(s, 17).
letter_factor(h, 19).
letter_factor(r, 23).
letter_factor(d, 29).
letter_factor(l, 31).
letter_factor(c, 37).
letter_factor(u, 41).
letter_factor(m, 43).
letter_factor(w, 47).
letter_factor(f, 53).
letter_factor(g, 59).
letter_factor(y, 61).
letter_factor(p, 67).
letter_factor(b, 71).
letter_factor(v, 73).
letter_factor(k, 79).
letter_factor(j, 83).
letter_factor(x, 89).
letter_factor(q, 97).
letter_factor(z, 101).
chars_product([], 1).
chars_product([H|T], Product):-
letter_factor(H, Factor),
chars_product(T, Product0),
Product is Factor * Product0.
word_product(Word, Product):-
atom_chars(Word, Chars),
chars_product(Chars, Product).
organize([]):-
findall(Words, bagof(Word, word_product(Word-_), Words), WordsList),
write(WordsList).
organize([H|T]):-
word_product(H, P),
assertz(word_product(H-P)),
organize(T).
main:-
Anagrams = [opt, bat, saw, tab, pot, top, was],
organize(Anagrams), nl,
halt.
```

### Sample Run

```
$ gplc ch-1.p
$ ./ch-1
[[bat,tab],[opt,pot,top],[saw,was]]
```

### Notes

This is a Prolog version of Perl code I wrote for the same problem. In translating the approach of using the Fundamental Theorem of Arithmetic I tried to make sure the result was idiomatic Prolog as much as I could. I think I pulled that off!?!? In Prolog there is sometimes an over temptation to end up writing a bunch of mostly functional code. That is, code that could just as easily be written in, say, Haskell or ML. The `organize/1`

predicate is pretty much that sort of functional code but it is very short and wraps Prolog concepts such as `assertz/1`

, `findall/3`

, and `bagof/3`

.

I am using the same mathematical trick that I have used for anagrams in the past, starting with Challenge 005. The By the

**Fundamental Theorem of Arithmetic***every integer greater than 1 is either a prime number itself or can be represented as the unique product of prime numbers.*We use that to our advantage by having a prime number associated with each letter. Each word is a product of these numbers and words with the same product are anagrams.In this way we

`assert/1`

Word-Product pairs for all the given anagrams and once done use`findall/3`

with`bagof/3`

to collect the solutions for printing as desired. This method of collecting solutions is given a nice description here.The choice of letters and prime numbers is based on the

*Lewand Ordering*and it isn’t at all necessary (it stems from an early, unnecessary, design decision) but it does little harm so I left it in anyway.

## Part 2

*You are given a binary tree. Write a script to represent the given binary tree as an object and flatten it to a linked list object. Finally, print the linked list object.*

### Solution

```
:-initialization(main).
dfs(Node, Graph, [Node|Path]):-
dfs(Node, Graph, Path, []).
dfs(_, _, [], _).
dfs(Node, Graph, [AdjNode|Path], Seen) :-
member(r(Node, Adjacent), Graph),
member(AdjNode, Adjacent),
\+ memberchk(AdjNode, Seen),
dfs(AdjNode, Graph, Path, [Node|Seen]).
unseen_nodes(Nodes, NodeList, Unseen):-
unseen_nodes(Nodes, NodeList, [], Unseen).
unseen_nodes([], _, Unseen, Unseen).
unseen_nodes([H|T], NodeList, UnseenAccum, Unseen):-
\+ memberchk(H, NodeList),
unseen_nodes(T, NodeList, [H|UnseenAccum], Unseen).
unseen_nodes([H|T], NodeList, UnseenAccum, Unseen):-
memberchk(H, NodeList),
unseen_nodes(T, NodeList, UnseenAccum, Unseen).
paths_list(Paths, List):-
paths_list(Paths, [], List).
paths_list([], List, List).
paths_list([H|T], ListAccum, List):-
unseen_nodes(H, ListAccum, Unseen),
append(ListAccum, Unseen, ListAccum0),
paths_list(T, ListAccum0, List).
print_list([H|[]]):-
format("~d~n", [H]).
print_list([H|T]):-
format("~d -> ", [H]),
print_list(T).
main:-
findall(Path, dfs(1,[r(1,[2]),r(1,[3]),r(2,[4,5]),r(5,[6,7])], Path), Paths),
paths_list(Paths, List),
print_list(List), halt.
```

### Sample Run

```
$ gplc ch-2.p
-bash-5.0$ ./ch-2
1 -> 2 -> 4 -> 5 -> 6 -> 7 -> 3
```

### Notes

Some of this code is re-used from last week

The idea of a (singly) Linked List in Prolog is fundamental. We might consider any ordinary list in this way since Prolog allows direct access to the head and tail of

lists. This solution, therefore ends up being simpler than even the Perl solution to the same problem.The code above simply does a DFS of the given tree and then adds new nodes to a list in the order in which they are seen.

## References

posted at: 12:09 by: Adam Russell | path: /prolog | permanent link to this entry

### 2021-01-03

The Weekly Challenge 093 (Prolog Solutions)

## Part 1

*You are given set of co-ordinates @N. Write a script to count maximum points on a straight line when given co-ordinates plotted on 2-d plane.*

### Solution

```
:-initialization(main).
triangle_area(Points, Area):-
[[X1, Y1], [X2, Y2], [X3, Y3]] = Points,
Area is (X1 * (Y2 - Y3)) + (X2 * (Y3 - Y1)) + (X3 * (Y1 - Y2)).
collinear_points(Points, CollinearPoints):-
member(A, Points),
member(B, Points),
member(C, Points),
A \== B, A \== C, B \== C,
triangle_area([A, B, C], Area),
Area == 0,
CollinearPoints = [A, B, C].
main:-
N = [[5,3], [1,1], [2,2], [3,1], [1,3]],
collinear_points(N, CollinearPoints),
write(CollinearPoints), nl,
halt.
```

### Sample Run

```
$ gplc ch-1.p
$ ch-1
[[2,2],[3,1],[1,3]]
```

### Notes

Keep in mind that any two points determine a line. Therefore to consider all possible non-trivial lines we need to review all triples of points.

In determining collinearity I calculate the area of a triangle using the triple of points. If the area is zero we know that all the points lay on the same line.

## Part 2

*You are given a binary tree containing only the numbers 0-9. Write a script to sum all possible paths from root to leaf.*

### Solution

```
:-initialization(main).
dfs(Node, Graph, [Node|Path]):-
dfs(Node, Graph, Path, []).
dfs(_, _, [], _).
dfs(Node, Graph, [AdjNode|Path], Seen) :-
member(r(Node, Adjacent), Graph),
member(AdjNode, Adjacent),
\+ memberchk(AdjNode, Seen),
dfs(AdjNode, Graph, Path, [Node|Seen]).
sum_paths(Paths, Sum):-
sum_paths(Paths, 0, Sum).
sum_paths([], Sum, Sum).
sum_paths([H|T], PartialSum, Sum):-
sum_list(H, ListSum),
S is PartialSum + ListSum,
sum_paths(T, S, Sum).
path_lengths([], _).
path_lengths([H|T], [L|Lengths]):-
length(H, L),
path_lengths(T, Lengths).
partial_path(_, _, []).
partial_path(Path, MaxPathLength, [H|T]):-
length(Path, PathLength),
length(H, HLength),
(PathLength < MaxPathLength ; (subtract(Path, H, Remaining), length(Remaining, 0))),
partial_path(Path, MaxPathLength, T).
partial_path(Path, MaxPathLength, [H|_]):-
length(Path, PathLength),
length(H, HLength),
PathLength =< MaxPathLength,
subtract(Path, H, Remaining),
\+ length(Remaining, 0),
fail.
complete_paths(Paths, CompletePaths):-
path_lengths(Paths, PathLengths),
max_list(PathLengths, MaxPathLength),
complete_paths(Paths, Paths, MaxPathLength, [], CompletePaths).
complete_paths([], _, _, CompletePaths, CompletePaths).
complete_paths([H|T], Paths, MaxPathLength, CompletePathsAccum, CompletePaths):-
\+ partial_path(H, MaxPathLength, Paths),
complete_paths(T, Paths, MaxPathLength, [H|CompletePathsAccum], CompletePaths).
complete_paths([H|T], Paths, MaxPathLength, CompletePathsAccum, CompletePaths):-
partial_path(H, MaxPathLength, Paths),
complete_paths(T, Paths, MaxPathLength, CompletePathsAccum, CompletePaths).
main:-
findall(Path0, dfs(1,[r(1,[2]),r(2,[3,4])], Path0), Paths0),
complete_paths(Paths0, CompletePaths0),
sum_paths(CompletePaths0, Paths0Sum),
write(Paths0Sum), nl,
findall(Path1, dfs(1,[r(1,[2,3]), r(3,[5,6]), r(2,[4])], Path1), Paths1),
complete_paths(Paths1, CompletePaths1),
sum_paths(CompletePaths1, Paths1Sum),
write(Paths1Sum), nl, halt.
```

### Sample Run

```
$ gplc ch-2.p
$ prolog/ch-2
13
26
```

### Notes

The depth first search is pretty idiomatic Prolog and the `dfs/3`

code was something I grabbed from SO (see references). That code is straightforward enough, it finds all paths in a depth first manner. We are only concerned with the complete *root to leaf* paths and so a bit of effort goes into filtering out the partial paths. Once that is done the paths are summed and we are done.

## References

posted at: 16:36 by: Adam Russell | path: /prolog | permanent link to this entry

### 2020-12-24

# Advent of Code 2020

## AoC 2020 in Prolog

I had never heard of Advent Of Code before this year although it seems to have been around for six years running now! On Twitter I noticed some related posts about a week into this year’s event. After checking out what this was all about I decided this would be a fun way to continue honing my Prolog skills. I have been making a serious attempt to master Prolog and part of that has been doing a variety of programming challenges to get experience with different capabilities and nuances of the language.

Between the late start, wrapping up work projects before the end of year, and personal tasks related to the upcoming holidays (such as they will be this year) I got about halfway through the tasks. Now that I am aware that this event exists I’ll make sure to start off on day one next year!

Below are links to my solutions. The problem statements are all fairly lengthy so refer to the AoC 2020 website for details on what these are supposed to be doing.

- The solution to Part 2 should finish running sometime in late 2021. My initial approach was to create a tree of all possible adapters and then just count the leaves. It runs in just milliseconds for test cases but for the data set given … Yikes! … the run time will be very long. This is what happens when you write code too late at night! After sleeping on it I realized that something along the lines of using memoization to count and sum all the branches of the adapter tree should run reasonably quickly. However, I was using a slightly older version of SWI-Prolog that seemed to have an incomplete tabling implementation so I had to upgrade, which is easy enough I suppose, but after doing so I just decided to move on to Day 11 since I already had a late start as it was.

2020 Advent of Code Day 11 (Part 1)

2020 Advent of Code Day 11 (Part 2)

2020 Advent of Code Day 12 (Part 1)

2020 Advent of Code Day 12 (Part 2)

### Notes

I kept the solutions about as “vanilla” (i.e. no libraries) as can be. As a result all the solutions have some boilerplate code for reading and preparing the input data. Each of the solutions are otherwise fairly compact, a testament to the power of Prolog.

## References

posted at: 15:35 by: Adam Russell | path: /prolog | permanent link to this entry

### 2020-12-13

The Weekly Challenge 090 (Prolog Solutions)

## Part 1

*Write a script to print the nucleiobase count in the given DNA sequence. Also print the complementary sequence where Thymine (T) on one strand is always facing an adenine (A) and vice versa; guanine (G) is always facing a cytosine (C) and vice versa.*

### Solution

```
:- initialization(main).
nucleotide_pair('A', 'T').
nucleotide_pair('C', 'G').
compliment([H|T], [Compliment|RestOfCompliment]):-
var(Compliment),
atom_chars(A, [H]),
(nucleotide_pair(A, X); nucleotide_pair(X, A)),
atom_chars(X, [Compliment]),
compliment(T, RestOfCompliment).
compliment([H|T], [Compliment|RestOfCompliment]):-
(nucleotide_pair(A, Compliment); nucleotide_pair(Compliment, A)),
atom_chars(A, [H]),
compliment(T, RestOfCompliment).
compliment([], _).
compliment(_, []).
main:-
Sequence = 'GTAAACCCCTTTTCATTTAGACAGATCGACTCCTTATCCATTCTCAGAGATGTGTTGCTGGTCGCCG',
atom_chars(Sequence, SequenceChars),
length(SequenceChars, SequenceLength),
format("Sequence length is ~d.~n", [SequenceLength]),
length(Compliment, SequenceLength),
compliment(SequenceChars, Compliment),
atom_chars(ACompliment, Compliment),
compliment(OriginalSequence, Compliment),
atom_chars(AOriginalSequence, OriginalSequence),
format("Original sequence is ~a.~n", [AOriginalSequence]),
format("Complimentary sequence is ~a.~n", [ACompliment]),
halt.
```

### Sample Run

```
$ gplc ch-1.p
$ ch-1
Sequence length is 67.
Original sequence is GTAAACCCCTTTTCATTTAGACAGATCGACTCCTTATCCATTCTCAGAGATGTGTTGCTGGTCGCCG.
Complimentary sequence is CATTTGGGGAAAAGTAAATCTGTCTAGCTGAGGAATAGGTAAGAGTCTCTACACAACGACCAGCGGC.
```

### Notes

- The predicate
`compliment/2`

will work with either of the arguments instantiated. In other words it is a*bimodal predicate*.This is maybe a little silly in this case in particular because of the nature of the mappings between base pairs it is not necessary. Still, getting predicates to be bimodal requires some care and skill and I thought I’d take a crack at it here. - I am using Gnu Prolog here.
`gplc`

is the Gnu Prolog compiler. I love how it compiles Prolog to a single native executable. Beautiful!

## Part 2

*You are given two positive numbers $A and $B. Write a script to demonstrate Ethiopian Multiplication using the given numbers.*

### Solution

```
:- initialization(main).
ethiopean_multiplication(Operands, Product):-
ethiopean_multiplication(Operands, [], Product).
ethiopean_multiplication([1, _], IntermediateTerms, Product):-
sum_list(IntermediateTerms, Product).
ethiopean_multiplication(Operands, IntermediateTerms, Product):-
[A0, B0] = Operands,
A is A0 div 2,
B is B0 * 2,
M is A mod 2,
M == 1,
ethiopean_multiplication([A, B], [B|IntermediateTerms], Product).
ethiopean_multiplication(Operands, IntermediateTerms, Product):-
[A0, B0] = Operands,
A is A0 div 2,
B is B0 * 2,
M is A mod 2,
M == 0,
ethiopean_multiplication([A, B], IntermediateTerms, Product).
main:-
[A, B] = [14, 12],
ethiopean_multiplication([A, B], Product),
format("Product of ~d x ~d (via Ethiopean Multiplication) is ~d.~n", [A, B, Product]),
halt.
```

### Sample Run

```
$ gplc ch-2.p
$ ch-2
Product of 14 x 12 (via Ethiopean Multiplication) is 168.
```

### Notes

I think this is a fairly straightforward recursive Prolog solution. At each recursive step the two elements are either halved or doubled until the base step is hit and then the accumulated terms (the even elements from only the odd/even pairs) are summed.

## References

posted at: 18:34 by: Adam Russell | path: /prolog | permanent link to this entry

### 2020-12-06

Perl Weekly Challenge 089 (Prolog solutions)

## Part 1

*You are given a positive integer $N. Write a script to sum GCD of all possible unique pairs between 1 and $N.*

### Solution

```
range_list(0, [0]).
range_list(N, List):-
range_list(N, [], List).
range_list(0, List, List).
range_list(N, ListAccum, List):-
N0 is N - 1,
range_list(N0, [N|ListAccum], List).
unique_pairs(List, Pairs):-
unique_pairs(List, List, [], Pairs).
unique_pairs([], [], Pairs, Pairs).
unique_pairs([_|T0], [], PairsAccum, Pairs):-
unique_pairs(T0, T0, PairsAccum, Pairs).
unique_pairs([H0|T0], [H1|T1], PairsAccum, Pairs):-
\+ member([H0, H1], PairsAccum),
\+ member([H1, H0], PairsAccum),
H0 \= H1,
unique_pairs([H0|T0], T1, [[H0, H1]|PairsAccum], Pairs).
unique_pairs([H0|T0], [_|T1], PairsAccum, Pairs):-
unique_pairs([H0|T0], T1, PairsAccum, Pairs).
write_gcd_pairs(Pairs):-
write_gcd_pairs(Pairs, 0).
write_gcd_pairs([[I,J]|[]], Sum):-
Sum0 is Sum + gcd(I, J),
format("gcd(~d, ~d) = ~d ~n", [I, J, Sum0]).
write_gcd_pairs([[I,J]|T], Sum):-
format("gcd(~d, ~d) + ", [I, J]),
Sum0 is Sum + gcd(I, J),
write_gcd_pairs(T, Sum0).
main:-
range_list(4, L),
unique_pairs(L, Pairs),
write_gcd_pairs(Pairs),
halt.
```

### Sample Run

```
$ gprolog
GNU Prolog 1.4.5 (32 bits)
Compiled Dec 3 2020, 00:37:14 with gcc
By Daniel Diaz
Copyright (C) 1999-2020 Daniel Diaz
| ?- ['prolog/ch-1.p'].
compiling /home/adamcrussell/perlweeklychallenge-club/challenge-089/adam-russell/prolog/ch-1.p for byte code...
/home/adamcrussell/perlweeklychallenge-club/challenge-089/adam-russell/prolog/ch-1.p compiled, 36 lines read - 6945 bytes written, 52 ms
(26 ms) yes
| ?- main.
gcd(3, 4) + gcd(2, 4) + gcd(2, 3) + gcd(1, 4) + gcd(1, 3) + gcd(1, 2) = 7
```

### Notes

This mostly follows the approach of the Perl solution for this problem. The problem is broken down into the following pieces

- Generate all pairs of numbers between 1 and N. This is done here with
`range_list/2`

and`unique_pairs/2`

. - The pairs then have their GCDs computed, summed, and printed. Here this is all done via
`write_gcd_pairs/1`

.

## Part 2

*You are given m x n matrix of positive integers. Write a script to print spiral matrix as a list.*

### Solution

```
all_unique(_, []).
all_unique(L, [V|T]) :-
fd_exactly(1, L, V),
all_unique(L, T).
pwc_matrix(R1,R2,R3) :-
R1 = [A, B, C],
R2 = [D, E, F],
R3 = [G, H, I],
/* each element is a number from 1 to 9 */
fd_domain([A, B, C], 1, 9),
fd_domain([D, E, F], 1, 9),
fd_domain([G, H, I], 1, 9),
/* ensure each element is unique */
all_unique([A, B, C, D, E, F, G, H, I], [1, 2, 3, 4, 5, 6, 7, 8, 9]),
/* row constraints */
A + B + C #= 15,
D + E + F #= 15,
G + H + I #= 15,
/* column constraints */
A + D + G #= 15,
B + E + H #= 15,
C + F + I #= 15,
/* diagonal constraints */
A + E + I #= 15,
C + E + G #= 15,
/* label all variables to instantiate one solution */
fd_labeling([A, B, C, D, E, F, G, H, I]).
write_rows([]).
write_rows([H|T]):-
format('[ ~d ~d ~d ]~n', H),
write_rows(T).
write_solutions([]).
write_solutions([H|T]):-
write_rows(H),
nl,
write_solutions(T).
main :-
findall([R1,R2,R3], pwc_matrix(R1,R2,R3), Solutions),
write_solutions(Solutions).
```

### Sample Run

```
$ gprolog
GNU Prolog 1.4.5 (32 bits)
Compiled Dec 3 2020, 00:37:14 with gcc
By Daniel Diaz
Copyright (C) 1999-2020 Daniel Diaz
| ?- ['prolog/ch-2.p'].
compiling /home/adamcrussell/perlweeklychallenge-club/challenge-089/adam-russell/prolog/ch-2.p for byte code...
/home/adamcrussell/perlweeklychallenge-club/challenge-089/adam-russell/prolog/ch-2.p compiled, 42 lines read - 6581 bytes written, 50 ms
(31 ms) yes
| ?- main.
[ 2 7 6 ]
[ 9 5 1 ]
[ 4 3 8 ]
[ 2 9 4 ]
[ 7 5 3 ]
[ 6 1 8 ]
[ 4 3 8 ]
[ 9 5 1 ]
[ 2 7 6 ]
[ 4 9 2 ]
[ 3 5 7 ]
[ 8 1 6 ]
[ 6 1 8 ]
[ 7 5 3 ]
[ 2 9 4 ]
[ 6 7 2 ]
[ 1 5 9 ]
[ 8 3 4 ]
[ 8 1 6 ]
[ 3 5 7 ]
[ 4 9 2 ]
[ 8 3 4 ]
[ 1 5 9 ]
[ 6 7 2 ]
(5 ms) yes
```

### Notes

This is really not too tremendously different than a Sudoku, which just came up a couple of weeks ago. Obviously the constraints are different but the general approach is very similar. Here I am using Gnu Prolog’s finite domain (FD) constraint solver functions. In the past I primarily used SWI-Prolog and for this sort of problem I’d specifically make use of SWI-Prolog’s library(clpfd). Conceptually, the implementations are virtually identical which slightly different predicates. Gnu Prolog seems much faster, but I have neither confirm nor denied that with any sort of benchmarks.

posted at: 20:10 by: Adam Russell | path: /prolog | permanent link to this entry

### 2020-11-29

Perl Weekly Challenge 088 (Prolog solutions)

## Part 1

*You are given an array of positive integers @N. Write a script to return an array @M where $M[i] is the product of all elements of @N except the index $N[i].*

### Solution

```
/*
You are given an array of positive integers @N.
Write a script to return an array @M where $M[i] is
the product of all elements of @N except the index $N[i].
*/
product_a_b(A, B, P):-
P is A*B.
list_product([], 1).
list_product([H|T], P) :-
foldl(product_a_b, T, H, P).
list_products(List, Products):-
length(List, L0),
L is L0 - 1,
list_products(List, L, [], Products).
list_products(_, -1, Products, Products).
list_products(List, Index, ProductsAccum, Products):-
nth0(Index, List, _, Remainder),
list_product(Remainder, Product),
NewIndex is Index - 1,
list_products(List, NewIndex, [Product|ProductsAccum], Products).
main:-
list_products([5, 2, 1, 4, 3], Products),
write(Products),
halt.
```

### Sample Run

```
$ swipl -s prolog/ch-1.p -g main
[24,60,120,30,40]
```

### Notes

A problem like this really underscores the nature of lists in Prolog. That is, if you are programming in most other languages you might think of lists and arrays as interchangeable concepts for the most part. In Prolog that is very clearly not the case: lists are lists and obtaining an element at a certain position in the list is not as straightforward as simply accessing it with the usual square bracket notation. Not to say it is all that hard to do in Prolog, it’s just … different. Or, at least, requires a different mindset. People with experience in pure functional languages have this mindset of thinking of lists in terms of recursions and maps but fluency in these techniques is, sadly, less common these days.

## Part 2

*You are given m x n matrix of positive integers. Write a script to print spiral matrix as a list.*

### Solution

```
/*
You are given m x n matrix of positive integers.
Write a script to print spiral matrix as a list.
*/
write_remove_top(Matrix, UpdatedMatrix):-
nth0(0, Matrix, Top),
atomic_list_concat(Top, ",", TopString),
write(TopString),
nth0(0, Matrix, _, UpdatedMatrix).
write_remove_last([], UpdatedMatrix, UpdatedMatrix).
write_remove_last([H|T], RemainderAccum, UpdatedMatrix):-
length(H, L),
Last is L - 1,
nth0(Last, H, Right),
write(Right),
write(","),
nth0(Last, H, _, UpdatedRow),
write_remove_last(T, [UpdatedRow|RemainderAccum], UpdatedMatrix).
write_remove_right(Matrix, UpdatedMatrix):-
write_remove_last(Matrix, [], UpdatedMatrix).
write_remove_first([], UpdatedMatrix, UpdatedMatrix).
write_remove_first([H|T], RemainderAccum, UpdatedMatrix):-
nth0(0, H, Left),
write(Left),
write(","),
nth0(0, H, _, UpdatedRow),
write_remove_first(T, [UpdatedRow|RemainderAccum], UpdatedMatrix).
write_remove_left(Matrix, UpdatedMatrix):-
write_remove_first(Matrix, [], UpdatedMatrix).
write_remove_bottom(Matrix, UpdatedMatrix):-
length(Matrix, L),
Last is L - 1,
nth0(Last, Matrix, Bottom),
reverse(Bottom, ReverseBottom),
atomic_list_concat(ReverseBottom, ",", BottomString),
write(BottomString),
nth0(Last, Matrix, _, UpdatedMatrix).
spiral(Matrix):-
spiral(Matrix, _).
spiral(Matrix, UpdatedMatrix):-
write_remove_top(Matrix, UpdatedMatrix),
write(","),
write_remove_right(UpdatedMatrix, RightRemainder),
reverse(RightRemainder, RemainderRight),
write_remove_bottom(RemainderRight, BottomRemainder),
write(","),
reverse(BottomRemainder, RemainderBottom),
write_remove_left(RemainderBottom, LeftRemainder),
spiral(LeftRemainder, _).
spiral(_, []):-
write("\b").
main:-
spiral([
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]
]), halt.
```

### Sample Run

```
$ swipl -s prolog/ch-2.p -g main
1,2,3,6,9,8,7,4,5
```

### Notes

The spiral print works in a repeated pattern from the outside in: top row, right column, bottom row, left column. My solution puts each write/remove step of this pattern in their own predicates. A few things worth pointing out

- The matrix is a 2d list: a list with inner lists.
- For the spiral effect we need to print bottom up and right to left. In those cases I just use reverse/2 on the list being printed.
- I use a simple trick to remove a trailing comma in the final output; I print a backspace to the terminal which deletes it!
- Like Part 1 above I use nth0/4 as well as nth0/3 to directly access specific elements.

posted at: 20:23 by: Adam Russell | path: /prolog | permanent link to this entry

### 2020-11-15

Perl Weekly Challenge 086

## Part 1

*You are given an array of integers @N and an integer $A. Write a script to find find if there exists a pair of elements in the array whose difference is $A. Print 1 if exists otherwise 0.*

### Solution

```
:- use_module(library(optparse)).
/*
You are given an array of integers @N and an integer $A.
Write a script to find find if there exists a pair of elements
in the array whose difference is $A.
Print 1 if exists otherwise 0.
*/
opts_spec(
[
[opt(numbers),
default([10, 8, 12, 15, 5]),
longflags([numbers])],
[opt(a),
default(7),
longflags([a])]
]).
ch_1(L, N):-
member(A, L),
member(B, L),
A =\= B,
D is A - B,
N = D,
writeln(1).
ch_1(_, _):-
writeln(0).
main:-
opts_spec(OptsSpec),
opt_arguments(OptsSpec, [numbers(L), a(A)], _AdditionalArguments),
ch_1(L, A),
halt.
```

### Sample Run

```
$ swipl -s ch-1.p -g main --numbers="[10, 30, 20, 50, 40]" --a=15
0
$ swipl -s ch-1.p -g main --numbers="[10, 8, 12, 15, 5]" --a=7
1
```

### Notes

My approach here is very similar to what I did last week. Please do check that out if you’re interested in a slightly longer discussion of the use of `optparse`

to handle the command line arguments.

## Part 2

*You are given Sudoku puzzle (9x9). Write a script to complete the puzzle*

### Solution

```
:- use_module(library(clpfd)).
sudoku(Puzzle, Solution) :-
Solution = Puzzle,
Puzzle = [S11, S12, S13, S14, S15, S16, S17, S18, S19,
S21, S22, S23, S24, S25, S26, S27, S28, S29,
S31, S32, S33, S34, S35, S36, S37, S38, S39,
S41, S42, S43, S44, S45, S46, S47, S48, S49,
S51, S52, S53, S54, S55, S56, S57, S58, S59,
S61, S62, S63, S64, S65, S66, S67, S68, S69,
S71, S72, S73, S74, S75, S76, S77, S78, S79,
S81, S82, S83, S84, S85, S86, S87, S88, S89,
S91, S92, S93, S94, S95, S96, S97, S98, S99],
ins(Puzzle, 1..9),
Row1 = [S11, S12, S13, S14, S15, S16, S17, S18, S19],
Row2 = [S21, S22, S23, S24, S25, S26, S27, S28, S29],
Row3 = [S31, S32, S33, S34, S35, S36, S37, S38, S39],
Row4 = [S41, S42, S43, S44, S45, S46, S47, S48, S49],
Row5 = [S51, S52, S53, S54, S55, S56, S57, S58, S59],
Row6 = [S61, S62, S63, S64, S65, S66, S67, S68, S69],
Row7 = [S71, S72, S73, S74, S75, S76, S77, S78, S79],
Row8 = [S81, S82, S83, S84, S85, S86, S87, S88, S89],
Row9 = [S91, S92, S93, S94, S95, S96, S97, S98, S99],
Column1 = [S11, S21, S31, S41, S51, S61, S71, S81, S91],
Column2 = [S12, S22, S32, S42, S52, S62, S72, S82, S92],
Column3 = [S13, S23, S33, S43, S53, S63, S73, S83, S93],
Column4 = [S14, S24, S34, S44, S54, S64, S74, S84, S94],
Column5 = [S15, S25, S35, S45, S55, S65, S75, S85, S95],
Column6 = [S16, S26, S36, S46, S56, S66, S76, S86, S96],
Column7 = [S17, S27, S37, S47, S57, S67, S77, S87, S97],
Column8 = [S18, S28, S38, S48, S58, S68, S78, S88, S98],
Column9 = [S19, S29, S39, S49, S59, S69, S79, S89, S99],
SubBox1 = [S11, S12, S13, S21, S22, S23, S31, S32, S33],
SubBox2 = [S41, S42, S43, S51, S52, S53, S61, S62, S63],
SubBox3 = [S71, S72, S73, S81, S82, S83, S91, S92, S93],
SubBox4 = [S14, S15, S16, S24, S25, S26, S34, S35, S36],
SubBox5 = [S44, S45, S46, S54, S55, S56, S64, S65, S66],
SubBox6 = [S74, S75, S76, S84, S85, S86, S94, S95, S96],
SubBox7 = [S17, S18, S19, S27, S28, S29, S37, S38, S39],
SubBox8 = [S47, S48, S49, S57, S58, S59, S67, S68, S69],
SubBox9 = [S77, S78, S79, S87, S88, S89, S97, S98, S99],
valid([Row1, Row2, Row3, Row4, Row5, Row6, Row7, Row8, Row9,
Column1, Column2, Column3, Column4, Column5, Column6, Column7, Column8, Column9,
SubBox1, SubBox2, SubBox3, SubBox4, SubBox5, SubBox6, SubBox7, SubBox8, SubBox9]).
valid([]).
valid([H|T]) :-
all_different(H),
valid(T).
```

### Sample Run

```
$ swipl -s prolog/ch-2.p
Welcome to SWI-Prolog (threaded, 64 bits, version 8.2.2)
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software.
Please run ?- license. for legal details.
For online help and background, visit https://www.swi-prolog.org
For built-in help, use ?- help(Topic). or ?- apropos(Word).
?- sudoku([_,_,_,2,6,_,7,_,1,6,8,_,_,7,_,_,9,_,1,9,_,_,_,4,5,_,_,8,2,_,1,_,_,_,4,_,_,_,4,6,_,2,9,_,_,_,5,_,_,_,3,_,2,8,_,_,9,3,_,_,_,7,4,_,4,_,_,5,_,_,3,6,7,_,3,_,1,8,_,_,_],Solution).
Solution = [4, 3, 5, 2, 6, 9, 7, 8, 1|...].
?- halt.
```

### Notes

How many Prolog programmers saw this part of the challenge and thought “finally my training has paid off!” For the unfamiliar, Sudoku is a natural fit for a language like Prolog and quite a few variations are possible. My approach here is to make some use of `clpfd`

but favored a more verbose style so that what is happening should be fairly clear. The clpfd manual includes a much more terse example.

posted at: 23:49 by: Adam Russell | path: /prolog | permanent link to this entry

### 2020-11-08

# Command Line Arguments with SWI_Prolog’s library(optparse)

SWI-Prolog has a nice library which allows you to pass arguments on the command line to your Prolog programs. Here are a couple of example of it’s use done with the two parts of Perl Weekly Challenge 085 implemented in Prolog.

## Part 1

*You are given an array of real numbers greater than zero. Write a script to find if there exists a triplet (a,b,c) such that 1 < a+b+c < 2. Print 1 if you succeed otherwise 0.*

### Solution

```
:- use_module(library(optparse)).
/*
You are given an array of real numbers greater than zero.
Write a script to find if there exists a triplet (a,b,c)
such that 1 < a+b+c < 2. Print 1 if you succeed otherwise 0.
*/
opts_spec(
[
[opt(numbers),
default([1.2, 0.4, 0.1, 2.5]),
longflags([numbers])]
]).
ch_1(L):-
member(A, L),
member(B, L),
member(C, L),
A =\= B,
B =\= C,
A =\= C,
D is A + B + C,
D > 1,
D < 2,
writeln(1).
ch_1(_):-
writeln(0).
main:-
opts_spec(OptsSpec),
opt_arguments(OptsSpec, [numbers(N)], _AdditionalArguments),
ch_1(N),
halt.
```

### Sample Run

```
$ swipl -s ch-1.p -g main --numbers="[1.0, 0.2, 3.4, 0.1]"
1
$ swipl -s ch-1.p -g main --numbers="[1.0, 1.2, 3.4, 0.1]"
0
```

### Notes

You can see from the Sample Run that we are passing a Prolog list as a command line argument. To avoid conflicts with the shell trying to interpret our square brackets and commas we put the list itself in quotes. But how does Prolog know what `--numbers`

means?

To use `library(optparse)`

you must define a specification for each of your command line arguments that you expect to take. Here we just have one. The line `opt(numbers)`

specifies the term that will be used to obtain the value in your program, `default([1.2, 0.4, 0.1, 2.5])`

provides a default value if the argument is not used on the command line. `longflags([numbers])`

indicates that to look for a flag of the form `--numbers`

. we could have also used `shortflags([numbers])`

instead if we would prefer to use `-numbers`

. I don’t know of any strong arguments for either one but find, personally, that the long form is more intuitive.

The values passed on the command line are extracted using `opt_arguments`

which will follow the specification you’ve provided. `_AdditionalArguments`

refers to any arguments passed on the command line without dashes but here we do not have any. `[numbers(N)]`

is a list of the parsed key(value) pairs from the command line. So for this example `N`

is the list value we entered on the command line and it is then used as needed.

## Part 2

*You are given a positive integer $N. Write a script to find if it can be expressed as a ** b where a > 0 and b > 1. Print 1 if you succeed otherwise 0.*

### Solution

```
:- use_module(library(clpfd)).
:- use_module(library(optparse)).
/*
You are given a positive integer $N.
Write a script to find if it can be expressed
as a ^ b where a > 0 and b > 1.
Print 1 if you succeed otherwise 0.
*/
opts_spec(
[
[opt(number),
default(0),
longflags([number])]
]).
/*
Ok, I'll admit, this is a pretty silly use of clpfd when
a simple logarithm calculation would do the job! Still clpfd
is more fun.
*/
ch_2(N) :-
N0 is N -1,
A in 0 .. N0,
B in 1 .. N0,
N #= A ^ B,
label([A,B]),
writeln(1).
ch_2(_) :-
writeln(0).
main:-
opts_spec(OptsSpec),
opt_arguments(OptsSpec, [number(N)], _AdditionalArguments),
ch_2(N),
halt.
```

### Sample Run

```
$ swipl -s ch-2.p -g main --number=7
0
$ swipl -s ch-2.p -g main --number=8
1
```

### Notes

For an extra bit of fun I decided to use`library(clpfd)`

for this although it is most definitely a bit of over engineering! This can be done rather simply using logarithms. Here we see the same pattern as done in Part 1: define the specification, extract the values from the command line, and then use the values. Here the single argument is just a single value and so there is no need to wrap the value in quotes, however, if you add quotes anyway it will have no effect. For example:
```
$ swipl -s ch-2.p -g main --number="100"
1
```

## Reference

posted at: 17:11 by: Adam Russell | path: /prolog | permanent link to this entry