# RabbitFarm

### 2022-08-07

The Weekly Challenge 176 (Prolog Solutions)

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

## Part 1

*Write a script to find the smallest integer x such that x, 2x, 3x, 4x, 5x and 6x are
permuted multiples of each other.*

### Solution

```
permuted(X, Y):-
number_chars(X, XChars),
number_chars(Y, YChars),
msort(XChars, XCharsSorted),
msort(YChars, YCharsSorted),
XCharsSorted == YCharsSorted.
permuted_multiple(Permuted):-
current_prolog_flag(max_integer, MAX_INTEGER),
between(1, MAX_INTEGER, X),
X2 is 2 * X,
X3 is 3 * X,
X4 is 4 * X,
X5 is 5 * X,
X6 is 6 * X,
permuted(X, X2), permuted(X2, X3),
permuted(X3, X4), permuted(X4, X5), permuted(X5, X6),
Permuted = X.
```

### Sample Run

```
| ?- permuted_multiple(Permuted).
Permuted = 142857 ?
(2150 ms) yes
```

### Notes

I implemented solutions for this problem in multiple languages and compared side by side on the same system this Prolog solution ran the fastest. I think that's because in Prolog the logic can be (naturally!) expressed very succinctly and so the underlying instructions getting executed are most efficient. Other implementations seem to require a bit more overhead, such as when deconstructing an integer into a list of digits for example.

The task here is for us to generate the smallest integer with this permutable property.
Technically the code here will generate all such numbers, however in search of other
solutions it seems there is just the one found before we exceed the bounds of
`MAX_INTEGER`

.

## Part 2

*Write a script to find out all Reversible Numbers below 100.*

### Solution

```
all_odd([]).
all_odd([H|T]):-
number_chars(Digit, [H]),
M is mod(Digit, 2),
M == 1,
all_odd(T).
reversible(X):-
number_chars(X, XChars),
reverse(XChars, XCharsReversed),
number_chars(R, XCharsReversed),
Sum is X + R,
number_chars(Sum, SumChars),
all_odd(SumChars).
reversible_under_n(N, Reversible):-
between(1, N, X),
reversible(X),
Reversible = X.
```

### Sample Run

```
$ gprolog --consult-file prolog/ch-2.p
| ?- reversible_under_n(100, Reversibles).
Reversibles = 10 ? a
Reversibles = 12
Reversibles = 14
Reversibles = 16
Reversibles = 18
Reversibles = 21
Reversibles = 23
Reversibles = 25
Reversibles = 27
Reversibles = 30
Reversibles = 32
Reversibles = 34
Reversibles = 36
Reversibles = 41
Reversibles = 43
Reversibles = 45
Reversibles = 50
Reversibles = 52
Reversibles = 54
Reversibles = 61
Reversibles = 63
Reversibles = 70
Reversibles = 72
Reversibles = 81
Reversibles = 90
(10 ms) no
```

### Notes

Here we also use Prolog's ability to identify multiple solutions to generate all solutions
less than `100`

, as required.

## References

posted at: 11:54 by: Adam Russell | path: /prolog | permanent link to this entry