Hamish Campbell: Polyominoes - An Exploration in Problem Solving with Python

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Hamish Campbell:
Polyominoes - An Exploration in Problem Solving with Python
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@ Kiwi PyCon 2013 - Saturday, 07 Sep 2013 - Track 2
http://nz.pycon.org/

**Audience level**

Intermediate

**Description**

A "Polyomino" is a geometric figure of adjacent squares. But just how many "n" polyominos are there? Solving this question is one of our favourite interview questions and the techniques that can employed to solve it are instructive for a range of problems. In this talk, we discuss the potential solutions and offer prizes for the best solutions submitted by the end of the conference.

**Abstract**

A "Polyomino" is a geometric figure of adjacent squares. But just how many "n" polyominos are there?

This question has been one of our favourite interview questions. It's a deceptively simple problem with opportunities for innovative solutions and disastrous missteps. In this talk, we'll discuss the benefits of a number of approaches, demonstrate a few common pitfalls and challenge the audience to provide their own solutions, with prizes on offer at the end of the conference.

This is a talk for everyone: beginners will be introduced to a problem that can be solved using primitives, experts will see opportunities to deploy advanced optimizations.

**YouTube**

http://www.youtube.com/watch?v=HXQW3MCOECU

Transcript

1. Kiwi PyCon 2013 Hamish Campbell @polemic
2. Polyominoes an exploration in problem solving
3. Software Development Art Science Engineering
4. What is a Polyonomino? Monomino Domino Tromino
5. Let's be specific Fixed Distinct when none is a translation of another... One-sided And none is a rotation of another... Free And none is a reflection / glide reflection of another.
6. 4-Ominoes “In the wild”
7. How Many One-Sided N-ominos?
8. Define a Polyomino class Polyomino: def __init__(self, iterable): self.squares = tuple(iterable) p = Polyomino(((0,0), (0,1), (1,1)))
9. Identifying Candidates Enumerating search space – or – Building from lower order
10. Enumerating from Search Space (n = 6)
11. Enumerating from Search Space n2 = 36 squares “From n choose r “ import gmpy gmpy.comb(36, 6) >> 1947792 GMPY: https://code.google.com/p/gmpy/ Multiple-precision arithmetic for Python
12. Enumerating from Search Space n(n+1)/2 = 21 gmpy.comb(21, 6) >> 54264
13. Enumerating from Search Space 1 +
14. Enumerating from Search Space 1 + (2 x 5) nCr 6 +
15. Enumerating from Search Space 1 + (2 x 5) nCr 6 + (3 x 4) nCr 6 1 + \ gmpy.comb(2*5, 6) +\ gmpy.comb(3*4, 6) >> 1135
16. Enumeration def count_enumerations(d): import gmpy return 1 + sum( gmpy.comb(i * (1 + d ­ i), d) for i in range(1 + d // 2, d) )
17. Enumeration for i in range(5, 10): print i, count_enumerations(i) 5 183 6 1,135 7 18,668 8 172,732 9 3,655,850
18. Building from Lower Order (n = 1) 1
19. Building from Lower Order (n = 2) 1 x (n - 1) x 4 (worse case)
20. Building from Lower Order (n = 3) (n - 2) x 4 x (n -1) x 4 n = 4 (n – 3) x 4 x (n – 2) x 4 x (n – 1) x 4
21. Building from Lower Order Assumes 4 edges! In reality we can pick a lower number, e.g. 1 import math for i in range(5, 10): print i, math.factorial(i) 5 120 8 40,320 6 720 9 362,880 7 5,040
22. We have a winner! 3,655,850 vs 409,114
23. Identifying a Polyomino Tuples are hashable, that's handy. def __hash__(self): return hash(self.squares)
24. Identifying a Polyomino Consistently hashable. Order matters. def __init__(self, iterable): self.squares = tuple( sorted(iterable) )
25. Identifying a Polyomino But remember the rules def __hash__(self): p = self, k = self.key() for _ in range(3): p = self.rotate().translate() k = min(k, p.key()) return k def key(self): return hash(self.squares)
26. Eliminating Duplicates What's the easiest way to store an array of unique items?
27. Using the set helpers How does the set `in` operator work? hash(a) == hash(b) and (a is b or id(a) == id(b))
28. Identifying a Polyomino def __eq__(self, other): return hash(self) == hash(other)
29. Unique Polyominoes len(set([ Polyomino(((0,0), (0,1))), Polyomino(((1,0), (0,0))), Polyomino(((1,1), (1,1))), ])) == 1
30. What's left? Raising the order Building the result set. We'll stop here.
31. Challenge
32. Challenges 1. Fastest 2. Lowest memory usage 3. Most esoteric, obfuscated or shortest solution See us at the Treehouse, 11am - 1pm Sunday ThinkGeek vouchers and spot prizes Reference implementation: http://bit.ly/1enn1ny (numpy arrays? pypy vs others? threading?)
33. import sys, itertools class Polyomino(tuple): def __hash__(_):return(getattr(_,'',''))or[setattr(_,'',setattr(_,'_',map(min,zip( *(setattr(_,'p',_.__class__((c[1],m­c[0]­1)for(c)in(_.p))if'p'in(_.__dict__)else(_)) or(getattr(_,'p'))))))or(min(getattr(_,'',hash(tuple(sorted(setattr(_,'p',_.__class__( [(x­_._[0],y­_._[1])for(x,y)in(_.p)])if(_._[0]or(_._[1]))else(_.p))or(_.p))))), hash(tuple(sorted(_.p))))))for(m)in(itertools.repeat(max(max(_,key=max)),4))]and(getattr(_,'')) def __eq__(*_):return(1&len(set(map(hash,_)))) def __call__(_,__):return[_.__class__(((__,__),))]if(not(1%__))else(set(itertools.chain(*[set([_.__class__ (_+tuple([a]))for(a)in(set((c[0]+x,c[1]+y)for(c,(x,y))in(itertools.product( _,((­1,0),(1,0),(0,­1),(0,1)))))).difference(_)])for(_)in(_(__­1))]))) print(len(Polyomino()(int(sys.argv.pop())))) Thanks for listening @polemic hamish.campbell@koordinates.com