Solving Problems

The theme for November’s Code Share is Solving Problems. The programmers job is to solve problems with code.

Robert L Glass in Software Conflict 2.0 (1990)

Why do we write code? To solve problems. The best code is the code that solves the most important problems.

Leo Brodie in the preface to Thinking Forth (2004)

We need to be careful, because sometimes we inadvertently create new ones. Those that follow us can spend more time solving the problems we created rather than moving forward the solutions we provided.

Michael Jackson in Problem Frames (2001)

When we read code we need to ask ourselves some important questions:

• What problem is being solved? How important is the problem being solved?
• How effectively is the problem reflected in the code? Does the code show the concerns that need to be addressed to deliver results in the real world?
• How clear does the code make the technical solution? Does it bury the important aspects of the design below layers of abstraction?

November challange: Solving Soduku

The complexity of the Soduku puzzle is not incalculable, but the creation of soduku solvers has much to teach us .

One again we are going to use a python solution by Peter Norvig as a starting point:

Solve Every Sudoku Puzzle

See

http://norvig.com/sudoku.html]9

Throughout this program we have:

r is a row, e.g. ‘A’

c is a column, e.g. ‘3’

s is a square, e.g. ‘A3’

d is a digit, e.g. ‘9’

u is a unit, e.g. ‘A1’,’B1’,’C1’,’D1’,’E1’,’F1’,’G1’,’H1’,’I1’

grid is a grid,e.g. 81 non-blank chars, e.g. starting with ‘.18…7…

values is a dict of possible values, e.g.

def cross(A, B): “Cross product of elements in A and elements in B.” return a+b for a in A for b in B

digits = ‘123456789’ rows = ‘ABCDEFGHI’ cols = digits squares = cross(rows, cols) unitlist = (cross(rows, c) for c in cols + cross(r, cols) for r in rows + cross(rs, cs) for rs in (‘ABC’,’DEF’,’GHI’) for cs in (‘123’,’456’,’789’)) units = dict((s, u for u in unitlist if s in u) for s in squares) peers = dict((s, set(sum(unitss,))-set(s)) for s in squares)

Unit Tests

def test(): “A set of tests that must pass.” assert len(squares) == 81 assert len(unitlist) == 27 assert all(len(unitss) == 3 for s in squares) assert all(len(peerss) == 20 for s in squares) assert units‘C2’ == ‘A2’, ‘B2’, ‘C2’, ‘D2’, ‘E2’, ‘F2’, ‘G2’, ‘H2’, ‘I2’, ‘C1’, ‘C2’, ‘C3’, ‘C4’, ‘C5’, ‘C6’, ‘C7’, ‘C8’, ‘C9’, ‘A1’, ‘A2’, ‘A3’, ‘B1’, ‘B2’, ‘B3’, ‘C1’, ‘C2’, ‘C3’ assert peers‘C2’ == set(‘A2’, ‘B2’, ‘D2’, ‘E2’, ‘F2’, ‘G2’, ‘H2’, ‘I2’, ‘C1’, ‘C3’, ‘C4’, ‘C5’, ‘C6’, ‘C7’, ‘C8’, ‘C9’, ‘A1’, ‘A3’, ‘B1’, ‘B3’) print ‘All tests pass.’

Parse a Grid

def parse_grid(grid): ””“Convert grid to a dict of possible values, {square: digits}, or return False if a contradiction is detected.””” ## To start, every square can be any digit; then assign values from the grid. values = dict((s, digits) for s in squares) for s,d in grid_values(grid).items(): if d in digits and not assign(values, s, d): return False ## (Fail if we can’t assign d to square s.) return values

def grid_values(grid): “Convert grid into a dict of {square: char} with ‘0’ or ’.’ for empties.” chars = c for c in grid if c in digits or c in ‘0.’ assert len(chars) == 81 return dict(zip(squares, chars))

Constraint Propagation

def assign(values, s, d): ””“Eliminate all the other values (except d) from valuess and propagate. Return values, except return False if a contradiction is detected.””” other_values = valuess.replace(d, ”) if all(eliminate(values, s, d2) for d2 in other_values): return values else: return False

def eliminate(values, s, d): ””“Eliminate d from valuess; propagate when values or places <= 2. Return values, except return False if a contradiction is detected.””” if d not in valuess: return values ## Already eliminated valuess = valuess.replace(d,”) ## (1) If a square s is reduced to one value d2, then eliminate d2 from the peers. if len(valuess) == 0: return False ## Contradiction: removed last value elif len(valuess) == 1: d2 = valuess if not all(eliminate(values, s2, d2) for s2 in peerss): return False ## (2) If a unit u is reduced to only one place for a value d, then put it there. for u in unitss: dplaces = s for s in u if d in valuess if len(dplaces) == 0: return False ## Contradiction: no place for this value elif len(dplaces) == 1: # d can only be in one place in unit; assign it there if not assign(values, dplaces0, d): return False return values

Display as 2-D grid

def display(values): “Display these values as a 2-D grid.” width = 1+max(len(valuess) for s in squares) line = ’+’.join(’-’(width3)3) for r in rows: print ”.join(valuesr+c.center(width)+(’|’ if c in ‘36’ else ”) for c in cols) if r in ‘CF’: print line print

Search

def solve(grid): return search(parse_grid(grid))

def search(values): “Using depth-first search and propagation, try all possible values.” if values is False: return False ## Failed earlier if all(len(valuess) == 1 for s in squares): return values ## Solved! ## Chose the unfilled square s with the fewest possibilities n,s = min((len(valuess), s) for s in squares if len(valuess) > 1) return some(search(assign(values.copy(), s, d)) for d in valuess)

Utilities

def some(seq): “Return some element of seq that is true.” for e in seq: if e: return e return False

def from_file(filename, sep=’\n’): “Parse a file into a list of strings, separated by sep.” return file(filename).read().strip().split(sep)

def shuffled(seq): “Return a randomly shuffled copy of the input sequence.” seq = list(seq) random.shuffle(seq) return seq

System test

import time, random

def solve_all(grids, name=”, showif=0.0): ””“Attempt to solve a sequence of grids. Report results. When showif is a number of seconds, display puzzles that take longer. When showif is None, don’t display any puzzles.””” def time_solve(grid): start = time.clock() values = solve(grid) t = time.clock()-start ## Display puzzles that take long enough if showif is not None and t > showif: display(grid_values(grid)) if values: display(values) print ‘(%.2f seconds)\n’ % t return (t, solved(values)) times, results = zip(time_solve(grid) for grid in grids) N = len(grids) if N > 1: print “Solved %d of %d %s puzzles (avg %.2f secs (%d Hz), max %.2f secs).” % ( sum(results), N, name, sum(times)/N, N/sum(times), max(times))

def solved(values): “A puzzle is solved if each unit is a permutation of the digits 1 to 9.” def unitsolved(unit): return set(valuess for s in unit) == set(digits) return values is not False and all(unitsolved(unit) for unit in unitlist)

def random_puzzle(N=17): ””“Make a random puzzle with N or more assignments. Restart on contradictions. Note the resulting puzzle is not guaranteed to be solvable, but empirically about 99.8% of them are solvable. Some have multiple solutions.””” values = dict((s, digits) for s in squares) for s in shuffled(squares): if not assign(values, s, random.choice(valuess)): break ds = valuess for s in squares if len(valuess) == 1 if len(ds) >= N and len(set(ds)) >= 8: return ”.join(valuess if len(valuess)==1 else ’.’ for s in squares) return random_puzzle(N) ## Give up and make a new puzzle

grid1 = ‘003020600900305001001806400008102900700000008006708200002609500800203009005010300’ grid2 = ‘4…..8.5.3……….7……2…..6…..8.4……1…….6.3.7.5..2…..1.4……’ hard1 = ’…..6….59…..82….8….45……..3……..6..3.54…325..6………………’

if name == ‘__main__’: test() solve_all(from_file(“easy50.txt”, ‘========’), “easy”, None) solve_all(from_file(“top95.txt”), “hard”, None) solve_all(from_file(“hardest.txt”), “hardest”, None) solve_all(random_puzzle() for _ in range(99), “random”, 100.0)

References used:

http://www.scanraid.com/BasicStrategies.htm]10

http://www.sudokudragon.com/sudokustrategy….]11

http://www.krazydad.com/blog/2005/09/29/an-…]12

http://www2.warwick.ac.uk/fac/sci/moac/curr…]13

Please have a go at creating your own soduku solver. Whatever you do, though, don’t just translate the above code into another language. Take the time to understand the problem and possible solutions, then express your own solution in code as elegantly as you can.

 

What’s Going To Happen?

 

On Monday 31st October, a couple of days before the share, we’ll send out an email to everybody who has signed up. If you have any code to contribute please send it in a reply to that email.

On Wednesday 2nd November, the Code Share itself, we will all have a couple of short presentations. This will be followed by breaking out into groups to cast a critical eye over the code for some solutions. We will print these out, so you won’t have to bring a laptop. Afterwards we will come back together as a group to discuss what we have learned.

After the event we will be heading to the Half Moon, 213-223 Mile End Road, Mile End, Greater London, E1 4AA - http://www.jdwetherspoon.co.uk/home/pubs/th… - for drinks/networking.

ThoughtWorks are delighted to be sponsors of the November Code Share! 

This event is hosted in collaboration with the London Java Community . We expect many programmers across the range from junior to experienced to be participating, so it’s a great chance to see how code is viewed by an experienced programmer.