Dynamic Programming with Python dataclasses and joblib

Starting in Python 3.7, the module dataclasses introduces a decorator that allows us to create immutable structures (like tuples) but with their own batteries-included methods. I wanted to give dataclasses a try with some non-trivial workloads. Having always been interested the coin change problem where the solution involves dynamic programming, this seemed to be a good test of dataclasses.

Instead of the classic coin change problem (finding the minimum number of coins that add up to a given amount), however, here I want to keep track of all possible combinations. The end result (at the bottom of this post) is an interactive plot to explore the results.

Note: You can get the original Jupyter notebook for this post here.

from dataclasses import dataclass, field
from joblib import Memory

from bokeh.io import output_notebook, show
from bokeh.plotting import figure

Loading BokehJS ...

Here we'll use joblib to cache the results, so we don't need to redo computations we already have completed:

cachedir = "/tmp/memo1"
memory = Memory(cachedir, verbose=0)
WARNING:root:[Memory(location=/tmp/memo1/joblib)]: Flushing completely the cache

Now, let's define the universe of possible coins.

# coins = [1, 2, 5, 10, 20, 50, 100]  # European coins
# coins = [1, 5, 10, 25] # US Coins in wide circulation
coins = [1, 5, 10, 25, 50, 100] # All US Coins circulation

Here's where we create our dataclass—this will hold a specific collection of coins (the purse variable). We'll also define how to add coins from one purse to another (__add__) and set some helper methods allowing us to sort (__hash__), compare (__eq__, __le__, __lt__) and print (__repr__).

class change:
    purse: list
    value: int = field(init=False)
    ncoins: int = field(init=False)
    sort_index: int = field(init=False, repr=True)

    def __post_init__(self):
        self.value = sum(map(lambda x: x[0]*x[1], zip(coins, self.purse)))
        self.ncoins = sum(self.purse)
        self.sort_index = self.ncoins

    def n(self, coin):
        return self.purse[coins.index(coin)]

    def __add__(self, other):
        return change([x+y for x,y in zip(self.purse,other.purse)])

    def __eq__(self, other):
        return tuple(self.purse) == tuple(other.purse)

    def __le__(self, other):
        return self.ncoins <= other.ncoins

    def __lt__(self, other):
        return self.ncoins < other.ncoins

    def __hash__(self):
        return hash(tuple(self.purse))

    def __repr__(self):
        return f"({self.value}) {dict(zip(coins,self.purse))}  {self.ncoins}"

Next we'll define the dynamic programming bit, which recursively looks for ways to make change, saving the results for a certain value, and reusing previously computed values.

empty_purse = [0]*len(coins)

def get_change(val):

    if val < coins[0]:
        return [change(empty_purse)]
    if val == coins[0]:
        purse = empty_purse.copy()
        purse[0] = 1
        return [change(purse)]

    final_list = []
    for i, coin in enumerate(coins):
        if val < coin:

        purse = empty_purse.copy()
        purse[i] = 1
        to_add = change(purse)
        final_list += [prev + to_add for prev in get_change(val - coin)]

    return list(set(final_list))
%time a = get_change(75)
CPU times: user 391 ms, sys: 74.3 ms, total: 466 ms
Wall time: 528 ms

That's pretty quick given all the possibilities (as we'll see). Let's compute all possibilities up to 100 cents.

max_n = 101
raw = [(x, sorted(get_change(x))) for x in range(1,max_n)]
points = []
uniques = []
min_coins = []
min_coins_scale = []

max_n = 0
for val, purse_list in raw:
    ncoins = [x.ncoins for x in purse_list]
    if val == 1:
        uniques.append((val, len(ncoins)))
        if len(ncoins) != uniques[-1][1]:
            uniques.append((val, len(ncoins)))

    for nc in set(ncoins):
        max_n = max(max_n, ncoins.count(nc))
        combos = []
        for x in [y for y in purse_list if y.ncoins == nc]:
            combos.append("-".join([str(x) for x in x.purse]))
        combos_str = "<br>".join(combos)
        points.append((val, nc, ncoins.count(nc), combos_str))


The value max_n is the most number of combinations for a certain fixed value size. For instance, if we want to make change for 97 cents with exactly 25 coins there are 7 different ways to do that:


Now let's make an interactive plot:

from bokeh.models import ColumnDataSource,  LabelSet, Label
from bokeh.transform import linear_cmap
from bokeh.models import HoverTool

source = ColumnDataSource(data={
    'x' : [x[0] for x in points],
    'y' : [x[1]/(x[0]) for x in points],
    'y_nat': [x[1] for x in points],
    'n' : [max(2,20*x[2]/max_n) for x in points],
    'n_nat': [x[2] for x in points],
    'str': [x[3] for x in points]

hover = HoverTool(point_policy='snap_to_data', line_policy='nearest')

hover.tooltips = [
    ("(val,n_coins)", "(@x, @y_nat)"),
    ("n_combos", "@n_nat"),
    ("combos", "@str{safe}")

source_label = ColumnDataSource(data=dict(x=[x[0] for x in uniques],
                                    y=[1 for x in uniques],
                                    lab=[str(x[1]) + '→' for x in uniques]))
labels = LabelSet(x='x', y='y', text='lab', level='glyph',
              x_offset=-1, y_offset=0.01, source=source_label, render_mode='canvas', text_font_size="8pt")
p = figure(plot_width=975, plot_height=500)
p.xaxis.axis_label = 'Total Amount'
p.yaxis.axis_label = 'Fraction of coins needed (out of max)'

p.line([x[0] for x in raw], min_coins_scale, line_width=2, color="red", alpha=0.8, legend_label="min coins")

p.circle('x', 'y', size='n', color=linear_cmap('n', 'Viridis256', 2, 20),
         fill_alpha=0.6, source=source)

p.legend.location = "bottom_left"



This should be a dynamic plot – you can zoom in to see some of the interesting structures. Hover over circles to see the coin combinations. I color coded and sized the circles so you can see where there is a lot of non-unique combinations. For example, we can see that 30 cents is the lowest value where there is a non-unique number of coin combinations (you can get 30 with 6 coins in 2 different ways). The numbers at the top of the plot show the total number of unique solutions. Try this out on your own; for example, what differences do you see with the Euro cent combination?

Joshua Bloom
Professor of Astronomy

Astrophysics Prof at UC Berkeley, former Wise.io cofounder (acquired by GE); Inventor; Dad, Tennis everything. Anti #TransparentMoon. Check out his group activities at ml4science.org.