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(December 2011, on Hacker News)

I had some free time these past few days, so I got back to Project Euler: I like fooling around with the problems there, particularly because the process reminds me of what it felt like when I begun working with computers; the ecstasy of "pure" problem solving, without the nasty side-effects of dealing with clients :‑)

Hey, on occasion I am even known to pretend to be a scientist :‑)

The problem I've enjoyed most so far, is problem 31:

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation: 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

It is therefore possible to make £2 in the following way:
1x£1, 1x50p, 2x20p, 1x5p, 1x2p, 3x1p
How many different ways can £2 be made using any number of coins?

Hmm...

Being the trigger-happy engineer that I am, I quickly tried to brute-force:

#include <stdio.h>
#include <signal.h>

// We want to accumulate 200 cents (2 pounds)
#define TARGET 200

int coins[] = {1, 2, 5, 10, 20, 50, 100, 200};
long long total = 0LL;

int newCoin(int s)
{
    if (s == TARGET) {
	total ++; // landed right on 200 cents, increase total
	return 1; // and report "break out of the loop"
    } else if (s>TARGET) {
	// we overflowed, go back and report "break out of the loop"
	return 1; 
    } else {
	// recurse, adding each available coin in turn
	for(int idx=0; idx<sizeof(coins)/sizeof(coins[0]); idx++) {
	    int earlyAbort = newCoin(s+coins[idx]);
	    // if the callee reported non-zero, we break
	    if (earlyAbort)
		return 0;
	}
    }
    return 0;
}

void pr(int i)
{
    // SIGINT (Ctrl-C) triggered reports
    printf("\n%lld\n",  total);
}

int main()
{
    signal(SIGINT, pr);
    newCoin(0); // start the game
    printf("%lld\n",  total);
}


The process starts with newCoin(0), called from main. This function will recursively call itself, until it either grows the accumulated sum so far (its parameter) beyond 200 cents, or it will reach exactly 200 cents - at which point the total counter will record it as a successful combination.

Sure, sure, I am not handling the signals properly, printf is not re-entrant - but who cares, I just want to quickly see if this solves the problem or not; Ctrl-C will print how high total is, so trying it out...

ttsiod@elrond:tmp$ g++ -O2 problem31.brute.cc
ttsiod@elrond:tmp$ ./a.out
^C
32378168
^C
51764088
^C
81886592
^C
168584448
^C
225451092
^Z
[1]+  Stopped                 ./a.out
ttsiod@elrond:tmp$ kill %1
[1]+  Killed                 ./a.out
Ooops. This thing grows rapidly (too rapidly)!

And it doesn't seem to finish. After 1 minute of running...

To make matters worse, the speed at which total goes up makes it clear that I am being an idiot: the blind recursion will obviously measure 1p+2p as a different combination to 2p+1p. Order is not supposed to count; the problem statement says "how many different ways", and this is clearly violating it.

Breaking it down, target: 1 cent

OK, thinking this through... Re-applying past experience, I build a table...
Target in centsUsing only 1pUsing <= 2pUsing <= 5p...
11...
If I wanted to accumulate just 1 cent (column "Target in cents"), how many ways are there to do it, using coins less than or equal to 1p?

Only one way - so, fill up the cell of column "Using only 1p" with 1.

Moving on - what if I could use coins less than or equal to 2p?

Target in centsUsing only 1pUsing <= 2pUsing <= 5p...
111...
There is still only one way - using 1p. The 2p coins can't be used, since our target (a total of 1 cent for this line of the table) is less than 2p.

So all the first line fills up with 1s:

Target in centsonly 1p<= 2p<= 5p<= 10p<= 20p<= 50p<= 100p<= 200p
111111111
Only one way to reach the target of 1 cent, no matter what coins I use.

Breaking it down, target: 2 cents

What if we move to a target of 2 cents?
Target in centsonly 1p<= 2p<= 5p<= 10p<= 20p<= 50p<= 100p<= 200p
111111111
21
Well, using only 1p there is only one way: 1p+1p.

When using coins less than or equal to 2p, however, we can also do it via a single 2p coin - so there are now 2 ways:

Target in centsonly 1p<= 2p<= 5p<= 10p<= 20p<= 50p<= 100p<= 200p
111111111
212222222
Additional coins don't help any - the total number of ways remains 2 (1p+1p,2p) throughout the rest of the line.

Breaking it down, target: 3,4,5,... cents

Thinking the same way:
Target in centsonly 1p<= 2p<= 5p<= 10p<= 20p<= 50p<= 100p<= 200p
111111111
212222222
312222222
413333333
513444444
So, now that we've done this process manually, what do we notice? For example, on line 5 (target: 5 cents) the highlighted cell of column "<=2p" is formed as 1+2: This lookup is key - we basically reuse previous calculations with a single lookup.

OK, we can now write it down in Python...

#!/usr/bin/env python

# the 8 coins correspond to 8 columns
coins = [1, 2, 5, 10, 20, 50, 100, 200]

TARGET=200

matrix = {}
for y in xrange(0, TARGET+1):
    # There is only one way to form a target sum N
    # via 1-cent coins: use N 1-cents!
    matrix[y, 0] = 1  # equivalent to matrix[(y,0)]=1

for y in xrange(0, TARGET+1):
    print y, ":", 1,
    for x in xrange(1, len(coins)):
        matrix[y, x] = 0
        # Is the target big enough to accomodate coins[x]?
        if y>=coins[x]:
            # If yes, then the number of ways to form
            # the target sum are obtained via:
            #
            # (a) the number of ways to form this target
            #     using ONLY coins less than column x
            #     i.e. matrix[y][x-1]
            matrix[y, x] += matrix[y, x-1]
            # plus
            # (b) the number of ways to form this target
            #     when USING the coin of column x
            #     which means for a remainder of y-coins[x]
            #     i.e. matrix[y-coins[x]][x]
            matrix[y, x] += matrix[y-coins[x], x]
        else:
            # if the target is not big enough to allow
            # usage of the coin in column x,
            # then just copy the number of ways from the
            # column to the left (i.e. with smaller coins)
            matrix[y, x] = matrix[y, x-1]
        print matrix[y, x],
    print
...and marvel at seeing it run infinitely faster than brute-force (finishes in less than 30 milliseconds):
ttsiod@elrond:tmp$ python code/problem31.py
0 : 1 1 1 1 1 1 1 1
1 : 1 1 1 1 1 1 1 1
2 : 1 2 2 2 2 2 2 2
3 : 1 2 2 2 2 2 2 2
4 : 1 3 3 3 3 3 3 3
5 : 1 3 4 4 4 4 4 4
6 : 1 4 5 5 5 5 5 5
7 : 1 4 6 6 6 6 6 6
8 : 1 5 7 7 7 7 7 7
9 : 1 5 8 8 8 8 8 8
10 : 1 6 10 11 11 11 11 11
11 : 1 6 11 12 12 12 12 12
...
195 : 1 98 1980 14140 43450 62156 65934 65934
196 : 1 99 2000 14350 44275 63500 67425 67425
197 : 1 99 2020 14560 45100 64844 68916 68916
198 : 1 100 2040 14770 45925 66188 70407 70407
199 : 1 100 2060 14980 46750 67532 71898 71898
200 : 1 101 2081 15211 47696 69118 73681 73682
my level
The right-most column tells us that there are 73682 ways to form a total of 2 pounds.

The above method, in case you've never seen it before, is called dynamic programming - and just as was done here, can be used to quickly solve problems that are impossible to brute force.

OK, project Euler... one more down, three hundred to go! :‑)


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