Pouring Tanks (the bc(1) version)

README

sh pour.sh 8 5 5
Filling right when left has 0
Immediate step #1
sh pour.sh 5 8 2
Filling left when right has 0
Pouring left to right gives 5
Filling left when right has 5
Pouring left to right gives 2
Solved at step #4
sh pour.sh 3 4 2
Filling left when right has 0
Pouring left to right gives 3
Filling left when right has 3
Pouring left to right gives 2
Solved at step #4
sh pour.sh 18 33 9
Filling right when left has 0
Pouring right to left gives 15
Emptying left when right is 15
Pouring right to left gives 15
Filling right when left has 15
Pouring right to left gives 30
Emptying left when right is 30
Pouring right to left gives 12
Emptying left when right is 12
Pouring right to left gives 12
Filling right when left has 12
Pouring right to left gives 27
Emptying left when right is 27
Pouring right to left gives 9
Solved at step #14
sh pour.sh 6 10 3
Filling right when left has 0
Pouring right to left gives 4
Emptying left when right is 4
Pouring right to left gives 4
Filling right when left has 4
Pouring right to left gives 8
Emptying left when right is 8
Pouring right to left gives 2
Emptying left when right is 2
Pouring right to left gives 2
Filling right when left has 2
Pouring right to left gives 6
Emptying left when right is 6
Pouring right to left gives 0
Filling right when left has 6
Emptying left when right is 10
Pouring right to left gives 4
Emptying left when right is 4
Failed at step #18

Makefile

all: pour.sh
	sh $< 8 5 5
	sh $< 5 8 2
	sh $< 3 4 2
	sh $< 18 33 9
	sh $< 6 10 3

clean:
	-rm -f *.tmp

pour.sh

script='
define p(x) {
	auto a, r

	if (x) {
		"Pouring left to right gives "
	} else {
		"Pouring right to left gives "
	}

	a = m[x] - t[x]
	if (a > t[!x]) {
		t[x] += t[!x]
		t[!x] = 0
		r = t[x]
	} else {
		t[!x] -= a
		t[x] = m[x]
		r = t[!x]
	}

	r
	return (r)
}

define e(x) {
	t[x] = 0
	if (x) {
		"Emptying right when left is "
	} else {
		"Emptying left when right is "
	}
	t[!x]
	return (t[!x])
}

define f(x) {
	t[x] = m[x]
	if (x) {
		"Filling right when left has "
	} else {
		"Filling left when right has "
	}
	t[!x]
	return (t[x])
}

define s(l, r, q, d) {
	auto i, c, u, v

	m[0] = l
	m[1] = r

	while (1) {
		if (!t[!d]) {
			i += 1
			v = f(!d)
			if (q == v) {
				"Immediate step #"
				return (i)
			}
		}

		if (t[d] == m[d]) {
			i += 1
			v = e(d)
			if (c == v) {
				"Failed at step #"
				return (i)
			}

			if (!c) {
				c = v
			}
		}

		i += 1
		v = p(d)
		if (q == v) {
			"Solved at step #"
			return (i)
		}
	}
}
'

echo "$script s($1, $2, $3, 0)" | bc >left.tmp
echo "$script s($1, $2, $3, 1)" | bc >right.tmp

left=`wc -l <left.tmp`
right=`wc -l <right.tmp`

if [ "$left" -lt "$right" ]; then
	cat left.tmp
else
	cat right.tmp
fi

rm -f {left,right}.tmp