| Brian
Harvey University of California, Berkeley |
| Download PDF version |
| Back to Table of Contents |
| BACK chapter thread NEXT |
| MIT Press web page for Computer Science Logo Style |
|
|
|
Program file for this chapter: pour
You have probably seen puzzles like this one many times:
You are at the side of a river. You have a three-liter pitcher and a seven-liter pitcher. The pitchers do not have markings to allow measuring smaller quantities. You need two liters of water. How can you measure two liters?
These puzzles are used in some IQ tests, so many people come across them in schools. To solve the problem, you must pour water from one pitcher to another. In this particular problem, there are six steps in the shortest solution:
This example is a relatively hard pitcher problem, since it requires six steps in the solution. On the other hand, it doesn't require pouring water back into the river, and it doesn't have any unnecessary pitchers. An actual IQ test has several such problems, starting with really easy ones like this:
You are at the side of a river. You have a three-liter pitcher and a seven-liter pitcher. The pitchers do not have markings to allow measuring smaller quantities. You need four liters of water. How can you measure four liters?
and progressing to harder ones like this:
You are at the side of a river. You have a two-liter pitcher, a five-liter pitcher, and a ten-liter pitcher. The pitchers do not have markings to allow measuring smaller quantities. You need one liter of water. How can you measure one liter?
The goal of this project is a program that can solve these problems. The program should take two inputs, a list of pitcher sizes and a number saying how many liters we want. It will work like this:
? pour [3 7] 4 Pour from river to 7 Pour from 7 to 3 Final quantities are 3 4 ? pour [2 5 10] 1 Pour from river to 5 Pour from 5 to 2 Pour from 2 to river Pour from 5 to 2 Final quantities are 2 1 0
How do people solve these problems? Probably you try a variety of special-purpose techniques. For example, you look at the sums and differences of the pitcher sizes to see if you can match the goal that way. In the problem about measuring four liters with a three-liter pitcher and a seven-liter pitcher, you probably recognized right away that 7−3=4. A more sophisticated approach is to look at the remainders when one pitcher size is divided by another. In the last example, trying to measure one liter with pitchers of two, five, and ten liters, you might notice that the remainder of 5/2 is 1. That means that after removing some number of twos from five, you're left with one.
Such techniques might or might not solve any given pitcher problem. Mathematicians have studied ways to solve such problems in general. To a mathematician, a pitcher problem is equivalent to an algebraic equation in which the variables are required to take on integer (whole number) values. For example, the problem at the beginning of this chapter corresponds to the equation
In this equation, x represents the number of times the three-liter pitcher is filled and y represents the number of times the seven-liter pitcher is filled. A positive value means that the pitcher is filled from the river, while a negative value means that it's filled from another pitcher.
An equation with two variables like this one can have infinitely many solutions, but not all the solutions will have integer values. One integer-valued solution is x=3 and y = -1. This solution represents filling the three-liter pitcher three times from the river (for a total of nine liters) and filling the seven-liter pitcher once from the three-liter pitcher. Since the seven-liter pitcher is bigger than the three-liter pitcher, it has to be filled in stages. Do you see how this analysis corresponds to the sequence of steps I gave earlier?
An equation with integer-valued variables is called a Diophantine equation. In general, a Diophantine equation will have infinitely many solutions, but they won't all be practical as solutions to the original problem. For example, another solution to the equation we've been considering is x = -4 and y=2. This solution tells us to fill the seven-liter pitcher from the river twice, and the three-liter pitcher from the seven-liter pitcher four times. Here's how that works out as a sequence of steps:
This solution works, but it's more complicated than the one I used in the first place.
One way to solve Diophantine equations is graphically. For example, consider the problem about measuring one liter of water with pitcher capacities two, five, and ten liters. It turns out that the ten-liter pitcher is not actually needed, so let's forget it for now and consider the simpler but equivalent problem of using just the two-liter and the five-liter pitchers. This problem gives rise to the equation
For the moment, never mind that we are looking for integer solutions. Just graph the equation as you ordinarily would. The graph will be a straight line; probably the easiest way to draw the graph is to find the x-intercept (when y=0, 2x=1 so x=1/2) and the y-intercept (when x=0, y=1/5).
Once you've drawn the graph, you can look for places where the line crosses the grid points of the graph paper. In this case, two such points of intersection are (-2,1) and (3,-1). The first of these points represents the solution shown earlier, in which the five-liter pitcher is filled from the river and then used as a source of water to fill the two-liter pitcher twice. The second integer solution represents the method of filling the two-liter pitcher from the river three times, then pouring the water from the two-liter pitcher to the five-liter pitcher each time. (On the third such pouring, the five-liter pitcher fills up after only one liter is poured, leaving one liter in the two-liter pitcher.) What about the original version of this problem, in which there were three pitchers? In this case, we have a Diophantine equation with three variables:
The graph of this equation is a plane in a three-dimensional coordinate system. An example of a solution point that uses all three pitchers is (-2,-1,1). How would you interpret this as a series of pouring steps?
By the way, not all pitcher problems have solutions. For example, how could
you measure one liter with a two-liter pitcher and a ten-liter pitcher? The
answer is that you can't; since both pitchers hold an even number of liters,
any amount of water measurable with them will also be even.*
*You can
find a computational algorithm to solve (or show that there are no solutions
to) any linear Diophantine equation with two variables on page 50 of Courant
and Robbins, What Is Mathematics? (Oxford University Press,
1941).
My program does not solve pitcher problems by manipulating Diophantine equations. Instead, it simply tries every possible sequence of pouring steps until one of the pitchers contains the desired amount of water. This method is not feasible for a human being, because the number of possible sequences is generally quite large. Computers are better than people at doing large numbers of calculations quickly; people have the almost magical ability to notice the one relevant pattern in a problem without trying all the possibilities. (Some researchers attribute this human ability to "parallel processing"--the fact that the human brain can carry on several independent trains of thought all at once. They are beginning to build computers designed for parallel processing, and hope that these machines will be able to perform more like people than traditional computers.)
The possible pouring steps for a pitcher problem form a tree. The
root of the tree is the situation in which all the pitchers are empty.
Connected to the root are as many branches as there are pitchers; each
branch leads to a node in which one of the pitchers has been filled from the
river. Each of those nodes has several branches connected to it,
corresponding to the several possible pouring steps. Here is the beginning
of the tree for the case of a three-liter pitcher and a seven-liter pitcher.
Each node is represented in the diagram by a list of numbers indicating the
current contents of the three-liter pitcher and the seven-liter pitcher; for
example, the list [3 4] means that the three-liter pitcher is
full and the seven-liter pitcher contains four liters.
Actually, I have simplified this tree by showing only the meaningful pouring steps. The program must consider, and of course reject, things like the sequence
and individual meaningless steps like pouring from a pitcher into itself, pouring from an empty pitcher, and pouring into a full pitcher. For a two-pitcher problem there are three possible sources of water (the two pitchers and the river) and three possible destinations, for a total of nine possible pouring steps. Here is the top of the full tree:
At each level of the tree, the number of nodes is multiplied by nine. If we're trying to measure two liters of water, a six-step problem, the level of the tree at which the solution is found will have 531,441 nodes! You can see that efficiency will be an important consideration in this program.
In some projects, a tree is represented within the program by a Logo list. That's not going to be the case in this project. The tree is not explicitly represented in the program at all, although the program will maintain a list of the particular nodes of the tree that are under consideration at a given moment. The entire tree can't be represented as a list because it's infinitely deep! In this project, the tree diagram is just something that should be in your mind as a model of what the program is doing: it's searching through the tree, looking for a node that includes the goal quantity as one of its numbers.
Many programming problems can be represented as searches through trees. For example, a chess-playing program has to search through a tree of moves. The root of the tree is the initial board position; the second level of the tree contains the possible first moves by white; the third level contains the possible responses by black to each possible move by white; and so on.
There are two general techniques for searching a tree. These techniques are called depth-first search and breadth-first search. In the first technique, the program explores all of the "descendents" of a given node before looking at the "siblings" of that node. In the chess example, a depth-first search would mean that the program would explore all the possible outcomes (continuing to the end of the game) of a particular opening move, then go on to do the same for another opening move. In breadth-first search, the program examines all the nodes at a given level of the tree, then goes on to generate and examine the nodes at the next level. Which technique is more appropriate will depend on the nature of the problem.
In a
programming language like Logo, with recursive procedures and local
variables, it turns out that depth-first search leads to a simpler program
structure. Suppose that we are given an operation called
children that takes a node as input and gives us as its output
a list of all the children (one level down) of that node. Suppose we also
are given a command called process that takes a node as input
and does whatever the program needs to do for each node of the tree. (You
can just use print in place of process if you want
to see what's in the tree.) Here is how to do a depth-first search:
to depth.first :node process :node foreach (children :node) "depth.first end
In this program, the structure of the tree is reflected in the
structure of recursive invocations of depth.first.
It might be worthwhile to consider a specific example of how this program works. One of the suggested activities in Chapter 11 was to write a program that takes a telephone number as input and prints out all possible spellings of that number as letters. (Each digit can represent any of three letters. To keep things simple, I'm going to ignore the problem of the digits zero and one, which don't represent any letters on telephone dials in the United States.) Here is a partial picture of the tree for a particular telephone number. Each node contains some letters and some digits. (In the program, a node will be represented as a Logo list with two members, a word of letters and a word of digits.) The root node is all digits; the "leaf" nodes will be all letters.
The operation children must output a list of three nodes, selecting
each of the three possible letters for the first remaining digit. If the
input to children is a leaf node (one with all letters), it must
output the empty list to indicate that there are no children for that node.
to children :node
if emptyp last :node [output []]
output map [child (first :node) ? (butfirst last :node)] ~
letters first last :node
end
to letters :digit
output item :digit [[] abc def ghi jkl mno prs tuv wxy]
end
to child :letters :this :digits
output list (word :letters :this) :digits
end
?show children [tnt 9827]
[[tntw 827] [tntx 827] [tnty 827]]
The top-level procedure has to turn a number into a root node and invoke a depth-first search:
to spell :number depth.first list " :number end
What about the process command? The program wants to print only leaf
nodes:
to process :node if emptyp last :node [print :node] end
» Try this program. To get the tree illustrated above, use the instruction
spell 8689827
Then try again, but investigate the order in which the program
searches the nodes of the tree by using a different version of process:
to process :node print :node end
This will let you see the order in which the program encounters the nodes of the tree.
Writing a breadth-first search is a little more complicated because the program must explicitly arrange to process all the nodes of a given level before processing those at the next level. It keeps track of the nodes waiting to be processed in a queue, a list in which new nodes are added at the right and the next node to be processed is taken from the left. Here is the program:
to breadth.first :root
breadth.descend (list :root)
end
to breadth.descend :queue
if emptyp :queue [stop]
process first :queue
breadth.descend sentence (butfirst :queue) ~
(children first :queue)
end
This breadth-first search program uses the same children and
process subprocedures as the depth-first version. You can try a
breadth-first listing of telephone number spellings simply by changing the
top-level spell procedure to invoke breadth.first
instead of depth.first. What you'll find is that (with the
version of process that only prints leaf nodes) the two
versions produce the same results, but the depth-first program trickles the
spellings out one by one, while the breadth-first version prints nothing for
a long time and then spits out all the spellings at once. If you use the
version of process that prints all the nodes, you can see why.
The telephone number speller is an unusual example of a tree-search program for two reasons. First, the tree is finite; we know in advance that it extends seven levels below the root node, because a telephone number has seven digits. Second, the goal of the program requires searching the entire tree. It's more common that the program is looking for a solution that's "good enough" in some sense, and when a solution is found, the program stops looking. For example, in the pitcher problem program, once we find a sequence of steps to measure the desired amount of water, we don't care if there is also a second way to do it.
For the pitcher problem solver, I decided that a breadth-first search is appropriate. The main reason is that I wanted to present the shortest possible solution. To do that, first I see if any one-step sequences solve the problem, then I see if any two-step sequences solve it, and so on. This is a breadth-first order.
At first, I thought that I would represent each node of the tree as a list of numbers representing the contents of the pitchers, as in the diagram I showed earlier. I called this list of quantities a state. This information is enough to be able to generate the children of a node. Later, though, I realized that when I find a winning solution (one that has the goal quantity as one of the quantities in the state list) I want to be able to print not only the final quantities but also the sequence of pouring steps used to get there. In a depth-first search, this information is implicitly contained in the local variables of the procedure invocations leading to the winning solution. In a breadth-first search, however, the program doesn't keep track of the sequence of events leading to a given node. I had to remember this information explicitly.
The solution I chose was to have an extra member in the list representing a state, namely a list of pourings. A pouring is a list of two numbers representing the source and the destination of the water being poured. Zero represents the river; numbers greater than zero are pitcher numbers. (A pitcher number is not the same as the size of the pitcher. If you enter the instruction
pour [2 5 10] 1
then the two-liter pitcher is pitcher number 1, the five-liter is
number 2, and the ten-liter is number 3.) The list of pourings is the first
member of the expanded state list; pourings are added to that list at the
front, with fput. For example, in the interaction
?pour [3 7] 4 Pour from river to 7 Pour from 7 to 3 Final quantities are 3 4
the extended state information for the final solution state is
[[[2 1] [0 2]] 3 4]
In this list, the sublist [0 2] represents pouring water
from the river into pitcher number 2, which is the seven-liter pitcher.
The sublist [2 1] represents pouring water from pitcher number 2 into
pitcher number 1.
Up to this point I've continued to call this expanded data
structure a state. That's what I did in the program, also, until I
found that certain procedures needed the new version, while other procedures
dealt with what I had originally considered a state, with only the final
quantities included in the list. As a result, my program had local
variables named state in several procedures, some of which contained
the old kind of state, and some the new kind. I thought this might be
confusing, so I did what I should have done in the first place: I invented a
new name for the expanded data structure. It's now called a path;
when you read the program you can confidently assume that :state
represents a list like
[3 4]
while :path represents a list like
[[[2 1] [0 2]] 3 4]
The trouble with using a list of lists of lists in a program is that it can
become very complicated to keep track of all the uses of selectors like
first and constructors like fput. For example,
suppose the value of the variable oldpath is a path, and we
decide to pour water from pitcher number :from to pitcher
number :to. We now want to construct a new path, which will
include a new state (computed from the old state and the two pitcher
numbers) and a new list of moves, with the new move added to the existing
ones. We'd end up saying
make "newpath fput (fput (list :from :to) first :oldpath) ~
(newstate butfirst :oldpath :from :to)
assuming that we have a procedure newstate that computes the
new state. This instruction is hard to read! The two invocations of
fput have quite different purposes. One adds a new move to a
list of moves, while the other connects a list of moves to a state in order
to form a path. We can clarify instructions like this one if we make up
synonyms for procedures like first and fput to be
used in particular contexts. For example, we make a new path using
fput, but we'll call it make.path when we're using
it for that purpose. Just as fput is a constructor, and
first a selector, for lists, we can invent constructors and
selectors for abstract data types (ones that we make up, rather
than ones built into Logo) such as paths:
to make.path :moves :state output fput :moves :state end to path.moves :path output first :path end to path.state :path output butfirst :path end
That unreadable instruction shown earlier would now be written this way:
make "newpath make.path (fput (list :from :to) path.moves :oldpath) ~
(newstate (path.state :oldpath) :from :to)
At first glance this may not seem like much of an improvement, since the new
names are longer and less familiar than the old ones. But we can now read
the instruction and see that it calls a constructor make.path
with two inputs, one that seems to have to do with moves, and the other that
seems to have to do with states. If we remember that a path has two parts,
a list of moves and a state, this makes sense.
» Invent a constructor and selectors for a move data type.
Sentence as a CombinerThe general breadth-first search program I showed earlier contains this procedure:
to breadth.descend :queue if emptyp :queue [stop] process first :queue breadth.descend sentence (butfirst :queue) (children first :queue) end
The most common use of sentence is in generating English
sentences. In that use, the input and output lists are sentences or
flat lists. You're supposed to think, "Sentence takes two words or
sentences as inputs; its output is a sentence containing all the words of
the inputs." In this program, we're using sentence in a different
way, more like what is called append in Lisp. Here you're supposed to
think, "Sentence takes two lists as inputs; its output is a list
containing the members of the inputs." Those members could be words or
lists, but in this case they'll be lists, namely paths.
Recursive procedures that manipulate non-flat lists generally use
fput as the combiner. That wouldn't work here for two
reasons. First, the queue structure that we need to implement breadth-first
search requires that we add new entries at the opposite end of the list from
where we look for the next node to process. If we use first to
select a node and fput to add new candidate nodes, then instead
of a queue we'd be using a stack, in which the newest entries are
processed first instead of the oldest ones first. That would give us a
depth-first tree search algorithm. We could solve that problem by using
lput as the combiner, but the second reason for choosing
sentence is that we don't generate new entries one at a time.
Instead, children gives us several children to add to the queue
at once. That means we must append the list output by children
to the list that represents the nodes already queued.
Pour is going to work essentially by invoking breadth.first on a
root node containing zeros for all the current quantities. But in this case
we want to pick a single node that satisfies the conditions of the problem,
so we must modify breadth.first to make it an operation that
outputs the first such node:
to breadth.first :root
output breadth.descend (list :root)
end
to breadth.descend :queue
if emptyp :queue [output []]
if winnerp first :queue [output first :queue]
output breadth.descend sentence (butfirst :queue) ~
(children first :queue)
end
The predicate winnerp will output true if its input is
a node that satisfies the problem conditions:
to winnerp :path output memberp :goal path.state :path end
If breadth.first runs out of nodes without finding a
solution, it returns an empty list to indicate failure.
Here is a
simplified version of pour:
to pour :sizes :goal win breadth.first make.path [] all.empty :sizes end to all.empty :list output map [0] :list end
All.empty is an operation that outputs a state in which all
of the values are zeros. The number of zeros in the list is equal to the
number of members in its input, which is the number of pitchers. Pour
combines this initial state with an empty list of moves to produce the
first path.
To allow breadth.first to work, we must have an operation
called children that outputs a list of the children of a node.
Starting from a particular state, what are the possible outcomes of a single
pouring? As I mentioned earlier, the source of a pouring can be the river
or any of the pitchers, and the destination can also be the river or any of
the pitchers. If there are n pitchers, then there are n+1
sources, n+1 destinations, and therefore (n+1)2
possible pourings. Here is how the program structure reflects this. I'm
assuming that we've created (elsewhere in the program) a variable called
pitchers whose value is a list of all the integers from zero to
n.
to children :path
output map.se [children1 :path ?] :pitchers
end
to children1 :path :from
output map.se [child :path :from ?] :pitchers
end
to child :path :from :to
output (list make.path (fput (list :from :to) path.moves :path)
(newstate (path.state :path) :from :to))
end
The version of child presented here is simpler than the one
in the actual project, but the other procedures are the real versions.
We'll see later how child is expanded. The immediately important
point is to see how children and children1 ensure that every
possible source (:from) and destination (:to) from zero to the
number of pitchers are used.
You should be wondering, at this point, why children1 uses
sentence as a combiner. (That's what it means to use
map.se rather than map.) It makes sense for
children to combine using sentence because, as I
discussed earlier, the things it's combining are lists of nodes, the outputs
from invocations of children1. But children1 is
not combining lists of nodes; it's combining the outputs from invocations of
child. Each invocation of child computes a single
child node. It would be more straightforward to write the program this way:
to children1 :path :from ;; simplified
output map [child :path :from ?] :pitchers
end
to child :path :from :to ;; simplified
output make.path (fput (list :from :to) path.moves :path) ~
(newstate (path.state :path) :from :to)
end
This also eliminates the use of list in child, needed
in the other versio