CAI~IFORNIA
STATE UNIVERSITY, NORTHRIDGE
LATTICE POINTS:
AP~CATIONS
IN PROBLEM SOLVING
A thesis submitted in partial satisfaction of the
requirements for the degree of Master of Science in
Mathematics
by
Carol Mercer Cocker
.
I
t
June, 1977
·~·---~--·-------·····-
·-·
·-----------------·------·------·-------------------------- ----- - - - -------
-
'
'
The thesis of Carol Mercer Cocker is approved:
E. Muriel .J. Wright, chairperson
California State University, Northridge
~-----~---~-~~--- ~---------- ---~-~-~----~- ---------~-------------------~~----------------
ii
-----------------
ACKNOWLEDGEMENT
My sincere thanks and appreciation go to Dr. Muriel
Wright for her kind encouragement and special guidanceJ and
to Dr. Joel Zeitlin and Dr. v1lilliam Watkins for their
constructive critic ism and manuscript sugge_stions.
My
thanks also go to my husband, Dave, for his patience and
understanding during the past months as well as the more
than
50
diagrams located throughout the text.
Paul and Wilma
~liercer,
My parents,
deserve a special thanks for their
support of this paper shown by the many volunteer hours
of babysitting they have done the past months.
L_______ -----------
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------
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iii
------------------ - - - - - - - - -
---
TABLE OF CONTENTS
ACKNOWLEDGEI\lliNT~ .
...•.•........•..••..•....•........... iii·
TABLE OF CONTENTS ..........•...........•.•.•.•.•.•..... i v
ABSTRACT • .••• a • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • vi
Chapter
page
I..
LATTICE POINTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Definition . ................................. 1
Visible Points on a Lattice ..•.•.•..•..••..• 2
Diagonals on a Lattice ..................•.•. 3
Approximating the Value of 1{2......•.•.•.... 4
Billiards and Lattice Points ...•...•.•.•••.. 5
IT.
AREA OF SilVIPLE POLYGONS ..........•.•...•.•.•.. 9
Def'inition of Simple Polygon ....•........... 9
Pick's Theorem ............................... 9
Squared verses Sheared Lattices ...•.•...... 12
Hexagonal Grids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
Area of' Simple Polygons with "Holes" ...•... 15
Area of Simple Closed Curves ..•.......•.... 18
Approximating the Value of Pi ............•. 20
III ..
A "DIFFICULT CROSSING" PROBLEM .•...•.•....... 22
Statement of Problem ....•.•.•...•.•.•.•..•. 22
Solution by Tree Diagram .•.•...•.•...••.•.. 24
Solution by Lattice Points .••....•.•...•... 26
Conditions Which Make Problem Unsolvable .•. 29
Conditions Which Always Give Solution ...... 31
Additional Problems ........•.....•.•..•.... 32
IV.
THE THREE JUG PROBLEM .•.•.•.•.•.•.•.•.•.•.... 34
Statement of Problem .•.•.•...•.•.•.....••.. 34
Solution by Trial and Error .....•.•..•.•... 35
Solution by Trilinear Coordinate System .... 39
Conclusions Reached by Graphing Solution ... 42
Additional Examples ....•.•.•...•.•...•....• 44
Problem With No Solution ..•.•...•.•••...•.. 48
-----------------
------------------
- - --------------- -
iv
--------------------
V.
THE THREE JUG PROBLEM- CONTINUED ........•... 51
Specific Jug Problem to be Analyzed ......•. 51
Discussion of Points on the Graph ...••.• ... 52
Discussion of Paths on the Graph ......•.•.. 53
Formula for Number of Accessible Points .•.. 64
FOOTNOTES~ . ~
i
BIBLIQGRAP}IY
ft
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67
170
.,---~----------···-------------------------·-
ABSTRACT
LATTICE POINTS:
APPLICATIONS IN PROBLEM SOLVING
by
Carol Mercer Cocker
Master of Science in Mathematics
The purpose of this paper is to investigate specific
problems in mathematics which have solutions on a set of
lattice points.
I
I
In each problem, the introduction of a set!
of points simpl_i:fies the solution and gives a means of
discovering a general solution which can be applied to
I
I
similar :problems.
The paper begins with a study of the lattice points as integers.
Chapter I also gives a method of approximating
irrational numbers.
Later a finite lattice is applied to
the game of billiards.
Chapter I I is a discussion of the area of simple
polygons.
It .includes a proof of Pick's Theorem, Pick's
Theorem adapted to polygons with "holes", as well as two
[methods of approximating pi using·lattice points.
vi
1
'
~------C-h~pters--III,-IV-~andV-consider
!puzzles from recreational mathematics.
the solution of two
The "difficult
!crossing" problem uses a square lattice as one method of
!solution.
I
the
1
II
The "pouring" problem uses a triangular lattice.
In the discussion of these puzzles questions such as
~ollowing
1.
are answered.
Is there a systematic way of finding possible
'solutions?
I
2.
Is there a general method for finding the solution
!which can be used when the conditions of the problem are
changed?
J.
Under what circumstances will there be no
solution?
L
vii
CHAPTER I
LATTICE POINTS
Lattice points can be used to solve problems in many
' areas of mathematics.
Topics in number theory, probability,
combinatorial analysis and geometry, as well as specific
i
puzzles and games in recreational mathematics can be
!
[discussed by the use of lattice points.
Topics taught at
; the elementary and secondary school level can be emphasized
!
and simplified through the introduction of a lattice.
A complete lattice may be regarded as a system of
ordered pairs of numbers or vectors.
If P and Q are the
points {x 1 ,y1 ) and (x 2 ,y 2 ) and 0 the origin then the coordinates of any point S of the lattice based on OP and OQ
:are x
= a 1 x 1 + a 2x 2 and y = a 1 y 1 + a 2y 2 where a 1 and a 2
• are integers.
In general a lattice is any subset of
(a 1v 1 + a 2 v 2 + ... + anvnl ai are integersJ
·For most situations, v 1 = (1,0) and v 2
(0,1) so the
, lattice is the set of integer points.
=
In Chapter I the points of a lattice are discussed.
1The lattice will be the Cartesian coordinate system where
:x
~
0 and y Z:. 0 and x and y are equal to integers.
Figure 1
'shows a section of this lattice which is bounded on the
left side and bottom.
Let~
be the point (0,0).
Let A' be any lattice point
A be the first lattice point along ray OA'.
1
A is
called a visible point because there is no lattice point
between 0 and A to hide A from an observer at 0.
A necessary and sufficient condition for (x,y) to be
•
''
•
•
•
''
•
•
•
•
•
• • • •
•
/§'
•
1-'
•
/3
•
•
•
II
/()
g
•
•
•
8
•
•
•
•
7
6
s
•
•
•
~
s
''
•
•
•
2
•
•
/
0
/
2
9
4
D'
6'
8
9
Figure 1
/4 /5
/{3
/7···
-
-
·visible is that the integers x andy are relatively prime;
:that is that the greatest common factor of x and y is one.
·Each invisible .point has coordinates which have a common
• factor greater than one.
The enlarged points in Figure 1
:show the visible points from (0,0).
In an article by Martin.Gardner 1 observations con: cerning lattice points
~re
made.
After shading the visible
1
· points. certain diagonal paths can be seen.
Two diagonals
from (0,1) and (1,0) which extend up and to the right
'
:contain visible points only.
The reason is because two
: numbers which differ by 1 are always relatively prime.
'
i
(See Figure 1).
Another observation is that there are
certain diagonals running from upper left to lower right
which,. except for the ends, consist entirely of visible
:points.
All these diagonals cut the coordinate axes at
i prime numbers.
Each of the coordinates of the points along
these diagonals have a sum equal to the prime indicated on
the axes at the end of the diagonal.
This indicates that
two numbers whose sum is prime must be relatively prime.
For if they had a common factor, that factor would also
evenly divide their sum.
So far only rays have been discussed which contain a
visible point.
Yet there is an infinite number of rays
from (0,0) which do not have any lattice points.
These
rays would only contain points which have coordinates that
are incommensurable with each other.
'-·--------·--·--·
---------- ------------ -----··---- -·--------------------------------- ----
Such a ray would have
------
-------- ----- ------- -----·-- -- ------ --------------------
r;--sl;p;-;ith a~- irr-~tion~~~i;;~---For example a-ray-can___l
be constructed to represent the
12
I
by constructing an angle
between the ray·and the x axis whose tangent is equal to
112.
Since the slope is equal to the tangent of the angle,
the slope will have an irrational value.
I
Figure 2
I
In Figure 2, the line representingV;Z is shown. The
Ilattic~ points near the line can give a good approximation
I to 1[b_ First write the lattice points as fractions, y/x. II
l
I Then the. points below the line and close to 12 are fractio~
l
smaller than
1'2.
All the points above the line and close
I
to it are fracti-~nsiargerthan
12.
As the ray ..v;f
approaches infinity the points become closer and closer
: approximations of
W.
One way to expressr2 is by the continued fraction
1 + 1
2 + 1
2 + 1
2 + ••.
Starting from the top and forming the partial sums 1, 1 +
I
J1 + 1/C 2 +
1.; ,
I the sums are
1,
1- + 1
I
c2 +
t) ~
1 + 1/ [ 2 + 1/ ( 2 + 1
3/2, 4/3, 7/5, 10/7'".. •
fl ... ,
These are the
I same fractions which are formed by the coordinates of the
lattice points in Figure 2.
As stated before, there is an infinite number of rays
which do not contain lattice points.
If a man stood at
i
i (OjO)
and his line of sight was determined in a random
:way, the probability of his line of vision containing a
visible point would be zero.
This is shovm from the fact
· tha.t the only rays which contain lattice points have
rational slope.
If instead of an infinte array of lattice points,
finite rectangular regions of different sizes are used, an
: application to the game of billiards can be made.
Assume
'
i
the height and widths of the rectangles are integral and
that a billiard ball is placed at (0,0).
Also assume the
, ball will-travel along diagonals which are at 45° angles
sides of the rectangle
an~
will roll forever or
!,
.
-·-·-
-
-- ---
until it hits a corner.
-
-- ---- - --Then whatever the dimensions of
~
.
,the rectangle the ball will hj_ t one of the table's other
three corners after a finite number of rebounds.
In fact
the length of the total path can be calculated as well. 2
Whatever the dimensions of the rectangular table a
;square can be formed by placing a finite number of conigruent rectangles side by side. See Figure J. This is
l
possible by reflecting the rectangle in one and then another
tof its sides.
The smallest square formed by these reflect-
;
:ions will have a length that is the least common multiple
'
iof the rectangles two sides.
I
The path of the rebounding
:ball is replaced by a straight line if the rectangles are
I
!
;thought as a mirror reflection of each rectangle adjacent
i
!
ito
•
1.
t
9
See Figure J.
r- ---- ---------------
-
-----
--- - -
--
-
--
The length of AD is the hypotenuse of a right isoscelffi
:triangle, so the measure of AD is the least common
n.
'multiple of the sides times
It is even possible to predict which corner the ball
will hit without drawing a picture each time.
If both
.sides are odd it will strike the diagonal corner from the
: ( 0, 0) point.
This can be shown by shading the ( 0, 0) point
:and every second lattice point.
The ball will only pass
:through these points and since only one corner is shaded,
:it must be the one the ball will eventually reach.
The
diagonal corner is always shaded when both sides are odd.
'see Figure 4.
·"
•
•
•
•
/
/ "
"
."
/
/
.
0
''
•
~
''
•
•
"
' '·
•
"
/.
•
''
/
Figure 4
If one side is even and one is odd it will strike an
adjacent corner to (0,0) and will be the corner on the
:even side of the rectangle.
This can be shown as above.
!See Figure 5.
•
I
----·
•
•
0
•
0
..
•
•
•
•
•
l
Figure 5
---~-------
-----·-------
----~~---~-----
---
·--~-~-------------------------~---- ~----------------
_____
_;
~------
--~-
--------
--··
·-
----
----
If both sides are even the same argument will not
.hold.
When the possible points are shaded in, all three
•corners are shaded.
This means the ball could hit any one
'of them depending on the length of the sides.
See Figure
l6.
.
•
..
•
•
.
•
Figure 6
However if the numbers are first divided by the greatest
common factor, the rules above will then apply-.
The result
from the division will be either two odd numbers or one
even and one odd and the rules above can be used.
See
Figure 7.
Figure 7
!_ _ _
---
---~-- -~
--~ -~---------------- -------------~----
--------
-----------------------
---
CHAPTER II
AREA OF SIMPLE POLYGONS
•
I
Figure 1
I
Lattice points can be used to find the. area of a
I
By definition a simple polygon is a closed;
simple polygon.
figure, each side a straight line, with no sides having any
other points in common with any other side of the polygon
except vertices.
The vertices must be lattice points.
The;
I
iarea can be estimated quickly by counting the squares and
parts of squares.
I
However in 1899, G. Pick discovered a
theorem which can be used to find the area.
Pick's Theorem:
I
If b is the number of points on the
boundary and i is the number of points 1n the interior of
the polygon, then the area is given by A = tb + i - 1. 3
In Figure 1 above, b = 11, i = 6' so A = tot.
Proof:
Pick 1 s Theorem is easy to show for a rectangle.
I
I
If\
a is equal to the length of the rectangle and w equals the /
width,
I of
th~n
the area A is equal to aw.
The boundary (b)
the rectangle has 2a + 2w lattice points.
~o~nt~_~n__e_ae~
The number
row of the interior is (a - 1) and the
I
1
of~
_j
lnumber of poi~t~-i;-~ach~ol~~ of___ti1-;-lnt~~ior
lSo
i;-(;--=-i)l
-the number of points in the interior is (a - 1) ( w - 1).
I'
Substituting into Pick's Theorem of tb + i - 1,
~(2a
+ 2w) + (a- 1)(w
1)
a + w + aw - a - w + 1 - 1
1
=
=
aw
!which is the area of the rectangle.
If a parallelogram is
placed on a lattice of points the number of lattice points
on the altitude of the parallelogram equals the number on
the width.
So the same argument for a rectangle holds for
Therefore Pick's Theorem is true for a
a parallelogram.
parallelogram.
To complete the proof of Pick's Theorem some assumptions must be made.
1.
Each polygon has its vertices at lattice points.
2.
The polygon can be divided into fundamental tri-
angles.
definition:
A fundamental polygon is one which has no
interior lattice points and no lattice points along the
.
e d ge excep t ver t 1ces.
4
See Figure 2 for one possible
division of a polygon into fundamental triangles .
•
'------~--
_j
Figure 2
- - - - -
.
---·-- --- ·-·-
J.
--·
--·-
--··-.
-- -
-
.
.
-
-
The area of any fundamental triangle is
A
~.
--
-·--·
---·
fundamental parallelogram is made up of two fundamental
triangles.
From Pick's Theorem the area of a fundamental
parallelogram is equal to 1.
Therefore the fundamental
~.
triangle has area of
To complete the proof: .Assume P (polygonal figure) is
partitioned into fundmental triangles.
Using some results
of Descartes, add the angles inside all the fundamental
triangles to get
~F
radians, where F is the number of
.fundamental triangles.
Adding the same angles vertex by
vertex
?r'F = '7((B - 2) + 27(I.
7(
(B - 2) is the sum of the interior angles of a polygon
with B sides and 2 7( I comes from the fact that each inter.br
vertex is surrounded by fundamental triangles.
Then
F = B - 2 + 21.
Since F is equal to twice the area of P,
~F
= ~B
+ I - 1
which is Pick's Theorem.5
There are other proofs of Pick's Theorem. See
Coxeter, 6 Grossman, 7 Kelley, 8 and Niven and Zuckerman.9
It is interesting to note that Pick's Theorem is valid
-
· for lattice points in a plane which are not square.
If a
; parallelogram is used for a unit (a lattice based on
. nonperpendieular vectors) then the lattice points are
. sheared.
j
L_ ____ ·--···---~-----------··---,-- - - - - - - - - - - -
--~-----------~-----~---- -~--
--- ------ - - - - - - - - - -
,------------Figure
1 In
3 the area of the polygon is
~(11)
.
.. .
+ 9 - 1
I
..
•
•
•
·a.
Figure 3
•
walter and
Marlon
s tephen
10 .
ln an ar t•lc 1 e on geo+.
Brown
boards leave several CJ.Uestions about lattice points in a
1
pl_ane unanswered.
IQ!!<l.
I
Two sets of lattice points are given,
1
SQUare, one sheared, with the same unit chosen along
lthe axes.
!
The sheared lattice is placed on top of the
square one so that the bottom points coincide.
1
I
Then they
ask:
1.
Will any of the other points coincide?
2.
Does the answer depe.nd on the angle of the shear? i
I
Gall the angle between the x and y axis on the sheared;
lattice
e.
If the tangent of
e
I
is irrational then none of I
the coordinate points are rational and the points can neveri
I
coincide with the integral coordinates of the square
I
lattice.
However not all tangents of
e
eCJ.ual to a rational!
number give rational coordinates for the sheared lattice.
For example if the shear is one of 45° the coordinates of
the square lattice are on the same ray as the sheared
lattice.
However all the points on the sheared lattice
!
I
'
I
I
'
i
lhave irrational coordinates in terms of the SC[Uare lattice .1
I
A
B
Figure 4
In Figure 4 the "' stand for points on the sheared lattice
and x stand for points on the square lattice.
All distancel
are i:p terms of the.square lattice.
Then the distance AC,
(denoted d ( AC j is 1 and d (BC) = d (AB) .
By the Pythagorean
Theorem d(AB) = d(BC)
'
to ( 1";.._
;).
II lattice
1
1
,
~~)
r/;t.
.
=.J;..
So the coordinate Cis equal
-vi:
1,1
1.
i
All the points along this ray of the sheared i
will be multiples of ( .,2;.
nate must be
;
i
, ~'2. ) •
At D the y coordi-!
i
so again all the points along this ray will
v:l.
1
be multiples of an irrational number.
Therefore all the
points of the sheared lattice system will have irrational
values when superimposed on the square lattice system.
A sheared lattice must be found which has rational
values before it can have any points cqinciding with a
square lattice.
One can be constructed. which will have
points coinciding with a square lattice.
On any of the
sheared lattices a right triangle can be made using a
1
I
Ic oordi nate
( x_,y_)_a_n_d_t_h_e_x_a_n_d_y
__a_x_e_s_.__r_f_t_h_e__d_i_s_t_a_n_c_e_s__:____J
d(BC) a;d- d(AB)areb-~th---~atio-~al-~;d.-s;ti-;;;fy the ~qu~tio;;[
x2 + y 2
= 1 then the sheared lattice will have points
which coincide with the square lattice.
1
One such lattice
II
j
is shown in Figure 5 with tane = 4/3, d(AB) = 3/5, and
d(BC)
= 4/5.
Jt
)(
X
J(
J(
•
X
•
•
)(
.
•
){
'X
'){
){
•
Figure 5
Other tessellating figures can used to develop a grid
!similar to a lattice.
For example a grid of regular
bexagons
can be constructed.
t-
Several questions can then be
asked concerning polygonal figures formed on it. 1 ~
1..
Will Pick's Theorem be valid for finding the area
of the polygonal figures?
2..
If not, can the formula be modified so that the
area can be found using the boundary and interior points
Jof the figlire :?
'L
In Figure 6, there are 10 boundary points, 0 interior
- ·
- - - - - - - ; - - - - - : - - - - - - '
-----~
•
..
•
SJ
.·. <! .
•
.
.. .
•
•
•
.
-
•
• •
Figure 6
1
I
points and the area is 2 unit hexagons.
!(10) + 0 - 1
By Pick's Theorem,!
= 4 I 2,
so Pick's Theorem does not hold.
In fact the area of a
polygonal figure on a regular hexagon lattice is not a
function of' the boundary and interior points alone.
. .
Figure 7
The three figures in Figure 7 each have 4 boundary points,
and_ 0 interior points but each has a different area.
If
a b.exa,;gon is the unit area, then the first has area!, the
second 2/3 and the third
1/J.
So a formula based only on
interior and boundary points is not valid.
Pick"s Theorem can be adapted to simple polygons
which contain one or more "holes" in them.
These "holes"
.LV
-.---
--··-
~--
-·-··--·· ··-·-
-----.-
--·-
must also be simple polygons.
:
~
Figure 8
To find the area of the shaded section, the area of
ithe
l~rge
polygon and small polygon can be found by Pick's
IJTheorem.
Then the difference of the two can be calculated.
!However a formula can be found to find it directly. 12
'
Let b
1
and b
2
be the number of points on the bound-
and i
2
1
be the number of interior points of the two polygons. Let
'aries of the larger and smaller polygons . . J.Jet i
'
iB
and I be the number of boUndary and interior points,
respectively, of the shaded area.
1I = i
,~b
1
1
- i
+ i1
Then B
=
+ b and
1
2
The area of the larger polygon is
- b •
2
- 1 and the smaller polygon is ~b
2
b
2 +
i
2
- 1.
:Then the area of the shaded section is
tb1 + i1
j
'------- - - - - - --
-
1
-
(~b2 + i 2 • - 1)
= !b1
~b2 + 11
i2
= !b1
tb2 + i1
i2
1 + 1
--- - -- - - - - - · - -------------------------------------------
---
-----~-
-------------
ll
I
--------------------------------------------------,
=
-!b 1 + -!b 2 + I
= i(b + b ) + I ·
2
1
I
= iB +I
If there are two holes in the polygon, B
and I = i1
-
l2 - iJ
-
b2 - bJ.
-
ib1 + l1 - 1
(ib2 + i2
-
-
1 + ib3 + iJ
-
ib1
ib2
j,_b
=
ib1
ib 2
ib3 + I + b2 + b3 + 1
=
ib1 + ib 2 + ibJ + I + 1
=
iB + I + 1
J + i1
+ b2 + b
Then
-·
2
= b1
i2
-
1)
i3 + 1
In general the formula is iB + I + n - 1 where n is the
number of holes.
Semi-holes can be of two types.
Figure 9
The first type actually changes the original simple polygon
to a new polygon.
area.
Pick's theorem can be used to find the
See Figure 10.
Figure 1.0
·J.U
r:::h:~h~s:~~on; t t~:\:
I
I of'
semi-holes.
ling
lit
I
Combining the two gives a formula for find-1
I
the area of a polygon with "holes" and "semi-holes."
is iB +I+ !m + n- 1. 1 3
Pick 1 s Theorem can also be useful in finding the area
1
II
:o:m:i:ms~ill~i:::~: :h::o~:ea::::er
of' simple closed curves· such as a circle or ellipse.
To
get a bet-ter approximation of the area of these curved
I figures
I square
finer grids are useful.
However to keep the unit
constant some modifications of Pick's Theorem must
be made.
11}
.........
D
... .
a
b
..........
........
c
I
Figure 11
If 11(a) represents a unit square the area must be onei
and ·by Pick's Theorem 4/2 + 0 - 1 = 1.
grid of
l
In FigurTehe11s(bqu)arae
has been imposed on the unit square.
.J'
stil.l has an area of one but by Pick 1 s Theorem
· 8/2 + 1 - 1
= 4
=/=
1.
Pick's Theorem must be modified so
that the area for (b) is found by*(~+ i - 1).
I same
..
l
In the
way if the grid is ~ the area of (c) is found by
1
I, T6
!
·(b
.
If a grid of 1/n is introduced on the unit
· 2 + ·~ - 1 ) •
square then the formula for finding the area would be
L
A=
-.-2.....,::(~=-----1~) t~
+ i - 1]
--------------------------------------------------------------~
19
rr-;xa~ple
the--area-of a.--q:u~arter--
!units is equal to~.
circle-- o-T rad.ius
two
Using finer grids gives a closer
approximation to pi.
unit square
!(5) +
1 - 1
~ [t(9)
= 2.5
2.875
+ 8
!I
1/16 [
~ ( 21)
+ 41 -
~
=
3.156
1/64
Figure 12
c~(39)
-
+ 183
I
~=J .14B
-- ... ---- --------
---
---··
-
-
Gauss also used lattice points to find approximations
:for the value of
g.
Given a circle with the center at a
' lattice point, ·radius r, r an integer, and f(r) the number
: of points on the boundary and interior of the circle.
• the lim
= /(.
15
. .
. .
.
/
Then
v
•
.
r---
.,........
. . . . . .
. .
~
.
.
""' \ \ .
........
/
I
.
\..
.
7 .
/
..
_. /. . . .
I
. .·""'.~
Figure 13
(
Discussion:
Every square has unit area so f(r) is equal
to the number of squares whose lower left-hand corners are
inside or on the boundary of the circle.
If A(r) is the
area of all the squares cut by the boundary of the circle,
l(r2 the area of the circle, then
lf{r) - 7Tr21
~
liid
£A..(cl
r2
r2
- lr'
A(r)
'All the squares cut by the boundary of the circle are in an
. area between two circles of radii r +
;
l_. ----·· - ------ --- --· ----- - .
1'2 and r - 12. 1/2
21
.
-
···----~-
---.
.
: i~ thG maximum
: ~q~ar~,
This
-.
di~t~n9G
are~
:be
petween any two points on a unit
equ~~
to
'?((~ + 112) 2 - 7r(r - V'2) 2
__ '11 {:r~ + 2f2r + 2 - r 2 + 2V2r - 2)
B(r) __
4 ~7(r
I
from
!~J
(.411277'
r
:r -·
thi~ ~ limitin~ p~QQ~ee
gives lim
r~oe
Usin~ v~tues
fo:r
the f9llQwing are
r=
f(r) found by Gauss empirically,
~ ~ng
e,:pp~Q:}{:bmations ~o
f{?-)=
tr •
f(r)/r 2=
H>
]1/
3·17
~0
!~~?
3.1425
'0
~~iH
3·134
100
J14tr
3.1417
~OQ
1~~€}g~
3.1407
jOQ
~~~g~'f
3.14107
i'-------------------~- ---
0/-
•
···--------------------~-----------------------~--------------------
-----------------------
------·---------- - - - - -
~------------------
CHAPTER III
l
A "DIFFICULT CROSSING" PROBLEM
A group consisting of three cannibals and three
missionaries seeks to cross a river.
which-will hold up to two people.
A boat is available
If the missionaries on
either side of the river are outnumbered at any time by
I
the cannibals on that side, even momentarily, the cannibals!
I
will do away with the unfortunate, out-numbered missionaries.
What schedule of crossings can be devised to permit
j
I
the ent.ire party to cross safely? 16
This· problem in recreational mathematics is one of a
group that puzzle enthusiatist Henry E. Dudeney labeled
"crossing· river pr,oblems".
This type puzzle dates back to
Abbott .Alcuim of the 6th century.
Later mathematicans
such as- Tartaglia and Bachet worked on general solutions
to these, puzzles.
In the solution given by Dudeney, he
states, "Tt would appear that no general rule can be given
for solving these river-crossing puzzles ... As in the case
I
of measuring puzzles we generally have to rely on individual
ingenuity ... t7
However since Dudeney's time, mathematicans !
have use·d methods to generalize these "difficult crossing"
puzzles.
and
In particular, articles by Schwartz,
Farle~.
Cooke, and Detrick
20
methods for solution.
22
18
I
!
Bellman, 19 1
have given some general
,
23
r----- -··-------·-----·--·----.
·-··---· --·---··----------------------·-·---1
The method is to translate the problem into a series
of" moves from one condition to another.
In the problem
above, each crossing of the river is an operation which
!changes the condition.
To further simplify the problem,
the number of missionaries and cannibals can be expressed
as an ordered pair.
If c equals the number of cannibals
on the first side and m·equals the number of missionaries
:
1,,1
I.
I
I
I
I
on ·the fir.st side, then the number of cannibals and
missionaries on the first side can be represented by ( c ,m).
For example (2,J) indicates 2 cannibals and 3 missionaries
on the first side.
It is important to note that at the
same time it also indicates one cannibal and zero
missionaries on the second side, since the total number of
m and c must be
I
J. 21
_using the information above the possible states can
·be listed.
They are:
(0,3)
( 1 3)
J
(2,3)
(3.3)
(0,2)
( 1, 2)
(2,2)
CL?)
(0,1)
( 1 '1)
(2,1)
(3,1)
(0,0)
( 1, 0)
(2,0)
( 3 '0).
These are the possible states but they are not all allowable.
In the puzzle there is the additional condition
I
I
·that the cannibals cannot outnumber the missionaries.
Therefore {J,2),
/ nated.
I
I second
Since
side,
(3,1), and (2,1) are automatically elimi-
(1,2) on the first side means (2,1) on the
(1,2) is not allowable.
Using the same
----------·------------------------------------
I
I
,--------------------·-··------·---·-------------------------------------
reasoning (0,2) and (0,1) are also eliminated.
,.
I
I
in population, of course, indicate which
,people are being ferried which way.
The trips across the
river alternate with a decrease and then an increase as
I
the people are ferried across the river and then return
I
to the init.ial side.
I
See Figure 1 for a complete tree
diagram showing all possibilities with the states not
allowed and returns to previous states crossed off.
II
J
The tree diagram translates into the following solutioj
I
1
I
I
1.
Two cannibals cross the river
2•
One cannibal returns
J.
Two cannibals cross the river
4.
One cannibal returns
.5·
6.
Two missionaries cross the river
7·
Two missionaries cross the river
8.
One cannibal returns
9·
Two ·cannibals cross the river
One carmibal and one missionary return
10.
One cannibal returns
11.
Two cannibals cross the river.
r·----
1
iboat is on _____ side
l
jfirst
Isecond
first
second
first
second
1
first
I
I
!second
I
i
i
1
first
I
I second
I
first
I
(2,0)
t~T§~
(1,1)
-~
(0,0)
i second
I =====
return to previous state
state not allowed
*not a previous state because boat is on first side of
river; in step 2 it is on second side
1-----
Figure 1
------------------------------------------------~--
If the puzzle can be translated into ordered pairs
then the tree diagram makes the problem easier to do.
Although it is still trial and error, it is a systematic
method of getting the answer.
The ordered pairs also
suggest plotting the allowable pairs (c,m) on a set of
lattice points.
0~ c
~
J, 0
~m~
Using a rectangular coordinate system 23
J , and c
=
m or m
=
0 or m
=
J.
0
Figure 2
The ten allowable states are shown in Figure 2.
The
object is to connect the points, called vertices, so that
the path leads from (J,J) to (0,0) within the limitations
given in the problem.
Each.path between two points
represents a crossing of the river, and is called an edge.
The following observations are important in finding the
paths possible for this problem.
1.
When the boat goes from the first to the second
side, c, m, or both c and m decrease.
is either one or two.
I graph which are
l triangle
.The total decrease
Therefore the only points of the
r~achable
from a given point lie in a right
with the given point at the upper right
27
vertex.
2.
When the boat goes from the second side back to
the first side c, m,. or both c and m will increase and
J
5imilarly the only reachable points are in a right triangle
with the given point at the lower left vertex.
For example, from the point· (3,3) the possible moves
are shown by the right triangle in Figure· 3·
The points
within the triangle are the allowable vertices.
So the
allowable paths are to (1 ,3), (2,3), and (2,2).
Drawing
the possible paths from each allowable point the graph
is constructed.
See Figure 4.
Figure 3
3·
Figure 4
Each path can be traveled in either direction
since all ferryings are reversible.
And since trips from
I
GO
--·-
the first to the second side must alternate with trip~---~
-~-----------·-----~--------------------·-
from the second to the first,· paths leading down or to the
left must alternate with paths leading up or to the right.
In-the words of Farley, "A method for finding a
solution to the puzzle can now be described in the followin ,
terms.
Starting at the upper right corner of the graph,
make a sequence of transitions to allowable points in
the reachable triangles, moving alternately down or to
the left and up or to the right, until the lower left
corner is reached." 24 After a little exper~mentation
the solution shown is:
Figure 5
The graphing method may not be much better than a
systematic trial and error or a tree diagram but the graph
I
can now be used to answer questions about this and other
puzzles.
1
I
1.
Is the solution the only one?
2.
In general if one starts with some other number of
cannibals and missionaries, is the puzzle. solvable?
3·
Which solution if there are several, requires
rthe- f'ewe-st
cr·o-ssingso-f' theri ver?_______
The f'ollowing are some statements, some with proof',
wh.ich will answer the questions above.
I.
There are f'our distinct solutions, all requiring
eleven crossings.
The retracing of' steps or returning to
j ~revious states are not allowable soluti.ons.
Proof':
categories:
First classif'y the allowable states into three
T f'or those along the top where m = 3, B f'or
those along the bottom where m
the diagonal where 0 ~ m = c =. J.
=
0, and D f'or those along
Since the boat holds
only two people it is impossible to go directly f'rom a
T to a B category.
So any path must include a D point.
From (2,3) or (3,3) to (2,2) the next step must go to (2,3)
~
or (3,3) and nothing is gained.
T state it must go to (1,1).
Jf'rom (1,3).
and (2,0).
When the path leaves the
The only way to do this is
From (1,1) to a B point is then to go to (2,2)
The solution path must contain the sequence
(1,3)--:,
(1,1)~
(2,2)- (2,0).
(1,3) can be reached f'rom (3,3) by
· ( 3 , 3) -
( 1 , 3) -
( 2 , 3) -
( 0 , 3) ~ ( 1 , 3)
(3,3)- (2,2) _,. (2,3)-- (0,3)
~
or
(1,3).
There are two ways to reach (0,0) f'rom (2,0).
Therefore
there are exactly four possible solutions each requiring
eleven crossings.25
II.
I
Four cannibals and four missionaries cannot be
ltaken across by a~oat holdi~ only two peoEle.
)V
Figure 6
Proof:
The graph for four cannibals and four
missionaries is constructed in Figure 6.
To leave the T
points without immediately returning one must go from
(2,4) to (2,2).
From (2,2) the path must- go back to (2,4)
or go to (J,J).
B points.
III.
From (J,J), there is no way to get to the
26
Therefore the puzzle has no solution.
If the boat holds three people instead of two,
up to five cannibals and five missionaries· can safely
cross.
Discussion:
For three persons in the boat the
reachable states change.
For trips across the river the
right triangle is shown in Figure ?(a).
For the returning
trips it 1s shown in Figure ?(b).
b.
a
Figure 7
::!
~b~ca::e t:: :::::b~:a:e:::a:~::::::s ~~: ~~:::::ary
Iin
·the boat.
u·sing this reachable diagram it is possible
to :find solutions for four and five persons, but not six.
The .impossibility proof for SlX is similar to that in
sta:tement I I above. 2 7
IV.
If the boat holds four people any number of
I
cannibals and equal number of missionaries can cross.
When four persons are allowed ln the boat
Proof:
th.e reachable states change.
See Figure 8 .
•
Figure 8
!Again classify the allowable states into three categories
IT,
I
I
~B
and D as . was done in the proof of I.
Since there
are two reachable points on the diagonal (D) it is now
possible to travel up and down the diagonal regardless of
the number of cannibals and missionaries.
A partial
gra_ph for m missionaries and cannibals is found in
Figur:e
9.
back to (m
A
path can go from (m,m) to (m - 2,m -2)
1 ,m
-
1) down to (m
Sa :a soluti,.on can always
next page •
L
- J,m -
J) and so on.
28
be found. See Figure 9 nn the
I
·•
..
.
, .,.---.
/
Figure 9
Other variations of this puzzle can be solved or
shown impossible by using the graphing technique.
Two
examples are as follows.
Problem 1~
Five missionaries and four cannibalspwerissh ntso.
cross a river using a boat which holds only.two
,~
0
The canni.bals cannot outnumber the missionaries at any
/
29 !
Draw a diagram and show the puzzle has a solution.
i
Solution: The allowable states are (0,5), (1,5), (2,5),
time.
(J,5)' (4,5), (J,4), (4,4)' (2,J), (J,J)' (1 ,2)' (2,2)'
(0,1), (1,1), (0,0), (1,0), (2,0), (J,O), (4,0).
The diagrams showing the paths and one possible solution
are shown in Figures 10 and 11
/f1'l/
~
e
Figure 11
33
·-----
-----~----~-----------------
Using m for missionary and c for cannibal the solutioni~
1.
One c and one m cross the· river; one c returns.
2.
One c and one m cross the river; one m returns.
3·
One c and one m cross the river; one c returns.
4.
One c and one m cross the river; one m returns.
5·
One c and one m cross the river; one c returns.
6.
One c and one m cross the river; one m returns.
7·
One c and orne m cross the river; one c returns.
8.
One c and one m cross the river.
Problem 2.
Suppose there are four missionaries and two
cannibals who want to cross a river using a boat which
holds only two persons.
The cannibals are stronger than
the missionaries to the extent that the missionaries are
safe only when they exceed the cannibals in number (on
shore and in the boat).
Draw a diagram and show the
puzzle does not have a solution.3°
~
1
Figure 12
The only way to get to (1,2) from aT state is from (1,4).
From (1,2) the next step must increase and would return to
(1,4).
The only way to get to a B state and possible
solution is through (1,2).
solution.
Therefore there is no
I
CHAPTER IV
THE THREE JUG PROBLEM
Two men have a jug filled with eight liters of wine
:which they wish to divide equally between them.
They have
empty jugs which are five and three liters.
!they accomplish this division?3 1
How can
,~wo
This puzzle as with most problems in recreational
•mathematics has certain unwritten assumptions in it.
For
this puzzle natural assumptions would be:
1.
The jugs are exactly eight, five, and three liters.
2.
When the wine is poured, none is spilled.
J.
No wine is lost to evaporation or to drinking.
4.
The jugs are unmarked and there are no other
~._-
measuring devices available.
This last assumption is
. important, for when pouring wine from one ·jug to another,
the pouring stops when one jug is empty or the other is
'full, whichever occurs first.
This decanting puzzle was first given by Nicola
:Fontana (Tartaglia) a 16th century Italian mathematican.
:Even in the early 20th century it was bel_ieved that it could
,only be solved by trial and error.
In.1917, Dudeney,
known for his books in recreational mathematics wrote,
"It
.is the general opinion that puzzles of this type can only
be solved by trial, but I think formula~ can be constructed
'for the solution.
It is a practically unexplored field
34
J...J
:------------··-----·---~_32--
I
o-:f investigation."
Since that time mathematicans have
found systematic ways of approaching this puzzle and other
rdecanting problems of the same type.
l
In searching for solutions to this puzzle several
questions about its solutions can be asked.
1.
Is there a method of· keeping track of the sequence
l
I
I
pourings?
i
2.
Is there a systematic way of finding possible
solutions?
J.
Is there a general method for finding the
I
solution I
of this type of puzzle?
4.
Does the solution give more information than just
the solution, such as, all possible solutions or those
cases where there are no solution?
5.
Is there a best solution?
6.
Can the methods of solution be used on other
puzzles unrelated to the jug problem?
These are the kinds of questions answered in this
chapter.
The first approach to solving this decanting puzzle
will be trial and error.
According to Schuh, trial is an
activity that is important and useful when solving puzzle
problems.
However as he says, even "if you are lucky
enough to f'ind a solution, you will not know whether this
is the only solution let alone how many solutions there
Iare. n)J
The method should really be call·ed systematic
L_________________________________________~
jO
,trial for it is not just a guessing and trying aga1n
:method but a systematic way of arriving at a solution.
Some type of table, list or diagram is kept so that as
;pourings are made you know exactly which cases have already
:been considered and which have not.
In a systematic trial,
'records should not only tell accurately which :possibilities
·have been Bxamined, but also which solutions have come from
:one :possibility and which from another.
Bellman suggests
i
:a tree diagram as his systematic procedure for constructing
!solutions to the :pouring puzzles.
This makes use of the
I
'fact that whatever the contents are at a given stage, only
!certain pourings are :possible.
When constructing the tree,
'
itriples (x,y, z) are used where x, y, z are the contents of
Jthe 8, 5, 3 jugs respectively.J 4
So (8,0,0) means 8 liters
•in the 8 liter container and no wine in the 5 or 3 liter
container.
As the branches develop, if a duplication or
previous state occurs those branches will stop.
Fi~re
See
1.
The tree must stop growing as there is a finite
[number of possible states for the three containers.
In
this case, the 8 liter container can contain 0, 1, 2D 3, 4,
5, 6, 7, or 8 liters of wine or 9 possibilities.
The 5
/liter has 6 possibilities and the 3 liter has 4 :possibil)ities, so it would have to repeat after 9 x 6 x 4 or 216
! :pourings. - In this particular example however, it is not
'
:possible to :pour more than
15 times without returning to
-----------
I
I
I
I
{5,0,3)
(3,5,0)
~·
~
(3,2,3) (8,0,0) (0,5,3)
~
~
(5,0,3)
(3,5,0) (5,0,3)
I
(8,0,0) (0,5,3) (5,3,0)
I
'
''
Figure 1
------~
J
,-----------~-
---~--
----
---
---·------·--
--~--------------.-
---~----------
Ia previous state.
-··---------------
Since the tre.e diagram represented the liquid in the
·three jugs as a triple (x,y,z) it suggests several possibil!ities for representing these triples geometrically.
One
way would oe to use them as coordinates of a point in a
three dimensional space.
This is introduced by Bellman in
the book Algorithms, GraJ2hs, and Com:Ruters.35
The set of
ipossible states correspond to a set of points.
From the
•conditions stated in the problem these coordinates are
·integers, nonnegative, x f: 8, y
~
5, z .:S 3 and x +. y + z = 8.
!Then all points (x,y,z) lie on the plane x + y + z
!the positive octant.
>the conditions.
=8
in
There are 24. triples which satisfy
In Figure 2 several of the triples are
i
;illustrated.
Since x + y + z = 8, if two points are
given, the third can be found.
So if x and y are recorded
. on a rectangular coordinate system z can b.e found from
:z
'
=
8 - x- y.
This 2 dimensional representation may
'
:be more convenient.
y
Figure 2
---
------------------------·-------------------~- -~-----
-------~-------------------------------·
---
·---
----
Looking at the intersections of the x, y, z planes,
x, y, z integers, we have the-trilinear coordinate system
that is introduced by Coxeter in the book Geometry Revisited.
'A trilinear coordinate system uses three sets of parallel
lines which divide the paper into a lattice of equilateral
;triangles.3 6 On this paper an equilateral triangle is
drawn whose altitude is that of the amount of liquid given.
Each side represents one of the three jugs.
From each of
:these sides the parallel lines are counted, one line for
'each liter of the measure of the jug.
The distance then
corresponds to the size of each container and a line is
idrawn parallel to the side at this distance.
The figure
I
i
:formed will be a convex polygon of 3 to 6 sides.
The trilinear coordinates of any point P are distances
x, y, z, of P. from the three sides of the triangle and
'can be represented by the coordinate (x,y,z).
Theorem:
Proof:
The trilinear coordinates have a constant sum.
Given triangle ABC with side a, altitude h.
The
)coordinate P(x.y,z) is the distances x, y, z of P from the
three sides BC,
I
AC~
•inside the triangle.
and AB.
This is positive because Pis
Then !ax + iay + iaz = area of Ll PBC +
area of ClPCA + area of
~PAB =
area of b. ABC = iah.
:Therefore x + y + z = h and is therefore a constant.
End
·of proof.37.
One o-f the coordinates stays fixed while the other
two vary with a constant sum.
Th~
point (x,y,z) moves
:along a line parallel to one side of the triangle so that
'the operation of pouring liquid from the second to the
third is represi:mted by a motion of the point (x,y,z) along
;a line where x is constant, y decreases, and z increases.
[Points corresponding to the possible contents for all the
'measures at any one time will·be on the boundary or the
interior of the figure formed.
1
Using the same problem
:(8,5,3) this method is illustrated.
Call the vessels a, b, c and the amount of liquid they
contain x, Ys z.
;will be x
!
:x and y
Then x + y + z
= 8, andy and z = 0.
= 4 and z = 0.
=
8.
The initial state
The final state will be
Using the trilinear paper an
'
I
:equilateral triangle of altitude 8 is drawn.
(The altitude
i
must be the amount of liquid).
'between x = 0, x = 8, y
=
The possible states lie
0, y = 5, z = 0, z = J.3 8 Figure
3 is a projection from the three dimensional coordinate
!
= 8.
:system (Figure 2) to the plane x + y + z
~the
In Figure J,
figure has been flipped to correspond to Coxeter's
lnotation.
X-::.0
~
,,
0
Figure 3
j~---------------------------·--·__:·--~-----------------~-----------
- - - - - - - - - - - - - ------- - - - - ----- - - - - - - - - - - - ----
T.L
1.:-
-~-------~-------------------
.
iWhen the water is transferred from one vessel to another,
the amount in the -third remains unchanged.
The effect.of
a single transfer is that the point travels along one of
the lines within the limits of the polygon.
Since the ves-
sel is completely emptied or filled, the transfer is
represented by a point on the boundary.
In this case the point (8,0,0) represents the initial
state.
The broken lines from this point represent the
two possible operations from (8,0,0).
For example _the path
to (3,5,0) represents filling the second jug from the
first.
See Figure
LJ-
Figure 4
The lines in Figure 5 show one of the various ways
of going from (8,0,0) to (4,4,0).
The
~ath
goes along
a direction parallel to one side of the triangle, until
it reaches a side.
At the point it reacts like a billiard
ball bounding back along another line.
1
I
--------·
Figure 5
Although the puzzle is solved us1ng this method,
several questions come to mind concerning the procedure.
Why was the particular :path chosen?
2.
Are there other paths which will give the solution ..
J.
Is this path the shortest?
4.
Will the procedure work on any jug problem?
helps to answer these questions.
with a 1 the initial point
Iis
I
I
~o
On the diagram mark
(8~0,0).
already marked with a number.
luntil
J
1.
Tweedie, in an article from the Mathematical Gazette
I
I
The points that can
Continue in this manner
f'urther labeling is possible or until the solution
-
reached.39 See Figure 6,
From this simple procedure the possible paths to
(4,4,0) are quickly seen.
More importantly conclusions
Figure 6
can be obtained that were not possible by studying the
i
!
tree diagram.
This procedure first es:tablishes the order
in which filling the containers is performed.
From this
it shows whether or not the problem is even solvable.
Once
it is established that there is a solution it gives all
possible solutions and which is the shortest or best
solution.
This simple graph gives even more information.
It
tells whether all measures ·from 1 to 8 can be measured.
It
also tells how many moves it would take to reach any
particular point and therefore which measurement would
:be
most difficult to obtain.
l
:
i
Notice that the figure obtained in the previous
example was a parallelogram.
Dependinp on the conditions
in the problem this figure can be an equilateral triangle,
a regular or irregular hexagon, a pentagon, a trapezoid,
, or a parallelogram.
L_ ________________________________ ~-------------------·-~--~--------------·----------------------___;
point will be on the inside of the
. figure if none of the jugs are completely full or empty.
'
I
It will be on the boundary if at least one is either full
j
or empty and on the vertex if two or three are either
~
full or empty.
This diagram also shows that only boundary points are
possible, making all interior points impossible.
These
inside points cannot .be reached because each pouring
operation stops when either the jug from which pouring is
t
empty or the one into which pouring is full.
At the
! conclusion of each step there is at least one jug which is
i
; either full or empty.
If ne = the number of empty jugs
land nf =the number of full jugs then ne + nf> 1. 40
i
!' Even
if the initial point is that of an interior coordinate,
• after one pouring, every state will be on a point of the
:boundary.
A few more examples with some explanation and comment
will show what a powerful tool this simple graph can be in
solving and analzying these problems.
The original problem of (8,5,3) was special in several
ways.
The amount of wine was equal to that of the largest
; container.
Also the sum of the two smaller jugs equalled
I
i the larger.
Here is a similar problem which does not have
!
)these conditions.
A marl has three jugs of 7, 6, and 3 ·liters and 8 liters
i
of liquid.
It is distributed to _the three jugs so that
l----------·---··--------------------·--------·-------·-·-·------~----------;
I.
lt·h~~e---~~-;3- li t~-~;-f~-the?-iiter -c~ntaine~~- i-~- the
container~
6 liter
1
wishes to divide the liquid into two equal parts.
1
and 2 in the 3
lite~
1
container.
He
41
l
i
''
1.·
!
I
This may sound more complicated than the first problemJ
1-.·ut
'.)
the g1..;aph quickly solves it for us.
First make an
I
'
I
!equilateral triangle of height 8 and draw the size of the
I•
'three containers parallel to the three sides. Locate the
I
!initial position (J,J,2) and the final position (4,4,0).
7
Figure 7
Using Tweedie's method1 number the points possible
from (J,J,2).
See Figure 8.
One possible solution is
then (J,J,2) - (5,3,0) - (?,1,0) - (4,1,3) - (4,4,0).
The figure gives more information than _just the solution.
It also says that in this case. all measllres between 1 and
7 can be measured.
It also gives all states which are not
possible, such as (1,5,2).
Remember that no interior
!point is possible~
--------------------------~--------------------~
' .
Figure ·8
In the last example, the figure formed was a hexagon.
I
In the next example, the figure formed has 5 sides.
A
I
I
!man has three jars whose capacities are 10, 6, and 5 liters
I!
The 5 liter jar is empty and the others contain 6 liters
each.
I
He wishes to get J liters in one jar, 4 in another
land 5 in the other.
I
1
1
What are the simplest ways of getting
th"1s d".
t "b u t"1on. 42
1s·r1
I
The six points
(5~.394), (5,4~J),
(4,J,5), (4,5,J),
(J,4,5), and (J,5,4) each satisfy the conditions required
lin the final answer but only (4,J,5) and (J,4,5) are
accessible from (6,6,0) since they are the only ones which '
lie on the boundary of the diagram.
Figure 9 shows the initial state and the possible
I final
states.
Figure 10 shows that (4,J,5) can be
jreached in 4 moves and (J,4,5) takes 8 moves.
I solution would
The best
be (6,6,0), (10,2,0), (10;0,2), (4,6,2),
'I
---
-·--·-
----
----
-·
--- -- --·- --- --- --
~
------ -----·--------,
(4,J,5).
/0
Figure 9
Figure 10
I
----~------
~
~
--
------------~~----------~--
--------------
-------
--
I
~-------·--·--------------·----
---·
·-·---·---------·-------,
.
/where all the measures are possible.
(All the boundary
jpoints were accessible from the initial point).
Icourse
solvable.
I "
'true.
Of
it would be interesting if for all problems, all
jmeasures Could be found.
1
I
All the examples given so far have produced cases
Then every jug puzzle would be
1
But one example will show this is not always
Take three jugs of measure 8, ?,
of liquid.
1
6 and 10 liters
Put 8 in the 8 liter, 1 in the 7 liter and
1 in the 6 liter.
.
1
I
1
Using the same method, number the points
accessible from (8,1,1).
Figure 11
is quickly seen that
not have a number.
(0,5,5), (5,0,5), and (5,5,0) do
In fact unless
5 liters is initially ini
5 canna t b e meas.ure d f rom any lnl
. . t"l alli
.
I
Notice that (0,5,5), (5,0,5), and (5,5,0) form a
'
.
one of the cont alners,
I state.
triangle which cannot be entered from another path.
ligure 12.
See
'/
---·-----~-
·----
{o,s;s).
Figure 12
In general if t is the total amount of liquid and a, b, c,
the size of the containers and t
4
d cannot be measured. 3
Proof:
= 2d 2 a> b "> c :;;>d, then
If d cannot be measured the points on the boundary
of the polygon equal to d must be inaccessible from other
points on the polygon.
This means they must be on a closed
curv:e; a curve not accessible from other paths formed from
the boundary points.
The coordinates (x,y,z) which contain d must be
(t,t,O), (t,O,t), (O,t,t).
2 2
2
2
2 2This is argued from the fact that on the boundary of the
polygon one of the coordinates must be 0.
Since d
= t, and 1I
2
the sum of the coordinates is equal to t, the other two
coordinates must each be t.
2
I
r---------~----;----------------~--~------
I
1
---------------------------:-----l
The three coord1nates (t,t,O), (t,O,t), (O,!,t) form
2 2
2
2
2 2
an equilateral .triangle.
a closed curve.
The. equilateral triangle forms
I
Because all the. possible paths reflect at a
60 degree angle, there
~s
no way to enter the equilateral
triangle from another path.
measured.
Therefore d cannot be
End of proof.
Another case where all measures are not possible is
when the containers are of a size 8, 6, 4.
The problem
is that the measures of the containers are not relatively
prime.
In fact the only measures possible are multiples
. .
44
In general then the jugs measure
of th e common d1v1sor.
·must be relatively prime before it is relevant to consider
whether every measure is possible.
in Chapter V ~
[
Since 2d2 a"?> b ':7 c 7 d, the vertices of the '
!triangle must be along the boundary of the polygon, not
at a vertex.
i
I
This will be discussed
. r··-·
____ ..
---~
----~--------
-----·---
··---~
-· -------- -·-----·-
---~--·
------ --·-----
CHAPTER V
THE THREE JUG PROBLEM - CONTINUED
The original problem ( 8, 5, 3) is a; special case of the
pouring problem, for the total is equal to the capacity of
:the larger jug and the sum of the two smaller jugs is
;equal to the largest.
Through the use of the trilinear
'
:coordinate system this special case will be analyzed
.. -----·--···
~.~-------
algebtGl.i.Q?-lly.
.
:-~,..~'·"·····"'-'""_....
This analysis follows that for the Sawyer
·-· .-,.._,,, ___ ,__...
graphs introduced by Bellman in the book Algorithms, Graphs
'
4
and Computers. 5
The purpose of this analysis is to find
some general results for pouring problems.
In particular
we find when all the points of the figure are accessible.
Since general results are the goal, the containers
must be of arbitrary sizes.
Call the containers A, B, and
C and the amount of liquid each can hold a, b, and c.
total amount of liquid is t.
The
The assumptions are:
1.
a, b, and c are positive integers.
2.
a = b + c
A typical state is the trilinear coordinate (x,y,z)
where x, y, and z indicate the amount in A, B, and C.
Since a, b, c do not have definite values the diagram
: is drawn symbolically.
L_ ---------
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - --- --
51
---------
---~-
---- ---- - - -
------------ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - · - - - ,
Figure 1
The interior points are inaccessible as they represent
states with no container full or empty.
The points on the
parallelogram are the ones that are possibly accessible;
there are 2a such points since 2b + 2c = 2a.
Each of these
2a points is in one of three sets.
I.
or empty.
The points where every container is either full
These are the 60 degree vertices of the
parallelogram or (a,O,O) and (O,b,c).
There are two paths
from these coordinates and neither is a interior path.
II.
The points where exactly two containers are
either full or empty.
These are the 120 degree vertices of
the parallelogram or {c,b,O) and (b,O,c).
There are three
paths from these coordinates. One is an interior path.
III.
The points where exactly one container is
either full or empty.
These are the coordinates of the
sides of the parallelogram.
There are four paths from
these points. Two of these paths are interior paths.
These three sets can be seen by checking the diagrams.
'Def~ini.:t.:hQ_l'J:
A path is called a reversible path if given
two states it is possible to pour ,from one to the other
and then return to the previous state. 46
k_,_
/i
A reversible path is possible when the jug from
which~
)
l_pouring is full or the one receiving is empty.
:Theorem:
/
/ .
From each state in set III exactly four
;pourings are possible, two reversible and two not
:reversible.
'Proof:
I
To make a transition from one state to another
'there must be two containers used;
one to pour into.
one to pour from and
'rhere are six ways to make this choice.
I
-·
;However if a jug is empty it is impossible to pour from
i
•
: l.
!
t
,
or if a jug is full it is impossible to pour into
One of these cases occurs in the set III state so the
• number of outgoing paths is four.
• is either full or empty.
' reversible operations.
There is one jug that
Therefore 'there must be two
Either pouring from the full
• container into each of the other two or the pouring into
: the empty from each of the other two jugs. This completes
the proof. 4 7
Using the diagram it is easy to see that four paths
; are possible.
The two reversible paths are those leading
to another side of the parallelogram (the interior paths)
or a point in set III.
The two nonreversible paths are
those leading to points (b,O,c) and (O,b,c), a vertex.
-------~-~------------------·------------------------------- \_
_____
-~-----
----------- ---
They are points of set I or set II.
See Figure 2.
(o, b,c)
Figure 2
Suppose we begin at (c,b,O).
There are three possible
pouring operations, resulting in (a,O,O), (O,b,c) and
(c,b-c,c).
(a,O,O) and (O,b,c) are in set I and (c,b-c,c)
; is in set III.
Since there are two reversible operations
• from each point of set III, a path is formed between
; points in set III.
i
'-··--·-- ----~-~~ ~ ---- ---~----
- - - - - - - - - - - - - - - - - - - - - ~--------------------- --
---
~heorem:
There is an unique reversible interior path
beginning at (c·,b,O) and ending at (b,O,c).
'Proof:
The boundary points are in one of three sets.
;Since set I, (a,O,O) and (O,b,c) have no interior
·reversible paths, the points ·cannot be part of any interior
/path beginning at (c,b,O).
Points of set II (c,b,O) and
· (b,O,c), have one reversible interior path; the path
;will begin or end at this type of point.
;the path begins at (c,b,O).
We assume that
Set III contains points
which have two reversible interior paths.
: it and one to leave it.)
(One to reach
Both paths are used as the path
goes to and away from the point.
The path cannot pass
through it again unless it re-hits a previous point so
1
there is no first repetition.
1
repetition until it ends.
The path continues without
It must end for there are·= a
finite number of boundary points.
Therefore it must end
at (b,O,c), the only point not used in set II.
This
completes the proof.
However it does not follow that this path contains
all of the points of set III.
There may be one or more
separate paths not connected to the others.
i if' the problem is
For example
(10,6,4) the path consists of (4,6,0),
{4,2,4), (8,2,0), (8,0,2), {2,6,2), (2,4,4), (6,4,0), and
{6,0,4).
{6,0,4) is equal to (b,O,c).
'----------------------·· - - - - - - - -
However other
----~------·------·-----~-----
-
-------~-
----
------
,--------------------·-----~---,--,---,.---
---
points of set III such as (1,5,4) are not included.
I
· ·
d oes t he pa t h 1nc
· 1 u d e a 11 po1"nts.!.;
Under what cond1t1ons
jof set III, making all points accessible?
This is the
!question the following remarks will answer.
However beforel
Ianswering 'the question it is convenient to subdivide the
Ipoints of set III. Set III contains the· points where one
I•
I container is full or empty and the other two contain
lliquid but are not full or empty.
subsets.
There are four disjoint
In the first two subsets, one jug is empty, and
no jug is full.
(1).
(x,O,a-x) where xis b
~
x< a.
x·-,.b because a - x (c and since a - c = b, x)'b.
a - x
<c
( 2) •
is true because the third container is not full.
( x, a-x, 0) where x is c < x < a .
x > c because a - x < b, or a - b < x.
Since a - b = c
X)' C.
A third case where jug A is empty is not necessary
because (O,b,c) is of set r·.
In the second two subsets
one jug is full and no jug is empty.
(3).
(x,b,c-x) where x is 0 < x< c.
(4).
(x,b-x,c) where xis O<x<.b.
Through the remainder of this chapter the subsets will be
denoted (1), (2), (J), (4). 48
The four subsets each refer to one side of the
!parallelogram of the general trilinear diagram.
I
[Figure
J.
I
See
1
1
I
I,
Figure 3
Using these subsets the analysis of the pouring
that began at (c,b,O) continues
\Theorem:
algebraical~y.
The unique reversible path beginning at (c,b,O)
will contain a sequence of states of types (4) and (2).
Proof:
The first step is (c,b-c,c).
From (c,b..:..c,c) we
jean return to original (c,b,O) or go to (2c,b-c,O).
lis possible since 2c < b + c =a.
This
The next reversible
I
step is to (2c ,b~2c ,c).
If b - 2c > 0 then the path
continues until b - 2c <
o.
When b - 2c
~
0 then
(2c,b-2c,c) equals (2c,O,c) which is set I.
steps
set - subset
(c,b,O)
II
( c, b-e, c)
III
(4)
(2c,b-c,O)
III
(2)
(2c,b-2c,c)
III
(4)
if b - 2c
7
0
(Jc,b-c,O)
III
(2)
if b - 2c
<.
0 then
(2c,O,c)
set I
lEach time a new pair of set III state is added to the path
---·--··---··--··--------- · - - - - -
the content of jug A goes up by c and jug B goes down by
c.
These steps can be carried out as long as the content
of the first jug (jug A) does not exceed a and the amount
!withdrawn from jug B does not exceed b.
Figure 4
Or in general
steps
set - subset
( c ,b ,0)
II
(c .b-e ~6)
III
(4)
(2c,b-c,O)
III
(2)
( (n-1)c, b-(n-1)c, c)
III
(4)
(nc, b-(n-1)c, 0)
III
(2)
.
where n is the largest integer such that nc6 a.
End of
proof •.49
So far it has been shoWh that a path from (c,b,O) to
(c.b-c,c) begins as a series of points which alternate
!between points of subset (2) and (4).
i
I
•
As the path
lcont1.nues, one of two things will occur.
Either the path
~
i
I
I
point from
This should be self-evident from the figure
ion the previous page.
I
i
How the path continues depends on
lthe relationship between a and c.
!Theorem:
If a\c is an integer then the path stops at
I
i{b,O,c) and only includes points of type (2) and (4).
I
~Proof:
Given that n is ·the largest integer such that
I
Inc ~a and the ( 4) to ( 2) path continues to (A) which equals
l
I< (n-1)c, b-(n-l)c,c) to
Iprevious theorem.
(B)~
Since a
(nc, b-(n-l)c, 0) from the
= b + c, nc ~ b + c or ( n-1) c 6 b.
I
!If aJc is an integer nc = a or (n - 1)c =b. Then (A)
1becomes (b,b~b,c) or (b,O,c) which is the vertex of set II
1
and (B) is (nc,b-b,O) or (a,O,O).
Therefore the path
1
stops.
End of proof. 50
From this it is also seen that there are (n - 2)
pairs of points of subset (4).and (2) and therefore 2n
.points on this connected path.
'
An
ex~!fiple
,.-.""' ''''"~--~·'·'""""'=,.,.""'"-"~-·--~'>'-";:l
will help understand the previous theorems.
Take jugs measuring (12,10,2).
Then the path that begins
at (2,10,0) to (2,8,2) is the following.
(2,10~0)
(c,b,O)
(2,8,2)
{c,b-c,c)
(4,8,0)
(2c,b-c,c)
(4,6,2)
(2c,b-2c,c)
(6,6,0)
(Jc,b-2c,O)
I (6,4,2)
(Jc,b-Jc,c)
I
'
(8~~4,0)
i
·
~
~- ~
---~~-·~c
4c
~b ~
3c, o)
~~-
~
(8,2,2)
( 4c , b -4c , .o )
(10,2,0)
(5c,b-4c,O)
(10,0,2)
(5c,b-5c,c) or ((n- 1)c,b -(n- 1)c,6)
(12,0,0)
(6c,b-5c,O) or (nc,b - {n - 1)c,O)
Figure 5
·There are (n-2) pairs of (4) to (2) points or in this.
example, 4 pairs or 8 points.
path or 12.
There are 2n points on the
(The eleven above plus the other vertex (0, 8,2)
:which can be reached from (b, 0, c).)
The second possibility mentioned above is that ale is
·not an integer.
'Theorem:
If ale is not an integer then the path beginning
lat (c,b,O) to (c,b-c,c) will begin with a series of points
i
:from subset (4) and (2), contain a point of subset (1),
,contain a point 0f subset (J)' and then begin a series of
i
7 •
:(4) to (2) points again.
In other words the path will have
·the general formula of
·(c,b,O)
II
:cc,b-c,c)
III {4)
i (
2c , b -c , 0)
III (2) .
.
'{{n-1)c,b-(n-1)c,c)
III (4)
(nc,b-{n-1)c,O)
III (2)
' {nc, 0, b- (n-1) c)
III (1)
(nc-b ,b ,b-·(n-1)c)
III (J)
:(nc-b, 2b-nc, c)
III (4)
:CCn + 1)c-b,2b-nc,O)
III (2)
:((n + 1 ) c -·b , 2b-n ( n + 1)c,c)
III (4)
;
:((n + 2)c-b,2b-(n + 1)c,O)
Proof:
Then
If
ale
III (2)
is not an integer then nc(a <(n + 1)c.
((n-1)c,b-(n~1)c,c)
and (nc,b-(n-1)c,O) are of subset
. (4) and (2) since b - (n- 1)c :.>0 and nc <.a.
However it
, is not possible to withdraw a full c amount from jug B.
The next step consists of emptying jug B into jug C.
Then
(nc,b-(n-1)c,O) becomes (nc,O,b-(n-1)c) which is a set III,
, subset (1) point.
From a subset (1) point, a reversible
. step is one that goes into jug B.
We can pour from
• jug A or jug C, but if we pour from jug C we return to
(nc,b-(n-1)c,O).
So the step which gives a new point
. is to pour from jug A.
Jug B will be filled and the new
. state is (nc-b,b,b-(n-1)c).
L--·-~------·--
[Why will B be filled?
Since a ( ( n + 1) c or a
<nc
or a - c = b, nc >·b J
This new state is of subset (J).
+ c or nc
> a - c.
Since a = b+c
i
The next step is pouring from the full jug into jug C
which is (nc-b,2b-nc,c).
((n+1)c-b,.2b-nc,O).
(subset (4):f
(subset (2)]
=b
therefore b - (nc
nc + c - b.
~nc
2b - nc =
Now c "'7 nc - b because
- (nc - b))
c ;;-nc - b _,. nc < b + c
It is possible because
b < a . [Why?.
jug B, 2b-nc )' 0 and (n+1) c
b - nc + b
The next step is
<a which is true.
b =
\
(n + 1) c - b < a + c - b.
Since I
b) ,. b - c .,. 0.
Since nc <a,
Since c > nc -b J
(n + 1) c
b>c, a+ c- b <a and therefore (n + 1)c- b<.a.
There-:1.
fore there will again have a sequence of points of subset
1
(4) and (2). End of proof.5
The path of points from the previous theorem continues
until jug B contains an amount y with 0
~ y!:
c.
If y is a
multiple of c then, as it did when a\c, it· will become
empty and the path will end.
For example if some multiple
of c is exactly 2b say (m -· 1)c
= 2b.
Then the path ends
with ((m- 1)c- b,2b- (m- 1)c,c) or (b,O,c) and then
(me- b,2b - (m- 1)c,O) or (a,O,O).
The total pairs of
points on the path is (m - 1) + 1 + (m - n + 1)
The total number of points
l.S
2(m + 1)
= m + 1.
= 4a/c.
If, as in the case where ale is not an integer, no
multiple of c is 2b then let m be the largest integer such
I that ( m - 1 ) c < 2b < me or
I(me
0
< 2t. - (m - 1 ) c < c .
- b,2b - (m -·1)c,O) the next step will be
Then from
!
i
Arguing as before I
r(mc - b,O ,2b :.. (m - 1)c) or subset (1).
'the following steps will be (mc-2b,b,2b-(m-1)c) (subset (J)J
(mc-2b,Jb-mc,c} (subset (4)), ((m + 1)c-2b,Jb-mc,O)
(subset (2)).
So the path beginning at (c,b,O) to (c,b-c,c)
(n - 1) pairs of points, the second (m - n) pairs.
I the
Adding I
pairs of (1), (J) points the number of pairs of points I
I down to the second (4) to (2) path is n - 1 + 1 + m - n = ml
.I
The purpose of this discussion is to see if there
I
I
are patterns in the pouring problems which will lead to
some general conclusions.
For example a formula giving
the number of points on a path from (c,b,O) to (b,O,c)
would be useful in checking if all possibly accessible
I
1
l
1
points are accessible from. (c.,b ,0).
The last two points on the path were (mc-2b,Jb-mc,c)
land ((m + 1)c-2b,Jb-mc,O) of subset (4) and (2).
This
next path of (4) to (2) points will be
(mc-2b Jb-mc, c)
It
..
((m + 1)c-2b,Jb-mc,O)
I ((k-1)? - 2b,Jb-(k-1)c,c)
I (kc - 2b • Jb - ( k -1 ) c , 0 )
III (4)
III (2)
III (4)
III (2)
I
!where k is the greatest integer such that 0 <Jb-(k-1)c<c.
!This can also be written (k-1)c<Jb <kc.
Iof
pairs of points will be m + 1 + k- m
The total number
=k
+ 1.
The nextl
'
(4) to (2) path will end with ((j-1)c-Jb,4b-(j-1)c,c),
(jc-Jb,4b-(j-1)c,O) where j is the greatest integer such
that 0
< 4b-(j-1)c <c or (j-1)c < 4b < jc.
The number of
points will be j + 2.
Now suppose N is the smallest nonnegative integer
such that c divides (N + 1)b.
Then the final (4) to (2)
path will be
(A)= ((p-1)c- Nb,(N + 1)b- (p-1)c,c)
(B)
= (pc -
Nb, (N + 1)b
(p - 1)c,O)
(A) will be of set II and equal (b,O,c) and (B) will be of
set I and equal (a,O,O).
Then (N + 1)b- (p- 1)c =
or (p - 1)c = (N + 1)b.
The total number- of pairs of
( 4) , ( 2) , ••• ( 1) , ( 3) , ( 4) , ( 2) , • • • is p + N - 1.
0
[
Adding
I
the original (c,b,O) and {O,b,c) to this number the follow-:
ing theorem is proved.5 2
Theorem:
Let N be the smallest nonnegative integer such
that c divides (N + 1)b, arid let p be defined as (p - 1)c =
(N + 1)b, from above.
Then· the total number of accessible
points is 2(p + N) = 2(N + 1)a . 53
c
An example will show this result.
(27,15,12).
Then a= 27, b = 15, c
Take three jugs of
= 12.· First find
the smallest nonnegative integer such that (N + 1) 15
12
an integer.
I
In this case, N
= J.
(3 + 1)(15) or p- 1 = (15){4)
12
I number
is
Then (p - 1)12 =
and p = 6·. · Then the
of accessible points or the number of points on the
-
ipath from (c,b,O) is 2(p + N) = 18 or 2(4)(2?) = 18.
12
'There are 2a possibly accessible points or 54, so 18 are
'accessible from this path.
:corollary:
If b and c have no common factor then all points
'are accessible.
Proof:
From the theorem
1)c = (N + 1)b
(p
pc - c = {N + 1)b
pc = c + (N + 1)b
p
=
c + {N + 12b
c
!
iSince b and c have no common factor, and p must be an
!integer, (N + 1)
N
=
= c.
The smallest value_ of N is then
c - 1.
p
= c + ((c-12 + 1)b
c
p
= 1 +
or p = 1 +b.
cb
c
:using the formula of the theorem
2((1 +b) + (c - 1)) = 2(b + c)
= 2a.
2a is the number of all possibly accessible points so the
path contains all accessible points.5 4 End of proof.
In the (8,5,3) problem, a
and p = 6.
! 2a
= 8,· b = 5, c = 3, N = 2,
The number of points is 2(p + N) or 16.
= 16.
Corollary:
If b and c have a common factor then not all
;points are accessible.
Proof:
If b and c are not relatively prime there exists
----~--~·-··-~----------
---~--
- - - - - - - - - - - - - - - - - - - - - - - · - - - - - - - - - - - - - - - - - - - - -------I
vv
an integer k 71 such that k is the greatest common factor
of b and c..
= kb 1 and c = kc 1 .
Then b
From the theorem 1
, N is the smalle.st nonnegative integer such that c divides
'
+ 1)kb 1
I
kc1
Then
:(N+1)b.
(N
and for c 1 to divide
( (N + 1)b , c 1 = N + 1. Then .N = c - 1.
1
1
From the theorem, the total number of accessible points is
2{p + N) or 2(N + 1)a .
c
.Substituting c 1
=N
+ 1 into 2(N + 1)a
accessible points is
;' 2c a
1
c
=
the number of
c
=
Now c
2a
k
= kc 1 so
<. 2a.
. Since 2a is the number of possible accessible points, the
!
number of accessible points is less than the possible
: number of accessible points. ·End of proof.
-
~--------------~
-------- - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - -
'i
I
FOOTNOTES
1.
Martin Gardner, "The Lattice of Integers," Scientific
American, 212 (May 1965), 120-126.
2.
Ibid., 121.
J.
H. Steinhaus, Mathematical Snapshots (New York: Oxford i
University Press, 1969), 96.
Duane DeTemple and Jack Robertson, "The Equivalence
of Euler's and Pick's Theorem," Mathematics Teacher,
67 (March 1974), 222.
5.
Ibid., 224-225.
6.
H.S.M. Coxeter, Introduction to Geometry (New York:
John Wiley and Sons,: 1969), 209.
7.
Howard Grossman, "Fun With Lattice Points," Scri:Qta
Mathematica, 12 (1946), 87.
8.
David Kelley, "Areas of Simple Polygons," The
Pentagon, 20 (Fall 1960), J-11.
9.
Ivan Niven and H.S. Zuckerman, "Lattice Points and
Polygonal Area, " American Mathematics Monthly, 74
(1967), 1195-1200.
.
10.
Marion Walter and Stephen Brown, "What If Not?,"
Mathematics Teaching, 46 (Spring 1969), 41.
11.
Ibid., 45.
12.
A.G. Marshall, "Pick~ With Holes," Mathematics
Teaching, 50 (Spring 70), 67.
1J.
Ibid., 68.
14.
J.B. Harkin, "The Limit Concept on the Geoboard,"
Mathematics Teacher, 65 (January 1972), 13-17.
15.
D.Hilbert and S. Cohn-Vossen, Geome.try and the
Imagination (New York: Chelsea Publishing Co, 1952),
JJ.
I
I
I
bts
-------------------------
16.
I
Robert Fraley, Kenneth Cooke, Peter Detrick, "Graphical!
Solution of Difficult Crossing Puzzles," Mathematics
Magazine,39 (May 1966), 151.
I
1
17.
Henry Dudeney, Amusements in Mathematics (New York:
Dover Publications, 1970), 236.
18.
Benjamin Schwartz, "An Analytic Method for the
"Difficult Crossing" Puzzles," Mathematics Magazine,
34 (1961), 187.
19.
Richard Bellman, Kenneth Cooke, JoAnn Lockett,
AlgQrithms, Graphs, and Computers {New York: Academic
Press , 197 0 ) , 196 •
20.
Farley, op. cit., 1.51-157·
21.
Ibid. , 151.
22.
Bellman, op. cit., 197·
23.
Ibid., 198.
24.
Farley, op. cit., 153·
25.
Ibid., 153·
26.
Ibid. , 153·
27.
Ibid. , 154.
28.
Ibid. , 154.
-~
29 •. Bellman, op. cit., 203.
30.
Ibid., 204.
31.
Ibid. , 142.
32.
Dudeney, op. cit., 109.
33·
Fred Schuh, The Master Book of Mathematical
Recreations (New York: Dover Publishing 1968), 6.
34.
Bellman, op. cit., 145.
J5.
Ibid._, 149.
36.
H.S.M. Coxeter, Geometry Revisited (New York: Random
House, 1967), 89.
.
I
r---1
1
37.
Ibid., 89.
38 ..
T.H. O'Beirne, Puzzles and Paradoxes (New York: Oxford i
University-Press, 1965), 51.
M.C.K. Tweedie, "A Graphical Method of Solving
Tartaglian Measuring Puzzles," The Mathematical
Gazette, 23 (July, 1939), 279.
40.
Bellman, op. cit., 167.
41.
Coxeter, op. cit., .90.
42 ..
Tweedie, op. cit. , 281.
4J.,
Coxeter, op. cit., 91 ..
44 ..
Ibid. , 92.
45.
Bellman, op. cit., 174.
46.
Ibid., 178.
47.
Ibid., 179·
48.
Ibid. , 180.
49-
Ibid. , 181.
50-
Ibid., 182.
51 ..
Ibid., 183.
52 ..
Ibid., 185.
53·
Ibid.
54-
Ibid. , 186.
-~
L _____
---,----___1
U';t
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