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BINARY AND TERNARY
INVESTIGATIONSDr Chris Pritchard
The Mathematical Association Annual Conference 2016
University of Oxford
Binary Application 1: Cuisenaire rods
4
1 + 3Take a 4 rod and make the
same length using a
combination of other rods.
Task 1: Find the other six possibilities? 1 + 1 + 1 + 1
1 + 1 + 2
1 + 2 + 1
2 + 1 + 1
2 + 2
3 + 1
To cut or not to cut,
that is the question!
4
1 + 3
Using a 1 for a cut and a 0
for the lack of a cut we can
produce a binary coding.
100
000
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 11
0
1 1
0 0
0 1
1 0
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 11
0
1 1
0 0
0 1
1 0
Length Length Length Length
2 3 4 5
Discuss
Number of
patterns for each
length
Symmetries in rods
and binary codes
Abandon the physical model:
Length 6, i.e. 5 cuts (32 possibilities)
11111 11011 10111 10011 01111 01011 00111 00011
11110 11010 10110 10010 01110 01010 00110 00010
11101 11001 10101 10001 01101 01001 00101 00001
11100 11000 10100 10000 01100 01000 00100 00000
Task 3: Consider the rods up to length 5. Identify the patterns arising
from using rods of length 1 and 2 only.
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 11
0
1 1
0 0
0 1
1 0
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 11
0
1 1
0 0
0 1
1 0
Fibonacci
Sometimes
presented
as a steps
problem
Length 6: where are the Fibonacci steps now?
00000 00001 00010 00011 00100 00101 00110 00111
01000 01001 01010 01011 01100 01101 01110 01111
10000 10001 10010 10011 10100 10101 10110 10111
11001 11001 11010 11011 11100 11101 11110 11111
Remove any binary codes with consecutive zeros.
00000 00001 00010 00011 00100 00101 00110 00111
01000 01001 01010 01011 01100 01101 01110 01111
10000 10001 10010 10011 10100 10101 10110 10111
11001 11001 11010 11011 11100 11101 11110 11111
01010 01011 01101 01110 01111
10101 10110 10111
11010 11011 11101 11110 11111
13 possibilities
Binary Application 2: Bubbles
Here are two bubbles
drawn in different ways:
Task 4: Jot down the different ways of showing 3 bubbles.
Moving towards binary
Crop top and bottom
Interpret as a sequence o brackets ( ( ) ) ( )
Code an opening bracket as a 0
and a closing bracket as a 1. 0 0 1 1 0 1
Since each such binary string starts
with a 0 and ends with a 1 we
could remove these two digits.
0 1 1 0
Task 5: Now find the binary codes for the other three
configurations to complete the set.
1010 0110 0101 0011
Why does each code have the same number of zeros and ones?
Task 6: There are two more such codes. Find them,
draw the bubbles diagrams and interpret your answers.
1001 ( ) ( ( ) )
1100 ( ) ) ( ( )
Now we can go on to generate all possible patterns
with 4, 5, 6, … bubbles starting with full sets of binary
codes of the correct length.
Binary Application 3: Folds and creases
Reference: ATM’s Points of Departure 1 (1997).
Take a strip of paper and fold it repeatedly over from right to left.
0 0 1
0
After folding once there is
an ‘in fold’, coded 0.
After folding twice there are
three folds (in, in, out),
coded 001
Task 7: Jot down the binary code we get after 3 folds. Is there
an easy way of generating the next binary code each time?
0 0 1 0 0 1 1
For the next design each time, put in a new pivot at the end
and then reflect the previous sequence changing all creases
(like taking complements). In folding on the pivot each 0 hits a
1 and vice versa, each out fold tucks into an in fold.
0 0 1 0 0 1 1 0 0 0 1 1 0 1 1
Work in progress: folding in two directions
UNDER-UNDER
UNDER-OVER
OVER-UNDER
OVER-OVER
UL UR DL DR LU RU LD RD
1
0
0
0
0010
Example: over-over up left
Binary Application 4: Nim
Nim is an ancient game, probably originating in China and
coming into Europe in the early sixteenth century.
In its standard form, there are three piles of counters or
stones or matches. Players take turns to remove any
number of the objects from one pile, with the winner being
the player who takes the last.
The name ‘Nim’ was assigned by Harvard mathematician,
Charles Leonard Boulton (1869-1922).
In 1901, he analysed the game using binary.
Example: begin with piles of size 9, 12 and 11
which in binary are 1001, 1100 and 1011.
1001
1100
1011
1110
Sum without carrying
Winning strategy is to always reduce the sum to zeros.
1001
1100
1011
1110
1001
10
1011
0000
Reduce the middle row from
12 to 2 by removing ten
matches from the middle pile.
Example of a move in a winning strategy:
From Binary to Ternary: Weighing
Task 7: We have a pan balance which we intend using to
weigh some object. The object is placed on one pan
while the weights go only on the other pan. The smallest
weight is of 1 unit and there is only one such weight. All
other weights are multiples of the basic unit weight and
there is only one of each. Which weights should we
choose to produce an efficient system of weighing?
We need a 1 and a 2, but not a 3 because 3 = 2 + 1 …
1 = 1
2 = 2
3 = 2 + 1
4 = 4
5 = 4 + 1
6 = 4 + 2
7 = 4 + 2 + 1
8 = 8
9 = 8 + 1
1 x 1
1 x 2
1 x 2 + 1 x 1
1 x 4
1 x 4 + 1 x 1
1 x 4 + 1 x 2
1 x 4 + 1 x 2 + 1 x 1
1 x 81 x 8 + 1 x 1
1
10
11
100
101
110
111
1000
1001
Binary
4 2
181625
25 is 11001 in binary
But what if we were allowed to place weights on both pans?
We could also weigh 25 by moving the 1 onto the left-hand
pan and adding the 2 to the right-hand pan.
A binary weights system is not now the most efficient.
Task 8: What is the best system now?
1 = 1
2 = 3 – 1
3 = 3
4 = 3 + 1
5 = 9 – 3 – 1
6 = 9 – 3
7 = 9 – 3 + 1
8 = 9 – 1
9 = 9
10 = 9 + 1
11 = 9 + 3 – 1
1 = 1
2 + 1 = 3
3 = 3
4 = 3 + 1
5 + 3 + 1 = 9
6 + 3 = 9
7 + 3 = 9 + 1
8 + 1 = 9
9 = 9
10 = 9 + 1
11 + 1= 9 + 3
1 x 1 1
1 x 3 + (-1) x 1 1 11 x 3 1 0
1 x 3 + 1 x 1 1 1
1 x 9 + (-1) x 3 + (-1) x 1 1 1 1
1 x 9 + (-1) x 3 1 1 0
1 x 9 + (-1) x 3 + 1 x 1 1 1 1
1 x 9 + (-1) x 1 1 0 11 x 9 1 0 0
1 x 9 + 1 x 1 1 0 1
1 x 9 + 1 x 3 + (-1) x 1 1 1 1
Signed ternary
Chris Pritchard ‘A balanced approach to number bases’,
Mathematics in School, May 2016
£14.99 to MA members