Round2: Maths Challenge organised by ‘rep2rep’ research group at Sussex&Cambridge universities

Full name 1: Mohammed Alasmar
Email: M.ALASMAR@SUSSEX.AC.UK
GENERAL INFORMATION
Problem & Approaches to Solutions
MATHS CHALLENGE1
http://users.sussex.ac.uk/~gg44/rep2rep/1
The problems that have been solved:
Problem 1 - Two Integers
Problem 3 - City Grid
1
09/12/2018
Full name 2: Alexander Jeffery
Email: A.P.Jeffery@SUSSEX.AC.UK

2
Problem 1 - Two Integers
Two different integers between 1 and 10 are chosen at random. What is the
probability that they are consecutive?
Problem 3 - City Grid
You and your friend live in a grid-like city. Every edge of the grid is a road. The distance from
your place to your friend's place is n roads north and m roads east. You and your friend
decide to visit each other but both of you leave at the same time and walk at the same
speed. If both of you take 'optimal paths' (minimising the length of the paths), what's the
probability that you will meet on your way?

4
Index Number 1 Number 2
1 1 2
2 1 3
3 1 4
4 1 5
5 1 6
6 1 7
7 1 8
8 1 9
9 1 10
10 2 1
11 2 3
12 2 4
13 2 5
14 2 6
15 2 7
16 2 8
17 2 9
18 2 10
19 3 1
20 3 2
21 3 4
22 3 5
23 3 6
24 3 7
25 3 8
26 3 9
27 3 10
28 4 1
29 4 2
30 4 3
31 4 5
32 4 6
33 4 7
34 4 8
35 4 9
36 4 10
37 5 1
38 5 2
39 5 3
40 5 4
41 5 6
42 5 7
43 5 8
44 5 9
45 5 10
46 6 1
47 6 2
48 6 3
49 6 4
50 6 5
51 6 7
52 6 8
53 6 9
54 6 10
55 7 1
56 7 2
57 7 3
58 7 4
59 7 5
60 7 6
61 7 8
62 7 9
63 7 10
64 8 1
65 8 2
66 8 3
67 8 4
68 8 5
69 8 6
70 8 7
71 8 9
72 8 10
73 9 1
74 9 2
75 9 3
76 9 4
77 9 5
78 9 6
79 9 7
80 9 8
81 9 10
82 10 1
83 10 2
84 10 3
85 10 4
86 10 5
87 10 6
88 10 7
89 10 8
90 10 9
Case 1:
Permutations without replacement i.e., the two numbers are
different and order is considered, for example (1,2) and (2,1) should
be taken in the outcome.
We define the consecutive numbers as the numbers which follow
each other in order, without gaps, from smallest to largest.
Picking two different number between [1,10] gives 90 possible
outcomes, where 9 of them represent the cases of getting
consecutive numbers, as shown in Fig.1 (the yellow cells).
Hence, the probability that the two numbers are consecutive can be
found as follows:
! =
9
90
= #. %
Fig.1

5
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
Case 1:
We define the consecutive numbers
as the numbers which follow each
other in order, without gaps, from
smallest to largest.
Fig.2(a) shows are the possible
outcomes from picking two
different number between [1,10].
These are represented by the black
squares in the figure, which are 90
squares. The cases of consecutive
numbers are represented by the
yellow squares in Fig.2(b), which are
9 squares.
Hence, the probability that the two
numbers are consecutive can be
found as follows:
All outcomes (the black squares = 90) The cases of consecutive numbers
(the yellow squares = 9)
! =
number of yellow squares
number of black squares
=
9
90
= #. %
Fig.2
Fig.2(a)
Fig.2(b)
Number 1 Number 1
Number2
Number2

6
2
3
4
5
1 6
7
8
9
10
1
3
4
5
2 6
7
8
9
10
1/9
1/10
1/10
1
2
4
5
3 6
7
8
9
10
1
2
3
5
4 6
7
8
9
10
1/10
1/10
1
2
3
4
5 6
7
8
9
10
1
5
3
4
6 5
7
8
9
10
1/10
1/10
1
2
3
4
7 5
6
8
9
10
1
2
3
4
8 5
6
7
9
10
1/10
1/10
1
2
3
4
9 5
6
7
8
10
1
2
3
4
10 5
6
7
8
9
1/10
1/10
1/9
1/9
1/9
1/9
1/9
1/9
1/9
1/9
Case 1:
The tree diagram is used in this solution (see below). The first stage in this tree (picking the first number at random) contains ten outcomes: [1:10], so the probability of each one of these
outcomes is 1/10. In the second stage, 9 possible numbers can be selected at random. Only one of these 9 numbers is a consecutive number to the number that was selected at the first stage
except for number 10 which does not have any consecutive number.
Thus, the probability that the two numbers are consecutive can be found as follows: ! = #×
1
10
×
%
#
=
1
10
= &. %

7
The Permutation is defined as a selection of numbers in which the order of the numbers matters, and there is no
replacement. This is the case of this question i.e., we solve it with order and without replacement. Hence, we can use
the permutation formula to find the total number of possible outcomes, as follows:
!(#, %) = # % =
#!
# − % !
10 2 =
10!
10 − 2 !
= 90
Now, the number of choices with consecutive numbers = # − 1 = 10 − 1 = 9
This is because each number has a consecutive but not the last number (this is based on our definition of consecutive
numbers, i.e., the consecutive numbers are the numbers which follow each other in order, without gaps, from smallest
to largest)
Therefore, the probability of choosing consecutive numbers is
Case 1:
! =
# − 1
#!
# − % !
=
10 − 1
10!
10 − 2 !
=
9
90
= 0.1

8
If we want to pick 2 different numbers from
the set of numbers [1:10], then there are 10
options to pick from for the first number and
9 options for the second number, as shown
in Fig.3.
Thus, the total number of selections =
number of available values for number 1 x
number of available values for number 2 =
10 × 9 = 90
Now, for the first picked number there is
only one consecutive number expect for the
largest number, which does not have any
consecutive number (see Fig.4). This means
that there are 9 possibilities of getting
consecutive numbers.
Therefore, the probability of choosing
consecutive numbers is
& =
9
90
= '. )
10 9 8 7 6 5 4 3 2 1
123456789
Case 1:
Fig.3 the number of available numbers for picking
10 available
Pick number 1 Pick number 2
9 available
Fig.4 each number has a consecutive but not 10
No consecutive

9
clc,close all
x = 1:10;
y = 1:10;
counterAll = 0;
counterCons = 0;
for i=1:10
for j = 1:10
if y(j)==x(i)+1
plot(x(i),y(j),'rs','LineWidth', 15)
counterCons = counterCons+1;
end
hold on
if x(i)~=y(j)
plot(x(i),y(j),'bd','LineWidth', 4)
counterAll = counterAll+1;
end
end
end
grid on,xlabel('number 1'), ylabel('number 2')
set(gca,'fontsize',15)
legend('consecutive','all outcomes')
prob = counterCons/counterAll
prob = 0.1000
Running the code gives the probability value and Fig.5.
By using the following Matlab code.
Fig.5

10
The second number is bigger than the first. If the first number is k (which happens with probability 1/n), then
the probability the second number is 1/(n-1).
Case 1
! = #
$%&
'(&
1
*
×
1
* − 1
! = #
$%&
&-(&
1
10
×
1
9
= 9×
1
10
×
1
9
=
1
10
= 0.1

11
Case 2:
Solving the problem using combination, i.e.,
there is no replacement and order is not
important, for example, we just take one of
these groups (1,2) and (2,1).
We define the consecutive numbers as the
numbers which follow each other in order,
without gaps, from smallest to largest.
Picking two different number between
[1,10] gives 90 possible outcomes, where 9
of them represent the cases of getting
consecutive numbers, as shown in Fig.6 (the
yellow cells).
numbers are consecutive can be found as
follows:
! =
9
45
= #. %
Fig.6
Index Number 1 Number 2
1 1 2
2 1 3
3 1 4
4 1 5
5 1 6
6 1 7
7 1 8
8 1 9
9 1 10
10 2 3
11 2 4
12 2 5
13 2 6
14 2 7
15 2 8
16 2 9
17 2 10
18 3 4
19 3 5
20 3 6
21 3 7
22 3 8
23 3 9
24 3 10
25 4 5
26 4 6
27 4 7
28 4 8
29 4 9
30 4 10
31 5 6
32 5 7
33 5 8
34 5 9
35 5 10
36 6 7
37 6 8
38 6 9
39 6 10
40 7 8
41 7 9
42 7 10
43 8 9
44 8 10
45 9 10

12
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10
Case 2:
We define the consecutive numbers
as the numbers which follow each
other in order, without gaps, from
smallest to largest.
Fig.7(a) shows are the possible
outcomes from picking two different
number between [1,10].
These are represented by the black
squares in the figure, which are 45
squares. The cases of consecutive
numbers are represented by the
yellow squares in Fig.7(b), which are
9 squares.
numbers are consecutive can be
found as follows:
All outcomes (the black squares = 45) The cases of consecutive numbers
(the yellow squares = 9)
! =
number of yellow squares
number of black squares
=
9
45
= #. %
Fig.7
Fig.7(a)
Fig.7(b)
Number 1 Number 1
Number2
Number2
1 2 3 4 5 6 7 8 9 10
1
2
3
4
5
6
7
8
9
10

13
2
3
4
5
1 6
7
8
9
10
3
4
5
2 6
7
8
9
10
4
5
3 6
7
8
9
10
5
4 6
7
8
9
10
5 6
7
8
9
10
6
7
8
9
10
7
8
9
10
8
9
10
9
10
Case 2:
The tree diagram is used in this solution. The first stage in this tree (picking the first number at random) contains ten outcomes: [1:10], so the probability of each of these outcomes is 1/10. In the
second stage, 9 possible numbers can be selected at random. Only one of these 9 numbers is a consecutive number to the number that was selected at the first stage except for number 10 which
does not have any consecutive number. Now, the probability that the two numbers are consecutive can be found as follows:
! =
number of consecutives
total number of possible outcome
=
3
45
= 6. 8

14
The Combination is defined as a selection of numbers in which the order of the numbers does not matter,
and there is no replacement. This is the case of this question i.e., without order and without replacement.
Hence, we can use the Combination formula to find the total number of possible outcomes, as follows:
! ", $ =
"
$
=
"!
" − $ ! $!
10
2
=
10!
10 − 2 ! 2!
= 45
Now, the number of choices with consecutive numbers = " − 1 = 10 − 1 = 9
This is because each number has a consecutive but not the last number (this is based on our definition of
consecutive numbers, i.e., the consecutive numbers are the numbers which follow each other in order,
without gaps, from smallest to largest)
Therefore, the probability of choosing consecutive numbers is
Case 2:
. =
" − 1
"
$
=
" − 1
"
2
=
10 − 1
10
2
=
9
10!
10 − 2 ! 2!
=
9
45
= 0.2

15
If we want to pick 2 different numbers from
the set of numbers [1:10], then there are 10
options to pick from for the first number
and 9 options for the second number, as
shown in Fig.8.
Thus, the total number of selections =
10
2
= 45
Now, for the first picked number there is
only one consecutive number expect for the
largest number, which does not have any
consecutive number (see Fig.9). This means
that there are 9 possibilities of getting
consecutive numbers.
Therefore, the probability of choosing
consecutive numbers is
' =
9
45
= 0.2
10 9 8 7 6 5 4 3 2 1
123456789
Case 2:
Fig.8 the number of available numbers for picking
10 available
Pick 2 numbers randomly without replacement
Fig.9

16
clc,close all
x = 1:10; y = 1:10;
counterAll = 0; counterCons = 0;
for i=1:10
for j = i:10
if y(j)==x(i)+1
plot(x(i),y(j),'rs','LineWidth', 15)
counterCons = counterCons+1;
end
hold on
if x(i)~=y(j)
plot(x(i),y(j),'bd','LineWidth', 4)
counterAll = counterAll+1;
end
end
end
grid on,xlabel('number 1'), ylabel('number 2')
set(gca,'fontsize',15)
legend('consecutive','all outcomes')
prob = counterCons/counterAll
prob = 0.2000
Running the code gives the probability value and Fig.10.
Fig.10

18
• My friend and I each walk towards each other’s houses on an optimal path.
• Since we walk at the same speed, we will always meet at a point equidistant from each of our houses.
• We call such a point a ‘midpoint’.
• There are many possible paths we each could choose, but as long as we choose paths that have a midpoint, we will
meet each other.
• So to calculate the probability, we must know:
1. how many midpoints there are for all optimal paths (k)
2. how many ways there are to reach each midpoint
• The probability we meet can then be given by:
! =
δ
$
=
number of ways to meet at the midpoints
total number of ways to all midpoints

19
• We continue with an example, where n = 4 and m = 2.
• In this grid, there are three midpoints {A, B and C}.
• By using tree diagram, we can show all the possible paths for
both persons.
• Person P1 walks either to the right (R) or up (U), while person P2
walks either to the left (L) or down (D)
• At each step we get one of these options for both persons:
(P1,P2) = {(Up,Left), (Up,Down), (Right,Left) or (Right,Down)}
Left
Down
Right
Up

For the first person P1:
• There are 3 ways to reach A from P1 • There is 1 way to reach C from P1• There are 3 ways to reach B from P1
Now, the total ways P1 can go to reach the midpoints is given as:
3 + 3 + 1 = %
Example: solution 1
20

21
• And the total ways P2 can go to reach the midpoints is given as:
1 + 3 + 3 = 7
Similarly, for the second person P2:
• There is 1 way to reach A from P2 • There are 3 ways to reach B from P2 • There are 3 ways to reach C from P2
• Therefore, the total number of ways both persons can reach the midpoints is & = 49 (i.e. 7*7).
• And the number of those ways in which the two persons choose the same midpoint (i.e.,
number of ways to meet at the midpoints ) is given by (see the next slide):
• δ = 9:;< =>?@ A1 B? C × 9:;< =>?@ A2 B? C + 9:;< =>?@ A1 B? F × 9:;< =>?@ A2 B? F + 9:;< =>?@ A1 B? G × 9:;< =>?@ A2 B? G = 3×1 + 3×3 + 1×3 = 15
• Hence, the probability we meet can then be given by:
A =
δ
&
=
=
15
49
= 0.3061

22
A
A
B
B
C
C
!"#$ %&'( )1 +' , × !"#$ %&'( )2 +' , + !"#$ %&'( )1 +' 0 × !"#$ %&'( )2 +' 0 + !"#$ %&'( )1 +' 1 × !"#$ %&'( )2 +' 1
3 × 1 + 3 × 3 + 1 × 3 = 15 = δ

23
both persons.
Left
Down
Right
Up
Up è Up è Right
Up è Right è Up
Right è Up èUp
Up è Right è Right
Right è Right è Up
Right è Up è Right
Right è Right èRight
Left è Left è Left
Left è Left è Down
Left è Down è Left
Down è Left è Left
Down è Down è Left
Down è Left è Down
Left è Down è Down
Example: solution 2

24
Up è Up è Right
Up è Right è Up
Right è Up èUp
We can generate a tree diagram from these possible outcomes, as follows:
They meet
at A
They don’t
meet
1 out of 7
Subtree1

25
Up è Up è Right
Up è Right è Up
Right è Up èUp
They meet
at A
They don’t
meet
1 out of 7
Subtree2

26
Up è Up è Right
Up è Right è Up
Right è Up èUp
They meet
at A
They don’t
meet
1 out of 7
Subtree3

27
Up è Up è Right
Up è Right è Up
Right è Up èUp
They meet
at B
They don’t
meet
They don’t
meet
3 out of 7
Subtree4

28
Up è Up è Right
Up è Right è Up
Right è Up èUp
They meet
at B
They don’t
meet
They don’t
meet
3 out of 7
Subtree5

29
Up è Up è Right
Up è Right è Up
Right è Up èUp
They meet
at B
They don’t
meet
They don’t
meet
3 out of 7
Subtree6

30
Up è Up è Right
Up è Right è Up
Right è Up èUp
They meet
at C
They don’t
meet
They don’t
meet
3 out of 7
Subtree7

31
Hence, from this tree diagram we can calculate the total number of outcomes
1. The number of ways they meet at a midpoint in Subtree1 equals: 1 out of 7
Thus, the number of ways they meet at the midpoints is given as: 15 out 49
The probability they meet can is given by:
! =
#
$
=
%&'()* +, -./0 1+ '))1 .1 12) '345+3%10
1+1.6 %&'()* +, -./0 1+ .66 '345+3%10
= 15/49 = 0.3061

32
both persons.
Left
Down
Right
Up
Up è Up è Right
Up è Right è Up
Right è Up èUp
Example: solution 3
The following table
shows all the possible
outcomes from these
options. It is clear that
we will get 7*7 = 49
outcomes.

33
Step 1 Step 2 Step 3
P1 P2 P1 P2 P1 P2
P1 P2 P1 P2 P1 P2
P1 P2 P1 P2 P1 P2
The yellow cells represent the cases
when P1 and P2 meet at a midpoint,
which are 15 out of 49 outcomes. Thus,
the probability they meet is 15/49 =
0.3061

34
Generalising the solution
to work for any grid size
• The probability we meet is given by:
=
?
?
56789: ;< =>?@ A; 799A >A AB9 7CDE;C5A@
F =
G
H
=

35
Generally speaking we can use the
combination formula to calculate the
number of ways we can meet at a
midpoint.
For a grid of size m by n the number of
ways we can meet at a midpoint is given
by:
In the example of 2*4 grid,
6
4
=
6!
2! 4!
= 15
( + *
*
6!
4! 2!
+,-./0 12 3456 71 -//7 47 78/ -9:;19+76
+,-./0 12 3456 71 -//7 47 78/ -9:;19+76 =
δ
= δ

36
We now turn to generalising the total number of ways to all midpoints. We first need to define the following
parameters:
S = number of steps to midpoint, which can be found as:
! =
# + %
2
k = number of midpoints, which is calculated as:
' = min #, % + 1
In our 2*4 grid example, ! =
-./
/
= 3 12341, and ' = min 2,4 + 1 = 3
3
= 3
3
1
= 3
3
0
= 1
Note that all the
equations here work
with a gird of an even
sum of its dimension,
e.g., 2*4 not 2*3. We
will explain the case of
the odd sum later.
E

37
Hence, the total number of ways one can reach a midpoint is given by:
!
"#$
%&'
(
)
= !
"#$
+&'
3
)
=
3
0
+
3
1
+
3
2
= 1 + 3 + 3 = 7
3
2
3
1
3
0
This is the same value for the second person P2.
3
0
3
1
3
2
1
3
3

38
The same applies for the other person. We therefore square this value to get the total number of ways we
cab each go to a midpoint without necessarily meeting.
!
"#$
%&'
(
)
*
total number of ways to all midpoints =
• The probability we meet is given by:
=
= + ?
?
∑A#B
C&D E
A
F
GHIJKL MN OPQR SM IKKS PS STK IUVWMUGSR
….. (*)
X =
Y
Z
=
= [

39
Now, let’s run some examples using equation(*).
Example 1:
! = # $%& % = '
( =
) + +
2
=
6
2
= 3
/ = min ), + + 1 = 2 + 1 = 3
The probability we meet =
! + %
%
∑678
9:; <
6
#
=
=
'
∑678
# >
6
# =
;?
'@
= 8. >8=;

40
Example 2:
! = # $%& % = '
( =
) + +
2
=
10
2
= 5
0 = min ), + + 1 = 4 + 1 = 5
! + %
%
∑789
:;< =
7
>
=
<9
'
∑789
# ?
7
> =
><9
@'<
= 9. ><B?

41
Example 3:
! = # $%& % = '
( =
) + +
2
=
10
2
= 5
0 = min ), + + 1 = 3 + 1 = 4
! + %
%
∑89:
;<= >
8
?
=
=:
'
∑89:
# @
8
? =
=?:
A'A
= :. =''@

42
Example 4:
! = # $%& % = '#
( =
) + +
2
=
20
2
= 10
/ = min ), + + 1 = 5 + 1 = 6
! + %
%
∑789
:;' <
7
=
=
=9
'#
∑789
# '9
7
= =
'##9>
>9?9>>
= 9. 9AB'

43
Example 5:
! = #$ %&' & = ($
) =
* + ,
2
=
50
2
= 25
0 = min *, , + 1 = 20 + 1 = 21
! + &
&
∑78$
9:; <
7
(
=
=$
($
∑78$
($ (=
7
( =
>. @;(AB + ;#
;. ;(>AB + ;=
= $. $>;A

44
Note that all the
equations here work
with a gird of an even
sum of its dimension,
e.g., 2*4 not 2*3. We
will explain the case of
the odd sum later.
Now, we show how to
deal with a grid that has
an odd sum of its
dimension.
• Again, we need to find the probability we meet:
= ?

45
2*3 (odd sum)
Example, !′= number of midpoints = 5
!′ = 2×min ), + + 1 = 4 + 1 = 5
1
2
1
2
2
1
1
In this example, there are 5 midpoints and each walker needs 2.5 steps to reach a midpoint.
The figure below shows the number of ways that walker 1 needs to reach each midpoint.

46
2*3 (odd sum)2*4 (even sum)
!"#$%& '( )*+, -' #%%- *- -.% #/01'/!-, =
= 3×5 + 3×3 + 3×5 = 57
Or
; + <
<
=
=
>
= 57
3
3
1
1 1
1
2
2
2
1
1
2
1
= 5×5 + ?×3 + ?×? + 5×? + 5×5 = 5@
Or
; + <
<
=
7
3
= 5@
δ

Hence, the formula
! + #
#
is valid for both cases (even and odd sums), which
is used to find the $%&'() *+ ,-./ 0* &((0 -0 01( &234*2$0/.
The challenging part is finding total number of ways to all midpoints. Instead
of deriving a new formula for the odd sum, we can observe that the β value is
same for both 2*4 and 2*3, as shown below.
β = 3 + 3 + 1 × 1 + 3 + 3 = 7×7 = MN β = 1 + 2 + 2 + 1 + 1 × 1 + 1 + 2 + 2 + 1 = 7×7 = MN

48
The β value of m*n grid with odd sum m+n equals the β value of m*(n+1), which is an even sum. This
means we still can use the derived formulas of the even sum for the odd sum but we need to add one to the
n value. Note that is just to find β not ".
Let’s apply this conclusion to the same example that we use here:
Even sum(m,n)
"
=
$ + &
&
Number of steps to
midpoints:
' =
$ + &
(
Number of midpoints:
) = *+, $, & + . / = 0
1=2
)−.
'
1
(
Example: m = 2, n =
4
" =
4
5
= .6
7 =
2 + 4
2
= 3
; = 2 + 1 = 3
/ = 5=
> =
"
/
=
15
49
Odd sum(m,n) A =
* + ,
,
Number of steps to
midpoints:
B =
* + ,
(
Number of midpoints:
)C = (×$1& $, & + .
We still need the k value for the
summation:
) = $1& $, & + .
/ = 0
+=2
E−.
B
+
(
Example: m = 2, n =
3
" =
6
F
= .2
7 =
2 + 3
2
= 3
;′ = 2×2 + 1 = 5
; = 2 + 1 = 3 / = 5=
> =
"
/
=
10
49

49
Example 1:
! = # $%& % = '
( =
) + +
#
= #. - = 3
K’ = number of midpoints, which is calculated as:
/′ = #×)2+ ), + + 4 = #×# + 4 = -
/ = )2+ ), + + 4 = # + 4 = '
! + %
%
∑678
9:4 ;
6
#
=
-
'
∑678
# '
6
# =
48
<=
= 8. '8>4

50
Example 1:
! = # $%& % = '
( =
) + +
#
= ,. ' = .
/′ = #×)2+ ), + + 4 = #×# + 4 = '
/ = )2+ ), + + 4 = # + 4 = ,
! + %
%
∑678
9:4 ;
6
#
=
<
'
∑678
# .
6
# =
#4
4#4
= 8. 4<,=

51
Example 2:
! = # $%& % = '
( =
) + +
,
= #. ' = '
./ = ,×)1+ ), + + 3 = ,×# + 3 = 4
. = )1+ ), + + 3 = # + 3 = '
! + %
%
∑678
9:3 ;
6
,
=
4
'
∑678
# '
6
, =
3,<
4<3
= 8. 3=33

52
Example 3:
! = # $%& % = '
( =
) + +
,
= -. / = /
01 = ,×)3+ ), + + 5 = ,×# + 5 = 6
0 = )3+ ), + + 5 = # + 5 = -
! + %
%
∑89:
;<5 =
8
,
=
>
'
∑89:
# /
8
, =
?-
'6'
= :. 5,-#

53
Example 4:
! = # $%& % = '(
) =
* + ,
-
= .. # = '0
1′ = -×*4, *, , + ' = -×# + ' = ''
1 = *4, *, , + ' = # + ' = 6
! + %
%
∑890
:;' <
8
-
=
'.
'(
∑890
# '0
8
- =
''6-=
(0>0((
= 0. 0-=6

54
Example 5:
! = #$ %&' & = #(
) =
* + ,
#
= #-. / = #/
0′ = #×*3, *, , + 5 = #×#( + 5 = -5
0 = *3, *, , + 5 = #( + 5 = #5
! + &
&
∑78(
9:5 ;
7
#
=
-$
#(
∑78(
#5 #/
7
# =
#. <#=<> + 5?
5. 5#-$> + 5/
= (. (#/5

55
Example 6:
! = # $%& % = '
( =
) + +
,
= ,. . = /
0′ = ,×)3+ ), + + # = ,×# + # = /
0 = )3+ ), + + # = # + # = ,
! + %
%
∑678
9:# ;
6
,
=
.
'
∑678
# /
6
, =
.
#<
= 8. /#,.

56
%% city grid - meeting at midpoint
% @Mohammed Alasmar 2018
clc, clear,close all
n = 5;
m = 1;
if ( mod((m+n),2) == 0 )
numSteps = ceil((m+n)/2);
numMidPoints = min(m,n) + 1;
halfStep = 1;
for i = 1:numMidPoints
scores(i) = nchoosek(numSteps , i-1) ;
end
else
numSteps = ((m+n)/2);
numMidPoints =2*min(m,n) + 1;
halfStep = 0.5;
scores = zeros(1,numMidPoints);
for i = 1: min(m,n) + 1
scores(i) = nchoosek(ceil((m+n)/2) , i-1);
end
end
x = 1:m+1; y = 1:n+1; [X,Y] = meshgrid(x,y);
figure, plot(X,Y, '-ob', 'MarkerSize',10), hold on , plot(X',Y', '-ob', 'MarkerSize',10)
view(2), set(gca,'xtick',[]) , set(gca,'ytick',[])
if m<=n
ax = 1;
ay = numSteps+1;
xStep = 1;
yStep = -1;
end
if m>n
ax = numSteps+1;
ay = 1;
xStep = -1;
yStep = 1;
end
for i = 1:numMidPoints
hold on
plot(ax,ay, 'ro' ,'MarkerSize',14 , 'MarkerFaceColor' , 'r')
score = num2str(scores(i));
text(ax-0.07,ay-0.07,score, 'FontSize',11);
ax = ax + xStep*halfStep;
ay = ay + yStep*halfStep;
end
chancesOfMeetingAtMidpoint = nchoosek(m+n ,n)
totalNumOfWaysToMidpoint = sum(scores)^2
prob = chancesOfMeetingAtMidpoint /totalNumOfWaysToMidpoint
% figure, bar3(prob), xlabel('dim m'), ylabel('dim n'), zlabel('probToMeet')
Change the values m and n, then run the program to get
the required probability

57
For fun, we run the program the we to find the probability to meet at different dimensions m = 1:30
and n = 1:30
This will give 30*30 = 900 probability values, as shown in the figure.
It is noticed that if the two friends would like to increase the chances of meeting on the way, then
they need to live in a city grid of equal dimensions.

Round2: Maths Challenge organised by ‘rep2rep’ research group at Sussex&Cambridge universities

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