The Maths of Sorting Things Out

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John D Barrow
Gresham College
DAMTP Cambridge
The Maths of Sorting Things Out
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Complexity is All Around Us
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389965026819938 5569 389965026814369
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Number of ways an even number can be written as
the sum of two primes (4 lt n lt 1000)
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Number of ways an even number can be written as
the sum of two primes (4 lt n lt 106)
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Computer-aided Mathematics Computer
Proofs Experimental mathematics Is Anything
Beyond Computers?
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Four Colours Suffice
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Tractability Easy problems N parts Calculation
time grows as N Hard problems N
parts Calculation time grows as 2N
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Monkey Puzzles
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6-city Travelling Salesman
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6-city Travelling Salesman
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8 years of parallel computing
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Some More Hard ProblemsWhere the best solution
cant be predicted by a general recipe

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Packing Problems
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Examples

• Fitting music tracks on CDs
• Fitting packages in boxes
• Scheduling computer or machine tasks
• Building shelves
• Adverts in TV breaks
• Data storage on disk
• Fitting your holiday stuff in a suitcase

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The Problem
25 data files to fit onto disks that each hold a
maximum of 10 GB What is the smallest number of
disks you could use? How do you pack these files
to do it?
6,6,5,5,5,5,4,3,2,2,3,7,6,5,4,3,2,2,4,4,5,8,2,7,1
Total volume to be fitted on disks is 106
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Strategy 1 Next Fit 6,6,5,5,5,5,4,3,2,2,3,7,6,
5,4,3,2,2,4,4,5,8,2,7,1

• Put first track on first disk, then next in the
  same disk until it wont fit.
• Then start new disk.
• You cant go back and fill gaps in earlier disks.
• 14 disks used. 3 full.
• 32 out of 140 slots empty
• 108/140 77 filled

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Other Strategies to Try

• First Fit
• You can go back and put tracks on the first disk
  in the line that still has room
• Worst Fit
• Put next track on the disk that currently has
  most space available
• Best Fit
• Put next track on the disk that leaves the least
  room on the disk after it is added

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Strategy 2 First Fit 6,6,5,5,5,5,4,3,2,2,3,7,
6,5,4,3,2,2,4,4,5,8,2,7,1

• Put first track on first disk which has space,
• Then start new disk.
• You can go back and fill gaps in earlier disks.
• 12 disks used. 6 full.
• 14 slots empty
• 106/120 88 filled

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Sorting Helps

• Sort the files into descending size order
• 8,7,7,6,6,6,5,5,5,5,5,5,4,4,4,4,3,3,3,2,2,2,2,2,1
• Now try sorted Next Fit and First Fit
  strategies
• What was the best strategy?
• Is it the best possible for this problem?

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Strategy 3 Sorted Next Fit 8,7,7,6,6,6,5,5,5,5
,5,5,4,4,4,4,3,3,3,2,2,2,2,2,1

• Put first track on first disk, then next in the
  same disk until it wont fit.
• Then start new disk.
• You cant go back to fill gaps in earlier disks.
• 14 disks used. 4 full.
• 33 slots empty
• 107/140 76 filled

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Strategy 4 Sorted First Fit
8,7,7,6,6,6,5,5,5,5,5,5,4,4,4,4,3,3,3,2,2,2,2,2,1

• Put first track on first with disk with space
• Then start new disk.
• 11 disks used. 10 full.
• 4 slots empty. Cant do better
• 106/110 96 filled
• THIS IS THE BEST POSSIBLE PACKING

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Chicken McNugget Numbers
6, 9 and 20
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What is the Largest Non-McNugget Number
Any sufficiently large number can me written as
a linear sum of 6s, 9s and 20s
44 69920 45 99999 46 62020 47
99920 48 669999 49 92020
Make all larger numbers by adding a 6 to any one
of these and so on
The largest non-McNugget number is 43 6s and 9s
only make multiples of 3. Using a 20 or two
doesnt help. All the non-McNugget numbers
are 1,2,3,4,5,7,8,10,11,13,14,16,17,19,22,23,25,28
,31,34,37,43.
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The Happy Meal 4-McNugget Box (1979) makes it
mathematically far less interesting
!
The numbers 4,6,9,20 now allow only six
non-McNugget numbers 1,2,3,5,7 and 11
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Whats the smallest total stamp value that cant
fit on the envelope?
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The Postage Stamp Problem
An envelope can only hold a maximum number of
stamps ? h Stamps have values An a1,
a2,..an Sort in ascending order so that 1 a1
lt a2 lt.. lt an We want the smallest integer that
cant be written as N(h, An) F1a1 F2a2 ..
Fnan where F1 F2 .. Fn lt h
The number of consecutive possible postage
amounts that will fit on the envelope is n(h,
An) N(h, An) - 1
Exact solution only known for n 2 n(h, A2)
(h 3 a2)a2 2 for h ? a2 - 2 n(h,2) (h2
6h 1)/4 is integer part, so 3.14 3
The first values are 2,4,7,10,14,18,23,28,34,40,.
N(h,4) gt 2.008 ? (h/4)4 O(h3)
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The Spare Change Problem
St Gaudens 20 Gold Double Eagle
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Smallest set of coins that will make all amounts
of change from 1 to 100
Europe 1, 2, 5, 10, 20, 50 denominations USA
1, 5, 10, 25 Minimum number of coins given by
the handy Greedy Algorithm Use as many of the
largest value at each step 76 pence 50 20 5
1 (4 coins) 76 cents 50 25 1 (3 coins)
Old money in the UK didnt have the handy
property ½ , 1 , 3 , 6 , 12 , 24 , 30 old pence 4
shillings 48d 30 12 6 24 24 This
happens when we double a coin value (gt 2d) eg
with a groat 4d 8d 4 4 6 1 1
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Shallits Improved American System
Average number of coins needed to make 1 to 100
cents with 1, 5, 10, 25 denominations is
4.7 Worst case is with only a 1 cent coin need
99 coins to make 99c etc Average number needed is
49.5
What is the best case using only 4 values? 1, 5,
18, 29 and 1, 5, 18, 25 are both better! They
need an average of 3.89 coins to make all 1 ?
100 cent amounts Best to use the 1, 5, 18, 25
values And replace the dime with a new 18c piece
18c
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Improved UK or Euro system ?
1, 2, 5, 10, 20, 50, 100, 200p coins
Average number of coins to make all amounts up to
500 is 4.6 Add a 133p or 137p coin and the
average number falls to 3.92
137p
133p
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A Further Improvement?

• Lots of currencies use the pattern of
  denominations
• 1,2,5, and 10,20, 50, and 100, 200, 500
• They would do (about 6) better and use less
  change with
• 1,3,4, and 10, 30, 40, and 100, 300, 400
• 68 50 10 5 2 1 40 10 10 4 4
• 134 100 20 10 2 2 100 30 4
• 243 200 20 20 2 2 200 40 3
• 279 200 50 20 5 2 2 200 40
  30 3 3 3
• Greedy algorithm has one failure because 6 33
  is better than 6 411
• Best to do away with the 1c or 1p coin though!