Computing Science 1P

Computing Science 1P

Large Group Tutorial: Lab Exam & Class Test

Simon GayDepartment of Computing Science

University of Glasgow

2006/07

2006/07 Computing Science 1P Tutorial on Lab Exam / Class Test - Simon Gay 2

Remember the lab exam…

A sequence of items, each of weight between 1 and 20, are tobe packed into bins, each of capacity 20.

As each item appears, work out which bin to put it in:- using the “first fit” algorithm- using the “best fit” algorithm

This is an example of a bin-packing problem, with manyapplications of practical importance. In general, working outthe best packing given all of the items is an example of anNP-complete problem. No-one knows a better way thantrying all possible packings.

As we are allocating each item as it appears, it’s an online bin-packing problem.


Bin-packing

The question tells us to define a function

findBin

which is given:

a list of the weights of the bins in use

the weight of the next item

and returns:

either: the number of the bin to put the next item inor: -1, if it won’t fit in any of the bins


Bin-packing

The function definition will start with

def findBin(bins,weight):

and it must return an integer value, including the possibility of -1

There will be two versions of findBin, one for the first-fitalgorithm and one for the best-fit algorithm.

BEWARE: does findBin return the number of the bin (in therange 1…) or its index position (in the range 0…) ?

I will use the index position.


Bin-packing

Examples:

findBin([15,17],2)

should return 0 or 1, depending on whether we are usingfirst-fit or best-fit.

findBin([15,17],6)

should return -1.

Assume that we have findBin.


Bin-packing: main program

Obviously we need a variable to store the list of bins beingused, with the weight of each bin.

Example: [ 10, 5, 16 ]

The values in this list will never be 0, because we won’tstart a new bin unless there is something to put in it.

Initially this list will be empty.

bins = []

It’s worth asking whether we need to store the weights ofall of the items.


Do we need a variable to store a list of the weights of all the items?

• Yes• No• Don't know



All we need to do is consider each item in turn, as it is entered;there is no need to store them. The program has a familiarstructure:

bins = []weight = input(“Enter weight: ”)while weight != 0:

## do something with weight#weight = input(“Enter weight: ”)



What do we do with weight ? We have to callfindBin(bins,weight) to find out which bin to put it in.If the result is -1 then that’s a special case: a new bin.

bins = []weight = input(“Enter weight: ”)while weight != 0:

bin = findBin(bins,weight)if bin == -1:

bins = bins + [weight]else:

bins[bin] = bins[bin] + weightweight = input(“Enter weight: ”)



To output the report of the latest allocation:

bins = []item = 0weight = input(“Enter weight: ”) while weight != 0:

item = item + 1bin = findBin(bins,weight)if bin == -1:

bin = len(bins)bins = bins + [weight]

else:bins[bin] = bins[bin] + weight

print “Item”, item, “weight”, weight, “bin”, bin+1weight = input(“Enter weight: ”)


Do you prefer bins = [ ] or bins = [ 0 ] ?

• bins = [ ]• bins = [ 0 ]


How many bins are needed to pack 0 items?

• None• One, but it is empty• Two, but they are both empty• Some other number• Don't know


Bin-packing: final output

The very easy way:

print “Bin weights:”, bins

But it prints square brackets, which we don’t want.



The easy way without the square brackets:

print “Bin weights:”,for i in bins:

print i,

But it doesn’t print commas between the numbers.



With commas but not quite right:

print “Bin weights:”,for i in bins:

print i, “,”,

It prints an unwanted comma after the final number.



The right way:

print “Bin weights:”,for i in range(len(bins)-1):

print str(bins[i]) + “,”,if len(bins) > 0:

print bins[len(bins)-1]


Bin-packing: almost everythingbins = []item = 0weight = input(“Enter weight: ”)while weight != 0:

item = item + 1bin = findBin(bins,weight)if bin == -1:

bin = len(bins)bins = bins + [weight]

else:bins[bin] = bins[bin] + weight

print “Item”, item, “weight”, weight, “bin”, bin+1weight = input(“Enter weight: ”)

print “Bin weights:”,for i in range(len(bins)-1):

print str(bins[i]) + “,”,if len(bins) > 0:

print bins[len(bins)-1]


Defining findBin for first-fit

def findBin(bins,weight):b = 0while b < len(bins):

if bins[b]+weight <= 20:return b

else:b = b + 1

return -1


Defining findBin for best-fit

def findBin(bins,weight):b = 0best = -1 # index of best so farwhile b < len(bins):

if bins[b]+weight <= 20:if best == -1 or bins[b] > bins[best]:

best = bb = b + 1

return best


Class Test: Question 5

Input a string and output the number of occurrences of eachletter, in a tabular format.

this is the day of the cs1p class test

a: 3 b: 0 c: 2 d: 2e: 3 f: 1 g: 0 h: 2 i: 1 j: 0 k: 0 l: 1 m: 0 n: 0 o: 2 p: 1 q: 0 r: 0 s: 5 t: 5 u: 0 v: 0 w: 0 x: 0 y: 2 z: 0


Part (a)

Define a function which takes a character and a string asparameters and returns the number of times that thecharacter appears in the string.

def count(c,s):n = 0for nextC in s:

if nextC == c:n = n + 1

return n

Exam technique: the function should DO WHAT THE QUESTION ASKED


Part (a)

Alternatively:

def count(c,s):n = 0i = 0while i < len(s):

if s[i] == c:n = n + 1

i = i + 1return n

A good example of when for … in is simpler: we don’t need to do anything with the position within the string, just thecharacter at each position.


Part (b)

A simple way to do this is to loop over a string (or a list) of all the letters in the alphabet.

for c in “abcdefghijklomnopqrstuvwxyz”:


Part (b)

Of course we need to input the string to analyze:

text = raw_input(“Enter the string: ”)

for c in “abcdefghijklomnopqrstuvwxyz”:


Part (b)

Now we can use the function from part (a):


for c in “abcdefghijklomnopqrstuvwxyz”:print c+“:”, count(c,text), “ ”,

This prints the output on one long line:

a: 3 b: 0 c: 2 d: 2 e: 3 ...


Part (b)

So we need to keep track of which column we are on, andstart a new line when necessary:


column = 0for c in “abcdefghijklomnopqrstuvwxyz”:

print c+“:”, count(c,text), “ ”,column = column + 1if column = 4:

column = 0print


Part (b)

Alternatively:


column = 0for c in “abcdefghijklomnopqrstuvwxyz”:

print c+“:”, count(c,text), “ ”,column = column + 1if column % 4 == 0:

print


Part (b)

Another approach avoids using a string of the whole alphabet.It uses the functions ord and chr, which you might not know.


Do you know what ord and chr do?

• Yes• No


Part (b)

The function chr is given an integer and returns thecorresponding character. chr(97) = ‘a’ and so on.

The function ord does the opposite: ord(‘a’) = 97

To avoid having to remember the magic number 97, we canuse the following expression to give us the n’th letter of thealphabet, starting the numbering from 0:

chr(n+ord(‘a’))

Of course we could wrap this up in a function (exercise…)

Look up ASCII


Part (b)

This gives an alternative solution:


column = 0for i in range(26):

c = chr(i + ord(‘a’))print c+“:”, count(c,text), “ ”,column = column + 1if column % 4 == 0:

print

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Computing Science 1P