Word autocorrect can be a lot like emacs abbrev-mode

Tired of typing the same stock phrases over and over again in Word? You can use a stand-in word and global search and replace when your document is finished but it's a lot more fun to watch Word automatically type for you.

For example, if you add an entry to your autocorrect dictionary as shown above, every time you type the key qnato, Word will instantly replace it with North American Treaty Organization just like emacs abbrev-mode.

If you have different groups of abbreviations for different kinds of writing you can prefix them all with the same letter or number and they will stay grouped and separate in the Word autocorrect dictionary.

Some general ideas for using this would be:

   @gg = @gmail.com
   @yy = @yahoo.com
   @me = your.name@bigcorp.com

You can also highlight an entire block of text and the entire block including all formatting can be assigned to an autocorrect  key. This would be useful for having various signature blocks, addresses and boilerplate sentences and paragraphs for business letters.

And the last thing you can do is select a photo or picture in a Word document and add an autocorrect key for it. For example if you paste a picture of The Dude into your document and add the key dude for it, every time you type dude, a picture of The Dude will be inserted.

Teachers can use this for easy clip art insertion. Any bit of art that is regularly being inserted into your documents can be assigned to a key.

You can build a dictionary with an image for every number and letter if you want for puzzle construction.

If it turns out that this a seldom used feature of autocorrect I would worry about normal.dot corruption if the images are large, especially since the art is probably stored in normal.dot.

Samsung SyncMaster 204B Power Supply Repair

Samsung SyncMaster 204B PSU PCB

The SyncMaster 204B is a very nice 20.1" UXGA (1600x1200) panel circa 2006.
I really like this aspect ratio but everything has gone widescreen and a monitor like this is basically no longer available.

The CapXon caps bulge and fail and before complete failure the monitor flickers more and more since the caps are on the switching power supply that powers the backlights. I think only the three caps after the chopper were bulging on this unit.

See: http://en.wikipedia.org/wiki/Capacitor_plague

Opening the monitor is a PITA and you don't want to do it twice so replace all the caps the first time. Use flush cutting side cutters to cut away as much of the leads and solder as you can. Solder wick rubbed with flux and pressed into what remains will loosen the caps to the extent that they will almost fall off the PCB.

More info:

http://tinymicros.com/wiki/Samsung_SyncMaster_204B_LCD_Repair

A proper parts list is a stumbling block for some people. These parts are perfect for my board rev 0.1 as of December 2011. These are not the cheapest possible parts but the parts with the longest lifetime ratings.

Digi-key part numbers

Prox on, Prox off: Windows Proxy On, Windows Proxy Off with PowerShell

These four files will give you the ability to switch between
having a proxy server set in Windows and not having one set 
without having to navigate a bunch of windows each time. 

Because Windows won't allow a PowerShell script to execute by 
default the scripts are called using batch files so the execution 
policy can be altered before and after calling the script.

proxyon.ps1:

set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name MigrateProxy -value 1
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyEnable -value 1
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyHttp1.1 -value 0
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyServer -value 'http://10.1.10.109:3128'
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyOverride -value '<local>'

proxyoff.ps1:

set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name MigrateProxy -value 1
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyEnable -value 0
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyHttp1.1 -value 0
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyServer -value ''
set-itemproperty 'HKCU:\Software\Microsoft\Windows\CurrentVersion\Internet Settings' -name ProxyOverride -value '<local>'

poff.bat:
 
powershell {Set-ExecutionPolicy Unrestricted}
powershell "& C:\proxyoff.ps1"
powershell {Set-ExecutionPolicy Restricted}

pon.bat:

powershell {Set-ExecutionPolicy Unrestricted}
powershell "& C:\proxyon.ps1"
powershell {Set-ExecutionPolicy Restricted}

Set Windows Network Settings From The Command Line Using Net Shell (netsh.exe)

If you are changing network settings repeatedly on a test system it gets old having to point and click each time. Net Shell (netsh.exe) can be called as follows from the command line:

   netsh interface ipv4 set address name="Local Area Connection" dhcp HOST_IP
   SUBNETMASK GATEWAY_IP


For example:

   netsh interface ipv4 set address name="Local Area Connection" 10.0.0.2   
   255.255.255.0 10.0.0.1

Typing this over and over will get old as well so you can make a simple batch file called
ip.bat containing:

   netsh interface ipv4 set address name="Local Area Connection" %1 %2 %3 %4

Now you can type:

   ip [dhcp | static] HOST_IP SUBNETMASK GATEWAY_IP

For example:

   ip dhcp 10.0.0.2 255.255.255.0 10.0.0.1


Of course you can just create individual batch files for each test scenario.

Dell Laptop Power Jack Repair

A common problem with laptops is that the power receptacle on the mainboard stops making a reliable connection to the barrel connector on the power brick.  Often this leads to intermittent battery charging unless the connector is in just the "right" position. Eventually the connector fails completely and the laptop will not power on.

The entire job took about three hours. Getting to the connector to desolder it is a bear. A complete disassembly was required. Thus the mountain of parts in the right side of the photo.

Some newer laptops have daughterboards containing the power receptacle so that desoldering is not necessary. Not having good desoldering equipment made it difficult to get the through holes in the mainboard clean and ready for a new connector but a cheap solder sucker and solder wick eventually did the trick.

If your laptop is worth less than the cost of a professional repair, give it a try. If your laptop is worth much more I would leave this repair to the professionals.

MIT 6.00 Problem Set 8 - Problem 2

The problem statement does not mention sorting. But sorting the entire dict would be helpful. And the comparator could be used with sorted() but since there is no mention of sorting the data set I leave it unsorted.  So this may be a completely wrong approach but I take two tuples at a time and whichever is best is added to the solution set where best is judged by the comparator.

Python Code:

def greedyAdvisor(subjects, maxWork, comparator):
    schedule = {} # initial solution
    firstSubject = {}
    odd = True
    work = 0    
    test = subjects.copy()
    for key in subjects.keys():
        if odd == True:
            firstSubject = subjects[key]
            odd = False
        else:
            odd = True
            if comparator(firstSubject, subjects[key]):
                if work + firstSubject[WORK] <= maxWork:
                    schedule[key] = firstSubject
                    work += firstSubject[WORK]
            else:
                if work + subjects[key][WORK] <= maxWork:
                    schedule[key] = subjects[key]
                    work += subjects[key][WORK]

    if test != subjects:
        print "SUBJECTS MUTATED"

    return schedule

Output:

cmpValue
Course  Value   Work
======  ====    =====
12.05   7       1
14.01   6       1
22.12   10      18
6.09    8       4
7.11    7       4
7.15    3       3
7.17    10      1
7.19    10      5
8.10    4       6
8.14    3       7
8.18    4       10

Total Value:    72
Total Work:     60

cmpWork
Course  Value   Work
======  ====    =====
12.03   2       1
14.01   6       1
22.12   8       15
6.09    8       4
7.11    7       4
7.15    3       3
7.17    10      1
7.19    10      2
8.10    4       6
8.14    3       7
8.16    2       6
8.18    4       10

Total Value:    67
Total Work:     60

cmpRatio
Course  Value   Work
======  ====    =====
12.03   2       1
12.05   7       1
14.01   6       1
22.12   10      18
6.09    8       4
7.02    3       2
7.11    7       4
7.15    3       3
7.17    10      1
7.19    10      2
8.14    3       7
8.16    2       6
8.18    4       10

Total Value:    75
Total Work:     60

MIT 6.00 Problem Set 8 - Problem 1

Python Code:

def loadSubjects(filename):
    catalog = {}
    inputFile = open(filename)
    for line in inputFile:        
        line = string.strip(line)
        lineList = string.split(line, ',')

        lineList[1] = int(lineList[1])
        lineList[2] = int(lineList[2])
        catalog[lineList[0]] = lineList[1:]
    return catalog

MIT 6.00 Lecture 8 - Stirling's approximation

Approximating large factorials:

import math

def stirling(n):
    return math.sqrt(2*math.pi*n) * (n/math.e)**n

if __name__ == '__main__':
    for i in range(1,20):
        s = stirling(i)
        f = math.factorial(i)
        print "stirling: %21.2f  factorial: %21d  error: %3.2f%%" % (s, f, abs(((f-s)/s)*100.0))

Output:

stirling:                  0.92  factorial:                     1  error: 8.44%
stirling:                  1.92  factorial:                     2  error: 4.22%
stirling:                  5.84  factorial:                     6  error: 2.81%
stirling:                 23.51  factorial:                    24  error: 2.10%
stirling:                118.02  factorial:                   120  error: 1.68%
stirling:                710.08  factorial:                   720  error: 1.40%
stirling:               4980.40  factorial:                  5040  error: 1.20%
stirling:              39902.40  factorial:                 40320  error: 1.05%
stirling:             359536.87  factorial:                362880  error: 0.93%
stirling:            3598695.62  factorial:               3628800  error: 0.84%
stirling:           39615625.05  factorial:              39916800  error: 0.76%
stirling:          475687486.47  factorial:             479001600  error: 0.70%
stirling:         6187239475.19  factorial:            6227020800  error: 0.64%
stirling:        86661001740.60  factorial:           87178291200  error: 0.60%
stirling:      1300430722199.47  factorial:         1307674368000  error: 0.56%
stirling:     20814114415223.14  factorial:        20922789888000  error: 0.52%
stirling:    353948328666101.12  factorial:       355687428096000  error: 0.49%
stirling:   6372804626194313.00  factorial:      6402373705728000  error: 0.46%
stirling: 121112786592294176.00  factorial:    121645100408832000  error: 0.44%

As n increases the error decreases:

n=165 error: 0.0505177446%
n=166 error: 0.0502133452%
n=167 error: 0.0499125922%
n=168 error: 0.0496154204%
n=169 error: 0.0493217663%

MIT 6.00 Lecture 8 Notes on exp()

In Lecture 8 a recursive algorithm for finding a to the b is shown. The algorithm flip-flops between two cases (even, odd) until it reaches the base case of b == 0. For odd cases the problem is reduced by one, for even cases the problem is reduced by half.

I think it can be hard to see where the final result comes from at first. What can exp() return? It can return 1, a or *a. 1 is returned for the edge case b == 0 so returning a or *a are the typical results. Any particular result is going to be of the form:

result = a*a*a .  .  .


For 2**15:

    exp3(2,15)
2*exp3(2,14)
    exp3(4,7)
4*exp3(4,6)
    exp3(16,3)
16*exp3(16,2)
    exp3(256,1)
b == 1
a: 256
result: 32768


result = 2*4*16*256 => 32768

Also notice that in the below code that the even test is (b%2)*2 == 0 not == b.

Python Code:

def exp(a,b,mult):
    print "Call %d*exp3(%d,%d)" % (mult,a,b)
    if b == 0:
        print 'b == 0'
        return 1

    if b == 1:
        print 'b == 1'
        print "a: %d" % (a)
        return a

    if (b%2)*2 == 0:
        print '\tif (b%2)*2 == 0: ',
        print "[b: %d]" % (b)
        return exp(a*a, b/2, 1)
    else:
        print "\telse              [b: %d]" % (b)
        return a*exp(a, b-1, a)

if __name__ == '__main__':
    for i in range(0,16):
        print "===> 2^%d" % (i)
        print "result: %d\n" % exp(2,i,1)

'''
exp(2,4)
  exp(2*2, 4/2)
    exp(4*4, 2/2)
    b == 1
    return 16

==> 16

exp(2,5)
  2*exp(2, 5-1)
    exp(2*2, 4/2)
      exp(4*4, 2/2)
        b == 1
        return 16

==> 16 * 2 = 32

exp(2,6)
  exp(2*2, 6/2)
    4*exp(4, 3-1)
      exp(4*4, 2/2)
        b == 1
        return 16

==> 16 * 4 = 64          
'''

Output:

===> 2^0
Call 1*exp3(2,0)
b == 0
result: 1

===> 2^1
Call 1*exp3(2,1)
b == 1
a: 2
result: 2

===> 2^2
Call 1*exp3(2,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(4,1)
b == 1
a: 4
result: 4

===> 2^3
Call 1*exp3(2,3)
        else              [b: 3]
Call 2*exp3(2,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(4,1)
b == 1
a: 4
result: 8

===> 2^4
Call 1*exp3(2,4)
        if (b%2)*2 == 0:  [b: 4]
Call 1*exp3(4,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(16,1)
b == 1
a: 16
result: 16

===> 2^5
Call 1*exp3(2,5)
        else              [b: 5]
Call 2*exp3(2,4)
        if (b%2)*2 == 0:  [b: 4]
Call 1*exp3(4,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(16,1)
b == 1
a: 16
result: 32

===> 2^6
Call 1*exp3(2,6)
        if (b%2)*2 == 0:  [b: 6]
Call 1*exp3(4,3)
        else              [b: 3]
Call 4*exp3(4,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(16,1)
b == 1
a: 16
result: 64

===> 2^7
Call 1*exp3(2,7)
        else              [b: 7]
Call 2*exp3(2,6)
        if (b%2)*2 == 0:  [b: 6]
Call 1*exp3(4,3)
        else              [b: 3]
Call 4*exp3(4,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(16,1)
b == 1
a: 16
result: 128

===> 2^8
Call 1*exp3(2,8)
        if (b%2)*2 == 0:  [b: 8]
Call 1*exp3(4,4)
        if (b%2)*2 == 0:  [b: 4]
Call 1*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 256

===> 2^9
Call 1*exp3(2,9)
        else              [b: 9]
Call 2*exp3(2,8)
        if (b%2)*2 == 0:  [b: 8]
Call 1*exp3(4,4)
        if (b%2)*2 == 0:  [b: 4]
Call 1*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 512

===> 2^10
Call 1*exp3(2,10)
        if (b%2)*2 == 0:  [b: 10]
Call 1*exp3(4,5)
        else              [b: 5]
Call 4*exp3(4,4)
        if (b%2)*2 == 0:  [b: 4]
Call 1*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 1024

===> 2^11
Call 1*exp3(2,11)
        else              [b: 11]
Call 2*exp3(2,10)
        if (b%2)*2 == 0:  [b: 10]
Call 1*exp3(4,5)
        else              [b: 5]
Call 4*exp3(4,4)
        if (b%2)*2 == 0:  [b: 4]
Call 1*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 2048

===> 2^12
Call 1*exp3(2,12)
        if (b%2)*2 == 0:  [b: 12]
Call 1*exp3(4,6)
        if (b%2)*2 == 0:  [b: 6]
Call 1*exp3(16,3)
        else              [b: 3]
Call 16*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 4096

===> 2^13
Call 1*exp3(2,13)
        else              [b: 13]
Call 2*exp3(2,12)
        if (b%2)*2 == 0:  [b: 12]
Call 1*exp3(4,6)
        if (b%2)*2 == 0:  [b: 6]
Call 1*exp3(16,3)
        else              [b: 3]
Call 16*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 8192

===> 2^14
Call 1*exp3(2,14)
        if (b%2)*2 == 0:  [b: 14]
Call 1*exp3(4,7)
        else              [b: 7]
Call 4*exp3(4,6)
        if (b%2)*2 == 0:  [b: 6]
Call 1*exp3(16,3)
        else              [b: 3]
Call 16*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 16384

===> 2^15
Call 1*exp3(2,15)
        else              [b: 15]
Call 2*exp3(2,14)
        if (b%2)*2 == 0:  [b: 14]
Call 1*exp3(4,7)
        else              [b: 7]
Call 4*exp3(4,6)
        if (b%2)*2 == 0:  [b: 6]
Call 1*exp3(16,3)
        else              [b: 3]
Call 16*exp3(16,2)
        if (b%2)*2 == 0:  [b: 2]
Call 1*exp3(256,1)
b == 1
a: 256
result: 32768

MIT 6.00 Problem Set 6 - Problem 5

LPG Delivery, Seoul

Assume worst case. pick_best_word() is going to produce 13699 candidate words to be checked against the dictionary. pick_best_word_faster() is going to produce 127 candidate words to be checked against the dictionary.

The original word list is length 83667 while the rearranged word list is length 69091. The reduction in length is from the collapse of sets of words made from the same letters into single words.

I think the bulk of the performance improvement is from the reduction in candidate words, not dictionary length or access methods.


Python Output:


>>> len(list(itertools.permutations('abcdefg', 7)))
5040
>>> len(list(itertools.permutations('abcdefg', 6)))
5040
>>> len(list(itertools.permutations('abcdefg', 5)))
2520
>>> len(list(itertools.permutations('abcdefg', 4)))
840
>>> len(list(itertools.permutations('abcdefg', 3)))
210
>>> len(list(itertools.permutations('abcdefg', 2)))
42
>>> len(list(itertools.permutations('abcdefg', 1)))
7
>>> 5040+5040+2520+840+210+42+7
13699

>>> len(list(itertools.combinations('abcdefg', 7)))
1
>>> len(list(itertools.combinations('abcdefg', 6)))
7
>>> len(list(itertools.combinations('abcdefg', 5)))
21
>>> len(list(itertools.combinations('abcdefg', 4)))
35
>>> len(list(itertools.combinations('abcdefg', 3)))
35
>>> len(list(itertools.combinations('abcdefg', 2)))
21
>>> len(list(itertools.combinations('abcdefg', 1)))
7
>>> 1+7+21+35+35+21+7
127

MIT 6.00 Problem Set 6 - Problem 4 Answer

Anhinga, Florida Everglades

This code is roughly twice as fast as the code from Problem 3. Two changes are responsible for the increased speed.

The first change is the switch from itertools.permutations() to itertools.combinations(). In Problem 3 every permutation of n letters is checked against the dictionary. Since order doesn't mater with a sorted dictionary key permutations would be redundant.


Python Code Answer:

def pick_best_word_faster(hand, rearrange_dict):
    score = 0
    high_word = ""
    perms = []
    attempts = 0
    local_hand = []

    # itertools.permutations is iterating for letters
    # that are zero count in the hash, dump the zeros
    # convert dict to list
    for k in hand.keys():
        if hand[k] > 0:
            for i in range(hand[k]):
                local_hand.append(k)

    print "local_hand: "
    print "\t",
    print local_hand
    print

    take = len(local_hand)
    
    # try for a bingo, then one less than a bingo, etc.
    while take != 0:
        perms = list(itertools.combinations(local_hand, take))
        take -= 1
    
        for letter_list in perms:
            attempts += 1
            sorted_word = ''.join(sorted(letter_list))

            if rearrange_dict.get(sorted_word, 0):         
                word = rearrange_dict[sorted_word]       
                if score < points_dict[word]:
                    score = points_dict[word]
                    high_word = word
                    print "word: %s  score: %d" % (high_word, score)

    print "------------------------------------------------------------"
             
    print "attempts: %d" % (attempts)
       
    if score < 1:
        return '.'
    else:
        return high_word



def get_word_rearrangements(word_list):
    d = {}

    for word in word_list:
        d[''.join(sorted(word))] = word

    return d

Output:


d g i i z m t
calling with:
        {'d': 1, 'g': 1, 'i': 2, 'z': 1, 'm': 1, 't': 1}
local_hand:
        ['d', 'g', 'i', 'i', 'z', 'm', 't']

word: digit  score: 7
word: timid  score: 8
word: ditz  score: 14
------------------------------------------------------------
attempts: 127
0.0350000858307
Score: 400.00  Total Score: 400.00
Turn took 0.04 seconds, 0.31 seconds remaining

g i m
calling with:
        {'d': 0, 'g': 1, 'i': 1, 'z': 0, 'm': 1, 't': 0}
local_hand:
        ['g', 'i', 'm']

word: mig  score: 6
------------------------------------------------------------
attempts: 7
0.0209999084473
Score: 285.72  Total Score: 685.71
Turn took 0.02 seconds, 0.29 seconds remaining

Final Score: 685.71



a c e j m n n
calling with:
        {'a': 1, 'c': 1, 'e': 1, 'j': 1, 'm': 1, 'n': 2}
local_hand:
        ['a', 'c', 'e', 'j', 'm', 'n', 'n']

word: nance  score: 7
word: mace  score: 8
word: jean  score: 11
word: jam  score: 12
------------------------------------------------------------
attempts: 127
0.0339999198914
Time limit exceeded.
Score: 0.00  Total Score: 0.00

c e n n
calling with:
        {'a': 0, 'c': 1, 'e': 1, 'j': 0, 'm': 0, 'n': 2}
local_hand:
        ['c', 'e', 'n', 'n']

word: ne  score: 2
------------------------------------------------------------
attempts: 15
0.0210001468658
Time limit exceeded.
Score: 0.00  Total Score: 0.00

c n
calling with:
        {'a': 0, 'c': 1, 'e': 0, 'j': 0, 'm': 0, 'n': 1}
local_hand:
        ['c', 'n']

------------------------------------------------------------
attempts: 3
0.018000125885
No legal combination possible.

MIT 6.00 Problem Set 6 Answers - Problem 3

I don't think the intent of this exercise was to write a permutation routine or to use itertools but I used itertools.  pick_best_word() is pretty much the whole assignment.

Python Code Answer:

def pick_best_word(hand, points_dict):
    score = 0
    length = 0
    high_word = ""
    perms = []
    local_hand = []
    attempts = 0

    # itertools.permutations is iterating for letters
    # that are zero count in the hash, dump the zeros
    # convert dict to list
    for k in hand.keys():
        if hand[k] > 0:
            for i in range(hand[k]):
                local_hand.append(k)
        
    print "local_hand: "
    print "\t",
    print local_hand
    print

    take = len(local_hand)
    
    # try for a bingo, then one less than a bingo, etc.
    while take != 0:
        perms = list(itertools.permutations(local_hand, take))
        take -= 1
    
        for letter_list in perms:
            attempts += 1
            word = ""
            for c in letter_list:
                word += c

            if points_dict.has_key(word):
                if score < points_dict[word]:
                    score = points_dict[word]
                    high_word = word
                    print "word: %s  score: %d" % (word, score)

    print "------------------------------------------------------------"
             
    print "attempts: %d" % (attempts)
       
    if score < 1:
        return '.'
    else:
        return high_word

For each seven letter hand 13699 permutations are checked
in the first pass:

>>> import itertools
>>> perms = list(itertools.permutations('1234567', 7))
>>> len(perms)
5040
>>> perms = list(itertools.permutations('1234567', 6))
>>> len(perms)
5040
>>> perms = list(itertools.permutations('1234567', 5))
>>> len(perms)
2520
>>> perms = list(itertools.permutations('1234567', 4))
>>> len(perms)
840
>>> perms = list(itertools.permutations('1234567', 3))
>>> len(perms)
210
>>> perms = list(itertools.permutations('1234567', 2))
>>> len(perms)
42
>>> perms = list(itertools.permutations('1234567', 1))
>>> len(perms)
7
>>> 5040+5040+2520+840+210+42+7
13699

Output:


s u u f y j m
calling with:
        {'s': 1, 'u': 2, 'f': 1, 'y': 1, 'j': 1, 'm': 1}
local_hand:
        ['s', 'u', 'u', 'f', 'y', 'j', 'm']

word: fumy  score: 12
------------------------------------------------------------
attempts: 13699
Score: 184.62  Total Score: 184.62
Turn took 0.06 seconds, 1.75 seconds remaining

s u j
calling with:
        {'s': 1, 'u': 1, 'f': 0, 'y': 0, 'j': 1, 'm': 0}
local_hand:
        ['s', 'u', 'j']

word: jus  score: 10
------------------------------------------------------------
attempts: 15
Score: 400.00  Total Score: 584.61
Turn took 0.03 seconds, 1.73 seconds remaining

Final Score: 584.61


p s t z m o o
calling with:
        {'p': 1, 's': 1, 't': 1, 'z': 1, 'm': 1, 'o': 2}
local_hand:
        ['p', 's', 't', 'z', 'm', 'o', 'o']

word: pomos  score: 9
word: zooms  score: 16
------------------------------------------------------------
attempts: 13699
Score: 275.86  Total Score: 275.86
Turn took 0.06 seconds, 1.59 seconds remaining

p t
calling with:
        {'p': 1, 's': 0, 't': 1, 'z': 0, 'm': 0, 'o': 0}
local_hand:
        ['p', 't']

------------------------------------------------------------
attempts: 4
No legal combination possible.



s s u d g i t
calling with:
        {'s': 2, 'u': 1, 'd': 1, 'g': 1, 'i': 1, 't': 1}
local_hand:
        ['s', 's', 'u', 'd', 'g', 'i', 't']

word: disgust  score: 9
------------------------------------------------------------
attempts: 13699
Score: 1053.57  Total Score: 1053.57
Turn took 0.06 seconds, 1.36 seconds remaining

Final Score: 1053.57

MIT 6.00 Problem Set 6 Answer - Problems 1 & 2

Python Code Answer:

def play_hand(hand, word_list):
    total_score = 0.0
    min_elapsed = 0.010
    hand_score = 0.0

    time_limit = raw_input('Enter Time Limit: ')
    
    if time_limit.isdigit():
        time_limit = float(time_limit)
    else:
        print "Enter Valid Time Limit"
        return

    while True:        
        print

        num_letters = 0
        for key in hand.keys():
            for i in range(hand[key]):
                num_letters += 1
                print key,

        if num_letters < 1:
            break

        start = time.time()
        word = raw_input('Word: ')
        stop = time.time()
        
        elapsed = stop - start
        time_limit -= elapsed

        # prevent divide by zero error, force min time
        if elapsed < min_elapsed:
            elapsed = min_elapsed

        if word == '.':
            break

        if is_valid_word(word, hand, word_list):
            update_hand(hand, word)
            if time_limit > 0.0:
                hand_score = get_word_score(word, num_letters) / elapsed
                total_score += hand_score
            else:
                print "Time limit exceeded."
            print "Score: %0.2f  Total Score: %0.2f" % (hand_score, total_score)
        else:
            print "Invalid word. Try again."

        if time_limit >= 0:
            print "Turn took %0.2f seconds, %0.2f seconds remaining" % (elapsed, time_limit)
    
    print "Final Score: %0.2f" % (total_score)

Output:

Loading word list from file...
   83667 words loaded.
Enter n to deal a new hand, r to replay the last hand, or e to end game: n
Enter Time Limit: 10

q e t h j o n Word: note
Score: 0.44  Total Score: 0.44
Turn took 9.05 seconds, 0.95 seconds remaining

q h j Word: .
Final Score: 0.44

Enter n to deal a new hand, r to replay the last hand, or e to end game:
Invalid command.
Enter n to deal a new hand, r to replay the last hand, or e to end game: n
Enter Time Limit:
Enter Valid Time Limit

Enter n to deal a new hand, r to replay the last hand, or e to end game: n
Enter Time Limit: asdf
Enter Valid Time Limit

Enter n to deal a new hand, r to replay the last hand, or e to end game: n
Enter Time Limit: 10

a q v p j j o Word:
Invalid word. Try again.

a q v p j j o Word: .
Final Score: 0.00

Enter n to deal a new hand, r to replay the last hand, or e to end game: n
Enter Time Limit: 10

p u t i h z n Word: putz
Score: 3.28  Total Score: 3.28
Turn took 4.57 seconds, 5.43 seconds remaining

i h n Word: hin
Score: 1.51  Total Score: 4.79
Turn took 3.98 seconds, 1.45 seconds remaining

Final Score: 4.79

Enter n to deal a new hand, r to replay the last hand, or e to end game:

MIT 6.00 Problem Set 4 - Problem 5 Answer

Key Largo

I spent a bunch of time thinking about how to do an exhaustive search with a given fragment by appending every letter of the alphabet all the way up to the maximum word length in the dictionary to see if the fragment was the beginning of a dictionary word. I thought there had to be a better way.


This code below is looking for an exact match. If fragment_str is the beginning of a word that is in the wordlist but not a standalone word then this test will return False.

if fragment_str in wordlist:

The fragment of length n should be compared with the first n letters of each dictionary word. If there is a match then there is a word containing that fragment. The search itself is brute force. Words that are too short are checked. Words that don't even begin with the same first letter are checked. So the efficiency can be improved but the game play is correct. 

Python Code Answer:

def fragment_starts_word(fragment_str, wordlist):
    fragment_str_len = len(fragment_str)

    for word in wordlist:
        if word[0:fragment_str_len] == fragment_str:
            if len(word) > len(fragment_str):
                print "Fragment (%s) is start of at least one longer word (%s)" % (fragment_str, word)
                return True


def other(player):
    if player == 1:
        return 2
    else:
        return 1

          
def ghost():
    wordlist = load_words()
    fragment = []
    player = 1
    fragment_str = ""

    while True:
        print "\nFragment: " + fragment_str
        turn = raw_input("Player " + str(player) + " Turn: ")

        if len(turn) == 1 and turn in string.ascii_letters:
            fragment.append(turn.lower())
            fragment_str = "".join(fragment)

            if fragment_str in wordlist:
                print fragment_str + " is a word."
                if len(fragment) > 3:
                    print "Player " + str(player) + " loses."
                    print "Player " + str(other(player)) + " wins!"
                    sys.exit(0)
            elif fragment_starts_word(fragment_str, wordlist) != True:
                    print "Fragment is not beginning of any word."
                    sys.exit(0)
        else:
            print "Input must be a single letter."
            break

        player = other(player)

if __name__ == '__main__':
    ghost()

Output:

Loading word list from file...
   83667 words loaded.

Fragment:
Player 1 Turn: p
Fragment (p) is start of at least one longer word (pa)

Fragment: p
Player 2 Turn: e
pe is a word.

Fragment: pe
Player 1 Turn: a
pea is a word.

Fragment: pea
Player 2 Turn: f
Fragment (peaf) is start of at least one longer word (peafowl)

Fragment: peaf
Player 1 Turn: o
Fragment (peafo) is start of at least one longer word (peafowl)

Fragment: peafo
Player 2 Turn: w
Fragment (peafow) is start of at least one longer word (peafowl)

Fragment: peafow
Player 1 Turn: l
peafowl is a word.
Player 1 loses.
Player 2 wins!

******************************************************

Fragment:
Player 1 Turn: p
Fragment (p) is start of at least one longer word (pa)

Fragment: p
Player 2 Turn: y
Fragment (py) is start of at least one longer word (pya)

Fragment: py
Player 1 Turn: t
Fragment (pyt) is start of at least one longer word (python)

Fragment: pyt
Player 2 Turn: h
Fragment (pyth) is start of at least one longer word (python)

Fragment: pyth
Player 1 Turn: o
Fragment (pytho) is start of at least one longer word (python)

Fragment: pytho
Player 2 Turn: n
python is a word.
Player 2 loses.
Player 1 wins!

******************************************************

Fragment:
Player 1 Turn: q
Fragment (q) is start of at least one longer word (qabala)

Fragment: q
Player 2 Turn: z
Fragment is not beginning of any word.

MIT 6.00 Problem Set 5 - Problems 1 - 4 Answers

Juvenile Gull, Australia

Problem set 5 has a framework of provided code to complete the assignment within. Below is the code that needs to be added to the framework to complete the assignment.

Python Code Answer:

# Problem #1: Scoring a word
def get_word_score(word, n):
    score = 0

    for c in word:
        score += SCRABBLE_LETTER_VALUES[c]

    if len(word) == HAND_SIZE:
        score += 50

    return score



# Problem #2: Update a hand by removing letters

def update_hand(hand, word):
    for c in word:
        hand[c] -= 1

    return hand


# Problem #3: Test word validity
def is_valid_word(word, hand, word_list):
    local_hand = hand.copy()

    for c in word:
        if local_hand.get(c, 0) == 0:
            return False
        else:
            local_hand[c] -= 1
    
    if word.lower() in word_list:
        return True
    else:
        return False
            

# Problem #4: Playing a hand
def play_hand(hand, word_list):
    total_score = 0

    while True:        
        print

        num_letters = 0
        for key in hand.keys():
            for i in range(hand[key]):
                num_letters += 1
                print key,

        if num_letters < 1:
            break

        word = raw_input('Word: ')

        if word == '.':
            break

        if is_valid_word(word, hand, word_list):
            update_hand(hand, word)
            hand_score = get_word_score(word, num_letters)
            total_score += hand_score
            print "Score: %d  Total Score: %d" % (hand_score, total_score)
        else:
            print "Invalid word. Try again."
    
    print "Final Score: %d" % (total_score)

Output:

Loading word list from file...
   83667 words loaded.
Enter n to deal a new hand, r to replay the last hand, or e to end game: n

u y h j o n n Word: joy
Score: 13  Total Score: 13

u h n n Word: hun
Score: 6  Total Score: 19

n Word: .
Final Score: 19

Enter n to deal a new hand, r to replay the last hand, or e to end game: n

r r u v k l o Word: .
Final Score: 0

Enter n to deal a new hand, r to replay the last hand, or e to end game: n

a r u w y h j Word: yaw
Score: 9  Total Score: 9

r u h j Word: .
Final Score: 9