Self Extinction Events

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GREAT FILTERS

We as a global culture and civilization face what many call “A Great Filter” in our future (This century and the next).

In the past “Great Filters” have littered the fossil record of our planet with extinction events. Climate Change brought on by asteroid strikes and so forth.

ONE OF THE FIRST GREAT FILTERS

One of the very first “Great Filters” was of a species of bacteria (anaerobic) that took in carbon dioxide and as a waste product produced oxygen. It bloomed all over the planet when atmospheric carbon dioxide was the more prominent than oxygen as an atmospheric gas. That bacteria wiped itself out with just a few examples remaining. This species executed a “SELF-EXTINCTION-EVENT

A CURRENT GREAT FILTER

There is another species doing this by digging-up, pumping-up sequestered hydrocarbons and turning them into energy and producing carbon dioxide in great quantities as a waste product. This is leading to climate change that could very likely take this species extinct along with many others. This species is in the process of attempting a “SELF-EXTINCTION-EVENT“.

The above is an argument against why we as a species must without delay move away from our ‘Hydrocarbon Economy”.  We know how to do it.  We have the tools and the technology.  All that remains is a global cooperative effort to do it.

GUN VIOLENCE AND MODERN CIVILIZATION

Additionally this second species now has a population and technology big enough to demand a global cooperative society. However there are certain members of the species who resist that global cooperative civilization and cling to their evolutionary excess baggage in the form of tribal behaviors. They drag down the very important move toward global civilization.

We all know what species we are talking about here.

I’m an ‘old-man’ and I got my fill of guns back in 1968-1969 in Southeast Asia.

There is NO PLACE for devices designed to kill other humans in a modern civilized society.

And we all know how tribal behaviors manifest themselves. The above is an argument against why we as a species do not need devices to kill each other.

What Are The Odds You Already Have My Phone Number

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09 NOV 2018
538 Riddler Express
What Are The Odds
You Already Have My Phone Number

Assume the phone company assigns seven digit phone numbers in a random way. What are the chances that when you get a second phone then the number it will be assigned will contain the same digits as your first phone number, including any repetitions of digits.  Such numbers are said to be ‘scrambles’ of each other.

An example of a ‘scramble’ is the pair 608-2338 and 830-2386.  Also suppose that all the phone numbers from 000-0000 up to and including 999-9999 are are up for grabs. (Note there are many different seven digit numbers that the phone company actually never assigns, and we ignore this).

NOTE: if your first phone’s number is like 000-0000 or 111-1111 or other number with all seven digits the same then you would be guaranteed to get a number that is not a ‘scramble’ of your first phone when you get the number for your second phone since all ‘scrambles’ of such numbers come out the same. There is only ONE of each of these types.  So your chances of getting a scramble would be: ZERO. So we do not consider these.

So here is how to figure this out:  Think of a line of buckets each filled    with numbers each number in a any bucket is a scrambles of any other in that bucket.  If you pick two numbers from any of those buckets you will be guaranteed they will be scrambles of each other.

What are the chances that after randomly picking a ball from one of ‘m’ buckets.  You again choose a bucket at random and that is the same bucket you picked your first ball from.

Easy: 1/m — you have ‘one-chance-in-m’ of getting the same bucket.

Here is the interesting part:  How many buckets are there?

For each number from 000 0000 all the way to 999 9999 take that number and reduce it to a number whose digits are ‘in-order’.

For example the number 608 2338 would be reduced to 023 3688.  Scratch that number on the bucket and toss in the original number.  Do this and track that numbers from 000 0000 to 999 9999 when the number you come up with after ordering the digits already exist then just toss your number in the bucket.  If there is no bucket just scratch the ordered number on the bucket and toss it in.

Some buckets have a few numbers and some buckets have a lot of numbers.  The buckets with the fewest numbers in them have only 7 numbers in them.  The buckets with the most numbers in them have 5,040 numbers in them. And the total number of buckets is 11,430 buckets.

So grab a number from a bucket then you have a 1 in 11,430 chance of putting your hand in the same bucket.  But if you do then you get a ‘scramble’. 

Here is one way to count the number of buckets:

import sys
import random

n_dct = {}

def set_itm_val_in_dct(n_raw):
    """
        Update The Dictionary:
        Using the input number 'n_raw'
        create a new value/pair for the dictionary 'n_dct'
        or increment the value part of an existing value/pair
    """
    n_str = str(n_raw)
    n_lst = list(n_str)
    n_len = len(n_lst)
    for _ in range(7-n_len):
        n_lst.insert(0,"0")
    n_lst.sort()

    """ Ignore Numbers That Have All The Same Digits """
    if n_lst[0] == n_lst[6]:
        return

    n_lst = "".join(n_lst)

    if n_lst in n_dct:
        n_dct[n_lst] += 1
    else:
        n_dct[n_lst] = 1

def n_dct_fill(n_max):
    """
        Build The Dictionary
        Create n_dct using the values 0 ... n (n_max-1)
    """
    for n in range(1, n_max):
        set_itm_val_in_dct(n)

def dct_print():
    """
        Print The Dictionary
    """
    print len(n_dct)
    count = 0
    n_tup = n_dct.items()
    n_tup.sort(key=lambda elem: elem[1])
    for tup in n_tup:
        count += 1
        if count % 10 == 0:
            print("")
        prt_str = '{: >7}:{: >5}'.format(tup[0], tup[1])
        print (prt_str),
    print count

# Create The Dictionary
if __name__ == "__main__":
    """ From 000-0000 to 999-9999 """
    n_dct_fill(10000000)
    dct_print()

 

Is Today Your Birthday

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"""
    Is Today Your Birthday?
    Written By Zach Cox (http://www.zachdcox.name)
    On the occosion of the anniversary his nephew's birth.
    Happy Birthday James!
    Note:  Sadly It was not finished for this year's anniversary
           But you can adjust command line argument "yyyy_mm_dd"
           To see what different inputs yield.

"""
import datetime
import sys
import os
import traceback

class Is_Today_Your_Birthday:
    usage_messages = ["Usage:  python Is_Today_Your_Birthday.py yyyy-mm-dd",
                      "Where:  yyyy_mm_dd is the date of your birth",
                      "       'yyyy' 'mm' 'dd' are each numeric character strings",
                      "        of length 4, 2, and 2 and seperated by the dash character.",
                      "        yyyy is the year, mm is the month and dd is the day of the month."]
    birthday_msgs = ["You Are One Decade Old!",
                     "You Are Twenty Percent Dead!",
                     "You Are Flirty Shirty Thirty ",
                     "Lordy, Lordy, You Are Forty!",
                     "You Are Nifty, Thrifty Fifty!",
                     "You Are Sketchy, Sexy Sixty!",
                     "You Are Sensibly, Steadily Seventy!",
                     "You Are Weighty, Eighty Matey!",
                     "You Are Shiny, Briny Ninety!",
                     "Dang, Ain't You Dead Yet?",
                     "Congradulations!"]
    arg_list = None
    # Today
    now = None
    # Date Of Birth
    dob = None
    # Birthday Anniversary
    bda = None

    def __init__(self):
        """ set up the today/now vars """
        self.now = datetime.date.today()
        return

    def speak(self, message, do_speak):
        if do_speak:
            ret_val = os.system("say " + message)

    def print_usage(self,error_message,do_speak=False):
        """ Print Error Message And  Usage Rules """
        print error_message
        self.speak(error_message, do_speak)
        for msg in self.usage_messages:
            print msg
            self.speak(msg, do_speak)

    def check_command_line_arguments(self):
        """
            set up the date of birth and this year's birthday vars
			return True when date
            check for a valid command line argument
        """
        ret_value = False
        self.arg_list = sys.argv
        if len(self.arg_list) >= 2:
            # date of birth
            try:
                dtobj = datetime.datetime.strptime(self.arg_list[1], '%Y-%m-%d')
            except ValueError:
                self.print_usage("INVALID ARGUMENT")
                return(False)

            self.dob = dtobj.date()
            # this year's birthday anniversary
            self.bda = self.dob.replace(year=self.now.year)
            return True
        else:
            self.print_usage("MISSING CMD LINE ARGUMENT")
            return False

    def dob_vs_now(self):
        # Born On This Very Day (WOW!)?
        if self.now == self.dob:
            print("WOW! Today Is The Day Of Your Birth!")
            print("YOU ARE BRAND NEW!")
            print("Welcome To the 'outside' World!")
            return
        # Born Sometime In The Future?
        if self.dob > self.now:
            print("WOW! You Are From The FUTURE.")
            print("TIME TRAVEL!  Who Knew?")
            return
        # Today Is Your Birthday
        if self.bda == self.now:
            print("Happy Birthday!")
            years_old = int(self.now.year)-int(self.dob.year)
            print("You Are %d Years Old Today" % (years_old) )
            msg_idx = len(self.birthday_msgs)-1
            if years_old > 90:
                msg_idx = len(self.birthday_msgs)-2
            elif years_old % 10 == 0:
                msg_idx = (years_old / 10)-1
            print(self.birthday_msgs[msg_idx])
            return
        # You Have Already Had Or Will Have A Birthday This Year
        print("Today Is NOT Your Birthday")
        days = abs((self.now - self.bda).days)
        days_str = "Days"
        if days == 1:
            days_str = "Day"
        if self.bda > self.now:
            extra_msg = "Your Birthday Will Be %d %s From Now." % (days, days_str)
            print(extra_msg)
            years_old = int(self.now.year)-int(self.dob.year)
            message = "You Will Be %d Years Old That Day!" % (years_old)
        else:
            extra_msg = "Your Birthday Was %d %s Ago." % (days, days_str)
            print(extra_msg)
            years_old = int(self.now.year)-int(self.dob.year)
            message = "You Turned %d Years Old That Day!" % (years_old)

        print(message)

def main():
    """
        Instanciate the object
        Invoke the check command line arguments method on the object
        Invoke the analysis of date of birth to today's date method on the object
    """
    print " "
    is_today_your_birthday = Is_Today_Your_Birthday()
    bool_wtf = is_today_your_birthday.check_command_line_arguments()
    if bool_wtf:
        is_today_your_birthday.dob_vs_now()
    print " "

"""
    Python Artifact
"""
if __name__ == '__main__':
    main()
<span id="mce_SELREST_start" style="overflow:hidden;line-height:0;">&#65279;</span>

538 Puzzles – The Maths Anarchists

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Here is the 538 Blog Riddle Page For Last Week:

When Will The Arithmetic Anarchists Attack

I started this yesterday and I think I successfully solved the riddle.  However I got slightly different answers than the solution published today.  I do not know where the error is.  Was the solution published in the 538 Blog incorrect?  Or is my solution incorrect?

Their solution was that there were 212 events and the longest gap was 1,096 days.  Their solution tagged the most events for any one year to be 7 for the year 2012 which did match my solution.

My solution was:

For Each Date From 01 JAN 2001 to 31 DEC 2099
Consider dd/mm/yy where dd*mm == yy
Then Answer The Following Questions:

[Q1] How Many Of These Events Are There?
[A1] There Are 210 Events In the 21st Century.

[Q2] Which Year Has The Most?
[A2] The Year 2024 Will Have 7 Events.

[Q3] What Is The Longest Gap Between Any Two?
[A3] The Longest Gap = 1097 Days And Occurs Between 03/19/2057 And 03/20/2060

I figured this out using a Python script which is given below.

I’m still looking at my code in hopes of finding the problem (assuming it is my problem).

"""
    When Will The Maths Anarchists Attack
    For Each Year From 2000-2100:
        Consider "mm/dd/yy"
        When mm*dd == yy
        Then For That (mm, dd, yy) Combination
            An Attack Will Be Launched
    [1] How Many Statues Will Be Destoryed?
    [2] Which Year(s) Will Have The Most Statues Destroyed
    [3] What Will Be The Longest Gap Between Statue Destruction
"""
import datetime
import calendar

class Havoc_From_The_Maths_Anarchists:
    """
        Answer Questions 1,2,3 Above
    """
    def __init__(self):

        self.list_of_triples = []
        self.q1 = "[Q1] How Many Of These Events Are There?"
        self.q2 = "[Q2] Which Year Has The Most?"
        self.q3 = "[Q3] What Is The Longest Gap Between Any Two?"
        self.a1 = "[A1] "
        self.a2 = "[A2] "
        self.a3 = "[A3] "

    def build_list_of_dates(self, debug=True):
        """
            For each month number and each day of the month number
            When month number times day of the month number is not less than or equal to 99 skip and consider next date
            If this day of the month is greather than the number of days in this month skip and consier the next date
            If this year, month, day of the month already found skip and consider the next date
            Add this year, month, day of the month to the list dates and consider the next date
        """
        self.list_of_triples = []

        for idx in range(1,13):
            for jdx in range(1,32):
                this_mo = idx
                this_da = jdx
                this_yr = this_mo*this_da
                # SKIP IF YEAR NOT IN RANGE 01 ... 99
                if this_yr &gt; 99:
                    continue
                # SKIP OUT OF RANGE DAYS
                if this_da &gt; calendar.monthrange(this_yr,this_mo)[1]:
                    if debug:
                        print "[BIG_da] mo = ",this_mo," da = ", this_da, " yr = ",this_yr
                    continue
                # ADD THIS DAY TO ALLOWED DAYS
                this_one = (2000+this_yr, this_mo, this_da)
                if debug:
                    print "[NEW_da] mo = ", this_mo, " da = ", this_da, " yr = ",this_yr
                self.list_of_triples.append( this_one )

    def find_answers(self, debug = True):
        self.list_of_triples.sort()
        current_year = 2001
        current_list = []
        list_of_years = []
        list_of_dates = []
        for idx in range(0, len(self.list_of_triples)):
            list_of_dates.append(datetime.date(self.list_of_triples[idx][0],self.list_of_triples[idx][1],self.list_of_triples[idx][2]))
            if self.list_of_triples[idx][0] != current_year:
                current_year = self.list_of_triples[idx][0]
                list_of_years.append(current_list)
                current_list = []
            this_one = datetime.date(self.list_of_triples[idx][0],self.list_of_triples[idx][1],self.list_of_triples[idx][2])
            current_list.append(this_one)

        # Find Answer 1 and Answer 2
        total_dates = 0
        max_events = 0
        max_year = ""

        for a_list in list_of_years:
            if debug:
                for a_date in a_list:
                    print a_date.strftime('%m/%d/%Y'),
                print
            this_years_events = len(a_list)
            total_dates += this_years_events
            if this_years_events &gt; max_events:
                max_events = this_years_events
                max_year = a_list[0].strftime('%m/%d/%Y').split('/')[2]
        self.a1 += "There Are "+str(total_dates)+" Events In the 21st Century."
        self.a2 += "The Year "+max_year + " Will Have " + str(max_events) + " Events."

        # Find Answer 3
        date_then = datetime.date(2000,12,31)
        date_now  = date_then
        delta     = int((date_now - date_then).days)
        max_gap   = delta
        gap_dates = (date_now,date_then)

        for a_date in list_of_dates:
            date_then = date_now
            date_now = a_date
            delta = int((date_now - date_then).days)

            if delta &gt; max_gap:
                max_gap = delta
                gap_dates = [date_now, date_then]
                if debug:
                    print "MAX_GAP = " ,max_gap," Occured Between ",gap_dates[1].strftime('%m/%d/%Y')," and ",gap_dates[0].strftime('%m/%d/%Y')

        self.a3 += "The Longest Gap = "+ str(max_gap)+ " Days And Occurs Between "+gap_dates[1].strftime('%m/%d/%Y')+" And "+gap_dates[0].strftime('%m/%d/%Y')

    def print_answers(self, debug = True):
        print
        print
        print "For Each Date From 01 JAN 2001 to 31 DEC 2099"
        print "Consider dd/mm/yy where dd*mm == yy"
        print "Then Answer The Following Questions:"
        print
        print self.q1
        print self.a1
        print
        print self.q2
        print self.a2
        print
        print self.q3
        print self.a3

def main():
    answer_questions = Havoc_From_The_Maths_Anarchists()
    answer_questions.build_list_of_dates(False)
    answer_questions.find_answers(False)
    answer_questions.print_answers(False)

if  __name__ =='__main__':main()<span id="mce_SELREST_start" style="overflow:hidden;line-height:0;"></span>

 

Integer Multiplication Explained

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When I first left undergraduate school I took a job teaching 8th and 9th grade math. My 8th graders were introduced to integer multiplication for the first time at that level. They certainly knew natural number multiplication and even knew integer addition which includes subtraction by the time we took up integer multiplication.

When teaching integer addition I presented it as addition of vectors directed along the number line. So +2 + -4 would be an arrow starting at zero of length 2 pointing to the right then append an arrow of length 4 pointing to the left. The tip of the second arrow wound up at the point -2 on the number line.

To extend that idea to integer multiplication I introduced items I called ‘Number Line Frogs’. Each frog had a number associated with it and a direction. So a +2 frog could jump two places at a time and it pointed to the right. The number -3 was represented by a frog that could jump three places and pointed to the left.

I then asked the students to think of integer multiplication problems in two parts. The problem a x b was to be visualized as an ‘a-frog’ being asked to do something and that something was to perform ‘b’ jumps. The problem always started at the origin or ‘0’ on the number line. The sign of the number ‘b’ indicated which direction the frog was to jump. The absolute value of the number ‘b’ indicated how many jumps the frog was to take. The sign of the number ‘a’ indicated which direction the frog was pointing and the absolute value of the number ‘a’ indicated how far the frog could jump.

In this way of thinking about it the problem +2 x -3 was the same as saying ‘Allow a +2 frog to jump three jumps backward (-3)’. I stressed that the problem -3 x +2 was the same problem as could be verified by saying ‘Allow a -3 frog to take two jumps forward (+2)’ In both cases the end result was the number -6.

The rule for ‘positive-a times positive-b’ was to say ‘let a postive-a-frog take b-jumps-forward’ which came out to be the same problem as saying ‘let a positive-b-frog take a-jumps-forward’.

The rule for ‘positive-a times negative-b’ was to say ‘let a positive-a-frog take b-jumps-backward, while the rule for ‘negative-b times positive-a’ was to say ‘let a negative-b frog take a-jumps-forward’.

Finally the rule for ‘negative-a times negative-b’ was to say ‘let a ‘negative-a frog take ‘b-jumps-backward’.

When viewed this way and taking into account the ‘frog’ always started at zero the sign table for integer multiplication was justified. Additionally the commutative property of multiplication was illustrated by noting any number that could be decomposed into ‘b-groups of length a’ was equivalent to the observation that the same number could be decomposed into ‘a-groups of length b’.

I will say that the 8th grade group I had were sorted in an inhomogeneous manner with all the top students put into one class and that class was assigned to me. I did worry about this sorting process and felt that it was wrong on multiple levels but given there was nothing a first year math teacher could do to change the situation I made the most of it and presented many mathematical concepts using non standard methods (i.e. presented not-by-the-book). I think that math class did enjoy my presentations.

Can I Get Your Digits! (Except for one of them)

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I follow the blog “Math Jokes 4 Mathy Folks“.  A recent post gave a game/riddle you can play with kids or even post a set of questions on some social media site and play with your friends.  The post is “Can I Get Your Digits“.

So, here are the questions you can ask:

[Step-1] Pick a number with more than four digits (keep it secret).
[Step-2] Add the digits of the number together (keep it secret).
[Step-3] Subtract this sum from the original number (keep it secret).
[Step-4] From [Step-3] give me all the digits except for one of them (keep that digit secret).
[Step-5] I will ‘magically’ guess the missing (secret) digit!

The ‘magic-guess’ is: Add up the digits given to you and then add what ever number it takes to get to the nearest multiple of nine and that number is the missing digit.

So for example: the number 87694. When the digits are added you get 34.  When you subtract 36 from 87694 you get 87660.  Below is a table for each of the possible missing digits from number 87660.

Missing the ‘0’ … 8+7+6+6 = 27 (The missing digit is ZERO)
Missing the ‘6’ … 8+7+6+0 = 21 (The missing digit is SIX)
Missing the ‘7’ … 8+6+6+0 = 20 (The missing digit is SEVEN)
Missing the ‘8’ … 7+6+6+0 = 19 (The missing digit is EIGHT)

Now here is an explanation of why this always works.

The way numbers are represented is in what is called a ‘Place Value System’.  In this system a number can be thought of as a sum of the digits of the number times a power of 10 (for a base 10 number).  The power of 10 depends on the ‘Place’ of the digit.  The ‘least-significant’ digit (the on on the right end of a number) can be thought of as that digit times 10^{0}.  The next ‘Place’ is called the 10’s place and can be thought of as that digit times 10^{1}, with the next ‘Place’ being the 100’s place or the digit times 10^{2} and so forth.

So for example:

87694\quad =\quad 8\times {10}^{4}+7\times {10}^{3}+6\times {10}^{2}+9\times {10}^{1}+4\times {10}^{0}

And this representation is equivalent to the following one when you replace the powers of 10 with the corresponding addition problem:

8\left( 1+9999 \right) +7\left( 1+999 \right) +6\left( 1+99 \right) +9\left( 1+9 \right) +4

It is now clear how this ‘magic guess’ works. When you distribute each digit over the corresponding sum you get the following:

8+8\left( 9999 \right) +7+7\left( 999 \right) +6+6\left( 99 \right) +9+9\left( 9 \right) +4

Regrouping you get the following very suggestive formula:

(8+7+6+9+4)\quad +\quad 8\left( 9999 \right) +7\left( 999 \right) +6\left( 99 \right) +9\left( 9 \right)

This is clearly the sum of the digits of the original number plus another number that is clearly a multiple of nine:

\sum { digits\_ of(87694) } \quad +\quad 9\left[ 8\left( 1111 \right) +7\left( 111 \right) +6\left( 11 \right) +9\left( 1 \right)  \right]

The main thing to realize here is that when you subtract \sum { digits\_ of(87694) } from the original number you are left with a number that (whatever its actual digits are) is a multiple of nine.

Finally it is the case that any number that is a multiple of nine also has the property that the number resulting from summing its digits is also a multiple of nine. This final fact is what you use to find the missing digit. Leaving out a single digit will ‘disturb’ the sum so it is no longer a multiple of nine and you only need sum up what you are given and find that number that causes the sum to be a multiple of nine to deduce the missing digit.

Notice: When the dropped digit is either a ‘0’ or a ‘9’ your magic ‘guess’ should be: “You left out either a 9 or a 0.”

Notice: This explanation can be extended to an arbitrary number of digits and thus it would be a proof.

Notice: That a multiple of nine also has the properties that the sum of its digits is also a multiple of nine can be explained in exactly the same way as the explanation above goes. So assume some number is a multiple of nine:

X\quad =\quad 9x

Now write X as a sum of its digits with each one times its corresponding power of ten:

X\quad =\quad \sum _{ i=0 }^{ n }{ { a }_{ i } } { 10 }^{ i }

Notice that we can write each term of the sum as a digit of the the original number times the quantity \left( 1+9\cdots 9 \right) where the symbol “9\cdots 9 ” is 9 or 99 or 999 or 9999 and so forth.

X\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1+9\cdots 9 \right)

This way of writing the number makes it clear what happens when each digit is distributed over the its corresponding sum and the the sum regrouped so that the sum of the digits of the original number are distinct from the rest of the number.

X\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } +\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 9\cdots 9 \right)

Now it is clear that each term of the rightmost sum can be factored by the number nine giving the formula below. The symbol “1\cdots 1 ” is 1 or 11 or 111 or 1111 and so forth.

X\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } +\quad 9\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1\cdots 1 \right)

Recalling that the original number is a multiple of nine we get the following:

9x\quad =\quad { a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } +\quad 9\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1\cdots 1 \right)

Finally subtracting the rightmost sum from both sides of the formula and factoring out the number nine we get the following which does show that the sum of the digits of the original number is a multiple of nine.

9\left[ x-\sum _{ i=1 }^{ n }{ { a }_{ i } } \left( 1\cdots 1 \right)  \right] ={ a }_{ 0 }+\sum _{ i=1 }^{ n }{ { a }_{ i } } \quad

Fizz Buzz (Beep?)

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I subscribe to a blog called Quora.  You can set filters on what sort of questions are presented. I’ve set them to show math, physics, and computer science posts.

Today I saw a post that asked if someone with an advanced degree in computer science could fail the “Fizz Buzz Test“.

I had no idea what this test was so I did a Google Search on the words “fizzbuzz test” and came up with a wikipedia page dedicated to the game which also mentions the test (here is the page).

After reading the article section named “Other Uses” and messing around with a python script today I can safely say I do fall into the category of someone who has used this game as a “Software Kata

Here is what I’ve done so far.  And, in my defense, I did start out with a VERY SIMPLE version of this game and it sort of “grew-like-topsy” to what you see below.

I’ve used the WordPress tags to embed the code below.  So this blog post not only talks about the Fizz Buzz Test but also gives a demonstration of how source code embedding works on WordPress.

# Fizz Buzz (beep) Script
import os

def sayIt( anIndex, aCount, phrase):
  indexString = "The "+str(aCount)
  if aCount >= 11 and aCount <= 20:
    indexString += 'th'
  elif aCount % 10 == 1:
    indexString += 'st'
  elif aCount % 10 == 2:
    indexString += 'nd'
  elif aCount % 10 == 3:
    indexString += 'rd'
  else:
    indexString += 'th'
  indexString += ' element is '+str(anIndex)+' and we say '+phrase
  os.system("say " +indexString)

def findPhrase( index, fizzNumber, buzzNumber, beepNumber ):
  newNumber = index
  thePhrase = ''
  if fizzNumber != 0:
    while newNumber % fizzNumber == 0:
      thePhrase += 'FIZZ '
      newNumber = newNumber / fizzNumber
  if buzzNumber != 0:
    while newNumber % buzzNumber == 0:
      thePhrase += 'BUZZ '
      newNumber = newNumber / buzzNumber
  if beepNumber != 0:
    while newNumber % beepNumber == 0:
      thePhrase += 'BEEP '
      newNumber = newNumber / beepNumber
  return thePhrase

def fizzBuzzBeep(maxCount, fizzNumber, buzzNumber, beepNumber):
  count = 0
  index = 1
  while (count<maxCount):
    aPhrase = findPhrase( index, fizzNumber, buzzNumber, beepNumber )
    if len(aPhrase) != 0:
      count +=1
      print("{0:3d} - {1:3d} {2}".format(count, index, aPhrase))
      sayIt( index, count, aPhrase)
    index +=1

if __name__ == '__main__':
  fizzBuzzBeep(100, 3, 5, 7)

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