CS 1000 H. James de St. Germain

Problem Solving

Writing a Program to Approximate Pi (or why

3.141592653589793238462643383279502884197169399 is better than

3.141592653589793238462643383279)

CS 1000

Problem

• We need pi…

• The computer needs pi…– At least for geometric, mathematical, physics,

mechanical, civil, etc. computations

• The computer doesn’t know pi– Okay, it does know pi, but someone had to tell

it. – What if you were that someone?

CS 1000

Option 1: Don’t use a Symbolic Name, just Memorize the digits…

3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679 8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196 4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273 7245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094 3305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912 9833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132 0005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235 4201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859 5024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303 5982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989 3809525720106548586327886593615338182796823030195203530185296899577362259941389124972177528347913151 5574857242454150695950829533116861727855889075098381754637464939319255060400927701671139009848824012 8583616035637076601047101819429555961989467678374494482553797747268471040475346462080466842590694912 9331367702898915210475216205696602405803815019351125338243003558764024749647326391419927260426992279 6782354781636009341721641219924586315030286182974555706749838505494588586926995690927210797509302955 3211653449872027559602364806654991198818347977535663698074265425278625518184175746728909777727938000 8164706001614524919217321721477235014144197356854816136115735255213347574184946843852332390739414333 4547762416862518983569485562099219222184272550254256887671790494601653466804988627232791786085784383 8279679766814541009538837863609506800642251252051173929848960841284886269456042419652850222106611863 0674427862203919494504712371378696095636437191728746776465757396241389086583264599581339047802759009 9465764078951269468398352595709825822620522489407726719478268482601476990902640136394437455305068203 4962524517493996514314298091906592509372216964615157098583874105978859597729754989301617539284681382 6868386894277415599185592524595395943104997252468084598727364469584865383673622262609912460805124388 4390451244136549762780797715691435997700129616089441694868555848406353422072225828488648158456028506 0168427394522674676788952521385225499546667278239864565961163548862305774564980355936345681743241125 1507606947945109659609402522887971089314566913686722874894056010150330861792868092087476091782493858 9009714909675985261365549781893129784821682998948722658804857564014270477555132379641451523746234364 5428584447952658678210511413547357395231134271661021359695362314429524849371871101457654035902799344 0374200731057853906219838744780847848968332144571386875194350643021845319104848100537061468067491927 8191197939952061419663428754440643745123718192179998391015919561814675142691239748940907186494231961 5679452080951465502252316038819301420937621378559566389377870830390697920773467221825625996615014215 0306803844773454920260541466592520149744285073251866600213243408819071048633173464965145390579626856 1005508106658796998163574736384052571459102897064140110971206280439039759515677157700420337869936007 2305587631763594218731251471205329281918261861258673215791984148488291644706095752706957220917567116 7229109816909152801735067127485832228718352093539657251210835791513698820914442100675103346711031412 6711136990865851639831501970165151168517143765761835155650884909989859982387345528331635507647918535 8932261854896321329330898570642046752590709154814165498594616371802709819943099244889575712828905923 2332609729971208443357326548938239119325974636673058360414281388303203824903758985243744170291327656 1809377344403070746921120191302033038019762110110044929321516084244485963766983895228684783123552658 2131449576857262433441893039686426243410773226978028073189154411010446823252716201052652272111660396 6655730925471105578537634668206531098965269186205647693125705863566201855810072936065987648611791045 3348850346113657686753249441668039626579787718556084552965412665408530614344431858676975145661406800 7002378776591344017127494704205622305389945613140711270004078547332699390814546646458807972708266830

CS 1000

• The current unofficial (as of October 4 2006) world record is 100,000 decimal places, and was set by a Japanese mental health counselor named Akira Haraguchi, who is currently 59 years of age…

Perhaps not the best use of Time?

CS 1000

Why Approximate Pi

• Pi is:– Irrational (X/Y ~= pi)– Infinite and Varied

• Computers cannot represent– Irrational numbers– Infinite numbers

• AH HA! You said in the previous slides that the computer knows pi! You were WRONG! The computer must only know an approximation of pi!– Perhaps we can do better than the Matlab

Programmers?

CS 1000

Your “Assignment” (or word problems are fun)

The magical number pi (3.141927…) can be calculated by a sequence of subtractions and additions (e.g., 4 - 4/3 + 4/5 -

4/7 + 4/9 - 4/11 + 4/13 - 4/15 + etc). Write a program to calculate pi. First ask the user how many steps she wants to take. Make sure the inputted number is positive. Print out the final value.

CS 1000

Solving the Pi Problem

• Your Objective:– Turn Problem Statement into Code

• Process– Separate out and Identify (in high level

English) the Steps necessary to accomplish the goal!

– Convert each step into Pseudocode – Convert Pseudocode into Code

CS 1000

Things to Remember

• Computer programs execute ONE statement at a time and thus you have to think in a series of small steps while programming.

• While DESIGNING you should think in high level concepts, such as:– Get input– Calculate values (or use a loop to calc values)– Plot data

• Design First!

CS 1000

Programming Problem

Calculate Pi Program

This is DAUNTING!

This is TOO Much!

Must Not Panic. Must Not Panic!

Lets break this into smaller easier steps!

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Identify High Level Goals

• First re-read the problem statement:The magical number pi (3.141927…) can be

calculated by the following sequence of subtractions and additions.:• 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + etc

Write a program to calculate pi. First ask the user how many steps she wants to take. Make sure the inputted number is positive. Print out the final value.

CS 1000

Break Down the Problem

• First lets break this down:– “Write a program to calculate pi. First ask the user

how many steps she wants to take. Make sure the number is positive. Print out the value.”

• This becomes1. The user must input how many steps in the

calculation. – This “number of steps” must be positive.

2. The program calculates an approximation to pi.– We have a formula for this

3. Print out value of pi.

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Program Decomposition Graph

Calculate Pi Program

Ask user for non negativenumber of steps

Use formula tocalculate pi

Output theanswer

Much Better!

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Or Flow Chart

End

Start

Get Sequence Length

Calculate Pi

Print Output

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Next: Analyze the Formula

• We are calculating the value of pi.

• Don’t panic, we are given the formula.– 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + etc – Step 1 2 3 4 5 6 ...

CS 1000

Oh my! It’s Math!

• Many people experience what I call the, “Oh my, its math!” phenomenon.

• Try not to let the idea that you are doing “Math” scare you into not thinking about the problem.

• The computer is going to do the work for you.

• Your job is to tell the computer how

CS 1000

So, first:Solve the Problem by Hand…

• … for a few simple examples:– If the user inputs the number zero,

• you ask again.– If the user inputs 1 you would print out that pie is

• 4 – If the user inputs 2 you would print out that pie is

• 2.6666 (4 – 4/3)– If the user inputs 3 you would print out that pie is

• 3.4666 (4 – 4/3 + 4/5)– etc, etc, etc

Formula: 4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 - 4/15 + etcStep: 1 2 3 4 5 6 …

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Looking for Patterns!

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Write a For loop!

• for i = 1 : 1000 % for index = start:cnt:end– fprintf(‘%d ‘, i);

• end

• fprintf(‘\n’);

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Write a While loop!

• value = 1

• while ( value < 1000 ) % while ( true )– fprintf(‘%d ‘, value);– value = value + 1;

• end

• fprintf(‘\n’);

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Write a For Loop!

• for i=2:2:1000– fprintf(‘%d ‘, i);

• end

• fprintf(‘\n’);

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Write a While Loop!

• value = 2;

• while (value < 10000)– fprintf(‘%d ‘, value);– value = value + 2;

• end

• fprintf(‘\n’);

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Write a Program. Use a While loop!

• current_digit = 1;• while ( true )

– fprintf(‘%d ‘, current_digit);– if (current_digit == 1)

• current_digit = 0;

– else• current_digit = 1;

– end

• end

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Write a program. Use a For loop!

• for i = 1:10000– if ( rem ( i, 2) == 0)

• fprintf(‘1 ‘);

– else• fprintf(‘0 ‘);

– end

• end

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Write a Loop

• Can you figure this one out?

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Back to Pi: Looking for Patterns!

• The formula is:

4 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + 4/13 – …

• What patterns do you see? – Write them down on your notes:

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Does re-writing the formula help?

• Lets rewrite this on its side, step by step:4

- 4 / 3

+ 4 / 5- 4 / 7

+ 4 / 9

- 4 / 11

…etc …

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Making the pattern consistent.

• Lets make the pattern consistent:+ 4 / 1

- 4 / 3

+ 4 / 5- 4 / 7

+ 4 / 9

- 4 / 11

…etc …

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Seek the PatternWhat do we divide by?

STEP Division 1 4 / 1

2 4 / 3

3 4 / 5

4 4 / 7

--------------------------------

i 4 / ?

Symbolic of all possible values

? = some math with i

Which is often written:

? = f(i) % some function of i

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step divisor 1 1

2 3

3 5

4 7

5 9

6 11

7 13

• Write down a math equation to compute the “divisor” value from the “step” variable!

• divisor = 2 * step – 1• or ( ( 2 * i ) – 1)

Seek the PatternCompute the Divisor?

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Seek the PatternWhen do we Add?

When do we Subtract?

STEP Add/Sub1 +

2 -

3 +

4 -

---------------------------------------------------------------------------

i ?

Write your Pseudocode in your notes:

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When do we Add?When do we Subtract?

What is the pattern?If step number “i” is odd we add

If step number “i” is even we subtract

In Matlab:if ( odd(i) )

add the new value

else

subtract the new value

end

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Thus we have the following patterns:

STEP Add/Sub Value1 + 4 / 1

2 - 4 / 3

3 + 4 / 5

4 - 4 / 7

---------------------------------------------------------------

i add when i is odd 4 / (2*i-1)

CS 1000

Time to Start Programming

• We know we are supposed to:1. Query the user for how many iterations (how many

steps, how big (i) gets) in our loop– If the number is not positive repeat step 1

2. Calculate the “current” approximation to pi using our formula

3. Output answer

Calculate Pi Program

Ask user for non negativenumber of steps

Use formula tocalculate pi

Output theanswer

CS 1000

Breaking the steps down

• We know we are supposed to:1. Query the user for how many iterations (how many steps) in

our loopnumber_of_steps = input(‘How many steps: ‘);

2. If the number is not positive repeat step 1How do we repeat an unknown number of times?

WHILE LOOP

3. Calculate the value of pi using our formulaWhat are we doing? Repeating something a pre-defined

number of times. What programming element do we use?

FOR LOOP

4. Ouput the value

CS 1000

Getting the number of steps from the user

• We can combine our ideas into the following code:

number_of_steps = input(‘How many steps: ‘);while (number_of_steps < 1)

number_of_steps = input(‘How many steps: ‘);end

– This is a DESIGN PATTERN that you will see many times in your programming career.

CS 1000

While Loop Design Pattern

• Ask for data

• While data is incorrect– Re-ask for data

• End while

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Now for the calculation

• If we are calculating something, we had better have a variable for it:

our_pi = 0;

Note 1: Even though we are using the variable name our_pi, what we really mean is:

• “our approximation to the value of pi over the previous X steps” (but that would be too long a variable name)

Note 2: we initialize the value of our variable to 0 to represent that no work has yet been done.

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Now for the calculation

our_pi = 0; % (our pi value starts at 0).

• Here is our formula (at STEP=i, compute 4 / ( i*2 -1 ), and add to our_pi )

– our_pi = our_pi + ( 4 / ( (i * 2) – 1 ) ;

• remember: sign alternates but initially lets just use ‘+’

• Thus we need a Loop

for i = 1 to number_of_stepsdo calculation

end

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Calculation continued:

• Lets combine our steps:– Remember

• Initialize variable

• Write loop

• Insert formula into loop

our_pi = 0; % (our pi value starts at 0).

for i = 1 : number_of_steps

our_pi = our_pi + ( 4 / ( (i * 2) – 1 ) );

end

% read this to be, “Our approximation of Pi becomes equal to the old approximation of Pi plus (or minus) the new computation.”

CS 1000

Sign Alternates!

• We need to take into account the fact that the sign alternates. There are many options:

1. Mathish expression– add if odd ( i )

2. Another Math expression– sign = ( -1 ^ (i+1) );

3. A boolean flag (called a toggle):sign_is_positive = true; % then change to false.sign_is_positive = ~sign_is_positive; % inside loop

4. Others are possible

CS 1000

Alternating Sign using odd even

• if ( odd(i) )

• our_pi = our_pi + % PLUS formula

• else % i must be even

• our_pi = our_pi - % MINUS formula

• end

CS 1000

Alternating Sign using a Boolean

if (time_for_a_negative)

our_pi = our_pi - % MINUS formulatime_for_a_negative = ~time_for_a_negative;

else % only alternative is time for a positive

our_pi = our_pi + % PLUS formulatime_for_a_negative = ~time_for_a_negative;

end

CS 1000

Cleaning up Duplicate Code

if (time_for_a_negative) our_pi = out_pi - % MINUS formula

else % only alternative is time for a positive our_pi = our_pi + % PLUS formula

end

time_for_a_negative = ~time_for_a_negative;

time_for_a_negative = ~time_for_a_negative;

CS 1000

Alternating Sign using Math

our_pi = … our_pi + (formula * -1^(i+1));

Advantage: No IF statementDisadvantage: Perhaps Compute

Intensive

CS 1000

Insert alternating sign code• Old code:

our_pi = 0; % (our pi value starts at 0).for i = 1 : number_of_steps

our_pi = our_pi ? ( 4 / ( (i * 2) – 1 ) );end

• New code using “boolean flag” to alternate add/subour_pi = 0; % (our pi value starts at 0).sign_is_positive = true;for i = 1 : number_of_steps

if (sign_is_positive == true)

our_pi = our_pi + ( 4 / ( (i * 2) – 1 ) );else

our_pi = our_pi - ( 4 / ( (i * 2) – 1 ) );endsign_is_positive = ~sign_is_positive; % must switch each time

end

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Finally: Print out the Answer

• This step is easy! (I hope!)fprintf(‘the value of pi over %d iterations is approximated

as %f\n’, number_of_steps, our_pi);

• %d – print integer number (whole number)

• %f – print float number (with decimal point)

• \n – print a newline

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Summary

• What we did:– Analyzed program from the TOP DOWN.– Found major components of program.– For each major component, wrote

pseudocode – For each section of pseudocode we wrote

actual code

CS 1000

What does the final program look like?%% Program by : H. James de St. Germain% This program calculates the value of pi, using the formula% pi = 4 – 4/3 + 4/5 – 4/7 + 4/9 + . . . number_of_steps = input('How many steps: ');while (number_of_steps < 1) number_of_steps = input('How many steps: ');end our_pi = 0; sign_is_positive = true;for i = 1 : number_of_steps if ( sign_is_positive == true ) our_pi = our_pi + ( 4 / ( ( i * 2 ) - 1 ) ); else our_pi = our_pi - ( 4 / ( ( i * 2 ) - 1 ) ); end sign_is_positive = ~sign_is_positive;end fprintf('the value of pi over %d iterations is %f\n', number_of_steps, our_pi);

CS 1000

Lets Convert to C

CS 1000

Questions?