︠8c18a7b9-ae80-471d-978c-eedeec1be08f︠
### Programming == make the machine to do a routine work
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︠4d5ad766-d427-4000-98fa-3c324a37ecdc︠
### Python is case-sensative
a=1;
A
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︠c49a28ea-b2d1-48a8-98ad-a520e191e9f9︠
### Python cares about line break (unlike LaTeX)
a
=
1
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︠330ba9e5-1f5d-407d-a526-c5187e217bee︠
### Python cares about indents
if "Z">"B":
print "sign the contract"

### Use tab and ctrl+tab to do indents
### convention: 1 tab = 4 spaces
### On CoCalc, change your setting at “Account” and check the box of “Spaces instead of tabs”.
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︠53b849ff-e05b-4492-affb-6e63fe9e1ddf︠
### if statement
### this code translates a percentage grade to a letter grade
### try changing the value of score and run again

score = 90;
if score >= 80 and score <= 100:
    print "A";
elif score >= 70 and score < 80:
    print "B";
elif score >= 60 and score < 70:
    print "C";
elif score >= 0 and score < 60:
    print "D";
else:
    print "Input score not valid";
︡396ccea7-760c-4f80-b5e2-9aefa03d9ed8︡
︠311f0390-ac1a-4f6e-8dba-63cee89a52b9︠
### for loop
### this code prints all positive integers less than or equal to 100
###  that is a multiple of 5 or 7

for i in range(1,101):
    if i%5==0 or i%7==0:
        print i;
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︠43b16e39-94a4-4d66-b9e6-4d20145f456a︠
### while loop
### this code is a primitive way to find
###  the least common multiple of 5 and 7

i=1;
while True:
    if i%5==0 and i%7==0:
        print i;
        break;
    else:
        i=i+1;
︡afe028b3-d9bb-4e3a-872c-189bf72866d9︡
︠c2f401c2-f41f-43dc-9654-37b22a5ef460︠
### Exercise:
###    Print all integers from 1 to 100
###    that is a multiple of 3, 5, or 7.
###    How many of them you get?  (Don't count by hand...)
count=0;
for i in range(1,101):
    ???

print count;
︡18153486-596e-4f92-aef7-124c843094a2︡
︠6877d3b5-43cb-4c42-a505-6f485fff945d︠
### Exercise:
###     Find the sum of k^3 for k=1,...,10.
total=0;
for ???:
   ???
print total
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︠6b4df72b-bf8c-47c8-a430-3221102bc7cf︠
### Exercise:
###     Count how many pairs (a,b) with gcd(a,b)==1
###     for a=1,...,10 and b=1,...,10.
for a in ???:
    for b in ???:
        ???
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︠13119363-51de-4109-a48a-b33546df7d8as︠
### How to call a submatrix?

M = matrix(
[[1,4,9],
[4,5,6],
[7,8,9]])

print M;
print "---"
print M[[1,2],[0,1]];


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### Recursion: you may call a function in the definition of the function.

def det(M):
    m,n = M.dimensions();
    if m != n:
        return False; # determinant only works on square matrices
    if n == 0:
        return 1; # this is a vacuous definition
    if n == 1:
        return M[0,0];
    total = 0;
    ### expand on the first row
    for j in range(n):
        rows = range(n);
        rows.remove(0);
        cols = range(n);
        cols.remove(j);
        total += (-1)^(0+j)*M[0,j]*det(M[rows,cols]);
    return total;

print M;
print M.determinant(), det(M);
︡e08c49cb-6f1d-4bfc-89a6-19ff4bc01fd5︡
︠9f20d4a5-6626-47fd-9308-d8da5f58503di︠
%md

## Project 1
Let $a_0=3$, $a_1=2$, $a_2=14$ and $a_n-3a_{n-1}+0a_{n-2}+4a_{n-3} = 0$.

Write a function to find $a_n$ through the recurrence relation.
︡1491331a-ff0e-4231-ae95-9c18acf6be1f︡{"done":true,"md":"\n## Project 1\nLet $a_0=3$, $a_1=2$, $a_2=14$ and $a_n-3a_{n-1}+0a_{n-2}+4a_{n-3} = 0$.\n\nWrite a function to find $a_n$ through the recurrence relation."}
︠e8a40d27-ed26-4b3c-81aa-4bc1b950f580s︠
def an_rec(n):

    ### Your answers:
    if n == 0:
        return 3;
    if n == 1:
        return 2;
    if n == 2:
        return 14;
    return 3*an_rec(n-1) - 4*an_rec(n-3);
    ### this is not an efficient algorithm, but easy to program.
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︠9521f063-083a-4bb6-b448-aef313add05ci︠
%md
## Project 2
Use the answer you found in Problem 1 of the homework to write another function that compute $a_n$ directly by the formula.

Then test your answers by comparing `an_rec(n)` and `an_for(n)` for $n=1,\ldots,10$.
︡bc1a6203-7757-4240-9efd-494b052c3ed8︡{"done":true,"md":"## Project 2\nUse the answer you found in Problem 1 of the homework to write another function that compute $a_n$ directly by the formula.\n\nThen test your answers by comparing `an_rec(n)` and `an_for(n)` for $n=1,\\ldots,10$."}
︠19071467-d0cd-4343-9c78-e4b15712975c︠
def an_for(n):

    ### You answers:
    return 2*(-1)^n + 2^n + n*2^n;

### test;
for n in range(1,11):
    print n,an_rec(n),an_for(n);
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︠6a6551c2-dba9-4378-809b-1fc31e17c0aa︠