Tag Archives: Algorithm
Project Euler Question 15
Posted by Shawn Zhang
on December 23, 2012
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Question:
Starting in the top left corner of a 2*2 grid, there are 6 routes (without backtracking) to the bottom right corner.How many routes are there through a 20*20 grid?
Solution:
This is not a particular question for programming ,it seems like a more mathematics problem for me . for moving from top left to right down , only consist two direction "down/ right", all path is a combination of 20 down and 20 right moves, so ,solution is just number of combination 20 of 40 instruction on move .
For Read more [...]
Project Euler Problem 26
Posted by Shawn Zhang
on December 23, 2012
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Question
A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2
=
0.5
1/3
=
0.(3)
1/4
=
0.25
1/5
=
0.2
1/6
=
0.1(6)
1/7
=
0.(142857)
1/8
=
0.125
1/9
=
0.(1)
1/10
=
0.1
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d 
1000 for which 1/d contains Read more [...]
1000 for which 1/d contains Read more [...] Project Euler Question 14
Posted by Shawn Zhang
on December 20, 2012
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Question:
The following iterative sequence is defined for the set of positive integers:
n 
n/2 (n is even)
n 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 40 20 10 5 16 8 4 2 1
It can be seen that this sequence (starting at 13 and finishing Read more [...]
n/2 (n is even)
n 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following sequence:
13 40 20 10 5 16 8 4 2 1
It can be seen that this sequence (starting at 13 and finishing Read more [...] Project Euler Question 29
Posted by Shawn Zhang
on December 19, 2012
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Question:
Consider all integer combinations of ab for 2 
a 5 and 2 b 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 a 100 Read more [...]
a 5 and 2 b 5:
22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125
If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
How many distinct terms are in the sequence generated by ab for 2 a 100 Read more [...] Project Euler Question 31
Posted by Shawn Zhang
on December 18, 2012
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Question 31:
In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).x
It is possible to make £2 in the following way:
1X£1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p
How many different ways can £2 be made using any number of coins?
Solution :
actually this is a linear algebra question : Find all possible 
that satisfy below equation
Algorithm Read more [...]
that satisfy below equation
Algorithm Read more [...] Python Solution for Project Euler(Problem 007 - 009 )
Posted by Shawn Zhang
on March 20, 2012
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Question 007
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
Solution : 104743
Thanks to Prime_number_theorem(link: http://en.wikipedia.org/wiki/Prime_number_theorem ) There is approximate formula to figure the Nth number of prime number . The answer is actually programming language independent.
It must be a joke in J, I do serious considering learning J !
p: 10000
Question 8:
Find the Read more [...]
Python & R Solution for Project Euler(Problem 004 - 006)
Posted by Shawn Zhang
on March 18, 2012
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A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.
Solution 906609
Python: , just setup a loop of combination of 2 3-digit numbers ,and test if product of them is a palindrome .
def palindrome():
rlt = 0
for i in range (999, 99, -1):
for j in range (999, 99, -1):
prod = str(i * j);
prod_rev Read more [...]
Python & R Solution for Project Euler(Problem 001 - 003)
Posted by Shawn Zhang
on March 17, 2012
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1. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.Find the sum of all the multiples of 3 or 5 below 1000.
Solution: 233168
Python, powerful range function generate list numbers via step
sum(range(0,1000,3)) + sum(range(0,1000,5)) - sum(range(0,1000,15))
R, we use seq() to generate this value list .
> sum(seq(from=0, to=999, by=3)) + sum(seq(from=0,to=999,by=5)) -sum(seq(from=0,to=999,by=15))
[1] 233168
2. Read more [...]
A Discrete Probability Models in R
Posted by Shawn Zhang
on March 10, 2012
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There is an interesting Example in real world (From <mathematical modeling 3th>,Ch07):
An electronics manufacturer produces a variety of diodes. Quality control engineers attempt to insure that faulty diodes will be detected in the factory before they are shipped . It is estimated that 0.3% of the diodes produced will be faulty . It is possible to test each diode individually. It is also possible to place a number of diodes in series and test the entire group . If this test fails Read more [...]
Find Equal Sum in Two Number Set
Posted by Shawn Zhang
on February 28, 2012
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Accounting Nightmare
In account book, it is really nightmare if you try to find out a even group of transactions within a larger group .(by even group, we mean sum of credit equals to sum of debit)
It is exhausting to find out 15 Debit + 25 Debit equals to 30 Credit + 10 Credit in a large transactions scope. Thanks to latest development of Python(2.7 with itertools module) , it is easy to implement a algorithm to find all possible groups in two data set.
Abstraction
Lets back to Read more [...]