Wednesday, November 28, 2018
Friday, November 16, 2018
The Book Empire Strikes Back
This is my updated book list Starting from June 2018
1. 7th Heaven (James Patterson)
1. 7th Heaven (James Patterson)
2. The Book Jumper (Methchild Glaser)
3. Listen to your heart (Kasie West)
4. Nice and Mean (Jessica Leader)
5. Save the Date (Morgan Matson)
6. Warriors: SkyClan’s Destiny (Erin Hunter)
7. Pivot Point (Kasie West)
8. Warriors: A Vision of Shadows- Shattered Sky (Erin
Hunter)
9. Pretty Little Liars: Flawless (Sara Shepard)
10. Warriors: Crookedstar’s Promise (Erin Hunter)
11. Little Monsters (Kara Thomas)
12. Postcards from Venice (Dee Romito)
13. The Last to Let Go (Amber Smith)
14. Warriors: A Vision of Shadows- Darkest Night (Erin
Hunter)
15. To All the Boys I’ve Loved Before (Jenny Han)
16. Pretty Little Liars: Perfect (Sara Shepard)
17. The List (Siobhan Vivian)
18. Three Dark Crowns (Kendare Blake)
19. Royals (Rachel Hawkins)
20. Catwoman: Soulstealer (Sarah J. Maas)
21. Losing Brave (Bailee Madison & Stefne Miller)
22. Love, Life, and the List (Kasie West)
23. The Clique (Lisi Harrison)
24. Pretty Little
Liars: Unbelievable (Sara Shepard)
25. Warriors: A Vision of Shadows- River of Fire (Erin
Hunter)
26. That's Not What Happened (Kody Keplinger)
27. Warriors: Tigerheart's Shadow (Erin Hunter)
28. This Savage Song (Victoria Schwab)
29. Legendary (Stephanie Garber)
30. Warriors: Crowfeather's Trial (Erin Hunter)
31. The Winner's Curse (Marie Rutoski)
32. One of Us is Lying (Karen M. McManus)
33. Lucky in Love (Kasie West)
34. The Cheerleaders (Kara Thomas)
35. Once and for all (Sarah Dessen)
36. Lady Midnight (Cassandra Clare)
37. An Ember in the Ashes (Sabaa Tahir)
38. Red Queen (Victoria Aveyard)
39. Reboot (Amy Tintera)
40. Warriors: Tallstar's Revenge (Erin Hunter)
41. Renegades (Marissa Meyer)
42. Throne of Glass (Sarah J. Maas)
43. Pretty Little Liars: Wicked (Sara Shepard)
44. A Study in Charlotte (Brittany Cavallaro)
45.
Friday, November 2, 2018
Pigeon-Hole Principle
What is a Pigeon-Hole Principle?
If I ask you the following questions "Are there two people in NYC who have the same number of hairs on their head?" or "Are there two homeschoolers in Long Island who have the same birthday?"
What will you answer? The answer to both questions is "YES".
How? The proof uses the Pigeon-Hole Principle a simple concept yet a powerful tool.
Lets say that there are 6 pigeons in a cold winter day and they have to shelter in 5 holes. If all pigeons should have shelter, then at least two pigeons have to fit in a hole. This is generalized as the pigeon-hole principle.If there are x pigeons and y holes, and x>y and each hole must be occupied, at least two pigeons will have to fit in a hole.
This looks like common knowledge. Why is it important in mathematics to be called a "Principle"?
Let us look at how we will answer the questions asked above? A human on an average has 150,000 hairs. There are around 2 million people in NYC. Think of the number of hairs as holes and people as pigeons. So by Pigeon Hole Principle there are at least two people who have the same number of hairs.
I read the following proof recently in the book A Walk Through Combinatorics and it really fascinated me.
Look at the sequence: 7, 77, 777, 7777,...
Claim: There is an element of the sequence which is divisible by 2003.
How to prove? By Contradiction.
On the contrary that none of the first 2003 terms are divisible by 2003. The possible remainders are 1,2,3,...2002.
There are 2003 members of the sequence. They are like pigeons.
There are 2002 remainders. They are like holes.
By Pigeon-Hole principle, there should be at least two pigeons (two members of the sequence) in a home (remainder).
Thus at least two members share the same remainder.
Name the two numbers of the sequence as x and y.
x = d(2003) + r
y = f(2003) + r.
Since x and y are in the sequence, we can assume that x > y.
x-y = (d-f)2003
This implies that x-y is divisible by 2003.
If we show that x-y is a member of the sequence 7, 77, 777,... then we have showed that our assumption is false.
But why should x-y be of the form 77...7
We know that x = 7777...777
y = 7...777
Hence x-y = 777 x
Since 2003 divides x-y, and 2003 does not divide
2003 must divide 77...777 < x.
Hence we have found a number in the sequence that is divisible by 2003 which renders our assumption false. Hence 2003 divides at least one of the first 2003 terms.
If I ask you the following questions "Are there two people in NYC who have the same number of hairs on their head?" or "Are there two homeschoolers in Long Island who have the same birthday?"
What will you answer? The answer to both questions is "YES".
How? The proof uses the Pigeon-Hole Principle a simple concept yet a powerful tool.
Lets say that there are 6 pigeons in a cold winter day and they have to shelter in 5 holes. If all pigeons should have shelter, then at least two pigeons have to fit in a hole. This is generalized as the pigeon-hole principle.If there are x pigeons and y holes, and x>y and each hole must be occupied, at least two pigeons will have to fit in a hole.
This looks like common knowledge. Why is it important in mathematics to be called a "Principle"?
Let us look at how we will answer the questions asked above? A human on an average has 150,000 hairs. There are around 2 million people in NYC. Think of the number of hairs as holes and people as pigeons. So by Pigeon Hole Principle there are at least two people who have the same number of hairs.
I read the following proof recently in the book A Walk Through Combinatorics and it really fascinated me.
Look at the sequence: 7, 77, 777, 7777,...
Claim: There is an element of the sequence which is divisible by 2003.
How to prove? By Contradiction.
On the contrary that none of the first 2003 terms are divisible by 2003. The possible remainders are 1,2,3,...2002.
There are 2003 members of the sequence. They are like pigeons.
There are 2002 remainders. They are like holes.
By Pigeon-Hole principle, there should be at least two pigeons (two members of the sequence) in a home (remainder).
Thus at least two members share the same remainder.
Name the two numbers of the sequence as x and y.
x = d(2003) + r
y = f(2003) + r.
Since x and y are in the sequence, we can assume that x > y.
x-y = (d-f)2003
This implies that x-y is divisible by 2003.
If we show that x-y is a member of the sequence 7, 77, 777,... then we have showed that our assumption is false.
But why should x-y be of the form 77...7
We know that x = 7777...777
y = 7...777
Hence x-y = 777 x
Since 2003 divides x-y, and 2003 does not divide
2003 must divide 77...777 < x.
Hence we have found a number in the sequence that is divisible by 2003 which renders our assumption false. Hence 2003 divides at least one of the first 2003 terms.
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