Proving

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Hi.:)

[Question]
Here's what we had to do:

http://img85.imageshack.us/img85/7103/puzcopy3qk.png

There were 4 match sequences (see top right of image) and we had to find the formula for each the number of matches in the nth patern in each one. That was simple.

H being height, n being number in sequence.

Then we had all of the formulas and had to think of a formula to get the amount of matches in any sequence.

[Difficulty]
I managed to do that and get (2h+1)n+h but that was just by looking at it and couldn't really show any working out.

Could you advise me on what working out to show and then also prove it.

[Thoughts]
Well, I'm just a bit confused about showing working out and proving, and I currently don't have any ideas as to what to show.



Thank you for your time!:)
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Due to me wording it quite crap up there I'll have another go:

Each line is a match, we had 4 sequences and we had to figure out the formula for each sequence.

http://img85.imageshack.us/img85/6161/hpim08387up.jpg

E.G.

In the first sequence we notice it goes up by 3 each time. We notice if we times n by 3 and add 1 each time we get the number of matches in that sequence ONLY.

Then once we had all our formulas for each sequence we had to look at them and figure out a formula to find the number of matches in any sequence.

E.G.

When the height is one the formula is 3n+1.
When the height is two the fomrula is 5n+2.
When the height is three the formula is 7n+3.
When the height is four the formula is 9n+4.

So what can we write fot a formula for the number of matches in any sequence?

Call the height h and the sequence number n.

Notice the amount on n is 2 x the height + 1 in every formula.

So we get (2h+1).

Now because this figure is being multiplied by n, we multiply by n (2h+1)n.

Finally notice the number being added on at the end is the same as the height.

So the formula is (2h+1)n+h.

We can test it for example putting in:

Height of 2, pattern 2, (see diagram, 2x2).

(2x2+1)2+2

5x2+2=12

http://img85.imageshack.us/img85/8163/hpim08370gd.jpg

So we know the formula works, now I just need to show some "mathematical" working out and also "prove" it.

Do you understand what I mean now? Is there a specific area? Please say you undestand now.:D Thanks.:)
 

dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
"n" can be referred to as columns (up-down). Similarily, "h" can be referred to as rows (left-right).

Examine a single column by itself, if we say that the first row of that column requires 3 matches (forming the C shape), then every row after the first will require 2 matches (forming the L shape). In general, the number of matches in a single column is 3+2(h-1).

Single Column (C=3 matches, L=2 matches)
C
L
L

If you have multiple columns, the general formula is n[3+2(h-1)] because for each column you will have 3+2(h-1) matches.

Multiple Columns (C=3 matches, L=2 matches)
CC CC
L L L L
L L L L

You will also need 1 match for every row to "close off" the opening at the end. So in general, "h" additional matches are needed.

Complete (C=3 matches, L=2 matches, I=1 match)
CC CCI
L L L LI
L L L LI

Putting that all together, you get the general formula for the number of matches which is n[3+2(h-1)]+h.

n[3+2(h-1)]+h
=n(3+2h-2)+h
=n(2h+1)+h
=(2h+1)n+h

I'm hoping that will satisfy your question. :D
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Argh, I'm confused, I was told h was height and n was the sequence number (or rows).... :cry:

Also for the first bit you posted for the one column bit; 3+2(h-1), I put one in for what you said h was and it didn't come out right.... :(
 

dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
Rows go from left to right so when you increase "h" by 1, you are actually adding another row.

h1 (row1) --->
h2 (row2) --->
h3 (row3) --->

Columns go from top to bottom so when you increase "n" by 1, you are actually adding another column.

n1 (column1) n2 (column2) n3 (column3)
........|.................|.......................|
........V................V......................V

(ignore the dots)

3+2(h-1) is the general formula for a column when not including the additional 1 match per row needed to close off the openings at the end, that's why when you plug in 1 for "h", you get 3 matches instead of 4.

Doing it this way allows every column to have the same general formula which allows you to write n[3+2(h-1)] as the general formula for any number of columns.

Then all you need to do is add "h" because you need 1 match per row to close off all the openings at the end. The actual general formula for the total number of matches is n[3+2(h-1)]+h, not 3+2(h-1).

Hope that clarifies it. :D
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Sorry but I'm terribly confused....Sorry for all these questions.

1) So we cannot use the first formula bit because it's not finished?

I thought the first sequence was like this:

http://img90.imageshack.us/img90/1687/hpim08394qs.jpg

Do you mean it like this?:

http://img286.imageshack.us/img286/5766/hpim08415hp.jpg

2) What if the first part of the sequence was just a square then it wouldn't be 3 +2....

3) Also can I just ask, what do these mean [ ]...? So I know what I'm writing.

4) Finally if you say h is rows:

"You will also need 1 match for every row to "close off" the opening at the end. So in general, "h" additional matches are needed."

CC CCI
L L L LI
L L L LI

Isn't that 4 rows yet you've only added 3 matches so how can it be add h?

5)Finally finally :p, would it be possible to do this problem a different way to prove it? As that's what my teacher told me to do.

I appreciate all this. :)
 

dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
1) I was just showing you how to get to the general formula of n[3+2(h-1)]+h which is equal to (2h+1)n+h. First you have 3+2(h-1) of matches per column. Then you multiply by the number of columns, "n", so you have n[3+2(h-1)]. Finally, you add a match for every row, so you have n[3+2(h-1)]+h.

You got it. That is exactly what I mean. Columns go up and down though, but by the way those numbers are positioned, I think you just scanned it sideways.

2) If there is only 1 row in the column, then you would have 3+2(1-1) matches per column.

CCCI - 3 matches per column + 1 per row

3) [ ] is the same thing as ( ).

4) You are getting rows and columns confused. Remember, rows go left to right, columns go up and down. There are 3 rows and 4 columns.

row 1 CC CCI
row 2 L L L LI
row 3 L L L LI

5) There might be but I do not know it. :D
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Today my teacher told me definately h was height and n was width....

http://img133.imageshack.us/img133/3003/help8uv.png

Would it be possible to show it like that? :D

Like in the first formua:

3+2(h-1).

I was told by my teacher h should be the height but now I don't have a clue where to go from!

Thanks. :)
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Sorry for d.p. just wanted to bump it up the page.

So how did you get (h-1) that's what's got me confused.

And is n:

<--------->

And is h:


^
¦
¦
¦
\/

The first formula: 3+2(h-1) represents

C
L
L

So if I put 3+2(3-1); how come it comes out with 10 and not 7?

Thanks.
 

dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
h represents the number of rows, so h-1 represents the number of rows beyond the first. There are always 3 matches in the first row of a column so that is a constant. There are 2 more matches for every row beyond the first, so that is represented by 2(h-1). 3+2(h-1) represents the number of matches in a single column.

When n=1, you have 1 column
When n=2, you have 2 columns
When n=3, you have 3 columns...etc.
From that, we can say that n=# of columns

When h=1, you have 1 row
When h=2, you have 2 rows
When h=3, you have 3 rows...etc.
From that, we can say that h=# of rows
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
dagr8est said:
h represents the number of rows, so h-1 represents the number of rows beyond the first. There are always 3 matches in the first row of a column so that is a constant. There are 2 more matches for every row beyond the first, so that is represented by 2(h-1). 3+2(h-1) represents the number of matches in a single column.

When n=1, you have 1 column
When n=2, you have 2 columns
When n=3, you have 3 columns...etc.
From that, we can say that n=# of columns

When h=1, you have 1 row
When h=2, you have 2 rows
When h=3, you have 3 rows...etc.
From that, we can say that h=# of rows
Aha, I think I'm getting it now, so 3+2 is being timsed by (h-1) as that is the number of rows after the first?

Finally:p:

The first formula: 3+2(h-1) represents

C
L
L

So if I put 3+2(3-1); how come it comes out with 10 and not 7?

I'm getting there! :lol:
 

dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
So if I put 3+2(3-1); how come it comes out with 10 and not 7?
Your order of operations is not correct.

3+2(h-1)
=3+2(3-1)
=3+2(2)
=3+4
=7
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
Bump. Come on almost finished! One more question (see above)!:D
 

dagr8est

Junior Member
Joined
Nov 2, 2004
Messages
128
Sorry, I haven't checked this forum in a couple of days. What you have written down looks correct although I hope that's not your final copy. Also you didn't explain why you added "h". Just say you need 1 match for every row to close off the opening at the end, so in general you need "h" additional matches.
 

Monkeyseat

Full Member
Joined
Jul 3, 2005
Messages
298
dagr8est said:
Sorry, I haven't checked this forum in a couple of days. What you have written down looks correct although I hope that's not your final copy. Also you didn't explain why you added "h". Just say you need 1 match for every row to close off the opening at the end, so in general you need "h" additional matches.
Why is it a bit messy?:p Should i just write it up neater?

I did write that kinda under the 2nd diagram.

Thanks for all the help, I think I've finally got my head around it.
 
Top