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Please Stop Triple Spacing


DonRocks
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Actually, if anyone on the board sees a triple-spaced post, would you please cut-and-paste the URL of the post, and PM the poster? Thanks.

And I'm sorry if I'm cranky, but I'm working on a problem right now. In fact, I could use some help.

If anyone can get me the answer, I'll take them out to dinner at Maestro.

Anyway, I need f (4, 2) resolved to an integer value for the following function:

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

f (x, y) =

y + 1 ........................ {when x = 0}

f (x-1, 1) ................... {when x > 0 and y = 0}

f (x-1, f (x, y-1)) ........ {when x > 0 and y > 0)

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

It seems pretty simple, so probably won't take too long to resolve. Nevertheless, I can't seem to get the answer, and am willing to pay accordingly to the first person who can get it solved.

First person to PM me the answer gets it.

Thanks!

Rocks

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Actually, if anyone on the board sees a triple-spaced post, would you please cut-and-paste the URL of the post, and PM the poster? Thanks.

And I'm sorry if I'm cranky, but I'm working on a problem right now. In fact, I could use some help.

If anyone can get me the answer, I'll take them out to dinner at Maestro.

Anyway, I need f (4, 2) resolved to an integer value for the following function:

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

f (x, y) =

y + 1 ........................ {when x = 0}

f (x-1, 1) ................... {when x > 0 and y = 0}

f (x-1, f (x, y-1)) ........ {when x > 0 and y > 0)

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

It seems pretty simple, so probably won't take too long to resolve. Nevertheless, I can't seem to get the answer, and am willing to pay accordingly to the first person who can get it solved.

First person to PM me the answer gets it.

Thanks!

Rocks

I was told that there would be no math in this debate ...

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Actually, if anyone on the board sees a triple-spaced post, would you please cut-and-paste the URL of the post, and PM the poster? Thanks.

And I'm sorry if I'm cranky, but I'm working on a problem right now. In fact, I could use some help.

If anyone can get me the answer, I'll take them out to dinner at Maestro.

Anyway, I need f (4, 2) resolved to an integer value for the following function:

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

f (x, y) =

y + 1 ........................ {when x = 0}

f (x-1, 1) ................... {when x > 0 and y = 0}

f (x-1, f (x, y-1)) ........ {when x > 0 and y > 0)

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

It seems pretty simple, so probably won't take too long to resolve. Nevertheless, I can't seem to get the answer, and am willing to pay accordingly to the first person who can get it solved.

First person to PM me the answer gets it.

Thanks!

Rocks

I suck at math. What are the letters doing in there? :lol:

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f(x, y) =

y + 1 ........................ {when x = 0}

f(x - 1, 1) ................... {when x > 0 and y = 0}

f(x - 1, f(x, y - 1)) ........ {when x > 0 and y > 0)

f(0, y) = 1 + y

Using induction to prove f(1, y) = y +2:

Base case:

f(1, 0) = f(0, 1) = 2 = 0 + 2

Assume for particular but arbitrarily chosen (pbac) k that it is true for (k - 1) then show for k:

f(1, k) = f(0, f(1, k - 1)) = 1+ f(1, k - 1) = 1 + [(k - 1) + 2] = k + 2 QED.

Using induction to prove f(2, y) = 2y + 3:

Base case:

f(2, 0) = f(1, 1) = 3

Assume for pbac k that it is true for (k - 1) then show for k:

f(2, k) = f(1, f(2, k - 1)) = 2 + f(2, k - 1) = 2 + 2(k-1) + 3 = 2 + 2k - 2 + 3 = 2k + 3 QED.

Using induction to prove f(3, y) = 23 + y - 3:

Base case:

f(3, 0) = f(2, 1) = 5 = 23 + 0 - 3

Assume for pbac k that it is true for (k - 1) then show for k:

f(3, k) = f(2, f(3, k - 1)) = f(2, 23 + k - 1 - 3)

= 2 * (23 + k - 1 - 3) + 3

= 2 * (2k + 2 - 3) + 3 = 2k + 3 - 6 + 3 = 2k + 3 - 3 QED.

Calculating f(4, 2):

f(4, 0) = f(3, 1) = 13

f(4, 1) = f(3, f(4, 0)) = f(3, 13) = 216 - 3 = 65533

f(4, 2) = f(3, f(4, 1))

= f(3, 65533) = 265536 - 3

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2^65536 - 3

Thanks. Now I'll just need you to PM me the integer (not an equation) and I'll take you to Maestro.

And I'll buy you dessert if you PM me the answer to f (4, 3).

Of course, there are more digits in that answer than there are particles in the known universe. And the only number I know of that's larger than this is THE NUMBER OF POSTS THAT I'VE EDITED IN THE PAST 21 MONTHS THAT HAVE BEEN TRIPLE-SPACED.

Please stop triple-spacing!

And if you're curious, click here to learn more about the imposing Ackermann function.

(Oh, and Jake, I have to say ... pretty impressive.)

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Thanks. Now I'll just need you to PM me the integer (not an equation) and I'll take you to Maestro.
Binary: 265536 is a 1 with 65536 zeroes after, 265536 - 1 is 65536 ones, so 265536 - 3 is 65534 ones, then a zero, then a one.

Hex: 265536 is 1616384 or a 1 with 16384 zeroes after, so 265546 - 3 is 16383 F's, then one D.

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well, vienna sausages in a can.
This one time I was at the 7-11 by Foggy Bottom Metro, and there was a guy standing outside with a bag and a soda. He produced a can of Vienna sausages, popped it open, and proceeded to drink the liquid from the can. That accomplished, he ate the sausages.
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f (x, y) =

y + 1 ........................ {when x = 0}

f (x-1, 1) ................... {when x > 0 and y = 0}

f (x-1, f (x, y-1)) ........ {when x > 0 and y > 0)

f(0, y) = 1 + y

Using strong induction to prove f(1, y) = y +2:

Base case:

f(1, 0) = f (0, 1) = 2 = 0 + 2

f(1, 1) = f (0, f (1, 0)) = f(0, f(0, 1)) = f (0, 2) = 3 = 1 + 2

Assume for all k = 2 to y-1 that it is true then show for y:

f(1, y) = f(0, f(1, y-1)) = 1+ f(1, y-1) = 1 + [(y - 1) + 2] = y + 2 QED.......................f(4, 2) = f(3, f(4, 1)) = f(3, 65533) = 2^65536 - 3

I stand by 11.
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Is that what shut the site down?

I just spent ten dollars on additional bandwidth to get us through January. That's an app. Not an applet; an appetizer. Fucking Parrott has no life, chewed up his employer's CPU all afternoon computing a 19,000-digit number, and cost me an order of mozzarella sticks.

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I just spent ten dollars on additional bandwidth to get us through January. That's an app. Not an applet; an appetizer. Fucking Parrott has no life, chewed up his employer's CPU all afternoon computing a 19,000-digit number, and cost me an order of mozzarella sticks.

Next time I see you, I'll buy you some!

ETA: I was right to be afraid of all those numbers and letters!

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I just spent ten dollars on additional bandwidth to get us through January. That's an app. Not an applet; an appetizer. Fucking Parrott has no life, chewed up his employer's CPU all afternoon computing a 19,000-digit number, and cost me an order of mozzarella sticks.

I presume those are some of Fabio's Maestro mozzarella sticks.

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I just spent ten dollars on additional bandwidth to get us through January. That's an app. Not an applet; an appetizer. Fucking Parrott has no life, chewed up his employer's CPU all afternoon computing a 19,000-digit number, and cost me an order of mozzarella sticks.
It's the old addage, be careful what you ask for. And also, rocks needed to spend $10 to get us through January, which ends tomorrow. I hope they had internet access at whatever bar he was in when he got word that the site was down.

ETA: And I can just see the shit eating grin that must have been on Jake's face when he hit the send button on that PM.

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It's the old addage, be careful what you ask for. And also, rocks needed to spend $10 to get us through January, which ends tomorrow. I hope they had internet access at whatever bar he was in when he got word that the site was down.

It's tomorrow. And the answer to life, the universe, and everything is actually 42.

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f (x, y) =

y + 1 ........................ {when x = 0}

f (x-1, 1) ................... {when x > 0 and y = 0}

f (x-1, f (x, y-1)) ........ {when x > 0 and y > 0)

f(0, y) = 1 + y

Using strong induction to prove f(1, y) = y +2:

Base case:

f(1, 0) = f (0, 1) = 2 = 0 + 2

f(1, 1) = f (0, f (1, 0)) = f(0, f(0, 1)) = f (0, 2) = 3 = 1 + 2

Assume for all k = 2 to y-1 that it is true then show for y:

f(1, y) = f(0, f(1, y-1)) = 1+ f(1, y-1) = 1 + [(y - 1) + 2] = y + 2 QED.

Using strong induction to prove f(2, y) = 2y + 3

Base cases:

f(2, 0) = f(1, 1) = 3

f(2, 1) = f(1, f(2, 0)) = f(2, 0) + 2 = 5 = 2(1) + 3

Assume for all k = 2 to y-1 that it is true then show for y:

f (2, y) = f(1, f(2, y-1)) = 2 + f(2, y-1) = 2 + 2(y-1) + 3 = 2 + 2y - 2 + 3 = 2y + 3 QED.

Using strong induction to prove f(3, y) = 2^(3+y) - 3 [where 2^x denotes 2 raised to the power of x]:

Base case:

f(3, 0) = f (2, 1) = 5 = 2^(3+0) - 3

f(3, 1) = f (2, f(3, 0)) = f (2, 5) = 13 = 2^(3+1)-3

Assume for all k=0 to y-1 that f (3, k) = 2^(3+k) - 3. Now prove for y.

f(3, y) = f(2, f(3, y-1)) = f(2, 2^(3+y-1) - 3) = 2{[2^(3+y-1)] - 3] + 3

= 2* {[2^(y+2)] - 3} + 3 = 2^(y+3) - 6 + 3 = 2^(y+3) - 3 QED.

f(4, 0) = f(3, 1) = 13

f(4, 1) = f(3, f(4, 0)) = f(3, 13) = (2^16)-3 = 65533

f(4, 2) = f(3, f(4, 1)) = f(3, 65533) = 2^65536 - 3

Uhhhh. Que???!!!

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And I can just see the shit eating grin that must have been on Jake's face when he hit the send button on that PM.
Actually, I was quite amused that such a tool existed!
Fucking Parrott has no life
Guilty. Or are you talking about my long-lost second cousin, eleven times removed?
chewed up his employer's CPU all afternoon computing a 19,000-digit number
Actually, it took about a minute on a Dell Latitude D505 laptop.
cost me an order of mozzarella sticks.
I wonder. What would Vincent Feraud propose alongside some nice fritti da mozzarella?
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If this dish is on the menu six months from now at $20, and has the same quantity and quality of ingredients, I'll run down K Street with no clothes on during rush hour.

Cheers,

Rocks.

Go Jake! Enjoy your dinner at Maestro (go sooner rather than later, I would advise.) Aren't you glad he wagered a meal at Maestro, and not the above? BTW, has anybody checked the menu at Blue Duck Tavern recently?
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This whole thing goes well beyond the Algebra 2 level of education I attained in high school.

e^i(pi) = -1

I have a file on my computer of pi to 10 million digits. I like to see how many times my phone number appears in it (twice), and friends' birthdays.

Mmmmmmm.... pie.....

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ETA: There are 19,729 digits in that number. It takes up 6 pages in a Word document.
I thought I had posted that earlier, but it was removed in a fit of pique--the moderator spilled Vietnamese beef stew on his coarsely-knit polo shirt.

Note, of course, that you can calculate the number of digits without writing out the number by taking the base-10 logarithm:

log10265536 = 65536 * log102 ~= 65536 * 0.30103 ~= 19728.3
and rounding up.
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THE NUMBER OF POSTS THAT I'VE EDITED IN THE PAST 21 MONTHS THAT HAVE BEEN TRIPLE-SPACED.

Please stop triple-spacing!

OK, I'm confused. We were PM'd that we were guilty of triple spacing. Are we talking actual triple spaces, or are we talking ellipsis'? I know I'm a big ellipsis user. Is this what is causing all the grief?

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OK, I'm confused. We were PM'd that we were guilty of triple spacing. Are we talking actual triple spaces, or are we talking ellipsis'? I know I'm a big ellipsis user. Is this what is causing all the grief?
Answer: Rocks himself ( :unsure: ) triple-spaced in post #31. It fascinates me that NOBODY, including Rocks his ownself, picked up on this. Thus making this whole subject completely irrelevant.

ETA: WELCOME, RRSMITH, to our merry, if somewhat confused, band. :lol:

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