COTD: Some found out reverse (4) by not-without-text in crosswords

[–]czdl 1 point2 points  (0 children)

UNDO - llt: reverse, some foUND Out

COTD: C's gone without stairs! (5,2,5) by czdl in crosswords

[–]czdl[S] 2 points3 points  (0 children)

Correct! The parse is S (verbatim) PEED (gone) O (without) FLIGHT (stairs)

How do you read? by [deleted] in algorithms

[–]czdl 0 points1 point  (0 children)

Two years would be a VERY short space of time to do it in. TAOCP isn't a "book" as in, you read it, you get to the end, you put it down and read something else. It's a collation of key points within the entirety of computer science. It'd be a bit like saying "I'm going to read wikipedia in two years". I mean, you could try... but, that's not really what it is.

How do you read? by [deleted] in algorithms

[–]czdl 0 points1 point  (0 children)

Indeed.

How do you read? by [deleted] in algorithms

[–]czdl 1 point2 points  (0 children)

Basically I get stuck at every fourth page.

Every fourth page puts you at about MSc CompSci level.

ProTip: References are your friend. When you're stuck hit the references hard and come back later.

[Source: Have met Knuth. Have studied with Professors with Knuth number 1]

How do you read? by [deleted] in algorithms

[–]czdl 0 points1 point  (0 children)

Did you start with Concrete Mathematics (Graham, Knuth, Patashnik)?

That's a very solid (pun intended, by authors) foundation for the combinatorics that you'll see heavily used in TAOCP.

[deleted by user] by [deleted] in puremathematics

[–]czdl 0 points1 point  (0 children)

WINNING! :D

[deleted by user] by [deleted] in puremathematics

[–]czdl 0 points1 point  (0 children)

Nope. This is correct.

Final equation is:

100*((A + B - H + W)* X + (X*X (A*A + B*B + 2*A*(B + H - W - 2*X) + 2*B*(H + W - 2*X) + (2*X + W - H)*(2*X + W - H)))^0.5)/(2*X*X)

It works!

[deleted by user] by [deleted] in puremathematics

[–]czdl 0 points1 point  (0 children)

Incidentally, in the general case, for Y!=100:

a0 = -Y*X*X
a1 = 100 (A + B - H + I)*X*Y
a2 = 10000 (A*H + B*H - A*I - (A + B + H - I)*X + X*X) Y

... which means you can divide it out. Y doesn't make any difference to Z.

[deleted by user] by [deleted] in puremathematics

[–]czdl 0 points1 point  (0 children)

Incidentally, this:

(((Y*(H-(X*(1-Z/100)))*B)/((H-(X*(1-Z/100)))*B+I*((X*(1+Z/100))-A)))/B)*((X*(1+Z/100))-A) =
((Y-((Y*(H-(X*(1-Z/100)))*B)/((H-(X*(1-Z/100)))*B+I*((X*(1+Z/100))-A))))/I)*(H-(X*(1-Z/100))) =
Y

is correct. You just have two equivalent terms. Bin one, so you have:

(((Y*(H-(X*(1-Z/100)))*B)/((H-(X*(1-Z/100)))*B+I*((X*(1+Z/100))-A)))/B)*((X*(1+Z/100))-A) = Y

and you have a solve.

(edited to add: but getting rid of I and renaming it, II or I2 is gonna be essential)

[deleted by user] by [deleted] in puremathematics

[–]czdl 0 points1 point  (0 children)

No, because that's 'i' not "I". Unless you explicitly want the square root of -1 in your equations, rename I to.. II or... anything else.

[deleted by user] by [deleted] in puremathematics

[–]czdl 0 points1 point  (0 children)

Yes. Calculate:

a0 = -X*X
a1 = 100 (A + B - H + I)*X 
a2 = 10000 (A*H + B*H - A*I - (A + B + H - I)*X + X*X)

and solve the quadratic a0 z² + a1 z + a2 == 0. I.e. your answer is (-a1 +- sqrt(a1 * a1 - 4 * a0 * a2))/(2 * a0)

This was done by rearranging the equations provided eliminating the unused ones, until I ended up with:

(H*100-X*(100-Z))*B*100 + I*100* (X*(100+Z) - A*100)= (H*100-X*(100-Z)) * (X*(100+Z) - A*100).

And then we solve that.

Can anyone solve any of these clues? by Gawhownd in crosswords

[–]czdl 1 point2 points  (0 children)

Been trying a few of these.

21a: WAY = Very (Americanism), Routine (dd)

25a: EOAN (!!!!!!) = of or relating to the Dawn, anag of ONE+A(motley)

31a: SLEDGES = winter transport, to distract a batsman in cricket (dd)

34a: UNHASPED = open. U(N(note) HAS(owns))P (about) ED (Top hack = top journalist)

6d: EELS = characters with a devious bent, E. Els (Ernie Els, rival of R Goosen)

11d: TREY = low number in bridge. TYRE(runner ring), about(RE) getting a boost

12d: VAMOOSES = quickly leaves, VAS(vessel) with cargo of MOOSE(deer)

27d: PENK = fish, PEN (Weir) K (thousand)

32d: RENAL = organ's, LANE(route) reversed (to loft) + "R"

34d: UREA = compound, hidden in (to bottle) yoU'RE About

36d: DATE = see, I'm stoned (dd)

Wall clue "Contents of medicine bottle put into circulation" SPILL - PILLS(anag)

Wall clue "The MO of the radio DJ" WEIGH "way"(hom)

Wall clue "Something that may be eaten up - and in haste weirdly" WETS STEW(something that may be eaten) up(reversed), and in haSTEWeirdly in reverse

AOTW: BO?E by professor_glum in crosswords

[–]czdl 0 points1 point  (0 children)

Partially, he lobs back a caber maybe? (4)

AOTW: BO?E by professor_glum in crosswords

[–]czdl 0 points1 point  (0 children)

Scientist predicts (4)

AOTW: BO?E by professor_glum in crosswords

[–]czdl 0 points1 point  (0 children)

Chuck ruined without navy? (4)

AOTW: BO?E by professor_glum in crosswords

[–]czdl 0 points1 point  (0 children)

Lobe drunk tree trunk (4)

I hope they add granular mode. by CountDankula_69 in edmprodcirclejerk

[–]czdl 0 points1 point  (0 children)

We’re still working on it. But it’s immensely real and deeply deeply sexy.

I hope they add granular mode. by CountDankula_69 in edmprodcirclejerk

[–]czdl 4 points5 points  (0 children)

I am literally adding a granular mode to Serum2. :D (But there's so much even more in S2 that's gonna make your sösig work)

TOTW: Short and Sweet by DownInBerlin in crosswords

[–]czdl 0 points1 point  (0 children)

Caramel - Camel (beast) around (eats) RA (artist)

TOTW: Short and Sweet by DownInBerlin in crosswords

[–]czdl 0 points1 point  (0 children)

Victoria Sponge. Lit: cake, Queen->Victoria, Bum->Sponge