Top #breakout dweets

function u(t) {

c.width=1920 // clear the canvas
for(i=0;i<9;i++)
x.fillRect(400+i*100+S(t)*300,400,50,200) // draw 50x200 rects

} // 122/140

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    u(t) is called 60 times per second.
    t: Elapsed time in seconds.
    S: Shorthand for Math.sin.
    C: Shorthand for Math.cos.
    T: Shorthand for Math.tan.
    R: Function that generates rgba-strings, usage ex.: R(255, 255, 255, 0.5)
    c: A 1920x1080 canvas.
    x: A 2D context for that canvas.

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pavel

remix of d/13495 by u/pavel

function u(t) {

for(i=t?c.width|=C=_=>u[j=q+n&-n|p/n]=-!u[j]|1:p=q=r=s=571;i--;u[i]||x.fillRect(i?i%n*n:p+=r*=C(p+=r),i?i&-n:q+=s*=C(q+=s),z=i?n:15,-z))n=64

Apr 11 2020 10:31 PM UTC

}// 140/140

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#breakout

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u/pavel

Fixed the tilted look, the missing upper left block, and increased the size of the ball.
u/pavel

Why does changing 571 break the code for most values? :)
u/pavel

472, 582, 593 work well though
u/iznax

The code seems to work with most starting values if you change it to "C=_=>(p|q)<0?-1:u[j=q+n&-n|p/n]=-!u[j]|1" and handle negative coordinates as blockers?
u/KilledByAPixel

👍
u/pavel

Its shorter to init positions and velocities to same number but makes velocities too fast. Another problem is that init needs to happen once and shortest way to do that is t?<spot>:init which creates a spot to do something every non-first frame. I solve both problems by naming my collision detection function C and defining it on every non-first frame. On the first frame C is Math.cos so the velocity update code does r = r*C(p+=r) = 571*Math.cos(571+571) = 17.7 which is a reasonable speed.
u/pavel

The ball needs to be updated once per frame. Since I am iterating with a single index its easiest to check for index 0. This corresponds to the upper left block which is why it was missing in the original. A short fix is to shift everything up one block by using negative height for the blocks and adding a row to ball position encoding. The +n in q+n&-n|p/n.
u/pavel

I encode and decode 2d co-ordinates into one number. The block coordinates are (i%n,i/n|0) and their width is n making their real positions (i%n*n,(i/n|0)*n). Lets simplify the second one. If we choose n=2**k then (i/n|0)*n = (i/2**k|0)*2**k = (i>>k)<<k so we are zeroing out the last k binary digits. But if n=2**k then -n is a bunch of ones followed by k zeroes (https://en.wikipedia.org/wiki/Two%27s_comple…)! Therefore (i/n|0)*n = i&-n.
u/pavel

Need to check if ball pos (p,q) touches a new block. Ball is always in some block, find that blocks pos index. Dividing ball position by n gives the virtual coords and blocks are 1x1 so truncating gives 2d block coordinates (p/n|0,q/n|0). Want to encode this into index j where (x,y)=(j%n,j/n|0). Therefore j=(q/n|0)*n+(p/n|0). If n=2**k then in binary j=[32-k binary digits y-coord][k binary digits of x-coord] so we can use bitwise-or which also auto-truncates so j=(q/n|0)*n|p/n. And as we saw (q/n|0)*n=q&-n so j=q&-n|p/n! Real code uses j=q+n&-n|p/n to add one row to ball pos to account for shift up to hide missing (0,0)` block.
u/pavel

p+=r*=C(p+=r) encodes an unbelievable trick that is perfect for collision detection. For detection we need to independently try advancing vertically and horizontally, check for collision, and reverse course if hit. The temporary state involved in trying can cost a lot of chars. p+=r*=C(p+=r) does it all and C doesn't even read its argument! When C is invoked the global p it sees was already incremented by r. However, the outer p+= will eventually discard that increment. Magic!
u/rep_movsd

Un-effing-believable
u/hvianna

Wow, this is insane!
u/jylikangas

Masterclass in code golfing!
u/13thptr

awesome, looks like the ball escapes without breaking everything after a while though. will read the explanation more thoroughly.
u/pavel

No chars left for bounds checks! It freaks out with negative positions. If I find a way to shorten I will add bounds.
u/ersagun

I need to read and reread that explanation on a loop until I get this. This is incredible.
u/Qwitter

Just wow.

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KilledByAPixel

function u(t) {

eval(unescape(escape`琿挮睩摴桼㴰㨨䔽㔰ⱘ㵙㴵㌰ⱁ㴷ⱂ㴴ⱄ㵛崩㭆㴨圬娩㴾⡊㵄孇㴨圯䕼〩⭚ⵚ╅崬䑛䝝㴱ⱊ籼ⴱ⤻硛昽❦楬汒散琧崨堫㵁⨽䘨堫䄬天ⱙ⬽䈪㵆⡘ⱙ⭂⤬㤬㤩㭦潲⡩㴰㭩㰲攳㬩硛晝⡩╅⩅Ⱪ⵩╅Ⱑ䑛椫⭝⩅ⱅ⤻`.replace(/u(..)/g,"$1%")))

Apr 26 2019 7:59 PM UTC

}// 139/140

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#Breakout tribute

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u/pavel

Nice!
u/Jezza

This is so satisfying. Brilliant!
u/aleamb

great! relaxing
u/katkip

How
u/KilledByAPixel

See pavel's uncompressed version, it is much cleverer.
u/katkip

#compressed

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pavel

remix of d/13487 by u/KilledByAPixel

function u(t) {

for(i=t?c.width|=d=_=>u[j=q&-64|p>>6]=-!u[j]|1:(p=q=540,r=7,s=4);i--;)u[i]||x.fillRect(i?i%64*64:p+=r*=d(p+=r),i?i:q+=s*=d(q+=s),z=i?64:9,z)

Apr 27 2019 4:26 AM UTC

}// 140/140

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#breakout #tribute

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u/pavel

Had to make some visual sacrifices to get this #uncompressed
u/pavel

I discovered p+=r*=d(p+=r);q+=s*=d(q+=s) as part of messing around with this. Why does it work?
u/pavel

:)
u/KilledByAPixel

Oh wow, nice! That's the code that updates the position, velocity and applies collision for x and y. It needs to call the d function to check its colliding and also set the block to 1 if it does. It needs to check x and y separately so it knows which side to bounce off
u/pavel

Notice how d only takes one argument, and it ignores it.
u/KilledByAPixel

I see what you did there. So many great tricks in this one.
u/pavel

p+=r*=d(p+=r) the inner p+=r increments p and d ends runs with that value, however the outer p+= caches the old value of p so when d resolves the inner p+=r is silently ignored!
u/pavel

#physics

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KilledByAPixel

remix of d/22668 by u/KilledByAPixel

function u(t) {

eval(unescape(escape`𩡯𬠨𪐽𨰮𭱩𩁴𪁼🐰𛁴🱸𛡦𪑬𫁒𩑣𭀨𦀫👁𚠽𡠨𦀫👁𚐬𦐫👂𚠽𡠨𦐫👂𚐬𞐬𞐩𞠨𦠽𜰴𛁅🐶𜀬𦀽𞀶𜀬𦐽𝰰𜀬𠐽𝐬𠠽𜰩𞱩𛐭𞱸𛡦𪑬𫁒𩑣𭀨𪐥𡐪𦠬𚁩𛱅𯀰𚐪𦠬𭀿𘑵𦱩𧐪𦠺𭑛𪑝🐡𚁩𙑅𙡩𛱅𚐬𦠩𚑆👟🐾𭑛𡰽𚁘𛱚𯀰𚐫𚁙𛱚𯀰𚐪𡑝🐭𘑵𦱇𧑼𜐻`.replace(/u../g,'')))

Apr 30 2021 11:00 PM UTC

}// 139/140

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Breakinski #sierpinski #breakout

u/Nascretis

That is really cool, as always!
u/KilledByAPixel

Thank you. ✌️😊

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metamero

function u(t) {

c.width|=onmousemove=e=>X=e.x
t?M.map((e,i)=>e||x.fillRect(i?8+(i-1>>3)*136:X*1920/innerWidth-60,i?8+(i%8)*44:999,120,40)):M=[...Array(113)]

May 14 2019 8:05 PM UTC

}// 140/140

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#breakout #tribute #interactive come to discord chat to draw the ball and test collisions

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