TedTest

From the Ted Talk by Alex Rosenthal: The Tower of Epiphany | Think Like A Coder, Ep 7

Unscramble the Blue Letters

Ethic and Hedge are on the ground floor of a massive tower. Barriers of energy separate them from their quest’s second goal: the Node of Creation. To reach it, Ethic must use three energy streams to cmilb the tweor. As soon as she steps forward a timer will begin counting down from 60 seconds. At the back of the room there’s a basin made of invisible towers that can hold energy between them. After one minute, a terrnot of energy will pour down from above, flniilg one unit at a time, with a force field preventing it from spliinlg out the fnrot or back. During the 60 calm sdeoncs, Ethic and Hedge must decide exactly how many units of energy will fall. For each of the three celalhnges, they must cohose the amount that will fill the basin exactly. If they do so, the energy will poprel them further upwards. But if they get the amount at all wrong, the energy lift will fail, dropping them. dgmriaas on the wlals illustrate some examples. This configuration will capture exactly 2 units of energy. This configuration will capture 4— 3 here, and 1 here. And this one will also capture 4, because any energy on the right would spill out. The energy will rain down in such a way that it’ll only overflow if there’s no space that could hold it. Hedge can make one tower of blocks vblisie at a time and count how tall it is, but he can’t look at the whole structure all at once. How does Ethic program Hedge to fugrie out exactly how much energy each basin can hold? Pause now to figure it out for yourself. Here’s one way of thinking about what’s happening: each uecponucid cell will hold energy if and only if there is a wall eventually to its left, and a wall eventually to its right. But it would take a long time for Hedge to check this for each individual cell. So what if he were to consider a whole column of blocks at a time? How many untis of energy can this hold, for instance? Pause now to figure it out for yourself. Let’s aalynze the problem by looking at our example. There are 5 cmlnuos of bolkcs here. The leftmost one can’t hold any energy, because there’s nothing higher than it. The 2nd stack can have 3 units above it, as they would be trapped between these two 4 block saktcs. We get 3 units by taking the height where the energy would level off— 4, and subtracting the hhgiet of the stack— so that’s 4 minus 1. The 3rd stack is similar— 4 to the left, 4 to the right, and it’s 3 high, so it’ll hold 4 minus 3 equals 1 unit. The 4th stack and 5th stacks have nothing higher than them to the right, so they can’t hold any egrney. We can adapt this idea into an algorithm. Considering one column at a time as the point of reference, hgdee can look to the left stack by stack to find the height of the tallest one, look to the right to find the height of the tallest one, and take the smaller of the two as the height the energy can fill up to. If the result is higher than the column in question, subtract the height of the original culomn, and the result will be the number of units that column can hold. If it's equal to or below the level of the column in question, the energy would spill off. Hedge can apply that to an entire basin with a loop that starts on the left-most column and moves right, one column at a time. For each column, he’ll run the same steps— look all the way left for the tallest, do the same to the right, take the lower height of the two, satburct the original column height, and increase the grnad total if that number is positive. His loop will repeat as many times as there are columns. That will work, but it’ll take a long time for a large basin. At every step Hedge repeats the action of looking left and looking right. If there are N stacks, he’ll look at all N stacks N times. Is there a faster way? Here’s one time saver: before doing anything else, Hedge can start on the left, and keep a running tally of what the highest stack is. Here that would be 2, 2 again, since the first was higher, then 4, 4, 4. He can then find the highest right-most stacks by doing the same going right-to-left: 1, 3, 4, 4, 4. In the end he’ll have a table like this in his memory. Now, Hedge can take one more pass to calculate how much energy there will be above every sactk with the same equation from before: take the smaller of the stored left and right values, and subtract the height of the current tower. Instead of looking at N stacks N times, he’ll look at N stacks just 3 times— which is what’s called laenir time. There are ways to optimize the solution even further, but this is good enough for our heroes. ehtic and Hedge work as one. The first cascade is a breeze, and they rise up the tower. The second is a little tougher. The third is huge, with dozens of stacks of blocks. The temir ticks down towards zero, but Ethic’s program is fast. She gets the wheel in piiootsn just in time, and the energy lifts them to the Node of Creation. Like the first, it reveals a vision: memories of years gone by. The wlrod machine chagend everything, and Ethic, in her position as chief robotics engineer, grew troubled by what she saw. When the Bradbarrier went up to keep the people in, she knew something was seriously wnorg. So she created three artifacts with the abiitly to rtersoe people’s power, creativity, and memory, and smuggled them to three communities. Before she could tell people how to use them, the gmreveonnt discovered her efforts and sent bots to arrest her and the other programmers. The last thing Ethic used the world machine to create was a robot that would protect the ancient device from the fcreos of ignorance by enclosing it in a giant maze. She nmead her ctoriean Hedge. Without warning, the energy lift felcriks, then fizzles out.

Open Cloze

Ethic and Hedge are on the ground floor of a massive tower. Barriers of energy separate them from their quest’s second goal: the Node of Creation. To reach it, Ethic must use three energy streams to _____ the _____. As soon as she steps forward a timer will begin counting down from 60 seconds. At the back of the room there’s a basin made of invisible towers that can hold energy between them. After one minute, a _______ of energy will pour down from above, _______ one unit at a time, with a force field preventing it from ________ out the _____ or back. During the 60 calm _______, Ethic and Hedge must decide exactly how many units of energy will fall. For each of the three __________, they must ______ the amount that will fill the basin exactly. If they do so, the energy will ______ them further upwards. But if they get the amount at all wrong, the energy lift will fail, dropping them. ________ on the _____ illustrate some examples. This configuration will capture exactly 2 units of energy. This configuration will capture 4— 3 here, and 1 here. And this one will also capture 4, because any energy on the right would spill out. The energy will rain down in such a way that it’ll only overflow if there’s no space that could hold it. Hedge can make one tower of blocks _______ at a time and count how tall it is, but he can’t look at the whole structure all at once. How does Ethic program Hedge to ______ out exactly how much energy each basin can hold? Pause now to figure it out for yourself. Here’s one way of thinking about what’s happening: each __________ cell will hold energy if and only if there is a wall eventually to its left, and a wall eventually to its right. But it would take a long time for Hedge to check this for each individual cell. So what if he were to consider a whole column of blocks at a time? How many _____ of energy can this hold, for instance? Pause now to figure it out for yourself. Let’s _______ the problem by looking at our example. There are 5 _______ of ______ here. The leftmost one can’t hold any energy, because there’s nothing higher than it. The 2nd stack can have 3 units above it, as they would be trapped between these two 4 block ______. We get 3 units by taking the height where the energy would level off— 4, and subtracting the ______ of the stack— so that’s 4 minus 1. The 3rd stack is similar— 4 to the left, 4 to the right, and it’s 3 high, so it’ll hold 4 minus 3 equals 1 unit. The 4th stack and 5th stacks have nothing higher than them to the right, so they can’t hold any ______. We can adapt this idea into an algorithm. Considering one column at a time as the point of reference, _____ can look to the left stack by stack to find the height of the tallest one, look to the right to find the height of the tallest one, and take the smaller of the two as the height the energy can fill up to. If the result is higher than the column in question, subtract the height of the original ______, and the result will be the number of units that column can hold. If it's equal to or below the level of the column in question, the energy would spill off. Hedge can apply that to an entire basin with a loop that starts on the left-most column and moves right, one column at a time. For each column, he’ll run the same steps— look all the way left for the tallest, do the same to the right, take the lower height of the two, ________ the original column height, and increase the _____ total if that number is positive. His loop will repeat as many times as there are columns. That will work, but it’ll take a long time for a large basin. At every step Hedge repeats the action of looking left and looking right. If there are N stacks, he’ll look at all N stacks N times. Is there a faster way? Here’s one time saver: before doing anything else, Hedge can start on the left, and keep a running tally of what the highest stack is. Here that would be 2, 2 again, since the first was higher, then 4, 4, 4. He can then find the highest right-most stacks by doing the same going right-to-left: 1, 3, 4, 4, 4. In the end he’ll have a table like this in his memory. Now, Hedge can take one more pass to calculate how much energy there will be above every _____ with the same equation from before: take the smaller of the stored left and right values, and subtract the height of the current tower. Instead of looking at N stacks N times, he’ll look at N stacks just 3 times— which is what’s called ______ time. There are ways to optimize the solution even further, but this is good enough for our heroes. _____ and Hedge work as one. The first cascade is a breeze, and they rise up the tower. The second is a little tougher. The third is huge, with dozens of stacks of blocks. The _____ ticks down towards zero, but Ethic’s program is fast. She gets the wheel in ________ just in time, and the energy lifts them to the Node of Creation. Like the first, it reveals a vision: memories of years gone by. The _____ machine _______ everything, and Ethic, in her position as chief robotics engineer, grew troubled by what she saw. When the Bradbarrier went up to keep the people in, she knew something was seriously _____. So she created three artifacts with the _______ to _______ people’s power, creativity, and memory, and smuggled them to three communities. Before she could tell people how to use them, the __________ discovered her efforts and sent bots to arrest her and the other programmers. The last thing Ethic used the world machine to create was a robot that would protect the ancient device from the ______ of ignorance by enclosing it in a giant maze. She _____ her ________ Hedge. Without warning, the energy lift ________, then fizzles out.

Solution

changed
forces
analyze
propel
tower
ability
spilling
linear
front
government
height
hedge
grand
wrong
visible
energy
flickers
climb
columns
figure
position
stacks
units
restore
diagrams
world
seconds
filling
timer
ethic
creation
unoccupied
challenges
choose
blocks
column
torrent
stack
walls
subtract
named

Original Text

Ethic and Hedge are on the ground floor of a massive tower. Barriers of energy separate them from their quest’s second goal: the Node of Creation. To reach it, Ethic must use three energy streams to climb the tower. As soon as she steps forward a timer will begin counting down from 60 seconds. At the back of the room there’s a basin made of invisible towers that can hold energy between them. After one minute, a torrent of energy will pour down from above, filling one unit at a time, with a force field preventing it from spilling out the front or back. During the 60 calm seconds, Ethic and Hedge must decide exactly how many units of energy will fall. For each of the three challenges, they must choose the amount that will fill the basin exactly. If they do so, the energy will propel them further upwards. But if they get the amount at all wrong, the energy lift will fail, dropping them. Diagrams on the walls illustrate some examples. This configuration will capture exactly 2 units of energy. This configuration will capture 4— 3 here, and 1 here. And this one will also capture 4, because any energy on the right would spill out. The energy will rain down in such a way that it’ll only overflow if there’s no space that could hold it. Hedge can make one tower of blocks visible at a time and count how tall it is, but he can’t look at the whole structure all at once. How does Ethic program Hedge to figure out exactly how much energy each basin can hold? Pause now to figure it out for yourself. Here’s one way of thinking about what’s happening: each unoccupied cell will hold energy if and only if there is a wall eventually to its left, and a wall eventually to its right. But it would take a long time for Hedge to check this for each individual cell. So what if he were to consider a whole column of blocks at a time? How many units of energy can this hold, for instance? Pause now to figure it out for yourself. Let’s analyze the problem by looking at our example. There are 5 columns of blocks here. The leftmost one can’t hold any energy, because there’s nothing higher than it. The 2nd stack can have 3 units above it, as they would be trapped between these two 4 block stacks. We get 3 units by taking the height where the energy would level off— 4, and subtracting the height of the stack— so that’s 4 minus 1. The 3rd stack is similar— 4 to the left, 4 to the right, and it’s 3 high, so it’ll hold 4 minus 3 equals 1 unit. The 4th stack and 5th stacks have nothing higher than them to the right, so they can’t hold any energy. We can adapt this idea into an algorithm. Considering one column at a time as the point of reference, Hedge can look to the left stack by stack to find the height of the tallest one, look to the right to find the height of the tallest one, and take the smaller of the two as the height the energy can fill up to. If the result is higher than the column in question, subtract the height of the original column, and the result will be the number of units that column can hold. If it's equal to or below the level of the column in question, the energy would spill off. Hedge can apply that to an entire basin with a loop that starts on the left-most column and moves right, one column at a time. For each column, he’ll run the same steps— look all the way left for the tallest, do the same to the right, take the lower height of the two, subtract the original column height, and increase the grand total if that number is positive. His loop will repeat as many times as there are columns. That will work, but it’ll take a long time for a large basin. At every step Hedge repeats the action of looking left and looking right. If there are N stacks, he’ll look at all N stacks N times. Is there a faster way? Here’s one time saver: before doing anything else, Hedge can start on the left, and keep a running tally of what the highest stack is. Here that would be 2, 2 again, since the first was higher, then 4, 4, 4. He can then find the highest right-most stacks by doing the same going right-to-left: 1, 3, 4, 4, 4. In the end he’ll have a table like this in his memory. Now, Hedge can take one more pass to calculate how much energy there will be above every stack with the same equation from before: take the smaller of the stored left and right values, and subtract the height of the current tower. Instead of looking at N stacks N times, he’ll look at N stacks just 3 times— which is what’s called linear time. There are ways to optimize the solution even further, but this is good enough for our heroes. Ethic and Hedge work as one. The first cascade is a breeze, and they rise up the tower. The second is a little tougher. The third is huge, with dozens of stacks of blocks. The timer ticks down towards zero, but Ethic’s program is fast. She gets the wheel in position just in time, and the energy lifts them to the Node of Creation. Like the first, it reveals a vision: memories of years gone by. The world machine changed everything, and Ethic, in her position as chief robotics engineer, grew troubled by what she saw. When the Bradbarrier went up to keep the people in, she knew something was seriously wrong. So she created three artifacts with the ability to restore people’s power, creativity, and memory, and smuggled them to three communities. Before she could tell people how to use them, the government discovered her efforts and sent bots to arrest her and the other programmers. The last thing Ethic used the world machine to create was a robot that would protect the ancient device from the forces of ignorance by enclosing it in a giant maze. She named her creation Hedge. Without warning, the energy lift flickers, then fizzles out.

Frequently Occurring Word Combinations

ngrams of length 2

collocation	frequency
hold energy	2
energy lift	2
wall eventually	2
long time	2
world machine	2

Important Words

ability
action
adapt
algorithm
amount
analyze
ancient
apply
arrest
artifacts
barriers
basin
block
blocks
bots
bradbarrier
breeze
calculate
called
calm
capture
cascade
cell
challenges
changed
check
chief
choose
climb
column
columns
communities
configuration
count
counting
create
created
creation
creativity
current
decide
device
diagrams
discovered
dozens
dropping
efforts
enclosing
energy
engineer
entire
equal
equals
equation
ethic
eventually
examples
fail
fall
fast
faster
field
figure
fill
filling
find
fizzles
flickers
floor
force
forces
front
giant
good
government
grand
grew
ground
hedge
height
heroes
high
higher
highest
hold
huge
idea
ignorance
illustrate
increase
individual
instance
invisible
knew
large
left
leftmost
level
lift
lifts
linear
long
loop
machine
massive
maze
memories
memory
minute
moves
named
node
number
optimize
original
overflow
pass
pause
people
point
position
positive
pour
power
preventing
problem
program
programmers
propel
protect
question
rain
reach
reference
repeat
repeats
restore
result
reveals
rise
robot
robotics
room
run
running
seconds
separate
smaller
smuggled
solution
space
spill
spilling
stack
stacks
start
starts
step
steps
stored
streams
structure
subtract
subtracting
table
tall
tallest
tally
thinking
ticks
time
timer
times
torrent
total
tougher
tower
towers
trapped
troubled
unit
units
unoccupied
values
visible
wall
walls
warning
ways
wheel
work
world
wrong
years