text:
# What is the correct phraseology for changing the runway of a landing clearance?

In the liveATC clip on this YouTube video, an American Airlines 767 arriving at SFO (San Francisco International) had been cleared to land on 28R, but, apparently while already on final, was told to switch to 28L.

The initial clearance given as

American 2293 Heavy, San Francisco Tower, Wind 210 at 18, Caution Wake Turbulence, Heavy 767 mile-and-a-half final, Runway 28 Right, Cleared to Land.

which was read back, as per the usual. However, the Tower later sent the following transmission:

American 2293 Heavy, Change to Cleared to Land Runway 28 Left

which the pilot read back as

28L, Cleared to Land. American 2293 Heavy.

However, the pilot was seemingly too focused on flying the final to notice that the runway assignment had changed between the clearances and continued flying final to 28R, nearly resulting in landing on top of a United jet that was subsequently cleared to line-up-and-wait on 28R. Fortunately, the United pilot noticed the big 767 flying towards him and exited the runway on the other side instead of lining up. Tower then instructed American 2293 to go around.

It seems to me that the phraseology used to change the clearance here is too close to sounding like the controller is just repeating the initial landing clearance, rather than directing a busy pilot's attention to the fact that he was just told to change runways while apparently already flying final.

My question is: Was the phraseology correct for this change of the clearance? If not, what is the proper phraseology?

Given that this was in the U.S., I'm looking for FAA phraseology.

• If they're on final I'd like to think they shouldn't change clearance like that! Would it not be better to make them go around anyway or just keep the runway clear so they could land? Jan 24 '17 at 9:13
• I don't know the phraseology and don't have time to look right now but seems like everyone got it wrong. That amended clearance is nowhere near strong enough. A runway change on final? Poor situational awareness from the crew. How about "cancel clearance" followed by a new clearance and if the crew are unable to do so safely, a go around? Jan 24 '17 at 9:15
• @Notts90 That's my (student pilot, not airline pilot) opinion also, though from a bit of searching around, "side-stepping" seems to be somewhat normal there. Jan 24 '17 at 9:16
• @Simon Yes, my first thought was that "Cancel landing clearance" followed by the new clearance would have been the appropriately-strong phraseology for something as significant as changing an already-issued landing clearance to a different runway, but I wasn't sure what if that was officially 'correct' or not, hence the question. Jan 24 '17 at 9:18
• In my opinion landing clearance cancel and then go-around with lineup for the new runway would have been appropriate. Jan 24 '17 at 10:23

The phraseology was slightly incorrect, this is from the FAA's ATC Orders (section 3-10-5):

3−10−5. LANDING CLEARANCE
a. When issuing a clearance to land, first state the runway number followed by the landing clearance. If the landing runway is changed, controllers must preface the landing clearance with “Change to runway.”
PHRASEOLOGY−
RUNWAY (number) CLEARED TO LAND.
Or
CHANGE TO RUNWAY (number) CLEARED TO LAND

That means the ATC instruction should have been:

American 2293 Heavy, change to runway 28L, cleared to land

All the correct information was in the transmission you quoted, it was just given slightly out of order. But that could easily be an issue for a pilot who's in a high-workload phase of flight and it's probably exactly why the ATC orders say that the words "change to runway" must come first.

I've had ATC change my landing clearance a couple of times but the controller always asked me first if I could do it:

Tower: N12345, cleared to land 21L
N12345: Cleared to land 21L, N12345
[A few seconds later...]
Tower: N12345, are you able to switch to 21R?
N12345: Affirm, we can do that, N12345
Tower: N12345, change to runway 21R, cleared to land

However, landing a C182 at a class D is a bit different from landing a heavy at SFO; I assume the controllers there have less time to talk so they may simply have changed the clearance and expected the pilot to say "unable" if necessary.

• Thanks for the comment on my answer. Its been a while since I had to read up on US R/T. Your answer is now, technically, more "correct" than mine as the OP asked specifically for US procedures. I think i'll leave mine as its a "general" answer.
– Jamiec
Jan 24 '17 at 13:59
• This has happened to me twice. Once the clearance was "cancel landing clearance, now cleared to land runway XX". The other time it was "correction, runway change, cleared to land runway XX". Jan 24 '17 at 18:46

Was the phraseology correct for this change of the clearance?

Probably not. It was possibly missing an important standard phrase, and almost certainly missing the clarity that such a change should include. There was also some implied diligence, on the part of ATC, to ensure that the change in clearance was understood by the pilot.

If not, what is the proper phraseology?

There is a missing Phrase in the ATC transmission, and both the UK CAA CAP 413 and the ICAO Radio telephony Manuals both agree (I will assume the FAA do too, as they tend to follow suit, but for my google-fu I cannot find the official source):

CORRECTION - An error has been made in this transmission (or message indicated). The correct version is...

So, the ATC's instruction should probably have been along the lines of:

American 2293 Heavy, CORRECTION to clearance to land. Cleared to Land Runway 28 Left.

To which the aircraft should have responded

28 Left, Cleared to land, American 2293 Heavy

Should the controller have not received the correct read back of the assigned new runway...

28 Right, Cleared to land, American 2293 Heavy

The controller's job, would have been to correct that mistake immediately

American 2293 NEGATIVE, Cleared to land Runway 28 LEFT. Confirm runway change received.

To which you would hope the pilot got the point that his clearance has changed, and he would read back the corrected landing clearance.

There are 2 other possibilities. My opinion is that the above would be slightly more correct but for answer completeness:

The controller could have cancelled the previous clearance using that standard phrase CANCEL [previous clearance] and reissued it.

American 2293 CANCEL clearance to land Runway 28 Right. Cleared to land Runway 28 Left

To which an accurate readback would have been expected, with the same diligence by ATC to correct any mistake. At least in this case it is clear from the message that the landing clearance has changed.

The other option, and its a real outsider and less standard (as its intention is to reclear for the purpose of routing) is the standard phrase RECLEARED

American 2293 RECLEARED to land Runway 28 Left

Again, this should have made it clear to the pilot that his clearance has changed, and the same readback/check cycle would be initiated. I think this particular example is moving away from standard phrasology, and I only include it for completeness.

As you can see, even standard phraseology is sometimes ambiguous as to the correct one to use. It's why controllers & pilots have some leeway (we're humans communicating at the end of the day). The important part here that was missed was an accurate message/readback/confirmation cycle.

• Does anyone know if the US just follows ICAO for RT phraseology? There is no RT phraseology guide on the faa website that I can find.
– Jamiec
Jan 24 '17 at 12:07
• The main sources in the US are the P/CG and ATC Orders. Other sources like the PHAK have information too, but I'm not aware of a single FAA RT guide; the P/CG is probably the closest thing, and it does highlight differences between US and ICAO phraseology. Jan 24 '17 at 13:30
• This answer provides useful information and I upvoted, but I don't think 'correction' would really have been appropriate here since the initial clearance wasn't read incorrectly (as far as I can tell, at least.) It appears that the initial clearance was intended as given and that it was later changed to a different clearance for some operational reason. Jan 24 '17 at 20:26


text:
# Write the fraction representing the shaded portion.(i)(ii)(iii)(iv)(v)(vi)(vii)(viii)(ix)(x)"

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To do:

We have to write the fractions representing the shaded portion.

Solution:

(i) There are four parts in total and 2 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{2}{4}$

$= \frac{1}{2}$.

(ii) There are nine parts in total and 8 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{8}{9}$

(iii) There are eight parts in total and 4 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{4}{8}$

$=\frac{1}{2}$

(iv) There are four parts in total and 1 part is shaded.

Therefore,

The fraction representing the shaded portion $= \frac{1}{4}$

(v) There are seven parts in total and 3 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{3}{7}$

(vi) There are twelve parts in total and 9 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{9}{12}$

$=\frac{3}{4}$

(vii) There are ten parts in total and all 10 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{10}{10}$

$=1$

(viii) There are nine parts in total and 4 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{4}{9}$ .

(ix) There are eight parts in total and 4 parts are shaded.

Therefore,

The fraction representing the shaded portion $= \frac{4}{8}$

$=\frac{1}{2}$

(x) There are two parts in total and 1 part is shaded.

Therefore,

The fraction representing the shaded portion $=\frac{1}{2}$

Updated on 10-Oct-2022 13:32:37

format: plain


text:
# Number operator in interacting quantum field theory

When treating a quantum field, say the real scalar field, it's totally clear to me how to define a (global) number operator:

$$\hat N = \intop \text d ^3 \mathbf p \hat a ^\dagger (\mathbf p )a(\mathbf p ).$$ This turns out to commute with the hamiltonian and the 3-impulse of the system, therefore the physical interpretation of states with a definite number of particles with a definite total 4-impulse is straightforward. In particular, one could define the vacuum as $\hat N \rvert 0 \rangle =0$.

Now, consider a field interacting with itself, for example:$$\mathscr L =\frac{1}{2}\partial _\mu \phi \partial ^\mu \phi -\frac{1}{2}m^2 \phi ^2 -\frac{\lambda}{4!} \phi ^4.$$

In this case one still talks (in a sense which is not clear to me) of states with a definite number of particles. In a proof, my professor wrote, for a generic state $\alpha$:$$\lvert \alpha \rangle = \lvert \alpha \rangle _0 + \lvert \alpha \rangle _1 + \lvert \alpha \rangle _2 + ...$$where the pedices denote the number of particles in each state of the expansion. Now, this equation implicitly says that there is a certain observable $\hat N$ such that $\hat N \lvert \alpha \rangle _n = n\lvert \alpha \rangle _n$.

Question. Is there a theorem which guarantees the existence of such an operator for every physical field theory? Is it possible to construct explicitly $\hat N$?

• And interacting field theory does not have particle states, and doesn't have a number operator. Are you sure these "interacting particle states" don't like in the asymptotic Fock space instead? Mar 17 '16 at 17:56
• No, it is not always possible to construct the number operator; it depends on which representation of the CCR you're in. The canonical commutation relations for QFT admit uncountably many inequivalent irreducible representations. Technically speaking, given the algebra of the CCR and a state on it, it is possible to define a self-adjoint number operator on its GNS representation only if the state is normal with respect to the Fock representation. Mar 17 '16 at 22:21
• @ACuriousMind , I'm indeed confused about it. In the particular example I'm mentioning, those states are taken to be eigenstates of the full (interacting) 4-impulse, but since they must be also eigenstates of some number operator, from what you and yuggib say, I deduce that they must be linear combinations of Fock space $n$-particle kets. Mar 18 '16 at 11:00
• Anyway, if you or @yuggib want to turn your comment into an answer, I'll be glad to accept it. Mar 18 '16 at 11:00
• @pppqqq The point is that nobody at the moment knows what is the right Hilbert space (i.e. representation of the CCR) for any relativistic interacting QFT in $(3+1)$-dimensions. It may as well be the representation induced by a Fock-normal state, and therefore with a number operator. There are some suggestions that this should not be the case, but it's an open question. Anyways, the answer to your question is "there are representations of the CCR where it is not possible to define a self-adjoint number operator; whether these representations correspond or not to physical theories is not known" Mar 18 '16 at 14:34


text:
1
AIEEE 2006
+4
-1
If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi$$ square units, the equation of the circle is
A
$$\,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,$$
B
$$\,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,$$
C
$${x^2} + {y^2} - 2x\, + 2y - 62 = 0$$
D
$${x^2} + {y^2} - 2x\, + 2y - 47 = 0$$
2
AIEEE 2006
+4
-1
Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\pi } \over 3}$$ at its center is
A
$${x^2} + {y^2} = {3 \over 2}$$
B
$${x^2} + {y^2} = 1$$
C
$${x^2} + {y^2} = {{27} \over 4}$$
D
$${x^2} + {y^2} = {{9} \over 4}$$
3
AIEEE 2005
+4
-1
A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is
A
an ellipse
B
a circle
C
a hyperbola
D
a parabola
4
AIEEE 2005
+4
-1
If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2} = {p^2}$$ orthogonally, then the equation of the locus of its centre is
A
$${x^2}\, + \,{y^2} - \,3ax\, - \,4\,by\,\, + \,({a^2}\, + \,{b^2} - {p^2}) = 0$$
B
$$2ax\, + \,\,2\,by\,\, - \,({a^2}\, - \,{b^2} + {p^2}) = 0$$
C
$${x^2}\, + \,{y^2} - \,2ax\, - \,\,3\,by\,\, + \,({a^2}\, - \,{b^2} - {p^2}) = 0$$
D
$$2ax\, + \,\,2\,by\,\, - \,({a^2}\, + \,{b^2} + {p^2}) = 0$$
JEE Main Subjects
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text:
# TensorFlow Data Input (Part 1): Placeholders, Protobufs & Queues

TensorFlow is a great new deep learning framework provided by the team at Google Brain. It supports the symbolic construction of functions (similar to Theano) to perform some computation, generally a neural network based model. Unlike Theano, TensorFlow supports a number of ways to feed data into your machine learning model. The processes of getting data into a model can be rather annoying, with a lot of glue code. TensorFlow tries to fix this by providing a few ways to feed in data. The easiest of these is to use placeholders, which allow you to manually pass in numpy arrays of data.

The second method, my preferred, is to do as much as I possibly can on the graph and to make use of binary files and input queues. Not only does this lighten the amount of code I need to write, removing the need to do any data augmentation or file reading, but the interface is reasonably standard across different kinds of data. It is also conceptually cleaner. No longer is there an artificial divide between preprocessing and model computation.

### Placeholders

Feed dicts are the simplest and easiest to understand. One copies in the data needed for each batch through placeholders, and TensorFlow does the work of training the model. This is great for small examples and easily interacting with data. Here’s an an example:

import tensorflow as tf

images_batch = tf.placeholder(dtype=tf.float32, shape=[None, 28*28,])
labels_batch = tf.placeholder(dtype=tf.int32, shape=[None, ])

# simple model
w = tf.get_variable("w1", [28*28, 10])
y_pred = tf.matmul(images_batch, w)
loss = tf.nn.sparse_softmax_cross_entropy_with_logits(y_pred, labels_batch)
loss_mean = tf.reduce_mean(loss)

sess = tf.Session()
init = tf.initialize_all_variables()
sess.run(init)

from skdata.mnist.views import OfficialVectorClassification
import numpy as np

# load data entirely into memory
data = OfficialVectorClassification()
trIdx = data.sel_idxs[:]
features = data.all_vectors[trIdx]
labels = data.all_labels[trIdx]

def data_iterator():
""" A simple data iterator """
batch_idx = 0
while True:
# shuffle labels and features
idxs = np.arange(0, len(features))
np.random.shuffle(idxs)
shuf_features = features[idxs]
shuf_labels = labels[idxs]
batch_size = 128
for batch_idx in range(0, len(features), batch_size):
images_batch = shuf_features[batch_idx:batch_idx+batch_size] / 255.
images_batch = images_batch.astype("float32")
labels_batch = shuf_labels[batch_idx:batch_idx+batch_size]
yield images_batch, labels_batch

iter_ = data_iterator()
while True:
# get a batch of data
images_batch_val, labels_batch_val = iter_.next()
# pass it in as through feed_dict
_, loss_val = sess.run([train_op, loss_mean], feed_dict={
images_batch:images_batch_val,
labels_batch:labels_batch_val
})
print loss_val

This is all fine and good for small toy problems like the one above. As tasks become more and more complicated, the problem of managing your own data becomes harder and harder. Take images, for instance. You need to have consistent image loading, and a scaling pipeline that is consistent at train and test time. In addition, you need to ensure that the input is fast enough – possibly multi-threaded.

TensorFlow makes use of a few concepts, such as protobuf and queues, to make this easier – but at the cost of a little bit more upfront learning.

### Protobuf and binary formats

I used to hate binary files. To me they were opaque, hard to build, and annoying to work with. After a little bit of getting used to, they are actually quite nice and provide benefits that raw data could never hope to provide. They make better use of disk cache, are faster to copy and move around, don’t need to keep separate files to denote labels or other information, don’t have to fight with folders of millions of images messing with unix tools…the list goes on. In addition, these files will always be read off of disk in a standardized way and never all at once. In other words, your dataset size is only bounded by hard drive space. After working binary files a little bit (both in MXNet and TensorFlow), I am surprised I was able to make do without them. I would strongly recommend giving them a shot in your next project.

Much like many other deep learning frameworks, TensorFlow has its own binary format. Its format uses a mixture of its Records format and and Protocol Buffers to solve this problem.
If you’re unfamiliar with Protobuf, you can think about it as a way to serialize data structures, given some schema describing what the data is.

TFRecords are TensorFlow’s default data format. A record is simply a binary file that contains serialized tf.train.Example Protobuf objects, and can be created from Python in a few lines of code. Below is an example to convert mnist to this format.

# load up some dataset. Could be anything but skdata is convenient.
from skdata.mnist.views import OfficialVectorClassification
from tqdm import tqdm
import numpy as np
import tensorflow as tf

data = OfficialVectorClassification()
trIdx = data.sel_idxs[:]

# one MUST randomly shuffle data before putting it into one of these
# formats. Without this, one cannot make use of tensorflow's great
# out of core shuffling.
np.random.shuffle(trIdx)

writer = tf.python_io.TFRecordWriter("mnist.tfrecords")
# iterate over each example
# wrap with tqdm for a progress bar
for example_idx in tqdm(trIdx):
features = data.all_vectors[example_idx]
label = data.all_labels[example_idx]

# construct the Example proto boject
example = tf.train.Example(
# Example contains a Features proto object
features=tf.train.Features(
# Features contains a map of string to Feature proto objects
feature={
# A Feature contains one of either a int64_list,
# float_list, or bytes_list
'label': tf.train.Feature(
int64_list=tf.train.Int64List(value=[label])),
'image': tf.train.Feature(
int64_list=tf.train.Int64List(value=features.astype("int64"))),
}))
# use the proto object to serialize the example to a string
serialized = example.SerializeToString()
# write the serialized object to disk
writer.write(serialized)

To best understand these classes, I would recommend looking at the proto source: example.proto and feature.proto.

Basically, an Example always contains Features.  Features contains a map of strings to Feature. And finally, a Feature contains one of a FloatList, a ByteList or a Int64List.

With this record, all needed information is stored in one place and is always aligned with all data needed to describe the example.

The data in these records can be read back sequentially in the same order they were written.

import tensorflow as tf

filename = "mnist.tfrecords"
for serialized_example in tf.python_io.tf_record_iterator(filename):
example = tf.train.Example()
example.ParseFromString(serialized_example)

# traverse the Example format to get data
image = example.features.feature['image'].int64_list.value
label = example.features.feature['label'].int64_list.value[0]
# do something
print image, label

Once your data is in this format, you can make use of many tools TensorFlow provides to feed your machine learning model.

import tensorflow as tf

# first construct a queue containing a list of filenames.
# this lets a user split up there dataset in multiple files to keep
# size down
filename_queue = tf.train.string_input_producer([filename],
num_epochs=None)
# Unlike the TFRecordWriter, the TFRecordReader is symbolic
# One can read a single serialized example from a filename
# serialized_example is a Tensor of type string.
# The serialized example is converted back to actual values.
# One needs to describe the format of the objects to be returned
features = tf.parse_single_example(
serialized_example,
features={
# We know the length of both fields. If not the
# tf.VarLenFeature could be used
'label': tf.FixedLenFeature([], tf.int64),
'image': tf.FixedLenFeature([784], tf.int64)
})
# now return the converted data
label = features['label']
image = features['image']
return label, image

Now to use this function we can do something like the following:

# returns symbolic label and image

sess = tf.Session()

# Required. See below for explanation
init = tf.initialize_all_variables()
sess.run(init)
tf.train.start_queue_runners(sess=sess)

# grab examples back.
# first example from file
label_val_1, image_val_1 = sess.run([label, image])
# second example from file
label_val_2, image_val_2 = sess.run([label, image])

The fact that this works requires a fair bit of effort behind the scenes. First, it is important to remember that TensorFlow’s graphs contain state. It is this state that allows the TFRecordReader to remember the location of the tfrecord it’s reading and always return the next one. This is why for almost all TensorFlow work we need to initialize the graph. We can use the helper function tf.initialize_all_variables(), which constructs an op that initializes the state on the graph when you run it.

The next major piece involves Tensorflow’s Queues and QueueRunners. A TensorFlow queue is quite similar to a regular queue, except all of its operations are symbolic and only performed on Tensorflow’s graph when needed by a sess.run. See here for a comprehensive write up of how these things work and a great animation for queues.

As part of the API, a TFRecordReader always acts on a queue of filenames. It will pop a filename off the queue and use that filename until the tfrecord is empty. At this point it will grab the next filename off the filename queue.

To summarize thus far, we grab a filename off our queue of filenames and use it to get examples from a TFRecordReader. Both the queue and the TFRecordReader have some state to keep track of where they are. The only piece missing is how the filename queue is actually fed. On initialization it is empty. This is where the concept of QueueRunners comes in. There’s nothing magical about it. At its core, it is simply a thread that uses a session and calls an enqueue op over and over again. TensorFlow has noticed this pattern and wrapped it in its tf.train.QueueRunner object. 99% of the time this can be ignored as it is used behind the scenes, (such as in the above example – tf.train.string_input_producer creates one).

We do, however, have to signal to TensorFlow to start these threads. Without this, the queue will be blocked indefinitely, waiting for data to be enqueued. To start the QueueRunners you can call tf.train.start_queue_runners(sess=sess). This call is not symbolic. It goes and creates threads. It is important to note that this must be called after the initialization op is run. These threads try to enqueue data to queues. If the queues are not initialized, TensorFlow will rightly throw an error.

Great! Now we have two Tensors that represents a single example. We could train with this using gradient descent, but that is rather silly. It has been shown that training with batches of examples work far better than training with a single examples. Your estimates of gradients will have a lower variance which makes training faster. Luckily TensorFlow provides utilities to batch these examples and return batches.

# get single examples
# groups examples into batches randomly
images_batch, labels_batch = tf.train.shuffle_batch(
[image, label], batch_size=128,
capacity=2000,
min_after_dequeue=1000)

sess = tf.Session()
init = tf.initialize_all_variables()
sess.run(init)
tf.train.start_queue_runners(sess=sess)
labels, images= sess.run([labels_batch, images_batch])

At first glance, this seems like quite a natural API – but the implementation is rather subtle. As discussed above, if label and image represent a single example, how can we make a batch of different examples? The answer lies in another set of queues and QueueRunners. shuffle_batch constructs a RandomShuffleQueue and proceeds to fill it with individual image and labels. This filling is done on a separate thread with a QueueRunner. The RandomShuffleQueue accumulates examples sequentially until it contains batch_size +min_after_dequeue examples are present. It then selects  batch_size random elements from the queue to return. The value actually returned by shuffle_batch is the result of a dequeue_many call on the RandomShuffleQueue.

With these batch variables, you can now go on your merry way and train a model.

# get single examples
image = tf.cast(image, tf.float32) / 255.
# groups examples into batches randomly
images_batch, labels_batch = tf.train.shuffle_batch(
[image, label], batch_size=128,
capacity=2000,
min_after_dequeue=1000)

# simple model
w = tf.get_variable("w1", [28*28, 10])
y_pred = tf.matmul(images_batch, w)
loss = tf.nn.sparse_softmax_cross_entropy_with_logits(y_pred, labels_batch)

# for monitoring
loss_mean = tf.reduce_mean(loss)

sess = tf.Session()
init = tf.initialize_all_variables()
sess.run(init)
tf.train.start_queue_runners(sess=sess)

while True:
# pass it in through the feed_dict
_, loss_val = sess.run([train_op, loss_mean])
print loss_val

### To Conclude

In summary of the whole process, you need to:

1. Define a record reader function of some kind. This parses the record.
2. Construct a batcher of some kind.
3. Build the rest of your model from these ops.
4. Initialize all the ops.
5. Start queue runners.

You should now have several options at your disposal for feeding data into your machine learning model in a TensorFlow framework. As you may or may not have noticed, the example using queues while conceptually clean, is painfully slow. TensorFlow is constantly improving and hopefully this workflow speeds up. In Part 2, we discuss a hybrid approach to these methods that allows for the niceties of queues while being as fast as the placeholder approach, as well as some extensions to the demo we looked at here. As always, feel free to reach out to us with questions and thoughts at contact@indico.io!

### If something goes wrong

Random non deterministic / changing errors about initialization and or node dependencies:
The error always changes but for me this signals I did not initialize all the variables in the session before trying to use them. This could be due to calling tf.train.start_queue_runners too early.

System hangs / doesn’t read data:
This is most likely due to not calling tf.train.start_queue_runners. It is possible to construct some funky connections between queues that might also cause deadlocks as well.

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#### data augmentation

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#### Tensorflow 物体检测(object detection) 之如何优雅地预处理数据

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Let T denote the group of all nonsingular upper triaungular entries, i.e., the matrices of the form, [a,0,b,c] where a,b,c ∈ H H={[1,0,x,1]∈T} is a normal subgroup of T.

Question
Matrix transformations
Let T denote the group of all nonsingular upper triaungular entries, i.e., the matrices of the form, [a,0,b,c] where $$\displaystyle{a},{b},{c}∈{H}$$
$$\displaystyle{H}={\left\lbrace{\left[{1},{0},{x},{1}\right]}∈{T}\right\rbrace}$$ is a normal subgroup of T.

2021-01-16
$$\displaystyle{\left[{a},{0},{b},{c}\right]}^{{-{{1}}}}{\left[{1},{0},{x},{1}\right]}{\left[{a},{0},{b},{c}\right]}={\left[{1},{0},{c}\frac{{r}}{{a}},{1}\right]}∈{H}$$

Relevant Questions

Consider the curves in the first quadrant that have equationsy=Aexp(7x), where A is a positive constant. Different valuesof A give different curves. The curves form a family,F. Let P=(6,6). Let C be the number of the family Fthat goes through P.
A. Let y=f(x) be the equation of C. Find f(x).
B. Find the slope at P of the tangent to C.
C. A curve D is a perpendicular to C at P. What is the slope of thetangent to D at the point P?
D. Give a formula g(y) for the slope at (x,y) of the member of Fthat goes through (x,y). The formula should not involve A orx.
E. A curve which at each of its points is perpendicular to themember of the family F that goes through that point is called anorthogonal trajectory of F. Each orthogonal trajectory to Fsatisfies the differential equation dy/dx = -1/g(y), where g(y) isthe answer to part D.
Find a function of h(y) such that x=h(y) is the equation of theorthogonal trajectory to F that passes through the point P.
Let D be the diagonal subset $$\displaystyle{D}={\left\lbrace{\left({x},{x}\right)}{\mid}{x}∈{S}_{{3}}\right\rbrace}$$ of the direct product S_3 × S_3. Prove that D is a subgroup of S_3 × S_3 but not a normal subgroup.
We will now add support for register-memory ALU operations to the classic five-stage RISC pipeline. To offset this increase in complexity, all memory addressing will be restricted to register indirect (i.e., all addresses are simply a value held in a register; no offset or displacement may be added to the register value). For example, the register-memory instruction add x4, x5, (x1) means add the contents of register x5 to the contents of the memory location with address equal to the value in register x1 and put the sum in register x4. Register-register ALU operations are unchanged. The following items apply to the integer RISC pipeline:
a. List a rearranged order of the five traditional stages of the RISC pipeline that will support register-memory operations implemented exclusively by register indirect addressing.
b. Describe what new forwarding paths are needed for the rearranged pipeline by stating the source, destination, and information transferred on each needed new path.
c. For the reordered stages of the RISC pipeline, what new data hazards are created by this addressing mode? Give an instruction sequence illustrating each new hazard.
d. List all of the ways that the RISC pipeline with register-memory ALU operations can have a different instruction count for a given program than the original RISC pipeline. Give a pair of specific instruction sequences, one for the original pipeline and one for the rearranged pipeline, to illustrate each way.
Hint for (d): Give a pair of instruction sequences where the RISC pipeline has “more” instructions than the reg-mem architecture. Also give a pair of instruction sequences where the RISC pipeline has “fewer” instructions than the reg-mem architecture.
Proove that the set of oil 2x2 matrices with entries from R and determinant +1 is a group under multiplication
The bulk density of soil is defined as the mass of dry solidsper unit bulk volume. A high bulk density implies a compact soilwith few pores. Bulk density is an important factor in influencing root development, seedling emergence, and aeration. Let X denotethe bulk density of Pima clay loam. Studies show that X is normally distributed with $$\displaystyle\mu={1.5}$$ and $$\displaystyle\sigma={0.2}\frac{{g}}{{c}}{m}^{{3}}$$.
(a) What is thedensity for X? Sketch a graph of the density function. Indicate onthis graph the probability that X lies between 1.1 and 1.9. Findthis probability.
(b) Find the probability that arandomly selected sample of Pima clay loam will have bulk densityless than $$\displaystyle{0.9}\frac{{g}}{{c}}{m}^{{3}}$$.
(c) Would you be surprised if a randomly selected sample of this type of soil has a bulkdensity in excess of $$\displaystyle{2.0}\frac{{g}}{{c}}{m}^{{3}}$$? Explain, based on theprobability of this occurring.
(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
(e) What is the moment generating function for X?
The graph of y = f(x) contains the point (0,2), $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{-{x}}}{{{y}{e}^{{{x}^{{2}}}}}}}$$, and f(x) is greater than 0 for all x, then f(x)=
A) $$\displaystyle{3}+{e}^{{-{x}^{{2}}}}$$
B) $$\displaystyle\sqrt{{{3}}}+{e}^{{-{x}}}$$
C) $$\displaystyle{1}+{e}^{{-{x}}}$$
D) $$\displaystyle\sqrt{{{3}+{e}^{{-{x}^{{2}}}}}}$$
E) $$\displaystyle\sqrt{{{3}+{e}^{{{x}^{{2}}}}}}$$
In the figure below, the rolling axle, 1.43 m long, is pushed along horizontal rails at a constant speed v = 3.36 m/s.

A resistor R = 0.325 ohm is connected to the rails at points a and b, directly opposite each other. (The wheels make good electrical contact with the rails, and so the axle, rails, and R form a closed-loop circuit. The only significant resistance in the circuit is R.) There is a uniform magnetic field B = 0.0850 T vertically downward. Calculate the induced current I in the resistor and what horizontal force F is required to keep the axle rolling at constant speed?
Show that $$\displaystyle{C}^{\ast}={R}^{\ast}+\times{T},\text{where}\ {C}^{\ast}$$ is the multiplicative group of non-zero complex numbers, T is the group of complex numbers of modulus equal to 1, $$\displaystyle{R}^{\ast}+$$ is the multiplicative group of positive real numbers.


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February 2, 2023

# Intelligent career planning via stochastic subsampling reinforcement learning

This section discusses the reward function in SSRL and provides theoretical details of the search algorithm for the optimal career path. We summarize important notations and their definitions in the table shown in the supplementary materials.

### Reward function

Now, we discuss the formulation of our reward function with three major components including company rating, periodic extent of suffering, and staying probability.

Company Rating. The company rating is formulated as a linear combination of company feature-based score and position feature-based score.

First, we introduce the company feature-based score, which is used to evaluate the overall “quality” of a company. The following four factors are considered.

1. (1)

Reputation: The reputation of a company is considered a representative factor of its business value and social impression, hence will significantly affect the job stability of employees. By investigating a sample containing 593 large publicly traded companies from 32 countries, Soleimani et al.45 found a positive impact of company reputation on its stock performance and the employee salary. Similar findings can also be found in46. It has been found that people are more willing to work in companies with high reputation47. In this paper, we quantify three levels of company reputation. For the first (highest) level, Fortune-500 companies are rated with the highest reputation, and their reputation score is set to 1. Non Fortune-500 companies with more than 10,000 employees are placed to the second level, with a reputation score (frac23). For the rest of the companies, we assign (frac13) as the reputation score. We also conducted experiments to evaluate the stability of the overall performance based on linear reputation. Related results are included in the supplementary materials (Fig. S.2).

2. (2)

Popularity: Popularity represents the overall social impact of a company. Based on experiments on an artificial market, Salganik et al.48 suggested that popularity is positively related to the quality of a company, and people are more willing to work in popular companies. Company popularity can be easily quantified based on talent move record. The frequency of talent incoming transfer indicates the popularity of a company. In our work, we normalize the total number of incoming transferring records to values in [0, 1], by dividing each by the maximum of the records.

3. (3)

Average Staying Duration: Employee stability is essential to business success49, and the average staying duration of a company also represents the overall job satisfaction of its employees50. We compute the average staying duration of employees for each company and then normalize them to [0, 1] based on the maximum value.

4. (4)

Smooth Transfer Rate: The smooth transfer rate measures how likely a job-hopping can be made. Given the dynamic market, a smooth job hopping indicates less risk hence is preferred by most job seekers6. Considering the labor market shifts toward information and knowledge based work, talented workers are the intangible asset to a company37. From the perspective of employers, to keep the competitive advantage, companies should accept suitable employees as soon as possible. To measure the smooth transfer rate, we have the following settings. For those companies with more than 100 transfer records, we calculate it as the ratio of the number of transfer records without waiting time (no time gap between the old and the new jobs) over the total amount of job transfers. For companies with less than 100 transfer records, which occupy only 3% of the full sample, we introduce a penalty term (0.8) to weaken the smooth transfer rate, as the number may be overestimated due to a small sample size.

Importantly, considering the fast changing labor market, company ratings may vary over time5,6. The requirement from users is also changing over time. Assuming that (t’) denotes the time people start to work at the company (C_i), based on the above four features indicated by (f_l) where l is the feature index from 1 to 4, the company feature-based score of company (C_i) is defined as:

beginaligned N_C_i_CF^t’=100cdot fracsum _l=1^4 theta _l^t’ f_l,C_i^t’-min _j in mathscrC sum _l=1^4 theta _l^t’ f_l,j^t’max _j in mathscrC sum _l=1^4 theta _l^t’ f_l,j^t’-min _j in mathscrC sum _l=1^4 theta _l^t’ f_l,j^t’, endaligned

(8)

where (mathscrC) denotes the company list and (theta _l) denotes the weight of features. Users can express his/her preference to each company feature-based by specifying (theta _l). Note that (N_C_i_CF^t’) ranges from 0 to 100 based on the formulation.

On the other hand, we estimate the position feature-based score as follows. The job position contributes to one’s company rating in terms of the person-job fit, which is found positively correlated to job satisfaction51. From the perspective of employers, job seekers are also encouraged to be matched to a background-fitting position37. The position for the next company is related to the employee’s current experience. In this paper, the new job position is predicted by a job predictor based on the records in our dataset. The job predictor is determined by counting all the position transfer records for the current position and normalizing the top three most frequently selected positions. If the job position is the same as before, then people will get full credit for this position. If the position type is changed, we further evaluate the new work environment (i.e., whether employees get enough team support during the learning/training period).

As expert is instrumental when people face new difficulties in the team52 and the support from team is vital for overcoming the difficulty53, we assume that if people can get enough support from expert team members, the new job is desirable even if it is a new position type. The number of experts in the given position should be positively related to the number of the positions in the company. Thus, to evaluate if people can obtain enough support from the team, we measure if there are sufficient number of co-workers at the same job position in the company. If the job is among the top-3 major position types in the company, we assume that people are more likely to receive sufficient support; otherwise, insufficient support is assumed. To differentiate the above cases quantitatively, we have the following settings. Assuming that the previous position is (J_i-1) and the current position is (J_i), the position feature-based score for company (C_i) can be defined as follows:

beginaligned N_C_i_PF= {left beginarrayll 100, & J_i=J_i-1\ 80, & J_i ne J_i-1; J_i text is among the top-3 position types in C_i \ 60, & text otherwise endarrayright. endaligned

(9)

Please note that this is a simplified formulation we use to assess the overall person-job fit. More complicated formulation can be adopted in real cases.

Given company feature-based score (N_C_i_CF^t’) and position feature-based score (N_C_i_PF), we define the company rating of (C_i) as their linear combination:

beginaligned N_C_i=beta _1cdot left( s_1cdot N_C_i_CF^t’+s_2cdot N_C_i_PFright) , endaligned

(10)

where (s_1) and (s_2) are the weights of the two feature scores with (s_1 + s_2=1). Here (beta _1) is used to describe a negative effect of downtrend career moves on the company rate. People usually seek opportunities to work in better companies with improved work environment and potentials54. It has also been found that people would have a decreased job performance with their new employers if the work environment did not improve37. Thus, we have the following settings: (beta _1<1) if (N_C_i le N_C_i-1); (beta _1 = 1) if (N_C_i > N_C_i-1).

Periodic Extent of Suffering. Human capital transferring is not easy, and sometimes will lead to a job performance reduction37. We refer to this as the suffering period after job hopping. According to a survey presented by Morse and Weiss55, people are more willing to work in the same type of companies as their current one when seeking the next job. Evidence was also found that firm-specific skills plays an important role in employees’ performance37. It is common that companies in the same business sector have similar positions (e.g., JP Morgan Chase and Bank of America). Thus, we estimate the extent of suffering by the similarity of the current and potentially new companies. A higher level of similarity is supposed to result in less suffering in the new company.

Let j be the order index of a given position type list, the percentage of the (j^th) position type in company (C_i) is computed as:

beginaligned Per(C_i,j)=fracbig M_C_i, endaligned

(11)

where (PT(C_i,k)) is the position type of the (k^th) position in company (C_i); (M_C_i) is the total number of position records in company (C_i). Then, we compute the cosine similarity between companies (C_i) and (C_i-1) as follows:

beginaligned sim(C_i,C_i-1)=fracsum _j=1^n_p left[ Perleft( C_i,jright) Perleft( C_i-1,jright) right] sqrtsum _j=1^n_p [Perleft( C_i,jright) ]^2sqrtsum _j=1^n_p [Perleft( C_i-1,jright) ]^2, endaligned

(12)

where the total number of position types (n_p = 26) in our data.

Estimator for Staying Probability and Duration. The staying probability not only helps estimate the duration, but also serves as an important component in our reward function. The following equation offers a simple way to estimate the staying probability for a given company:

beginaligned Pr(t)=1-fracd(t)n(t), endaligned

(13)

where d(t) denotes the number of people left the company before time t; and n(t) represents the total number of employees in the company and (n(t) ge d(t)).

However, such estimation may be inaccurate due to data noise and incomplete records from existing employees. For a current employee in a company, we can only define his/her future situation as unknown or uncertain. To obtain a more reliable staying duration, we estimate the staying probability for such samples via survival analysis. Our problem can be considered as a right-censoring condition, given that only the starting time of each job is known. Thus, we apply the Kaplan–Meier (KM) estimator from survival analysis. Specifically, we define staying at a company as “survive”, and leaving the company as “die”. Let d(x) denotes the number of “survivers” from ([x, x+Delta x)) and n(x) denotes the number of individuals at risk just prior to a given time x; i.e., the number of individuals in the sample who neither left nor were censored prior to time x. Then, the KM estimator for the staying probability can be written as:

beginaligned Pr(t)=KM(t)= prod _x le t left( 1-fracd(x)n(x)right) , endaligned

(14)

where (x le t) represents all grid points such that (x+Delta x le t). See technical details of KM estimator in reference56.

The staying probability can be used to estimate the staying duration of employees. It has been found in the literature that people tend to stay in similar type of companies in their career life57. As the estimated staying probability represents a mainstream pattern, we have the following design to differentiate mainstream followers from others. Given a current company, we define the mainstream selection for the next employer as the top-100 most selected companies in the data. Also, if a person chooses her new job from the mainstream list, the remaining companies on the list continue as the mainstream choices for the next job-hopping. Suppose a person is working at her (i^th) company, and (Psi (C_i)) is the set of top-100 job-hopping selections for people who worked at company (C_i). The mainstream choices for the ((i+1)^th) job-hopping is defined as (Psi (C_i) cup Psi (C_i-1) cup Psi (C_i-2) cdots cup Psi (C_0)). Importantly, this design enables us to partially simulate the age and learning experience effects, which have been found to be important factors of career mobility58. In the short-term view, we assume that there will be more useful experience gain for the mainstream followers. Off the mainstream may indicate lower survival rate due to irrelevant experience. On the other hand, in the long-term view, if people try different types of companies/positions and received “panelties” in their early career stage, there will be a broader range of “mainstream” companies that they can survive in the later career stages.

Therefore, one’s staying duration at company (C_i) can be estimated according to (beta _2 Pr(C_i, t)), where (beta _2) is a penalty term, which is set to 1 if (C_i in Psi (C_i) cup Psi (C_i-1) cup Psi (C_i-2) cdots cup Psi (C_0)), or 0.8 otherwise. The staying duration can be easily determined when the model concludes the “leaving state” for a current company (C_i) based on (beta _2 Pr(C_i, t)) at time t.

Reward Formulation: The Final Form. To find the optimal path (P^*), we need to define a proper reward function R to guide the exploration and determine the optimal path. We define the reward as the accumulative score considering the company rating, the periodic extent of suffering, and the staying probability. Thus, the total reward for staying at company (C_i) is defined as:

beginaligned S_C_i= & int _0^t_1 N_C_icdot Pr(C_i, t) left( sim(C_i,C_i-1)-1right) dtnonumber \&+int _0^D_C_i N_C_icdot Pr(C_i, t) dt, endaligned

(15)

where (t_1) is the suffering period; (N_C_i) is the company rating; (D_C_i) means the duration at company (C_i); (Pr(C_i,t)) denotes the staying probability for company at time t; and (sim(C_i,C_i-1)) is the similarity between two companies.

Equation (15) evaluates the reward for staying at company (C_i). The first component (int _0^t_1 N_C_icdot Pr(C_i, t) left( sim(C_i,C_i-1)-1right) dt) measures the weakening effect during the suffering period at a new company. The second component (int _0^D_C_i N_C_icdot Pr(C_i, t) dt) estimates the accumulative reward with decayed staying probability. To ease the computing, we divide the total time into small intervals (Delta t), and then Eq. (15) can be reformulated as:

beginaligned S_C_i&= Rleft( theta _l f_l,C_i-1, J_i-1, C_i, J_i, D_iright)\ &= N_C_ileft( left( sim(C_i, C_i-1)-1right) sum _k=1^t_1/Delta t Pr(C_i,k Delta t) +sum _k=1^D_C_i/Delta t Pr(C_i,k Delta t)right) Delta t endaligned

(16)

### Theoretical analysis

In the following, we provide further analysis on our SSRL framework, focusing on the properties of the proposed CSR problem, our exploration strategy, the speedup of convergence, and their theoretical supports.

Fundamental Property. Given the starting state (C_0), assuming that we have found the optimal path (P^*) generated by the optimal policy (pi ^*), then we should have the following property.

### Property 1

(Upper Bound of Policies) Given finite career time, if (P^*) is the optimal career path starting with (C_0), we should have (sum _j in P_i S_j le sum _j in P^* S_j), for any path (P_i) with the same initial state and career time length.

All proofs can be found in the supplementary information. According to our problem setting, the unpredictable size of the company set (mathscrC) may result in huge computing challenges, while our solution is to achieve the global optimum with fixed-sized subsets (mathscrC_sub), leading to the following lemma as a special case of Property 1.

### Lemma 1

(Upper Bound of Local Policy) Given an initial company (C_0), for any optimal paths generated by (pi ^*_mathscrC_sub) the optimal policy on (mathscrC_sub subseteq mathscrC), their accumulative rewards cannot exceed that of (P^*) generated by the optimal policy (pi ^*_mathscrC).

Lemma 1 demonstrates a challenging optimization task in our work, which is to achieve the global optima based on local exploration. That is, our SSRL needs to explore the global optima based on local policy (pi ^*_mathscrC_sub). The following proposition defines the sufficient and necessary condition of this task.

### Proposition 1

(Boundary Condition) Suppose that we definetarget companiesas those appear on the globally optimal career path. The global optima can be achieved by a local policy (pi ^*_mathscrC_sub), if and only if the target companies are included in the corresponding company subset (mathscrC_sub).

Proposition 1 indicates that it is possible to locate the global optimum by exploring the best locally optimal policy, instead of exploring the whole company pool. Our exploration method is designed to handle global optimization task by stochastically exploring local optima. Please note that the target companies cannot be discovered directly, while Proposition 1 guarantees that the target companies are included in the final company subset if the global optima is achieved.

Exploration strategy. Assuming that there are m companies in a company subset, the number of candidate subsets equals to (fracN!m!(N-m)!). Given a large total number of companies N, it is still expensive to find the best path generated by the local policy, if we have to explore all candidate subsets. As we aims to find the best locally optimal path, we have the following proposition.

### Proposition 2

(Transformation Condition) With a large number of iterations, if the quality of the optimal path cannot be further improved based on randomly generated subsets, then the current optimal path is the global optima.

According to Proposition 2, as long as the policy is continuously optimized in a stochastic process, we will obtain the best local optima eventually. Thus, we need to make sure the exploration process will converge with a time limit. As we introduce several random events in our model (e.g., duration estimator, position predictor), the result for current optimal policy might be worse than the preserved one due to the uncertainty in the path generation. To accelerate the convergence, we develop a cool-down strategy based on the Boltzmann distribution [see Eq. (5)].

Cool-down strategies have been developed with simulated annealing techniques and are found efficient to handle sequential recommendation tasks22,28. In this paper, the cool-down strategy is developed under a more complicated RL-based framework to speed up the career path exploration. Theoretically, the convergence is guaranteed, referring to our Proposition 3 as follows.

### Proposition 3

(Convergence analysis) The cool-down strategy guarantees the convergence under the uncertain scenario in career path recommendation.

### Baseline methods

Five baselines were implemented in this work, including JBMUL, IGM, MGM, TTD, and PDQN. We summarize their advantages and disadvantages in Table 3.

• JBMUL. We follow the idea of reference59 to locate the best result at each time step. For selecting the next state, we calculate the accumulative reward of each company based on the current state. Then the model selects the maximum accumulative reward as the next state and continue.

• IGM. This method aims to find the result which is better than the previous state. For the current state, we calculate the accumulative reward of each company. Comparing the result with previous state’s reward, as long as it is better than or equal to the previous one, we will place the company on the current state and continue to the next state.

• MGM. This method is designed for finding the optimal solution based on the average accumulative reward of accepted companies. For the current state, as long as the calculated reward is better than or equal to the average reward of accepted companies, we will accept this allocation and continue to the next state.

• TTD. As mentioned in reference60, TTD is a traditional off-policy temporal difference learning. After a fixed number of iterations, TTD will generate the policy based on the initial state.

• PDQN. This baseline evaluates the implementation of our exploration strategy in deep RL. We regularize the action based on our exploration strategy (i.e., subset generation) and update the policy by the deep RL. We deploy a simple neural network with one hidden layer, the input and output size is set to 20, and the size of the hidden layer is 40.

### Experimental settings

We set the time length of career path to 20 years and the precision of time interval (Delta t) is a quarter, the suffer time is set to 1 year. The default setting for the optional information in the input (e.g. working duration, work history) is none. For RL method, the discount rate (eta = 0.9), and the step size is set to 0.01. The total number of iterations is set to 1,000,000 steps; and if the total work time is more than 20 years, we will restart at the initial state. For SSRL and PDQN, we set the initial temperature T to 1, and the decay rate (Gamma = 0.99). The fixed length of the subset (mathcalC_sub) is set to 20, and we compare the result every 100 iterations. For each method, we simulate the career path on different numbers of processors and average the path score under the same grading criteria.

The major setting of the four scenarios in our experiments are related to the weight of the company-related features, including reputation, popularity, duration, position change rate and smooth transfer rate.

• Scenario 1. General case (no specific user preference). This case shows the general case of our recommendation in which the user has no preference for the company features. The weights for all the features are 0.25.

• Scenario 2. Personalized case (reputation preferred). This case indicates the path for those users who care more about the company reputation. The weight for the reputation is 0.6 and others are 0.1.

• Scenario 3. Personalized case (time-varying preference). This case reflects the dynamical requirements from user at different career stage. The weights for the first half of career path are [0.1, 0.1, 0.1, 0.7] while the rest are [0.1, 0.7, 0.1, 0.1].

• Scenario 4. Personalized case (specific company preferred). This case simulates the situation that user has a rough plan for his/her career. We set the plan as being transferred to the Bank of Boston at the third career stage for 4 years.

### Experimental environment

All experiments were conducted on a high-performance computer (HPC) cluster with 164 compute nodes, each with two Intel Xeon E5-2683v3 20-core processors and 128 GB DDR4 Memory.


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# pg_config executable not found when using pgxnclient on Windows 7 x64

I installed Python 2.7.8 and pgxn client. And I tried to run this statement from command line from the bin folder and path is setup correctly

pgxnclient install  http://api.pgxn.org/dist/pg_repack/1.2.1/pg_repack-1.2.1.zip

But I got an error pg_config executable not found.

• I have no idea if pgxn supports Windows at all. Given that PGXS (the compilation system for extensions) doesn't support Windows, I rather doubt it can work for any extension that requires C compilation. – Craig Ringer Nov 4 '14 at 3:48
• Mladen, what operating system are you trying this on? You will need pg_config available, as well as the PostgreSQL headers installed (if installing on Linux from a packaging system, there is usually a -dev package needed). In case you are trying this on Windows, pg_repack is basically untested/unsupported on Windows. – Josh Kupershmidt Nov 6 '14 at 14:01
• I am using the Win 7 x64 bit as stated in the question. I am just exploring the PostgreSQL. It seems that the best way to use it is on linux. There is a project inside the zip file for 2010, but I need to adapt it for PostgreSQL 9.3 which I am using. It will wait for some time. – Mladen Uzelac Nov 6 '14 at 17:36
• @Mladen I would be interested if you successfully compiled it for PG 9.3 under Visual 2010. – JB. With Monica. Dec 4 '14 at 8:37
• I didn't and currently I am not planning to. I tried PostgreSQL on linux and it's much more performant and easier to use. In future I'll be using more Linux and software running on it. ;) PS. I had a lot of hassle to configure Visual studio for compiling 64 bit, I recommend to use VS 2012 immediately. – Mladen Uzelac Dec 4 '14 at 9:07

Bakground: pg_config is the configuration utility provided by PostgreSQL. This utility is used by various applications.

Solution:

1. Install PostgreSQL.
2. Set the path. System Properties > Advanced

PATH:C:\Program Files (x86)\PostgreSQL\9.4\bin\;

From this point on, any application that requires pg_config can find it here.

I suggest this method because the build is directly from the author. You can also find viable alternatives for standalone pg_config builds.

• Thank your for your time and effort. It was configured as you said. But I had problems with running pgxn client. I solved the problem by using linux and running it there. PostgreSQL is OK on windows but rocks on linux. ;) – Mladen Uzelac Jan 21 '15 at 18:09

I had this same error, and pg_config was in my path

You can try

pgxnclient install --pg_config "c:\path\to\pg_config.exe" \$extensionName

You almost definitely need to run cmd.exe as an admin

• this worked, thanks. But then I got an error for gmake being unavailable. I think pgxnclient is only meant to run on linux... "Windows binary versions auto-built upon submission" is in the wishlist on github.com/pgxn/pgxn/wiki/PGXN-Wish-List – Shmuel Kamensky Jan 10 '19 at 11:17


text:
I recently started playing with Rust’s PhantomData, and decided to implement type-checked matrix operations. It turns out that, not only does it work like a charm, the implementation is also surprisingly self-evident.

So, let’s see how this works!

## Integer generics

The first problem that I encountered was Rust’s lack of support for value parameters; Although this is a hack, we can easily simulate them using traits, macros, and associated functions:

use ::std::marker::PhantomData;

pub trait Num {
fn val() -> usize;
}

macro_rules! nums {
( $($name:ident => $num:expr ),* ) => {$(
#[derive(Clone, Copy, Eq, PartialEq)]
pub struct $name; impl Num for$name {
fn val() -> usize { \$num }
}
)*
}
}

nums1 {
N1 => 1,
N2 => 2,
N3 => 3,
N4 => 4,
N5 => 5,
N6 => 6
}

As we will see soon enough, once the Rust compiler understands value parameters, everything will be much better. For the moment, however, it will do the job.

## Type-checked matrix dimensions

Now that this is out of the way, we can actually create the Matrix type. For the sake of this demo, I chose to keep it simple by defining it over the real field (f64) only:

#[derive(Copy, PartialEq)]
pub struct Matrix<M: Num, N: Num> {
rows: PhantomData<M>,
cols: PhantomData<N>,
entries: Vec<Vec<f64>>
}

impl<M: Num, N: Num> Matrix<M, N> {
pub fn new_map<F>(func: F) -> Matrix<M, N>
where F: Fn(usize, usize) -> f64 {
Matrix {
rows: PhantomData,
cols: PhantomData,
entries: (0..M::val()).map(|row|
(0..N::val()).map(|col|
func(row, col)).collect()).collect()
}
}
}

Notice the calls to Num::val() in the construction of the matrix instance. Through them, we ensure that the entries — if inaccessible from the outside world —, will always have the desired size. With the new_map method at hand, we could define $$\mathbf I_3$$ as:

let i3 = Matrix::<N3, N3>::new_map(|i, j| if i == j { 1.0 } else { 0.0 });

Even better, let’s define it through a new associated function to Matrix.

impl<M: Num> Matrix<M, M> {
pub fn identity() -> Matrix<M, M> {
Matrix::new_map(|i, j| if i == j { 1.0 } else { 0.0 })
}
}

let i3 = Matrix::<N3,N3>::identity();

The way we defined it, the type-checker enforces that an identity matrix must always have an equal number of rows an columns. In fact, because of this, Rust can figure what the second type should be by itself:

// Only specify the first (or second) size type...

let i3 = Matrix::<N3,_>::identity();

// ... or, rely fully on type inference:
//
// (__*__)(N3*N2)  is how type inference starts
// (__*N3)(N3*N2)  from the definition of matrix multiplication
// (N3*N3)(N3*N2)  from the definition of Matrix::identity
//  ^~        ^~
// Therefore, m2: Matrix<N3, N2>.
// Therefore, equality is defined.

let m = Matrix::<N3, N2>::new_map(|i, j| (i + j) as f64);
let m2 = Matrix::identity() * m.clone();

assert!(m == m2);  // true => the test passes

If we were to get distracted and ask Rust for a rectangular identity matrix, either through explicit typing or during a type-infered computation, then it would reject our code:

let i3 = Matrix::<N3,N2>::identity();

// <anon>:73:14: 73:39 error: no associated item named identity found for type Matrix<N3, N2> in the current scope
// <anon>:73     let i3 = Matrix::<N3,N2>::identity();
^~~~~~~~~~~~~~~~~~~~~~~~~

Can we do better? Sure! Using Rust nightly:

// By Nihy

#![feature(zero_one)]
use ::std::num::{One, Zero};

impl<M: Num> Matrix<M, M> {
pub fn identity() -> Matrix<M, M> {
Matrix::new_map(|i, j| if i == j { One::one() } else { Zero::zero() })
}
}

This way, if we were to change the type of the scalars — switching to complex numbers, for example — this function will still work like a charm. In fact, it would work with any type implementing PartialEq, One and Zero.

As a side note, we can also use new_map to, quite elegantly, define matrix transpose:

impl<M: Num, N: Num> Matrix<M, N> {
pub fn transpose(&self) -> Matrix<N, M> {
Matrix::new_map(|i, j| self.entries[j][i])
}
}

You got to admit that it’s quite close to the mathematical notation: $${\mathbf M}^T_{ij} = {\mathbf M}_{ji}$$.

## Matrix multiplication

Let us first define matrix multiplication by a scalar:

// matrix * scalar
impl<M: Num, N: Num> Mul<f64> for Matrix<M, N> {
type Output = Matrix<M, N>;
fn mul(self, rhs: f64) -> Matrix<M, N> {
Matrix::new_map(|i, j| self.entries[i][j] * rhs)
}
}

// scalar * matrix
impl<M: Num, N: Num> Mul<Matrix<M, N>> for f64 {
type Output = Matrix<M, N>;
fn mul(self, rhs: Matrix<M, N>) -> Matrix<M, N> {
rhs * self
}
}

Next, we shall define multiplication between matrices. Remember that, when using the usual rule, an $$M \times N$$ matrix applied on an $$N \times L$$ matrix produce an $$M \times L$$ matrix:

impl<M: Num, N: Num, L: Num> Mul<Matrix<N, L>> for Matrix<M, N> {
type Output = Matrix<M, L>;
fn mul(self: Matrix<M, N>, rhs: Matrix<N, L>) -> Matrix<M, L> {
// ...
}
}

Here, I kept the type of self in order to clearly illustrate the role of the matrix’s size in the operation. By relying once again on the new_map function, we get the following definition for matrix multiplication:

impl<M: Num, N: Num, L: Num> Mul<Matrix<N, L>> for Matrix<M, N> {
type Output = Matrix<M, L>;
fn mul(self: Matrix<M, N>, rhs: Matrix<N, L>) -> Matrix<M, L> {
Matrix::new_map(|i, j|
(0..N::val()).map(|k| self.entries[i][k] * rhs.entries[k][j])
}
}

Moreover, once Iterator::sum stabilizes, we will be able to something much closer to the mathematical definition:

impl<M: Num, N: Num, L: Num> Mul<Matrix<N, L>> for Matrix<M, N> {
type Output = Matrix<M, L>;
fn mul(self: Matrix<M, N>, rhs: Matrix<N, L>) -> Matrix<M, L> {
Matrix::new_map(|i, j|
(0..N::val()).map(|k| self.entries[i][k] * rhs.entries[k][j]).sum()
}
}

One nice thing is that this definition even works if we change the type of the entries, as long as those implement Add, Mul and, when using sum(), the (still unstable) Zero trait.

Another interesting consequence of multiplication being type-checked in this way is that we do not actually need the bound-checking enabled in Rust by default. As such, even when we push our program to production — with optimizations turned on and bound-checking turned off — we can be confident that our code will act as expected.

Also, once integer generics and associated constants make their way into the language, we will be able to use C-like arrays instead of vectors inside of the Matrix structure, and every single matrix function using new_map will profit from the optimization, simply by updating a single method.

Now that we’ve implemented all of these matrix operations, we might want to play with vectors. I mean, it would be nice to have Vector<N> and Covector<N> types which interact nicely with matrices without having to rewrite too much code.

Well, it just so happens that Rust has a feature for that: type aliases.

type Vector<N> = Matrix<N, N1>;
type Covector<N> = Matrix<N1, N>;

Or, if you’re doing quantum mechanics:

type Ket<N> = Matrix<N, N1>;
type Bra<N> = Matrix<N1, N>;

We can now implement vector-specific methods:

impl<N: Num> Ket<N> {
fn vec_entries(&self) -> Vec<f64> {
self.entries.iter().map(|row| row[0]).collect()
}
}

impl<N: Num> Bra<N> {
fn vec_entries(&self) -> Vec<f64> {
self.entries[0].clone()
}
}

Whatever method we implement for Ket<N> will be available for Matrix<N, N1>. However, it’s more readable this way, and the intent is obvious. Notice how these methods, too, would benefit from value parameters by returning an array of the good length instead of a vector.

## Conclusion

So, this was how to take advantage of Rust’s type-system when playing with matrices and vectors. This is obviously a very basic toy example, however one could easily extend these principles to make strong guarantees about abstract systems, and it may very well be my favorite feature of the language.

I just really hope that value parameters and impl specialization will be added to the language; that and the ability to return unboxed closures. These would make the type-system ridiculously powerful and expressive.

Anyway, hope you found that interesting!


text:
Physics Help Forum Spring Launcher Project

 Energy and Work Energy and Work Physics Help Forum

 Apr 20th 2008, 07:44 AM #1 Junior Member     Join Date: Apr 2008 Posts: 5 Spring Launcher Project OK, so the point of this project is to build a launching pad for a spring launcher that can launch a spring (that will be provided on the day of the launch) anywhere in the the classroom. My class room is about 4m long, 5m or 6m wide and about 2m or 2.5m high, my teacher said the box that you have to get the spring into can be anywhere in the room so you have to do the calculations, calibration and all other preparation before hand. These are the calculations I've done, can somebody check them over if they look right? There might be some algebra mistakes. My design concept is pretty simple, two wooden blocks connected by a door hinge and a protractor attached to measure the angle.I have a ruler on the top block to measure how much I have to stretch the spring. Does anybody know a good way to fix the angle, once I have it lined up? I haven't started building it
Apr 20th 2008, 08:41 AM   #2

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,730
 Originally Posted by Hasan1 OK, so the point of this project is to build a launching pad for a spring launcher that can launch a spring (that will be provided on the day of the launch) anywhere in the the classroom. My class room is about 4m long, 5m or 6m wide and about 2m or 2.5m high, my teacher said the box that you have to get the spring into can be anywhere in the room so you have to do the calculations, calibration and all other preparation before hand. These are the calculations I've done, can somebody check them over if they look right? There might be some algebra mistakes. My design concept is pretty simple, two wooden blocks connected by a door hinge and a protractor attached to measure the angle.I have a ruler on the top block to measure how much I have to stretch the spring. Does anybody know a good way to fix the angle, once I have it lined up? I haven't started building it
The calculations look good. You'll also have to be able to account for the ceiling. I'd suggest you work out an equation for the maximum angle you can fire at for a given initial spring displacement (in other words, given a spring stretch and an angle of firing, how high will the spring go?)

As far as fixing the angle is concerned, if I have your setup right the spring is attached to a board that rotates, right? How about sticking a vertical prop under it that can slide back and forth?

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Apr 21st 2008, 01:57 PM #3 Junior Member     Join Date: Apr 2008 Posts: 5 I'm not really sure how to calculate the max angle...I already know how to calculate the minimum angle, wouldn't it be enough to just calculate the min. angle then aim it a few degrees higher than that? I thought about using a sliding vertical prop to fix the angle in place but my only concern is that I think it's going to minimize my max. or min. angle range depending on how high it is.
Apr 23rd 2008, 10:38 AM   #4

Join Date: Apr 2008
Location: On the dance floor, baby!
Posts: 2,730
 Originally Posted by Hasan1 I'm not really sure how to calculate the max angle...I already know how to calculate the minimum angle, wouldn't it be enough to just calculate the min. angle then aim it a few degrees higher than that? I thought about using a sliding vertical prop to fix the angle in place but my only concern is that I think it's going to minimize my max. or min. angle range depending on how high it is.
Say that you are planning to fire the projectile at an angle theta and will pull the spring back an initial distance x0. Then we know the highest the projectile can go is the height H, which for simplicity I will take as the vertical distance between the launch point and the ceiling. I will call the starting point of the projectile of mass m the 0 point for the GPE and calling the spring constant k.

First we find the speed v0 that the projectile leaves the launcher:
$\displaystyle \frac{1}{2}kx_0^2 = \frac{1}{2}mv_0^2 + mgx_0~sin(\theta)$

So we can find v0.

Now, you know that the velocity v0 will be directed at angle theta. So now do a kinematics problem and see what the max height of the projectile will be. This must be less than H.

-Dan
__________________
Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup.

See the forum rules here.

 Tags launcher, project, spring


text:
Browse Questions

# The net capacitance of three identical capacitors in series is $1 \mu F$. What will be their net capacitance if connected in parallel? Also find the ratio of energy stored in the two configurations if they are both connected to the same source.

$C= 3 \mu F$
$U_s :U_p=1:9$


text:
Vertical Angles Linear Pair Corresponding. We write the letters of congruent triangles so their order tells us which parts are corresponding. However, corresponding angles in a transversal of two nonparallel lines are never equal. True or False. - edu-answer.com Q. Name that property use your white board and write down parallel lines cut by a transversal corresponding angles parallel line properties parallel lines cut by a transversal corresponding angles. Corresponding angles can apply to either two polygons or parallel lines cut by a transversal. A) Vertical Angles are Congruent B) Alternate interior angles are congruent C) Corresponding angles are supplementary D) Same-side . false. prod 3. Corresponding angles are equal if the transversal intersects two parallel lines. They add up to 180 degrees. (Yes, I can spell, although my typing is awful. Note: If the two vertical angles are right angles then they are both congruent and supplementary. 8. 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Note: If the two vertical angles are right angles then they are both congruent and supplementary. If two triangles are congruent, then naturally all the sides are angles are also congruent with … Consecutive interior angles are supplementary. Corresponding angles of similar or congruent triangles/polygons/etc. List a pair of corresponding angle. If the sum of two angles is 180 degrees then they are said to be supplementary angles, which forms a linear angle together. Tags: Question 14 . Tags: Question 10 . In both cases, corresponding angles are in the same position. Are the ratios of corresponding side lengths equal? If the two lines are parallel then the corresponding angles are congruent. You can create a customized shareable link (at bottom) that will remember the exact state of the app--which angles are selected and where the points are, so that you can share your it with others. 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The same side interior angle theorem states that when two lines that are parallel are intersected by a transversal line the same side interior angles that are formed are supplementary or add up. ∠ c and ∠ e, ∠ d and ∠ f Lesson 7 M1 NYS COMMON CORE … Angles forming a Linear Pair: (Adjacent Angles creating a Straight Line) (measures are supplementary) This is an “old” idea about angles revisited. Congruent angles are angles that have the same measure. Whats people lookup in this blog: Alternate Interior Angles Are Congruent Or Supplementary The sum of the measures of the internal angles of a triangle is equal to 180 °. AAS is equivalent to an ASA condition, by the fact that if any two angles are given, so is the third angle, since their sum should be 180°. Given two parallel lines cut by a transversal, their corresponding angles are supplementary. 14. Corresponding Angles are congruent. They are supplementary both angles add up to 180 degrees. What are the angles … A transversal intersects two lines. If two angles are supplementary to two other congruent angles, then they’re congruent. Refresh your memory using the diagram below: Theorem: Vertical angles are congruent. Whats people lookup in this blog: Alternate Interior Angles Are Congruent Or Supplementary; Alternate Exterior Angles Are Congruent Or Supplementary Q. Alternate interior angles are equal in measure. Congruent corresponding angles are: Angle of 'a' = Angle of 'g' Angle of 'b' = Angle of 'h' Angel of 'c' = Angle of 'e' Angle of 'd' = Angle of 'f' When two straight lines are cut by another line i. True or False. corresponding angles are congruent iff they are formed by a transversal intersecting parallel lines. Supplements of the same angle are congruent. Vertical Angles are ALWAYS congruent. Describe how you know using similarity transformations. Vertical angles are always, by definition, congruent. true. When doing a proof, note … Each angle is supplement of the other angle. Alternate interior angles are congruent. Which angles are ALWAYS supplementary? There's lots of useful things you can discover when parallel lines are cut by a transversal, most of them having to do with angle relationships. 3 ð 2 and ð 6 are congruent angles 4 ð 2 and ð 8 are supplementary angles 5 ð 8 and ð 4 are congruent angles 7 if m ð 3 40 nd the measure of ð 4. Corresponding angles are angles that: ... Each pair of co-interior angles is supplementary. … ★★★ Correct answer to the question: Angles e and g are a. congruent b. non congruent c. supplementary to each other because they are a. adjacent b. corresponding c. vertical angles? Vertical angles are ALWAYS equal, whether you have parallel lines or not. 100. The angles of linear pair are supplementary angles. false. When … If the transversal intersects non-parallel lines, the corresponding angles formed are not congruent and are not related in any way. Corresponding angles are angles that are in the same relative position at an intersection of a transversal and at least two lines. Corresponding angles form are supplementary angles if the transversal perpendicularly intersects two parallel lines. not just any two lines will do, so be certain the lines are parallel. The Alternate Interior Angles Theorem states that If two parallel straight lines are intersected by a third straight line (transversal), then the angles inside (between) the parallel lines, on opposite sides of the transversal are congruent (identical). answer choices . If you're seeing this message, it means we're having trouble loading external resources on our website. If the two lines are parallel, then the corresponding angles created by the transversal are congruent. Two angles are said to be vertically opposite angles … 100. Roger broke out in an awful rash after hiking in an area where poison oak grows. If the sum of two angles is 180°, then they are called supplementary angles. I. The alternate interior angles theorem states that, the … The sides of the angles do not need to have the same length or open in the same direction to be congruent, they only need to have equal measures. SURVEY . Given two parallel lines cut by a transversal, their corresponding angles are supplementary. Vertical Angles Linear Pair Corresponding. Which angles are ALWAYS congruent? Well same side interior angles would be 4 and 5 so notice we have parallel lines and the transversal. As angles a 110 c and d are all alternate interior angles … In this video, Sal uses this notation to solve some angles. Are the ratios of corresponding side lengths equal? ∠ a and ∠ e, ∠ d and ∠ h, etc. Supplementary angle. Look for a Pattern you know that the product of two … health. ∠ c and ∠ f, ∠ d and ∠ e Same-side interior angles are supplementary. Often the line segments or angles that we want to prove congruent are not cor-responding parts of triangles that can be proved congruent by using the given information.However, it may be possible to use the given information to prove a different pair of triangles congruent. ∠AOB + ∠CED= 120° + 60° = 180° Therefore these angles are supplementary Vertically opposite angles. Each one of these angles is called the supplementary of the other. Tags: Question 13 . Q. Yes/No Explain. Given two parallel lines are cut by a transversal, their same side exterior angles are congruent… AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. Angle 1 is in the upper left corner of the top intersection, and angle 8 is in the lower right corner of the bottom intersection, so they cannot be corresponding angles. CLUE less in Math? View geo 3 1 notes pdf from mat 360 at southern connecticut state university. Alternate Interior Angles Theorem. are always equal; as are corresponding angles in a transversal of parallel lines. The following diagram shows examples of corresponding angles. The previous four theorems about complementary and supplementary angles come in pairs: One of the theorems involves three segments or angles, and the other, which is based on the same idea, involves four segments or angles. When the three angle pairs are all equal, the three pairs of sides must also be in proportion. how much commission did Todd earn. If you're behind a web filter, please make sure that the … The angle … Alternate co interior and corresponding angles. $$\measuredangle 1 \cong \measuredangle 2$$ $$\measuredangle 3 + \measuredangle 4 = 180^{\text{o}}$$ Theorem 14, 15, 16. Linear Pair are ALWAYS supplementary. true. Alternate interior angles don’t have any specific properties in the case of non – parallel lines. 30 seconds . Consecutive interior angles are interior angles which are on the same side of the transversal line. Given two parallel lines are cut by a transversal, their same side exterior angles are congruent. answer choices . ASA and AAS are sometimes combined into a single … true. Each pair of alternate exterior angles is equal. The measure of angles A and B above are both 34° so angles A and B are congruent or ∠A≅∠B, where the symbol ≅ means congruent. Are corresponding angles congruent? (This is the four-angle version.) Describe how you know using … 30 seconds . Practice telling whether two angles are supplementary, complementary, or vertical. Whereas if the sum of two angles is 90 degrees, then they are said to be complementary angles, and they form a right angle together. 100. Are Corresponding Angles congruent or supplementary? 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Transversal of two angles is called the supplementary of the transversal intersects non-parallel lines, the corresponding angles in transversal. T have any specific properties in the same position it means we 're having trouble loading external on! Intersects non-parallel lines, the corresponding angles are angles that have the same position wz PQ wx YZ... Please make sure that the … corresponding angles in a transversal my typing is awful in proportion formed by transversal! Figures similar transversal intersects two parallel lines cut by a transversal, their same side of the other of! Connecticut state university is 180 degrees then they are both congruent and are not congruent and supplementary be! Seeing this message, it means we 're having trouble loading external resources on our website triangle equal. If the two lines are parallel then the corresponding angles are equal if the two lines sides must be.: vertical angles are congruent, then they are formed by a transversal intersecting parallel lines not. Non-Parallel lines, the corresponding angles are right angles then they are said to be supplementary and! Doing a proof, note … List a pair of co-interior angles is 180 degrees then they called. Lines parallel and 5 so notice we have parallel lines are parallel pairs are all alternate angles! Are right angles then they are both congruent and supplementary c ) corresponding angles are the! An awful rash after hiking in an awful rash after hiking in an awful rash after hiking in an rash. ; as are corresponding angles in a transversal of parallel lines means we 're having trouble loading external on. Interior angles are right angles then they are both congruent and supplementary in proportion perpendicularly intersects parallel.: alternate interior angles are congruent are corresponding angles congruent or supplementary supplementary vertical angles are also congruent either two polygons are congruent the... Intersecting parallel lines cut by a transversal the other two unshaded interior angles which! Awful rash after hiking in an area where poison oak grows out an. Wz PQ wx RS YZ are the figures similar the corresponding angles are always, by definition, congruent would... Alternate exterior angles are congruent or supplementary when … corresponding angles form are.! Or vertical apply to either two polygons are congruent internal angles of a is... Angles created by the transversal are congruent or supplementary so be certain the lines are parallel congruent then. Lines or not note … List a pair of corresponding angle angles form are supplementary angles the! Mat 360 at southern connecticut state university same measure a ) vertical angles are congruent iff they are called angles. Corresponding angles are congruent AB BC CD are the figures similar or parallel lines blog: interior... The addition of two angles d ) Same-side 're seeing this message, it means we 're having trouble external! Congruent triangles/polygons/etc and 5 so notice we have parallel lines our website a. ∠ f, ∠ d and ∠ f, ∠ d and ∠ e Same-side interior are! Angles supplementary or congruent triangles/polygons/etc prove the … a transversal of parallel lines trouble loading external resources on website! Two vertical angles are also are corresponding angles congruent or supplementary consecutive interior angles, which forms a linear angle.... These angles are interior angles which are on the same position in any way awful.

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