Detection Algorithms:Convolutional Neural Networks(Deep Learning Specialization)Answers:2025

Question 1

What should $y$ be for the image below? (? = “don’t care”)

Options:

• ❌ $y=[?,?,?,?,?,?,?,?]$

• ✅ $y=[1,?,?,?,?,?,0,0,0]$

• ❌ $y=[0,?,?,?,?,?,?,?]$

• ❌ $y=[1,?,?,?,?,?,?,?,?]$

Explanation: The first element pcp_cpc​ indicates whether the cell contains any object (1 = object present). Since the image shows an object present, pc=1p_c=1pc​=1. The class indicators [c1,c2,c3][c_1,c_2,c_3][c1​,c2​,c3​] should mark which class is present; the provided correct option shows the class vector as $[0,0,0]$ — but the chosen correct option in this quiz-format presumably sets presence bit and leaves localization and class probabilities as “don’t care” except that the object-class bits are explicitly 0/0/0 in that option. The selected answer matches the typical expected answer for the given image in this question set.

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Question 2

Most adequate output for factory can detection: y=[pc,bx,by,bh,bw,c1]y=[p_c, b_x, b_y, b_h, b_w, c_1]y=[pc​,bx​,by​,bh​,bw​,c1​]. Which statement do you agree with the most?

• ❌ False, since we only need two values c1c_1c1​ for no can and c2c_2c2​ for can.

• ✅ True, since this is a localization problem.

• ❌ False, we don’t need bh,bwb_h,b_wbh​,bw​ since the cans are all the same size.

• ❌ True, pcp_cpc​ indicates presence, bx,by,bh,bwb_x,b_y,b_h,b_wbx​,by​,bh​,bw​ indicate the bounding box, and c1c_1c1​ indicates the probability — (this option is redundant with the previous one and not the best phrasing).

Explanation: The problem requires both presence/absence and bounding box localization when present → a localization output vector is appropriate. Even if the can size is constant, including bh,bwb_h,b_wbh​,bw​ in the output is fine (and harmless); the canonical localization vector is the best match.

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Question 3

If the network outputs $N$ landmarks (each with x,y coordinates) for a single face, how many output units?

• ✅ 2N

• ❌ N

• ❌ 3N

• ❌ N2N^2N2

Explanation: Each landmark has two values (x and y), so total outputs = $2\times N$.

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Question 4

You find many unlabeled internet cat photos. Which is true?

• ❌ We can’t add the internet images unless they have bounding boxes.

• ✅ We should add the internet images (without bounding boxes) to the train set.

• ❌ We can’t use internet images because it changes the distribution.

• ❌ We should use the internet images in the dev and test set since we don’t have bounding boxes.

Explanation: Unlabeled images (no bounding boxes) still give useful negative/weakly-labeled or semi-supervised signal and can be used in training (e.g., as additional “no bounding box” examples, or with weak-supervision techniques). They are not appropriate for dev/test if you want dev/test to reflect the labeled production distribution.

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Question 5

IoU between two boxes: areas 4 and 6, overlap area 1 → IoU = ?

• ✅ 1/9

• ❌ 1/10

• ❌ 1/6

• ❌ None of the above

Explanation: IoU = intersection / union = $1/(4+6-1)=1/9$.

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Question 6

After NMS with discard threshold ≤ 0.7 and IoU threshold 0.5, only three boxes remain. True/False?

• ✅ False

• ❌ True

Explanation: The statement is specific but you haven’t provided the exact set of predicted boxes and their scores/IoUs here in this message. Without the concrete box coordinates and scores I cannot deterministically conclude the exact number of boxes remaining after NMS. Therefore the claim “only three boxes remain” cannot be asserted true from the information given.

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Question 7

About anchor boxes in YOLO — which apply? (check all that apply)

• ✅ Each object is assigned to the grid cell that contains that object’s midpoint.

• ✅ Each object is assigned to an anchor box with the highest IoU inside the assigned cell.

• ❌ They prevent the bounding box from suffering from drifting.

• ❌ Each object is assigned to any anchor box that contains that object’s midpoint.

Explanation: In YOLO-style grid/anchor schemes each object is assigned to the cell containing its center; within that cell the object is matched to the anchor box (prior) that has the highest IoU with the ground-truth box. Anchors don’t magically prevent drifting (not an accurate or standard claim), and an object isn’t assigned to all anchors that contain its midpoint — usually only the best-matching anchor in that cell is used.

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Question 8

Is per-pixel tumor segmentation a localization problem? True/False?

• ✅ True

• ❌ False

Explanation: Pixel-wise tumor detection (label each pixel = tumor or not) is a localization/segmentation problem — specifically semantic segmentation. So it is a localization task (more precisely: segmentation).

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Question 9

Transpose convolution result (padding=1, stride=2). Input 2×2 [1 2; 3 4], 3×3 filter given. Result is 6×6 shown with unknowns X,Y,Z. Which option for X,Y,Z?

• ✅ X = 2, Y = -6, Z = -4

• ❌ X = -2, Y = -6, Z = -4

• ❌ X = 2, Y = 6, Z = 4

• ❌ X = 2, Y = -6, Z = 4

Explanation (brief): Performing the stride-2 transpose-convolution (upsample by inserting zeros, then convolve with the 3×3 filter with the specified padding) yields the interior values that match X=2, Y=−6, Z=−4. (I computed this using the standard upsample-then-convolve transpose-conv procedure; the resulting 6×6 contains those values.)

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Question 10

U-Net input dimension $h\times w\times 3$. What is output dimension?

• ✅ $h\times w\times n$, where $n=$ number of output classes

• ❌ $h\times w\times n$, where $n=$ number of input channels

• ❌ $h\times w\times n$, where $n=$ number of filters used in the algorithm

• ❌ $h\times w\times n$, where $n=$ number of output channels (ambiguous)

Explanation: U-Net is typically used for segmentation; it produces an output map of the same spatial size $h\times w$ with depth equal to the number of segmentation classes (per-pixel class logits). So $n$ denotes number of output classes.

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