Question

تذكر الوقوف بي يدي الله والحساب وحقوق العباد
the histogram of an image is given next. perform histogram equalization and give the new histogram.

original histogram
gray level i | 0 | 1 | 2 | 3 | 4 | 5
count n_i | 43 | 0 | 48 | 29 | 15 | 17

new histogram
gray level i | 0 | 1 | 2 | 3 | 4 | 5
count n_i | 0 | 43 | 0 | 48 | 44 | 17

Explanation:

Step1: Calculate Total Pixels

First, find the total number of pixels \( N \) by summing all \( n_i \).
\( N = 43 + 0 + 48 + 29 + 15 + 17 = 152 \)

Step2: Compute Cumulative Distribution Function (CDF)

For each gray level \( i \), compute \( CDF(i) = \sum_{k=0}^{i} n_k \).
- \( CDF(0) = 43 \)
- \( CDF(1) = 43 + 0 = 43 \)
- \( CDF(2) = 43 + 0 + 48 = 91 \)
- \( CDF(3) = 43 + 0 + 48 + 29 = 120 \)
- \( CDF(4) = 43 + 0 + 48 + 29 + 15 = 135 \)
- \( CDF(5) = 43 + 0 + 48 + 29 + 15 + 17 = 152 \)

Step3: Histogram Equalization Transformation

The transformation for histogram equalization is \( s_i = \lfloor \frac{(L - 1)}{N} \cdot CDF(i) floor \), where \( L = 6 \) (number of gray levels, 0–5).
- For \( i = 0 \): \( s_0 = \lfloor \frac{5}{152} \cdot 43 floor \approx \lfloor 1.414 floor = 1 \)
- For \( i = 1 \): \( s_1 = \lfloor \frac{5}{152} \cdot 43 floor = 1 \) (same as \( i=0 \), since \( CDF(1)=CDF(0) \))
- For \( i = 2 \): \( s_2 = \lfloor \frac{5}{152} \cdot 91 floor \approx \lfloor 2.987 floor = 2 \)
- For \( i = 3 \): \( s_3 = \lfloor \frac{5}{152} \cdot 120 floor \approx \lfloor 3.947 floor = 3 \)
- For \( i = 4 \): \( s_4 = \lfloor \frac{5}{152} \cdot 135 floor \approx \lfloor 4.447 floor = 4 \)
- For \( i = 5 \): \( s_5 = \lfloor \frac{5}{152} \cdot 152 floor = 5 \)

Step4: Count New Histogram

Count how many original gray levels map to each new gray level:
- New gray level 0: No mappings (since all \( s_i \geq 1 \)) → \( n_0 = 0 \)
- New gray level 1: \( i=0,1 \) → \( n_1 = 43 + 0 = 43 \)
- New gray level 2: \( i=2 \) → \( n_2 = 48 \) (Wait, correction: Wait, \( i=2 \) maps to 2, so \( n_2 = 48 \)? Wait, no—wait, original \( n_2 = 48 \), and \( s_2 = 2 \), so \( n_2 \) in new is 48? But the given new histogram has 0 for gray level 2. Wait, maybe the initial new histogram in the image is incorrect. Let's recalculate properly.

Wait, re-express: The transformation is \( s_i = \text{round} \) or \( \lfloor \) of \( \frac{(L-1)}{N} CDF(i) \). Let's recheck \( L \): gray levels 0–5, so \( L = 6 \), \( L-1 = 5 \).

CDF(0) = 43, CDF(1)=43, CDF(2)=91, CDF(3)=120, CDF(4)=135, CDF(5)=152.

For \( i=0 \): \( (5/152)*43 ≈ 1.414 \) → floor to 1.
For \( i=1 \): same as i=0, so 1.
For \( i=2 \): (5/152)*91 ≈ 2.987 → floor to 2.
For \( i=3 \): (5/152)*120 ≈ 3.947 → floor to 3.
For \( i=4 \): (5/152)*135 ≈ 4.447 → floor to 4.
For \( i=5 \): (5/152)*152 = 5 → floor to 5.

Now, count the number of pixels mapping to each new gray level:
- New level 0: 0 (no \( s_i = 0 \))
- New level 1: \( i=0 \) (n=43) and \( i=1 \) (n=0) → total 43 + 0 = 43
- New level 2: \( i=2 \) (n=48) → 48
- New level 3: \( i=3 \) (n=29) → 29? Wait, no! Wait, original \( n_i \) is the count for gray level \( i \). So when \( i=3 \), \( n_i = 29 \), and \( s_3 = 3 \), so new \( n_3 = 29 \). But the given new histogram has 48 for gray level 3. This suggests a mistake in the initial new histogram.

Wait, the problem’s given new histogram has:
Gray Level 0: 0
Gray Level 1: 43
Gray Level 2: 0
Gray Level 3: 48
Gray Level 4: 44
Gray Level 5: 17

But according to correct calculation, let's sum the new counts: 0 + 43 + 0 + 48 + 44 + 17 = 152, which matches \( N \). So let's see how the mappings lead to this:

Alternative approach: Maybe the transformation is \( s_i = \frac{(L-1)}{N} CDF(i) \) rounded or adjusted. Let's check the sum of new counts: 0 + 43 + 0 + 48 + 44 + 17 = 152, which is correct.

Let's see which original \( n_i \) map to new gray levels:

- New gray level 1: original gray levels with \( s_i = 1 \). \( s_0 = 1 \), \( s_1 = 1 \) → \( n_0 + n_1 = 43 + 0 = 43 \) (matches new \( n_1 = 43 \))
- New gray level 3: original gray level 2 (n=48) → \( s_2 = 3 \)? Wait, (5/152)*91 ≈ 2.987, floor to 2, but if we round, it's 3. Ah, maybe rounding instead of flooring. Let's try rounding:

\( s_i = \text{round}( \frac{5}{152} CDF(i) ) \)

- \( i=0 \): (5*43)/152 ≈ 1.414 → round to 1
- \( i=1 \): same as i=0 → 1
- \( i=2 \): (5*91)/152 ≈ 2.987 → round to 3
- \( i=3 \): (5*120)/152 ≈ 3.947 → round to 4? No, 3.947 rounds to 4? Wait, 3.947 is closer to 4.
- \( i=4 \): (5*135)/152 ≈ 4.447 → round to 4
- \( i=5 \): (5*152)/152 = 5 → 5

Now:
- New level 1: i=0,1 → 43 + 0 = 43
- New level 3: i=2 → 48 (so \( s_2 = 3 \))
- New level 4: i=3,4 → 29 + 15 = 44 (matches new \( n_4 = 44 \))
- New level 5: i=5 → 17 (matches new \( n_5 = 17 \))
- New level 0: 0
- New level 2: 0 (no mappings)

Yes! That works. So:
- \( s_0 = 1 \), \( s_1 = 1 \) → new level 1: 43 + 0 = 43
- \( s_2 = 3 \) → new level 3: 48
- \( s_3 = 4 \), \( s_4 = 4 \) → new level 4: 29 + 15 = 44
- \( s_5 = 5 \) → new level 5: 17
- New level 0: 0
- New level 2: 0

Thus, the new histogram is:

| Gray Level i | 0 | 1 | 2 | 3 | 4 | 5 |
|--------------|---|---|---|---|---|---|
| Count \( n_i \) | 0 | 43 | 0 | 48 | 44 | 17 |

Answer:

The new histogram after equalization is:

Gray Level \( i \)	0	1	2	3	4	5