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Discrete mathematics / English

Calculate Greatest Common Divisors via the Euclidean Algorithm

SoftwareEngenius · YouTube · 5:01

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The explanation, unpacked.

Reviewed learning material · Video analysis · English
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The lesson develops the Euclidean algorithm from positive common divisors, divisibility and congruence, then presents a recursive implementation and its call-count bound. Integer GCD inputs are not both zero. For ordered positive inputs a≥b>0a\ge b>0, write a=bq+ra=bq+r with 0≤r<b0\le r<b; the identity gcd⁡(a,b)=gcd⁡(b,r)\gcd(a,b)=\gcd(b,r) preserves the answer. When the divisor reaches zero, return the preceding positive divisor. Editorial notes complete the reverse common-divisor argument that is not separately shown in the source, and correct a zero-quotient typo on the complexity slide. The O(log⁡2(N))O(\log_2(N)) bound, with N=a+bN=a+b, counts remainder calls under unit-cost arithmetic rather than total bit-level running time.

Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.

Chapters

0:00Introduction to Fast GCD0:07What is GCD?0:24The Brute Force Approach0:31Introducing the Euclidean Algorithm0:35The Three-Part Lemma1:14Proof of the Congruence Property1:41Lemma list for gcd facts1:47.5Proof that b∣a−cb\mid a-c preserves the gcd2:23.5Transition to the Euclidean algorithm2:30.5First division step and gcd reduction2:59.5Second division step and continued reduction3:22Euclidean Algorithm Derivation4:01Java Implementation4:17Time Complexity Proof4:46Conclusion

Learning script

Generated from the video's visuals and explanation; not verbatim speech.

The video opens by posing a practical problem: how to efficiently calculate the Greatest Common Divisor (GCD) of two large numbers, specifically 1071 and 462.

Before tackling the efficient computation, the fundamental concept is defined. The GCD of two integers is the largest positive integer that divides both numbers evenly. Simple examples like gcd⁡(7,21)=7\gcd(7, 21) = 7 and gcd⁡(24,30)=6\gcd(24, 30) = 6 illustrate this. The case where gcd⁡(7,9)=1\gcd(7, 9) = 1 also introduces the term 'coprime' for numbers sharing no common divisors other than 1.

For positive integer inputs, a direct search tests candidates from the smaller number down to 1. In the worst case the number of divisibility tests grows linearly with that smaller input; this comparison counts arithmetic tests rather than bit-level processing time.

To overcome this inefficiency, the Euclidean Algorithm is introduced as the superior method for computing the GCD.

The theoretical foundation of the Euclidean Algorithm is presented through a three-part lemma. First, it establishes the commutative property: gcd⁡(a,b)=gcd⁡(b,a)\gcd(a, b) = \gcd(b, a). Second, it covers the trivial case where one number divides the other: if a>0a > 0 and a∣ba \mid b, then gcd⁡(a,b)=a\gcd(a, b) = a. Third, and most importantly for the algorithm's recursive nature, it links modular arithmetic to the GCD: if a≡c(modb)a \equiv c \pmod b, then gcd⁡(a,b)=gcd⁡(c,b)\gcd(a, b) = \gcd(c, b).

For a positive modulus bb, the relation a≡c(modb)a\equiv c\pmod b means b∣(a−c)b\mid(a-c). Thus an integer yy exists with by=a−cby=a-c. These equations set up the common-divisor argument that follows.

The board lists basic GCD properties. The focus is the reduction property under the hypothesis b∣a−cb\mid a-c.

From that hypothesis, the writer introduces an integer yy with by=a−cby=a-c, then rearranges it to c=a−byc=a-by. This sets up the proof that replacing aa by the remainder-like quantity cc does not change the gcd with bb.

Next, the proof takes an integer dd dividing both aa and bb. In the spoken explanation this dd is identified with the greatest common divisor context, although the displayed line explicitly records only the divisibility assumptions d∣ad\mid a and d∣bd\mid b.

Because dd divides aa and bb, the video argues that dd must also divide the combination a−bya-by. The justification given aloud is that one is effectively using a multiple of bb, which remains divisible by dd.

Since c=a−byc=a-by, the same statement becomes d∣cd\mid c. At this point the board has linked any common divisor of aa and bb to a divisor of cc as well.

The board concludes gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c). Editorial completion: every common divisor of bb and cc divides a=c+bya=c+by, giving the reverse implication needed for equality. Thus the two pairs have the same positive common divisors.

This invariant turns the divisibility argument into a method for calculating the greatest common divisor.

For integer inputs a≥b>0a\ge b>0, write a=bq1+r1a=bq_1+r_1 with 0≤r1<b0\le r_1<b using division with remainder.

Applying the reduction lemma to this division step gives gcd⁡(a,b)=gcd⁡(b,r1)\gcd(a,b)=\gcd(b,r_1). Thus the original pair is replaced by the smaller pair consisting of the divisor and the remainder.

If r1>0r_1>0, divide again: b=r1q2+r2b=r_1q_2+r_2 with 0≤r2<r10\le r_2<r_1. Then gcd⁡(b,r1)=gcd⁡(r1,r2)\gcd(b,r_1)=\gcd(r_1,r_2). If a remainder is zero, division stops instead.

The same reduction repeats with strictly decreasing positive integer remainders. This descent is finite, and the following explanation identifies the last positive divisor as the answer.

The remainders are nonnegative integers. Once a zero remainder appears, the preceding positive divisor is the GCD. Taking r0=br_0=b also covers exact divisibility on the first division; the answer then is bb. This is the stopping rule for the remainder chain.

For nonnegative integer arguments, not both zero, when b=0b=0 the recursive function returns aa. Otherwise it calls gcd⁡(b,a mod b)\gcd(b,a\bmod b), using a nonnegative remainder. The source displays the ordered positive-input case; arbitrary signed inputs require normalization rather than direct use of this code.

For integers a>b>0a>b>0, the new remainder is at most a/2a/2. After two recursive calls the larger parameter has fallen to that remainder, unless the process has already stopped. Thus the number of remainder calls is O(log⁡2(a+b))O(\log_2(a+b)) under unit-cost arithmetic. This counts numerical-value reduction, not bit-level running time. Editorial correction: the slide says zero quotient would give k=bk=b, but it would give k=ak=a; the correctly bounded quotient yields the same halving conclusion.

The lesson closes after the recursive implementation and complexity discussion.

Knowledge cards

01

Greatest common divisor

The Greatest Common Divisor (GCD) of two integers is the largest positive integer that divides both numbers without leaving a remainder. If the GCD is 1, the numbers are called coprime. The inputs are integers and are not both zero.

gcd⁡(a,b)\gcd(a, b)
02

Brute Force Inefficiency

For positive integer inputs, checking candidates down to 1 gives a linear worst-case number of divisibility tests in the smaller input under a unit-cost arithmetic model.

03

Euclidean Algorithm Lemma: Part 1 & 2

The Euclidean algorithm relies on key properties of the GCD. 1) Commutativity: The order of inputs doesn't matter (gcd⁡(a,b)=gcd⁡(b,a)\gcd(a,b) = \gcd(b,a)). 2) Divisibility: If a positive number aa divides bb, their GCD is simply aa.

gcd⁡(a,b)=gcd⁡(b,a);a>0, a∣b ⟹ gcd⁡(a,b)=a\gcd(a,b)=\gcd(b,a);\quad a>0,\ a\mid b\ \Longrightarrow\ \gcd(a,b)=a
04

Euclidean Algorithm Lemma: Part 3 (Modulo)

The crucial property enabling the Euclidean algorithm's reduction step: if two numbers aa and cc leave the same remainder when divided by bb (i.e., a≡c(modb)a \equiv c \pmod b), then their GCDs with bb are equal (gcd⁡(a,b)=gcd⁡(c,b)\gcd(a,b) = \gcd(c,b)). Here the modulus bb is a positive integer.

a≡c(modb) ⟹ gcd⁡(a,b)=gcd⁡(c,b)a\equiv c\pmod b\ \Longrightarrow\ \gcd(a,b)=\gcd(c,b)
05

Proof Start: Congruence implies Divisibility

The proof for the modulo property begins by translating the congruence relation into a divisibility statement. If a≡c(modb)a \equiv c \pmod b, by definition, bb divides the difference (a−c)(a - c). This means there exists an integer yy such that a−c=bya - c = by.

a≡c(modb)  ⟹  b∣(a−c)  ⟹  ∃y∈Z,by=a−ca \equiv c \pmod b \implies b \mid (a - c) \implies \exists y \in \mathbb{Z}, by = a - c
06

Three gcd lemmas used before the algorithm

The clip begins with a handwritten lemma list: (1) gcd⁡(a,b)=gcd⁡(b,a)\gcd(a,b)=\gcd(b,a); (2) if a>0a>0 and a∣ba\mid b, then gcd⁡(a,b)=a\gcd(a,b)=a; (3) a reduction fact used to justify the Euclidean algorithm. These are presented as the basic facts from which the method follows.

gcd⁡(a,b)=gcd⁡(b,a);a>0, a∣b⇒gcd⁡(a,b)=a\gcd(a,b)=\gcd(b,a);\quad a>0,\ a\mid b \Rightarrow \gcd(a,b)=a
07

Reduction lemma: subtracting a multiple preserves gcd

For integer a,ca,c and positive bb, the condition b∣(a−c)b\mid(a-c) preserves the GCD. The source shows the forward implication; the editorial converse uses a=c+bya=c+by to show that any divisor of bb and cc also divides aa.

b∣a−c⇒gcd⁡(a,b)=gcd⁡(b,c)b\mid a-c \Rightarrow \gcd(a,b)=\gcd(b,c)
08

Proof skeleton for the reduction lemma

From c=a−byc=a-by, divisors of a,ba,b also divide cc. Editorially, divisors of b,cb,c also divide a=c+bya=c+by. Both directions are needed to equate the positive common-divisor sets.

by=a−c, c=a−by, d∣a, d∣b⇒d∣cby=a-c,\ c=a-by,\ d\mid a,\ d\mid b \Rightarrow d\mid c
09

Euclidean algorithm as repeated gcd reduction

For positive divisors and nonnegative remainders, repeated division preserves the GCD. Divide by the next remainder only if it is positive; the algorithm stops at zero.

a=bq1+r1, r1<b⇒gcd⁡(a,b)=gcd⁡(b,r1)a=bq_1+r_1,\ r_1<b \Rightarrow \gcd(a,b)=\gcd(b,r_1)
10

Why the remainders matter

Standard integer division gives nonnegative remainders smaller than the current positive divisor. Their positive values strictly decrease, so iteration must terminate.

r1<b,r2<r1r_1<b,\quad r_2<r_1
11

Euclidean Algorithm Termination Condition

For positive input pairs, nonnegative remainders decrease until zero. The preceding positive divisor is the GCD; take r0=br_0=b to include exact first-step divisibility.

gcd⁡(a,b)=gcd⁡(b,a mod b)  ⟹  ⋯  ⟹  gcd⁡(rk−1,0)=rk−1\gcd(a,b) = \gcd(b, a \bmod b) \implies \dots \implies \gcd(r_{k-1}, 0) = r_{k-1}
12

Recursive Implementation Logic

For nonnegative integer inputs, not both zero, return the first argument when the second is zero; otherwise recurse on the divisor and nonnegative remainder. Signed inputs require normalization.

gcd⁡(a,b)={ab=0gcd⁡(b,a mod b)b>0\gcd(a,b)=\begin{cases}a&b=0\\\gcd(b,a\bmod b)&b>0\end{cases}
13

Logarithmic Time Complexity Proof

For ordered positive integer inputs, the new remainder is at most half the earlier dividend. Within two calls the larger parameter has therefore halved, giving a logarithmic number of remainder calls under unit-cost arithmetic. This is not bit-level running time.

O(log⁡2(N))O(\log_2(N))

Detailed learning notes

Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.

Symbols · 14

gcd(a,b)

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The introduction explains the greatest common divisor and illustrates it with integer pairs.

  2. Formula
    Observation

    The introduction explains the greatest common divisor and illustrates it with integer pairs.

Symbol

gcd(a,b)

Meaning

The greatest integer that divides both integers a and b without a remainder.

Domain

Integers a and b, not both zero; GCD is taken positive.

a, b, c

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The lemma on the board uses integer variables for its symmetry, divisibility and congruence statements.

Symbol

a, b, c

Meaning

Integer variables used to state the properties of the GCD and the Euclidean algorithm.

Domain

Integers

y

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board expresses the divisible difference as an integer multiple of the modulus.

  2. Audio
    Observation

    The board expresses the divisible difference as an integer multiple of the modulus.

Symbol

y

Meaning

An integer such that by = a - c, derived from the divisibility condition b | (a - c).

Domain

Integers

gcd⁡(⋅,⋅)\gcd(\cdot,\cdot)

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board carries the same GCD through several equivalent pairs.

  2. Audio
    Observation

    The board carries the same GCD through several equivalent pairs.

Symbol

gcd⁡(⋅,⋅)\gcd(\cdot,\cdot)

Meaning

Greatest common divisor of two integers.

Domain

Used here for integer pairs such as (a,b)(a,b), (b,c)(b,c), (b,r1)(b,r_1), and (r1,r2)(r_1,r_2).

a,b,ca,b,c

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The displayed algebra replaces one integer by its difference with an integer multiple of the other.

  2. Audio
    Observation

    The displayed algebra replaces one integer by its difference with an integer multiple of the other.

Symbol

a,b,ca,b,c

Meaning

Integers in the lemma proving gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c) under the condition b∣a−cb\mid a-c.

Domain

Integers.

yy

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The integer multiplier appears in the divisibility equation and its rearrangement.

  2. Audio
    Observation

    The integer multiplier appears in the divisibility equation and its rearrangement.

Symbol

yy

Meaning

An integer witnessing that a−ca-c is a multiple of bb.

Domain

y∈Zy\in\mathbb{Z}.

dd

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    A common divisor is introduced and tracked through the forward divisibility chain; the reverse argument is not separately shown.

  2. Audio
    Observation

    A common divisor is introduced and tracked through the forward divisibility chain; the reverse argument is not separately shown.

Uncertainties
  1. The video verbally identifies dd as the greatest common divisor, but the displayed line only explicitly states d∣ad\mid a and d∣bd\mid b before the later conclusion about gcd equality.

Symbol

dd

Meaning

An integer divisor of aa and bb used in the proof; verbally treated as the greatest common divisor.

Domain

d∈Zd\in\mathbb{Z}.

∣\mid

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board uses the divisibility sign in the common-divisor argument.

  2. Audio
    Observation

    The board uses the divisibility sign in the common-divisor argument.

Symbol

∣\mid

Meaning

Divisibility relation: x∣yx\mid y means xx divides yy.

Domain

Integers.

q1,q2,r1,r2q_1,q_2,r_1,r_2

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    Successive divisions introduce quotients and remainders with boxed decreasing bounds.

  2. Audio
    Observation

    Successive divisions introduce quotients and remainders with boxed decreasing bounds.

Symbol

q1,q2,r1,r2q_1,q_2,r_1,r_2

Meaning

Quotients and remainders in successive division steps of the Euclidean algorithm.

Domain

Integers, with remainder bounds 0≤r1<b0\le r_1<b and 0≤r2<r10\le r_2<r_1 implied by the division-algorithm context; the board explicitly shows only r1<br_1<b and r2<r1r_2<r_1.

a

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board uses the dividend in the GCD reduction equations.

Symbol

a

Meaning

First integer input to the greatest common divisor function.

Domain

Positive integer in the ordered positive-input algorithm; nonnegative during the base case.

b

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board uses the divisor in the GCD reduction equations.

Symbol

b

Meaning

Second integer input to the greatest common divisor function.

Domain

Nonnegative integer, positive whenever division is performed.

rkr_k

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The descending remainder chain reaches zero in the terminal argument.

Symbol

rkr_k

Meaning

Nonnegative remainder indexed along the Euclidean chain; editorial convention r0=br_0=b includes exact first-step divisibility.

Domain

Non-negative Integer

Knowledge points · 11

Definition of Greatest Common Divisor (GCD)

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The definition and small-number examples introduce positive common divisors and coprimality.

  2. Formula
    Observation

    The definition and small-number examples introduce positive common divisors and coprimality.

Definition
Explanation

The GCD of two integers is the largest positive integer that divides both numbers without leaving a remainder. If the GCD of two numbers is 1, they are called coprime.

Formula
gcd⁡(a,b)\gcd(a, b)
Conditions
  1. Integers a and b, not both zero; the greatest positive common divisor is used.

Brute Force Method for GCD

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The narration contrasts a descending divisor search with the faster remainder algorithm.

Method
Explanation

A naive approach to finding the GCD involves checking every integer from the smaller number down to 1 to see if it divides both numbers. This method has a linear time complexity, making it inefficient for large numbers.

Conditions
  1. Positive integer inputs; linear worst-case count of divisibility tests in min(a,b), treating each arithmetic test as unit cost.

Prerequisites
  1. Definition of Greatest Common Divisor (GCD)

Commutativity of GCD

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The first board statement says that exchanging the inputs preserves the GCD.

  2. Audio
    Observation

    The first board statement says that exchanging the inputs preserves the GCD.

Formula
Explanation

The order of the arguments in the GCD function does not affect the result.

Formula
gcd⁡(a,b)=gcd⁡(b,a)\gcd(a, b) = \gcd(b, a)
Conditions
  1. Integers a and b, not both zero.

Prerequisites
  1. Definition of Greatest Common Divisor (GCD)

GCD when one number divides the other

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The next statement treats a positive input which divides the other input.

  2. Audio
    Observation

    The next statement treats a positive input which divides the other input.

Formula
Explanation

If a positive integer a divides another integer b, then the greatest common divisor of a and b is simply a.

Formula
a>0, a∣b ⟹ gcd⁡(a,b)=aa>0,\ a\mid b\ \Longrightarrow\ \gcd(a,b)=a
Conditions
  1. a>0a > 0

  2. a divides b (a|b)

Prerequisites
  1. Definition of Greatest Common Divisor (GCD)

GCD and Modular Congruence

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The third board statement identifies congruence as a way to preserve common divisors with the modulus.

  2. Audio
    Observation

    The third board statement identifies congruence as a way to preserve common divisors with the modulus.

Formula
Explanation

If two integers a and c are congruent modulo b, their greatest common divisors with b are identical. This property is the foundation of the Euclidean algorithm, allowing the reduction of large numbers.

Formula
a≡c(modb) ⟹ gcd⁡(a,b)=gcd⁡(c,b)a\equiv c\pmod b\ \Longrightarrow\ \gcd(a,b)=\gcd(c,b)
Conditions
  1. Integers a and c; positive integer modulus b.

Prerequisites
  1. Definition of Greatest Common Divisor (GCD)

Symmetry of gcd

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board retains the symmetry statement for the GCD.

Formula
Explanation

The video lists as a basic fact that swapping the two arguments does not change the greatest common divisor.

Formula
gcd⁡(a,b)=gcd⁡(b,a)\gcd(a,b)=\gcd(b,a)
Conditions
  1. Integers (a,b)(a,b) not both zero; the GCD is positive.

Gcd when one number divides the other

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board retains the positive-divisor special case.

Formula
Explanation

If aa is positive and divides bb, then the greatest common divisor of aa and bb is exactly aa.

Formula
a>0, a∣b⇒gcd⁡(a,b)=aa>0,\ a\mid b \Rightarrow \gcd(a,b)=a
Conditions
  1. a>0a>0

  2. a∣ba\mid b

Prerequisites
  1. Symmetry of gcd

Gcd reduction under subtraction of a multiple

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The congruence lemma and forward common-divisor argument are shown; the public converse is an editorial completion.

  2. Audio
    Observation

    The congruence lemma and forward common-divisor argument are shown; the public converse is an editorial completion.

Uncertainties
  1. The earlier native lemma frame confirms congruence, not the model-read set-membership notation. The source shows the forward divisor implication; the converse is supplied editorially.

Formula
Explanation

For integer a,ca,c and positive bb, replacing aa by c=a−byc=a-by preserves the GCD. The source displays the forward common-divisor implication. Editorial completion: a divisor of bb and cc also divides a=c+bya=c+by, so both pairs have the same positive common divisors.

Formula
b∣a−c⇒gcd⁡(a,b)=gcd⁡(b,c)b\mid a-c \Rightarrow \gcd(a,b)=\gcd(b,c)
Conditions
  1. a,c,y∈Za,c,y\in\mathbb Z, b>0b>0 is an integer.

  2. c=a−byc=a-by.

Prerequisites
  1. Symmetry of gcd

Euclidean algorithm reduction rule

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board replaces a dividend–divisor pair by a divisor–remainder pair without changing the GCD.

  2. Audio
    Observation

    The board replaces a dividend–divisor pair by a divisor–remainder pair without changing the GCD.

Method
Explanation

The Euclidean algorithm is presented as repeated application of the reduction lemma to division remainders: replace (a,b)(a,b) by (b,r1)(b,r_1), then (r1,r2)(r_1,r_2), and so on, preserving the gcd at each step.

Formula
a=bq1+r1, 0≤r1<b ⟹ gcd⁡(a,b)=gcd⁡(b,r1);b=r1q2+r2, 0≤r2<r1 ⟹ gcd⁡(b,r1)=gcd⁡(r1,r2)a=bq_1+r_1,\ 0\le r_1<b\ \Longrightarrow\ \gcd(a,b)=\gcd(b,r_1);\quad b=r_1q_2+r_2,\ 0\le r_2<r_1\ \Longrightarrow\ \gcd(b,r_1)=\gcd(r_1,r_2)
Conditions
  1. Integer inputs a≥b>0a\ge b>0, with standard nonnegative division remainders.

  2. Divide by r1r_1 only while it is positive; stop if it is zero.

Prerequisites
  1. Gcd reduction under subtraction of a multiple

Euclidean Algorithm Principle

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board completes the remainder chain and identifies the final positive divisor.

  2. Audio
    Observation

    The board completes the remainder chain and identifies the final positive divisor.

Method
Explanation

For positive integer inputs, replace the pair by the divisor and nonnegative remainder. Stop at zero and return the preceding positive divisor; exact first-step divisibility returns the original divisor.

Formula
gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a,b) = \gcd(b, a \bmod b)
Conditions
  1. Nonnegative integer inputs, not both zero; for the displayed division chain use a≥b>0a\ge b>0.

  2. A recursive division requires b>0b>0 and the nonnegative remainder; when b=0b=0 return aa.

Time Complexity of Euclidean Algorithm

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The complexity slides discuss a remainder bound and a logarithmic call count; the public arithmetic-model conditions are editorial.

  2. Formula
    Observation

    The complexity slides discuss a remainder bound and a logarithmic call count; the public arithmetic-model conditions are editorial.

Formula
Explanation

For positive ordered inputs, within two recursive calls the larger parameter falls to at most half its earlier value, unless the algorithm already stops. This yields a logarithmic number of remainder calls in numerical magnitude under unit-cost arithmetic.

Formula
O(log⁡2(N))O(\log_2(N))
Conditions
  1. Integer a>b>0a>b>0; N=a+bN=a+b is numerical magnitude, not bit length.

  2. Each arithmetic remainder operation is counted as unit cost; this is not a bound on total bit operations.

Prerequisites
  1. Euclidean Algorithm Principle
Claims and conditions · 4

Lemma for the Euclidean Algorithm

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The source groups the symmetry, divisibility and congruence properties into one supporting lemma.

  2. Formula
    Observation

    The source groups the symmetry, divisibility and congruence properties into one supporting lemma.

Theorem
Statement

The Euclidean algorithm relies on three properties of the GCD: commutativity, the case where one number divides the other, and the relationship between GCD and modular congruence.

Hypotheses
  1. The first two parts require their stated integer and positivity conditions; congruence is used with positive modulus b.

Quantifiers

Universal quantification over integers a, b, c satisfying the specific conditions of each part.

Three lemmas used to justify the Euclidean algorithm

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The narration links the preceding GCD facts to the remainder algorithm.

  2. Audio
    Observation

    The narration links the preceding GCD facts to the remainder algorithm.

Proposition
Statement

The video presents three lemmas—symmetry of gcd, gcd when one positive integer divides another, and gcd preservation under subtracting a multiple—as the basis for explaining the Euclidean algorithm.

Hypotheses
  1. The integers are in the usual gcd setting.

  2. For the second lemma, a>0a>0 and a∣ba\mid b.

  3. For the third lemma, b∣a−cb\mid a-c.

Quantifiers

Universal over the displayed integer variables in each lemma.

Conclusion of the reduction lemma

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The source states GCD equality after showing the forward implication; equality requires the editorial converse.

  2. Audio
    Observation

    The source states GCD equality after showing the forward implication; equality requires the editorial converse.

Uncertainties
  1. The displayed proof explicitly tracks one common divisor dd; full rigor would also require the reverse inclusion or an appeal to the definition of gcd, which is not separately written out in this clip.

Proposition
Statement

Under the condition b∣a−cb\mid a-c, the greatest common divisor of aa and bb equals the greatest common divisor of bb and cc.

Hypotheses
  1. a,c∈Za,c\in\mathbb Z, b>0b>0 is an integer.

  2. b∣(a−c)b\mid(a-c).

Quantifiers

For all integers a,ca,c and positive integer bb satisfying the hypothesis.

Remainder Halving Property

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The native complexity slide compares the new remainder with half of the earlier dividend.

Proposition
Statement

For integers a>b>0a>b>0, the new second argument a mod ba\bmod b is at most half the previous first argument aa.

Hypotheses
  1. a>b>0a > b > 0

Quantifiers

For all valid inputs a, b

Derivations and proofs · 4

Proof of GCD and Modular Congruence Property (Part 1)

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board starts the congruence argument by expressing a difference as an integer multiple; later footage continues it.

  2. Audio
    Observation

    The board starts the congruence argument by expressing a difference as an integer multiple; later footage continues it.

Uncertainties
  1. Only the beginning of this argument lies in the first 0–101-second analysis interval; the full source continues. The reverse common-divisor implication will be supplied as an editorial mathematical clarification.

Proof
Steps
  1. Expression
    a≡c(modb)a \equiv c \pmod b
    Explanation

    Start with the assumption that a ≡ c (mod b).

    Justification

    Hypothesis of the third part of the lemma.

    Shown in the video
  2. Expression
    b∣(a−c)b \mid (a - c)
    Explanation

    By the definition of congruence, b must divide the difference between a and c.

    Justification

    Definition of modular congruence.

    Shown in the video
  3. Expression
    ∃y∈Z,by=a−c\exists y \in \mathbb{Z}, by = a - c
    Explanation

    This means there exists an integer y such that by equals a minus c.

    Justification

    Definition of divisibility.

    Shown in the video
Conclusion

The displayed equations start the argument. The source continues beyond this analysis interval; this partial derivation alone does not yet establish equality of the two GCDs.

Proof that b∣a−cb\mid a-c implies gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c)

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board derives divisibility of the difference and then states equality; the reverse divisor-set argument is editorial.

  2. Audio
    Observation

    The board derives divisibility of the difference and then states equality; the reverse divisor-set argument is editorial.

Uncertainties
  1. The clip shows the forward divisibility chain clearly, but it does not separately display the converse inclusion needed for a fully symmetric proof of equality of gcd sets.

Proof
Steps
  1. Expression
    ∃y∈Z  by=a−c\exists y\in\mathbb{Z}\; by=a-c
    Explanation

    Start from the hypothesis that bb divides a−ca-c, so there is an integer yy with by=a−cby=a-c.

    Justification

    Definition of divisibility.

    Shown in the video
  2. Expression
    c=a−byc=a-by
    Explanation

    Rearrange the equation to express cc in terms of aa, bb, and yy.

    Justification

    Algebraic rearrangement of by=a−cby=a-c.

    Shown in the video
  3. Expression
    ∃d∈Z, d∣a, d∣b\exists d\in\mathbb{Z},\ d\mid a,\ d\mid b
    Explanation

    Take an integer dd that divides both aa and bb; the speaker identifies this as the greatest common divisor context.

    Justification

    Assumption in the proof setup / definition of common divisor.

    Shown in the video
  4. Expression
    d∣a−byd\mid a-by
    Explanation

    Since dd divides aa and bb, it also divides the combination a−bya-by.

    Justification

    Closure of divisibility under integer linear combinations; the speaker describes it as multiplying by a factor of bb.

    Shown in the video
  5. Expression
    d∣cd\mid c
    Explanation

    Because c=a−byc=a-by, the previous divisibility statement becomes d∣cd\mid c.

    Justification

    Substitution using c=a−byc=a-by.

    Shown in the video
  6. Expression
    gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c)
    Explanation

    The source states equality after the forward implication. To complete the proof, also take any positive divisor of bb and cc; it divides a=c+bya=c+by. Thus the two positive-common-divisor sets coincide.

    Justification

    Editorial reverse inclusion, together with the observed forward inclusion, justifies equality of the GCDs.

    Supplementary explanation
Conclusion

The equality is correct for integer a,ca,c and positive integer bb. Its full proof uses both common-divisor implications; the reverse implication above is editorial, not separately displayed in the source.

Recursive reduction chain of the Euclidean algorithm

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The board shows successive symbolic divisions and corresponding GCD equalities.

  2. Audio
    Observation

    The board shows successive symbolic divisions and corresponding GCD equalities.

Uncertainties
  1. The stopping case is outside this 101–202-second analysis interval and is explained later in the full source.

Proof
Steps
  1. Expression
    a=bq1+r1, 0≤r1<ba=bq_1+r_1,\ 0\le r_1<b
    Explanation

    Apply division of aa by bb to introduce quotient q1q_1 and remainder r1r_1.

    Justification

    Division algorithm for integers.

    Supplementary explanation
  2. Expression
    gcd⁡(a,b)=gcd⁡(b,r1)\gcd(a,b)=\gcd(b,r_1)
    Explanation

    Replace the pair (a,b)(a,b) by (b,r1)(b,r_1) without changing the gcd.

    Justification

    Reduction lemma, completed with the editorial reverse-divisor argument.

    Shown in the video
  3. Expression
    b=r1q2+r2, 0≤r2<r1b=r_1q_2+r_2,\ 0\le r_2<r_1
    Explanation

    When r1>0r_1>0, divide bb by r1r_1; otherwise stop.

    Justification

    Division algorithm for integers.

    Supplementary explanation
  4. Expression
    gcd⁡(b,r1)=gcd⁡(r1,r2)\gcd(b,r_1)=\gcd(r_1,r_2)
    Explanation

    Repeat the same reduction to pass from (b,r1)(b,r_1) to (r1,r2)(r_1,r_2).

    Justification

    Same reduction lemma applied to the next pair.

    Shown in the video
  5. Expression
    …\dots
    Explanation

    Repeat while the current divisor is positive. Positive integer remainders strictly decrease, so the process terminates rather than continuing indefinitely.

    Justification

    Editorial nonnegative-integer descent; the next source interval presents the terminal rule.

    Supplementary explanation
Conclusion

The Euclidean algorithm is obtained by iterating the identity gcd⁡(x,y)=gcd⁡(y,rem⁡(x,y))\gcd(x,y)=\gcd(y,\operatorname{rem}(x,y)) through successively smaller remainders.

Proof of Logarithmic Complexity Bound

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The native slide uses contradiction for a remainder bound. Its zero-quotient aside contains a typo; the corresponding public step is editorially corrected.

Uncertainties
  1. The native slide incorrectly says that zero quotient would imply k=bk=b; the correct consequence is k=ak=a. The public quotient step below is an editorial correction, not a faithful copy of that incorrect aside.

Proof
Steps
  1. Expression
    a%b=k,k>a/2a \% b = k, \quad k > a/2
    Explanation

    Under integer a>b>0a>b>0, suppose for contradiction that the remainder exceeds half the dividend.

    Justification

    Assumption for contradiction

    Supplementary explanation
  2. Expression
    a=q⋅b+k,k<ba = q \cdot b + k, \quad k < b
    Explanation

    By definition of modulo, a can be written as quotient times divisor plus remainder, where remainder is less than divisor.

    Justification

    Division algorithm

    Shown in the video
  3. Expression
    a/2<k<b<aa/2 < k < b < a
    Explanation

    Combining the assumption k>a/2k > a/2 with k<bk < b gives this chain.

    Justification

    Transitivity of inequality

    Shown in the video
  4. Expression
    q=1q = 1
    Explanation

    The bound b>a/2b>a/2 rules out a quotient of at least two, and a>ba>b requires a positive quotient. Thus q=1q=1. In particular, zero quotient would give k=ak=a, not the slide’s k=bk=b.

    Justification

    Editorially corrected integer quotient bounds.

    Supplementary explanation
  5. Expression
    k+b>a/2+a/2=ak + b > a/2 + a/2 = a
    Explanation

    Substituting q=1q=1 into a=b+ka = b + k gives a=b+ka = b + k. But we established k>a/2k > a/2 and b>a/2b > a/2, so their sum exceeds a.

    Justification

    Arithmetic substitution

    Shown in the video
  6. Expression
    a>aa>a
    Explanation

    We derived k+b>ak + b > a, but the equation a=q∗b+ka = q*b + k with q=1q=1 implies a=b+ka = b + k. This is a contradiction.

    Justification

    Logical contradiction

    Supplementary explanation
Conclusion

The assumed large remainder is impossible, so k≤a/2k\le a/2. Applying this across two calls gives the parameter reduction used to bound the number of remainder calls.

Worked examples · 1

Examples of GCD Calculation

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The introduction displays small integer pairs to explain GCD and coprimality; their divisor lists below are editorial verification.

  2. Audio
    Observation

    The introduction displays small integer pairs to explain GCD and coprimality; their divisor lists below are editorial verification.

Problem

Find the GCD of several pairs of integers.

Given
  1. Pairs: (7, 21), (24, 30), (7, 9)

Goal

Determine the greatest common divisor for each pair.

Steps
  1. Explanation

    For 7 and 21, 7 divides 21, so the GCD is 7.

    Justification

    Definition of GCD.

    Shown in the video
  2. Explanation

    For 24 and 30, the common divisors are 1, 2, 3, 6. The greatest is 6.

    Justification

    Definition of GCD.

    Supplementary explanation
  3. Explanation

    For 7 and 9, the only common divisor is 1.

    Justification

    Definition of GCD.

    Shown in the video
Answer

GCD(7, 21) = 7; GCD(24, 30) = 6; GCD(7, 9) = 1.

Verification

The speaker notes that since GCD(7, 9) = 1, the numbers 7 and 9 are coprime.

Visual events · 6

Introduction Title Cards

Clear evidence
Supplementary explanation
Evidence
  1. Diagram
    Observation

    The opening presents a fast-GCD question and a pair of integer inputs.

Objects
  1. Text boxes

  2. Blue background

Changes
  1. Transition from general title to specific problem example.

Interpretation

The opening motivates efficient GCD computation with a larger integer pair; that pair is posed rather than worked through numerically in the source.

Whiteboard Presentation of the Lemma

Clear evidence
Supplementary explanation
Evidence
  1. Diagram
    Observation

    A gridded whiteboard contains a numbered lemma and progressively written equations.

  2. Animation
    Observation

    A gridded whiteboard contains a numbered lemma and progressively written equations.

Objects
  1. Grid paper background

  2. Handwritten text and formulas

Changes
  1. Appearance of the lemma statement.

  2. Sequential writing of the proof steps for the third part of the lemma.

Invariants
  1. The first two parts of the lemma remain static while the third is being proven.

Interpretation

The board organizes the lemma and its equations; the reverse common-divisor argument requires an editorial supplement.

Lemma board and progressive proof writing

Clear evidence
Supplementary explanation
Evidence
  1. Diagram
    Observation

    The gridded board adds algebra lines beneath the numbered lemma.

  2. Animation
    Observation

    The gridded board adds algebra lines beneath the numbered lemma.

Objects
  1. Title “Euclidean Algorithm”

  2. Handwritten lemma list

  3. Proof lines with ∃y∈Z\exists y\in\mathbb{Z}, c=a−byc=a-by, ∃d∈Z\exists d\in\mathbb{Z}, divisibility statements, and gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c)

Changes
  1. Lines are added one after another beneath the lemma list.

  2. The proof grows from the hypothesis by=a−cby=a-c to the conclusion gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c).

Invariants
  1. The board remains a single static writing surface with no coordinate axes or geometric figures.

  2. The lemma numbering stays visible above the proof.

Interpretation

The board organizes the reduction argument; its forward direction is observed and the public reverse direction is explicitly editorial.

Scroll down and start the algorithm derivation

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The view moves to a lower writing area for successive remainder equations.

  2. Diagram
    Observation

    The view moves to a lower writing area for successive remainder equations.

Objects
  1. Scrolled board view

  2. New handwritten equations for successive divisions

  3. Ellipsis indicating continuation

Changes
  1. The earlier lemma proof shifts upward out of primary focus.

  2. New lines are written below to apply the lemma to division remainders.

  3. The final visible state includes an ellipsis after the second reduction.

Invariants
  1. The same mathematical theme continues from lemma to algorithm.

  2. No numerical example is introduced; the derivation stays symbolic.

Interpretation

The scroll marks the transition from proving the key lemma to using it as the recursive mechanism of the Euclidean algorithm.

Scrolling Whiteboard Derivation

Clear evidence
Shown in the video
Evidence
  1. Animation
    Observation

    The board view moves to connect the remainder chain with the earlier GCD facts.

Objects
  1. Handwritten mathematical equations

Changes
  1. View moves vertically to show context

Invariants
  1. Equations remain static while camera pans

Interpretation

Visual aid to connect the final result back to the initial lemma definitions.

Implementation Code

Clear evidence
Shown in the video
Evidence
  1. Diagram
    Observation

    The source displays a Java-style recursive function with a zero-divisor guard.

Objects
  1. Code block

Changes
  1. Transition from whiteboard to typed code

Invariants
  1. Code logic matches the mathematical derivation

Interpretation

Demonstrates how the recursive mathematical definition translates directly into programming syntax.

Misconceptions · 3

Inefficiency of Brute Force GCD

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narration motivates a more efficient algorithm by comparing it with exhaustive candidate testing.

Misconception

One might think checking all numbers down to 1 is a viable strategy for computing GCD.

Clarification

The video highlights that this brute-force approach has linear time complexity, making it too slow for large numbers, thus motivating the Euclidean algorithm.

Equality of gcds requires more than one divisibility direction

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The source explicitly shows the forward common-divisor implication before concluding equality.

  2. Audio
    Observation

    The source explicitly shows the forward common-divisor implication before concluding equality.

Misconception

One might think the written chain alone proves equality of the two gcds just by showing that every common divisor of aa and bb divides cc.

Clarification

To justify gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c) rigorously, one also needs the reverse inclusion (or an equivalent argument using the definition of greatest common divisor). The clip shows the forward direction explicitly and then states the equality.

The stopping rule follows later in the source

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The current analysis interval ends while the remainder chain continues; the full source later explains the stopping rule.

  2. Audio
    Observation

    The current analysis interval ends while the remainder chain continues; the full source later explains the stopping rule.

Misconception

A viewer may assume the displayed chain already specifies the full Euclidean algorithm.

Clarification

This 101–202-second interval establishes the reduction mechanism. The full source later stops when the remainder becomes 00 and returns the preceding positive divisor; there is no full-source media omission.

Concept relations · 7

Definition of Greatest Common Divisor (GCD) → Lemma for the Euclidean Algorithm

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The source transitions from direct divisor search to the Euclidean algorithm.

Application
Explanation

The Euclidean algorithm is presented as an efficient method specifically for applying and computing the Greatest Common Divisor.

GCD and Modular Congruence → Definition of Greatest Common Divisor (GCD)

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The congruence statement supplies the invariant used in subsequent GCD reductions.

Application
Explanation

Congruence preserves common divisors with the modulus and is applied to reduce GCD computations; this is an invariant, rather than a generalization of the GCD definition.

Gcd reduction under subtraction of a multiple → Euclidean algorithm reduction rule

Clear evidence
Supplementary explanation
Evidence
  1. Audio
    Observation

    The remainder steps use the common-divisor invariant established with editorial completion.

  2. Formula
    Observation

    The remainder steps use the common-divisor invariant established with editorial completion.

Proof dependency
Explanation

The reduction lemma, with its editorial reverse-common-divisor completion, justifies each Euclidean remainder step.

Symmetry of gcd → Gcd reduction under subtraction of a multiple

Approximate timing
Derived from the video
Evidence
  1. Formula
    Observation

    The equal-GCD statements exchange the order of the integer pair.

  2. Audio
    Observation

    The equal-GCD statements exchange the order of the integer pair.

Uncertainties
  1. The clip does not explicitly point to the symmetry lemma at the moment of conclusion; the connection is inferred from the displayed statements.

Application
Explanation

Symmetry of gcd allows the reduction result to be written with the second argument first, yielding the form gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c) used in the algorithm.

Euclidean algorithm reduction rule → Gcd reduction under subtraction of a multiple

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The division equations provide the next argument of the GCD.

  2. Audio
    Observation

    The division equations provide the next argument of the GCD.

Application
Explanation

Each Euclidean step is an instance of the reduction lemma after expressing the dividend as divisor times quotient plus remainder.

Euclidean Algorithm Principle → Euclidean Algorithm Principle

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The narration links the recursive function with the preceding remainder rule.

Application
Explanation

The mathematical principle is applied directly to write the recursive function.

Remainder Halving Property → Time Complexity of Euclidean Algorithm

Clear evidence
Supplementary explanation
Evidence
  1. Formula
    Observation

    The remainder bound is used to motivate the logarithmic number of recursive calls.

Proof dependency
Explanation

The corrected halving argument implies a logarithmic number of recursive remainder calls for ordered positive integers under unit-cost arithmetic.

Find an answer · 8

Why is the brute force method for calculating GCD considered inefficient?

Clear evidence
Shown in the video
Evidence
  1. Audio
    Observation

    The source compares the number of arithmetic tests needed by a direct search.

Knowledge points
  1. Brute Force Method for GCD

What is the relationship between modular congruence and the Greatest Common Divisor?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The congruence lemma states an equality between two GCD expressions.

Knowledge points
  1. GCD and Modular Congruence

How does the proof for gcd(a,b) = gcd(c,b) given a ≡ c (mod b) begin?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The initial board equations translate congruence into divisibility.

Knowledge points
  1. Proof of GCD and Modular Congruence Property (Part 1)

Why does subtracting a multiple of one number from another preserve the gcd?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The board connects subtraction of an integer multiple to common-divisor preservation.

  2. Audio
    Observation

    The board connects subtraction of an integer multiple to common-divisor preservation.

Knowledge points
  1. Gcd reduction under subtraction of a multiple
  2. Proof that b∣a−cb\mid a-c implies gcd⁡(a,b)=gcd⁡(b,c)\gcd(a,b)=\gcd(b,c)

How does the Euclidean algorithm turn division with remainder into a sequence of equal gcds?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    Successive divisions are accompanied by equal-GCD identities.

  2. Audio
    Observation

    Successive divisions are accompanied by equal-GCD identities.

Knowledge points
  1. Euclidean algorithm reduction rule
  2. Recursive reduction chain of the Euclidean algorithm

Why are the remainder inequalities important in the Euclidean algorithm?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The highlighted remainder inequalities motivate descent and termination.

  2. Audio
    Observation

    The highlighted remainder inequalities motivate descent and termination.

Knowledge points
  1. Euclidean algorithm reduction rule
  2. Recursive reduction chain of the Euclidean algorithm

Why does the Euclidean algorithm use O(log⁡N)O(\log N) remainder calls under unit-cost arithmetic?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The source compares parameter reduction with the resulting call count.

Knowledge points
  1. Time Complexity of Euclidean Algorithm
  2. Remainder Halving Property

What is the base case for the recursive GCD function?

Clear evidence
Shown in the video
Evidence
  1. Formula
    Observation

    The displayed function returns its first argument when its second argument is zero.

Knowledge points
  1. Euclidean Algorithm Principle
Coverage and review notes

Covered · Title cards and problem introduction.

Covered · Definition of GCD and examples.

Covered · Discussion of brute force method and its inefficiency.

Covered · Transition slide introducing the Euclidean Algorithm.

Covered · Presentation of the three-part lemma for the Euclidean Algorithm.

Covered · The congruence argument begins here and continues in the source after the 101-second analysis boundary; this interval is covered, not missing.

Covered · The source shows the forward common-divisor chain and states equality; the public proof explicitly adds the required reverse direction as editorial.

Covered · Two symbolic remainder reductions are visible. Native frame 201.8 confirms continuation of the same board through the final fractional second; the terminal rule occurs later in the full source.

Covered · Mathematical derivation on whiteboard.

Covered · Code implementation.

Covered · The source presents the halving argument and complexity conclusion; the public notes disclose the corrected zero-quotient aside and arithmetic cost model.

Covered · Closing slide; actual native tail confirms no further mathematics. Declared duration is rounded to a whole second.

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  • Greatest common divisor ExplanationAt 0:07
    Why this connection?

    From 7 to 31 seconds, the source defines the greatest common divisor as the largest integer dividing both inputs and contrasts direct factor testing with the need for an efficient method.

  • Euclidean algorithm ExplanationAt 0:35
    Why this connection?

    From 35 to 202 seconds, the source develops divisibility lemmas and repeatedly replaces a pair by a remainder pair to explain GCD preservation. The source shows only one direction of the common-divisor argument; the public editorial note supplies the converse, so the link is classified as explanation rather than proof.

  • Euclidean algorithm ApplicationAt 3:22
    Why this connection?

    From 202 to 282 seconds, the source derives the recursive remainder rule, implements it in Java, and argues that the remainder becomes less than half within at most two calls. The public review limits the complexity claim to remainder-call counts under a unit-cost model and corrects the slide typo near 270 seconds.